Burton, David M.-a First Course in Rings and Ideals-Addison-Wesley (1970) (1)

A First Course in ;"

.

RINGS AND IDEALS

. DAVID

M.~BURTON

University ofNew Hampshire

.'

'Y 'Y

- ADDISON-WESLEY PUBLISmNG COMPANY Reading, Massachusetts . Menlo Park, California . London . Don MilIs, Ontario

A First Course in ;"

.

RINGS AND IDEALS

. DAVID

M.~BURTON

University ofNew Hampshire

.'

'Y 'Y

- ADDISON-WESLEY PUBLISmNG COMPANY Reading, Massachusetts . Menlo Park, California . London . Don MilIs, Ontario

.~

'{ I

1 '1

'.1 l·

I

í To my Father. Frank Howard Burton

This book .ís in the

ADDISON-WESLEY SERIES IN MATHEMATICS Consulting Editor: Lynn H. Loomis

1 i

1

Standard Book Number 201.00731.2 AMS 1968 Subject Classifieations 1610, 1620. Copyright © 1970 by Addison-Wesley Publishing Company, lnc. Philippines copyright 1970 by Addison~Wesley PubJishing Company, Ine. AII rights reserved. No part ofthis publieation may be reproduced. stored in a retrievaI system, or transmitted. in any form or by any means. electronie, mechanical, photocopying, recording. or otherwise, without the prior written permission of the publisher. Printed in Ihe United Sta tes , of Ameriea. Published simultaneous1y in Canada. Library·of Congress CataJog Card No. 73-100855.

.~

'{ I

1 '1

'.1 l·

I

í To my Father. Frank Howard Burton

This book .ís in the

ADDISON-WESLEY SERIES IN MATHEMATICS Consulting Editor: Lynn H. Loomis

1 i

1

Standard Book Number 201.00731.2 AMS 1968 Subject Classifieations 1610, 1620. Copyright © 1970 by Addison-Wesley Publishing Company, lnc. Philippines copyright 1970 by Addison~Wesley PubJishing Company, Ine. AII rights reserved. No part ofthis publieation may be reproduced. stored in a retrievaI system, or transmitted. in any form or by any means. electronie, mechanical, photocopying, recording. or otherwise, without the prior written permission of the publisher. Printed in Ihe United Sta tes , of Ameriea. Published simultaneous1y in Canada. Library·of Congress CataJog Card No. 73-100855.

----------------------~~

----,------------ .----_.-

..

_.-

----

----:-----.---------- --"----- _ . _ - - - - _ . ,

PREFACE :1'

.' ".

~' el

.~

,

As the title suggests, this volume is designed to 'serve as an introduction to the basic ideas and techniques of ring theory; it is intendedto be an expository textbook, rather than a treatise on the subject. Thé,mathematical background required for a proper understanding of the coi1tents i~ not extensive. We assume that the average reader 4as had SbITle prior contact with' abstract algebra, but is stíll relatively inexperienced";in this respect. In consequence, nearly everything herein can be read by a person familiar with basic group-theoretic concepts and having a nodding acquaintance with linear algebra; . The level ofmaterial should p'rove suitable-for advanced undergraduates and beginning graduate students. Indeed, a built-in flexibility perrnits the book to be used, either as the basic text or to be read independently by interested students, in a variety of situations. The reader whose main interest is in ideal theory, for instance, could chart a course through Chapters 2, 3, 5, 8, 11, 12, 13. Taken as a whole, the present work is more nearly a beginning than an end. Our hope is that it may serve as a n~tural point of departure for the study of th,e advanced treatises on ring theory and, in sorne aspects of the subject, the periodicalliterature. As regards treatment, ollr guiding principIe is the'sifong conviction that intelligibility should be given priority over coverage; that a deeper understanding of a few important topics is preferable to a superficial knowledge of many. This calIs for 'a presentation in which the pace is unhurried and which is complete in the detaiis of proof, particularIy of basic resuIts. By adhering to the "theorem-proof" ,5tyle ofwriting, we hope to achieve greater clarity (perhaps at the sacrifice of elegance). Apart from the general knowledge presupposed, an attempt has been made to keep the text technicalIy self-contained, even to the extent of including sorne material which is undoubtedly familiar. The mathematiéalIy sophisticated reader may prefer to skip the earlier chapters and refer to them only if the need arises. At the end of each chapter, there will be found a colIection of problems of varying degrees of difficulty. These constitute an integral part of the book and reqliire the reader's active participation. They introduce a variety

v

----------------------~~

----,------------ .----_.-

..

_.-

----

----:-----.---------- --"----- _ . _ - - - - _ . ,

PREFACE :1'

.' ".

~' el

.~

,

As the title suggests, this volume is designed to 'serve as an introduction to the basic ideas and techniques of ring theory; it is intendedto be an expository textbook, rather than a treatise on the subject. Thé,mathematical background required for a proper understanding of the coi1tents i~ not extensive. We assume that the average reader 4as had SbITle prior contact with' abstract algebra, but is stíll relatively inexperienced";in this respect. In consequence, nearly everything herein can be read by a person familiar with basic group-theoretic concepts and having a nodding acquaintance with linear algebra; . The level ofmaterial should p'rove suitable-for advanced undergraduates and beginning graduate students. Indeed, a built-in flexibility perrnits the book to be used, either as the basic text or to be read independently by interested students, in a variety of situations. The reader whose main interest is in ideal theory, for instance, could chart a course through Chapters 2, 3, 5, 8, 11, 12, 13. Taken as a whole, the present work is more nearly a beginning than an end. Our hope is that it may serve as a n~tural point of departure for the study of th,e advanced treatises on ring theory and, in sorne aspects of the subject, the periodicalliterature. As regards treatment, ollr guiding principIe is the'sifong conviction that intelligibility should be given priority over coverage; that a deeper understanding of a few important topics is preferable to a superficial knowledge of many. This calIs for 'a presentation in which the pace is unhurried and which is complete in the detaiis of proof, particularIy of basic resuIts. By adhering to the "theorem-proof" ,5tyle ofwriting, we hope to achieve greater clarity (perhaps at the sacrifice of elegance). Apart from the general knowledge presupposed, an attempt has been made to keep the text technicalIy self-contained, even to the extent of including sorne material which is undoubtedly familiar. The mathematiéalIy sophisticated reader may prefer to skip the earlier chapters and refer to them only if the need arises. At the end of each chapter, there will be found a colIection of problems of varying degrees of difficulty. These constitute an integral part of the book and reqliire the reader's active participation. They introduce a variety

v

vi

PREFACE

C;rl: 4 585

of topics not treated in the body of the text, as well as impart additional information about material covered earlier; sorne, especially in the later chapters, provide substantial extensions of the theory. We have, on the whole, resisted the temptation to use the exercises to develop results that will subsequently be needed (although this is not hard and fast). Those problems whose solutions do not appear straightforward are often accompanied by hints. The text is not intended to be encyc10pedic in nature; many fascinating aspects of this subject vie for inc1usion and sorne choice is imperative. To this end, we merely followed our own tastes, condensing or omitting entirely a number of topics that might have been encompassed by a book of the same tltle. Despite sorne notable omissions, the coverage should provide a firm foundation on which to build. A great deal of valuable criticism was received in the preparation of this work and ourmoments of complacence have admitted many improvements. Of those students who helped, consciously or otherwise, we should like particularly to mention Francis Chevarley, Delmon Grapes, Cynthia Kennett, Kenneth Lidman, Roy Morell, Brenda Phalen, David Smith, and John Sundstrom; we valued their critical reading of sections of the manuscript and incorporated a number of their suggestions into the texto It is a pleasure, likewise, to record our indebtedness to Professor James Clay of the University of Arizoria, who reviewed the final draft and offered helpful comments leading to its correction and improvement. We also profited from many conversations with our colleagues at the University of New Hampshire, especial1y Professors Edward Batho, Homer Bechtell, Robb Jacoby, and Richard Johnson. In this regard, special thanks are due Professor William Witthóft, who was kind enough to read portions of the galleys; his eagle-eyed attention saved us from embarrassment more than once. We enjoyed the'luxury of unusually good secretarial help and take this occasion to express our appreciation to Nancy Buchanan and Sola'nge Larochelle for their joint labors on the typescript. To my wife must go tbe largest .debt of gratitud e, not only for generous assistance with the text at all stages of development, but for her patience and understanding on those occasions when nOtlling would go as we wished. Finally, we should like to acknowledge the fine cooperation of the staff of Addison-Wesley and the usual high quality of their work. The author, needless tq say, must accept the full responsibility for any shortcomings and errors which remain. Durham, New Hampshire J anuary 1970

°o' °7 I

CONTENTS Chapter 1 Introductory Concepts . Chapter 2 Ideals and Their Operations

16

Chapter 3 The Classical Isomorphism Theorems .

39

Chapter 4 Integral Domains and Fields

52

Chapter 5 Maximal, Prime, and Primary Ideals

71

Chapter 6 Divisibility Theory in Integral Domains

90

Chapter 7 PolynomiaI Rings

112 ...

Chapter 8 Certain Radicals of a Ring .

157

Chapter 9 Two Classic Theorems

180

Chapter lQ

Direct Sums of Rings

204

Chapter 11

Rings with Chain Conditions

217

Chapter 12 Further Results on Noetherian Rings .

234

Chapter 13 Some Noncommutative Theory

262 287

AppendixA. Relations. AppendixB.

Zorn's Lernma

296 300

, Bibliography Index oC Special Symbols

303

Index .

305

D.M.B. vii

vi

PREFACE

C;rl: 4 585

of topics not treated in the body of the text, as well as impart additional information about material covered earlier; sorne, especially in the later chapters, provide substantial extensions of the theory. We have, on the whole, resisted the temptation to use the exercises to develop results that will subsequently be needed (although this is not hard and fast). Those problems whose solutions do not appear straightforward are often accompanied by hints. The text is not intended to be encyc10pedic in nature; many fascinating aspects of this subject vie for inc1usion and sorne choice is imperative. To this end, we merely followed our own tastes, condensing or omitting entirely a number of topics that might have been encompassed by a book of the same tltle. Despite sorne notable omissions, the coverage should provide a firm foundation on which to build. A great deal of valuable criticism was received in the preparation of this work and ourmoments of complacence have admitted many improvements. Of those students who helped, consciously or otherwise, we should like particularly to mention Francis Chevarley, Delmon Grapes, Cynthia Kennett, Kenneth Lidman, Roy Morell, Brenda Phalen, David Smith, and John Sundstrom; we valued their critical reading of sections of the manuscript and incorporated a number of their suggestions into the texto It is a pleasure, likewise, to record our indebtedness to Professor James Clay of the University of Arizoria, who reviewed the final draft and offered helpful comments leading to its correction and improvement. We also profited from many conversations with our colleagues at the University of New Hampshire, especial1y Professors Edward Batho, Homer Bechtell, Robb Jacoby, and Richard Johnson. In this regard, special thanks are due Professor William Witthóft, who was kind enough to read portions of the galleys; his eagle-eyed attention saved us from embarrassment more than once. We enjoyed the'luxury of unusually good secretarial help and take this occasion to express our appreciation to Nancy Buchanan and Sola'nge Larochelle for their joint labors on the typescript. To my wife must go tbe largest .debt of gratitud e, not only for generous assistance with the text at all stages of development, but for her patience and understanding on those occasions when nOtlling would go as we wished. Finally, we should like to acknowledge the fine cooperation of the staff of Addison-Wesley and the usual high quality of their work. The author, needless tq say, must accept the full responsibility for any shortcomings and errors which remain. Durham, New Hampshire J anuary 1970

°o' °7 I

CONTENTS Chapter 1 Introductory Concepts . Chapter 2 Ideals and Their Operations

16

Chapter 3 The Classical Isomorphism Theorems .

39

Chapter 4 Integral Domains and Fields

52

Chapter 5 Maximal, Prime, and Primary Ideals

71

Chapter 6 Divisibility Theory in Integral Domains

90

Chapter 7 PolynomiaI Rings

112 ...

Chapter 8 Certain Radicals of a Ring .

157

Chapter 9 Two Classic Theorems

180

Chapter lQ

Direct Sums of Rings

204

Chapter 11

Rings with Chain Conditions

217

Chapter 12 Further Results on Noetherian Rings .

234

Chapter 13 Some Noncommutative Theory

262 287

AppendixA. Relations. AppendixB.

Zorn's Lernma

296 300

, Bibliography Index oC Special Symbols

303

Index .

305

D.M.B. vii

.ONE

INTRODUCTORY CONCEPTS .,¡'

CONVENTIONS Rere we sha11 set forth certain conventions in notatio'i(,~nd terminology used throughout. the text: the standard sym bols of seí" theory will be e~ployed, namely, E, u, n, -, and 0 for the empty set. In particular, A - B = {xix E A and x!É B}. As regards inclusion, the symbols ~ ~~d ;;2 mean ordinary inclusion between sets (they do not exclude the posslbIllty of equality), whereas e and ::J indicate proper inclusion. When we ~eal with an indexed co11ection of sets, say {A¡li E IJ, the cumbersome notatlOns U {A¡liEI} and n {A¡liEI} will genera11y. be ~bbreviated to. u A¡ and n A¡; it being uIiderstood that the operabons are always over the fu11 domain on which the index is defined. Fo11owing custom, {a} denotes the set whose only member is a. Provided that there is no dsk of confusion, a one-element set will be identified with the element itself. . A function f (synonymous with mapping) is indicated by a strai?ht arrow going from domain to range, as in the case f: X .-+ Y, and the no.tatl~n always signifies thatfhas domain X. Under these cl~cumstan.ces,fls sald to be a function on X, or from X, into. Y. In representmg functlOnal values, we adopt the convention of writing the .function on the left, so that/~x), or occasiona11y fx, denotes the image of an element x E X. The restnctlOn of f to a subset A of X is the function flA from.1 into Y defined. by (fIA)(x) = f(x) for a11 x in A: F~r the compo.sltlOn of two func~lOns f: X -+ Yand g: Y -+ Z, we wIlI wnte g o f; that lS, g o f: X ~ Z .satlsfies (g o f)(x) = g(J(x)) for each x E X. (It is important to bear m mmd that our policy is to apply the functions from right to left.) Sorne knowledge of elementary number theory is assumed. We simply remark that the term "prime number" is taken to mean a positive prime; in other words, an integer n > 1 whose only divisors are ± 1 and ± n. Fina11y, let us reserve the symbol Zfor th~ set of all integers, Z~ for the set of positive integers, Q fo! the set of ratlOnal numbers, and R for the set of real numbers. viii

The present chapter sets the stag'~for much that fo11ows, by reviewing sorne of the basic elements of ring theory: I t al so serves as an appropriate vehicle for codifying certain notation and, technical vocabulary used throughout the text . With an eye to the b~,r~ning student (as well as .to minimize a sense of vagueness), we have also'l!1duded a ~umbel of pertinenrexamples of rings. The mathematica11y matúre reader who finds thepace'somewhat tedious may prefer to bypass this section, referring to it for terminology when necessary. As a starting point, it would seem appropriate formally to define the principal object of inten!st in this book, the notion of a ringo A ring is an ordered triple (R, +,.) consisting of a nonempty set R and two binary operations + and . defined on R such that

Definition 1-1.

1)

(R, +) is a cornmutative group, 2) (R,') is a semigroup, and 3) the operation . is distributive (on both sides) over the operation

+.

The reader should understand clearly that +' and . represent abstract, unspecified, operations and not ordinary addition and multiplication. For convenience, however, one invariably refers to the operation + as addition and to the operation . as multiplication. In the light of this terminology, it is natural then to speak of the commutative group (R, +) as the additive group of the ri.p.g and of (R, .) as the multiplicative semigroup of the. ringo By analogy with the integers, the unique identity element for addition is caBed the zero element of the ring and is denoted by the usual symbol O. The unique additive inverse of an element a E R will hereafter be written as - a. (See Problem 1 for justification of the adjective "unique".) In order to minimize the use of parentheses in expressions involving both operations, we shall stipulate that multiplication is to be performed befo re addition. Accordingly, the expression a'b + e stand s for (a'b) + e and not for a'(b + e). Because ofthe general associative law, parentheses

.ONE

INTRODUCTORY CONCEPTS .,¡'

CONVENTIONS Rere we sha11 set forth certain conventions in notatio'i(,~nd terminology used throughout. the text: the standard sym bols of seí" theory will be e~ployed, namely, E, u, n, -, and 0 for the empty set. In particular, A - B = {xix E A and x!É B}. As regards inclusion, the symbols ~ ~~d ;;2 mean ordinary inclusion between sets (they do not exclude the posslbIllty of equality), whereas e and ::J indicate proper inclusion. When we ~eal with an indexed co11ection of sets, say {A¡li E IJ, the cumbersome notatlOns U {A¡liEI} and n {A¡liEI} will genera11y. be ~bbreviated to. u A¡ and n A¡; it being uIiderstood that the operabons are always over the fu11 domain on which the index is defined. Fo11owing custom, {a} denotes the set whose only member is a. Provided that there is no dsk of confusion, a one-element set will be identified with the element itself. . A function f (synonymous with mapping) is indicated by a strai?ht arrow going from domain to range, as in the case f: X .-+ Y, and the no.tatl~n always signifies thatfhas domain X. Under these cl~cumstan.ces,fls sald to be a function on X, or from X, into. Y. In representmg functlOnal values, we adopt the convention of writing the .function on the left, so that/~x), or occasiona11y fx, denotes the image of an element x E X. The restnctlOn of f to a subset A of X is the function flA from.1 into Y defined. by (fIA)(x) = f(x) for a11 x in A: F~r the compo.sltlOn of two func~lOns f: X -+ Yand g: Y -+ Z, we wIlI wnte g o f; that lS, g o f: X ~ Z .satlsfies (g o f)(x) = g(J(x)) for each x E X. (It is important to bear m mmd that our policy is to apply the functions from right to left.) Sorne knowledge of elementary number theory is assumed. We simply remark that the term "prime number" is taken to mean a positive prime; in other words, an integer n > 1 whose only divisors are ± 1 and ± n. Fina11y, let us reserve the symbol Zfor th~ set of all integers, Z~ for the set of positive integers, Q fo! the set of ratlOnal numbers, and R for the set of real numbers. viii

The present chapter sets the stag'~for much that fo11ows, by reviewing sorne of the basic elements of ring theory: I t al so serves as an appropriate vehicle for codifying certain notation and, technical vocabulary used throughout the text . With an eye to the b~,r~ning student (as well as .to minimize a sense of vagueness), we have also'l!1duded a ~umbel of pertinenrexamples of rings. The mathematica11y matúre reader who finds thepace'somewhat tedious may prefer to bypass this section, referring to it for terminology when necessary. As a starting point, it would seem appropriate formally to define the principal object of inten!st in this book, the notion of a ringo A ring is an ordered triple (R, +,.) consisting of a nonempty set R and two binary operations + and . defined on R such that

Definition 1-1.

1)

(R, +) is a cornmutative group, 2) (R,') is a semigroup, and 3) the operation . is distributive (on both sides) over the operation

+.

The reader should understand clearly that +' and . represent abstract, unspecified, operations and not ordinary addition and multiplication. For convenience, however, one invariably refers to the operation + as addition and to the operation . as multiplication. In the light of this terminology, it is natural then to speak of the commutative group (R, +) as the additive group of the ri.p.g and of (R, .) as the multiplicative semigroup of the. ringo By analogy with the integers, the unique identity element for addition is caBed the zero element of the ring and is denoted by the usual symbol O. The unique additive inverse of an element a E R will hereafter be written as - a. (See Problem 1 for justification of the adjective "unique".) In order to minimize the use of parentheses in expressions involving both operations, we shall stipulate that multiplication is to be performed befo re addition. Accordingly, the expression a'b + e stand s for (a'b) + e and not for a'(b + e). Because ofthe general associative law, parentheses

2

INTRODUCTOR Y CONCEPTS

FIRST COURSE IN RINGS AND IDEALS

Example 1-2 Let X be a. given set and P(X) be the collection oC all subsets of X. The symmetric difference of two subsets A, B ~ X is the set A I:l B, where A I:l B =. (A B) u (B - A).

can also be otnÍtted when writing out sums and products of more than two elements. With these remarks in mind, we can now give a more elaborate definition of a ringo A ring (R, +, . ) consists of a nonempty set R together with two binary operations + and . of addition and multiplícation on R for which the following conditions are satisfied : 1) 2) 3) 4) 5) 6)

+ b = b + a, (a +' b) + C = a + (b +

c),

there exists an element Oin R such that i;¡ + O a for every a E R, for each a E R, there exists an element -a E R such that a + (-a) = O, (a'b)'c a'(b'c), and a'(b + c) = a'b + a'c and (b + c)'a b'a + c'a,

where it is understood that a, b, c represent arbitrary elements of R. A ring (R, +, .) is saidto be a finite ring if, naturally enough, the set R of its elements is a finite set. By placing restrictions oIÍ the multiplication operation, several other specialized types of rings are obtained. Definition 1-2. 1) A commutative ring is a ring (R, +,.) in which multiplication is a commutative operation < a' b = b· a for all a, b E R. (In case a'b = b'a for a particular pair a,b, we express this fact by saying that a and b commute.) 2) A ring wíth identity is a ring (R, +, .) in which there exists an identity element for the operation of multiplication, normally represented by the symboll, so that a'l l'a = a for aH a E R.

Given a ring (R, +, .) with identity 1, an element a E R is said to be invertible, or to be a unit, whenever a possesses a (two-sided) inv,er'se with respect to multiplication. The multiplicative inverse of a is uniqúe, whenever it exists, and will be denoted by a-l, so thata'a- l = a-l'a t'ln the future, the set of all invertible elements of the ring will be designateél by the symbol R*. It follo~s easily that the system (R*, .) forms a group~ hown· as the group 01 invertible elements. In this connection, notice tbat R* is certainly nonempty, for, ifnothing else, 1 and -l'belong to R*. (Qne must not assume, however, that 1 and -1 are necessarily distinct.) ,;' A consideration of several concrete examples will serve to brii:J.g these ideas into focus. Example 1-1. If Z, Q, R# denote the sets of integers, rational, and real numbers, respectively, then the systems (Z,+,·),

If we define addition and multiplication in P(X) by A

/

a

(Q,+,'),

+

A I:l B,

B

A· B = A n B,

then the system (P(X), +, .) forros a commutative ring with identity. The empty set0 serves as the zero element, whereas the multiplicative identity is X. Furthermore, each set in P(X) is its own additive in verse. It is interesting to note that if X is nonempty, then neither (P(X), u, n) nor (P(X), n, u) constitutes a ringo . Example 1-3. Given a ring (R, +, '), we may consider the set M,,(R) oC n x n matrices over R. If 1" {1,2, .. " n}, a typical member oC M,,(R) is a function 1: In X 1" --, R. In practice, one identifies such a Cunction with its values aij I(í,}), which are displayed as the n x n rectangular array

~: 11 (

...

~: 1" )

(aij E R).

Il"l , .. a""

For the sake oC simplicity, let us hereafter abbreviate the n x n matrix whose (i,}) entry is aij ~o (a¡), . The operations required to make {Mn(R), +, ,) a ring are provlded by the familiar forrpulas where c¡j

=

L" aik'b¡'j' k=l

i

(We sha11 oCten indulge in this harmless duplícation oC symbols whereby + and . are used with two different meanings.) The zero element oC the resulting ring is the n x n matrix all .of whose entries are O; and -(aij) (-aij)' The ring (Mn(R), +, .) fails to be commutative Cor n > 1. It is equally easy to show that if (R, +,.) has an identity element 1, then the matrix with l's down the main diagonal (that is, aH = 1) and O's elsewhere will act as identity Cor matrix multiplication. In terms of the Kronecker delta symbol Oij' which is defined by

(R#,+,')

are all examples of rings (here, + and . are taken to be ordinary addition and multiplication). In each oC these cases, the ring is commutative and has the integer 1 for an identity elemento

3

oij

=

J1

iC i 1,0 ifi

= j =1=

j

(i,j = 1,2, ... , n),

the identity matrix can be written concisely as (oij)'

2

INTRODUCTOR Y CONCEPTS

FIRST COURSE IN RINGS AND IDEALS

Example 1-2 Let X be a. given set and P(X) be the collection oC all subsets of X. The symmetric difference of two subsets A, B ~ X is the set A I:l B, where A I:l B =. (A B) u (B - A).

can also be otnÍtted when writing out sums and products of more than two elements. With these remarks in mind, we can now give a more elaborate definition of a ringo A ring (R, +, . ) consists of a nonempty set R together with two binary operations + and . of addition and multiplícation on R for which the following conditions are satisfied : 1) 2) 3) 4) 5) 6)

+ b = b + a, (a +' b) + C = a + (b +

c),

there exists an element Oin R such that i;¡ + O a for every a E R, for each a E R, there exists an element -a E R such that a + (-a) = O, (a'b)'c a'(b'c), and a'(b + c) = a'b + a'c and (b + c)'a b'a + c'a,

where it is understood that a, b, c represent arbitrary elements of R. A ring (R, +, .) is saidto be a finite ring if, naturally enough, the set R of its elements is a finite set. By placing restrictions oIÍ the multiplication operation, several other specialized types of rings are obtained. Definition 1-2. 1) A commutative ring is a ring (R, +,.) in which multiplication is a commutative operation < a' b = b· a for all a, b E R. (In case a'b = b'a for a particular pair a,b, we express this fact by saying that a and b commute.) 2) A ring wíth identity is a ring (R, +, .) in which there exists an identity element for the operation of multiplication, normally represented by the symboll, so that a'l l'a = a for aH a E R.

Given a ring (R, +, .) with identity 1, an element a E R is said to be invertible, or to be a unit, whenever a possesses a (two-sided) inv,er'se with respect to multiplication. The multiplicative inverse of a is uniqúe, whenever it exists, and will be denoted by a-l, so thata'a- l = a-l'a t'ln the future, the set of all invertible elements of the ring will be designateél by the symbol R*. It follo~s easily that the system (R*, .) forms a group~ hown· as the group 01 invertible elements. In this connection, notice tbat R* is certainly nonempty, for, ifnothing else, 1 and -l'belong to R*. (Qne must not assume, however, that 1 and -1 are necessarily distinct.) ,;' A consideration of several concrete examples will serve to brii:J.g these ideas into focus. Example 1-1. If Z, Q, R# denote the sets of integers, rational, and real numbers, respectively, then the systems (Z,+,·),

If we define addition and multiplication in P(X) by A

/

a

(Q,+,'),

+

A I:l B,

B

A· B = A n B,

then the system (P(X), +, .) forros a commutative ring with identity. The empty set0 serves as the zero element, whereas the multiplicative identity is X. Furthermore, each set in P(X) is its own additive in verse. It is interesting to note that if X is nonempty, then neither (P(X), u, n) nor (P(X), n, u) constitutes a ringo . Example 1-3. Given a ring (R, +, '), we may consider the set M,,(R) oC n x n matrices over R. If 1" {1,2, .. " n}, a typical member oC M,,(R) is a function 1: In X 1" --, R. In practice, one identifies such a Cunction with its values aij I(í,}), which are displayed as the n x n rectangular array

~: 11 (

...

~: 1" )

(aij E R).

Il"l , .. a""

For the sake oC simplicity, let us hereafter abbreviate the n x n matrix whose (i,}) entry is aij ~o (a¡), . The operations required to make {Mn(R), +, ,) a ring are provlded by the familiar forrpulas where c¡j

=

L" aik'b¡'j' k=l

i

(We sha11 oCten indulge in this harmless duplícation oC symbols whereby + and . are used with two different meanings.) The zero element oC the resulting ring is the n x n matrix all .of whose entries are O; and -(aij) (-aij)' The ring (Mn(R), +, .) fails to be commutative Cor n > 1. It is equally easy to show that if (R, +,.) has an identity element 1, then the matrix with l's down the main diagonal (that is, aH = 1) and O's elsewhere will act as identity Cor matrix multiplication. In terms of the Kronecker delta symbol Oij' which is defined by

(R#,+,')

are all examples of rings (here, + and . are taken to be ordinary addition and multiplication). In each oC these cases, the ring is commutative and has the integer 1 for an identity elemento

3

oij

=

J1

iC i 1,0 ifi

= j =1=

j

(i,j = 1,2, ... , n),

the identity matrix can be written concisely as (oij)'

4

FIRST COURSE IN RINGS AND IDEALS ,

INTRODUCTORY CONCEPTS

I

,

t ,

Example 1-4. To develop our next example, let X be an arbitrary (non- ' enipty) set and(R, +, .) be a ringo We adoptthe notation map(X, R} for the set consisting óf a11 mappings from ,X into R; in symbols,

equivalently, [a] = [rJ. Thus, there are' at most n different congruence . claSses in Zn, namely, [OJ, [lJ, ... , [n - 1]. But these n c1assesare them'selvesdistinct. For ifO :s;; b < a < n, then O < a - b < n and so a - b cannot be divisible by n, whence [aJ =1 [b]. Accordingly, Zn cqnsists of exactly n eIements:

map(X, R) :: {JI!: X ~ R}. (Foreaseofnotation, let usalso agree to write map R in place o~map(R, R):) Now, the elements ofmap(X, R) can be combined by performmg algebralc operatio:ps ontheir functional values. More specifically, the poiritwise sum and producto'r f and g, denoted by f + g, and f· g,. respectively, are the functions whiéh satisfy , (f

+

g)(X) = f(x)

+ g(x),

(f'g)(x) = f(x)'g(x),(x

E

,1

=

{a + knlkeZ}.

Of course, the same congrtience class may very well arise from another integer; any integer a' Cor which [a/J = [a J is said to be a representative of [a]. Qne final, purely notational, remark : the collection of a11 congruence classes oCintegers modulo n will be designated by Zn' It can be shown tbat the congruence cIasses [OJ, [lJ, ... , [n - 1J exhaust the elements of Z.. Given 'an arbitrary integer a, the division algorithm asserts tbat there exist uniqueq, re Z, with O :s;; r < n, such that a = qn + r. By the definition of congruence, a == r (rnod n), or

{'

= {[OJ, [lJ, ... , [n

- 1J}.

Thert::ader should J¡:eep in nlind that each congruence class li~ted.,above is determined by any one of members; a1l we have done is to f~present the cIass by its smallest nonnegative representative. .':' Our next step i5 to define the manner in which the members,ol Zn are to be added and multiplied, so tbat the.resulting system will forma ringo Tbe definitions are as follows: for each [aJ, [b J E Zn, ,¡, "

its

It is read~1i'·~erified thaJ the aboye definitions provide map(X, R) with the structuriola ringo We simply point out that tbe zero element of tbis ring is the c9~stant functión wbose sole value is 0, and the additive inverse -foffis cha,racterized by the rule (-1)(x) = -f(x). Notice that,'the aIgebraic properties of map(X, R) are determined by what happens in thering (R, +,.) (the set X fumishes only the points for the pointwise operations).' , For instance, if (R, +, .) has a multipli.cative identity 1, then the ring (map(X, R), +,.) likewise possesses an identity element; namely, the constant functit;>n defined by l(x) = 1 f6r all x E. X. '

[aJ = {xe Zlx E a (mod n)}

Zn

,

X).

Example 1-5.. Our final example i8 that of the ring oC integers modulo n, wbere n is a fixed positive integer. In order to describe tbis system, we first introduce thé notion of congruence: two integers a and b are said to be eongruent modulo, n, written a == b (mod n), if and only if the difference a b is divisible by n; in other words, a == b (mod n) if ánd only if a - b kn for some k E Z. We leave the reader to convince himself that the relation "congruent modulo n" defines an equivalence relation on the , set Z of integers. As such, it partitions Z into disjoint c1asses of congruent e1ements, caBed eongruenee classes. For each integer a, let the congníence class to which a belongs be denoted by [aJ:

5

[aJ +n [bJ

[a

+ bJ,

. [ab].

:.... fr.

In other words, the sum and product of two congruence c1asses[áliá'nd [b J . are the unique members of Zn which contain' the ordinary sumo a. eV' b and ordinary product ab;respectively. Before considering the algebraic properties ofthese operations, it is necessary to make certain tbat they are well-defined' and do not depend upon which representatives of the congruence c1asses are chosen. In regaro to multiplication, for instance, we want to satisfy ourselves that if [a'J= [aJ and [b'J = [bJ, then [a'h[b'J = [aln[bJ, or, [bJ, .rather, that [a'b'J == [ah]. Now, al E [a'J = [aJ and b' e [b'] which signifies that a' = a' + kn and b = b + jn for some k, j e Z. But then l

.\.

a'b

l

= (a +

kn)(b + jn)

ab

+ (aj + bk + kjn)n .

Hence, a'b' == ab (mod n) and so [a'b'] = [abJ, as desired. The proof that addition is unambiguously defined proceeds similarIy. We omit the detailed verification of the faet tbat (Z., +n' 'n) is a commutatíve ring with identity (tradítioIiaIly known as the ring of integers moaulo n), remarking only that the various ring axiom8 hold in Zn simply because they hold in Z. The distributive law, for instance, fo11ows in Zn from its validity in Z: ' [aL([bJ

+" [eJ)

= [aL[b

+

[ab + aeJ

eJ = [a(b

+ e)]

[abJ +n [aeJ = [aL[bJ +11 [aL [e]. =

Notice, too, that the congruence classes [OJ and [lJ serve as the zero element and multiplieative identity, respectively, whereas [-aJ i8 the additive in verse of [aJ in Zn' When no eonfusion is likely, we sha1l1eave off the brackets from the elements of Zn, thereby making no genuine distínctiori

4

FIRST COURSE IN RINGS AND IDEALS ,

INTRODUCTORY CONCEPTS

I

,

t ,

Example 1-4. To develop our next example, let X be an arbitrary (non- ' enipty) set and(R, +, .) be a ringo We adoptthe notation map(X, R} for the set consisting óf a11 mappings from ,X into R; in symbols,

equivalently, [a] = [rJ. Thus, there are' at most n different congruence . claSses in Zn, namely, [OJ, [lJ, ... , [n - 1]. But these n c1assesare them'selvesdistinct. For ifO :s;; b < a < n, then O < a - b < n and so a - b cannot be divisible by n, whence [aJ =1 [b]. Accordingly, Zn cqnsists of exactly n eIements:

map(X, R) :: {JI!: X ~ R}. (Foreaseofnotation, let usalso agree to write map R in place o~map(R, R):) Now, the elements ofmap(X, R) can be combined by performmg algebralc operatio:ps ontheir functional values. More specifically, the poiritwise sum and producto'r f and g, denoted by f + g, and f· g,. respectively, are the functions whiéh satisfy , (f

+

g)(X) = f(x)

+ g(x),

(f'g)(x) = f(x)'g(x),(x

E

,1

=

{a + knlkeZ}.

Of course, the same congrtience class may very well arise from another integer; any integer a' Cor which [a/J = [a J is said to be a representative of [a]. Qne final, purely notational, remark : the collection of a11 congruence classes oCintegers modulo n will be designated by Zn' It can be shown tbat the congruence cIasses [OJ, [lJ, ... , [n - 1J exhaust the elements of Z.. Given 'an arbitrary integer a, the division algorithm asserts tbat there exist uniqueq, re Z, with O :s;; r < n, such that a = qn + r. By the definition of congruence, a == r (rnod n), or

{'

= {[OJ, [lJ, ... , [n

- 1J}.

Thert::ader should J¡:eep in nlind that each congruence class li~ted.,above is determined by any one of members; a1l we have done is to f~present the cIass by its smallest nonnegative representative. .':' Our next step i5 to define the manner in which the members,ol Zn are to be added and multiplied, so tbat the.resulting system will forma ringo Tbe definitions are as follows: for each [aJ, [b J E Zn, ,¡, "

its

It is read~1i'·~erified thaJ the aboye definitions provide map(X, R) with the structuriola ringo We simply point out that tbe zero element of tbis ring is the c9~stant functión wbose sole value is 0, and the additive inverse -foffis cha,racterized by the rule (-1)(x) = -f(x). Notice that,'the aIgebraic properties of map(X, R) are determined by what happens in thering (R, +,.) (the set X fumishes only the points for the pointwise operations).' , For instance, if (R, +, .) has a multipli.cative identity 1, then the ring (map(X, R), +,.) likewise possesses an identity element; namely, the constant functit;>n defined by l(x) = 1 f6r all x E. X. '

[aJ = {xe Zlx E a (mod n)}

Zn

,

X).

Example 1-5.. Our final example i8 that of the ring oC integers modulo n, wbere n is a fixed positive integer. In order to describe tbis system, we first introduce thé notion of congruence: two integers a and b are said to be eongruent modulo, n, written a == b (mod n), if and only if the difference a b is divisible by n; in other words, a == b (mod n) if ánd only if a - b kn for some k E Z. We leave the reader to convince himself that the relation "congruent modulo n" defines an equivalence relation on the , set Z of integers. As such, it partitions Z into disjoint c1asses of congruent e1ements, caBed eongruenee classes. For each integer a, let the congníence class to which a belongs be denoted by [aJ:

5

[aJ +n [bJ

[a

+ bJ,

. [ab].

:.... fr.

In other words, the sum and product of two congruence c1asses[áliá'nd [b J . are the unique members of Zn which contain' the ordinary sumo a. eV' b and ordinary product ab;respectively. Before considering the algebraic properties ofthese operations, it is necessary to make certain tbat they are well-defined' and do not depend upon which representatives of the congruence c1asses are chosen. In regaro to multiplication, for instance, we want to satisfy ourselves that if [a'J= [aJ and [b'J = [bJ, then [a'h[b'J = [aln[bJ, or, [bJ, .rather, that [a'b'J == [ah]. Now, al E [a'J = [aJ and b' e [b'] which signifies that a' = a' + kn and b = b + jn for some k, j e Z. But then l

.\.

a'b

l

= (a +

kn)(b + jn)

ab

+ (aj + bk + kjn)n .

Hence, a'b' == ab (mod n) and so [a'b'] = [abJ, as desired. The proof that addition is unambiguously defined proceeds similarIy. We omit the detailed verification of the faet tbat (Z., +n' 'n) is a commutatíve ring with identity (tradítioIiaIly known as the ring of integers moaulo n), remarking only that the various ring axiom8 hold in Zn simply because they hold in Z. The distributive law, for instance, fo11ows in Zn from its validity in Z: ' [aL([bJ

+" [eJ)

= [aL[b

+

[ab + aeJ

eJ = [a(b

+ e)]

[abJ +n [aeJ = [aL[bJ +11 [aL [e]. =

Notice, too, that the congruence classes [OJ and [lJ serve as the zero element and multiplieative identity, respectively, whereas [-aJ i8 the additive in verse of [aJ in Zn' When no eonfusion is likely, we sha1l1eave off the brackets from the elements of Zn, thereby making no genuine distínctiori

6

INTRODUCTOR Y CONCEPTS

FIRST COURSE IN RINGS AND IDEALS

betwec3D a congruence c1ass and its smallest nónnegative representative; under this convention, Z" = {O, 1, ... , n - 1}. It is perhaps worth commenting that, since Z 1 = Z, a number of texts specifically exc1ude the value 1 for n. Although it is logically correct (and often convenient) to speak of a ring as an ordered triple, the notation becomes unwieldy as one progresses further into the theory. We shall therefore adopt the usual convention of designating a ring, say (R, +, '), simply by the set symbol R and assume that + and . are known. The reader should realize, however, that a given set may perfectly well be the underlying set of several different rings. Let us also agree to abbreviate a + (-b) as a - b and subsequently refer to tbis expression as the difference between a and b. As a final concession to brevity, juxtaposition without a median dot will be used to denote the product oftwo ring elements. . With these conventions on record, let us begin our formal development of ring theory. The material covered in the next several pages will probably be familiar to most readers and is inc1uded more to assure completeness than to present new ideas. Theorem 1-1. If R is a ring, then for any a, b, e E R 1) Oa = aO = O, 2) a(-b) = (-a)b = -(ab), 3) (-a)(-b) = ab, and 4) a(b - e) = ab - ae, (b - e)a = ba - ea.

Proof. These turn out, in,t.he m~in, to be simple consequences of the distributive laws. For instari2e, Irom O + O = O, it follows that Oa·=;(O

+ O)a =

Oa

+

Da.

Thus, by the cancellation law:ror the additive group (R, +), we have Oa = O. In a like manner, one obtains aO = O. The proof of (2) requires the fact that each element of R h~SI a unique additive inverse (Problem 1). Since b + (-b) = O, '"

ab

+

a("':'~) = a(b

+

(-b)) = aO = O,

which then implies that -(ab) = a( -b). The argument that (-a)b is also the additive inverse of ab proceeds similarly. Tbis leads immediately'to (3):

(-a)( -b) = -( -a)b = - (-(ab))

=

abo

The last assertion is all but obvious. . There is one very simple ring that consists only of the additive identity O, with addition and multiplication given by O + O = O, 00 = O; tbis ring is usually called the trivial ringo '

7

Corollary. Let R be a ring with identity 1. If R is noUhe trivial ring, then the elements O and 1 are distinct.

Proof. Since R =1= {O}, there exists sorne nonzero element a E R. If Oand 1 were equal, it would follow that a = al.. = aO = O, an obvious contradiction. CONVENTION: Let us assume, once and for all, that any ring with identity contains more than one·element. This will rule out the possibility that O and 1 coincide. We now make several remarksabout the concept of zero divisors (the term "divisors of zero" is also in common use): . Defuútion 1-3. If R is a ring and O =1= a E R, then a is called a left (right) zero divisor in R if there exists sorne b =1= O in R such that ab = O (ba = O). A zero divisor is any element of R that is either a left or right zero divisor. According to this definition, O is not a zero divisor, and if R contains an identity 1, then 1 is not a zero divisor nor is any element of R which happens to possess a multiplicative inverse. An obvious example of a riIig with zero divisors is Z., where the integer n > 1 is composite; if n = n 1 n 2 in Z (O < n 1 , n 2 < n), then the product n 1 ·.n 2 = O in Z •. For the most part, we shall be studying rings without zero divisors. In such rings it is possible to conc1ude from the equation ab = O that either a = O or b = O. One can express the property of being with or without zero divisors in the following useful way. ,Theorem 1-2. A ring R is without zero divisors if and only if it satisfies the cancellation laws for multiplication; that is, for all a, b, e E R, ab = ae and ba = ca, where a =1= O, implies b = e.

Proof. Suppose that R is without zero divisors and let ab = ae, a =1= O. Then, the product a(b - e) = O, which means that b - e = O and b = e. The argument is the same for the equation ba = ca. Conversely, let R satisfy the cancellation laws and assume that ab = O, with a =1= O. We then' have ab = aO, whence by cancellation b = O. Similarly, b =1= O implies a = O, proving that there are no zero divisors in R. ' By an integral domain is meant a commutative ring with identity which has no zero divisors, Perhaps the best-known example ofan integral domain is the ring ofintegers; hence the choice ofterminology. Theorem 1-2 shows that the cancellation laws for multiplication hold in any integral domain. The reader should be warned that many authors do not insist on the presence of a multiplicative identity when defining integral domains; and

6

INTRODUCTOR Y CONCEPTS

FIRST COURSE IN RINGS AND IDEALS

betwec3D a congruence c1ass and its smallest nónnegative representative; under this convention, Z" = {O, 1, ... , n - 1}. It is perhaps worth commenting that, since Z 1 = Z, a number of texts specifically exc1ude the value 1 for n. Although it is logically correct (and often convenient) to speak of a ring as an ordered triple, the notation becomes unwieldy as one progresses further into the theory. We shall therefore adopt the usual convention of designating a ring, say (R, +, '), simply by the set symbol R and assume that + and . are known. The reader should realize, however, that a given set may perfectly well be the underlying set of several different rings. Let us also agree to abbreviate a + (-b) as a - b and subsequently refer to tbis expression as the difference between a and b. As a final concession to brevity, juxtaposition without a median dot will be used to denote the product oftwo ring elements. . With these conventions on record, let us begin our formal development of ring theory. The material covered in the next several pages will probably be familiar to most readers and is inc1uded more to assure completeness than to present new ideas. Theorem 1-1. If R is a ring, then for any a, b, e E R 1) Oa = aO = O, 2) a(-b) = (-a)b = -(ab), 3) (-a)(-b) = ab, and 4) a(b - e) = ab - ae, (b - e)a = ba - ea.

Proof. These turn out, in,t.he m~in, to be simple consequences of the distributive laws. For instari2e, Irom O + O = O, it follows that Oa·=;(O

+ O)a =

Oa

+

Da.

Thus, by the cancellation law:ror the additive group (R, +), we have Oa = O. In a like manner, one obtains aO = O. The proof of (2) requires the fact that each element of R h~SI a unique additive inverse (Problem 1). Since b + (-b) = O, '"

ab

+

a("':'~) = a(b

+

(-b)) = aO = O,

which then implies that -(ab) = a( -b). The argument that (-a)b is also the additive inverse of ab proceeds similarly. Tbis leads immediately'to (3):

(-a)( -b) = -( -a)b = - (-(ab))

=

abo

The last assertion is all but obvious. . There is one very simple ring that consists only of the additive identity O, with addition and multiplication given by O + O = O, 00 = O; tbis ring is usually called the trivial ringo '

7

Corollary. Let R be a ring with identity 1. If R is noUhe trivial ring, then the elements O and 1 are distinct.

Proof. Since R =1= {O}, there exists sorne nonzero element a E R. If Oand 1 were equal, it would follow that a = al.. = aO = O, an obvious contradiction. CONVENTION: Let us assume, once and for all, that any ring with identity contains more than one·element. This will rule out the possibility that O and 1 coincide. We now make several remarksabout the concept of zero divisors (the term "divisors of zero" is also in common use): . Defuútion 1-3. If R is a ring and O =1= a E R, then a is called a left (right) zero divisor in R if there exists sorne b =1= O in R such that ab = O (ba = O). A zero divisor is any element of R that is either a left or right zero divisor. According to this definition, O is not a zero divisor, and if R contains an identity 1, then 1 is not a zero divisor nor is any element of R which happens to possess a multiplicative inverse. An obvious example of a riIig with zero divisors is Z., where the integer n > 1 is composite; if n = n 1 n 2 in Z (O < n 1 , n 2 < n), then the product n 1 ·.n 2 = O in Z •. For the most part, we shall be studying rings without zero divisors. In such rings it is possible to conc1ude from the equation ab = O that either a = O or b = O. One can express the property of being with or without zero divisors in the following useful way. ,Theorem 1-2. A ring R is without zero divisors if and only if it satisfies the cancellation laws for multiplication; that is, for all a, b, e E R, ab = ae and ba = ca, where a =1= O, implies b = e.

Proof. Suppose that R is without zero divisors and let ab = ae, a =1= O. Then, the product a(b - e) = O, which means that b - e = O and b = e. The argument is the same for the equation ba = ca. Conversely, let R satisfy the cancellation laws and assume that ab = O, with a =1= O. We then' have ab = aO, whence by cancellation b = O. Similarly, b =1= O implies a = O, proving that there are no zero divisors in R. ' By an integral domain is meant a commutative ring with identity which has no zero divisors, Perhaps the best-known example ofan integral domain is the ring ofintegers; hence the choice ofterminology. Theorem 1-2 shows that the cancellation laws for multiplication hold in any integral domain. The reader should be warned that many authors do not insist on the presence of a multiplicative identity when defining integral domains; and

8.

FIRST COURSE IN RINGS AND IDEALS

in this case the term "integral domain"· would merely indicate a commutative ring without zeto divisors. We change direction somewhat to deal with the situation where a subset of a ring again constitutes a ringo Formally speaking,

INTRODUCTORY CONCEPTS

Example 1-7. The set.Z. of of integers, for

This odefinition is adequate, but unwieldy, siIice all the aspects of the definition of a ring must be checked in deteimining whether a given subset is a subring. In seeking a simpler criterion, noticethat (S, +, .) is a subring of (R, +, .) provided that (S, +) is a °subgroup of (R, +), (S, .) is a subsemi-: group of (R, '), and the two distributive laws are .satisfied in S. But the distributive and associative laws hold automaticaJly for elements of S as a consequence of their validity in R. Since these laws are inherited from R, there is no necessity of requiring them in the definition of a subring. Taking our cuefrom these remarks, a subring could just as well be defined as follows. The system (S, +, o) forms a subring of the ring (R, +, .) if and only if .

To add th~ final touch, even this definition can be improved upon; for the reader versed in group theory will recall that (S, +) is a subgroup of the group (R, +) provided that a - b E S whenever a, bES. By these observations we are led to·a set of c10sure conditions wruch make it somewhat easier to verify tbat a particular subset is actually a subring. Theorem 1-3. Let R be a ring and 0 =1= S 5; R. Theri, S i8 a subring of R if and only if 1) a, b E S imply a - b E S .(closure under differences), (closure under multíplication). 2) a, b E S imply ab E S If S is a subring of the ring R, then the zero element of S is that of R and, moreover, the additive inverse of an element of the subring S is the same as its inverse as a member of R, Verification of these assertions is left as an exercise.

=

2(2nm)

E

E

Z.,

Z •.

Prior to stating our next theorem, let us define the center of a ring R, denoted by cent R, to be thé set 1'"

cent R=' {a

E

RJar

=

ra for all r ER}.

Phrased otherwise, cenot R;:consists of those elements which conimute with every member of R. It s~6uld be apparent that a ring R is commutative if and only if cent R = R."::::. ',.',

Theoreinl-4. For ahy.ring R, cent R is a subring of R..~~I' Proof. To be conscientiouiabout details,first observe that 'tentRis nonempty; for, at the very least, the zero element O E R. Now pick any two elements a, b in cent R. By the definition of center, we know that ar = ra and br :;= rb for every choice ofr E R. Thus, for arbitrary rE R, (a - b)r

=

ar - br = ra - rb

=

r(a - b),

which implies that a - b E cent R. A similar argumentaffirms that the product ab also lies in cent R. In the lightof Theorem 1-3, theseare sufficient conditions for the centet to be a subring of R. It has aIread y been remarked that, when a ring has an identity, this need not be true of its subrings. Other interesting situations may arise,

1) Sorne subfing has a multiplitative identity, but the entire ring does noto 2) Boththe ring and one ofits subrings possess identity elements, but they are distinct. . In each of the cited cases the identity for the subring is necessarily a divisor of zero in the larger ringo To justify this claim, let l' =1= O denote the identity element of the subririg S; we assume further that l' does not act as an identity for the whole ring R. Accordingly, there exists sorne element a E R for which al'. =1= a. It is dear that (al')l'

Example 1-6. Every ring R has two obvious subrings, namely, the set {O}, consisting only of the zero element, and R itself. These two subrings are usually referred to as the trivial subrings of R; all other subrings (if any exist) are called nontrivial. We shall use the term proper subring to mean a subring which is different from R.

°

- m)

This example al so illustrates a fact worth bearing in mind: in a ring with identity, a subring need nof contain the identity elemento

o

1) S ·is a nonempty subset of R, 2) (S, +) is a subgroup of(R, +), and 3) the set S is closed under multiplication.

(2n) (2m)

integers forms a subring of the ring Z

= 2(n

2n -:- 2m o

Definition 1-4. Let (R, +, .) be a ring and S 5; R be a nonempty subset of R. Ifthe system (S, +, .) is itselfa ring (using the induced operations), then (S, +, .) is said to be a subring of (R, +, ').

ev~n

9

=

a(l'l')

=

al',

or (al' - a)l' = O. Since rieither al' -' a nor l' is zero, the ring R has zero divisors, and in particular l' is a zerp divisor. Example 1-8. to present a simple illustration of a ring in which the seoond . of the aforementioned possibilities occurs, consider the set R = Z x Z,

8.

FIRST COURSE IN RINGS AND IDEALS

in this case the term "integral domain"· would merely indicate a commutative ring without zeto divisors. We change direction somewhat to deal with the situation where a subset of a ring again constitutes a ringo Formally speaking,

INTRODUCTORY CONCEPTS

Example 1-7. The set.Z. of of integers, for

This odefinition is adequate, but unwieldy, siIice all the aspects of the definition of a ring must be checked in deteimining whether a given subset is a subring. In seeking a simpler criterion, noticethat (S, +, .) is a subring of (R, +, .) provided that (S, +) is a °subgroup of (R, +), (S, .) is a subsemi-: group of (R, '), and the two distributive laws are .satisfied in S. But the distributive and associative laws hold automaticaJly for elements of S as a consequence of their validity in R. Since these laws are inherited from R, there is no necessity of requiring them in the definition of a subring. Taking our cuefrom these remarks, a subring could just as well be defined as follows. The system (S, +, o) forms a subring of the ring (R, +, .) if and only if .

To add th~ final touch, even this definition can be improved upon; for the reader versed in group theory will recall that (S, +) is a subgroup of the group (R, +) provided that a - b E S whenever a, bES. By these observations we are led to·a set of c10sure conditions wruch make it somewhat easier to verify tbat a particular subset is actually a subring. Theorem 1-3. Let R be a ring and 0 =1= S 5; R. Theri, S i8 a subring of R if and only if 1) a, b E S imply a - b E S .(closure under differences), (closure under multíplication). 2) a, b E S imply ab E S If S is a subring of the ring R, then the zero element of S is that of R and, moreover, the additive inverse of an element of the subring S is the same as its inverse as a member of R, Verification of these assertions is left as an exercise.

=

2(2nm)

E

E

Z.,

Z •.

Prior to stating our next theorem, let us define the center of a ring R, denoted by cent R, to be thé set 1'"

cent R=' {a

E

RJar

=

ra for all r ER}.

Phrased otherwise, cenot R;:consists of those elements which conimute with every member of R. It s~6uld be apparent that a ring R is commutative if and only if cent R = R."::::. ',.',

Theoreinl-4. For ahy.ring R, cent R is a subring of R..~~I' Proof. To be conscientiouiabout details,first observe that 'tentRis nonempty; for, at the very least, the zero element O E R. Now pick any two elements a, b in cent R. By the definition of center, we know that ar = ra and br :;= rb for every choice ofr E R. Thus, for arbitrary rE R, (a - b)r

=

ar - br = ra - rb

=

r(a - b),

which implies that a - b E cent R. A similar argumentaffirms that the product ab also lies in cent R. In the lightof Theorem 1-3, theseare sufficient conditions for the centet to be a subring of R. It has aIread y been remarked that, when a ring has an identity, this need not be true of its subrings. Other interesting situations may arise,

1) Sorne subfing has a multiplitative identity, but the entire ring does noto 2) Boththe ring and one ofits subrings possess identity elements, but they are distinct. . In each of the cited cases the identity for the subring is necessarily a divisor of zero in the larger ringo To justify this claim, let l' =1= O denote the identity element of the subririg S; we assume further that l' does not act as an identity for the whole ring R. Accordingly, there exists sorne element a E R for which al'. =1= a. It is dear that (al')l'

Example 1-6. Every ring R has two obvious subrings, namely, the set {O}, consisting only of the zero element, and R itself. These two subrings are usually referred to as the trivial subrings of R; all other subrings (if any exist) are called nontrivial. We shall use the term proper subring to mean a subring which is different from R.

°

- m)

This example al so illustrates a fact worth bearing in mind: in a ring with identity, a subring need nof contain the identity elemento

o

1) S ·is a nonempty subset of R, 2) (S, +) is a subgroup of(R, +), and 3) the set S is closed under multiplication.

(2n) (2m)

integers forms a subring of the ring Z

= 2(n

2n -:- 2m o

Definition 1-4. Let (R, +, .) be a ring and S 5; R be a nonempty subset of R. Ifthe system (S, +, .) is itselfa ring (using the induced operations), then (S, +, .) is said to be a subring of (R, +, ').

ev~n

9

=

a(l'l')

=

al',

or (al' - a)l' = O. Since rieither al' -' a nor l' is zero, the ring R has zero divisors, and in particular l' is a zerp divisor. Example 1-8. to present a simple illustration of a ring in which the seoond . of the aforementioned possibilities occurs, consider the set R = Z x Z,

10

FIRST COURSE IN RINGS ANO IDEALS

INTROOUCTORY CONCEPTS

consisting of ordered pairs of integers. One converts R into a ring by defining addition and rnultiplicatiori componentwise : (a, b)

+

(e, d) = (a

+ e, b +

d),

(a,b)(e, d) = (ae, bd).

A routíne calculation will show that Z x {O} = {(a, O)la E Z} forms a subring with .identíty element (1, O). This obviously differs from the identity of the entire ring R, which turns out to be the ordered pair (1, 1). By our prevíous rernarks, (1, O)rnustbeazerodivisorinR;infact,(l, 0)(0,1) = (O, O), where (O, O) serves as the zero element of R. If R is an arbitrary ring and n a positive integer, then the nth power a" of an element a E R is defined by the inductíve condítions al = a and a" = a"-la. Frorn tbis the usuallaws of exponents follow at once: a"a m = an+ m, (a"t = a"m (n, m E Z+).

To establish these rules, fu m and proceed by induction on n. Observe also that iftwo elements a, bE R happen to cornrnute, so do all powers of a and b, whence (ab)" = a"b" for each positive integer n. In the event that R possesses an identíty element 1 and a- l exists, negaÚve powers of a can be introduced by intelJ)retíng a-" as (a- l )", where n > O. With the definition aO = 1, the symbol a" now has a well-defined meaning for every integer n (at least when attention is restricted to invertible elernents). Parallelíng the exponent notation for powers, there is the notation of integral multiples of an e1ement a E R. For each positive integer n, we define the nth natural multiple na recursively as follows: la = a and na = (n - l)a

+ a, when n >

1.

If ít is also agreed to let Oa = O and ( - n)a = - (na), then the definition of na can be extended to all íntegers. Integral multiples satisfy several identities which are easy to establish:

(n

+ m)a

=

na

+ ma,

(nm)a = n(ma), n(a

+ b)

= /la

+ nb,

for a, b E R and arbitrary integers n and m. In addition to these rules, there are two further properties resulting frorn the distributive law, namely, n(ab) = (na)b = a(nb),

and

(na)(mb) = (nm)(ab).

Experience impels us to emphasize that the expression na should not be regarded as a ring product (indeed, the integer n 'may not even be a member of R); the entire symbol na is just a convenient way of indicating

11

a certain sum of elements of R. However, when there is an identity for rnultiplication, it is possible to represent na as a product oftwo ring elements, namely, na = (n1)a. To proceed further with our límited program, we must first frame a definítion. Definition 1-5. Let R be an arbítrary ringo If there exists a posítive integer n such that na = Ofor all a E R, then the srnaIlest positive integer wíth this property is called the eharaeteristie of the ringo If no such positive integer exists (that is, n = Ois the only integer for which na =. O for all a in R), then R is said to be of eharaeteristie zero. We shall wnte char R for the characteristíc of R. The rings of integers, rational numbers, and real numbers are all standard exarnples of systerns having characteristíc zero (sorne· writers prefer the expression "characteristic infinity"). On the other hand, the ring P(X) of subsets of a fixed set X is of characteristic 2, since 2A = A A A = (A - A) u (A - A)

=

rp

for every subset A S; X. Although the definition of characteristíc makes an assertion about every element of the ring, in rings with identíty the characteristic is completely determined by the identity elernent. . We reach tbis conc1usion below. Theorem 1-5. If R is any ring wíth identity 1, then R has characteristic n > O if and only if n is the least posítive integer for which nI = O. Proo/. If char R = n > O, then na = Ofor every a E R and so, in particular, nI = O. Were ml = O, where O < m < n, jt wciuld necessarily follow that ma

= m(la) = (m1)a = Oa

=

O

for every elernent a E R. The implícatíon is that char R < n, which is impossible. One establishes the converse in rnuch the 'sarne way. As we have seen, multiplícation exerts a strong infiuence on the addítive structure of a ring through the distributive law. The following corollary to Theorem 1-5 shows that by sufficiently restricting the multiplícation in a ring R it is possible to reach sorne interestíng conc1usions regarding the characteristic of R. Corollary 1. In an integral dornain R all the nonzero elernents have the sarne addítive order; this order is the characteristic of the domain when char R > O and infinite when char R = o. Proo/. To verify this assertion, suppose first that char R = n > O. According to the definítion of characteristic, each element O =1= a E R will then possess a finíte additive order m, wíth m :::;; n. (Recall that for an element

10

FIRST COURSE IN RINGS ANO IDEALS

INTROOUCTORY CONCEPTS

consisting of ordered pairs of integers. One converts R into a ring by defining addition and rnultiplicatiori componentwise : (a, b)

+

(e, d) = (a

+ e, b +

d),

(a,b)(e, d) = (ae, bd).

A routíne calculation will show that Z x {O} = {(a, O)la E Z} forms a subring with .identíty element (1, O). This obviously differs from the identity of the entire ring R, which turns out to be the ordered pair (1, 1). By our prevíous rernarks, (1, O)rnustbeazerodivisorinR;infact,(l, 0)(0,1) = (O, O), where (O, O) serves as the zero element of R. If R is an arbitrary ring and n a positive integer, then the nth power a" of an element a E R is defined by the inductíve condítions al = a and a" = a"-la. Frorn tbis the usuallaws of exponents follow at once: a"a m = an+ m, (a"t = a"m (n, m E Z+).

To establish these rules, fu m and proceed by induction on n. Observe also that iftwo elements a, bE R happen to cornrnute, so do all powers of a and b, whence (ab)" = a"b" for each positive integer n. In the event that R possesses an identíty element 1 and a- l exists, negaÚve powers of a can be introduced by intelJ)retíng a-" as (a- l )", where n > O. With the definition aO = 1, the symbol a" now has a well-defined meaning for every integer n (at least when attention is restricted to invertible elernents). Parallelíng the exponent notation for powers, there is the notation of integral multiples of an e1ement a E R. For each positive integer n, we define the nth natural multiple na recursively as follows: la = a and na = (n - l)a

+ a, when n >

1.

If ít is also agreed to let Oa = O and ( - n)a = - (na), then the definition of na can be extended to all íntegers. Integral multiples satisfy several identities which are easy to establish:

(n

+ m)a

=

na

+ ma,

(nm)a = n(ma), n(a

+ b)

= /la

+ nb,

for a, b E R and arbitrary integers n and m. In addition to these rules, there are two further properties resulting frorn the distributive law, namely, n(ab) = (na)b = a(nb),

and

(na)(mb) = (nm)(ab).

Experience impels us to emphasize that the expression na should not be regarded as a ring product (indeed, the integer n 'may not even be a member of R); the entire symbol na is just a convenient way of indicating

11

a certain sum of elements of R. However, when there is an identity for rnultiplication, it is possible to represent na as a product oftwo ring elements, namely, na = (n1)a. To proceed further with our límited program, we must first frame a definítion. Definition 1-5. Let R be an arbítrary ringo If there exists a posítive integer n such that na = Ofor all a E R, then the srnaIlest positive integer wíth this property is called the eharaeteristie of the ringo If no such positive integer exists (that is, n = Ois the only integer for which na =. O for all a in R), then R is said to be of eharaeteristie zero. We shall wnte char R for the characteristíc of R. The rings of integers, rational numbers, and real numbers are all standard exarnples of systerns having characteristíc zero (sorne· writers prefer the expression "characteristic infinity"). On the other hand, the ring P(X) of subsets of a fixed set X is of characteristic 2, since 2A = A A A = (A - A) u (A - A)

=

rp

for every subset A S; X. Although the definition of characteristíc makes an assertion about every element of the ring, in rings with identíty the characteristic is completely determined by the identity elernent. . We reach tbis conc1usion below. Theorem 1-5. If R is any ring wíth identity 1, then R has characteristic n > O if and only if n is the least posítive integer for which nI = O. Proo/. If char R = n > O, then na = Ofor every a E R and so, in particular, nI = O. Were ml = O, where O < m < n, jt wciuld necessarily follow that ma

= m(la) = (m1)a = Oa

=

O

for every elernent a E R. The implícatíon is that char R < n, which is impossible. One establishes the converse in rnuch the 'sarne way. As we have seen, multiplícation exerts a strong infiuence on the addítive structure of a ring through the distributive law. The following corollary to Theorem 1-5 shows that by sufficiently restricting the multiplícation in a ring R it is possible to reach sorne interestíng conc1usions regarding the characteristic of R. Corollary 1. In an integral dornain R all the nonzero elernents have the sarne addítive order; this order is the characteristic of the domain when char R > O and infinite when char R = o. Proo/. To verify this assertion, suppose first that char R = n > O. According to the definítion of characteristic, each element O =1= a E R will then possess a finíte additive order m, wíth m :::;; n. (Recall that for an element

12

FIRST COURSE IN RINGS AND IDEALS

+

PROBLEMS

+

where p is a prime, then we are able to deduce considerably more: each nonzero element of Zl is invertible. Before establishing this, first observe that the set ZI, regarded as an additive cyclic group of order p, consists of p distinct elements, namely, the p ~ums nI, where n = O, 1, ... , p 1. Now letnl beany nonzeroelementoCZ1 (O < n < p). Since nandp are relatively 'prime, there exist integers r and s Cor which rp + sn = 1. But then 1 = (rp + sn)1 = r(p1) + (s1)(nl). . ¡As p1 = O, we obtain the equation 1 = (sl)(nl), so that sI serves as the í multiplicative in verse of nI in ZL The value of these remarks wiIl have too áwait further developments (in particular, see Chapter 4).

a Oofthe group (R, +) to have order m mean s that ma = Oand ka O irO < k < m.) But the retation O == ma = (ml)a ímplies that mI = O, Cor R is assumed to be free of zero divisors. We therefore conclude from the theorem that n ::;; m, whence m and n are equal. In consequence, every nonzero element of R has additive order n. A somewhat similar argumen't can be employed when char R O. The equaqon ma = O would lead, as before, to m1 = O or m O. In this case every nonzero eIement a E R musí be oC infinite order. The last result seives to bring on record as . .

OUt

anotner use fuI point, which we place

CoroHary 2. An integral dornain R has positive characteristic ifand only iC na O foro s?me O a E R and some integer n E Z+.

+

;',.:PROBLEMS

Continuing in thls veln, let us next show tbat not any commutative group can serve as the additive group of an integral domain.

::.' ¡). Verify that the zero elernent of a ring R is unique, as ís the additive inverse of each.: element a E R.

.'

.

.

,

). Let R be an additive commutative group: If the product of eveJ'Y pair of elernents is defined to be zero, show tbat tbe resulting systern f~rmiifa'coxirmutaÚve ring " (this is sometimes called tbe zero ring). .

Theorem 1-6. The cbaracteristic of an integral domain is either zero or a prime number. '

Prooj. Let R be of positive characteristic n and assume that n is not a prime. Then, n has a nonírivial factorization n = n I n2 , with I < nI' n 2 < n. It follows tbat O = nl = (n 1 n2)1 = (n I n2 )12 = (n I l)(n 2 1).

3. Prove that any ring R in which the two operations are equal (that is, a for all a, b e R) must be the trivial ring R = {O}. .

b = ab

a) !he identity elemen! for rnultiplieation is unique, b) if a e R has a rnultiplieative inverse, thena -1 is unique, e) ifthe'elernent a is invertible, then so. also is -a, d) no divisor of zero can Po.ssess a multiplica ti ve IDverse in R.

5. 'a) Ifthe set X eontains more than one elernent, prove that every nonempty proper subset of X is a zeró divisor in the ring P(X). b) Show that, if n > 1, the matrix ring Mn(R) has zero divison even though the '_ ring R may not.

CorolIary.. If R is a finite integral domain, then char R = .p, a prime. Turning again to the general theory, let R be any ring with identity and , consider the set Zl of integral multiples of the ídentity; stated symbolically,

6. Suppose !hat R is a ring with identity 1'itnd having no divisor s ofzero. For a, bE R, verify that a) ab 1 if and only if ha = 1, b) if a2 1, then either a 1 or a = -1.

= {nlln E Z}.

From the relations nI - mI = (n - m)l,

+

4., In a ring R with identity, establish each of the following:

By supposition,.R is without zero divisors, so that either nll Oor n2 I O. Since both nI and n2 are less than n, this contradicts the choice of n as the . leastpositive integer for which n1 = O. We therefore concluoe that char R must be prime.

Zl

13

(nl)(ml) = (nm)l

7~ Let a,' b be two elements ofthe ring

R. Ji nE Z.¡: and a and b eommute, derive the

binomial expansion

one can easily infer that ZI itself [orms a (commutative) ring with identity. The order of the additive cyclic group (Z1, +) is simply the characteristic oC the given ring R. When R happens to be an integral domain, then Zl is a subdomain of R (that ¡s, Z1 is also an integral domain with respect to the operations in R). In fact, ZI is the smallest subdomain oC R, in the sense that it is contained in every other subdomain of R. If R is a domain of characteristic p,

(a

+ bY'

=

an + ('i.)an-1b

+ '" + (k)án-kb k + '"

(n~l)Qb·-l

+ b",

where

[knj "" k!(n nIis the usual binomial coefficÍent , I

k)!

..

12

FIRST COURSE IN RINGS AND IDEALS

+

PROBLEMS

+

where p is a prime, then we are able to deduce considerably more: each nonzero element of Zl is invertible. Before establishing this, first observe that the set ZI, regarded as an additive cyclic group of order p, consists of p distinct elements, namely, the p ~ums nI, where n = O, 1, ... , p 1. Now letnl beany nonzeroelementoCZ1 (O < n < p). Since nandp are relatively 'prime, there exist integers r and s Cor which rp + sn = 1. But then 1 = (rp + sn)1 = r(p1) + (s1)(nl). . ¡As p1 = O, we obtain the equation 1 = (sl)(nl), so that sI serves as the í multiplicative in verse of nI in ZL The value of these remarks wiIl have too áwait further developments (in particular, see Chapter 4).

a Oofthe group (R, +) to have order m mean s that ma = Oand ka O irO < k < m.) But the retation O == ma = (ml)a ímplies that mI = O, Cor R is assumed to be free of zero divisors. We therefore conclude from the theorem that n ::;; m, whence m and n are equal. In consequence, every nonzero element of R has additive order n. A somewhat similar argumen't can be employed when char R O. The equaqon ma = O would lead, as before, to m1 = O or m O. In this case every nonzero eIement a E R musí be oC infinite order. The last result seives to bring on record as . .

OUt

anotner use fuI point, which we place

CoroHary 2. An integral dornain R has positive characteristic ifand only iC na O foro s?me O a E R and some integer n E Z+.

+

;',.:PROBLEMS

Continuing in thls veln, let us next show tbat not any commutative group can serve as the additive group of an integral domain.

::.' ¡). Verify that the zero elernent of a ring R is unique, as ís the additive inverse of each.: element a E R.

.'

.

.

,

). Let R be an additive commutative group: If the product of eveJ'Y pair of elernents is defined to be zero, show tbat tbe resulting systern f~rmiifa'coxirmutaÚve ring " (this is sometimes called tbe zero ring). .

Theorem 1-6. The cbaracteristic of an integral domain is either zero or a prime number. '

Prooj. Let R be of positive characteristic n and assume that n is not a prime. Then, n has a nonírivial factorization n = n I n2 , with I < nI' n 2 < n. It follows tbat O = nl = (n 1 n2)1 = (n I n2 )12 = (n I l)(n 2 1).

3. Prove that any ring R in which the two operations are equal (that is, a for all a, b e R) must be the trivial ring R = {O}. .

b = ab

a) !he identity elemen! for rnultiplieation is unique, b) if a e R has a rnultiplieative inverse, thena -1 is unique, e) ifthe'elernent a is invertible, then so. also is -a, d) no divisor of zero can Po.ssess a multiplica ti ve IDverse in R.

5. 'a) Ifthe set X eontains more than one elernent, prove that every nonempty proper subset of X is a zeró divisor in the ring P(X). b) Show that, if n > 1, the matrix ring Mn(R) has zero divison even though the '_ ring R may not.

CorolIary.. If R is a finite integral domain, then char R = .p, a prime. Turning again to the general theory, let R be any ring with identity and , consider the set Zl of integral multiples of the ídentity; stated symbolically,

6. Suppose !hat R is a ring with identity 1'itnd having no divisor s ofzero. For a, bE R, verify that a) ab 1 if and only if ha = 1, b) if a2 1, then either a 1 or a = -1.

= {nlln E Z}.

From the relations nI - mI = (n - m)l,

+

4., In a ring R with identity, establish each of the following:

By supposition,.R is without zero divisors, so that either nll Oor n2 I O. Since both nI and n2 are less than n, this contradicts the choice of n as the . leastpositive integer for which n1 = O. We therefore concluoe that char R must be prime.

Zl

13

(nl)(ml) = (nm)l

7~ Let a,' b be two elements ofthe ring

R. Ji nE Z.¡: and a and b eommute, derive the

binomial expansion

one can easily infer that ZI itself [orms a (commutative) ring with identity. The order of the additive cyclic group (Z1, +) is simply the characteristic oC the given ring R. When R happens to be an integral domain, then Zl is a subdomain of R (that ¡s, Z1 is also an integral domain with respect to the operations in R). In fact, ZI is the smallest subdomain oC R, in the sense that it is contained in every other subdomain of R. If R is a domain of characteristic p,

(a

+ bY'

=

an + ('i.)an-1b

+ '" + (k)án-kb k + '"

(n~l)Qb·-l

+ b",

where

[knj "" k!(n nIis the usual binomial coefficÍent , I

k)!

..

14

FIRST COURSE IN RINGS AND IDEALS

8. An e1ement a of a ring R is said to be idempotent if a2 = a and nilpotent if a" = O for sorne n E Z+. Show that a) a nonzero idempotent element cannot be nilpotent, b) every nonzero nilpotent element is a zero divisor in R. 9. Given that R is an integral domain, prove that a) the only nilpotent element is the zero element of R, b) the multiplicative identity is the only nonzero idempotent elemento

11. A Boo/ean ring is a ring with identity every element of which is idempotent. Prov~ that any Boolean ring R is cornmutative. [Hin!: First show that a = -a for every aE R.] 12. Suppose the ring R contains an element a such that (1) a is idempotent and (2) a is not a zero divisor of R. Deduce that a serves as a multiplicative identity for R. 13. Let S be a nonempty subset of the finite ring R. Prove that S is a subring of R if and only if S is c10sed under both the operations of addition and multiplication. 14. Assume that R is a ring l/lld a E R. If C(a) denotes the set of all elements which commute with a,

=

{r E Rlar

= ra},

show that C(a) js a ~ubring of R. Also, verify the equality tent R

=

n.on C(a).

1S. Given a ring R, prove that a) if SI is an arbitrary (indexed) coHection of subrings of R, then their intersection n S¡ is also a subring of R; b) for a nonempty subset T of R, the set (T) = n {SIT

S;

a) if there exists an integer k such that ka = O for aH a E R, then k is divisible by char R; b) if char R :> O, then char S ~ char R for any subring S of R; c) if R is an integral domain and. S is a subdomain of R, then char S = char R. 19. L:t R be a ring with a finite number of elements, sayal> a2' ... , ano and let ni be the order of ai regarded as a member of the additive group of R. Prove that the characteristic of R is the least common multiple of the integers ni (i = 1,2, ... , n).

10. If a is a nilpotent element of R, a ring with identity, establish that 1 + a is invertible in R. [Hint: (1 + a)-l = 1 - a + a2 + ... + (-l)"-la"-l, where a" = O.]

C(a)

S; S is a subring of R}

.is the smaHest (in the sense of inc1usion) subring of R to contain T; (T) is called the subring generated by T. 16. Let S be a subring of R, a ring with identity. For an arbitrary element a rt S, the subring generated by the set S u {a} is represented by (S, a). If a E cent R, establish tha t

20. Suppose that R is a ring with identity such that char R = n > O. If n is not prime, show that R has divisors of zero. 21. If R is a rillg which has no nonzero nilpotent elements, deduce that aH the idempotent elements of R belong to cent R. [Hint: If a2 = a, then (ara - arV = (ara - ra)2 = O for aH r E R.] 22. Assume that R is a ring with the property that a2 + a E cent R for every element a in R. Prove that R is necessarily a cornmutative ringo [Hint: Utilize the expression (a + W + (a + b) to show first that ab + ba lies in the center for aH a, b E R.] 23. Let (G, +) be a commutative group and R be the set ofall (group) homomorphisms of G into itself. The pointwise sum f + g and composition f o g of two functions f, g E R are defined by the usual rules

(f + g)(x) = f(x)

+ g(x),

= {ro +

rla

+ ... +

r.a"ln E Z+; r¡

E

S}.

(f o g)(x) = f(g(x»)

(XE G).

+, o) forms a ring. At the same time determine

Show that the resulting system (R, the invertible elements of R.

24. Let(G,') bea finite group (writtenmultiplicatively), say with elementsx¡, x 2 , and let R be an arbitrary ringo Consider the set R(G) of all formal sums

¿"

••• ,

x"'

(r¡ER).

r¡x i

i=l

Two such expressions are regarded as equal if they have the same coefficients .. Addition and multiplication can be defined in R(G) by taking

±r¡x¡ + ±s¡x¡ = ±(r¡ + s¡)x¡ (.± r¡x¡) [.± S¡X¡) = .± t¡x i=l

i=l

and

i=l

j,

,=1

l=1

(S, a)

15

PROBLEMS

1=1

where ti

=

¿

rjsk •

17. Let R be an arbitrary ring apd n E Z+. Ifthe set Sft is defined by

XJXk=Xf

= {a E Rlnka =

(The meaning of the last-written sum is that the surnmation is to be extended over aH subscripts j and k for which xjxk = x¡.) Prove that, with respect to these operations, R(G) constitutes a ring, the so-called group ring of G over R.

S.

O for sorne k> O},

determine whether S" is a subring of R .. 18. Establish the foHowing assertions concerning the characteristic of a ring R:

14

FIRST COURSE IN RINGS AND IDEALS

8. An e1ement a of a ring R is said to be idempotent if a2 = a and nilpotent if a" = O for sorne n E Z+. Show that a) a nonzero idempotent element cannot be nilpotent, b) every nonzero nilpotent element is a zero divisor in R. 9. Given that R is an integral domain, prove that a) the only nilpotent element is the zero element of R, b) the multiplicative identity is the only nonzero idempotent elemento

11. A Boo/ean ring is a ring with identity every element of which is idempotent. Prov~ that any Boolean ring R is cornmutative. [Hin!: First show that a = -a for every aE R.] 12. Suppose the ring R contains an element a such that (1) a is idempotent and (2) a is not a zero divisor of R. Deduce that a serves as a multiplicative identity for R. 13. Let S be a nonempty subset of the finite ring R. Prove that S is a subring of R if and only if S is c10sed under both the operations of addition and multiplication. 14. Assume that R is a ring l/lld a E R. If C(a) denotes the set of all elements which commute with a,

=

{r E Rlar

= ra},

show that C(a) js a ~ubring of R. Also, verify the equality tent R

=

n.on C(a).

1S. Given a ring R, prove that a) if SI is an arbitrary (indexed) coHection of subrings of R, then their intersection n S¡ is also a subring of R; b) for a nonempty subset T of R, the set (T) = n {SIT

S;

a) if there exists an integer k such that ka = O for aH a E R, then k is divisible by char R; b) if char R :> O, then char S ~ char R for any subring S of R; c) if R is an integral domain and. S is a subdomain of R, then char S = char R. 19. L:t R be a ring with a finite number of elements, sayal> a2' ... , ano and let ni be the order of ai regarded as a member of the additive group of R. Prove that the characteristic of R is the least common multiple of the integers ni (i = 1,2, ... , n).

10. If a is a nilpotent element of R, a ring with identity, establish that 1 + a is invertible in R. [Hint: (1 + a)-l = 1 - a + a2 + ... + (-l)"-la"-l, where a" = O.]

C(a)

S; S is a subring of R}

.is the smaHest (in the sense of inc1usion) subring of R to contain T; (T) is called the subring generated by T. 16. Let S be a subring of R, a ring with identity. For an arbitrary element a rt S, the subring generated by the set S u {a} is represented by (S, a). If a E cent R, establish tha t

20. Suppose that R is a ring with identity such that char R = n > O. If n is not prime, show that R has divisors of zero. 21. If R is a rillg which has no nonzero nilpotent elements, deduce that aH the idempotent elements of R belong to cent R. [Hint: If a2 = a, then (ara - arV = (ara - ra)2 = O for aH r E R.] 22. Assume that R is a ring with the property that a2 + a E cent R for every element a in R. Prove that R is necessarily a cornmutative ringo [Hint: Utilize the expression (a + W + (a + b) to show first that ab + ba lies in the center for aH a, b E R.] 23. Let (G, +) be a commutative group and R be the set ofall (group) homomorphisms of G into itself. The pointwise sum f + g and composition f o g of two functions f, g E R are defined by the usual rules

(f + g)(x) = f(x)

+ g(x),

= {ro +

rla

+ ... +

r.a"ln E Z+; r¡

E

S}.

(f o g)(x) = f(g(x»)

(XE G).

+, o) forms a ring. At the same time determine

Show that the resulting system (R, the invertible elements of R.

24. Let(G,') bea finite group (writtenmultiplicatively), say with elementsx¡, x 2 , and let R be an arbitrary ringo Consider the set R(G) of all formal sums

¿"

••• ,

x"'

(r¡ER).

r¡x i

i=l

Two such expressions are regarded as equal if they have the same coefficients .. Addition and multiplication can be defined in R(G) by taking

±r¡x¡ + ±s¡x¡ = ±(r¡ + s¡)x¡ (.± r¡x¡) [.± S¡X¡) = .± t¡x i=l

i=l

and

i=l

j,

,=1

l=1

(S, a)

15

PROBLEMS

1=1

where ti

=

¿

rjsk •

17. Let R be an arbitrary ring apd n E Z+. Ifthe set Sft is defined by

XJXk=Xf

= {a E Rlnka =

(The meaning of the last-written sum is that the surnmation is to be extended over aH subscripts j and k for which xjxk = x¡.) Prove that, with respect to these operations, R(G) constitutes a ring, the so-called group ring of G over R.

S.

O for sorne k> O},

determine whether S" is a subring of R .. 18. Establish the foHowing assertions concerning the characteristic of a ring R:

.

'

IDEALS ANO THEIROPERATIONS

TWO

17

, Example 2-1. For each integer a E Z"let (a) r.epresent the set consisting of all integral muItiples of a; that is,

(a)

{naln E Z};

The following relations confirm (a) to be an ideal ofthe ring ofintegers:

na

ma

(n - m)a,

m(mi) = (mn)a,

IDEALS AND THEIR OPERATIONS

A1though it is possibleto obtain sorne interesting conclusions conceming subrings, this concept,lifunrestricted, is too general for most purposes. To derive certain highly'desirable results (for instance, the fundamental iso1l'l:0rphism theorems), ad~tional assumptions that go beyond Definition 1-4 . must be imposed. Thus,in the present chapter we narrow the field and focus attention on a class of subrings with a stronger type of multiplícative elosure, namely, closure under multiplícation by an arbitrary ring element. ' DefinidoR 2-1. A subring I of the ring R is said to be a twa-sided ideal of R if and ónly if rE R anct a El imply both ra E I and ar E I. Viewed otherwise, Definition 2-1 asserts that whe~ever one of thé factors in a product belongs to I, then the product itself must be in l. (This may be roughly summarized by saying tbat theset I "captures" products.) Taking stock of Theorem 1-3, which gives a mini mal set of conditions to be a subring, our current definition of a two-sided ideal may be reformu, lated as follows. DefinitioR 2.;.2. Let I be a nonempty subset of a ring R. Then I is a two-sided ideal of R if and only if 1) .a, b E I imply a - b E I, and 2) r E R and a E I imply both products ra, ar E I.

n, n EZ.

In particular, since (2) = Ze,the ring of even integers forms an {deal of Z. " Notice, too, that (O) = {O} and (1) = Z.,¡ ;'.,' Example 2-2. Another illustration is furnished by map (X, R), th~ ring of ¡.' For a fixed ,l.,~ , element x E X, we denote by Ix the set of all mappings which take on the ... ,value O at . x: ,,',; .. )~?ppings froIn the set X into'the ring R (see Example 1-4).

~'.-

Ix

{f E map(X, R)/f(x)

= O}.

Now, choose J, g E Ix and hE map(X, R). From the définHion of the ring operations in map(X, R),

(f -g)(x)

= f(x)-g(x) = O-O = O,

while

(fh)(x) = f(x)h(x)

Oh (x)

O,

and, in a similar manner, (hf)(x) O. Thus, f - g, fh and hf all belong to Ix, which implies that Ix is an ideal. More generally, if S is any nonempty subset of X, then I = {f E map(X, R)lf(x)

= O for all x E S}

comprises an ideal of map(X, R). Since I nxesI.., we have a situation where. the intersection of ideals is once again an ideal. (Theorem 2-2shows that tbis is no accident.) '

If condition (2) of the aboye definition is weakened so as to require only that the product ra belongs to I for every choice of r E R and a E I, we are led to the notion of a left ideal; right ideals are defined in a symmetric way. Needless to say, if the ring R happens to be commutative (the most important case so far as we shall be concerned), then there is no distinction between left, right, and two-sided ideals.

Before presenting our next example, we derive a fact which, despite its apparent simplicity, will be frequently applied in the sequel.

In what follows, let us agree that theterm "ideal", unmodified, wilJ always mean two-sided ideal.

Proa! Let I be an ideal of R and suppose that there is sorne member a =1= O of I such that a-lexists in R. (The theorem is trivial when I = {O}.) Since'

CONVENTION

Before proceeding further, we pause to examine tbis concept by means of several specificexamples. 16

Theonim 2-1. If 1 is a proper (right, left, two-sided) ideal of a ring R with identity, then no element of I possesses a multiplicative inverse; that is, I n R* = 0.

I is closed under multiplication by arbitrary ring elements, it follows that 1 = a-la E l. By tile same reasoning, I contains r = rl for every r in R;

.

'

IDEALS ANO THEIROPERATIONS

TWO

17

, Example 2-1. For each integer a E Z"let (a) r.epresent the set consisting of all integral muItiples of a; that is,

(a)

{naln E Z};

The following relations confirm (a) to be an ideal ofthe ring ofintegers:

na

ma

(n - m)a,

m(mi) = (mn)a,

IDEALS AND THEIR OPERATIONS

A1though it is possibleto obtain sorne interesting conclusions conceming subrings, this concept,lifunrestricted, is too general for most purposes. To derive certain highly'desirable results (for instance, the fundamental iso1l'l:0rphism theorems), ad~tional assumptions that go beyond Definition 1-4 . must be imposed. Thus,in the present chapter we narrow the field and focus attention on a class of subrings with a stronger type of multiplícative elosure, namely, closure under multiplícation by an arbitrary ring element. ' DefinidoR 2-1. A subring I of the ring R is said to be a twa-sided ideal of R if and ónly if rE R anct a El imply both ra E I and ar E I. Viewed otherwise, Definition 2-1 asserts that whe~ever one of thé factors in a product belongs to I, then the product itself must be in l. (This may be roughly summarized by saying tbat theset I "captures" products.) Taking stock of Theorem 1-3, which gives a mini mal set of conditions to be a subring, our current definition of a two-sided ideal may be reformu, lated as follows. DefinitioR 2.;.2. Let I be a nonempty subset of a ring R. Then I is a two-sided ideal of R if and only if 1) .a, b E I imply a - b E I, and 2) r E R and a E I imply both products ra, ar E I.

n, n EZ.

In particular, since (2) = Ze,the ring of even integers forms an {deal of Z. " Notice, too, that (O) = {O} and (1) = Z.,¡ ;'.,' Example 2-2. Another illustration is furnished by map (X, R), th~ ring of ¡.' For a fixed ,l.,~ , element x E X, we denote by Ix the set of all mappings which take on the ... ,value O at . x: ,,',; .. )~?ppings froIn the set X into'the ring R (see Example 1-4).

~'.-

Ix

{f E map(X, R)/f(x)

= O}.

Now, choose J, g E Ix and hE map(X, R). From the définHion of the ring operations in map(X, R),

(f -g)(x)

= f(x)-g(x) = O-O = O,

while

(fh)(x) = f(x)h(x)

Oh (x)

O,

and, in a similar manner, (hf)(x) O. Thus, f - g, fh and hf all belong to Ix, which implies that Ix is an ideal. More generally, if S is any nonempty subset of X, then I = {f E map(X, R)lf(x)

= O for all x E S}

comprises an ideal of map(X, R). Since I nxesI.., we have a situation where. the intersection of ideals is once again an ideal. (Theorem 2-2shows that tbis is no accident.) '

If condition (2) of the aboye definition is weakened so as to require only that the product ra belongs to I for every choice of r E R and a E I, we are led to the notion of a left ideal; right ideals are defined in a symmetric way. Needless to say, if the ring R happens to be commutative (the most important case so far as we shall be concerned), then there is no distinction between left, right, and two-sided ideals.

Before presenting our next example, we derive a fact which, despite its apparent simplicity, will be frequently applied in the sequel.

In what follows, let us agree that theterm "ideal", unmodified, wilJ always mean two-sided ideal.

Proa! Let I be an ideal of R and suppose that there is sorne member a =1= O of I such that a-lexists in R. (The theorem is trivial when I = {O}.) Since'

CONVENTION

Before proceeding further, we pause to examine tbis concept by means of several specificexamples. 16

Theonim 2-1. If 1 is a proper (right, left, two-sided) ideal of a ring R with identity, then no element of I possesses a multiplicative inverse; that is, I n R* = 0.

I is closed under multiplication by arbitrary ring elements, it follows that 1 = a-la E l. By tile same reasoning, I contains r = rl for every r in R;

18

FIRST COURSE IN RINGS ANO IDEALS

that is, R S;; J, whence the equality J = R. This contradicts the hypothesis that J is a proper subset of R. Notice that, en route, we have also established Corollary. In a ring with identíty, no proper (right, left, two-sided) ideal contaíns the identity elemento Example 2...3. Thís example is given to show that the ring Mn(R#) of n x n matrices over the real numbers has no nontrivial ideals. As a notational device, let us define Eij to be the n x n matrix having 1 as its ijth entry and zeroes elsewhere. Now, suppose that J =1= {O} is any ideal of the ring Mn(R#). Then J must contain sorne nonzero matrix (aij), with, say, rsth entry a,s =1= O. Since J is a two-sided ideal, the product

Err(b¡) (a¡)Ess is a member of J, where the matrix (b;) is chosen to have the element a;. 1 down its main diagonal and zeroes everywhere else. As a result of al! the zero entries in the various factors, it is easy to verify that this product is .equal to E,s' Knowing thís, the relation Eu

= EúE,sE.l

(i, j = 1,2, ... , n)

implies tbat all n2 of the matrices Eu are contained. in 1. The clinching point is that the identity matrix (ou) can be written as (ou)

= Ell + E 22 + ... + E nn ,

which leads to the conelusion that (Oij) E J and, appealing to the above corollary, thatl = Mn(R#). In other words, Mn(R#) possesses no nonzero proper ideals, as assertt;d... As a matter of definhíon, let us call a ring R =1= {O} simple if R has no two-sided ideals other tbAn {O} and R. In the light of Example 2-4, the matrix ring Mn(R#) is a simple ringo We now take up some of the standard methods for constructing new ideals from given ones. To begin with simpler things: Theorem 2-2. Leí {J¡} be an arbitrary collection of (right, left, twos.ided) ideals of the ring R, where i ranges over sorne index set Then n Ji is also a (right, left, two-sided) ideal of R.

Proof. We give the proof for the case in whích the ideals are two-sided. First, observe that the intersection n Ji is nonempty, for each of the ideals Ji must contain the zero element of the ringo Suppose that the elements a, b E n Ji and r E R. Then a and b are members of Ji' where i varies over the indexing set. Inasmuch as JI is assumed to be an ideal of R, it follows that a - b, ar and ra alllie in the set Ji' But this is true for every value of

IDEALS ANO THEIR OPERATIONS

19

i, whence the elements a - b, ar and ra belong to n Ji' making n Ji an ideal of R. Consider, for the moment, an arbitrary ring Ji and a nonempty subset S of R. By the symbol (S) we sháll mean the set (S)

= n {JI S

S;;

J; J is an ideal of R}.

The collection of all ideal s which contain S is not ernpty, since the entire ring itself is an ideal containing any subset of R; thus, the set (S) exists and satisfies the inclusion S S;; (S). By virtue ofTheorem 2-2, (S) forms an ideal of R, lrnown as the ideal generated by the set S. It is noteworthy that whenever J is any ideal of R with S s;; J, then necessarily (S) S;; 1. For tbis rcason, one often speaks of (S) as being the smallest ideal of R to contain the set S. It should be apparent that corresponding remarks apply to the o'he-sided ideals generated by S. If S consists of a finite number of elements, sayal' a2 , ••• , an , then, the l.,. idcal'which t4ey generate is customarily denoted by (al' a2, ... , aJo Such an ideal is said to be finitely generated with the given eIernents al as its generators. An ideal (a) generated by just one ring eIement is termed a

principal ideal. A natural undertaking is to determine the precise form of the members of the various ideals (right, left, two-sided) generated by a single element, saya, of an arbitrary ring R. The right ideal generated by a is caBed a principal right ideal and is denoted by (a),. Being closed with respect to multiplication on the right, (a), necessarily contains al! products ar (r E R), as well as the elements na (n an integer), and, hence, ineludes their sum ar + na. (As usual, the notation na represents the n-fold sum of a.) It is a fairly simple matter to check that the set of elements of the form ar + na constitutes a right iGeal of R. Observe, too, that the elernent a is a member of the ideal, .since a = aO + la. These remarks make it clear that

(a),

=

{ar

+ nalrER; nE Z}.

When there is an identity element present, the term na becomes superfluous, for, in this setting, we rnay write the expression ar + na more simply as

ar

+ na

= ar

+ a(nl) =

a(r

+

nI)

ar',

where r ' = r + nI is some ring elemento Thus, the set (a), consists of all right multiples of a byelements of R. If R is a ring with identity, we shall frequently employ the more suggestive notation aR in place of (a),; that is,

(a)r = aR Similar remarks

bya.

{arlr E R}.

~pply, of course, to the principal left ideal (a)¡ generated

18

FIRST COURSE IN RINGS ANO IDEALS

that is, R S;; J, whence the equality J = R. This contradicts the hypothesis that J is a proper subset of R. Notice that, en route, we have also established Corollary. In a ring with identíty, no proper (right, left, two-sided) ideal contaíns the identity elemento Example 2...3. Thís example is given to show that the ring Mn(R#) of n x n matrices over the real numbers has no nontrivial ideals. As a notational device, let us define Eij to be the n x n matrix having 1 as its ijth entry and zeroes elsewhere. Now, suppose that J =1= {O} is any ideal of the ring Mn(R#). Then J must contain sorne nonzero matrix (aij), with, say, rsth entry a,s =1= O. Since J is a two-sided ideal, the product

Err(b¡) (a¡)Ess is a member of J, where the matrix (b;) is chosen to have the element a;. 1 down its main diagonal and zeroes everywhere else. As a result of al! the zero entries in the various factors, it is easy to verify that this product is .equal to E,s' Knowing thís, the relation Eu

= EúE,sE.l

(i, j = 1,2, ... , n)

implies tbat all n2 of the matrices Eu are contained. in 1. The clinching point is that the identity matrix (ou) can be written as (ou)

= Ell + E 22 + ... + E nn ,

which leads to the conelusion that (Oij) E J and, appealing to the above corollary, thatl = Mn(R#). In other words, Mn(R#) possesses no nonzero proper ideals, as assertt;d... As a matter of definhíon, let us call a ring R =1= {O} simple if R has no two-sided ideals other tbAn {O} and R. In the light of Example 2-4, the matrix ring Mn(R#) is a simple ringo We now take up some of the standard methods for constructing new ideals from given ones. To begin with simpler things: Theorem 2-2. Leí {J¡} be an arbitrary collection of (right, left, twos.ided) ideals of the ring R, where i ranges over sorne index set Then n Ji is also a (right, left, two-sided) ideal of R.

Proof. We give the proof for the case in whích the ideals are two-sided. First, observe that the intersection n Ji is nonempty, for each of the ideals Ji must contain the zero element of the ringo Suppose that the elements a, b E n Ji and r E R. Then a and b are members of Ji' where i varies over the indexing set. Inasmuch as JI is assumed to be an ideal of R, it follows that a - b, ar and ra alllie in the set Ji' But this is true for every value of

IDEALS ANO THEIR OPERATIONS

19

i, whence the elements a - b, ar and ra belong to n Ji' making n Ji an ideal of R. Consider, for the moment, an arbitrary ring Ji and a nonempty subset S of R. By the symbol (S) we sháll mean the set (S)

= n {JI S

S;;

J; J is an ideal of R}.

The collection of all ideal s which contain S is not ernpty, since the entire ring itself is an ideal containing any subset of R; thus, the set (S) exists and satisfies the inclusion S S;; (S). By virtue ofTheorem 2-2, (S) forms an ideal of R, lrnown as the ideal generated by the set S. It is noteworthy that whenever J is any ideal of R with S s;; J, then necessarily (S) S;; 1. For tbis rcason, one often speaks of (S) as being the smallest ideal of R to contain the set S. It should be apparent that corresponding remarks apply to the o'he-sided ideals generated by S. If S consists of a finite number of elements, sayal' a2 , ••• , an , then, the l.,. idcal'which t4ey generate is customarily denoted by (al' a2, ... , aJo Such an ideal is said to be finitely generated with the given eIernents al as its generators. An ideal (a) generated by just one ring eIement is termed a

principal ideal. A natural undertaking is to determine the precise form of the members of the various ideals (right, left, two-sided) generated by a single element, saya, of an arbitrary ring R. The right ideal generated by a is caBed a principal right ideal and is denoted by (a),. Being closed with respect to multiplication on the right, (a), necessarily contains al! products ar (r E R), as well as the elements na (n an integer), and, hence, ineludes their sum ar + na. (As usual, the notation na represents the n-fold sum of a.) It is a fairly simple matter to check that the set of elements of the form ar + na constitutes a right iGeal of R. Observe, too, that the elernent a is a member of the ideal, .since a = aO + la. These remarks make it clear that

(a),

=

{ar

+ nalrER; nE Z}.

When there is an identity element present, the term na becomes superfluous, for, in this setting, we rnay write the expression ar + na more simply as

ar

+ na

= ar

+ a(nl) =

a(r

+

nI)

ar',

where r ' = r + nI is some ring elemento Thus, the set (a), consists of all right multiples of a byelements of R. If R is a ring with identity, we shall frequently employ the more suggestive notation aR in place of (a),; that is,

(a)r = aR Similar remarks

bya.

{arlr E R}.

~pply, of course, to the principal left ideal (a)¡ generated

20

IDEA,LS AND THEIR O,Í>ERATIONS

FIRST COURSE IN RINGS AND IDEALS

Then 11 + 12 + ... + In is likewise an ideal of R and is the smil.llest ideal of R which contains every 1,; phrased in another way, 11 + 12 + ... + In is the ideal generated by the union 11 u I2t U ... U In' In the special case of two ideals 1 and J, our definitiQn reduces to

As a general cornment, observe that the products ar (r E R) comprise the set of elements'of a right ideal of R even when the ring does not possess an identity. The difficulty, however, is that this ideal need not contain a itse1f. With regard to the two-sided ideal (a) generated by a, the situation is more complicated. Certainly the elements ras, ra, as and na must all belong to the ideal (a) for every choice of r, s E R and n E Z. In general, the sum of two elements ras and r'as' is no longer of the same form, so that, in order to have c10sure under addition, any finite sum rias i, where r" si E R, is also required tobe in (a). The reader will experience no difficulty in'showing that the principal ideal generated bya is given by ,',

=

{na

+ ra + as +

¿

More generally, let {J;} be an arbitrary indexed collection of ideals of 1, and is the ideal of R whose members are aH possible finite sumS of elements from the various ideals I i : \L '¿ 1, = {¿ afia, ~:jJ.

R. , The sum of this collection may be deno~ed by

.

r,as;!r,s,r"s,ER; nEZ}.,'

finite

+ a2 + ... + a"la,EIJ,

I i ; aH but a

,

+

Proof If 1 = {O}, the theorem is triviaHy true, since' the zero ideal {O} is the principal ideal generated by O. Suppose then tbat 1 does not consist ofthe zero element alone. Now, ifm E 1, -m also Iles in 1, so that the set 1 contains positive integers. Let n designate the least positive integer in 1. As 1 forms a~ ideal of Z, each integral multiple of n must belong to 1, whence

{al

E

¿

Theorem 2-3. The ring Z of integers is a principal ideal ring; in fact, if 1 is an ideal of Z, then 1 = (n) for sorne nonnegative integer n.

'=

{¿a,la,

=

terms. Just as n Ii can be interpreted as the largest ideal'of R contained in every I i, the sum Ii supplies the dual notion of the smallest ideal containing every Ii' U R = 11 12 + ... + In' then each element X E R can be expressed in the form x = al + a 2 + ... + an, where ai lies in J f . ' There is no guarantee, however, that tbis representation of x is unique. To eilsure that every member of R IS uniquely expressíble as a sum of elements from the ideals I i , an auxíliary definitíon ís required. .

The foHowing theorem furnishes an example of such rings.

+ ... + In

..

finíte¡~Ú.llnber ofthe a¡ ~~e O}, "':.;: where it ís understood that ¿ represents an arbitrar y sum whh orie or more ¿I,

Definítion 2-3. A ring R is said to be a principal ideal ring if eve~y ideal 1 of R is of the form 1 = (a) for sorne a E R.

11 +1 2

finite··

¿

¿

Let us noW describe certain binary operations on the set of aH ideals of R. (Similar consideratíons apply to the sets of right, and left ideal s, but for economy of effort we concentrate on two-sided ideals.) Given a finite number of ideals 11,12 , .,. ,1" of the ring R, one defines their sum in the natural way:

¿

The reader will take careto remember that, although {J,} may be an infinite family of ideals, only finíte sums of eleménts of R are involved in the definition above. An alternative descriptio~o'f 1, could be 'given by

In case R happens to haye an identity, this description of (a) reduées to the set of aH finite sums r , a s , . ' : , " A particularIy important typeof ring is a principal ideal ririg" wbich we now d e f i n e . : - '

~~L ' To establish the inc1usion 1 ~ (n), let k be an arbitrary element of 1. By the di:vision aigorithm there existintegers q and r for which k = qn + r, with O ~ r < n. Since k and qn are both members of 1, it foHows that r = k - qn E 1. Ifr > O, we would have a contradiction to the assumption that n is the smaHest positive integer in 1. Accordingly, r ~ O and k = qn E (n). Thus, only multiples of n belong to 1, implying that 1 ~ (n). The two inc1usions show that 1 = (n) and the argument is complete.

+ J = {a + b/aEI; bEJ}.

1

¿

(a)

21

. Definition ~-4. Let 11 , 12 , ... , In be ideals <;>fthe ring R. We caH R the internal direct sum off 1, 12 , ... , In' and write R = 11 EB 12 EB ... EB In' provided that .¡

a) R = 11 b) Ii n (JI

+ 12 + ... + In' and' + ... + I i- 1 + Ii+1 + ... + In) =

{O} for each i.

As was heralded by out remarks, we are now in a position to prove Theorem 2-4. Let 11 , 12 , ... , In be ideals of the ring R. following statements are equivalent:

Then the

1) R is the rntermiJ direct sum of 1 1 ,1 2 , ... , In' 2) Each element x of R is uniquely expressible in the form

x

=

al

+ a2 + ... + an, where ai E I¡.

Proof There is no loss in confining ourselves to thecase n = 2; the general arg~ment proceeds along similar lines. Webegin by assumíng that R = 11 EB 12 , Suppose further that an element x E R has two representatíons

x = al

+ b1 = a2 +

Q2

(a i EI 1, b¡EI 2 ).

20

IDEA,LS AND THEIR O,Í>ERATIONS

FIRST COURSE IN RINGS AND IDEALS

Then 11 + 12 + ... + In is likewise an ideal of R and is the smil.llest ideal of R which contains every 1,; phrased in another way, 11 + 12 + ... + In is the ideal generated by the union 11 u I2t U ... U In' In the special case of two ideals 1 and J, our definitiQn reduces to

As a general cornment, observe that the products ar (r E R) comprise the set of elements'of a right ideal of R even when the ring does not possess an identity. The difficulty, however, is that this ideal need not contain a itse1f. With regard to the two-sided ideal (a) generated by a, the situation is more complicated. Certainly the elements ras, ra, as and na must all belong to the ideal (a) for every choice of r, s E R and n E Z. In general, the sum of two elements ras and r'as' is no longer of the same form, so that, in order to have c10sure under addition, any finite sum rias i, where r" si E R, is also required tobe in (a). The reader will experience no difficulty in'showing that the principal ideal generated bya is given by ,',

=

{na

+ ra + as +

¿

More generally, let {J;} be an arbitrary indexed collection of ideals of 1, and is the ideal of R whose members are aH possible finite sumS of elements from the various ideals I i : \L '¿ 1, = {¿ afia, ~:jJ.

R. , The sum of this collection may be deno~ed by

.

r,as;!r,s,r"s,ER; nEZ}.,'

finite

+ a2 + ... + a"la,EIJ,

I i ; aH but a

,

+

Proof If 1 = {O}, the theorem is triviaHy true, since' the zero ideal {O} is the principal ideal generated by O. Suppose then tbat 1 does not consist ofthe zero element alone. Now, ifm E 1, -m also Iles in 1, so that the set 1 contains positive integers. Let n designate the least positive integer in 1. As 1 forms a~ ideal of Z, each integral multiple of n must belong to 1, whence

{al

E

¿

Theorem 2-3. The ring Z of integers is a principal ideal ring; in fact, if 1 is an ideal of Z, then 1 = (n) for sorne nonnegative integer n.

'=

{¿a,la,

=

terms. Just as n Ii can be interpreted as the largest ideal'of R contained in every I i, the sum Ii supplies the dual notion of the smallest ideal containing every Ii' U R = 11 12 + ... + In' then each element X E R can be expressed in the form x = al + a 2 + ... + an, where ai lies in J f . ' There is no guarantee, however, that tbis representation of x is unique. To eilsure that every member of R IS uniquely expressíble as a sum of elements from the ideals I i , an auxíliary definitíon ís required. .

The foHowing theorem furnishes an example of such rings.

+ ... + In

..

finíte¡~Ú.llnber ofthe a¡ ~~e O}, "':.;: where it ís understood that ¿ represents an arbitrar y sum whh orie or more ¿I,

Definítion 2-3. A ring R is said to be a principal ideal ring if eve~y ideal 1 of R is of the form 1 = (a) for sorne a E R.

11 +1 2

finite··

¿

¿

Let us noW describe certain binary operations on the set of aH ideals of R. (Similar consideratíons apply to the sets of right, and left ideal s, but for economy of effort we concentrate on two-sided ideals.) Given a finite number of ideals 11,12 , .,. ,1" of the ring R, one defines their sum in the natural way:

¿

The reader will take careto remember that, although {J,} may be an infinite family of ideals, only finíte sums of eleménts of R are involved in the definition above. An alternative descriptio~o'f 1, could be 'given by

In case R happens to haye an identity, this description of (a) reduées to the set of aH finite sums r , a s , . ' : , " A particularIy important typeof ring is a principal ideal ririg" wbich we now d e f i n e . : - '

~~L ' To establish the inc1usion 1 ~ (n), let k be an arbitrary element of 1. By the di:vision aigorithm there existintegers q and r for which k = qn + r, with O ~ r < n. Since k and qn are both members of 1, it foHows that r = k - qn E 1. Ifr > O, we would have a contradiction to the assumption that n is the smaHest positive integer in 1. Accordingly, r ~ O and k = qn E (n). Thus, only multiples of n belong to 1, implying that 1 ~ (n). The two inc1usions show that 1 = (n) and the argument is complete.

+ J = {a + b/aEI; bEJ}.

1

¿

(a)

21

. Definition ~-4. Let 11 , 12 , ... , In be ideals <;>fthe ring R. We caH R the internal direct sum off 1, 12 , ... , In' and write R = 11 EB 12 EB ... EB In' provided that .¡

a) R = 11 b) Ii n (JI

+ 12 + ... + In' and' + ... + I i- 1 + Ii+1 + ... + In) =

{O} for each i.

As was heralded by out remarks, we are now in a position to prove Theorem 2-4. Let 11 , 12 , ... , In be ideals of the ring R. following statements are equivalent:

Then the

1) R is the rntermiJ direct sum of 1 1 ,1 2 , ... , In' 2) Each element x of R is uniquely expressible in the form

x

=

al

+ a2 + ... + an, where ai E I¡.

Proof There is no loss in confining ourselves to thecase n = 2; the general arg~ment proceeds along similar lines. Webegin by assumíng that R = 11 EB 12 , Suppose further that an element x E R has two representatíons

x = al

+ b1 = a2 +

Q2

(a i EI 1, b¡EI 2 ).

22

FIRST COURSE IN RINGS ANO IDEALS

IDEALS ANO THEIR OPERATIONS

Then al - a 2 = b2 - bl' But the left-hand side of this last equation lies in 1 1 , while the right-hand side is in 12 , so that both sides be long to 11 n 12 = {O}. Itfollowsthata 1 - a 2 = b2 - b1 = 0,ora 1 = a2 ,b 1 = b2 • In other words, x is uniquely representable as a sum a + b, a E 1 l' b E 12 , Conversely, assume that assertion (2) holds and, that the element x E 11 n 12 , We may then express x in two different ways as the sum of an, element in 11 and an element in 12 ; namely, x = x + O (here x E 1 1 and OE 12 ) and x = O + x (here OE 1 1 and x E 12 ), The uniqueness assumption of(2) implies that x = O, in'consequence ofwhich 1 1 n 12 = {O}; hence, R = 11 EB 12 , This completes the proof of the theorem. We now come to a less elementary, but extremely useful,notion; namely, the product of ideals. Once again, assume that 1 and J are two ideals of the ring R. To be consistent with our earlier definition of the sum 1 + J, we should define fue product 'IJ to be the collection of all simple products ab, where a E 1 and b EJ. Unfortunately, the resulting set fails to form an ideal. (Why?) To counter this difficulty, we instead take the elements of IJ to be all possible ,finite sums of simple products; stated explicitly, IJ = O=a¡b¡la¡El; b¡EJ}. finile

With this definition IJ indeed becomes an ideal of R. For, suppose that x, y E IJ and r E R; then,

23

In this connection, it is important to observe that

1

::2

12

::2

13

::2 '"

::2

1"

::2

forms a decreasing chain of ideals.

Remark. If 1 is a right ideal and S a nonempty subset of the ring R, then ' SI

=

{L a¡r¡la¡ E S; r¡ El} finite

forms a right ideal of R. In particular, if S = {a}, then al (a notation we prefer to {a} 1) is gi ven by

al

=

{arlr E I}.

Analogous statements can be made when 1 is a left ideal of R, but not, of course, a two-sided ideal. The last ideal-theoretic operation which we wish to consider is that of the quotient (or residual), defined below.

Definition 2-5. Let 1 and J be two ideal s of the ring R. The right (left) quotient of 1 by J, denoted by the symbol 1 :r J (I :1 J), consists of all elements a E R such that aJ S;; 1 (:la S;; 1). In the event R is a commutative ring, we simply writel: J. It is by no means obvious that the set

where the a¡ and a; are in 1, and the b¡ and b; are in J. From this we obtain

x - y rx

=

=

a 1b 1 + ... + anb n + (-a'1)b'1' + ... (ra 1)b 1 + (ra 2 )b 2 + ... + (ra~~bp.

+

(-a~)b~,

Now, the elements -a; and raí necessarily lie in 1, so that x - y and rx E IJ; likewise, xr E IJ, making IJ an ideal of R. In point of fact, JJ is just the ideal generated by the set Qf all products ab, a E 1, bE J. There is no difficulty in extending the a1:ibve temarks to any finite number of ideals 11> 12 , ••• , In of the ring R. A'moment's thought shows that the product 11 12 ... In is the ideal consisting of finite sums of terms of the form a 1a2 ... an' with a¡ in 1¡. (It is perhaps appropriate to point out that, because of the assodative law for multiplication in R, the notation 1112 ,., In is unambiguous.) A special case irnmediately presents itself: namely, the situation where all the ideal s are alike, say equaJ to the ideal l. Rere, we see that 1" is the set of finite sums of products of n elements from 1:

1"

= {L..,¡ "

finite

a·11 a·12 ... a·In la.Ik El}.

1:.1 = {aERlaJ

S;;

I}

actually forms an ideal of R, whenever 1 and J are ideals. To verify this, suppose that the elements a, b El:, J and rE R. For any x E J, we clearly have (a - b)x = ax - bx E 1, since ax and bx both belong to 1 by definition. This establishes the inclusion (a - b)J S;; 1, which in turn signifies that a - bE 1 :r J. Likewise, the relations raJ S;; rl ~ 1 and arJ S;; aJ S;; 1 imply that ra, ar E r:r J. In consequence, 1:, J coinprises an ideal of R in its own right, and that 1 :1 J is also an ideal follows similarly. The purpose of the coming theorem is to point out the connection between the quotient ideal and the operations defined previously. This result, !llthough it might seem to be quite special, will serve us in good stead when we develop the theory of Noetherian rings. Theorem 2-5. The following relations hold for ideals in a ring R (capital letters indicate ideals of R): 1) (n 1¡}:.1 = n (li :.1), 2) 1 :, L J¡ = n (I :, J¡), 3) 1 :r(JK)'= (1 :rK) :rJ.

22

FIRST COURSE IN RINGS ANO IDEALS

IDEALS ANO THEIR OPERATIONS

Then al - a 2 = b2 - bl' But the left-hand side of this last equation lies in 1 1 , while the right-hand side is in 12 , so that both sides be long to 11 n 12 = {O}. Itfollowsthata 1 - a 2 = b2 - b1 = 0,ora 1 = a2 ,b 1 = b2 • In other words, x is uniquely representable as a sum a + b, a E 1 l' b E 12 , Conversely, assume that assertion (2) holds and, that the element x E 11 n 12 , We may then express x in two different ways as the sum of an, element in 11 and an element in 12 ; namely, x = x + O (here x E 1 1 and OE 12 ) and x = O + x (here OE 1 1 and x E 12 ), The uniqueness assumption of(2) implies that x = O, in'consequence ofwhich 1 1 n 12 = {O}; hence, R = 11 EB 12 , This completes the proof of the theorem. We now come to a less elementary, but extremely useful,notion; namely, the product of ideals. Once again, assume that 1 and J are two ideals of the ring R. To be consistent with our earlier definition of the sum 1 + J, we should define fue product 'IJ to be the collection of all simple products ab, where a E 1 and b EJ. Unfortunately, the resulting set fails to form an ideal. (Why?) To counter this difficulty, we instead take the elements of IJ to be all possible ,finite sums of simple products; stated explicitly, IJ = O=a¡b¡la¡El; b¡EJ}. finile

With this definition IJ indeed becomes an ideal of R. For, suppose that x, y E IJ and r E R; then,

23

In this connection, it is important to observe that

1

::2

12

::2

13

::2 '"

::2

1"

::2

forms a decreasing chain of ideals.

Remark. If 1 is a right ideal and S a nonempty subset of the ring R, then ' SI

=

{L a¡r¡la¡ E S; r¡ El} finite

forms a right ideal of R. In particular, if S = {a}, then al (a notation we prefer to {a} 1) is gi ven by

al

=

{arlr E I}.

Analogous statements can be made when 1 is a left ideal of R, but not, of course, a two-sided ideal. The last ideal-theoretic operation which we wish to consider is that of the quotient (or residual), defined below.

Definition 2-5. Let 1 and J be two ideal s of the ring R. The right (left) quotient of 1 by J, denoted by the symbol 1 :r J (I :1 J), consists of all elements a E R such that aJ S;; 1 (:la S;; 1). In the event R is a commutative ring, we simply writel: J. It is by no means obvious that the set

where the a¡ and a; are in 1, and the b¡ and b; are in J. From this we obtain

x - y rx

=

=

a 1b 1 + ... + anb n + (-a'1)b'1' + ... (ra 1)b 1 + (ra 2 )b 2 + ... + (ra~~bp.

+

(-a~)b~,

Now, the elements -a; and raí necessarily lie in 1, so that x - y and rx E IJ; likewise, xr E IJ, making IJ an ideal of R. In point of fact, JJ is just the ideal generated by the set Qf all products ab, a E 1, bE J. There is no difficulty in extending the a1:ibve temarks to any finite number of ideals 11> 12 , ••• , In of the ring R. A'moment's thought shows that the product 11 12 ... In is the ideal consisting of finite sums of terms of the form a 1a2 ... an' with a¡ in 1¡. (It is perhaps appropriate to point out that, because of the assodative law for multiplication in R, the notation 1112 ,., In is unambiguous.) A special case irnmediately presents itself: namely, the situation where all the ideal s are alike, say equaJ to the ideal l. Rere, we see that 1" is the set of finite sums of products of n elements from 1:

1"

= {L..,¡ "

finite

a·11 a·12 ... a·In la.Ik El}.

1:.1 = {aERlaJ

S;;

I}

actually forms an ideal of R, whenever 1 and J are ideals. To verify this, suppose that the elements a, b El:, J and rE R. For any x E J, we clearly have (a - b)x = ax - bx E 1, since ax and bx both belong to 1 by definition. This establishes the inclusion (a - b)J S;; 1, which in turn signifies that a - bE 1 :r J. Likewise, the relations raJ S;; rl ~ 1 and arJ S;; aJ S;; 1 imply that ra, ar E r:r J. In consequence, 1:, J coinprises an ideal of R in its own right, and that 1 :1 J is also an ideal follows similarly. The purpose of the coming theorem is to point out the connection between the quotient ideal and the operations defined previously. This result, !llthough it might seem to be quite special, will serve us in good stead when we develop the theory of Noetherian rings. Theorem 2-5. The following relations hold for ideals in a ring R (capital letters indicate ideals of R): 1) (n 1¡}:.1 = n (li :.1), 2) 1 :, L J¡ = n (I :, J¡), 3) 1 :r(JK)'= (1 :rK) :rJ.

24

IDEALS AND THEIR OPERA TIONS

FIRST COURSE IN RINGS AND IDEALS

Proo! Concerning (1), we have (n 1J:rJ

= =

{a

E

RlaJ ~ n 1¡}

=

{a

E

RlaJ ~ 1¡ for all

i}

n{aERlaJ ~ 1¡} c= n(l¡:r J )·

. With an eye to proving(2), notice that the inc1usion J¡ ~ ~ J¡ implies a~~ J;) ~ 1 if and only if aJ¡ ~ 1 for all i; thus,

25

often referred to as the pseudo-in verse of a. In the commutative case, ,the equation aa'a = a may, o'f course, be written as a2a' = a. . The result which we have in mind now follows. Theorem 2-6. Let 1 be an ideal of the regular ring R. Then any ideal J of 1 is likewise an ideal of R. .

. ,

Remark. Simila~ results,:h6Id for left quotients; the sole difference being that, instead of (3),'one riD\V has 1 :¡ (J K) = (1 :¡ J) :1 K.

Proof. To start, notice that 1 itself may be regarded as a regular ringo Indeed, if a EJ, then aa'a = a for sorne a' in R. Setting b = a'aa', the element b b~longs to 1 and has the property that ,"'. aba = a(a'aa')a = .(aa1a)a'a = aa'a = a. Our aim :is to show that whenever a E J ~ 1 and rE R, then both ar and ra lie in; We already know that ar El; hence, by the above,there exists an element x in 1 for which arxar = aro Sínce rxar is a member of 1 and J ifa:ssumed to be an idea16f 1, it follows that the product a(rxar) must belong,t6, J, or, equivalently, W'E J. A symmetric arg~ment confirms '. that ra E J.'\,

This may be a good pl~ce to observe that if 1 is an ide~l of. the ring R and J is an ideal of 1, then J need not be 'an ideal of the enhre rmg. R. For an illustration, we tum to the ring map R # and l~t R be the ~ubnng consisting of all continuous functions from, R # into ltself. Conslder the sets

Although Definition 2-6 appears to have a somewhat artificial air, we might remark that the set of alllinear transfonrtations on a finite dimensional vector space over a field forms a regular ring (Problem 20, Chapter 9). This in itself would amply justify the study of such rings.

1 = (filfE R;f(O) = O}, J = {ji2 + ni21fE R;f(O) = O; n E Z},

We now turn our attention to functions between rings arid, more specifically, to functions which preserve both the ring operations.

l: r ~,J¡

= fa E Rla(~ J;)~ 1} = {q:E RlaJ¡' ~ Ifor all i}

n (1 :r J¡).

.

Confirmation of the final' assertion follows from l: r (JK)

= {a~,Rla(JK) ~J} = {aERI(aJ)K ~ 1}' = {~ERlaJ ~ l: r K} = (1 :r K) :r J.

where i denotes the identity function on R # (that is, i(x) = x for all x E R #). A routine calculation verifies that J is an ideal of 1,which, in turn, forms an 2 2 ideal ofR. However, J fails to be an ideal of R, since i E J, while ti .~ J. (The symbol .1 is used in this setting to represent the constant funch.on 2 whose value a~ each real number is l) We as sume that ti E J and denve a contradiction. Then, ti 2 = fi 2 + ni 2 for a suitable choice of f E R and n ~ Z, with feO) = O. In consequence, fi2 = (t - n)i 2, implying that f(x) = t - n.=I= O for ;very O =1= x E R~; in other words, f is a nonzero constant functlOn on R - {O}. But tbis obviously violates the continuity off at O. . . A condition which will ensure that J is also an Ideal of R IS to take R to be a regular ring, a notion introduced by Von Neumann [52]. Definition 2-6. A ring R is said to be regular if for each element a E R there exists sorne a' E R such that aa' a = a.

If the ele~ent a happens to have a multiplicative i~verse, t~en t~e 1 regularity condition is satisfied by setting a' = a- ; in Vlew of thlS, a' IS

. ~

' I

r

Definition 2-7. Let R and R' be two rings. By a (ring) homomorphism, or homomorphic mapping, from R ínto R' is meant a functionf: R --+ R'

such that

f(a+ b) = fea)

+ f(b), f(ab)

= f(a)f(b)

,'" .'. for every pair of elements a, b E R. A homomorphísm which is also one-to-one as a map on the underlying sets is called an isomorphism. We emphasize that the + arid . occurring on the left-hand sides of the equations in Definition 2-7 are those of R, whereas the + and . occurring on the right-hand sídes are those of R'. This use of the same symbols for tbe operations of addition and multiplication in two different rings should cause no ambiguityif the reader attends c10sely to the context in which the notation is employed. If f is a homomorphism of R into R', then the image f(R) of R under f will be called the homomorphic image of R. When R = R', so that the two rings are the same, we say that f is a homomorphism of R into itself. In this connection, a homomorphism of R into itself is frequently referred to as an endomorphism of the ring R or, if an isomorphísm onto R, an automorphísm of R.

24

IDEALS AND THEIR OPERA TIONS

FIRST COURSE IN RINGS AND IDEALS

Proo! Concerning (1), we have (n 1J:rJ

= =

{a

E

RlaJ ~ n 1¡}

=

{a

E

RlaJ ~ 1¡ for all

i}

n{aERlaJ ~ 1¡} c= n(l¡:r J )·

. With an eye to proving(2), notice that the inc1usion J¡ ~ ~ J¡ implies a~~ J;) ~ 1 if and only if aJ¡ ~ 1 for all i; thus,

25

often referred to as the pseudo-in verse of a. In the commutative case, ,the equation aa'a = a may, o'f course, be written as a2a' = a. . The result which we have in mind now follows. Theorem 2-6. Let 1 be an ideal of the regular ring R. Then any ideal J of 1 is likewise an ideal of R. .

. ,

Remark. Simila~ results,:h6Id for left quotients; the sole difference being that, instead of (3),'one riD\V has 1 :¡ (J K) = (1 :¡ J) :1 K.

Proof. To start, notice that 1 itself may be regarded as a regular ringo Indeed, if a EJ, then aa'a = a for sorne a' in R. Setting b = a'aa', the element b b~longs to 1 and has the property that ,"'. aba = a(a'aa')a = .(aa1a)a'a = aa'a = a. Our aim :is to show that whenever a E J ~ 1 and rE R, then both ar and ra lie in; We already know that ar El; hence, by the above,there exists an element x in 1 for which arxar = aro Sínce rxar is a member of 1 and J ifa:ssumed to be an idea16f 1, it follows that the product a(rxar) must belong,t6, J, or, equivalently, W'E J. A symmetric arg~ment confirms '. that ra E J.'\,

This may be a good pl~ce to observe that if 1 is an ide~l of. the ring R and J is an ideal of 1, then J need not be 'an ideal of the enhre rmg. R. For an illustration, we tum to the ring map R # and l~t R be the ~ubnng consisting of all continuous functions from, R # into ltself. Conslder the sets

Although Definition 2-6 appears to have a somewhat artificial air, we might remark that the set of alllinear transfonrtations on a finite dimensional vector space over a field forms a regular ring (Problem 20, Chapter 9). This in itself would amply justify the study of such rings.

1 = (filfE R;f(O) = O}, J = {ji2 + ni21fE R;f(O) = O; n E Z},

We now turn our attention to functions between rings arid, more specifically, to functions which preserve both the ring operations.

l: r ~,J¡

= fa E Rla(~ J;)~ 1} = {q:E RlaJ¡' ~ Ifor all i}

n (1 :r J¡).

.

Confirmation of the final' assertion follows from l: r (JK)

= {a~,Rla(JK) ~J} = {aERI(aJ)K ~ 1}' = {~ERlaJ ~ l: r K} = (1 :r K) :r J.

where i denotes the identity function on R # (that is, i(x) = x for all x E R #). A routine calculation verifies that J is an ideal of 1,which, in turn, forms an 2 2 ideal ofR. However, J fails to be an ideal of R, since i E J, while ti .~ J. (The symbol .1 is used in this setting to represent the constant funch.on 2 whose value a~ each real number is l) We as sume that ti E J and denve a contradiction. Then, ti 2 = fi 2 + ni 2 for a suitable choice of f E R and n ~ Z, with feO) = O. In consequence, fi2 = (t - n)i 2, implying that f(x) = t - n.=I= O for ;very O =1= x E R~; in other words, f is a nonzero constant functlOn on R - {O}. But tbis obviously violates the continuity off at O. . . A condition which will ensure that J is also an Ideal of R IS to take R to be a regular ring, a notion introduced by Von Neumann [52]. Definition 2-6. A ring R is said to be regular if for each element a E R there exists sorne a' E R such that aa' a = a.

If the ele~ent a happens to have a multiplicative i~verse, t~en t~e 1 regularity condition is satisfied by setting a' = a- ; in Vlew of thlS, a' IS

. ~

' I

r

Definition 2-7. Let R and R' be two rings. By a (ring) homomorphism, or homomorphic mapping, from R ínto R' is meant a functionf: R --+ R'

such that

f(a+ b) = fea)

+ f(b), f(ab)

= f(a)f(b)

,'" .'. for every pair of elements a, b E R. A homomorphísm which is also one-to-one as a map on the underlying sets is called an isomorphism. We emphasize that the + arid . occurring on the left-hand sides of the equations in Definition 2-7 are those of R, whereas the + and . occurring on the right-hand sídes are those of R'. This use of the same symbols for tbe operations of addition and multiplication in two different rings should cause no ambiguityif the reader attends c10sely to the context in which the notation is employed. If f is a homomorphism of R into R', then the image f(R) of R under f will be called the homomorphic image of R. When R = R', so that the two rings are the same, we say that f is a homomorphism of R into itself. In this connection, a homomorphism of R into itself is frequently referred to as an endomorphism of the ring R or, if an isomorphísm onto R, an automorphísm of R.

26

FIRST COURSE IN RINGS AND IDEALS

For future use, we shalllabel the set of all hornornorphisrns frorn the ring R into the ring R' by the syrnbol horn(R, R'). In the event that R = R', the sirnpler notation horn R will be used in place of horn(R, R). (Sorne authors prefer to write end R, for endornorphisrn, in place of horn R; both notations have a certain suggestive power and it reduces to a rnatter of personal preference.) A knowledge of a few sirnple-rninded exarnples will help to fix ideas. Example 2-4. Let R and R' be arbitrary rings andf: R --> R' be the function which sends each elernent of R to the zero elernent of R'. Then, f(a

+ b) =

f(ab)

O = O + O = f(a)

= O = OO = f(a)

+ f(b),

(a, b E R),

f(b)

so thatfis a hornornorphic rnapping. This particular rnapping, the so-called trivial homomorphism, iS.the only constant function which satisfies Definition

2-7. Example 2-5. Consider the ring Z ofintegers and the ring Zn of integers modulo n. Definef: Z --> Zn by takingf(a) = [a]; that is, rnap each integer into the congruence class containing it. Thatfis a hornornorphisrn follows directly frorn the definition of the operations üi Zn: f(a + b) = [a + b] f(ab)

=

[ab]

=

=

[a] +n [b]

[aln[b]

=

f(a) +nf(b),

= f(a)·nf(b).

Example 2-6. In the ring rnap(X, R), define La to be the function which assigns to each fE rnap(X, R) its value at a fixed elernent a E X; in other words, La(J) =:= f(a): Then La is a hornornorphisrn frorn rnap(X, R) into R, known as the evaluation homomorphism at a. We need only observe that

Proo! Frorn f(O) = f(O + O) == f(O) + f(O), we obtain f(O) = O. The factthatf(a) +f(-a) =f(a + (-a») =f(O) = Oyieldsf(-a) = -f(a). As regards (3), let the elernent a E R satisfy f(a) = 1; then,j(l) = f(a)f(l) = f(a1) = f(a) = 1. Finally, the equationf(a)f(a- 1 ) = f(aa- 1 ) = f(l) = 1 shows thatf(a)-l = f(a- 1 ), whenever a E R has a rnultiplicative inverse.

Two cornrnents regarding part (3) of the aboye theorern are in order: First, it is evident that f(a)l

= (Jg)(a) = f(a)g(a)

f(a - b)

f(a)

=

f(a1) = f(a)f(l)

=

f(a)

+ f( -

b)

=

f(a) - f(b).

In short, any ring hornornorphisrn preserves differences as well as sums and products. The next theorem indicates the algebraic nature of direct and inverse images of subrings under homomorphisms. Among other things, we shall see that iffis a homomorphism from the ring R into the ring R', thenf(R) forms a subring of R'. The complete story is told below. Theorem 2-8. Letfbe a homomorphism from the ring R intotqe ring R'. Then, ... " 1) for each subring S of R,J(S) is a subring of R'; a n d ' . 2) for each subring S' of R',¡-.l(S') is a subring of R.

= Ta(J)~a(g).

l.

We now list sorne of the structural features preserved under hornornorphisrns. Theorem 2-7. Letfbe a hornomorphism frorn the ring R into the ring R'. Then the following hold: 1) f(O) = O,

Proo! To obtain the first part of the theorem, recall that, by definition, the imagef(S) = {f(a)la E S}. Now, suppose thatf(a) andf(b) are arbitrary elements of f{S). Then both a and b belong to th~ set S, as do a -:- b and ab (S being a subring of R). Hence, f(a) - f(b) = f(a - b) Ef(S)

and

2) f( - a) = - f(a) for all a E R.

If, in addition, R and R' are both rings with identity and f(R) = R', then 3) f(l) = 1, 4) f(a- 1) = f(a)-l for each in vertible elernent a

=

for any a in R. Knowing this, one rnight be ternpted to appeal (incorrectly) to the cancellation law to conclude thatf(l) = 1; what is actually required is the fact that rnultiplicative identities are unique. Second, ifthe hypothesis thatfrnap onto the set R' is ornitted, then it can only be inferred thatf(l) is the identity for the hornornorphic irnage f(R). The elernent f(l) need not serve as an identity for the en tire ring R' and, indeed, it rnay very well happen that f(l) =1= 1. We also observe, in passing, that, by virtue of staternent (2),

'7:a(j + g) = (J + g)(a) = f(a) + g(a) = La(J) + Ta(g), La(Jg)

27

IDEALS AND THEIR OPERATIONS

E

R.

f(a)f(b) = f(ab) Ef(S).

According to Theorem 1-3, these are sufficient conditions for f(S) to be a subring of R'. The proof of the second assertion proceeds similarly. First, remember thatf-l(S') = {a E Rlf(a) E S'}. Thus, if a, b Ef-l(S'), the imagesf(a) and

26

FIRST COURSE IN RINGS AND IDEALS

For future use, we shalllabel the set of all hornornorphisrns frorn the ring R into the ring R' by the syrnbol horn(R, R'). In the event that R = R', the sirnpler notation horn R will be used in place of horn(R, R). (Sorne authors prefer to write end R, for endornorphisrn, in place of horn R; both notations have a certain suggestive power and it reduces to a rnatter of personal preference.) A knowledge of a few sirnple-rninded exarnples will help to fix ideas. Example 2-4. Let R and R' be arbitrary rings andf: R --> R' be the function which sends each elernent of R to the zero elernent of R'. Then, f(a

+ b) =

f(ab)

O = O + O = f(a)

= O = OO = f(a)

+ f(b),

(a, b E R),

f(b)

so thatfis a hornornorphic rnapping. This particular rnapping, the so-called trivial homomorphism, iS.the only constant function which satisfies Definition

2-7. Example 2-5. Consider the ring Z ofintegers and the ring Zn of integers modulo n. Definef: Z --> Zn by takingf(a) = [a]; that is, rnap each integer into the congruence class containing it. Thatfis a hornornorphisrn follows directly frorn the definition of the operations üi Zn: f(a + b) = [a + b] f(ab)

=

[ab]

=

=

[a] +n [b]

[aln[b]

=

f(a) +nf(b),

= f(a)·nf(b).

Example 2-6. In the ring rnap(X, R), define La to be the function which assigns to each fE rnap(X, R) its value at a fixed elernent a E X; in other words, La(J) =:= f(a): Then La is a hornornorphisrn frorn rnap(X, R) into R, known as the evaluation homomorphism at a. We need only observe that

Proo! Frorn f(O) = f(O + O) == f(O) + f(O), we obtain f(O) = O. The factthatf(a) +f(-a) =f(a + (-a») =f(O) = Oyieldsf(-a) = -f(a). As regards (3), let the elernent a E R satisfy f(a) = 1; then,j(l) = f(a)f(l) = f(a1) = f(a) = 1. Finally, the equationf(a)f(a- 1 ) = f(aa- 1 ) = f(l) = 1 shows thatf(a)-l = f(a- 1 ), whenever a E R has a rnultiplicative inverse.

Two cornrnents regarding part (3) of the aboye theorern are in order: First, it is evident that f(a)l

= (Jg)(a) = f(a)g(a)

f(a - b)

f(a)

=

f(a1) = f(a)f(l)

=

f(a)

+ f( -

b)

=

f(a) - f(b).

In short, any ring hornornorphisrn preserves differences as well as sums and products. The next theorem indicates the algebraic nature of direct and inverse images of subrings under homomorphisms. Among other things, we shall see that iffis a homomorphism from the ring R into the ring R', thenf(R) forms a subring of R'. The complete story is told below. Theorem 2-8. Letfbe a homomorphism from the ring R intotqe ring R'. Then, ... " 1) for each subring S of R,J(S) is a subring of R'; a n d ' . 2) for each subring S' of R',¡-.l(S') is a subring of R.

= Ta(J)~a(g).

l.

We now list sorne of the structural features preserved under hornornorphisrns. Theorem 2-7. Letfbe a hornomorphism frorn the ring R into the ring R'. Then the following hold: 1) f(O) = O,

Proo! To obtain the first part of the theorem, recall that, by definition, the imagef(S) = {f(a)la E S}. Now, suppose thatf(a) andf(b) are arbitrary elements of f{S). Then both a and b belong to th~ set S, as do a -:- b and ab (S being a subring of R). Hence, f(a) - f(b) = f(a - b) Ef(S)

and

2) f( - a) = - f(a) for all a E R.

If, in addition, R and R' are both rings with identity and f(R) = R', then 3) f(l) = 1, 4) f(a- 1) = f(a)-l for each in vertible elernent a

=

for any a in R. Knowing this, one rnight be ternpted to appeal (incorrectly) to the cancellation law to conclude thatf(l) = 1; what is actually required is the fact that rnultiplicative identities are unique. Second, ifthe hypothesis thatfrnap onto the set R' is ornitted, then it can only be inferred thatf(l) is the identity for the hornornorphic irnage f(R). The elernent f(l) need not serve as an identity for the en tire ring R' and, indeed, it rnay very well happen that f(l) =1= 1. We also observe, in passing, that, by virtue of staternent (2),

'7:a(j + g) = (J + g)(a) = f(a) + g(a) = La(J) + Ta(g), La(Jg)

27

IDEALS AND THEIR OPERATIONS

E

R.

f(a)f(b) = f(ab) Ef(S).

According to Theorem 1-3, these are sufficient conditions for f(S) to be a subring of R'. The proof of the second assertion proceeds similarly. First, remember thatf-l(S') = {a E Rlf(a) E S'}. Thus, if a, b Ef-l(S'), the imagesf(a) and

28

FIRST COURSE IN RINGS AND IDEALS

f(b) must be members of S'. Since S' is asubring of R', it follows at once that fea - b) = fea) - f(b) E S' and f(ab) = f(a)f(b) E S'. . , This means that a - b and ab líe in f-l(S'), from which we conc1ude.that f-l(S') forms a subring of R. Left unresolved is the matter ofreplacing the term "subring" in Theorem 2-8 by "ideal". It is not difficult to show that part (2) of the th~orem remains true under such a substitution. More precisely: if /' is an ideal of R', then the subringf-l(!') is an ideal of R. For instance, suppo~e·that a Ef-l(!'), so thatf(a) E /', and let r be an arbitrary element of R. >.Then, f(ra) = f(r)f(a) E/';.in other words, the product ra is inf-l(/,). Likéwise, ar E f-l(/,), which helps to make f-l(!') an ideal of R . · .:.:.. . Without further restriction, it cannot be inferred that the image'f(l) will be an ideal of R', whenever 1 is an ideal of R. One would need to know that r'f(a) Ef(l) for all r' E R' and a E l. In general, there isno way of replacing r' by som~f(r) in order to exploit the fact that l is an ideal. The answer is o bvious : j ust take f to be an onto mapping. Surnmarizing these remarks, we may now state: CoroUary. 1) For each ideal /' of R', the subring f-l(/,) is an ideal of R. 2) Ifj(R) = R', then for each ideal l of R, the subringf(I) is an ideal ofR'. To go still further, we need to introduce a new idea. Definition 2-8. Let f be a homomorphism from the ring R into the ring R'. The kernel off, denoted by ker j,consistsof those elements in R which are mapped by f onto the zero element of the ring R': .

= {a E Elf(a) = O}. indicates that ker f is a nonempty ker f

Theorem 2-7 subset of R, since, if nothing else, O E ker f. Except for the case .of the trivial homomorphism, the kernelwill alwa,ys turn out to be a proper subset of R. As one might suspect, the kernel of a ring homomorphism forms an ideal. Theorem 2-9; The kernel ker Iof a homomorphism f from a ring R into a ring R' is an ideal'of R. Proof. We already know that the trivjal subring {O} forms an ideal of R'. Since ker f = f-l(O), the conclusion follows from the last corollary.

IDEALS AND THEIR OPERA TIONS

29

'The kernel of a hcimomorphism may be viewed as a measure of the extent to which the homomorphism fails to be one-to-one (hence, fails to be ~ isomorphism). In more concrete terms, we have Theorem 2-10. A homomorphism f from a ring R into a ring R' is an isomorphism if and only ifker f = {O} . Proof. First, iffis a one-to-one function andf(a) ,;;,. 0= feO), then a = O, whence ker f = {O}. On the other hand, supposéthat the kernel consists exactly of O. Uf(a) = f(b), theno.~., fea - b)

= fea) - f(b) =

O,';

which means that a - bE ker f. Since ker f = {O},\vemust have a - b = O, or a = b, making f a one-to-one function. ,:< : Two rings R and R' are said to be isomorpIÚ~;'-:denoted by R ~ R', if there exists an isomorphism fro'm the ring R ontP the ring R ':.. Although thisdefinition is unsymmetric in that it makes meil.t~on of a furistion from ~ne particular'ring to another, let us remark that iff: R --+ R'isaone~to-one, onto; homomorphic mapping, the function f -1: R" --+ Ralso enjoys these properties. We may therefore ignore the apparent lack of symmetry and merelyspeak of two rings R and R' as being isomorphic without specifying . one ring as isomorphic to the other; notationally, this situation is recognized by writing either R ~ R' or R' !:::: R. . Isomorphic rings are indistinguishable from the structural point of view, even though they may differ in the notation for and nature of their elements 'and operations. Two such rings, although not in gerieral formally identical,' are the same forall purposes; the underlying feature is the existence of a mapping which transports the algebraic structure of one ring to the other. Inpractice, we shall often identify isomorphic. rings without explicit mention. This seems to be a natural placeto insert an example. Example 2-7. Consider an arbitrary ring R with idelltity and the mapping f: Z --+ R given by f(n)·= nI: (At the risk ofbeing repetitious, let us again emphasize that nI means the n-fold sum of 1.) A simple computation shows that f, so defined, is a homomorphism from the ring Z of integers into the ring R: f(n '+ m) = (n + m)l= nI + mI = f(n) + f(m) and f(nm) = (nm)1 = n(ml) = (nI) (mI) = f(n)f(m). Since ker f constitutes an ideal of Z, a principal ideal ring, it follows that kerf = {ri

E

Zlnl = O} = (P)

for sorne nonnegative integer p. A moment's reflection should. convince the reader that the integer p is just the characteristic of R. In particular,

28

FIRST COURSE IN RINGS AND IDEALS

f(b) must be members of S'. Since S' is asubring of R', it follows at once that fea - b) = fea) - f(b) E S' and f(ab) = f(a)f(b) E S'. . , This means that a - b and ab líe in f-l(S'), from which we conc1ude.that f-l(S') forms a subring of R. Left unresolved is the matter ofreplacing the term "subring" in Theorem 2-8 by "ideal". It is not difficult to show that part (2) of the th~orem remains true under such a substitution. More precisely: if /' is an ideal of R', then the subringf-l(!') is an ideal of R. For instance, suppo~e·that a Ef-l(!'), so thatf(a) E /', and let r be an arbitrary element of R. >.Then, f(ra) = f(r)f(a) E/';.in other words, the product ra is inf-l(/,). Likéwise, ar E f-l(/,), which helps to make f-l(!') an ideal of R . · .:.:.. . Without further restriction, it cannot be inferred that the image'f(l) will be an ideal of R', whenever 1 is an ideal of R. One would need to know that r'f(a) Ef(l) for all r' E R' and a E l. In general, there isno way of replacing r' by som~f(r) in order to exploit the fact that l is an ideal. The answer is o bvious : j ust take f to be an onto mapping. Surnmarizing these remarks, we may now state: CoroUary. 1) For each ideal /' of R', the subring f-l(/,) is an ideal of R. 2) Ifj(R) = R', then for each ideal l of R, the subringf(I) is an ideal ofR'. To go still further, we need to introduce a new idea. Definition 2-8. Let f be a homomorphism from the ring R into the ring R'. The kernel off, denoted by ker j,consistsof those elements in R which are mapped by f onto the zero element of the ring R': .

= {a E Elf(a) = O}. indicates that ker f is a nonempty ker f

Theorem 2-7 subset of R, since, if nothing else, O E ker f. Except for the case .of the trivial homomorphism, the kernelwill alwa,ys turn out to be a proper subset of R. As one might suspect, the kernel of a ring homomorphism forms an ideal. Theorem 2-9; The kernel ker Iof a homomorphism f from a ring R into a ring R' is an ideal'of R. Proof. We already know that the trivjal subring {O} forms an ideal of R'. Since ker f = f-l(O), the conclusion follows from the last corollary.

IDEALS AND THEIR OPERA TIONS

29

'The kernel of a hcimomorphism may be viewed as a measure of the extent to which the homomorphism fails to be one-to-one (hence, fails to be ~ isomorphism). In more concrete terms, we have Theorem 2-10. A homomorphism f from a ring R into a ring R' is an isomorphism if and only ifker f = {O} . Proof. First, iffis a one-to-one function andf(a) ,;;,. 0= feO), then a = O, whence ker f = {O}. On the other hand, supposéthat the kernel consists exactly of O. Uf(a) = f(b), theno.~., fea - b)

= fea) - f(b) =

O,';

which means that a - bE ker f. Since ker f = {O},\vemust have a - b = O, or a = b, making f a one-to-one function. ,:< : Two rings R and R' are said to be isomorpIÚ~;'-:denoted by R ~ R', if there exists an isomorphism fro'm the ring R ontP the ring R ':.. Although thisdefinition is unsymmetric in that it makes meil.t~on of a furistion from ~ne particular'ring to another, let us remark that iff: R --+ R'isaone~to-one, onto; homomorphic mapping, the function f -1: R" --+ Ralso enjoys these properties. We may therefore ignore the apparent lack of symmetry and merelyspeak of two rings R and R' as being isomorphic without specifying . one ring as isomorphic to the other; notationally, this situation is recognized by writing either R ~ R' or R' !:::: R. . Isomorphic rings are indistinguishable from the structural point of view, even though they may differ in the notation for and nature of their elements 'and operations. Two such rings, although not in gerieral formally identical,' are the same forall purposes; the underlying feature is the existence of a mapping which transports the algebraic structure of one ring to the other. Inpractice, we shall often identify isomorphic. rings without explicit mention. This seems to be a natural placeto insert an example. Example 2-7. Consider an arbitrary ring R with idelltity and the mapping f: Z --+ R given by f(n)·= nI: (At the risk ofbeing repetitious, let us again emphasize that nI means the n-fold sum of 1.) A simple computation shows that f, so defined, is a homomorphism from the ring Z of integers into the ring R: f(n '+ m) = (n + m)l= nI + mI = f(n) + f(m) and f(nm) = (nm)1 = n(ml) = (nI) (mI) = f(n)f(m). Since ker f constitutes an ideal of Z, a principal ideal ring, it follows that kerf = {ri

E

Zlnl = O} = (P)

for sorne nonnegative integer p. A moment's reflection should. convince the reader that the integer p is just the characteristic of R. In particular,

I 30

any ring R with identity which is of characteristic zero will contain a subring isomorphic to the integers; more specifically, Z ~ Zl, where 1 is the identity of R. Suppose that f is a homomorphism from the ring R onto the ring R'. We have already observed that each ideal l of the ring R determines an ideal f(I) of the ring R'. It goes without saying that ring theory would be considerably simplified ifthe ideals of R were in a one-to-one correspondence with those of R' in this manner. Unfortunately, this need not be the case. The difficulty is refiected in the fact that if l and J are two ideals of R with l ~ J ~ l + kerJ, thenf(I) = f(J). The quickest way to see this is to notice tbat f(l) ~ f(J) ~ f(l

+

kerf) = f(I)

+ f(kerf)

31

IDEALS AND THEIR OPERA TIONS

FIRST COURSE IN RINGS AND IDEALS

= f(l),

from which we conclude that all the inclusions are actually equalities. In brief, distinct ideals of R may have the same image in R'. This disconcerting situation could be remedied by either demanding that kerf = {O} or else narrowing our view to coilsider only ideals l with ker f ~ l. In either event, it follows that l ~ J ~ l + ker f = l and, in consequence, l = J. The first of the restrictions just cited has the effect of making the function f one-to-one, in which case R and R' are isomorphic rings (and it then comes as no surprise to find their ideals in one-to-one correspondence). The second possibility is the subject of our next theorem. We tum aside briefiy to establish a preliminary lemma which will pro vide the key to later success. Lemma. Letfbe a homomorphism froro the ring R onto the ring R'. If l is any ideal of R such that ker f ~ 1, then l = f - 1 (J(l) ). Proof. Suppose that the element a E r1(j{l)), so that f{a} E f{l). Then f(a) = f(r) for sorne choice ofr in l. As a tesult, we will havef(a - r) = 0, or, what amounts to the same thing, a - rE ker f ~ l. This implies that a E l, yielding the inclusion f-1(J(I))~.-I. Since the reverse inclusion

such thatf(l) = /'. To·accomplish this, it is sufficient to take l = f-1(/,). By the corollary to Theorem 2-8,1 -1(/,) certainly forms an ideal of R and, since O E /', ker f = f-1(0)

f

l = f-1(J(I)) = f-1(J(J)) = J.

One finds in this way that fue correspondence l indeed one-to-one, completing the proof.

Proof. Our first concern is to show that the indicated correspondence

actually maps onto the set of all ideals of R'. In other words, starting with an ideal /' of R', we must produce sorne ideal l of the ring R, with ker f ~ l,

oH-

f(I), where kerf ~ l, is

Before announcing our next result, another definition is necessary. Definition 2-9. A ring R is said to be imbedded in a ring R' if there exists sorne subring S' of R' such that R ~ .S'.

In general if a ring R is imbedded in a ring R', then R' is referred to as an extension ~f R and we say that R can be extended to R'. The .most important cases are those in which one passes ~rom a given. ring R too an extension possessing sorne property not present In R. As a .sunple apph~a­ tion, let us prove that an arbitrary ring can be imbedded In an extenslon ring with identity. Theorem 2-12. (Dofroh Extension Theorein). Any ring R can be im-

bedded in a ring wiih identity. Proof. Consider the Cartesian product R x Z, where

:,RxZ= {(r,n)lrER;nEZ}. If addition and multiplication are defined by (a, n) + (b, m) = (a + b, n + m), (a, n)(b, m) = (ab + ma + nb, nm),

Here now is one of the main results of this section. from the ring R onto the ring R'. Then there is a one-to-one correspondel1ce between those ideals l of R such that ker f ~ l and the set of all ideal s /' of R'; specifically, /' is given by /' = f(l).

f- 1(/,).

Inasmuch as the function f is assumed to be an onto map, it also follows that f(I) = f(J-1(/,)) = /'. Next, we argue that this correspondence is one-to-one. To make things more specific, let ideals l and J of R be given, where ker f ~ l, ker ~ J, and satisfying f(I) = f(J). From the elementary lernma just estabhshed, we see that

always holds, the desired equality follows.

Theorem 2-11. (Correspondence Theorem). Let f be a homomorphism

~

then it is a simple matter to verify that R x Z forms a ring; we ~ea.ve t.he actual details as an exercise. Notice that this system has a multIphcatIve identity, namely, the pair (O, 1); for (a, n)(O, 1) = (aO

+

la

+

nO, nI) = (a, n),

and, similarly, (O, l)(a, n) = (a, n).

I 30

any ring R with identity which is of characteristic zero will contain a subring isomorphic to the integers; more specifically, Z ~ Zl, where 1 is the identity of R. Suppose that f is a homomorphism from the ring R onto the ring R'. We have already observed that each ideal l of the ring R determines an ideal f(I) of the ring R'. It goes without saying that ring theory would be considerably simplified ifthe ideals of R were in a one-to-one correspondence with those of R' in this manner. Unfortunately, this need not be the case. The difficulty is refiected in the fact that if l and J are two ideals of R with l ~ J ~ l + kerJ, thenf(I) = f(J). The quickest way to see this is to notice tbat f(l) ~ f(J) ~ f(l

+

kerf) = f(I)

+ f(kerf)

31

IDEALS AND THEIR OPERA TIONS

FIRST COURSE IN RINGS AND IDEALS

= f(l),

from which we conclude that all the inclusions are actually equalities. In brief, distinct ideals of R may have the same image in R'. This disconcerting situation could be remedied by either demanding that kerf = {O} or else narrowing our view to coilsider only ideals l with ker f ~ l. In either event, it follows that l ~ J ~ l + ker f = l and, in consequence, l = J. The first of the restrictions just cited has the effect of making the function f one-to-one, in which case R and R' are isomorphic rings (and it then comes as no surprise to find their ideals in one-to-one correspondence). The second possibility is the subject of our next theorem. We tum aside briefiy to establish a preliminary lemma which will pro vide the key to later success. Lemma. Letfbe a homomorphism froro the ring R onto the ring R'. If l is any ideal of R such that ker f ~ 1, then l = f - 1 (J(l) ). Proof. Suppose that the element a E r1(j{l)), so that f{a} E f{l). Then f(a) = f(r) for sorne choice ofr in l. As a tesult, we will havef(a - r) = 0, or, what amounts to the same thing, a - rE ker f ~ l. This implies that a E l, yielding the inclusion f-1(J(I))~.-I. Since the reverse inclusion

such thatf(l) = /'. To·accomplish this, it is sufficient to take l = f-1(/,). By the corollary to Theorem 2-8,1 -1(/,) certainly forms an ideal of R and, since O E /', ker f = f-1(0)

f

l = f-1(J(I)) = f-1(J(J)) = J.

One finds in this way that fue correspondence l indeed one-to-one, completing the proof.

Proof. Our first concern is to show that the indicated correspondence

actually maps onto the set of all ideals of R'. In other words, starting with an ideal /' of R', we must produce sorne ideal l of the ring R, with ker f ~ l,

oH-

f(I), where kerf ~ l, is

Before announcing our next result, another definition is necessary. Definition 2-9. A ring R is said to be imbedded in a ring R' if there exists sorne subring S' of R' such that R ~ .S'.

In general if a ring R is imbedded in a ring R', then R' is referred to as an extension ~f R and we say that R can be extended to R'. The .most important cases are those in which one passes ~rom a given. ring R too an extension possessing sorne property not present In R. As a .sunple apph~a­ tion, let us prove that an arbitrary ring can be imbedded In an extenslon ring with identity. Theorem 2-12. (Dofroh Extension Theorein). Any ring R can be im-

bedded in a ring wiih identity. Proof. Consider the Cartesian product R x Z, where

:,RxZ= {(r,n)lrER;nEZ}. If addition and multiplication are defined by (a, n) + (b, m) = (a + b, n + m), (a, n)(b, m) = (ab + ma + nb, nm),

Here now is one of the main results of this section. from the ring R onto the ring R'. Then there is a one-to-one correspondel1ce between those ideals l of R such that ker f ~ l and the set of all ideal s /' of R'; specifically, /' is given by /' = f(l).

f- 1(/,).

Inasmuch as the function f is assumed to be an onto map, it also follows that f(I) = f(J-1(/,)) = /'. Next, we argue that this correspondence is one-to-one. To make things more specific, let ideals l and J of R be given, where ker f ~ l, ker ~ J, and satisfying f(I) = f(J). From the elementary lernma just estabhshed, we see that

always holds, the desired equality follows.

Theorem 2-11. (Correspondence Theorem). Let f be a homomorphism

~

then it is a simple matter to verify that R x Z forms a ring; we ~ea.ve t.he actual details as an exercise. Notice that this system has a multIphcatIve identity, namely, the pair (O, 1); for (a, n)(O, 1) = (aO

+

la

+

nO, nI) = (a, n),

and, similarly, (O, l)(a, n) = (a, n).

.~

;1 ;1

32

FIRST COURSB IN RINGS AND IDEALS

Next, consider the subset R x {O} of R X Z consisting of aH pairs of the form (a, O). -Sin ce (a, O) (b O) (a, O)(b, O) e

(a - b, O),

(ab, O),

A point to be made in connection with the preceding theorem is that the imbedding process may be carried out even if the given ringhas. an identity to start with. Of course, in this case the construction has no particular merit; indeed, the original identity element only serves to introduce divisors of zero into the extended ringo Although Theorem 2-12 shows that we could confine our'study to rings with identity, it lS nonetheless desirable to develop as much Qf the theory as possible without the assumption of such an element. Thus,unless an explicit statement is rnade to the contrary, the subsequent discussions will not presuppose the existence of a multiplicative identity. _ We now talce a brief look at a different problem, namely, the problem of extending a function from a subring to the entire ringo In practice, one is usually con cerned with extensions which retain the characteristic features orthe given function. The theorem below, for instance, presents a situation in which it is possible to extend a homomorphism in such a way that the extended function also preserves both ring operations. Theorem 2-13. Let 1 be an ideal of the ring R andf a homomorphism from 1 onto R', a ring with identity. If 1 S;; cent R, then there is a unique homomorphic extension of f to all of R. Prooj As a start, we choose the element u E 1 so ihat f(u) = 1. Since 1 constitutes an ideal of R, the product au wi11lie in the set 1 for each choice of a E R. It is therefore possible to define a new function g: R -¡. R' by setting g(a) f(au) for all a in R. If the element a happens to belong to 1, then

f(au), = f(a)f(u)

=

f(a)1

fea),

showing that g actual1y extends the original functionj The next thing to confirm lS that both ring operations are _preserved by g. The case of addition is fairly obvious: if a, b E R, then g(a

+ b)

= f(a =

f(au)

As a preliminary step to demonstrating that g also preserves multiplication, notice that f(ab)u Z ) = f(abu)f(u) f(abu). From 1rus we are able to conclude that

it is evident that R X {O} constitutes a subring of R x Z. A straightforward caIculation, wruch we omit, shows that R x {O} is isomorphic to the given ring R under the mapping f: R -¡. R x {O} defined by fea) = (a, O). This process of extension therefore imbeds R in R x Z, a ring with .identity. .

g(a)

33

IDEALS AND THEIR OPERATIONS

+ b)u) f(au + bu) + f(bu) = g(a) + g(b).

g(ab)

f(abu)

= f(abu Z)

=

f(au)(bu»)

= f(au)f(bu) = g(a)g(b).

The crucial third equality is justified by the fact that u E cent R, hence, commutes with b. ... As regards the uniqueness assertion, let us assume that there is another homomorphic extension offto the set R'i9all it h. Sincefan'd h must agree on 1 and, more specifically, at the elemept u, h(u) = f(u) = -1. With this in mind, it follows that _,,:', ' h(a)

h(a)h(u) = h(au),=f(au) = g(a)

for all a E R and so h and g are the sa~~ 'fl;J,nction. Hence, tl1ere.i:;¡ one and only one way of extendingfhomomorphicálly from the ideallto'the whole . ring R.Before closing the present chapter, there is another type of direct sum whichdeserves mention. To this purpose, let R I, R z' ... , R" be a finite number of rings (not necessarily subrings of a common ring) and consider x R¡ consisting of all ordered n-tuples their Cartesian product R (al' az' Oo., a"), with al E Rí' One can easily convert R into a ring by performing the ring operations componentwise; in other words, if (al' az' : .. , a") and (b l , bz' ... , b,,) are two elements of R, simply define (al' az, : .. , all )

+ (b l , bz, ... ,b,,)

(al

+ b l , al + bz, ... , a" +

b,,)

and (al' az, ... , all)(bl,b z' ... , b,,) = (a l b l , azb z, ... , a"b"). The ring so obtained is caBed the external direct sum of R l , R z, , .. , R" and is convenientIy written R = RI Rz R". (Let us caution that the notation is not standard in this matter.) In brief, the situation is this: An external direct sum is a new ring constructed from a given set of rings, and an interna! direct sum is a representation of a given ring as a sum of certain of its ideals. The connection between these two types of direct sums will be made c1ear in the next paragraph. If R is the external direct sum of the rings R¡ (i = 1, 2, .,. , n), then the individual R¡ need not be subríngs, or even subsets, of R. However, there is an ideal of R which is the isomorphic image of Rí' A straightforward calculation will convillce the reader that the set

+ + ... +

I¡ = {(O, ... , O, al' 0, ... ,Olla; E RJ I1 !

I1

.~

;1 ;1

32

FIRST COURSB IN RINGS AND IDEALS

Next, consider the subset R x {O} of R X Z consisting of aH pairs of the form (a, O). -Sin ce (a, O) (b O) (a, O)(b, O) e

(a - b, O),

(ab, O),

A point to be made in connection with the preceding theorem is that the imbedding process may be carried out even if the given ringhas. an identity to start with. Of course, in this case the construction has no particular merit; indeed, the original identity element only serves to introduce divisors of zero into the extended ringo Although Theorem 2-12 shows that we could confine our'study to rings with identity, it lS nonetheless desirable to develop as much Qf the theory as possible without the assumption of such an element. Thus,unless an explicit statement is rnade to the contrary, the subsequent discussions will not presuppose the existence of a multiplicative identity. _ We now talce a brief look at a different problem, namely, the problem of extending a function from a subring to the entire ringo In practice, one is usually con cerned with extensions which retain the characteristic features orthe given function. The theorem below, for instance, presents a situation in which it is possible to extend a homomorphism in such a way that the extended function also preserves both ring operations. Theorem 2-13. Let 1 be an ideal of the ring R andf a homomorphism from 1 onto R', a ring with identity. If 1 S;; cent R, then there is a unique homomorphic extension of f to all of R. Prooj As a start, we choose the element u E 1 so ihat f(u) = 1. Since 1 constitutes an ideal of R, the product au wi11lie in the set 1 for each choice of a E R. It is therefore possible to define a new function g: R -¡. R' by setting g(a) f(au) for all a in R. If the element a happens to belong to 1, then

f(au), = f(a)f(u)

=

f(a)1

fea),

showing that g actual1y extends the original functionj The next thing to confirm lS that both ring operations are _preserved by g. The case of addition is fairly obvious: if a, b E R, then g(a

+ b)

= f(a =

f(au)

As a preliminary step to demonstrating that g also preserves multiplication, notice that f(ab)u Z ) = f(abu)f(u) f(abu). From 1rus we are able to conclude that

it is evident that R X {O} constitutes a subring of R x Z. A straightforward caIculation, wruch we omit, shows that R x {O} is isomorphic to the given ring R under the mapping f: R -¡. R x {O} defined by fea) = (a, O). This process of extension therefore imbeds R in R x Z, a ring with .identity. .

g(a)

33

IDEALS AND THEIR OPERATIONS

+ b)u) f(au + bu) + f(bu) = g(a) + g(b).

g(ab)

f(abu)

= f(abu Z)

=

f(au)(bu»)

= f(au)f(bu) = g(a)g(b).

The crucial third equality is justified by the fact that u E cent R, hence, commutes with b. ... As regards the uniqueness assertion, let us assume that there is another homomorphic extension offto the set R'i9all it h. Sincefan'd h must agree on 1 and, more specifically, at the elemept u, h(u) = f(u) = -1. With this in mind, it follows that _,,:', ' h(a)

h(a)h(u) = h(au),=f(au) = g(a)

for all a E R and so h and g are the sa~~ 'fl;J,nction. Hence, tl1ere.i:;¡ one and only one way of extendingfhomomorphicálly from the ideallto'the whole . ring R.Before closing the present chapter, there is another type of direct sum whichdeserves mention. To this purpose, let R I, R z' ... , R" be a finite number of rings (not necessarily subrings of a common ring) and consider x R¡ consisting of all ordered n-tuples their Cartesian product R (al' az' Oo., a"), with al E Rí' One can easily convert R into a ring by performing the ring operations componentwise; in other words, if (al' az' : .. , a") and (b l , bz' ... , b,,) are two elements of R, simply define (al' az, : .. , all )

+ (b l , bz, ... ,b,,)

(al

+ b l , al + bz, ... , a" +

b,,)

and (al' az, ... , all)(bl,b z' ... , b,,) = (a l b l , azb z, ... , a"b"). The ring so obtained is caBed the external direct sum of R l , R z, , .. , R" and is convenientIy written R = RI Rz R". (Let us caution that the notation is not standard in this matter.) In brief, the situation is this: An external direct sum is a new ring constructed from a given set of rings, and an interna! direct sum is a representation of a given ring as a sum of certain of its ideals. The connection between these two types of direct sums will be made c1ear in the next paragraph. If R is the external direct sum of the rings R¡ (i = 1, 2, .,. , n), then the individual R¡ need not be subríngs, or even subsets, of R. However, there is an ideal of R which is the isomorphic image of Rí' A straightforward calculation will convillce the reader that the set

+ + ... +

I¡ = {(O, ... , O, al' 0, ... ,Olla; E RJ I1 !

I1

34

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

therefore regard R as being an ideal of the ring R'. Our hypothesis now comes into play and asserts that R' = R E!1 J fór a suitable ideal J ~f R'. It is thus possible to choose an element (e, n) in J so that (O, -1) = (r, O) + (e, n), for sorne r E R. The last-written equation tells us that e = - r and n = -1; what is important is the resulting conc1usion that (e, -1) E J. For arbitrary rE R, the product (r, O){e, -1) = (re - r, O) will consequently be in both R and J (each being an ideal of R'). The fact that R n J = {O} forces (re - r, O) = (O, O); hence, re = r. In a like fashion, we obtain er = r, proving that R admits the element e as an identity.

(that is, the set consisting of aH n-tuples with zeroes in all places but the ith) forms an ideal of R naturally isomorphic to R¡ under the mapping which sends (O, ... , O, a¡, O, ... ,O) to the element a¡. Since (al' a 2 ,

••• ,

a n ) = (al' O, O, ... , O)

+

(O, a 2 , O, ... ,O)

+ '" +

(O, O, ... , O, a n),

it should also be c1ear that every member of R is uniquely representable as a sum of elements from the ideals l¡. Taking note of Theorem 2-4, this means that R is the internal direct sum of the ideals 1¡ and so RI ..j.. R 2 ..j.. ••• ..j.. Rn = 11 E!11 2 E!1'" E!1 In

(R¡ ~ 1J

In summary, the external direct sum R of the rings R I , R 2 , •• , , Rn is also the internal direct sum of the ideals 11 , 12 , ...• 1" and, for each i, R¡ and 1¡ are isomorphic. In view of the isomorphism just explained, we shall henceforth refer to the ring R as being a direct sum, not qualífying it with the adjective "internal" or "external", and rely exc1usively on the ®-notation. The term "internal" merely reflects the fact that the individual summands, and not isomorphic copies of them, líe in R. We take this opportunity to introduce the simple, but nonetheless useful, notion of a direct summand of a ringo In formal terms, an ideal 1 of the ring R is said to be a direct summand of R if there exists another ideal J of R such that R = 1 E!1 J. For future use, let us note that should the ideal 1 happen to have an identity element, say the e1ement e E 1, then it will automatically be a direct summand of R. Theargument proceeds as follows. For any choice of r E R, the product re E 1. The assumption that e serves as an identity for 1 then ensures that e(re) = re; At the same time (and for the samereasons), (er)e = ero Combining these pieces, we get re := ere = er, which makes it plain that the element e lies in the eenter of R. This is the key point in showing that the set J = {r - reir E R} forms an ideal oCR; the details are left to the reader. We contend that the ring R is actuaHy the direct sum of 1 and J. Certainly, eaeh element r of Rcmay be written as r = re + (r - re), where re E 1 and r - re E J. Since 1 n J = {O}, this is the only way r can be expressed as a sum of elements of 1 and J. (A moment's thoughtshowsthatifa El n J,saya = r - re,thena = ae = (r - re)e = r(e - e2 ) = O.) It is also true that the ideal 1 = eR = Re, but we did not need this faet here. As a further application of the idea of a direct summand, let us record Theorem 2-14. If the ring R is a direct summand in every extension ring containing it as an ideal, then R has an identity. Proof. To set this result in evidence, we first imbed R in the extension ring Then, R ~ R x {O}, where, as is easily verified, R x {O} constitutes an ideal of R'. We may

R' = R x Z in the standard way (see Theorem 2-12).

35

PROBLEMS 1. If 1 is a right ideal and J a left ideal of the ring R such that 1 n J that ab = O for aH a E 1, b E J.

= {O}, prove

2. Given an ideal 1 oC the ring R, define the set C(I) by C(I)

=

{r E Rlra - ar E 1 Cor all

QE

R}.

Verify that C(I) forms a subring of R.

I~

3. a) Show by example that if 1 and J are both ideals of the ring R, then 1 u J need not bean ideal oC R. b) If {1¡} (i = 1, 2, ...) is a colIection of ideal s ofthe ring R such thatI 1 ~ 12 ~ ... ~ In ~ ... , prove that u li is also an ideal of R. 4. Consider the ring Mn(R) oC n x n matrices over R, a ring with identity. A square matrix (alj) is said to be upper triqngular if aij = O for i > j and strictly upper triangular if a¡] = O Cor i ~ j. Let T,,(R) and T~(R) denote the sets of aH upper triangular.and strictIy upper triangular matrices in Mn(R), respectively. Prove each of the following: a) T,,(R) and T~(R) are both subrings of Mn(R). b) T:'(R) is an ideal of the ring T,,(R). e) A matrix (aij) E T" (R) is invertible in T" (R) if and only if aií is invertible in R Cor i = 1,2, ... , n. [Hint: Induct on the order n.] d) Any matrix (a¡j) E T~(R) is nilpotent; in particular, (a¡])n = O. 5. Let 1 b¡: an ideal of R, a commutative ring with identity. For an element a E R, the ideal generated by the set 1 u {a} is denoted by (1, a). Assurning that a rt 1, show that (1, a)

= {i +

rali

E

1, r

E

R}.

6. In the ring Z of integers consider the principal ideals (n) and (m) generated by the integers n and m. Using the notation of the previous probJem, verify that

((nj, m)

=

((m), n)

= (n) +

(m) = (n, m)

where d is the greatest common divisor of n and m.

= (d),

34

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

therefore regard R as being an ideal of the ring R'. Our hypothesis now comes into play and asserts that R' = R E!1 J fór a suitable ideal J ~f R'. It is thus possible to choose an element (e, n) in J so that (O, -1) = (r, O) + (e, n), for sorne r E R. The last-written equation tells us that e = - r and n = -1; what is important is the resulting conc1usion that (e, -1) E J. For arbitrary rE R, the product (r, O){e, -1) = (re - r, O) will consequently be in both R and J (each being an ideal of R'). The fact that R n J = {O} forces (re - r, O) = (O, O); hence, re = r. In a like fashion, we obtain er = r, proving that R admits the element e as an identity.

(that is, the set consisting of aH n-tuples with zeroes in all places but the ith) forms an ideal of R naturally isomorphic to R¡ under the mapping which sends (O, ... , O, a¡, O, ... ,O) to the element a¡. Since (al' a 2 ,

••• ,

a n ) = (al' O, O, ... , O)

+

(O, a 2 , O, ... ,O)

+ '" +

(O, O, ... , O, a n),

it should also be c1ear that every member of R is uniquely representable as a sum of elements from the ideals l¡. Taking note of Theorem 2-4, this means that R is the internal direct sum of the ideals 1¡ and so RI ..j.. R 2 ..j.. ••• ..j.. Rn = 11 E!11 2 E!1'" E!1 In

(R¡ ~ 1J

In summary, the external direct sum R of the rings R I , R 2 , •• , , Rn is also the internal direct sum of the ideals 11 , 12 , ...• 1" and, for each i, R¡ and 1¡ are isomorphic. In view of the isomorphism just explained, we shall henceforth refer to the ring R as being a direct sum, not qualífying it with the adjective "internal" or "external", and rely exc1usively on the ®-notation. The term "internal" merely reflects the fact that the individual summands, and not isomorphic copies of them, líe in R. We take this opportunity to introduce the simple, but nonetheless useful, notion of a direct summand of a ringo In formal terms, an ideal 1 of the ring R is said to be a direct summand of R if there exists another ideal J of R such that R = 1 E!1 J. For future use, let us note that should the ideal 1 happen to have an identity element, say the e1ement e E 1, then it will automatically be a direct summand of R. Theargument proceeds as follows. For any choice of r E R, the product re E 1. The assumption that e serves as an identity for 1 then ensures that e(re) = re; At the same time (and for the samereasons), (er)e = ero Combining these pieces, we get re := ere = er, which makes it plain that the element e lies in the eenter of R. This is the key point in showing that the set J = {r - reir E R} forms an ideal oCR; the details are left to the reader. We contend that the ring R is actuaHy the direct sum of 1 and J. Certainly, eaeh element r of Rcmay be written as r = re + (r - re), where re E 1 and r - re E J. Since 1 n J = {O}, this is the only way r can be expressed as a sum of elements of 1 and J. (A moment's thoughtshowsthatifa El n J,saya = r - re,thena = ae = (r - re)e = r(e - e2 ) = O.) It is also true that the ideal 1 = eR = Re, but we did not need this faet here. As a further application of the idea of a direct summand, let us record Theorem 2-14. If the ring R is a direct summand in every extension ring containing it as an ideal, then R has an identity. Proof. To set this result in evidence, we first imbed R in the extension ring Then, R ~ R x {O}, where, as is easily verified, R x {O} constitutes an ideal of R'. We may

R' = R x Z in the standard way (see Theorem 2-12).

35

PROBLEMS 1. If 1 is a right ideal and J a left ideal of the ring R such that 1 n J that ab = O for aH a E 1, b E J.

= {O}, prove

2. Given an ideal 1 oC the ring R, define the set C(I) by C(I)

=

{r E Rlra - ar E 1 Cor all

QE

R}.

Verify that C(I) forms a subring of R.

I~

3. a) Show by example that if 1 and J are both ideals of the ring R, then 1 u J need not bean ideal oC R. b) If {1¡} (i = 1, 2, ...) is a colIection of ideal s ofthe ring R such thatI 1 ~ 12 ~ ... ~ In ~ ... , prove that u li is also an ideal of R. 4. Consider the ring Mn(R) oC n x n matrices over R, a ring with identity. A square matrix (alj) is said to be upper triqngular if aij = O for i > j and strictly upper triangular if a¡] = O Cor i ~ j. Let T,,(R) and T~(R) denote the sets of aH upper triangular.and strictIy upper triangular matrices in Mn(R), respectively. Prove each of the following: a) T,,(R) and T~(R) are both subrings of Mn(R). b) T:'(R) is an ideal of the ring T,,(R). e) A matrix (aij) E T" (R) is invertible in T" (R) if and only if aií is invertible in R Cor i = 1,2, ... , n. [Hint: Induct on the order n.] d) Any matrix (a¡j) E T~(R) is nilpotent; in particular, (a¡])n = O. 5. Let 1 b¡: an ideal of R, a commutative ring with identity. For an element a E R, the ideal generated by the set 1 u {a} is denoted by (1, a). Assurning that a rt 1, show that (1, a)

= {i +

rali

E

1, r

E

R}.

6. In the ring Z of integers consider the principal ideals (n) and (m) generated by the integers n and m. Using the notation of the previous probJem, verify that

((nj, m)

=

((m), n)

= (n) +

(m) = (n, m)

where d is the greatest common divisor of n and m.

= (d),

---------~~--

--- -

-~-~-~--~----~-~~-

---------~

,1

1

"

36

FIRST COURSE iN RINGS AND IDEALS

I.

PROBLEMS

,37

7. Suppose that 1 is ~ left ideal and J a right ideal of the ring R. Consider the set 1J =

fE a¡b¡ja¡ El; b¡ E J},

where I: represents a finite sum of one or more terms. Establish that 1J is a twosided ideal of R and, whenever 1 and J are themselves two-sided, tbat IJ.¡;; 1 n J.

8. If S is any

giv~n

16. If 1 is a right ideal of R, a ring with identity, show tbat 1:/ R

nonempty subset of the ring R, then ann,S = {rE Rlar

a) Irl ¡;; J, then l:K ¡;; J:X and X:1 ;;¡ X:J. b) l:Jn+1 = (l:J"):J(l:J):J"foranynEZ+. e) l:J R ifand only if J ¡;; 1. d) l:J 1:(1 + J). {a

E

R IRa s; 1}

i8 the largest two-sided ideal of R contained in 1.

O for aIl a E S}

17. Given that f is a homomorphism from the ring R onto the ring R', prove that a) f(eent R) S; cent R'. . . b) If R 18 a principal ideal ríng, then the same is true of R'. [Hint: For any a E R,

is called the right ann,ihilator of S (in R); Similarly,

f(a»)

ann¡S = {r E Rlra = O for all a E S}

(f(a».]

.

,..,

,

e) Ir the element a E R is nilpotent, then its'image fea) is nilpotent in R'.

is the lefl annihilator of S. When R is a commutative ring, we simply,siJeak onhe annihi/ator of S and use the notation aní:l S. Prove the assertions below: . a) ann,S (ann/ S) is a 'right (left) ideal of R.

18. Let R be a ring witl:J identity. For eaeh invértible element a E R, show that the funetionf.: R ..... R defined by f.(x) = axa~'11s an automorphism of R.

b) If S is a right (left) ideal of R; tben ann,S (ann S) i8 an ideal of R. e) If S is an ideal of R, then ann,S and, ann I S are both ideals of R. d) When R has an identity elernent, ann,R = annlR = {O}.

19. Let f be a homomorphism from the ríng R Íj1to itself and S be Jhe set of elements that are left fixe;d by f; in symbols,

S

9. Le! 11,1 2 , ... ,1. be ideals ofthe ring R with R = 11 + 12 + ... + In' Show tbat this sum is direct if and only if al + a2 + ... + ah = O, with a¡ E 1¡, implies tbat each a¡ = O; 10. lf P(X) is the ring of all subsets of a given set X, prove tllat a) the collection of all finite subsets of X forms an ideal of P(X); b) for each subset Y ¡;; X, P(Y) ánd P(X ~ Y) are both principal ideals of P(X), with P(X) P(Y) Ea P(X ~ Y). 11. Suppose tbat R is a commutative ring with identity and that the element a E R is an idempotent dilferellt from Oor 1. Prove that R is the direct Sum of the principal a). ideals (a) and (1

1

+J

= (3),

1n J

(841

IJ

[Hint: In general, (a):(b) = (e), where e

= (21), then

(252),

l:J

= a}.

Establish that S forms a subring of R. 20. If f is a homomórphism from the ring R into the ríng R', wpere R has positi've . characterístie, verify that eharf(R) ~ char R.

21. Letfbe a homomorphism from the cornmutative rÍng R onto the ríng R'. If 1 and , Jare ídeals of R, verífy each ofthe following: a) b) e) d)

f(l + J) = f(1) + f(J); f(1J) f(1)f(1); f(l n J) f(1) n f(J), with equality if eíther 1 ;;¡ ker f or J ;;¡ ker f; f(l:J) S; f(l) :f(J), with equality if 1 ;;¡ ker f.

22. Sho'l'{ that the relation R ~ R' is añ equivalenee relation on any set üf rings.

12; Le! 1, J and X be ideaIs of the ring R. Prove that a) l(J + K) = 1J + IX, (1 + J)X = IX + JX; b) if 1 ;2 J, ihen 1 n (J + K) J + (1 n X). ' 13. Establish that in the ring Z, if 1 = (12) and J

{a E Rlf(a)

= (4),

J:1

= (7).

= afgcd (a, b).]

14. Given ideals 1 and J of the ring R, verify tbat . a) 0:,1 ann/1, and 0:/1 = ann,l (notation as in Problem 8); b) l:J (1:/ J) is tbe largest ideal of R with the property that (1 :,J)J ¡;; 1 (J(l:/ J) ¡;; 1). 15. Le! 1,J and X beideals of R, acornmutative ríng with identity. Pro ve thefollowing assertíons:

23. Let R be an arbitrary ringo For each fixed element a E R, define the left-multiplication function T,,: R ..... R by taking T,,(x) ax. If TR denotes th!; set of all such funetions, prove the following: a) T" is a (group) homomorphism of the additive group of R into itself; b) TI! forms a ríng, where multiplicatíon is taken to be funetional eomposition; e) Híe mappingf(a) = T" determines a homomorphism of R onto the ring TR ; d) the kernel offis the ideal annlR; . e) iUor each O =1= a E R, there exists so me b E R such that ab =1= O, then R ~ TR • (In particular, part(e) holds whenever R has an identity eIement.) 24. Let R be an arbitrary ring and R x Z be the exten8ion ríng eonstructed in Theorern 2-12. Establish that a) R)( {O} ii an ideal of R x Z; b) Z~ {O}xZ; e) if a is an idempotent element of R, then the pair (-a, 1) is idempotent in R x Z, while (a, O) is a zero divisor.

~---

---------~~--

--- -

-~-~-~--~----~-~~-

---------~

,1

1

"

36

FIRST COURSE iN RINGS AND IDEALS

I.

PROBLEMS

,37

7. Suppose that 1 is ~ left ideal and J a right ideal of the ring R. Consider the set 1J =

fE a¡b¡ja¡ El; b¡ E J},

where I: represents a finite sum of one or more terms. Establish that 1J is a twosided ideal of R and, whenever 1 and J are themselves two-sided, tbat IJ.¡;; 1 n J.

8. If S is any

giv~n

16. If 1 is a right ideal of R, a ring with identity, show tbat 1:/ R

nonempty subset of the ring R, then ann,S = {rE Rlar

a) Irl ¡;; J, then l:K ¡;; J:X and X:1 ;;¡ X:J. b) l:Jn+1 = (l:J"):J(l:J):J"foranynEZ+. e) l:J R ifand only if J ¡;; 1. d) l:J 1:(1 + J). {a

E

R IRa s; 1}

i8 the largest two-sided ideal of R contained in 1.

O for aIl a E S}

17. Given that f is a homomorphism from the ring R onto the ring R', prove that a) f(eent R) S; cent R'. . . b) If R 18 a principal ideal ríng, then the same is true of R'. [Hint: For any a E R,

is called the right ann,ihilator of S (in R); Similarly,

f(a»)

ann¡S = {r E Rlra = O for all a E S}

(f(a».]

.

,..,

,

e) Ir the element a E R is nilpotent, then its'image fea) is nilpotent in R'.

is the lefl annihilator of S. When R is a commutative ring, we simply,siJeak onhe annihi/ator of S and use the notation aní:l S. Prove the assertions below: . a) ann,S (ann/ S) is a 'right (left) ideal of R.

18. Let R be a ring witl:J identity. For eaeh invértible element a E R, show that the funetionf.: R ..... R defined by f.(x) = axa~'11s an automorphism of R.

b) If S is a right (left) ideal of R; tben ann,S (ann S) i8 an ideal of R. e) If S is an ideal of R, then ann,S and, ann I S are both ideals of R. d) When R has an identity elernent, ann,R = annlR = {O}.

19. Let f be a homomorphism from the ríng R Íj1to itself and S be Jhe set of elements that are left fixe;d by f; in symbols,

S

9. Le! 11,1 2 , ... ,1. be ideals ofthe ring R with R = 11 + 12 + ... + In' Show tbat this sum is direct if and only if al + a2 + ... + ah = O, with a¡ E 1¡, implies tbat each a¡ = O; 10. lf P(X) is the ring of all subsets of a given set X, prove tllat a) the collection of all finite subsets of X forms an ideal of P(X); b) for each subset Y ¡;; X, P(Y) ánd P(X ~ Y) are both principal ideals of P(X), with P(X) P(Y) Ea P(X ~ Y). 11. Suppose tbat R is a commutative ring with identity and that the element a E R is an idempotent dilferellt from Oor 1. Prove that R is the direct Sum of the principal a). ideals (a) and (1

1

+J

= (3),

1n J

(841

IJ

[Hint: In general, (a):(b) = (e), where e

= (21), then

(252),

l:J

= a}.

Establish that S forms a subring of R. 20. If f is a homomórphism from the ring R into the ríng R', wpere R has positi've . characterístie, verify that eharf(R) ~ char R.

21. Letfbe a homomorphism from the cornmutative rÍng R onto the ríng R'. If 1 and , Jare ídeals of R, verífy each ofthe following: a) b) e) d)

f(l + J) = f(1) + f(J); f(1J) f(1)f(1); f(l n J) f(1) n f(J), with equality if eíther 1 ;;¡ ker f or J ;;¡ ker f; f(l:J) S; f(l) :f(J), with equality if 1 ;;¡ ker f.

22. Sho'l'{ that the relation R ~ R' is añ equivalenee relation on any set üf rings.

12; Le! 1, J and X be ideaIs of the ring R. Prove that a) l(J + K) = 1J + IX, (1 + J)X = IX + JX; b) if 1 ;2 J, ihen 1 n (J + K) J + (1 n X). ' 13. Establish that in the ring Z, if 1 = (12) and J

{a E Rlf(a)

= (4),

J:1

= (7).

= afgcd (a, b).]

14. Given ideals 1 and J of the ring R, verify tbat . a) 0:,1 ann/1, and 0:/1 = ann,l (notation as in Problem 8); b) l:J (1:/ J) is tbe largest ideal of R with the property that (1 :,J)J ¡;; 1 (J(l:/ J) ¡;; 1). 15. Le! 1,J and X beideals of R, acornmutative ríng with identity. Pro ve thefollowing assertíons:

23. Let R be an arbitrary ringo For each fixed element a E R, define the left-multiplication function T,,: R ..... R by taking T,,(x) ax. If TR denotes th!; set of all such funetions, prove the following: a) T" is a (group) homomorphism of the additive group of R into itself; b) TI! forms a ríng, where multiplicatíon is taken to be funetional eomposition; e) Híe mappingf(a) = T" determines a homomorphism of R onto the ring TR ; d) the kernel offis the ideal annlR; . e) iUor each O =1= a E R, there exists so me b E R such that ab =1= O, then R ~ TR • (In particular, part(e) holds whenever R has an identity eIement.) 24. Let R be an arbitrary ring and R x Z be the exten8ion ríng eonstructed in Theorern 2-12. Establish that a) R)( {O} ii an ideal of R x Z; b) Z~ {O}xZ; e) if a is an idempotent element of R, then the pair (-a, 1) is idempotent in R x Z, while (a, O) is a zero divisor.

~---

38

FIRST COURSE IN RINGS AND IDEALS

25. Suppose that R is a ring of characteristic n. If addition and multiplication are defined in R x Z" = {(x, a)lx E R; a E Z,,} by (x, a)

+ (y,h) =

(x, a)(y, h)

THREE

+ y,a +" h), = (xy + ay + hx, a'"h), (x

prove that R x Z. is an extension ring of R of characteristic n. Also show that R x Z" has an identity element. 26. Leí R = R¡ Ef> R2 Ef> ... Ef> R" be the (external) direct sum of a finite number of rings R¡ with identity (i = 1, 2, '" , n). ' a) For fixed i, define the mapping "'¡: R -> R¡ a~ folIows: if a = (a¡, a2, ... , a,,), where aj E Rj , then 1l¡(a) = al' Prove that 1li is a homomorphism from the ring R onto Ri' b) Sh~w that every ideal l of R is of the form l = 1 ¡ Ef> l2 Ef> ... Ea l", with li an Ideal of R¡. [Hint: Take l¡ = 1l¡(l). If h¡ E l¡, then there exists sorne (h¡, oo., h¡, ... , h.) E l. It folIows that (h¡, ... , h¡, ... , h")(O, ... , 1, '" ,O) = (O, .oo , h¡, ... , O) E l.] 27. A nonempty subset A of a ring R is termed an adeal of R if (i) a, h E A imply a + h E A, (ii) /' E R and a E A imply both ar E A and ra E A. Prove that a) An adeal A of R is an ideal of R if for each a e A there is an integer n =1= O, dep~nding upon a, such that na E aR + Ra. (This condition is satisfied, in particular, if R has a multiplicative identity.) b) Whenever R is a commutative ring, the condition in part (a) is a necessary as weIl as sufficient condition for an adeal to be an ideal. [Hint: For any a'E R, the set A = {naln e Z +} + aR is an adeal of R; hence, an ideal of R.] 28. Let R be a ring with identity and M"(R) be the ring 6f n x n matrices over R. Prove the folIowing: a) If lis an idéal of the ring R, then M"(l) is an ideal of the matrix ring M"(R). b) Every ideal'of M,,(R) is of the fO rm M"(I), where, an ideal of R. [Hint: Let Fij(a) denote the matrix in M"(R) having a as its ijth entry and zeroes elsewhere. For a:ny ideal "ft in M"(R), let 1 be the set of elements in R which appear as entríes for the matrices in~. Given any a El, say a is the rsth entry of a matrix A E"ft, it foIlows that F¡ia) = F;,(1)AF.i1) e Jt.] c) If R is a simple ring, then so is the matrix ring M" (R).

lis

29. Let R be a ring with the prbperty that every subring of R is necessarily an ideal of R. (The ring Z, for instan ce, enjoys this property.) If R contains no divisors of zero, prove that muItiplication is commutative. [Hint: Given O =1= a e R, consider the subring S generated bya. For arbitrary h eR, ab = r e S, so that ar = ra.]

THE CLASSICAL ISOMORPHISM THE®REMS

In this chapter we shall discuss a nurnber of significant resuIts having to do with the relationship between homomorphisms and quotient rings (which we shall shortly define). Ofthese results perhaps the rnost crucial is Theorern 3-7, comrnonly known as the Fundamental Homomorphism Theorem for Rings. The importance of this result would be difficult to overemphasize, for 'it stand s 'as the cornerstone upon which much of the succeeding theory rests. , The notion of an ideal carries with it a natural equivalence relation. For, given an ideal 1 of the ring R, it is a routine matter to check that the relation defined by a == b if and only if a - b E 1 is actually an equivalence relation on R. As such, this relation induces a partition of R into equivalence classes, the exact nature of which is determined below. Theorem 3-1. If 1 is an ideal of the ring R, then the equivalence class of b E R for the relation == is the set b

+

1

=

+ il'ie l}. If a = b + i is any member of b + {b

Proo! Let [b] = {x E Rlx == b}. l, then a - b = i E l. By definition of ==, this implies that a E [b], and so b + 1 S;; [b]. On the other hand, if x E [b], we must have x. - b = i for sorne i in l, whence x = b +- i E b + l. Thus, the inclusion [b] S;; b + 1 also holds and equality follQws.

The usual practice is to speak of any set oL the form b + 1 as a coset of 1 in R, and to refer to the element b as a representative of b + l. For future reference we next li~t sorne of the basic. properties of cosets; these are well-known facts about equivalence classes (see Appendix A) translated into the present notation. Theorem 3-2. If 1 is an ideal of the ring R and ~, b E R, then each of the following is true: 1) a + 1 = 1 if and only if a E l. 2) a + 1 = b + 1 if and only if a - b E l. 3) Either a + 1 = b + 1 or else a + 1 and b + 1 are disjoint. 39

38

FIRST COURSE IN RINGS AND IDEALS

25. Suppose that R is a ring of characteristic n. If addition and multiplication are defined in R x Z" = {(x, a)lx E R; a E Z,,} by (x, a)

+ (y,h) =

(x, a)(y, h)

THREE

+ y,a +" h), = (xy + ay + hx, a'"h), (x

prove that R x Z. is an extension ring of R of characteristic n. Also show that R x Z" has an identity element. 26. Leí R = R¡ Ef> R2 Ef> ... Ef> R" be the (external) direct sum of a finite number of rings R¡ with identity (i = 1, 2, '" , n). ' a) For fixed i, define the mapping "'¡: R -> R¡ a~ folIows: if a = (a¡, a2, ... , a,,), where aj E Rj , then 1l¡(a) = al' Prove that 1li is a homomorphism from the ring R onto Ri' b) Sh~w that every ideal l of R is of the form l = 1 ¡ Ef> l2 Ef> ... Ea l", with li an Ideal of R¡. [Hint: Take l¡ = 1l¡(l). If h¡ E l¡, then there exists sorne (h¡, oo., h¡, ... , h.) E l. It folIows that (h¡, ... , h¡, ... , h")(O, ... , 1, '" ,O) = (O, .oo , h¡, ... , O) E l.] 27. A nonempty subset A of a ring R is termed an adeal of R if (i) a, h E A imply a + h E A, (ii) /' E R and a E A imply both ar E A and ra E A. Prove that a) An adeal A of R is an ideal of R if for each a e A there is an integer n =1= O, dep~nding upon a, such that na E aR + Ra. (This condition is satisfied, in particular, if R has a multiplicative identity.) b) Whenever R is a commutative ring, the condition in part (a) is a necessary as weIl as sufficient condition for an adeal to be an ideal. [Hint: For any a'E R, the set A = {naln e Z +} + aR is an adeal of R; hence, an ideal of R.] 28. Let R be a ring with identity and M"(R) be the ring 6f n x n matrices over R. Prove the folIowing: a) If lis an idéal of the ring R, then M"(l) is an ideal of the matrix ring M"(R). b) Every ideal'of M,,(R) is of the fO rm M"(I), where, an ideal of R. [Hint: Let Fij(a) denote the matrix in M"(R) having a as its ijth entry and zeroes elsewhere. For a:ny ideal "ft in M"(R), let 1 be the set of elements in R which appear as entríes for the matrices in~. Given any a El, say a is the rsth entry of a matrix A E"ft, it foIlows that F¡ia) = F;,(1)AF.i1) e Jt.] c) If R is a simple ring, then so is the matrix ring M" (R).

lis

29. Let R be a ring with the prbperty that every subring of R is necessarily an ideal of R. (The ring Z, for instan ce, enjoys this property.) If R contains no divisors of zero, prove that muItiplication is commutative. [Hint: Given O =1= a e R, consider the subring S generated bya. For arbitrary h eR, ab = r e S, so that ar = ra.]

THE CLASSICAL ISOMORPHISM THE®REMS

In this chapter we shall discuss a nurnber of significant resuIts having to do with the relationship between homomorphisms and quotient rings (which we shall shortly define). Ofthese results perhaps the rnost crucial is Theorern 3-7, comrnonly known as the Fundamental Homomorphism Theorem for Rings. The importance of this result would be difficult to overemphasize, for 'it stand s 'as the cornerstone upon which much of the succeeding theory rests. , The notion of an ideal carries with it a natural equivalence relation. For, given an ideal 1 of the ring R, it is a routine matter to check that the relation defined by a == b if and only if a - b E 1 is actually an equivalence relation on R. As such, this relation induces a partition of R into equivalence classes, the exact nature of which is determined below. Theorem 3-1. If 1 is an ideal of the ring R, then the equivalence class of b E R for the relation == is the set b

+

1

=

+ il'ie l}. If a = b + i is any member of b + {b

Proo! Let [b] = {x E Rlx == b}. l, then a - b = i E l. By definition of ==, this implies that a E [b], and so b + 1 S;; [b]. On the other hand, if x E [b], we must have x. - b = i for sorne i in l, whence x = b +- i E b + l. Thus, the inclusion [b] S;; b + 1 also holds and equality follQws.

The usual practice is to speak of any set oL the form b + 1 as a coset of 1 in R, and to refer to the element b as a representative of b + l. For future reference we next li~t sorne of the basic. properties of cosets; these are well-known facts about equivalence classes (see Appendix A) translated into the present notation. Theorem 3-2. If 1 is an ideal of the ring R and ~, b E R, then each of the following is true: 1) a + 1 = 1 if and only if a E l. 2) a + 1 = b + 1 if and only if a - b E l. 3) Either a + 1 = b + 1 or else a + 1 and b + 1 are disjoint. 39

40

FIRST COURSE IN RINGS AND IDEALS

THE CLASSICAL ISOMORPHISM THEOREMS

Given an ideal 1 of the ring R, let us employ the symbol Rjl to denote . the collection of all cosets of 1 in R; that is,

R.jl

{a -: IlaER}.

=

(a

+

+ 1) (a + l)(b + 1) 1)

+

(b

= (a = db

+ b) + + 1.

1,

One is faced with the usual problem of showing that these operatio~s are actually well-defined, so that thesum and product of the two cosets a + 1 and b + 1 do not depend on their particular representatives a and b. To tbis end, suppose that . ';' o"

~

ci+l=.a'+l

"

.. '

and

.b

+ 1 = b' + l.

Then a - a' = i l and b - b' = í2 for sorne il , i 2 .E l. From this we conclude that (a + b) - (a' +b' ) = (a ~. a') + (b - b') = i l + i 2 El, which, by Theorem 3-2, indicates that (a + b) + 1 = (a' + 6' ) + l. The net result is that (a + 1) + (b + 1) = (a' + 1) + (b' + 1). With regard to the multiplication of cosets, we observe that

ab - a'b'

a(b - b') + (a - a')b' ':'" ai 2 + ilb'El,

=

sincro both the products ai2and tI b' must be i~ 1. The implication, of course, is that ab + 1 = a'b' + 1; hence, our definition of multiplication in Rj1 is meaningful. The verifiGatiqn that Rjl, under the operations defined abo ve, forms a ring is easy and the details are left to the reader. To assure completeness, we simply state . Theorem 3-3. If1 is an ideal of the ring R, then Rj1 is also a ring, known as the quotient ring (or factor ring) of R by l. In Theorem 2-9 we saw that certain ideals occur as kernels of homo-. morphisms. Let us now demonstrate that every ideal does indeed arise in this manner. Theorem 3-4. Let 1 be an ideal of the ring R. Then the mapping nat f : R

--¡.

Rj1 defined by nat¡{a)

Proo! The fact that nat¡ is a homomorphism follows directly from the manner in which the operations aredefined in the quotient ring: nat¡(a

The set Rjl can be endowed with the structure of a ring in a natural way; all we need do is define addition and multiplication as follows:

=

a

+1

is a homomorphism of R onto the quotient ring Rjl; the kernel of natf . is precisely. the set l.

41

+ b)

=

+ b + 1 = (a + 1) + (b + + nat¡(b); ab + 1 = (a + l)(b + 1) a

= nat¡'(a)

nat¡(ab) = =

1) ¡'""

nat¡(a) nat¡(b).

That nat¡ carries R ont~ Rjl is all but obvious; indeed, every element of Rjl is a coset a + 1, with a E R, and so by definition nat¡(a) = a + 1. Inasmüch as the coset 1 =0 + 1 serves as the zero element for the ring Rj1, we necessarily have ker (nat¡) ~. {a . = . {a

E E

Rlnat¡(a) = l} Rla+ 1 = l}

=

1.

.

The last eq;~ality was achieved by invoking Theorem 3-2.__" ": . .. :. It is customary to speak of the function nat¡, which:~ssigns to each element of R the coset in Rjl of which it is the representativé,as the natural, or canonical, mapping of R onto the quotient ring Rj1. When there is no danger of confusion, we shall omit the subscript 1 in writing this mapping. There are two standard techniques for investigating the structure of a particular ringo One method calls for finding all the ideals of the ring, in the hope of gaining information about the ring through its local structure. The other approach is to deternrine all homomorphisms from the given ring hito a simpler ring; the idea here is that the homomorphi'c images will tend to reflect sorne of the algebraic properties of the original ringo (The reader is warned to proceed with sorne care, since, for example, it is quite possible for multiplication to be commutative in the image ring, without the given ring being commutative.) Although these lines of attack aim in different . directions, Theorems 2-9 and 3-4 show that for all practical purposes the'se . are the same; every ideal determines a homomorphism, and every homomorphism determines an ideal.

Example 3-1. A simple illustration to keep in mind when working with quotient rings is provided by the ring Z of integers and the principal ideal (n), where n is a positive integer. The cosets of (n) in Z take the form

a

+ (n)

=

{a

+ lenlle E Z}>

from which it is clear that the cosets of (n)are precisely the congruence classes modulo n. What we earlier described as the operations for congruence classes in Z" can now be vie:-ved as coset operations in Zj(n):

(a + (n)) + (b + (n)) (a + (n))(b + (n))

= =

a + b + (n), _ ab + (n).'

40

FIRST COURSE IN RINGS AND IDEALS

THE CLASSICAL ISOMORPHISM THEOREMS

Given an ideal 1 of the ring R, let us employ the symbol Rjl to denote . the collection of all cosets of 1 in R; that is,

R.jl

{a -: IlaER}.

=

(a

+

+ 1) (a + l)(b + 1) 1)

+

(b

= (a = db

+ b) + + 1.

1,

One is faced with the usual problem of showing that these operatio~s are actually well-defined, so that thesum and product of the two cosets a + 1 and b + 1 do not depend on their particular representatives a and b. To tbis end, suppose that . ';' o"

~

ci+l=.a'+l

"

.. '

and

.b

+ 1 = b' + l.

Then a - a' = i l and b - b' = í2 for sorne il , i 2 .E l. From this we conclude that (a + b) - (a' +b' ) = (a ~. a') + (b - b') = i l + i 2 El, which, by Theorem 3-2, indicates that (a + b) + 1 = (a' + 6' ) + l. The net result is that (a + 1) + (b + 1) = (a' + 1) + (b' + 1). With regard to the multiplication of cosets, we observe that

ab - a'b'

a(b - b') + (a - a')b' ':'" ai 2 + ilb'El,

=

sincro both the products ai2and tI b' must be i~ 1. The implication, of course, is that ab + 1 = a'b' + 1; hence, our definition of multiplication in Rj1 is meaningful. The verifiGatiqn that Rjl, under the operations defined abo ve, forms a ring is easy and the details are left to the reader. To assure completeness, we simply state . Theorem 3-3. If1 is an ideal of the ring R, then Rj1 is also a ring, known as the quotient ring (or factor ring) of R by l. In Theorem 2-9 we saw that certain ideals occur as kernels of homo-. morphisms. Let us now demonstrate that every ideal does indeed arise in this manner. Theorem 3-4. Let 1 be an ideal of the ring R. Then the mapping nat f : R

--¡.

Rj1 defined by nat¡{a)

Proo! The fact that nat¡ is a homomorphism follows directly from the manner in which the operations aredefined in the quotient ring: nat¡(a

The set Rjl can be endowed with the structure of a ring in a natural way; all we need do is define addition and multiplication as follows:

=

a

+1

is a homomorphism of R onto the quotient ring Rjl; the kernel of natf . is precisely. the set l.

41

+ b)

=

+ b + 1 = (a + 1) + (b + + nat¡(b); ab + 1 = (a + l)(b + 1) a

= nat¡'(a)

nat¡(ab) = =

1) ¡'""

nat¡(a) nat¡(b).

That nat¡ carries R ont~ Rjl is all but obvious; indeed, every element of Rjl is a coset a + 1, with a E R, and so by definition nat¡(a) = a + 1. Inasmüch as the coset 1 =0 + 1 serves as the zero element for the ring Rj1, we necessarily have ker (nat¡) ~. {a . = . {a

E E

Rlnat¡(a) = l} Rla+ 1 = l}

=

1.

.

The last eq;~ality was achieved by invoking Theorem 3-2.__" ": . .. :. It is customary to speak of the function nat¡, which:~ssigns to each element of R the coset in Rjl of which it is the representativé,as the natural, or canonical, mapping of R onto the quotient ring Rj1. When there is no danger of confusion, we shall omit the subscript 1 in writing this mapping. There are two standard techniques for investigating the structure of a particular ringo One method calls for finding all the ideals of the ring, in the hope of gaining information about the ring through its local structure. The other approach is to deternrine all homomorphisms from the given ring hito a simpler ring; the idea here is that the homomorphi'c images will tend to reflect sorne of the algebraic properties of the original ringo (The reader is warned to proceed with sorne care, since, for example, it is quite possible for multiplication to be commutative in the image ring, without the given ring being commutative.) Although these lines of attack aim in different . directions, Theorems 2-9 and 3-4 show that for all practical purposes the'se . are the same; every ideal determines a homomorphism, and every homomorphism determines an ideal.

Example 3-1. A simple illustration to keep in mind when working with quotient rings is provided by the ring Z of integers and the principal ideal (n), where n is a positive integer. The cosets of (n) in Z take the form

a

+ (n)

=

{a

+ lenlle E Z}>

from which it is clear that the cosets of (n)are precisely the congruence classes modulo n. What we earlier described as the operations for congruence classes in Z" can now be vie:-ved as coset operations in Zj(n):

(a + (n)) + (b + (n)) (a + (n))(b + (n))

= =

a + b + (n), _ ab + (n).'

42

FIRST COURSE IN RINGS AND IDEALS

THE CLASSICAL ISOMORPHISM THEOREMS

In short, the ring Zn of integers modulo n could just as well be interpreted as the quotient ring of Z by (n). As regards the incidence of ideals in a quotient ringí it should be noted that the Correspondence Theorem applies, in particular, to the case in which we start with an ideall of the ring R and take the hornomorphism fto be the natural mapping nat¡:R --+ Rjl. Since ker(nat¡) = l, the conclusion of the Correspondence Theorem is modified slightly. Theorem 3-5. Let l be an ideal ofthe dng R. Then there is a one-to-one correspondence between those ideals J of R such that l f; J and the set of all ideals J' of the quotient ring Rjl; specifically, J' is given by J' = nat¡J. Viewed otherwise, ,Theorem 3-5 asserts that the ideals of Rjl have the form Jjl, where J is an ideal of R containing l. In this context, Jjl and nat¡J are both used to designate the set {a + lla E J}. By way of an application ofthese ideas, consider the following statement: The ring Zn of integers modulo n has exactly one ideal for each positive divisor m of n, and no other ideals. In the first place, since Zn = Zj(n), Theorem 3-5 tells us that there is a one-to-one corresponden ce between those ideals of the ring Z which contain (n) and the set of ideal s of Z". But the ideal s of Z are just the principal ideals (m), where m ís a nonnegative ínteger. The outcome ís that there ís a one-to-one correspondence between the ídeals of Z" and those ídeals (m) of Z such that (m) ;;2 (n); thís last inclusion occurs íf and only if m dívides n. Theorem 3-6. (Factorízation of Homomorphísms). Let f be a homomorphism of the ríng R onto the ring R', and l be an' ídeal of R such that l f; ker! Then there exísts a uníque homomorphísmJ: Rjl --+ R' with the property thatf = lo nat¡. Proa! To start, we define a functíonJ: Rjl

--+

R', called the induced mapping,

by taking l(a

+

l)

=

f(a)

(a E R).

Thefirst question to be raised is whether or not 1 is actually well-defined. That is to say, we must establísh that this function has values which depend only upon the cosets of l and in no way on their particular representatives. In order to see this, let us assume a + l = b + l. Then a - b E l f; ker! This means that f(a)

=

f(a - b

+

b)

=

f(a - b)

+ f(b) =

f(b)

and, by the manner in which 1 was defined, that l(a + l) = J(b + l). Hence, the functionlis constant on the cosets of l, as we wished to demonstrate.

43

A routíne computation, involving the definitíon of the operations in Rjl, confirms thatlis indeed a homomorphism: l((a

+

l)

+

(b

+

l)) = J(a = f(a)

+ b + l) ~ f(a + b) _ + f(b) = f(a + l) + f(b +

l);

and, likewíse, l((a

+

l)(b

+

l))

= J(ab +

1) ~f(ab) _

= f(a)f(b)

= f(a +

l)f(b

+

l).

In this connection, notice that for each element a E R, f(a) = J(a

+

l) = l(nat¡ (a)) = (Jo nat¡ )(a)

whence the equality f = lo nat¡. It only remaíns to show that thís factorízation ís uníque. Suppose also thatf = g o nat¡ for sorne other function g: Rjl --+ R'. But then l(a

+

1)

=

f(a)

= (g o nat¡ Ha) =

g(a

+

l)

for all a in R, and so g = J. The índuced mappinglís thus the only functíon from the quotient ríng Rjl into R' satisfyíng the equatíon f = Jo nat¡. Corollary. The induced mapping 1 ís an isomorphísm íf and only íf kerf f; l. Proa! What is requíred here ís an explicit description of the kernel of J,

to wit

+ l~(a + l) = O} = {a + l f(a) = O} = {a .+ l a E ker f} = nat¡(ker f).

ker 1 = {a

With reference to Theorem 2-10 a necessary and sufficient condítíon for be an isomorphism is that' ker 1 = l. In the present settíng, thís amounts to the demand that nat¡(ker f) = l; which in tum'Ís equivalent to the ínclusíon ker f f; l;

1 to

In víew of the equalíty f = lo nat¡, the conclusíon of Theorem 3-Q is sometímes expressed by sayíng that the homomorphísm f can be factored through the quotient ríng Rjl or, alternatively, that f can be factored by nat¡. What we have proved, ín a technical sense, is that there exists one and only one function 1 whích makes the following diagram of maps commutative: R-LR'

-E~tl\

11

Rjl

(Speakíng informally, a "mapping díagram" is commutative íf, whenever there are two sequences of arrows in the díagram leading from one ríng to

42

FIRST COURSE IN RINGS AND IDEALS

THE CLASSICAL ISOMORPHISM THEOREMS

In short, the ring Zn of integers modulo n could just as well be interpreted as the quotient ring of Z by (n). As regards the incidence of ideals in a quotient ringí it should be noted that the Correspondence Theorem applies, in particular, to the case in which we start with an ideall of the ring R and take the hornomorphism fto be the natural mapping nat¡:R --+ Rjl. Since ker(nat¡) = l, the conclusion of the Correspondence Theorem is modified slightly. Theorem 3-5. Let l be an ideal ofthe dng R. Then there is a one-to-one correspondence between those ideals J of R such that l f; J and the set of all ideals J' of the quotient ring Rjl; specifically, J' is given by J' = nat¡J. Viewed otherwise, ,Theorem 3-5 asserts that the ideals of Rjl have the form Jjl, where J is an ideal of R containing l. In this context, Jjl and nat¡J are both used to designate the set {a + lla E J}. By way of an application ofthese ideas, consider the following statement: The ring Zn of integers modulo n has exactly one ideal for each positive divisor m of n, and no other ideals. In the first place, since Zn = Zj(n), Theorem 3-5 tells us that there is a one-to-one corresponden ce between those ideals of the ring Z which contain (n) and the set of ideal s of Z". But the ideal s of Z are just the principal ideals (m), where m ís a nonnegative ínteger. The outcome ís that there ís a one-to-one correspondence between the ídeals of Z" and those ídeals (m) of Z such that (m) ;;2 (n); thís last inclusion occurs íf and only if m dívides n. Theorem 3-6. (Factorízation of Homomorphísms). Let f be a homomorphism of the ríng R onto the ring R', and l be an' ídeal of R such that l f; ker! Then there exísts a uníque homomorphísmJ: Rjl --+ R' with the property thatf = lo nat¡. Proa! To start, we define a functíonJ: Rjl

--+

R', called the induced mapping,

by taking l(a

+

l)

=

f(a)

(a E R).

Thefirst question to be raised is whether or not 1 is actually well-defined. That is to say, we must establísh that this function has values which depend only upon the cosets of l and in no way on their particular representatives. In order to see this, let us assume a + l = b + l. Then a - b E l f; ker! This means that f(a)

=

f(a - b

+

b)

=

f(a - b)

+ f(b) =

f(b)

and, by the manner in which 1 was defined, that l(a + l) = J(b + l). Hence, the functionlis constant on the cosets of l, as we wished to demonstrate.

43

A routíne computation, involving the definitíon of the operations in Rjl, confirms thatlis indeed a homomorphism: l((a

+

l)

+

(b

+

l)) = J(a = f(a)

+ b + l) ~ f(a + b) _ + f(b) = f(a + l) + f(b +

l);

and, likewíse, l((a

+

l)(b

+

l))

= J(ab +

1) ~f(ab) _

= f(a)f(b)

= f(a +

l)f(b

+

l).

In this connection, notice that for each element a E R, f(a) = J(a

+

l) = l(nat¡ (a)) = (Jo nat¡ )(a)

whence the equality f = lo nat¡. It only remaíns to show that thís factorízation ís uníque. Suppose also thatf = g o nat¡ for sorne other function g: Rjl --+ R'. But then l(a

+

1)

=

f(a)

= (g o nat¡ Ha) =

g(a

+

l)

for all a in R, and so g = J. The índuced mappinglís thus the only functíon from the quotient ríng Rjl into R' satisfyíng the equatíon f = Jo nat¡. Corollary. The induced mapping 1 ís an isomorphísm íf and only íf kerf f; l. Proa! What is requíred here ís an explicit description of the kernel of J,

to wit

+ l~(a + l) = O} = {a + l f(a) = O} = {a .+ l a E ker f} = nat¡(ker f).

ker 1 = {a

With reference to Theorem 2-10 a necessary and sufficient condítíon for be an isomorphism is that' ker 1 = l. In the present settíng, thís amounts to the demand that nat¡(ker f) = l; which in tum'Ís equivalent to the ínclusíon ker f f; l;

1 to

In víew of the equalíty f = lo nat¡, the conclusíon of Theorem 3-Q is sometímes expressed by sayíng that the homomorphísm f can be factored through the quotient ríng Rjl or, alternatively, that f can be factored by nat¡. What we have proved, ín a technical sense, is that there exists one and only one function 1 whích makes the following diagram of maps commutative: R-LR'

-E~tl\

11

Rjl

(Speakíng informally, a "mapping díagram" is commutative íf, whenever there are two sequences of arrows in the díagram leading from one ríng to

,"1""

......

'. j

44

(

FIRST COURSE IN RINGS AND IDEALS

f

Tlieorem 3-7. (Fundamental Homomorphism Theorem). If f is a homomorphism from the ring R ont6 the ring R', then Rjker f ~ R'. Theorem 3-7 states that the images of R under ho~omorphisms can be duplicated (up to isomorphism) by quotient rings of R; to put it another way,every homomorpi?ism of R is "essentially" a natural mapping. Thus, the problem of determination of all homomorphic images of a ring has been 'reduced to the determination of its quotient rings. Let us use Theorem 3-7 to prove that any homomorphism onto the ring of integers is uniquely determined by its kernel. As the starting point of our endeavor, we establish a leIhma which is of independent interest. Lemma. The only nontrivial homomorphism from the ring Z of integersinto itself is the identity map izo

Proof. Because each po si ti ve integer n may be written as n = 1 + 1 + oo. + 1 (n summands), the operation-preserving nature of f implies that f(n) = nf(l), On the other hand, if n is an arbitrary negative integer, then -n EZ+ and so

=

-f(-n)

=

-(-n)f(1)

=

nf(1).

Plainly,f(O) = O = Of(1). The upshot is thatf(n) = nf(l) for every n in Z. Because f is not identically zero, we must have f(l) = 1; ·to 'see that this is so, sÍluply apply the cancellation láw to the relation f(m) = f(m1) = f(m)f(1), where f(m) =1= O. One finds in this way that f(n) = n == iz(n) Cor all n E Z, making f the identity map on Z. Corollary. There is at most one homomorphism under which an arbitrary ring R is isomorphic to the ring Z.

Proo.f Suppose that the rings R and Z are ísomorphic under two functions f, g: R -> Z. Then the composition .f o g - 1 is a homomorphic mapping from the ring Z onto itself. Knowing this, the ¡emma just pro ved implies thatfog-l = iz , orf= g. We now ha ve the necessary information to prove the following result. Theorem 3-8. Any homomorphism from an arbitrary ring R onto the ring Z ofintegers is uniquely determined by its kernel.

45

Proof. Letfand g be two homomorphisms from the ring R onto Z·with the property that kerf= ker g. Our aim, ofcourse, is to show thatfand g must be the same function. Now, by Theorem 3-7, the quotient rings Rjker f and Rjker g are both isomorphic to the ring of integers via the induced mappingsJand g, respectively. The assl,lmption thatfand g have a. common kernel, when oombined with the preceding corollary, forces J = g. It follows at once from the factorizatiQns .

another, the composition of mappings along these paths produces the same function.) A rather simple observation, with far-reaching implications, is that whenever 1 = ker f, so that both the Factorization Theorem and its Corollary are applicable, f induces a mapping J under which Rjl and R' are isomorphic rings. We surnmarize all this in the following theorem, a result which ~ll be invokedon many occasions in the sequel.

f(n) =f(-(-n))

THE CLASSICAL ISOMORPHISM THEOREMS

l'

=

.Jo natk~r J'

g =

g o natkerg

that the functions f and g are themselves identical.

. .~ r

The next two theorems are somewhat deeper result.s than usual and require the full force of our accumulated machinery. They c0mprise what are often called the First and Second Isomorphism Theorems and have important applications in the sequel. (The reader is caütioned' that there seems to be no universally accepted numbering for these,thecirems.) Theorem 3-9. Letfbe a·homomorphism of the rih~ R'cinto th~ ring R' and let 1 be an ideal of R. If ker f S;; 1, then Rjl ~ R' j f(1).

Proof. Before becoming in volved in the details of the proof, let us remark that the corollary to Theorem 2-8 implies that f(1) is an ideal of the ring R'; thus, it is meaningful to speak of the quotient ring R'j f(1). Let us now define the function g: R -> R'j f(1) by g = natJ(l) of, where nat J(l): R' -> R'j f(1) is the usual natural mapping. Thus, g merely assigns to each element a E R the coset fea) + f(1) in R'j f(1). Since the functions f and natJ(l) are both onto homomorphisms, their composition carries R homomorphically onto the quotient ring R' j f(1). The crux of the argument is to show thatker g = 1, for then the desired conclusion would be an immediate consequence of the Fundamental Homomorphism Theorem. Sin ce the zero element of R' j f(I) is just the coset f(I), the kernel of g consists of those members of R which are mapped . by g onto f(1): ker g = {a E Rlg(a)

= =

{a {a

E E

= f(l)} Rlf(a) + f(1) = f(1)} Rlf(a) Ef(/)} = f-l(f(1)).

The hypothesis that ker f S 1 allows us to appeal to the lemma preceding Theorem 2-11, from which it may be concluded that 1 = f-IU(I)). But then 1 = ker g, completing the argument. When applying this result, it is sometimes preferable to start with an arbitrary ideal in R' and utilize inverse images rather than direct images. The theorem can then be reformulated in the following way.

,"1""

......

'. j

44

(

FIRST COURSE IN RINGS AND IDEALS

f

Tlieorem 3-7. (Fundamental Homomorphism Theorem). If f is a homomorphism from the ring R ont6 the ring R', then Rjker f ~ R'. Theorem 3-7 states that the images of R under ho~omorphisms can be duplicated (up to isomorphism) by quotient rings of R; to put it another way,every homomorpi?ism of R is "essentially" a natural mapping. Thus, the problem of determination of all homomorphic images of a ring has been 'reduced to the determination of its quotient rings. Let us use Theorem 3-7 to prove that any homomorphism onto the ring of integers is uniquely determined by its kernel. As the starting point of our endeavor, we establish a leIhma which is of independent interest. Lemma. The only nontrivial homomorphism from the ring Z of integersinto itself is the identity map izo

Proof. Because each po si ti ve integer n may be written as n = 1 + 1 + oo. + 1 (n summands), the operation-preserving nature of f implies that f(n) = nf(l), On the other hand, if n is an arbitrary negative integer, then -n EZ+ and so

=

-f(-n)

=

-(-n)f(1)

=

nf(1).

Plainly,f(O) = O = Of(1). The upshot is thatf(n) = nf(l) for every n in Z. Because f is not identically zero, we must have f(l) = 1; ·to 'see that this is so, sÍluply apply the cancellation láw to the relation f(m) = f(m1) = f(m)f(1), where f(m) =1= O. One finds in this way that f(n) = n == iz(n) Cor all n E Z, making f the identity map on Z. Corollary. There is at most one homomorphism under which an arbitrary ring R is isomorphic to the ring Z.

Proo.f Suppose that the rings R and Z are ísomorphic under two functions f, g: R -> Z. Then the composition .f o g - 1 is a homomorphic mapping from the ring Z onto itself. Knowing this, the ¡emma just pro ved implies thatfog-l = iz , orf= g. We now ha ve the necessary information to prove the following result. Theorem 3-8. Any homomorphism from an arbitrary ring R onto the ring Z ofintegers is uniquely determined by its kernel.

45

Proof. Letfand g be two homomorphisms from the ring R onto Z·with the property that kerf= ker g. Our aim, ofcourse, is to show thatfand g must be the same function. Now, by Theorem 3-7, the quotient rings Rjker f and Rjker g are both isomorphic to the ring of integers via the induced mappingsJand g, respectively. The assl,lmption thatfand g have a. common kernel, when oombined with the preceding corollary, forces J = g. It follows at once from the factorizatiQns .

another, the composition of mappings along these paths produces the same function.) A rather simple observation, with far-reaching implications, is that whenever 1 = ker f, so that both the Factorization Theorem and its Corollary are applicable, f induces a mapping J under which Rjl and R' are isomorphic rings. We surnmarize all this in the following theorem, a result which ~ll be invokedon many occasions in the sequel.

f(n) =f(-(-n))

THE CLASSICAL ISOMORPHISM THEOREMS

l'

=

.Jo natk~r J'

g =

g o natkerg

that the functions f and g are themselves identical.

. .~ r

The next two theorems are somewhat deeper result.s than usual and require the full force of our accumulated machinery. They c0mprise what are often called the First and Second Isomorphism Theorems and have important applications in the sequel. (The reader is caütioned' that there seems to be no universally accepted numbering for these,thecirems.) Theorem 3-9. Letfbe a·homomorphism of the rih~ R'cinto th~ ring R' and let 1 be an ideal of R. If ker f S;; 1, then Rjl ~ R' j f(1).

Proof. Before becoming in volved in the details of the proof, let us remark that the corollary to Theorem 2-8 implies that f(1) is an ideal of the ring R'; thus, it is meaningful to speak of the quotient ring R'j f(1). Let us now define the function g: R -> R'j f(1) by g = natJ(l) of, where nat J(l): R' -> R'j f(1) is the usual natural mapping. Thus, g merely assigns to each element a E R the coset fea) + f(1) in R'j f(1). Since the functions f and natJ(l) are both onto homomorphisms, their composition carries R homomorphically onto the quotient ring R' j f(1). The crux of the argument is to show thatker g = 1, for then the desired conclusion would be an immediate consequence of the Fundamental Homomorphism Theorem. Sin ce the zero element of R' j f(I) is just the coset f(I), the kernel of g consists of those members of R which are mapped . by g onto f(1): ker g = {a E Rlg(a)

= =

{a {a

E E

= f(l)} Rlf(a) + f(1) = f(1)} Rlf(a) Ef(/)} = f-l(f(1)).

The hypothesis that ker f S 1 allows us to appeal to the lemma preceding Theorem 2-11, from which it may be concluded that 1 = f-IU(I)). But then 1 = ker g, completing the argument. When applying this result, it is sometimes preferable to start with an arbitrary ideal in R' and utilize inverse images rather than direct images. The theorem can then be reformulated in the following way.

46

THE CLASSICAL ISOMORPHISM THEOREMS

FIRST COURSE IN RINGS AND IDEALS

Corollary 1. Let f be a homomorphism from the ring R onto the rillg R'. If l' is any ideal of R', then R/ f -1(1') ~ R' / 1'. Proof In compliance with the corollary to Theorem 2-8,f - 1(1') forms an ideal of R. Furthermore, ker f S; f -1(1'), so that Theorern 3-9leads directly to the isomorphism R/r 1(I') ~ R'/f(J-1(I'))

=

R'/1'.

S;

l. Then

Proof As we know, if 1 is an ideal of the ring R and f is any hornomorphism of R, thenf(l) constitutes an ideal ofthe imagef(R). In the setting at hand, take f to be the natural. mapping natJ: R --+- RjJ; then l/J = nat J 1 forms an ideal ofthe quotient ring R' = RjJ. Since ker (natJ) = J S; 1, Theorern 3-9 implies tbat R/l is isomorphic to (RjJ)/(l/J) under the induced mapping g where g = nat¡!J o natJ.

The diagram displayed below may be of sorne help in visualizing the situation described by the last corollary: R

~R/l

~~R/J 1 ~ (R/J)/(l/J) 1' nat'/l

By virtue of our assumptions, there exists a (necessarily unique) isomorphism

g: R/l --+- (R/1)/(l/J) such that g o nat¡ =

nat¡jJ o natJ'

Let us now take up the second of our general isomorphism theorems. Theorem 3-10. If 1 and J are ideals of the ring R, then 1/(1 Í \ 1) ~ (1+ J)/J. Proof Reasoning as in Theorern 3-9, we seek a homornorphism f from 1 (regarded ~ a ring) onto the quotient ring (1 + 1)jJ such that ker f = 1 Í \ J. Our candidate for this functionfis defined by declaring thatf(a) = a + J, a E 1. A trivial, but useful, observation is that 1 S; 1 + J, whence f can be obtained bycomposing the injection map i¡: 1 --+- 1 + J with the natural mapping natJ: 1 + J _ (1 + 1)/J. To be quite explicit, f = nat J o i¡ or, in diagrammatic language,

1

1,

I~ (I

+

,1

+J

/nat, J)jJ

From this factorization, it is easy to see that f is a homomorphism with f(l) = (1 + J)/J. To confirm that the kernel ofjis precisely the set 1 Í \ J, notice that the coset J serves as the zero element of (1 + 1)/J, and so ker f

= =

{a {a

E

ll f (a) = J}

E

1a

+

J

=

J}

=

{a

E

lla E J}

=

1 Í \ J.

The asserted isomorphism should now be evident from the Fundamental Homomorphism Theorem.

Ariother special case, itself of interest, is the following. Corollary 2. Let 1 and J be two ideals of the ring R, with J ljJ is an ideal of R/J and (RjJ)/(l/J) ~ R/1.

47

We conclude tbis chapter with a brief excursion into the theory of nil and nilpotent ideals: a (right, left, two-sided) ideal 1 of the ring R is said to be a nil ideal if each element x in 1 is nilpotent; that is to say, if there exists a positive integer n for which x!' = O, where n depends upon the particular element x. As one might expect, the ideal 1 will be termed nilpotent provided 1" = {O} for sorne positive integer n. By definition, In denotes the set of all finite sums of products of n elements taken from 1, so that 1" = {O} is equivalent too requiring that for every choice of n elements al' a 2 , ... , a n E 1 (distinct or not), the product a 1a Z '" an = O; in particular, a n = O for all a in 1, whence every nilpotent ideal is autornatically a nil ideal. We speak of the ring R as being nil (ni!potent) if it is ni! (ni!potent) when regarded as an ideal. Notice, too, that any ideal containing a nonzero idempotent element cannot be nilpotent. With these definitions at our disposal, we can now prove two le,?mas. Lemma. 1) If R is a ni! (nilpotent) ring, then every subring and every homomorphic image of R is ni! (nilpotent). 2) If R contains an ideal 1 such that 1 and R/l are both ni! (ni!potent), then R is a pil (ni!potent) ringo Proof The proof of assertion (1) follows irnmediately from the definitions and Problem i-17.' To verify(2), assume that 1 and R/l are nil rings and that a E R. Then there exists sorne positive integer n for which the coset (a

+

l)n = a n

+1

= 1,

signifying that the élement a n E l. Inasrnuch as lis a nil ideal, (an)m = a nnl = O for sorne m E Z+. This implies that a is ni!potent as a member of R and, in consequence, R is a ni! ringo The remainder of the proof is left to the reader's careo Lemma. If NI and N 2 are two nil (nilpotent) ideals of the ring R, then their sum NI + N z is likewise a ni! (nilpotent) ideal. Proof With reference to Theorem 3-10, we have (NI + Nz)/N 1 ~ N z/(N 1 Í \ N 2)' The right-hand side (hence, the left-hand side) of this equation is a nil ring, being the hornomorphic image of the nil ideal N 2'

46

THE CLASSICAL ISOMORPHISM THEOREMS

FIRST COURSE IN RINGS AND IDEALS

Corollary 1. Let f be a homomorphism from the ring R onto the rillg R'. If l' is any ideal of R', then R/ f -1(1') ~ R' / 1'. Proof In compliance with the corollary to Theorem 2-8,f - 1(1') forms an ideal of R. Furthermore, ker f S; f -1(1'), so that Theorern 3-9leads directly to the isomorphism R/r 1(I') ~ R'/f(J-1(I'))

=

R'/1'.

S;

l. Then

Proof As we know, if 1 is an ideal of the ring R and f is any hornomorphism of R, thenf(l) constitutes an ideal ofthe imagef(R). In the setting at hand, take f to be the natural. mapping natJ: R --+- RjJ; then l/J = nat J 1 forms an ideal ofthe quotient ring R' = RjJ. Since ker (natJ) = J S; 1, Theorern 3-9 implies tbat R/l is isomorphic to (RjJ)/(l/J) under the induced mapping g where g = nat¡!J o natJ.

The diagram displayed below may be of sorne help in visualizing the situation described by the last corollary: R

~R/l

~~R/J 1 ~ (R/J)/(l/J) 1' nat'/l

By virtue of our assumptions, there exists a (necessarily unique) isomorphism

g: R/l --+- (R/1)/(l/J) such that g o nat¡ =

nat¡jJ o natJ'

Let us now take up the second of our general isomorphism theorems. Theorem 3-10. If 1 and J are ideals of the ring R, then 1/(1 Í \ 1) ~ (1+ J)/J. Proof Reasoning as in Theorern 3-9, we seek a homornorphism f from 1 (regarded ~ a ring) onto the quotient ring (1 + 1)jJ such that ker f = 1 Í \ J. Our candidate for this functionfis defined by declaring thatf(a) = a + J, a E 1. A trivial, but useful, observation is that 1 S; 1 + J, whence f can be obtained bycomposing the injection map i¡: 1 --+- 1 + J with the natural mapping natJ: 1 + J _ (1 + 1)/J. To be quite explicit, f = nat J o i¡ or, in diagrammatic language,

1

1,

I~ (I

+

,1

+J

/nat, J)jJ

From this factorization, it is easy to see that f is a homomorphism with f(l) = (1 + J)/J. To confirm that the kernel ofjis precisely the set 1 Í \ J, notice that the coset J serves as the zero element of (1 + 1)/J, and so ker f

= =

{a {a

E

ll f (a) = J}

E

1a

+

J

=

J}

=

{a

E

lla E J}

=

1 Í \ J.

The asserted isomorphism should now be evident from the Fundamental Homomorphism Theorem.

Ariother special case, itself of interest, is the following. Corollary 2. Let 1 and J be two ideals of the ring R, with J ljJ is an ideal of R/J and (RjJ)/(l/J) ~ R/1.

47

We conclude tbis chapter with a brief excursion into the theory of nil and nilpotent ideals: a (right, left, two-sided) ideal 1 of the ring R is said to be a nil ideal if each element x in 1 is nilpotent; that is to say, if there exists a positive integer n for which x!' = O, where n depends upon the particular element x. As one might expect, the ideal 1 will be termed nilpotent provided 1" = {O} for sorne positive integer n. By definition, In denotes the set of all finite sums of products of n elements taken from 1, so that 1" = {O} is equivalent too requiring that for every choice of n elements al' a 2 , ... , a n E 1 (distinct or not), the product a 1a Z '" an = O; in particular, a n = O for all a in 1, whence every nilpotent ideal is autornatically a nil ideal. We speak of the ring R as being nil (ni!potent) if it is ni! (ni!potent) when regarded as an ideal. Notice, too, that any ideal containing a nonzero idempotent element cannot be nilpotent. With these definitions at our disposal, we can now prove two le,?mas. Lemma. 1) If R is a ni! (nilpotent) ring, then every subring and every homomorphic image of R is ni! (nilpotent). 2) If R contains an ideal 1 such that 1 and R/l are both ni! (ni!potent), then R is a pil (ni!potent) ringo Proof The proof of assertion (1) follows irnmediately from the definitions and Problem i-17.' To verify(2), assume that 1 and R/l are nil rings and that a E R. Then there exists sorne positive integer n for which the coset (a

+

l)n = a n

+1

= 1,

signifying that the élement a n E l. Inasrnuch as lis a nil ideal, (an)m = a nnl = O for sorne m E Z+. This implies that a is ni!potent as a member of R and, in consequence, R is a ni! ringo The remainder of the proof is left to the reader's careo Lemma. If NI and N 2 are two nil (nilpotent) ideals of the ring R, then their sum NI + N z is likewise a ni! (nilpotent) ideal. Proof With reference to Theorem 3-10, we have (NI + Nz)/N 1 ~ N z/(N 1 Í \ N 2)' The right-hand side (hence, the left-hand side) of this equation is a nil ring, being the hornomorphic image of the nil ideal N 2'

PROBLEMS

48

49

FIRST COURSE IN RINGS AND IDEALS

Since (N I +, N 2)/N I and NI are both nil, it follows from the previous lemma that NI + N 2 is necessarily a nil ideal. Similar reasoning applies to the nilpotent case. Corollary. The sum of any finite number of ni] (nilpotent) ideals of the . ring R is again nil (nilpotent).

.after a certain, but not fixed, point. One may easily check that R constitut a subring ?f ringo S (in fact, R is not only a subring, but actually an of S): It IS III the nng R that we propose to construct our example of non-nilpotent nil ideal. ' a Le~ u~ den.ote b~ J the set o(sequences in R whose nth term belo'ngs to the pnnclpal Ideal III Zpn generated by p; in other words the sequence a E J if and only if it is of the form . ,

~he

ide:~

Havjng completed the necessary preliminaries, let us now establish Theorem 3-11. The sum I Ni bf all the nil ideals Ni of the ring R is a nil ideal.

Proa! If the element a E I Ni' then, by definition; a lies in some finite sum of nil ideal s of R; say, a E NI + N 2 + .... +. N n , wh<::re each N k is nil. By virtue of the last Gorollary, the sum N f' + N 2 + ... + N n must be a nil ideal; hence, the element a is nilpotent. This argument shows that I Ni is a nil ideal. It is possible to deduce somewhat moré; na~mely,

Corollary. The sum of all the nilpotent ideals of the ring R is a nil ideal. Proa!. Since each nilpotent ideal is a nil ideal, thesum N of all nilpotent i is itse1f ideals of R is contained in I Ni' the sum of all nil ideals. But a nil ideal, making N nil.

IN

=

a

(pr l ,pr2, ... ; prn' O, O, ... )

(r k E Zpk).

A routine ca1culation confirms that J is an ideal of R. Since each term of a is nilpotent in the appropriate ring, it follows that . ,'a" =

O = (O, O, O, ... )

for n large enough, .maki~g J a nil, ideal. (This also depends on the fact that a has only a fimte number of nonzero terms.) ." . At the present stage, it is still conceivable that 1 mighLbe a nilpotent Id~al of R. I:Iowever, we:can show that for each posit:ive.iiíteger n there eXIst elements (sequen ces) a E 1 for which' an f O. For instance defi b k' . ' ne a = .{ ak } Y ta lllg ak = P If k = 1, 2, .. , ,n + 1 and ak = O if k > n + l' ~~

,

a

=

(p, ... , p, p, O, : .. )

with

n

+ 1 p's.

One then obtains an = (O, ... , O, pn, O, ... ),

Example 3-2. For examples of nilpotent ideals~ let us turn to the rings Zpn, where p is a fixed prime and n > 1. By virtue of the remarks on page 42, Zpn has exacdy one ideal for each positive divisor of pn and no other ideals; these are simply the principal ideal s (l) = lz pn (O :::; k :::; n). For O < k :::; n, we ha ve '

where all the terms are zero except the (n + l)st, which is pn. Sin ce pn is a nnonzero elemento of the ring Zpn+l, the sequence an f O, implying that J. f {O}. As thls argument holds for any n E Z+, the ideal J carinot be mlpotent.

(I)n = (pnk) = (O) = {O},

W,e sh.all retu.rn. to these ide~s at the appropriate place in the sequel, at whlch tIme thelr Importance wI1l become dear.

so that each proper ideal of Zpn is nilpotent. Before leaving this chapter, we should present an example to show that, in generai, nil and nilpotent'are different concepts. Example 3-3. For a fixed prime p, let S be the collection of sequences a = {a } with the property that the nth term a",E Zpn (n ~ 1). S can be n made into a ring by performing the operations of addition and multiplication term by term :

{a,,} + {b,,}

=

{a n + bn},

{a,,}{b,,}

=

{a"b,,}.

The reader will find that the zero e1ement of this ring is just the sequence formed by the zero elements af the various Zpn and the negative of {a n} is { - a,,}. Now, consider the set R of all sequences in S which become zero

PROBLEMS

1. Le.t == be ~n equival.ence ~elation on the ring R. We say that ==, is eompatible (WI~ the nng operatlOns) If and only if a == b implies a + e == b + e, ae == be, ea =:' eb for all a, b, e E R. Prove that there is a one-to-one correspondence between the Ideals of R and the set of compatible equivalence relations on R.

'!. Ir R is an arbitrary ring and

11 E Z+, prove that a) the sets I" = {llaJa E R} and J" = {a E RJlla = O} are both ideals of R' b) char (R/ln) divides 11; , c) if char R =1= O, then char R divides 11 char (RfJ,,).

PROBLEMS

48

49

FIRST COURSE IN RINGS AND IDEALS

Since (N I +, N 2)/N I and NI are both nil, it follows from the previous lemma that NI + N 2 is necessarily a nil ideal. Similar reasoning applies to the nilpotent case. Corollary. The sum of any finite number of ni] (nilpotent) ideals of the . ring R is again nil (nilpotent).

.after a certain, but not fixed, point. One may easily check that R constitut a subring ?f ringo S (in fact, R is not only a subring, but actually an of S): It IS III the nng R that we propose to construct our example of non-nilpotent nil ideal. ' a Le~ u~ den.ote b~ J the set o(sequences in R whose nth term belo'ngs to the pnnclpal Ideal III Zpn generated by p; in other words the sequence a E J if and only if it is of the form . ,

~he

ide:~

Havjng completed the necessary preliminaries, let us now establish Theorem 3-11. The sum I Ni bf all the nil ideals Ni of the ring R is a nil ideal.

Proa! If the element a E I Ni' then, by definition; a lies in some finite sum of nil ideal s of R; say, a E NI + N 2 + .... +. N n , wh<::re each N k is nil. By virtue of the last Gorollary, the sum N f' + N 2 + ... + N n must be a nil ideal; hence, the element a is nilpotent. This argument shows that I Ni is a nil ideal. It is possible to deduce somewhat moré; na~mely,

Corollary. The sum of all the nilpotent ideals of the ring R is a nil ideal. Proa!. Since each nilpotent ideal is a nil ideal, thesum N of all nilpotent i is itse1f ideals of R is contained in I Ni' the sum of all nil ideals. But a nil ideal, making N nil.

IN

=

a

(pr l ,pr2, ... ; prn' O, O, ... )

(r k E Zpk).

A routine ca1culation confirms that J is an ideal of R. Since each term of a is nilpotent in the appropriate ring, it follows that . ,'a" =

O = (O, O, O, ... )

for n large enough, .maki~g J a nil, ideal. (This also depends on the fact that a has only a fimte number of nonzero terms.) ." . At the present stage, it is still conceivable that 1 mighLbe a nilpotent Id~al of R. I:Iowever, we:can show that for each posit:ive.iiíteger n there eXIst elements (sequen ces) a E 1 for which' an f O. For instance defi b k' . ' ne a = .{ ak } Y ta lllg ak = P If k = 1, 2, .. , ,n + 1 and ak = O if k > n + l' ~~

,

a

=

(p, ... , p, p, O, : .. )

with

n

+ 1 p's.

One then obtains an = (O, ... , O, pn, O, ... ),

Example 3-2. For examples of nilpotent ideals~ let us turn to the rings Zpn, where p is a fixed prime and n > 1. By virtue of the remarks on page 42, Zpn has exacdy one ideal for each positive divisor of pn and no other ideals; these are simply the principal ideal s (l) = lz pn (O :::; k :::; n). For O < k :::; n, we ha ve '

where all the terms are zero except the (n + l)st, which is pn. Sin ce pn is a nnonzero elemento of the ring Zpn+l, the sequence an f O, implying that J. f {O}. As thls argument holds for any n E Z+, the ideal J carinot be mlpotent.

(I)n = (pnk) = (O) = {O},

W,e sh.all retu.rn. to these ide~s at the appropriate place in the sequel, at whlch tIme thelr Importance wI1l become dear.

so that each proper ideal of Zpn is nilpotent. Before leaving this chapter, we should present an example to show that, in generai, nil and nilpotent'are different concepts. Example 3-3. For a fixed prime p, let S be the collection of sequences a = {a } with the property that the nth term a",E Zpn (n ~ 1). S can be n made into a ring by performing the operations of addition and multiplication term by term :

{a,,} + {b,,}

=

{a n + bn},

{a,,}{b,,}

=

{a"b,,}.

The reader will find that the zero e1ement of this ring is just the sequence formed by the zero elements af the various Zpn and the negative of {a n} is { - a,,}. Now, consider the set R of all sequences in S which become zero

PROBLEMS

1. Le.t == be ~n equival.ence ~elation on the ring R. We say that ==, is eompatible (WI~ the nng operatlOns) If and only if a == b implies a + e == b + e, ae == be, ea =:' eb for all a, b, e E R. Prove that there is a one-to-one correspondence between the Ideals of R and the set of compatible equivalence relations on R.

'!. Ir R is an arbitrary ring and

11 E Z+, prove that a) the sets I" = {llaJa E R} and J" = {a E RJlla = O} are both ideals of R' b) char (R/ln) divides 11; , c) if char R =1= O, then char R divides 11 char (RfJ,,).

50

FIRST COURSE IN RINGS AND IDEALS

3. Let l be an ideal of the ring R. Establish each of the following: a) R/I has no divisors ofzero ifand only if ab E 1 implies that either a or b belongs to l. b) R/I is commutative if and only if ab - ba E l for all a, b in R. c) R/I has an identity element if and only ir there is some e E R such that ae - a E 1 and ea - a E 1 for all a in R. d) Whenever R is a commutative ring with identity, Jhe~ so is the quotient ring

R/l. 4. Let R be a commutative ring with identity and let N denote the set of all nilpotent elements in R. Verify that a) The set N forms an ideal of R. [Hint: If d' = bm = O for integers n and In, consider (a - b)"h.J b) The quotient ring R/N has no nonzero nilpotent elements.

5. Prove the following generalization of the Factorization Theorem: Let fl and f2 be homomorphisms from the ring R onto the rings RI and R 2 , respectively. If ker fl S; ker f2' then there exists a unique homomorphism J: R 1 -+ R 2 satisfying h =]0 ft. [Hint: Mimic the argument of Theorem 3-6; that is, for any element fl(a) E R I, define ](JI(a») = f2(a).J 6. Let 1 be an ideal of the ring R. Assurne further that J and K are two subrings of R with 1 S; J, l S; K. Show that a) J S; K if and only if nat 1 J S; nat 1K b) nat1 (J 1\ K) = nat 1 J 1\ nat 1K. 7. If 1 is an ideal of the ring R, prove that . a) R/l is a simple ring if and only if there is no ideal J of R satisfying J e J e R; b) if R is a principal ideal ring, then so is the quotient ring R/l; in particular, Zn is a principal ideal ring for each n E Z+. [Hint: Problem 17, Chapter 2.J 8. a) Given a homomorphism f from the ring R onto the ring R', showthat U-I(b)\b E R'} constitutes a partition of R into the cosets of the ideal kerf [Hint: If b = f(a), then the coset a + kerf = f-l(b).J b) Verify that (up to isomorphism) the only homomorphic images of the ring Z of integers are the rings Zn' n > O, and {O}. 9. Suppose that S is a subring and 1 an ideal of the ring R. If S 1\ 1 = {O},prove that S is isomorphic to a subring of the quotient ring R/l. [Hint: Utilize the mappingf(a) = a + 1, where a E S.J 10. A commutatorin a ring R is defined to be any element oftheform [a, bJ = ab - ba. The commutator ideal of R, denoted by [R, RJ, is the ideal of R generated by the set of all commutators. Prove that a) Ris a commutative ringifand only if[R, RJ = {O} (in a sense, the size of[R, RJ provides a measure of the noncommutativity of R); b) for an ideal 1 of R, the quotient ring R/l is commutative if and only if [R, RJ S; l. 11. Assuming that f is a homomorphism from the ring R onto the commutative ring R', establish the assertions below:

PROBLEMS

51

a) [R, RJ S; kerf; b) f = lo nat(R RJ, where]is the induced mapping; c) if kerf s; [R, RJ, then R/[R, RJ "" R'/[R', R'J.

12. a) Suppose that l1 and 12 are ideills of the ring R for which R = 1 1 EB 1 2 • Prove that R/l1 "" 12, and R/12 "" 1 1. b) Let R be the direct sum of the rings R¡ (i = 1,2, ... , n). If 1¡ is an ideal of R¡ and 1 = 1 1 EB 12 EB ... EB 1", show that

[Hint: Find the kernel of the homomorphism f: R -+ L EB (RJl¡) that sends a = (a l ,a 2,···,an)tof(a) = (al + 11,a2 + 12,···,an + In)'] 13. For a proof of Theorem 3-9 that does not depend on the Fundamental Homomorphism Theorem, define the function h: R/l -+ R'/f(I) by taking h(a + 1) =

f(a)

+ f(l).

a) Show that h is a well-defined isomorphism onto R'/f(I); hence, R/l ~ R'/f(l). b) Establish that h is the unique mapping that makes the diagram below commutative:

R nat1

LR'=f(R)

1 R/l

1

nat f(I)

h

R'If(l)

14. Given integers m, n E Z+, establish that a) if m divides n, then Znj(m)/(n) "" Zm; b) if m and n are relatively prime, then Zmn

~

Zm EB Zn'

15. If 1 is an ideal of the ring R, prove that the matrix ring M n (R/l) is isomorphic to Mn(R)fMn(I). [Hint: Consider the mapping f: Mn(R) -+ M n(R/1) defined by f(a¡)) = (a¡) + l).J . 16. Let R be a ring without divisors of zero. Imbed E. in the ring R' = R x Z, as described in Theorem 2-12. (The case R = Ze illustrates that R' may contain zero divisors even though R does not.) Assuming that 1 denotes the left annihilator of R in R', 1 = {a ER'\ar = O for all r ER}, verify that a) l forms an ideal of R'. [Hint: R is an ideal of R'.J b) R'/l is a ring with identity which has no divisors of zero. c) R'/l contains a subring isomorphic to R. [Hint: Utilize Problem

9.J

50

FIRST COURSE IN RINGS AND IDEALS

3. Let l be an ideal of the ring R. Establish each of the following: a) R/I has no divisors ofzero ifand only if ab E 1 implies that either a or b belongs to l. b) R/I is commutative if and only if ab - ba E l for all a, b in R. c) R/I has an identity element if and only ir there is some e E R such that ae - a E 1 and ea - a E 1 for all a in R. d) Whenever R is a commutative ring with identity, Jhe~ so is the quotient ring

R/l. 4. Let R be a commutative ring with identity and let N denote the set of all nilpotent elements in R. Verify that a) The set N forms an ideal of R. [Hint: If d' = bm = O for integers n and In, consider (a - b)"h.J b) The quotient ring R/N has no nonzero nilpotent elements.

5. Prove the following generalization of the Factorization Theorem: Let fl and f2 be homomorphisms from the ring R onto the rings RI and R 2 , respectively. If ker fl S; ker f2' then there exists a unique homomorphism J: R 1 -+ R 2 satisfying h =]0 ft. [Hint: Mimic the argument of Theorem 3-6; that is, for any element fl(a) E R I, define ](JI(a») = f2(a).J 6. Let 1 be an ideal of the ring R. Assurne further that J and K are two subrings of R with 1 S; J, l S; K. Show that a) J S; K if and only if nat 1 J S; nat 1K b) nat1 (J 1\ K) = nat 1 J 1\ nat 1K. 7. If 1 is an ideal of the ring R, prove that . a) R/l is a simple ring if and only if there is no ideal J of R satisfying J e J e R; b) if R is a principal ideal ring, then so is the quotient ring R/l; in particular, Zn is a principal ideal ring for each n E Z+. [Hint: Problem 17, Chapter 2.J 8. a) Given a homomorphism f from the ring R onto the ring R', showthat U-I(b)\b E R'} constitutes a partition of R into the cosets of the ideal kerf [Hint: If b = f(a), then the coset a + kerf = f-l(b).J b) Verify that (up to isomorphism) the only homomorphic images of the ring Z of integers are the rings Zn' n > O, and {O}. 9. Suppose that S is a subring and 1 an ideal of the ring R. If S 1\ 1 = {O},prove that S is isomorphic to a subring of the quotient ring R/l. [Hint: Utilize the mappingf(a) = a + 1, where a E S.J 10. A commutatorin a ring R is defined to be any element oftheform [a, bJ = ab - ba. The commutator ideal of R, denoted by [R, RJ, is the ideal of R generated by the set of all commutators. Prove that a) Ris a commutative ringifand only if[R, RJ = {O} (in a sense, the size of[R, RJ provides a measure of the noncommutativity of R); b) for an ideal 1 of R, the quotient ring R/l is commutative if and only if [R, RJ S; l. 11. Assuming that f is a homomorphism from the ring R onto the commutative ring R', establish the assertions below:

PROBLEMS

51

a) [R, RJ S; kerf; b) f = lo nat(R RJ, where]is the induced mapping; c) if kerf s; [R, RJ, then R/[R, RJ "" R'/[R', R'J.

12. a) Suppose that l1 and 12 are ideills of the ring R for which R = 1 1 EB 1 2 • Prove that R/l1 "" 12, and R/12 "" 1 1. b) Let R be the direct sum of the rings R¡ (i = 1,2, ... , n). If 1¡ is an ideal of R¡ and 1 = 1 1 EB 12 EB ... EB 1", show that

[Hint: Find the kernel of the homomorphism f: R -+ L EB (RJl¡) that sends a = (a l ,a 2,···,an)tof(a) = (al + 11,a2 + 12,···,an + In)'] 13. For a proof of Theorem 3-9 that does not depend on the Fundamental Homomorphism Theorem, define the function h: R/l -+ R'/f(I) by taking h(a + 1) =

f(a)

+ f(l).

a) Show that h is a well-defined isomorphism onto R'/f(I); hence, R/l ~ R'/f(l). b) Establish that h is the unique mapping that makes the diagram below commutative:

R nat1

LR'=f(R)

1 R/l

1

nat f(I)

h

R'If(l)

14. Given integers m, n E Z+, establish that a) if m divides n, then Znj(m)/(n) "" Zm; b) if m and n are relatively prime, then Zmn

~

Zm EB Zn'

15. If 1 is an ideal of the ring R, prove that the matrix ring M n (R/l) is isomorphic to Mn(R)fMn(I). [Hint: Consider the mapping f: Mn(R) -+ M n(R/1) defined by f(a¡)) = (a¡) + l).J . 16. Let R be a ring without divisors of zero. Imbed E. in the ring R' = R x Z, as described in Theorem 2-12. (The case R = Ze illustrates that R' may contain zero divisors even though R does not.) Assuming that 1 denotes the left annihilator of R in R', 1 = {a ER'\ar = O for all r ER}, verify that a) l forms an ideal of R'. [Hint: R is an ideal of R'.J b) R'/l is a ring with identity which has no divisors of zero. c) R'/l contains a subring isomorphic to R. [Hint: Utilize Problem

9.J

INTEGRAL OOMAINS ANO FIELOS

53

ExampJe 4-2.
FOUR

+ (e, d) = (a + e, b . + d), (a, b)(~, d)' = (ae - bd, ad + be).

(a, b)

~he re~der may verify without difficulty that e, together with these operabons, IS a commutative ring with identity. In this setting, the pair (1 O) se~ves as the multiplicative identity, and (0, O) is the zero element ofthe ri~g. Glven any nonzero element (a, b) of e, either a =1= O or b =1= 0, so that a 2 + b 2 > O; thus, .

INTEGRAL DOMAINS AND FIELDS

In the preceding chapters a hierarchy of special rings has been establislied by impó~ing more and more restrictions on the muItiplicative semigroup of a riiig:: At first glance, one might be tempted to require that the multiplicativ,esemigroup actúalIy be a group; such an assumption would be far too demanding in that this situation can only take place in the trivial ring consisti~ ofzero alone. A less stringent conditionwould be the folIowing: . the nonzero elements comprise a group under multiplication. This leads to the notion of a field. . Definition 4-1. A ring F is said to be a field provided that the set F - {O} is a commutative group under the multiplication of F (the identity of this group wiI1 be written as 1). Definition 4-1 implicitly assumes that any field F contains at least one element different from zero, for F - {O} must be nonempty, serving as the set of elements of a group. It is also to be remarked that, since aO = O = Oa for any a é F, alI the members 'of F commute under multiplication and not merely the nonzero elements. Similarly, the relation 10 = 0= 01 implies that 1 is the identity for the entire ring F. Viewed otherwise: a fieJd 'is a commutative ring with identity in which each nonzero element possesses an inverse under multiplication. OccasionalIy, we shalI find it convenient to drop the requirement of commutativity in the consideration of a field, in which case the resulting system is called a division ring or skew fie/d. That is 1:0 say, a ring is a division ring if its nonzero elements form a group (not necessarily commutative) with respect to muItiplication. After this preamble, let us look at several examples. Example 4-1. Bere are some of the more standard illustrations of fields: the set Q of alI rational nUrribers, the set F = {a + b.J2la, b E Q}, and the set R# of alI real numbers. In each case the operations are ordinary addition and multiplication. 52

exists in

e and has the property that

(a b){ i'

¡

¡ ! :". . i

, \a

2

a

+

.

b2 '

. 2

,

2

- b ) _Lra +b a(-b) +ab»c " a2 + b2 - . 2 + b2 ' a2 + bf,: ;,~.(1·'9)·

\a

This shows that each nonzero member of e has an inverse 'under multi- . plication, thereby proving the system e to be a field. It is worth pointing out that the field e contains a subring isomorphic to the field ofreal numbers .. For, if . R# x {O} = {Ca, O)la E R#},

·i 1

! 11

,1

it follows that R # .~. R ~ x {O} via the mapping f defIned by fea) = (a, O). Inasmuch as the dlsbnctlOn between these systems is only one of notation we customarily identify the real ÍlUmber a with the corresponding ordered pair (a, O); in this sense, R# may be regarded as a subring of C.. Now, the definitión of the operations in e enables us to express an . arbitrary element (a, b) E e as . . . . ...... (a, b) = (a, O)

+

(b, 0)(0, 1),

. where the pair (O, 1) is such that (O, 1)2 = (O, 1)(0, 1) the symbol i as an abbreviation for (O, 1), we have (a, b)

=

(a, O)

+

=

(-1, O). Introducing

(b, O)í.

Finally, if it is agreed to replace pairs of the form (a, O) by the first component a ~this is justified by the precedíng paragraph), the dísplayed representatlOn becomes (a, b) = a

+

bi,

with

¡2 = -1.

I~ ot~er wor~s, the field e as defined initiaIly· is' nothing more than a . dlsgUlsed verSlOn ofthe fámilíar complex number system.

INTEGRAL OOMAINS ANO FIELOS

53

ExampJe 4-2.
FOUR

+ (e, d) = (a + e, b . + d), (a, b)(~, d)' = (ae - bd, ad + be).

(a, b)

~he re~der may verify without difficulty that e, together with these operabons, IS a commutative ring with identity. In this setting, the pair (1 O) se~ves as the multiplicative identity, and (0, O) is the zero element ofthe ri~g. Glven any nonzero element (a, b) of e, either a =1= O or b =1= 0, so that a 2 + b 2 > O; thus, .

INTEGRAL DOMAINS AND FIELDS

In the preceding chapters a hierarchy of special rings has been establislied by impó~ing more and more restrictions on the muItiplicative semigroup of a riiig:: At first glance, one might be tempted to require that the multiplicativ,esemigroup actúalIy be a group; such an assumption would be far too demanding in that this situation can only take place in the trivial ring consisti~ ofzero alone. A less stringent conditionwould be the folIowing: . the nonzero elements comprise a group under multiplication. This leads to the notion of a field. . Definition 4-1. A ring F is said to be a field provided that the set F - {O} is a commutative group under the multiplication of F (the identity of this group wiI1 be written as 1). Definition 4-1 implicitly assumes that any field F contains at least one element different from zero, for F - {O} must be nonempty, serving as the set of elements of a group. It is also to be remarked that, since aO = O = Oa for any a é F, alI the members 'of F commute under multiplication and not merely the nonzero elements. Similarly, the relation 10 = 0= 01 implies that 1 is the identity for the entire ring F. Viewed otherwise: a fieJd 'is a commutative ring with identity in which each nonzero element possesses an inverse under multiplication. OccasionalIy, we shalI find it convenient to drop the requirement of commutativity in the consideration of a field, in which case the resulting system is called a division ring or skew fie/d. That is 1:0 say, a ring is a division ring if its nonzero elements form a group (not necessarily commutative) with respect to muItiplication. After this preamble, let us look at several examples. Example 4-1. Bere are some of the more standard illustrations of fields: the set Q of alI rational nUrribers, the set F = {a + b.J2la, b E Q}, and the set R# of alI real numbers. In each case the operations are ordinary addition and multiplication. 52

exists in

e and has the property that

(a b){ i'

¡

¡ ! :". . i

, \a

2

a

+

.

b2 '

. 2

,

2

- b ) _Lra +b a(-b) +ab»c " a2 + b2 - . 2 + b2 ' a2 + bf,: ;,~.(1·'9)·

\a

This shows that each nonzero member of e has an inverse 'under multi- . plication, thereby proving the system e to be a field. It is worth pointing out that the field e contains a subring isomorphic to the field ofreal numbers .. For, if . R# x {O} = {Ca, O)la E R#},

·i 1

! 11

,1

it follows that R # .~. R ~ x {O} via the mapping f defIned by fea) = (a, O). Inasmuch as the dlsbnctlOn between these systems is only one of notation we customarily identify the real ÍlUmber a with the corresponding ordered pair (a, O); in this sense, R# may be regarded as a subring of C.. Now, the definitión of the operations in e enables us to express an . arbitrary element (a, b) E e as . . . . ...... (a, b) = (a, O)

+

(b, 0)(0, 1),

. where the pair (O, 1) is such that (O, 1)2 = (O, 1)(0, 1) the symbol i as an abbreviation for (O, 1), we have (a, b)

=

(a, O)

+

=

(-1, O). Introducing

(b, O)í.

Finally, if it is agreed to replace pairs of the form (a, O) by the first component a ~this is justified by the precedíng paragraph), the dísplayed representatlOn becomes (a, b) = a

+

bi,

with

¡2 = -1.

I~ ot~er wor~s, the field e as defined initiaIly· is' nothing more than a . dlsgUlsed verSlOn ofthe fámilíar complex number system.

54

FIRST COURSE IN RINGS AND lDEALS

Example 4-3. For an iJIustration of a division ring which is not a field, we turn to the ring of (Hamilton's) real quaternions. To introduce this ring, let the set H consist of all ordered 4-tuples of real nUll1bers:

It is easily verified that the product qij

(a, b, e, d) + (a', b', e', di) = (a + a', b + b', e + e', d + d'), (a, b, e, d)(a', b', e', di) = (aa' - bb' ee' dd', ab ' + ba' + ed' de', ae' - bd' + Ca' + db', ad' + be' - eb' + da'). A certaínamount of tedious, but nonetheless straightforward, calculation shows that the resulting system is a ring (known as the ring of real quaternions) in which (O, 0, 0, O) and (1, 0, 0, O) act as the zero and identity elements, respectively. Let us next introduce some special symbols by putting

k

= (0,0,0,1).

The elements 1, i, j, k have a number of distinctive properties; specifically, 1 is the ll1ultiplicative identity of H and

i2 =

i

ij = le, jk = i, kí = j, ji

k2

=

-1,

= - k,

lej = - i,

ik

al - (bi

+

cj

+ d/c)2 =

a2

+ b2 +

e2

+ d2 =t=

0,

+ (e, 0,0, Olí + (d,O, 0, O)k

Since the subring {(r, 0, 0, O)jr:ER#} is isomorphic to R#, the'notation can be rurther simplified on replacing (r, 0, 0, O) by the element r itself; adopting these conventions, the real quaternions may henceforth be regarded as the set H = {a +bi + ej + dkla,b,c,deR#}, with addition and multiplication performefl as for polynoll1ials (subject to the rules of the last paragraph). The reader versed in linear algebra should recognize that H comprises a four-dimensional vector space over R# having {1, i,j, le} as a basis. . The main point in our investigation is that any nonzero quatermon q = a + bi + ej + die (in other words, one of a, b, e, d must be different from zero) is a multiplicatively invertible element. By analogy with the complex numbers, each quaternion has a conjugate, defined as follows: q = a - bi - cj dk.

+ b2 +

c2

+ d2)-lq.

Incidentally, the totality of all members of H of the form (a, b, 0, O) = a + bi, the special quaternions, c:;onstitute a su bring isomorphic to e; as substitutes, one might also consider the set of all e1ements (a, 0, b, O) or all elements (a, O, 0, b). In tbis light, the real quaternions may be viewed as a suitable generalization of the complex numbers. The following theorem shows that any field is without divisors of zero, and consequently a syst'em in which the cancellation law for multiplication holds. Theorem 4-1. Every field F is an integral dornain.

Proof. Since every field is a commutative ring with identity, we need only prove that F .contains no zero divisors. To thls purpose, suppose a, b E F, with ab O. If the element a =t= 0, then it must possess a multiplicative inverse a -1 E F. But then the hypothesis that ab = yields

°=

=

These relations demonstrate that the commutative law for ll1ultiplication fails to hold in H, so that H definitely falls short of being a field. As in Example 4-2, the definition of the algebraic operations in H permits us to write each qua~ernion in the form (a, b, e: d) = (a, 0, 0,0)1.-1: (b, 0, 0, O)í

=

q-1 = (a 2

Addition and multiplication of the elements of H are defined by the rules

= (0,0,0, 1, O),

ijq

thus exhibiting that q has the multiplicative inverse

H = {(a, b, e, d)la, b, e, d E R#}.

(1,0,0, O), i = (0, 1,0, O), ' j

55

INTEGRAL DOMAINS AND FIELDS

°

a- 10 = a-1(ab) = lb

= b,

as desired. There obviously exist integral domains which are not fields; a prime example is the ring Z ofintegers. liowever, an integral dornain having only a finite ilUmber of elements must necessarily be a field. TheOÍ'em 4-2. Any integral domain R with only a finite number of ideals is a field.

Proof. Let a be any nonzero elernent of R. Consider the set of principal ideals (d'), where n e

(a n)

=

{r d'lr E R}.

Since R has only a finite number of distinct ideals, it follows that (a m ) = (a n) for certain positlve integers m, n with m < n. Now, am , as an element of (a"'), must lie in (a n). This being so, there exists some r E R for which a nl = r al!. By use of the cancellation law, 1 = ra n - m = (ra"-m-1)a. Because multiplication is commutative, we therefore have a- 1 = r an - m - 1 • This argument shows that every nonzero element of R is invertible; hence, R forrns a freid.

54

FIRST COURSE IN RINGS AND lDEALS

Example 4-3. For an iJIustration of a division ring which is not a field, we turn to the ring of (Hamilton's) real quaternions. To introduce this ring, let the set H consist of all ordered 4-tuples of real nUll1bers:

It is easily verified that the product qij

(a, b, e, d) + (a', b', e', di) = (a + a', b + b', e + e', d + d'), (a, b, e, d)(a', b', e', di) = (aa' - bb' ee' dd', ab ' + ba' + ed' de', ae' - bd' + Ca' + db', ad' + be' - eb' + da'). A certaínamount of tedious, but nonetheless straightforward, calculation shows that the resulting system is a ring (known as the ring of real quaternions) in which (O, 0, 0, O) and (1, 0, 0, O) act as the zero and identity elements, respectively. Let us next introduce some special symbols by putting

k

= (0,0,0,1).

The elements 1, i, j, k have a number of distinctive properties; specifically, 1 is the ll1ultiplicative identity of H and

i2 =

i

ij = le, jk = i, kí = j, ji

k2

=

-1,

= - k,

lej = - i,

ik

al - (bi

+

cj

+ d/c)2 =

a2

+ b2 +

e2

+ d2 =t=

0,

+ (e, 0,0, Olí + (d,O, 0, O)k

Since the subring {(r, 0, 0, O)jr:ER#} is isomorphic to R#, the'notation can be rurther simplified on replacing (r, 0, 0, O) by the element r itself; adopting these conventions, the real quaternions may henceforth be regarded as the set H = {a +bi + ej + dkla,b,c,deR#}, with addition and multiplication performefl as for polynoll1ials (subject to the rules of the last paragraph). The reader versed in linear algebra should recognize that H comprises a four-dimensional vector space over R# having {1, i,j, le} as a basis. . The main point in our investigation is that any nonzero quatermon q = a + bi + ej + die (in other words, one of a, b, e, d must be different from zero) is a multiplicatively invertible element. By analogy with the complex numbers, each quaternion has a conjugate, defined as follows: q = a - bi - cj dk.

+ b2 +

c2

+ d2)-lq.

Incidentally, the totality of all members of H of the form (a, b, 0, O) = a + bi, the special quaternions, c:;onstitute a su bring isomorphic to e; as substitutes, one might also consider the set of all e1ements (a, 0, b, O) or all elements (a, O, 0, b). In tbis light, the real quaternions may be viewed as a suitable generalization of the complex numbers. The following theorem shows that any field is without divisors of zero, and consequently a syst'em in which the cancellation law for multiplication holds. Theorem 4-1. Every field F is an integral dornain.

Proof. Since every field is a commutative ring with identity, we need only prove that F .contains no zero divisors. To thls purpose, suppose a, b E F, with ab O. If the element a =t= 0, then it must possess a multiplicative inverse a -1 E F. But then the hypothesis that ab = yields

°=

=

These relations demonstrate that the commutative law for ll1ultiplication fails to hold in H, so that H definitely falls short of being a field. As in Example 4-2, the definition of the algebraic operations in H permits us to write each qua~ernion in the form (a, b, e: d) = (a, 0, 0,0)1.-1: (b, 0, 0, O)í

=

q-1 = (a 2

Addition and multiplication of the elements of H are defined by the rules

= (0,0,0, 1, O),

ijq

thus exhibiting that q has the multiplicative inverse

H = {(a, b, e, d)la, b, e, d E R#}.

(1,0,0, O), i = (0, 1,0, O), ' j

55

INTEGRAL DOMAINS AND FIELDS

°

a- 10 = a-1(ab) = lb

= b,

as desired. There obviously exist integral domains which are not fields; a prime example is the ring Z ofintegers. liowever, an integral dornain having only a finite ilUmber of elements must necessarily be a field. TheOÍ'em 4-2. Any integral domain R with only a finite number of ideals is a field.

Proof. Let a be any nonzero elernent of R. Consider the set of principal ideals (d'), where n e

(a n)

=

{r d'lr E R}.

Since R has only a finite number of distinct ideals, it follows that (a m ) = (a n) for certain positlve integers m, n with m < n. Now, am , as an element of (a"'), must lie in (a n). This being so, there exists some r E R for which a nl = r al!. By use of the cancellation law, 1 = ra n - m = (ra"-m-1)a. Because multiplication is commutative, we therefore have a- 1 = r an - m - 1 • This argument shows that every nonzero element of R is invertible; hence, R forrns a freid.

56

Corollary. Any finite integral domain is a field. Because the aforementioned corollary is so basic a result, we offer a second proof. The "counting argument" involved in this latter proof adapts to a variety of situations in which the underIying ring is finite. The reasoning proceeds as follows. Suppose that al' a2 , ... , a" are the members of the integral domain R. For a fixed nonzero element a E R, we consi~er the n products aa l , aa 2 , ... ,aa". These products are all distinct, for.if aa¡ = aaj , the cancellation law (valid in any integral domain) would yie1d a¡ = ajO It follows that each e1ement of R must be of the form aa¡ for sorne choice of i. In particular, there exists sorne a¡E R such that aa¡ = 1. From the commutativity of multiplication, we infer thiit a- l = a¡, whence every nonzero element of R possesses a multiplicative iriverse. There are no fini.te division rings which are not fieÚ¡,s. To put it another way, in a finite systel11 in which all the field properti~s:except the cornmutativity of multiplicaticin are assumed, the multiplicati.on'must also be commutative. Proving this renowned result is far from being as elementary as the case of a finite integral domain and is deferred until Chaptet 9. For the moment, let us take a c10ser look at the multiplication structure of Z". It has been previously shown that, for each positive integer n, Z" comprises a commutative ring w!th identity. A reasonable question is: For precisely what yalues of n, if any at all, will this ring turn out to be a field? For a quick answer: n must be a prime number. (What could be simpler or more natural?) This fact is brought out by the coming theorem. Theorem 4-3. A nonzero element [aJE Z" is invertible in the ring Z" if and only if a and n are relatively prime integers (in the sense that gcd(a, n) = 1). Proo! If a and n are relatively prime, then there exist integers that al' + ns = 1. This implies that .

[lJ

=

[al'

+ nsJ =

[arJ

+"

= [arJ

+"

57

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE IN RINGS AND IDEALS'

l'

and s such

[nsJ [OJ = [aJ ." [rJ,

showing the congruence c1ass [1' J to be the multiplicative inverse of [a]. Now to the "only if" part. Assume [aJ to bea multiplicatively invertible elementofZ";say,withinverse[b]. Wethushave[abJ = [aJ'"[bJ = [lJ, so that there exists an integer k for which ab - 1 = kn. But then ab + n( -k) = 1; hence, a and n are relatively prime integers. Corollary., The zero divisors of Z" are precisely the nonzero elements of Z" which are not invertible. Proo! Naturally, no zero divisor of Z" can possess a multiplicative inverse. On the other hand, suppose that [aJ =1= [OJ is not invertible in Z", so that

gcd(a, n) = d, where 1 < d '< n. Then, a nonzero integers l' and s. This leads to

= rd and n = sd for suitable

[aJ ." [sJ = [asJ = [rdsJ = [rnJ = [O]. Since the defining properties of s rule oU,t the possibilitythat [sJ it follows that [aJ is a zero divisor of Z" . .,;

=

[OJ,

These results may be convenient1y submarized in the following statement. Theorem 4-4. The ring Z" of integer{!'\1odulo n is a field ,if and only if n is a prime number. If n is composit~,: then Z" is not an integral domain and the zero divisors of Z" are tho~~;!10nzero elements [aJ for which gcd(a, n) =1= 1. :j:i:. " Every field necessarily has at least twO: elements (1 being different from ' O); Theorem 4-4 indicates that there is ~>~eId having this miniT~ITI number as its number of elements, viz. Z2' <'éle:' ;~ .. As an interesting appÚca'tion of these ideas, considei' 'ihe following assertion: If there exists a homomorphism f: Z --+ F of the ring Z of integers onto a field F, then F is necessarily a finite field with a prime number of e1ements. For, by the Fundamental Homomorphism Theorem, Zjker f ~ F. But ker f = (n) for sorne positiveintegei n, since Zis a principal ideal domain. (In this connection, observe that n =1= O, for otherwise Z would be isomorphic to a field, an impossibility.) Taking stock ofthe fact that Zj(n) = Z", we are thus able to conc1ude that Z" ~ F, in consequence of which F has n elements. At this point Theorem 4-4 comes to our aid; since F, and in turn its isomorphic image Z", forms a field, n must be a prime nilmber. A useful counting function is the so-called Eufer phi1unCtion (totient), defined as follows: cjJ(l) =' 1 and, for each integer n > 1, cjJ(n) is the number of invertible e1ements in the ring Z". By virtue of Theorem 4-3~ cjJ(n) may al so be characterized as the number of positive integers < n which are relatively prime to n. For instance, cjJ(6) = 2, cjJ(9) = 6, and cjJ(12) = 4; it should be equally c1ear that whenever pis a primenumber, then cjJ(p) = p - 1. Lemma. If G" is the subset of Z" defined by G,; = {[aJ E Z"la is relatively prime to

n},

then (G", ',,) forms a finite group of order cjJ(n). Proo! In the light of the preceding remarks, (G", .") is simply the group of invertible elements of Z".

This leads at once to a c1assical result of Euler concerning the phifunction; the simplicity of the argument illustrates the advantage of the algebraic approach to number theory.

56

Corollary. Any finite integral domain is a field. Because the aforementioned corollary is so basic a result, we offer a second proof. The "counting argument" involved in this latter proof adapts to a variety of situations in which the underIying ring is finite. The reasoning proceeds as follows. Suppose that al' a2 , ... , a" are the members of the integral domain R. For a fixed nonzero element a E R, we consi~er the n products aa l , aa 2 , ... ,aa". These products are all distinct, for.if aa¡ = aaj , the cancellation law (valid in any integral domain) would yie1d a¡ = ajO It follows that each e1ement of R must be of the form aa¡ for sorne choice of i. In particular, there exists sorne a¡E R such that aa¡ = 1. From the commutativity of multiplication, we infer thiit a- l = a¡, whence every nonzero element of R possesses a multiplicative iriverse. There are no fini.te division rings which are not fieÚ¡,s. To put it another way, in a finite systel11 in which all the field properti~s:except the cornmutativity of multiplicaticin are assumed, the multiplicati.on'must also be commutative. Proving this renowned result is far from being as elementary as the case of a finite integral domain and is deferred until Chaptet 9. For the moment, let us take a c10ser look at the multiplication structure of Z". It has been previously shown that, for each positive integer n, Z" comprises a commutative ring w!th identity. A reasonable question is: For precisely what yalues of n, if any at all, will this ring turn out to be a field? For a quick answer: n must be a prime number. (What could be simpler or more natural?) This fact is brought out by the coming theorem. Theorem 4-3. A nonzero element [aJE Z" is invertible in the ring Z" if and only if a and n are relatively prime integers (in the sense that gcd(a, n) = 1). Proo! If a and n are relatively prime, then there exist integers that al' + ns = 1. This implies that .

[lJ

=

[al'

+ nsJ =

[arJ

+"

= [arJ

+"

57

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE IN RINGS AND IDEALS'

l'

and s such

[nsJ [OJ = [aJ ." [rJ,

showing the congruence c1ass [1' J to be the multiplicative inverse of [a]. Now to the "only if" part. Assume [aJ to bea multiplicatively invertible elementofZ";say,withinverse[b]. Wethushave[abJ = [aJ'"[bJ = [lJ, so that there exists an integer k for which ab - 1 = kn. But then ab + n( -k) = 1; hence, a and n are relatively prime integers. Corollary., The zero divisors of Z" are precisely the nonzero elements of Z" which are not invertible. Proo! Naturally, no zero divisor of Z" can possess a multiplicative inverse. On the other hand, suppose that [aJ =1= [OJ is not invertible in Z", so that

gcd(a, n) = d, where 1 < d '< n. Then, a nonzero integers l' and s. This leads to

= rd and n = sd for suitable

[aJ ." [sJ = [asJ = [rdsJ = [rnJ = [O]. Since the defining properties of s rule oU,t the possibilitythat [sJ it follows that [aJ is a zero divisor of Z" . .,;

=

[OJ,

These results may be convenient1y submarized in the following statement. Theorem 4-4. The ring Z" of integer{!'\1odulo n is a field ,if and only if n is a prime number. If n is composit~,: then Z" is not an integral domain and the zero divisors of Z" are tho~~;!10nzero elements [aJ for which gcd(a, n) =1= 1. :j:i:. " Every field necessarily has at least twO: elements (1 being different from ' O); Theorem 4-4 indicates that there is ~>~eId having this miniT~ITI number as its number of elements, viz. Z2' <'éle:' ;~ .. As an interesting appÚca'tion of these ideas, considei' 'ihe following assertion: If there exists a homomorphism f: Z --+ F of the ring Z of integers onto a field F, then F is necessarily a finite field with a prime number of e1ements. For, by the Fundamental Homomorphism Theorem, Zjker f ~ F. But ker f = (n) for sorne positiveintegei n, since Zis a principal ideal domain. (In this connection, observe that n =1= O, for otherwise Z would be isomorphic to a field, an impossibility.) Taking stock ofthe fact that Zj(n) = Z", we are thus able to conc1ude that Z" ~ F, in consequence of which F has n elements. At this point Theorem 4-4 comes to our aid; since F, and in turn its isomorphic image Z", forms a field, n must be a prime nilmber. A useful counting function is the so-called Eufer phi1unCtion (totient), defined as follows: cjJ(l) =' 1 and, for each integer n > 1, cjJ(n) is the number of invertible e1ements in the ring Z". By virtue of Theorem 4-3~ cjJ(n) may al so be characterized as the number of positive integers < n which are relatively prime to n. For instance, cjJ(6) = 2, cjJ(9) = 6, and cjJ(12) = 4; it should be equally c1ear that whenever pis a primenumber, then cjJ(p) = p - 1. Lemma. If G" is the subset of Z" defined by G,; = {[aJ E Z"la is relatively prime to

n},

then (G", ',,) forms a finite group of order cjJ(n). Proo! In the light of the preceding remarks, (G", .") is simply the group of invertible elements of Z".

This leads at once to a c1assical result of Euler concerning the phifunction; the simplicity of the argument illustrates the advantage of the algebraic approach to number theory.

58

FIRST COURSE IN RINGS AND IDEALS

Theorem 4-5. (Euler-Ferrnat). Un is a positive integer and a is relatively prime to n, then a",(n) 1 (mod n).

=

Proof. The congruence c1ass [a] can be viewed as an eIernent of the rnultiplicative group (G m ',,). Since this group has order
=

, There is an interesting relationship between fields and the lack of ideals; what we shaIl show is that fields have as trivial an ideal structure as possible. Theorem 4-6. Let R be a cornrnutative ring with identity. Then R is a field if and only if R has no non tri vial ideals. Proof. Assurne first that R is a field. We wish to show that the trivialideals

{O} and R are its only ideals. Let us suppose to the cOritrary that there exists sorne nontrivüll ideal 1 of R. By our assurnption, the subset 1 is such that 1 =1= {O} and 1 =1= R. This mean s that there exists sorne nonzero elernent a E l. Since R is taken to be a fieId, a has a rnultiplicative in verse a- 1 present in R. By the definition ofideal, we thus obtain 1 = a- l a E 1, which in turn implies that 1 = R, contradicting our choice of l. Conversely, suppose that the ring R has no nontrivial ideals. Given a nonzero eIernent a E R, consider the principal ideal (a) generated by a: (a)

=

{ralr

E

R}.

Now, (a) cánnot be the zero ideal, inasrnuch as a = a 1 E (a), with a =1= O. It foIlows frorn the hypothesis that the only other possibility is that (a) = R. In particular, since 1 E (a),there exists an elernent r E:R for whichr a = l. MuItiplication is cornrnutative, so that r = a-l. T~e;~fore, each nonzero elernent of R is rnultiplicativeIy in vertible and we are done. In view of this last result, the ring Z of integers faiIs to be a field sin ce it contains the nontrivial ideal Ze" ' Theorern 4-6 is useful in revealing the nature of hornornorphisrns between fieIds. We exploit it to prove i

'1

I

Theorem 4-7. Let f be a hornornorphisrn frorn the field Finto the field F'. Then eitherfis the trivial hornornorphism br elsefis one-to-one. , "

,

!I i , 1

II! 11

I

Proof. The proof consists of noticing that since ker fis an idea! of the field F, either ker f = {O} or else ker f = F. The condition ker f = {O} irnplies that f is a one-to-one function. On the other hand, if it happens that kerf = F, then each elernent of F is carried onto O; that is to say,Jis the trivial hornornorphisrn.

Corollary. Any hornornorphisrn of a fieId F onto itself is an autornorphisrn of F.

INTEGRAL DOMAINS AND FIELDS

59

Any ring with identity which is a subring of a field rnust of necessity be an integral dornain. Turning the situation around, one rnight ask whether. each integral dornain can be considered (apart frorn isornorphisrn) as a subríng of sorne fieId. More forrnaIly: Can a given integral dornain be irnbedded in a fieId? In the finite case there is plainly no difficulty, since any finite integral dornain already forrns a field. Our concern with this question arises frorn the natural desire to solve the linear equation ax = b, where a =1= O. A rnajor drawback to the notion of an integral dornain is that it does not always furnish a solution within the systern. (Of course, any such solution would ha ve to be unique, sin ce aX l = b = aX 2 irnplies that Xl = x 2 by the canceIlation law.) It hardly seerns necessary to point out that when the integral dornain happens to be a field, the equation ax = b (a =1= O) is always solvable, for one need only take x = a-lb. We begin our discussion of this problern with an obvious definition. Definition 4-2. By a subfield of a field F is rneant any subring P of F which is itself a field. For example, the ring Q of rational nurnbers is a subfield of the real field R # ; the sarne is true of the field F = {a + bJ2la, b E Q}. SureIy, the set P wilI be a subfield of the fieId F provided that (1) F' is a subgroup of the additive group of F and that (2) F' - {O} is a subgroup of the multiplicative group F - {O}. RecaIling our mínimal set of conditions for deterrnining s,ubgroups (see page 8), it foIlows that F' wilI be a subfield of F if and only if the foIlowing requirernents are rnet: 1) Pis a nonernpty subset of F containing at least one nonzero elernent, 2) a, b E F' irnply a - b E F', and 3) a, b E P, with b =1= O, irnply ab- t E F'. The corning theorern furnishes c1ue to the nature of the field in which we wish to irnbed a given integral dornain.

a

Theorem 4-8. Let the integral dornain R be a subring of the field F. If the set F' is defined by P

=

{ab-lla,bER;b =1=

O},

then P forrns a subfield of F with R 5; P; in fact, F' is the srnaIlest (in the sense of inc1usion) subfield of F containing R. Proof. Notice first that 1 = 11- 1 E P, so that F' =1= {O}. °Now consider two arbitrary elernents x, y of F'. With reference to the definition of P, we then have

58

FIRST COURSE IN RINGS AND IDEALS

Theorem 4-5. (Euler-Ferrnat). Un is a positive integer and a is relatively prime to n, then a",(n) 1 (mod n).

=

Proof. The congruence c1ass [a] can be viewed as an eIernent of the rnultiplicative group (G m ',,). Since this group has order
=

, There is an interesting relationship between fields and the lack of ideals; what we shaIl show is that fields have as trivial an ideal structure as possible. Theorem 4-6. Let R be a cornrnutative ring with identity. Then R is a field if and only if R has no non tri vial ideals. Proof. Assurne first that R is a field. We wish to show that the trivialideals

{O} and R are its only ideals. Let us suppose to the cOritrary that there exists sorne nontrivüll ideal 1 of R. By our assurnption, the subset 1 is such that 1 =1= {O} and 1 =1= R. This mean s that there exists sorne nonzero elernent a E l. Since R is taken to be a fieId, a has a rnultiplicative in verse a- 1 present in R. By the definition ofideal, we thus obtain 1 = a- l a E 1, which in turn implies that 1 = R, contradicting our choice of l. Conversely, suppose that the ring R has no nontrivial ideals. Given a nonzero eIernent a E R, consider the principal ideal (a) generated by a: (a)

=

{ralr

E

R}.

Now, (a) cánnot be the zero ideal, inasrnuch as a = a 1 E (a), with a =1= O. It foIlows frorn the hypothesis that the only other possibility is that (a) = R. In particular, since 1 E (a),there exists an elernent r E:R for whichr a = l. MuItiplication is cornrnutative, so that r = a-l. T~e;~fore, each nonzero elernent of R is rnultiplicativeIy in vertible and we are done. In view of this last result, the ring Z of integers faiIs to be a field sin ce it contains the nontrivial ideal Ze" ' Theorern 4-6 is useful in revealing the nature of hornornorphisrns between fieIds. We exploit it to prove i

'1

I

Theorem 4-7. Let f be a hornornorphisrn frorn the field Finto the field F'. Then eitherfis the trivial hornornorphism br elsefis one-to-one. , "

,

!I i , 1

II! 11

I

Proof. The proof consists of noticing that since ker fis an idea! of the field F, either ker f = {O} or else ker f = F. The condition ker f = {O} irnplies that f is a one-to-one function. On the other hand, if it happens that kerf = F, then each elernent of F is carried onto O; that is to say,Jis the trivial hornornorphisrn.

Corollary. Any hornornorphisrn of a fieId F onto itself is an autornorphisrn of F.

INTEGRAL DOMAINS AND FIELDS

59

Any ring with identity which is a subring of a field rnust of necessity be an integral dornain. Turning the situation around, one rnight ask whether. each integral dornain can be considered (apart frorn isornorphisrn) as a subríng of sorne fieId. More forrnaIly: Can a given integral dornain be irnbedded in a fieId? In the finite case there is plainly no difficulty, since any finite integral dornain already forrns a field. Our concern with this question arises frorn the natural desire to solve the linear equation ax = b, where a =1= O. A rnajor drawback to the notion of an integral dornain is that it does not always furnish a solution within the systern. (Of course, any such solution would ha ve to be unique, sin ce aX l = b = aX 2 irnplies that Xl = x 2 by the canceIlation law.) It hardly seerns necessary to point out that when the integral dornain happens to be a field, the equation ax = b (a =1= O) is always solvable, for one need only take x = a-lb. We begin our discussion of this problern with an obvious definition. Definition 4-2. By a subfield of a field F is rneant any subring P of F which is itself a field. For example, the ring Q of rational nurnbers is a subfield of the real field R # ; the sarne is true of the field F = {a + bJ2la, b E Q}. SureIy, the set P wilI be a subfield of the fieId F provided that (1) F' is a subgroup of the additive group of F and that (2) F' - {O} is a subgroup of the multiplicative group F - {O}. RecaIling our mínimal set of conditions for deterrnining s,ubgroups (see page 8), it foIlows that F' wilI be a subfield of F if and only if the foIlowing requirernents are rnet: 1) Pis a nonernpty subset of F containing at least one nonzero elernent, 2) a, b E F' irnply a - b E F', and 3) a, b E P, with b =1= O, irnply ab- t E F'. The corning theorern furnishes c1ue to the nature of the field in which we wish to irnbed a given integral dornain.

a

Theorem 4-8. Let the integral dornain R be a subring of the field F. If the set F' is defined by P

=

{ab-lla,bER;b =1=

O},

then P forrns a subfield of F with R 5; P; in fact, F' is the srnaIlest (in the sense of inc1usion) subfield of F containing R. Proof. Notice first that 1 = 11- 1 E P, so that F' =1= {O}. °Now consider two arbitrary elernents x, y of F'. With reference to the definition of P, we then have

60

Cor a suitable choice oC a, b, e, d E R, where b computation shows that

x

y = (ad

xy-l

i=

O, d

i=

O.

A simple

i=

= (ad)(eb)-l E F'.

~~"

a = al,

al- 1 E F '

for each a in!R, implying that R F ' . FinaHy; any subfield of F whiCh contruns R necéssarily'includes all products ab-l, with a, O i= bE R, and, hence, contaiús
classieal ring ofquotients of R is any ring Qc¡(R} satisCying the conditions 1) R S;;; Qc¡(R), 2} every element of Qc¡{R) has the form ah - 1, where a, b E R and b is a non-zero-divisor of R, and 3} every non-zero-divisor of R is invertible in Qc¡(R}.

Convention. An eIement a E R is termed a non-zero-divisor if ar i= 0, and ra i= for aH i= rE R; in particular, the phrase "non-zero-divisor"

°

°

exc1udes the zero element. As a starting point, let S denote the set of aH elements of R, a commutative ring, which are non-zero-divisors; we wíI1 assume hereafter that S i= 0· Needless to say, ifthere happens to be an identityelement 1 available, then 1 ES. NoHce, too, that the set S is dosed under multiplication. For, suppose that the elements S1' S2 E S and (s1s2)a = O. Then Sl(S2 a) = O and, since S1 is not a divisor of zero, it follows that S2a = O; this in turn implies thaf a O. Therefore,the product S1S2 is not a zero divisor oC R, whence S1S2 E S.

' {(a, s)la E R, s E S}.

RxS

A relation ~ may be introduced in R x S by taking

(a, s) "" (b, r) ir and only iC ar = bs.

O), we conc1ude that

By viitue of the remarks following Definition 4--2, this is sufficient to establish that the set F' is a subfield oC F.' Furthermore, . ,

Now consider the set of ordered pairs

'

be)(bd)-lEF'.

Also, if Y i8 nonzero (that is, whenever e

61

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE IN RINGS AND IDEALS

~We have in mind theprevious theorem, where ab- 1 lfad

= be.}'

=

ed- 1 if a~d ~>nly ',

, It is not difficult t~ verify tl:1at the relation "', thus defined, Ú- an equivalence relation in R x S. The transitive property is perhaps theléast obvious. To see this, assume that (a, s) "" (b, r) and (b,;') '" (e, t), so, ~hat

ar

=

sb,

bt

re.

'. ' '

'o·-·'!.'

Now multiply the first equation by t and the second by s '1.0 get

art

',;: -;-

= sbt,

sbt

sre.

Putting these'relations together, we obtain atr ser. $ince:r.js not a zero divisor, the canceIlation lawgives us at = se, which,is exactly the condition that (a, s) '" (e, t). Next, we label those elements of R xS which are equivalent to thepair (a, s) by thesymbol a/s; in other words, '

a/s

=

{(b, r)l(a, s) '" (b, r)} = sb}. .

= {(b, r) ar

The coIlection of all equivaIence cIasses a/s relative to '" will be denoted 'by Qc¡(R): Q.,¡(R). = {a/sla

E

R; s E S}.

From Theore¡n A-1, we knowthat the elements of Qcl(R) constitute a parti,tion of the set R x S. That is, the ordered pairs in Ji. x S faH into disjoint c1asses (calledformalfraetions), with each dass consisting of equivalent pairs, andnonequivalent pairs belonging to different cIasses. Further, two such c1asses a/s and b/r are identical if and only if ar = sb; in particular, all fractions ofthe form as/s, with s E S, are'equal. With these remarks in mind, let us introduce the operations of addition and niultiplication required to make Qc¡(R) into a ringo We do this by means of the formulas

a/s +, b/r (a/s)(b/r)

= =

(ar + sb)/sr, ab/sr.

Notice, incidentaIly, that since the set S is c10sed under multiplication ' ,the right-hand sides oC the defining equations are meaningful. , As usual, our first task is to justiry that these operations are well-defined ;

60

Cor a suitable choice oC a, b, e, d E R, where b computation shows that

x

y = (ad

xy-l

i=

O, d

i=

O.

A simple

i=

= (ad)(eb)-l E F'.

~~"

a = al,

al- 1 E F '

for each a in!R, implying that R F ' . FinaHy; any subfield of F whiCh contruns R necéssarily'includes all products ab-l, with a, O i= bE R, and, hence, contaiús
classieal ring ofquotients of R is any ring Qc¡(R} satisCying the conditions 1) R S;;; Qc¡(R), 2} every element of Qc¡{R) has the form ah - 1, where a, b E R and b is a non-zero-divisor of R, and 3} every non-zero-divisor of R is invertible in Qc¡(R}.

Convention. An eIement a E R is termed a non-zero-divisor if ar i= 0, and ra i= for aH i= rE R; in particular, the phrase "non-zero-divisor"

°

°

exc1udes the zero element. As a starting point, let S denote the set of aH elements of R, a commutative ring, which are non-zero-divisors; we wíI1 assume hereafter that S i= 0· Needless to say, ifthere happens to be an identityelement 1 available, then 1 ES. NoHce, too, that the set S is dosed under multiplication. For, suppose that the elements S1' S2 E S and (s1s2)a = O. Then Sl(S2 a) = O and, since S1 is not a divisor of zero, it follows that S2a = O; this in turn implies thaf a O. Therefore,the product S1S2 is not a zero divisor oC R, whence S1S2 E S.

' {(a, s)la E R, s E S}.

RxS

A relation ~ may be introduced in R x S by taking

(a, s) "" (b, r) ir and only iC ar = bs.

O), we conc1ude that

By viitue of the remarks following Definition 4--2, this is sufficient to establish that the set F' is a subfield oC F.' Furthermore, . ,

Now consider the set of ordered pairs

'

be)(bd)-lEF'.

Also, if Y i8 nonzero (that is, whenever e

61

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE IN RINGS AND IDEALS

~We have in mind theprevious theorem, where ab- 1 lfad

= be.}'

=

ed- 1 if a~d ~>nly ',

, It is not difficult t~ verify tl:1at the relation "', thus defined, Ú- an equivalence relation in R x S. The transitive property is perhaps theléast obvious. To see this, assume that (a, s) "" (b, r) and (b,;') '" (e, t), so, ~hat

ar

=

sb,

bt

re.

'. ' '

'o·-·'!.'

Now multiply the first equation by t and the second by s '1.0 get

art

',;: -;-

= sbt,

sbt

sre.

Putting these'relations together, we obtain atr ser. $ince:r.js not a zero divisor, the canceIlation lawgives us at = se, which,is exactly the condition that (a, s) '" (e, t). Next, we label those elements of R xS which are equivalent to thepair (a, s) by thesymbol a/s; in other words, '

a/s

=

{(b, r)l(a, s) '" (b, r)} = sb}. .

= {(b, r) ar

The coIlection of all equivaIence cIasses a/s relative to '" will be denoted 'by Qc¡(R): Q.,¡(R). = {a/sla

E

R; s E S}.

From Theore¡n A-1, we knowthat the elements of Qcl(R) constitute a parti,tion of the set R x S. That is, the ordered pairs in Ji. x S faH into disjoint c1asses (calledformalfraetions), with each dass consisting of equivalent pairs, andnonequivalent pairs belonging to different cIasses. Further, two such c1asses a/s and b/r are identical if and only if ar = sb; in particular, all fractions ofthe form as/s, with s E S, are'equal. With these remarks in mind, let us introduce the operations of addition and niultiplication required to make Qc¡(R) into a ringo We do this by means of the formulas

a/s +, b/r (a/s)(b/r)

= =

(ar + sb)/sr, ab/sr.

Notice, incidentaIly, that since the set S is c10sed under multiplication ' ,the right-hand sides oC the defining equations are meaningful. , As usual, our first task is to justiry that these operations are well-defined ;

62

INTEGRAL OOMAINS ANO FIELOS

FIRST COURSE IN RINGS ANO IDEALS

that is to say, it is necessary to show that the sum and product are independent of the particular elements of R used in their definition. Let us present the argument for addition in detail. Suppose, then, that a/s = a'/s' and b/r = b'/r'; we must show that

(ar + sb)/sr

(dr' + sb')/s'r'.

=

From what is given, it follows at once that

as'

= sa',

br' = rb'.

These equations imply

(ar + sb)(s'r') - (a'r' + s'b')(sr)

= =

(as' - sa')(rr') + (br' - rb')(ss') O(rr') + O(ss') = O.

By the definition of equality of equivalence classes, this amounts to saying that

(ar + sb)/sr

=

(a'r' + s'b')/s'r',

which proves addition to be well-defined. In much the same way, one can establish that

ab/sr

=

a'b'/s'r'.

The next lemma reveals the algebraic nature of Qc¡(R) under these operations. Lemma. The syst~m Qc¡(R) forms a commutative ring with identity.

Proof. It is an entirely -straightforward matter to confirm that Qc¡(R) is a

commutative ringo We leave the reader to make the necessary verifications at his leisure, and merely point out that O/s serves as the zero element, while -a/s is the negative oCa/s. That the equivalence class s'/s', where s' is any fixed non-zero-divi~or of R, constitutes themultiplicative identity is evidenced by the followmg computation: ': (a/s)(s'/s') = as'/ss = a/s "

for arbitrary a/s in Qc¡(R), since (as')s = (ss')a. Loosely speaking, common factors belonging to S may be .canceIled in a fracti
Proof. We begin by establishing that the ring Qc¡{R) contains a subring isomorphic to R. For this, consider the subset K of Qc¡(R) consisting of all

63

elements of the form aso/so, where So is a fixed non-zero-divisor of R (recall that the equivalence class aso/so depends only upon a, not upon the choice of so):

The reader can easily check that K is a subring of Qc¡(R). An obvious (onto) mappingf: R -+ K is defined by takingf(a) = aso/so. Since the condition aso/so = bso/so implies that as~ = bs~ or, after cancelling, that a = b, f will be a one-to-one function. Furthermore, it has the property of preserving both addition and multiplication:

fea + b) = (a + b)so/so = aso/so + bso/so = fea) + f(b) ¡(ab) = (ab)so/so = (ab)s~/s~ = (aso/so)(bso/so) = f(a)f(b). In this way, R can be isomorphically embedded in Qc¡(R). By identifying R with K, we may henceforth regard R as actually being contained in Qc¡(R). In practice, one simply replaces the fraction aso/so E K by the corresponding element a E R. We proceed to show that all the elements of S are invertible in Qc¡{R). Any non-zero-divisor s E S has, after identification, the formsso/s o' Now, the equivalence class so/ss o is also a member of Qc¡(R) (note the crucial use of the closure of S under multiplication) and satisfies the equation

(sso/so)(so/sso)

=

ss~/ss~

= so/so.

Since so/so plays the role of the identity element for Qc¡(R), we see at once that (ssO/SO)-l = so/sso' . All that remains to complete the proof is to verify that each member a/s of Q.¡(R) can be written as as-l. It should be clear that

a/s

=

(aso/so)(so/sso)

=

(aso/so)(ssO/SO)-l.

Replacing aso/so by a and sso/so by s, the displayed equation assumes the more familiar form a/s = as-l. The point is this: the set Qc¡(R) may now be interpreted as consisting of all quotients as- l , where a E R, s E S. Thus, Qc¡(R) satisfies Definition 4-3 in its entirety, thereby becoming a classical ring of quotients of R. Two comments are in order. In the first place, given any element s E R which is a non-zero-divisor, it follows that

(so

E

S).

Identifying sso/so with s and aso/so with a, we conclude from this that the equation sx = a always possesses a solution in Qc¡(R), namely, x = a/s = as-l. Second, notice that in Qc¡(R) multiplicative inverses exist not only for

62

INTEGRAL OOMAINS ANO FIELOS

FIRST COURSE IN RINGS ANO IDEALS

that is to say, it is necessary to show that the sum and product are independent of the particular elements of R used in their definition. Let us present the argument for addition in detail. Suppose, then, that a/s = a'/s' and b/r = b'/r'; we must show that

(ar + sb)/sr

(dr' + sb')/s'r'.

=

From what is given, it follows at once that

as'

= sa',

br' = rb'.

These equations imply

(ar + sb)(s'r') - (a'r' + s'b')(sr)

= =

(as' - sa')(rr') + (br' - rb')(ss') O(rr') + O(ss') = O.

By the definition of equality of equivalence classes, this amounts to saying that

(ar + sb)/sr

=

(a'r' + s'b')/s'r',

which proves addition to be well-defined. In much the same way, one can establish that

ab/sr

=

a'b'/s'r'.

The next lemma reveals the algebraic nature of Qc¡(R) under these operations. Lemma. The syst~m Qc¡(R) forms a commutative ring with identity.

Proof. It is an entirely -straightforward matter to confirm that Qc¡(R) is a

commutative ringo We leave the reader to make the necessary verifications at his leisure, and merely point out that O/s serves as the zero element, while -a/s is the negative oCa/s. That the equivalence class s'/s', where s' is any fixed non-zero-divi~or of R, constitutes themultiplicative identity is evidenced by the followmg computation: ': (a/s)(s'/s') = as'/ss = a/s "

for arbitrary a/s in Qc¡(R), since (as')s = (ss')a. Loosely speaking, common factors belonging to S may be .canceIled in a fracti
Proof. We begin by establishing that the ring Qc¡{R) contains a subring isomorphic to R. For this, consider the subset K of Qc¡(R) consisting of all

63

elements of the form aso/so, where So is a fixed non-zero-divisor of R (recall that the equivalence class aso/so depends only upon a, not upon the choice of so):

The reader can easily check that K is a subring of Qc¡(R). An obvious (onto) mappingf: R -+ K is defined by takingf(a) = aso/so. Since the condition aso/so = bso/so implies that as~ = bs~ or, after cancelling, that a = b, f will be a one-to-one function. Furthermore, it has the property of preserving both addition and multiplication:

fea + b) = (a + b)so/so = aso/so + bso/so = fea) + f(b) ¡(ab) = (ab)so/so = (ab)s~/s~ = (aso/so)(bso/so) = f(a)f(b). In this way, R can be isomorphically embedded in Qc¡(R). By identifying R with K, we may henceforth regard R as actually being contained in Qc¡(R). In practice, one simply replaces the fraction aso/so E K by the corresponding element a E R. We proceed to show that all the elements of S are invertible in Qc¡{R). Any non-zero-divisor s E S has, after identification, the formsso/s o' Now, the equivalence class so/ss o is also a member of Qc¡(R) (note the crucial use of the closure of S under multiplication) and satisfies the equation

(sso/so)(so/sso)

=

ss~/ss~

= so/so.

Since so/so plays the role of the identity element for Qc¡(R), we see at once that (ssO/SO)-l = so/sso' . All that remains to complete the proof is to verify that each member a/s of Q.¡(R) can be written as as-l. It should be clear that

a/s

=

(aso/so)(so/sso)

=

(aso/so)(ssO/SO)-l.

Replacing aso/so by a and sso/so by s, the displayed equation assumes the more familiar form a/s = as-l. The point is this: the set Qc¡(R) may now be interpreted as consisting of all quotients as- l , where a E R, s E S. Thus, Qc¡(R) satisfies Definition 4-3 in its entirety, thereby becoming a classical ring of quotients of R. Two comments are in order. In the first place, given any element s E R which is a non-zero-divisor, it follows that

(so

E

S).

Identifying sso/so with s and aso/so with a, we conclude from this that the equation sx = a always possesses a solution in Qc¡(R), namely, x = a/s = as-l. Second, notice that in Qc¡(R) multiplicative inverses exist not only for

¡l'

.'

64

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE lN' RINGS AND JDJ;ALS

members of S but for all elements of Qcl(R) which can be repiesented in the form r/s, where r, s are both non-zero-divisors; in fact, (r/s)(s/r) = rs/sr = so/so· When thering R is an integral domain, we may take the set S ofnon-zerodivisors as consisting of a1l the elements of R which are not Zefo. The last remark of the preceding paragraph then leads to the following:important theorem.' Theorem 4-10. For any integral domain R, the system Qc¡(R) forms a field, customari\y known as the field of quotients of R. Since an integral domain is (isomorphic to) a subring of lts field .of .... quotients, we also obtain Corollary. A ring is an integral domain if and only if it isasubring of a field:' . '. It should be pointed out that the hypothesis of commutativify:is essential to this last theorem; indeed, there exist noncommutative rings without divisors ofzero that cannot be imbedded in any division ringo The field of quotients constructed from the integral domain Z is, of course, the rational number field Q. Another fact of interest is that the field . of quotients is the smallest field in which an integral domain R ca~ be imbedded, in the sense that any field in which R is imbeddable contams a subfield isomorphic to QcI(R) (Problem 20). . The existence theorem for the classical ring of quotients can be supplemented by the following result, which shows that it is essentia11y unique.

Theorem 4-11. Let R and R' be two commutative rings, each containing at least'one non-zero-devisor. Then, any isomorphism of R onto R ' has a / únique extension to an isomorphism of QCI(R) onto Qcl(R );

Proo! To begin with, each member of Qcl(R) may be written in the form ab -1 where a b E R and bis a non-zero-divisor in R. Given an isomorphism 1 cp: R'-4 R' , the element cp(b) will be a no~-zero-divis~r of R' , s~ that cp(bt is present in Qcl(R /). Suppose that cp admlts an extenslOn to an Isomorphlsm 1 <1>: Qcl(R) -4 Qcl(R/). Sin ce a = (ab- )b, we would then ha ve . cp(a)

= (a)

= CD(ab- 1)(b)

=

(ab- 1)cp(b),

which, as a result, yields (ab- 1) = c!>(a)cp(b)-1. Thus, <1> is completely determined by the effect of cp on R and so determined uniquely, if it exists. at a11. . These remarks suggest that, in attempting to extend cp, we should consider the assignment: for a11

65

For a verification that· <1> is a well-defi~ed function, let ab -1 = er 1 in Qcl(R); that is to say, ad = be in R. Then. the equation c!>(a)c!>(d) = cp(b)cp(e) holds in R' ~ Qcl(R /) or, viewed otherwise, cp(a)cp(b)-1 = cp(e)cp(d)-1.But this means. (ab -1) = (er 1), .so that <1> do es not depend on how an element in Q~I(R) is expressed as a quotient. One verifies routinely that <1>, as defineci aboye, is a homomorphism of Q~I(R) into Qc¡(R /). . This homomorphism'certainly extends cp; indeed, if a is an arbitrary element of R and b is a .non-zero-divisor of R, <1> maps a = (db)b-1eititb (l., (a)

= cp(ab)ep(b)-1

= cp(a)~(b)cp(b)-1 .',

=

cp(a).

":

To see that· <1> is a one-tó-one function,we examine its kernel. Now, if (ab -1) = O, then cp(a) = O. But,cp being an isomorphism, this implies that .a· = 0, whence ab- 1 = O. Accordingly, ker == {O}, which forces <1> to be one-to-one. Without going into the'details, we also point out that <1> carries Qcl(R) onto Qcl(R /) (thi.s .stems frop?·;the fact that cpmaps onto R /). Therefore, <1> is the desired extensionof cp. '.' . A special case of particular importance occurs when R and R ' are the same ring and cp is taken to be the identity isomorphism on R. Coro]]ary. Any two quotient rings of a commutative ring R with at least one non-zéro-divisor are isomorphic by a unique mapping fixing all the elements of R. At this point we leave the theory of quotients and turn to prime fields. Cleady, any field F has at least one subfield, namely, F itself; a field which does not possess any proper subfields is ca11ed a prime field. Examp]e 4-4. The field Q bf rational numbers is the simplest example of a prime field. To see this, suppose that Fis any subfield of Q and let a E F be any nonzero element. Since F is a subfield of Q, it must contain the prod uct aa -1 = 1. In turn, n = nI E F for any n in Z; in other words, F contains all the integers. It then follows that every rational númber n/m =nm- 1 (m =1= O) also lies in F, so that F = Q. Example 4-5. For each priine p, the field Zp ofintegers modulo pis a prime field. The reasoning here depends on the fact that the additive group (Zp, +p) is a finite group ofprime order and therefore by Lagrange's theorem has no non-trivial subgroups. An observation which will not detain us long is that each field F contains a unique prime subfield. To make things more specific, let {F¡} be the collection of all subfields of F. Then the intersection (\ F¡ is also a subfield of F. Now, if F' is any subfield of the field (\ F¡, then F' E {F¡}, whence (\ F¡ ~ F' ; the implication is that F' = (\ F¡, forcing (\ F¡ to be a: prime

¡l'

.'

64

INTEGRAL DOMAINS AND FIELDS

FIRST COURSE lN' RINGS AND JDJ;ALS

members of S but for all elements of Qcl(R) which can be repiesented in the form r/s, where r, s are both non-zero-divisors; in fact, (r/s)(s/r) = rs/sr = so/so· When thering R is an integral domain, we may take the set S ofnon-zerodivisors as consisting of a1l the elements of R which are not Zefo. The last remark of the preceding paragraph then leads to the following:important theorem.' Theorem 4-10. For any integral domain R, the system Qc¡(R) forms a field, customari\y known as the field of quotients of R. Since an integral domain is (isomorphic to) a subring of lts field .of .... quotients, we also obtain Corollary. A ring is an integral domain if and only if it isasubring of a field:' . '. It should be pointed out that the hypothesis of commutativify:is essential to this last theorem; indeed, there exist noncommutative rings without divisors ofzero that cannot be imbedded in any division ringo The field of quotients constructed from the integral domain Z is, of course, the rational number field Q. Another fact of interest is that the field . of quotients is the smallest field in which an integral domain R ca~ be imbedded, in the sense that any field in which R is imbeddable contams a subfield isomorphic to QcI(R) (Problem 20). . The existence theorem for the classical ring of quotients can be supplemented by the following result, which shows that it is essentia11y unique.

Theorem 4-11. Let R and R' be two commutative rings, each containing at least'one non-zero-devisor. Then, any isomorphism of R onto R ' has a / únique extension to an isomorphism of QCI(R) onto Qcl(R );

Proo! To begin with, each member of Qcl(R) may be written in the form ab -1 where a b E R and bis a non-zero-divisor in R. Given an isomorphism 1 cp: R'-4 R' , the element cp(b) will be a no~-zero-divis~r of R' , s~ that cp(bt is present in Qcl(R /). Suppose that cp admlts an extenslOn to an Isomorphlsm 1 <1>: Qcl(R) -4 Qcl(R/). Sin ce a = (ab- )b, we would then ha ve . cp(a)

= (a)

= CD(ab- 1)(b)

=

(ab- 1)cp(b),

which, as a result, yields (ab- 1) = c!>(a)cp(b)-1. Thus, <1> is completely determined by the effect of cp on R and so determined uniquely, if it exists. at a11. . These remarks suggest that, in attempting to extend cp, we should consider the assignment: for a11

65

For a verification that· <1> is a well-defi~ed function, let ab -1 = er 1 in Qcl(R); that is to say, ad = be in R. Then. the equation c!>(a)c!>(d) = cp(b)cp(e) holds in R' ~ Qcl(R /) or, viewed otherwise, cp(a)cp(b)-1 = cp(e)cp(d)-1.But this means. (ab -1) = (er 1), .so that <1> do es not depend on how an element in Q~I(R) is expressed as a quotient. One verifies routinely that <1>, as defineci aboye, is a homomorphism of Q~I(R) into Qc¡(R /). . This homomorphism'certainly extends cp; indeed, if a is an arbitrary element of R and b is a .non-zero-divisor of R, <1> maps a = (db)b-1eititb (l., (a)

= cp(ab)ep(b)-1

= cp(a)~(b)cp(b)-1 .',

=

cp(a).

":

To see that· <1> is a one-tó-one function,we examine its kernel. Now, if (ab -1) = O, then cp(a) = O. But,cp being an isomorphism, this implies that .a· = 0, whence ab- 1 = O. Accordingly, ker == {O}, which forces <1> to be one-to-one. Without going into the'details, we also point out that <1> carries Qcl(R) onto Qcl(R /) (thi.s .stems frop?·;the fact that cpmaps onto R /). Therefore, <1> is the desired extensionof cp. '.' . A special case of particular importance occurs when R and R ' are the same ring and cp is taken to be the identity isomorphism on R. Coro]]ary. Any two quotient rings of a commutative ring R with at least one non-zéro-divisor are isomorphic by a unique mapping fixing all the elements of R. At this point we leave the theory of quotients and turn to prime fields. Cleady, any field F has at least one subfield, namely, F itself; a field which does not possess any proper subfields is ca11ed a prime field. Examp]e 4-4. The field Q bf rational numbers is the simplest example of a prime field. To see this, suppose that Fis any subfield of Q and let a E F be any nonzero element. Since F is a subfield of Q, it must contain the prod uct aa -1 = 1. In turn, n = nI E F for any n in Z; in other words, F contains all the integers. It then follows that every rational númber n/m =nm- 1 (m =1= O) also lies in F, so that F = Q. Example 4-5. For each priine p, the field Zp ofintegers modulo pis a prime field. The reasoning here depends on the fact that the additive group (Zp, +p) is a finite group ofprime order and therefore by Lagrange's theorem has no non-trivial subgroups. An observation which will not detain us long is that each field F contains a unique prime subfield. To make things more specific, let {F¡} be the collection of all subfields of F. Then the intersection (\ F¡ is also a subfield of F. Now, if F' is any subfield of the field (\ F¡, then F' E {F¡}, whence (\ F¡ ~ F' ; the implication is that F' = (\ F¡, forcing (\ F¡ to be a: prime

66

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

field. As regards the uniqueness assertion, suppose that K ¡ and K 2 are both prime subfields of F. Then K 1 n K 2 is a subfield of F as well as K ¡, with K¡ ;2 K¡ n K 2 • But K¡ can possess no proper subfields, which signifies thatK¡ = K¡ n K 2 • Likewise,K 2 = K¡ n K 2 , whenceK¡ = K 2 • We conc1ude this chapter by showing that, to within isómorphism, the rational number field and the fields Z pare the only prime fields. Theorem 4-12. Any prime field F is isomorphic either to Q, the field of'rational numbers, or to one of the fields Zp of integers modulo a prime p.

Proof. To begin, let 1 be the multiplicative identity of F and define the mapping f: Z ~ F by fin) = nI for any integer n. Then f is a homo~ morphism from Z onto the subring Zl of integral multiples of 1. In compliance with Theorem 3-7, we therefore have Z/ker f ~ Z1. But kerf is an ideal of Z, a principal ideal domain, whence ker f = (n) for sorne nonnegative integer n. The possibility that n = 1 can be ruled out, for otherwise 1 = f(l) = Oor, what amounts to the same thing, F = {O}. Notice further that if n =1= O, then n must in fact be a prime number. Suppose to the contrary that n = n¡n 2 ; where 1 < ni < n. Sin ce n E kerf, it follows that (n¡1)(n 2 1) = (n¡n 2 )1 = nI = O, yielding the contradiction that the field F has divisors of zero. (This result is not entirely unexpected, because the integer n is the characteristic of F and as such must be a prime, whenever n =1= O.) The preceding discussion indica tes that two possibilities arise: either 1) Zl ~ Z/(p) = Zp for sorne prime p, or 2) Zl ~ Z/(O) = Z. . Turning to a closer analysis of these cases, as sume flrst that Zl ~ Zp' with p prime. Inasmuch as the ring of integers modulO a prime forms a field, the subring Zl must itself be a field. But F, being a prime field, contains no proper subfields. Accordingly, Zl = F, 'wJ.:¡.ich leads to the ,_ isomorphism F ~ Z P' For the final stage of the proo( consider the situatiü\n where Zl ~ Z. Under these circumstances, the subring Zl is an integraldomain, but not a field. Taking stock of Theorem 4-8, as well as the hypothesis that F is a prime field, we conc1ude that F = {ab-¡Ia, bEZ1; b f O} = {(n1)(m1)-¡ln, m E Z; m

67

Since every field contains a unique prime subfield, the following subsidiary result is of interest. CoroUary l. Every field contains a subfield which is isomorphic either to the field Q or to one of the fields Z p' Theorem 4-12 also provides sorne information regarding field automorphisms. Corollary 2. If f is an automorphism of the field F, then f(a) = a for each element a in the prime subfield.of F (hence, a prime field has no automorphism except the identity).

Proof. The prime subfield of F is either F¡= {(n1)(m1)--¡ln,mEZ;m

=1=

O}

or

F2

=

{n1ln

=

0,1, ... ,p - 1},

according as the characteristic of F is O or a prime p. Since any automorphism of a field carries the identity 1 onto itself, the result should be c1ear.

PROBLEMS 1. a) Assuming that R is a division ring, show that cent R forms a field. b) Prove that every subring, with identity, ofa field is an integral domain. 2. Let R be an integral domain and consider the set Z1 of aH integral multiples of the identity eleme~t : Z1 = {n1ln e Z}. Establish that Z1 is a field if and onlyif R has positive characteristic. 3. In the field e, define a mapping f: e .:.. e by sending each complex number to its conjugate; that is,f(a + bi) = a --:-: ~i. Verify thatfis an automorphism of C. 4. FiiId the center of the quaternion rifig H. 5. Leí R be the subring of M 2( C) consisting of aH matrices of the form

(_~ ~)= (_~ 1~: ~ ~ ~~)

(a,b,e,deR#).

Prove that R is a division ring isomorphic to the division ring of real quaternions. =1=

O}.

It is now a purely routine matter to verify that the fields F and Q are isomorphic under the mapping g(n/m) = (n1)(m1)-1; we leave the details as an exercise.

6. By the quaternions over a field F is meant the set of aH q = a + bi + ej + dk, where a, b, e, d e F and where addition and multiplication are carried out as with real quaternions. Given that F is a field in which al + bl + e2 + d 2 = O if and only if a = b = e = d = O, establish that the quat~rnions over F form a division ringo

66

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

field. As regards the uniqueness assertion, suppose that K ¡ and K 2 are both prime subfields of F. Then K 1 n K 2 is a subfield of F as well as K ¡, with K¡ ;2 K¡ n K 2 • But K¡ can possess no proper subfields, which signifies thatK¡ = K¡ n K 2 • Likewise,K 2 = K¡ n K 2 , whenceK¡ = K 2 • We conc1ude this chapter by showing that, to within isómorphism, the rational number field and the fields Z pare the only prime fields. Theorem 4-12. Any prime field F is isomorphic either to Q, the field of'rational numbers, or to one of the fields Zp of integers modulo a prime p.

Proof. To begin, let 1 be the multiplicative identity of F and define the mapping f: Z ~ F by fin) = nI for any integer n. Then f is a homo~ morphism from Z onto the subring Zl of integral multiples of 1. In compliance with Theorem 3-7, we therefore have Z/ker f ~ Z1. But kerf is an ideal of Z, a principal ideal domain, whence ker f = (n) for sorne nonnegative integer n. The possibility that n = 1 can be ruled out, for otherwise 1 = f(l) = Oor, what amounts to the same thing, F = {O}. Notice further that if n =1= O, then n must in fact be a prime number. Suppose to the contrary that n = n¡n 2 ; where 1 < ni < n. Sin ce n E kerf, it follows that (n¡1)(n 2 1) = (n¡n 2 )1 = nI = O, yielding the contradiction that the field F has divisors of zero. (This result is not entirely unexpected, because the integer n is the characteristic of F and as such must be a prime, whenever n =1= O.) The preceding discussion indica tes that two possibilities arise: either 1) Zl ~ Z/(p) = Zp for sorne prime p, or 2) Zl ~ Z/(O) = Z. . Turning to a closer analysis of these cases, as sume flrst that Zl ~ Zp' with p prime. Inasmuch as the ring of integers modulO a prime forms a field, the subring Zl must itself be a field. But F, being a prime field, contains no proper subfields. Accordingly, Zl = F, 'wJ.:¡.ich leads to the ,_ isomorphism F ~ Z P' For the final stage of the proo( consider the situatiü\n where Zl ~ Z. Under these circumstances, the subring Zl is an integraldomain, but not a field. Taking stock of Theorem 4-8, as well as the hypothesis that F is a prime field, we conc1ude that F = {ab-¡Ia, bEZ1; b f O} = {(n1)(m1)-¡ln, m E Z; m

67

Since every field contains a unique prime subfield, the following subsidiary result is of interest. CoroUary l. Every field contains a subfield which is isomorphic either to the field Q or to one of the fields Z p' Theorem 4-12 also provides sorne information regarding field automorphisms. Corollary 2. If f is an automorphism of the field F, then f(a) = a for each element a in the prime subfield.of F (hence, a prime field has no automorphism except the identity).

Proof. The prime subfield of F is either F¡= {(n1)(m1)--¡ln,mEZ;m

=1=

O}

or

F2

=

{n1ln

=

0,1, ... ,p - 1},

according as the characteristic of F is O or a prime p. Since any automorphism of a field carries the identity 1 onto itself, the result should be c1ear.

PROBLEMS 1. a) Assuming that R is a division ring, show that cent R forms a field. b) Prove that every subring, with identity, ofa field is an integral domain. 2. Let R be an integral domain and consider the set Z1 of aH integral multiples of the identity eleme~t : Z1 = {n1ln e Z}. Establish that Z1 is a field if and onlyif R has positive characteristic. 3. In the field e, define a mapping f: e .:.. e by sending each complex number to its conjugate; that is,f(a + bi) = a --:-: ~i. Verify thatfis an automorphism of C. 4. FiiId the center of the quaternion rifig H. 5. Leí R be the subring of M 2( C) consisting of aH matrices of the form

(_~ ~)= (_~ 1~: ~ ~ ~~)

(a,b,e,deR#).

Prove that R is a division ring isomorphic to the division ring of real quaternions. =1=

O}.

It is now a purely routine matter to verify that the fields F and Q are isomorphic under the mapping g(n/m) = (n1)(m1)-1; we leave the details as an exercise.

6. By the quaternions over a field F is meant the set of aH q = a + bi + ej + dk, where a, b, e, d e F and where addition and multiplication are carried out as with real quaternions. Given that F is a field in which al + bl + e2 + d 2 = O if and only if a = b = e = d = O, establish that the quat~rnions over F form a division ringo

68

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

7. Establish the fo11owing faets eoneeming the Euler phi~funetion: a) If n andm are relatively prime integers, then .p(nm) = .p(n).p(m). b) For any prime P and n> O, ~(P') = p"(l - l/p) p' p.-¡. [Hint: The integers k sueh that O < k < p' and ged (le, p,) =1= 1 are p, 2p, ... , p"-lp.] e) If PI' P2' ... ,.p, are the distinct prime divisors of an ínteger n > 1, then .p(n) = n(1 l/p¡)(l 1/P2) ... (1 - l/p,). d) n = L.,I• .p(a). .,

16. Let F be a field of eharaeteristie p > O. Show that for fixed n E Z +} F' = {a E FlaP· = a is a subfield of F.

17. Let F be a field. F' a subfield of F, andfan automorphism of F. We say thatf flXes an element a E F in case fea) a. Prove the following assertíons: a) The.set of a11 automorphisms of F form a group (in which the binary operation is eomposition of funetions): . b) The.kt of aU automorphisms of F which fix eaehelement of F' eomprise a group. e) If G'is a group of automorphisms of F, then the set of all eIements of F that ar~'fixed by G (that is, the set F(G) = {a E Flf(a) = a for alifE G}) is a subfield'of F. known as the fixed field of G. .

'.

8. Let 1(~) denote the number of (dístinet) positive divisors of an integer n > 1. Prove that . , ' a) If n has the prime faetonzatíon n = pilpi' ... P'k" where the Pi are distinet primes and ni E Z+. then 1(n)'= (n¡ + 1)(n2 + 1) ... (n, + 1), b) The number of ideals of Z,i¡~t(n). e) "t"{n).p(n);;:: n, [Hint: II(n¡t1)II(1 - l/Pi) ;;:: 2'n(1/2)k.] ,; .

}

I

Th'iJfem:

10. a) Derive Fermat's Little If p is a prime number and a", O (mod p), then ar ¡ == 1 (mod p). ., b) If gcd (a, n) = 1, show thal the equation ax == b (mod n) has a unique solution modulo n. [Hint: AH solutions are given by x = bat!>(m-l)+ kn.]

di~lg¡'r of R. b) IfR-'has an identity and every non-zero-divisor of R ísÚ!"ertible iriR. then Rf=<.Qcl(R); in particular, F QcI(F) for any field F. e) Q.lni.l(R)) = Qc¡(R)~ . .. d) Ir R is finite, then R Qcl(R). [Hint: For any non-zero-divísor a E R, there is sorne bE R such that a2 b = a; ah is idempotent; thus, R has an identity element 1 and ab = 1 by Problem 12, Chapter 1.]

19. Utilize part (d) of the preeeding problern to give another proof that any finite integral domain ls a fieId.

JI. a) Prove thatevery field ís a principal ideal domain; . b) Show that the ring R = {a + b.J2Ia, b EZ} is not a field by exhibiting a nontrivial ideal of R.

20. Show that any field eontaining the integral domain R as a subriÍlg contains the field of quotients Q.I(R); in this sense, Q.¡(R) i8 the smallest field containing R.

12. Let f be a homomorphism from the ring R into the ring R' and suppose that R has a subring F whieh is a field. Establish that either F ~ ker f or else R' eontams a subring isomorphie to F. .

21. a) If R = {a + b.J2la, b E Z}, then R forms an integral domain under ordinary additíon and multiplication, but not a field. Obtain the field of quotients of R. Do the same fór the domain Ze' b) If K is a field of quotients of an integral domain R, prove that K is also a field of quotients of every subdomain of K containíng R.

13. Derive the fo11owing results: . a) The identity element of a subfield is the same as that of the field. .b) If {F¡}' is an index collection of subfields of the field F, then n F¡ is also a subfield of F. .. e) A subring F of a field F is a subfield of F if and only if F eontains at least one nonzero element and a- ¡ E F for every nonzero a E F', d) A subset F of a finite field F is a subfield of F if and only if F' eontains more than one element and is c10sed under addition and multiplieation.

22. Let R be an arbitrary ring (not necessarily commutative) with al least one non-zerodivisor. Prove that R possesses a c1assicaI ring of quotients if and only if it satisfies the so-ealled Ore eondition: for aU a, b E R. b being a non-zero-divisor, there exíst elements e, d E R, with d a non-zero-divisor such that ad be. 23. Prove tbat any automorpbism of an integral domain R admits a unique extension . to the field of quotients Qcl(R).

14. a) Consider the subset S of R # defined by S = {a

~ ~(

18. Let R he a commutative ring eontaining at least one non-zero-divisor. Prove that a) Aílé'lement ab- 1 is a non-zero-divisor of Q'I(R) if and only if a is a non-zero-

'

9. Given that the set H~= {[aJé Z,I[a] is not a zero divisor of Z.}, prove that (H" 'h) forms a finite gioup 'qforder .p(r¡).

69

24. Let F be a field and 21 the set of integral multiples of the identity.· Verify that the prime subfield of F coincides with Q,¡(Zl). [Hint: Problem 20.J

+ b.jPla, bE Q; p a fixed prime}.

Show that S is a subfield of R#. b) Prove that any subfield of the field R# must eontain the rational.numbers.

25. Establish the following assertion, thereby completing the proof of Theorem 4-12: If F is a field of eharaeteristie zero and

15. Prove that if the field F is of eharaeteristie p > O, then every subfield of F has eharacteristie p.

K "

i

1

= {(n1)(m1)-1In. m E Z; m =1= O)

is the prime subfieId of F, then K

C!!

Q via the mappingf(n/m)

(nl)(m1)-¡.

./..

68

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

7. Establish the fo11owing faets eoneeming the Euler phi~funetion: a) If n andm are relatively prime integers, then .p(nm) = .p(n).p(m). b) For any prime P and n> O, ~(P') = p"(l - l/p) p' p.-¡. [Hint: The integers k sueh that O < k < p' and ged (le, p,) =1= 1 are p, 2p, ... , p"-lp.] e) If PI' P2' ... ,.p, are the distinct prime divisors of an ínteger n > 1, then .p(n) = n(1 l/p¡)(l 1/P2) ... (1 - l/p,). d) n = L.,I• .p(a). .,

16. Let F be a field of eharaeteristie p > O. Show that for fixed n E Z +} F' = {a E FlaP· = a is a subfield of F.

17. Let F be a field. F' a subfield of F, andfan automorphism of F. We say thatf flXes an element a E F in case fea) a. Prove the following assertíons: a) The.set of a11 automorphisms of F form a group (in which the binary operation is eomposition of funetions): . b) The.kt of aU automorphisms of F which fix eaehelement of F' eomprise a group. e) If G'is a group of automorphisms of F, then the set of all eIements of F that ar~'fixed by G (that is, the set F(G) = {a E Flf(a) = a for alifE G}) is a subfield'of F. known as the fixed field of G. .

'.

8. Let 1(~) denote the number of (dístinet) positive divisors of an integer n > 1. Prove that . , ' a) If n has the prime faetonzatíon n = pilpi' ... P'k" where the Pi are distinet primes and ni E Z+. then 1(n)'= (n¡ + 1)(n2 + 1) ... (n, + 1), b) The number of ideals of Z,i¡~t(n). e) "t"{n).p(n);;:: n, [Hint: II(n¡t1)II(1 - l/Pi) ;;:: 2'n(1/2)k.] ,; .

}

I

Th'iJfem:

10. a) Derive Fermat's Little If p is a prime number and a", O (mod p), then ar ¡ == 1 (mod p). ., b) If gcd (a, n) = 1, show thal the equation ax == b (mod n) has a unique solution modulo n. [Hint: AH solutions are given by x = bat!>(m-l)+ kn.]

di~lg¡'r of R. b) IfR-'has an identity and every non-zero-divisor of R ísÚ!"ertible iriR. then Rf=<.Qcl(R); in particular, F QcI(F) for any field F. e) Q.lni.l(R)) = Qc¡(R)~ . .. d) Ir R is finite, then R Qcl(R). [Hint: For any non-zero-divísor a E R, there is sorne bE R such that a2 b = a; ah is idempotent; thus, R has an identity element 1 and ab = 1 by Problem 12, Chapter 1.]

19. Utilize part (d) of the preeeding problern to give another proof that any finite integral domain ls a fieId.

JI. a) Prove thatevery field ís a principal ideal domain; . b) Show that the ring R = {a + b.J2Ia, b EZ} is not a field by exhibiting a nontrivial ideal of R.

20. Show that any field eontaining the integral domain R as a subriÍlg contains the field of quotients Q.I(R); in this sense, Q.¡(R) i8 the smallest field containing R.

12. Let f be a homomorphism from the ring R into the ring R' and suppose that R has a subring F whieh is a field. Establish that either F ~ ker f or else R' eontams a subring isomorphie to F. .

21. a) If R = {a + b.J2la, b E Z}, then R forms an integral domain under ordinary additíon and multiplication, but not a field. Obtain the field of quotients of R. Do the same fór the domain Ze' b) If K is a field of quotients of an integral domain R, prove that K is also a field of quotients of every subdomain of K containíng R.

13. Derive the fo11owing results: . a) The identity element of a subfield is the same as that of the field. .b) If {F¡}' is an index collection of subfields of the field F, then n F¡ is also a subfield of F. .. e) A subring F of a field F is a subfield of F if and only if F eontains at least one nonzero element and a- ¡ E F for every nonzero a E F', d) A subset F of a finite field F is a subfield of F if and only if F' eontains more than one element and is c10sed under addition and multiplieation.

22. Let R be an arbitrary ring (not necessarily commutative) with al least one non-zerodivisor. Prove that R possesses a c1assicaI ring of quotients if and only if it satisfies the so-ealled Ore eondition: for aU a, b E R. b being a non-zero-divisor, there exíst elements e, d E R, with d a non-zero-divisor such that ad be. 23. Prove tbat any automorpbism of an integral domain R admits a unique extension . to the field of quotients Qcl(R).

14. a) Consider the subset S of R # defined by S = {a

~ ~(

18. Let R he a commutative ring eontaining at least one non-zero-divisor. Prove that a) Aílé'lement ab- 1 is a non-zero-divisor of Q'I(R) if and only if a is a non-zero-

'

9. Given that the set H~= {[aJé Z,I[a] is not a zero divisor of Z.}, prove that (H" 'h) forms a finite gioup 'qforder .p(r¡).

69

24. Let F be a field and 21 the set of integral multiples of the identity.· Verify that the prime subfield of F coincides with Q,¡(Zl). [Hint: Problem 20.J

+ b.jPla, bE Q; p a fixed prime}.

Show that S is a subfield of R#. b) Prove that any subfield of the field R# must eontain the rational.numbers.

25. Establish the following assertion, thereby completing the proof of Theorem 4-12: If F is a field of eharaeteristie zero and

15. Prove that if the field F is of eharaeteristie p > O, then every subfield of F has eharacteristie p.

K "

i

1

= {(n1)(m1)-1In. m E Z; m =1= O)

is the prime subfieId of F, then K

C!!

Q via the mappingf(n/m)

(nl)(m1)-¡.

./..

70

FIRST COURSE IN RINGS AND IDEALS

I

FIVE

In Problems 26-29, R is assumed to be a eommutative ringo 26. Let S be any multiplicatively c/osed subset of the ring R (that is, the produet of any two elements of S again lies in S) whieh eontains no zero divisor of R and O~ S. If the set Rs is defined by Rs

=

{ab- 1 EQc¡(R)laER,bES},

MAXIMAL, PRIME, AND PRIMARY IDEALS

prove that Rs is a subring of the ring of quotients Qc¡(R), known as the ring of qUQtients of R relative to S. 27. a) Show that the set S = {n E Zlp { n; p a' fixed prime} is multiplieatively c10sed and.determine Zs, the ring of quotients of Z relative to S. b) If R is any ring satisfying Z ~ R ~ Q, prove that R = Zs for a suitable multiplieatively c10sed subset S ~ Z. [Hint: Cónsider the set S = {m E Zlfor sorne n E Z, nlm E ,R; ged (n, m) = 1}.]

The present chapter is devoted to a study of certain special types of ideals, most notably maximal, prime, and primary ideals. On the whole, our hypothesis will restrict us to commutative rings with identity. The requirement is motivated to sorne extent by the fact that many of the standard examples of ring theory have this property. Another reason, which is perhaps more important from the conceptual point of view, is that the most satisfactory and complete results occur here. We begin our discussion with the following definition.

28. Let S be a multiplieat,ively c10sed subset of the ring R with O ~ S. Prove!he statements below: a) The set 1 = {a E Rlas = O for sorne s E S} is an ideal of R. b) Sil = nat¡S is a multiplieatively closed subset of the quotient ring RI1. e) No element of Sil is a zero divisor of RIl. (Thus, one can foqn !he ring of quotients of RIl relative to Sil; the result is eal1ed the generalized ring of

Definition 5-1. An ideal 1 of the ring R is said to be a maximal ideal provided that 1 =1= R and whenever J is an ideal of R with 1 e J ~ R, then J = R. Expressed somewhat loosely, an ideal is maximal if it is not the whole ring and is not properly contained in any larger proper ideal; the only ideal to contain a maximal ideal properly is the ring itself. It is usually quite awkward to pro ve that an ideal is maximal directIy from Definition 5-1. We therefore need several theorems which will help to determine whether or not a given ideal is actually maximal, but"wpich are, in genera}, easier to apply than Definition 5-1. One such re;¡¡.J!t is pn~sented b e l o w . :.. "

quotients of R relative to S.)

d) If S eontains nozero divisor of R. then (RI1)s!¡

= Rs.

29. Let S be a multiplieatively c10sed subset of the ring R whieh eontains no zero divisor of R nor zero. a) If 1 is an ideal of R. verify that the set IS-1 = {ab- 1 E Qc¡{R)la El, b E S} is anideal,ofQc¡(R). Conversely,eaehidealJ ofQc¡(R)isoftheformJ = (J n R)S-I. b) For ideals l. J of R. establish the identities

Theorem 5-1. Let 1 be a proper ideal of the ring R. Then 1 is a maximal ideal if and only if (J, a) = R for any element a ~ l. Here (J, a) dértotes the ideal generated by 1 u {a}. ,,' Proo!' First, notice that (J, a) satisfies 1 e (J, a) ~ R. These inc1~~ions imply that if 1. were a maximal ideal of R, we would necessarily. have (J, a) = R. On the other hand, assume that J is an ideal of the ring R with the property that 1 e J ~ R. If a is any element of J which does not lie in 1, then 1 e (1, a) ~ J. The requirement that (1, a) = R would thus force J to be all of R, and we could conc1ude that 1 is a maximal ideal. A knowledge of several moderately simple examples will provide some basis for understanding these ideas.

¡

71

70

FIRST COURSE IN RINGS AND IDEALS

I

FIVE

In Problems 26-29, R is assumed to be a eommutative ringo 26. Let S be any multiplicatively c/osed subset of the ring R (that is, the produet of any two elements of S again lies in S) whieh eontains no zero divisor of R and O~ S. If the set Rs is defined by Rs

=

{ab- 1 EQc¡(R)laER,bES},

MAXIMAL, PRIME, AND PRIMARY IDEALS

prove that Rs is a subring of the ring of quotients Qc¡(R), known as the ring of qUQtients of R relative to S. 27. a) Show that the set S = {n E Zlp { n; p a' fixed prime} is multiplieatively c10sed and.determine Zs, the ring of quotients of Z relative to S. b) If R is any ring satisfying Z ~ R ~ Q, prove that R = Zs for a suitable multiplieatively c10sed subset S ~ Z. [Hint: Cónsider the set S = {m E Zlfor sorne n E Z, nlm E ,R; ged (n, m) = 1}.]

The present chapter is devoted to a study of certain special types of ideals, most notably maximal, prime, and primary ideals. On the whole, our hypothesis will restrict us to commutative rings with identity. The requirement is motivated to sorne extent by the fact that many of the standard examples of ring theory have this property. Another reason, which is perhaps more important from the conceptual point of view, is that the most satisfactory and complete results occur here. We begin our discussion with the following definition.

28. Let S be a multiplieat,ively c10sed subset of the ring R with O ~ S. Prove!he statements below: a) The set 1 = {a E Rlas = O for sorne s E S} is an ideal of R. b) Sil = nat¡S is a multiplieatively closed subset of the quotient ring RI1. e) No element of Sil is a zero divisor of RIl. (Thus, one can foqn !he ring of quotients of RIl relative to Sil; the result is eal1ed the generalized ring of

Definition 5-1. An ideal 1 of the ring R is said to be a maximal ideal provided that 1 =1= R and whenever J is an ideal of R with 1 e J ~ R, then J = R. Expressed somewhat loosely, an ideal is maximal if it is not the whole ring and is not properly contained in any larger proper ideal; the only ideal to contain a maximal ideal properly is the ring itself. It is usually quite awkward to pro ve that an ideal is maximal directIy from Definition 5-1. We therefore need several theorems which will help to determine whether or not a given ideal is actually maximal, but"wpich are, in genera}, easier to apply than Definition 5-1. One such re;¡¡.J!t is pn~sented b e l o w . :.. "

quotients of R relative to S.)

d) If S eontains nozero divisor of R. then (RI1)s!¡

= Rs.

29. Let S be a multiplieatively c10sed subset of the ring R whieh eontains no zero divisor of R nor zero. a) If 1 is an ideal of R. verify that the set IS-1 = {ab- 1 E Qc¡{R)la El, b E S} is anideal,ofQc¡(R). Conversely,eaehidealJ ofQc¡(R)isoftheformJ = (J n R)S-I. b) For ideals l. J of R. establish the identities

Theorem 5-1. Let 1 be a proper ideal of the ring R. Then 1 is a maximal ideal if and only if (J, a) = R for any element a ~ l. Here (J, a) dértotes the ideal generated by 1 u {a}. ,,' Proo!' First, notice that (J, a) satisfies 1 e (J, a) ~ R. These inc1~~ions imply that if 1. were a maximal ideal of R, we would necessarily. have (J, a) = R. On the other hand, assume that J is an ideal of the ring R with the property that 1 e J ~ R. If a is any element of J which does not lie in 1, then 1 e (1, a) ~ J. The requirement that (1, a) = R would thus force J to be all of R, and we could conc1ude that 1 is a maximal ideal. A knowledge of several moderately simple examples will provide some basis for understanding these ideas.

¡

71

72

.,

FIRST COURSE IN RINGS AND IDEALS •

Example 5-1. We propose to show that in. the ring Z of integers the maximal ideáIs correspond to the prime numbers; more precisely: the principal ideal (n), n > 1, is maximal if and only- if n is a prime. Suppose tbat (n) is a maximal ideal of Z. If the integer n is not prime, then n = n l l1 2, where 1 < nl < n2 < n. This impliés that the ideals (nJ ~~~~~' (n) c: (n l ) c:; Z,

..- ' .

MAXIMAL, PRIME, AND PRIMAR Y IDEALS

73

'!I

(n) c: (n 2 ) c: Z,

,;

contrary to the maximality of (n). . ,'. For the opposite direction, ássume now that the integer n is priní~~ If tbe principal ideal (n) i8 not maximal in Z; then either (n) = Z or else~here exists sOD1e proper ideal (m) satisfying (n) c: (m) c:; Z. The first cáse is immediately ruled out. by the fact that 1 is not a multlple of any:¡:iiime number. The aIternativ~ possibility, (n) c: (m), means that n = km foi¡~9'me integer k > 1; tbis is equally untenable, since n is· prime, not comp~&~te. At any rate, we conclude that (n) must be a maximal ideal.~;.)," <:.

simply be their set-theoretic union. For this particular setting, Zorn 's Lemma may be formulated as follows: Let.91 be a nonempty family of subsets of some fixed nonempty' set with' the property tbat for each chain C(j in .91, the union u C(j also belongs . to.9l. Then.91 contains a set which is maximal in the sense that it is not properIy contained in any member of .91. i Because this may be the reader's first contact with Zorn's Lemma, we proceed'in somewhat leisurely fashion to establish ,'o,' Theorem 5-2. lf the ring R is finitely of R is contained in a maximal ideal.

generJi~d, then each proper ideal Y:\ ::;-,';":"

Proof Let 1 b~ any proper ideal of R, a fipüely gene¡:;ated ring; say, R ( a 1 , a2, ... ,an ). We define a family of ideal§ of R by taking d = {Jll ¡:; J; J Isa

proper;W~~l of R}.", : ····1:"(

Zorn's lemma. If (S, ~) is a partially ordered set with the property .. that every chain in S has an upper bound in S, then S possesses at least one maximal element

This family 18 obviously nonempty, for 1 itself belongs todo Now, consider an arbitrary chain {Ji} ofidealsln.9l. Our aim, of course, ís to establish that u li ís again a member of.9l. To this purpose, let the elements a, b E U li and rE R. Then there exist índices i and) for which a E l i , bE lJ' As the colIection {Ji} forms a chain, either li ¡:; lJ or eIse 1) ¡:; li' For definiteness, suppose that li ¡:; l j , so that both a, bE 1,. But 1j ís an ideal of R; hence, the difference a - b E J j ¡:; u 1 i' AI~o, the . products ar and ra E li ¡:; U li' AlI of this shows u li to be an ideal of R. Next, we must verify that u li i8 a pioper ideal of R. Suppose, to tbe contrary, that u li = R = (al' a2, ... ,an ). Then, each generator ak would belong to some ideal l/k of the chain {lJ. There being only finitely many l/k' one contains aIl others, calI it li" . Thus, al' a2' ... , an al11ie in this one.-....J li" In consequence, li' = R,which, is cIearIy impossible. ,FinalIy~ noti{e that 1 ¡:; u 1" whence the union u l¡ E .91. .\ Therefore, on the basis of Zorn's Lemma, the family .91 contains a maximal eIement M. It follows directIy from the definition of.91 that M is aproper ideal of the ring R with 1 ¡:; M. We a8ser.t that M is in fact a maximill ideal. To 8ee this, suppose that J is any ideal of R for which M c: J ¡:; R. . ,Since M is a maximal element of the family .91, J cannot belong to.9l. Accordingly, the ideal J must be improper, which is to say that J = R. We thus concIude that M is a maximal ideal of R, completing the proor.

ClearIy, some partial orderings are more useful than others in applications of Zorn's Lemma. In our later investigations we shall frequently take S to be a fami1y of subsets of a given set and tbe partíal ordering to be the usual inclusion relation; an upper bound of any chain of elements would

The significant point, of course, is that this theorem asserts the existence of certain maximal ideals, but gives no clue as to how actualIyto find them. The chiefvirtue ofTheorem 5-2 is that it Ieads immediately to the following " celebrated result. . '

Example 5-2. For an illustration of the practicality of Theorem 5-1,' we take R = mapR"', a commutative ring with identity (Example 4, Chapter 1). Consider the set M of alI functions which vanish at O:

M = {fERlf(O)

= O}.

EvidentIy, M forms an ideal ofthe ring R; we contend that it is actualIy a maximal ideal. Indeed, iff ~ M and i is the identity map on R"' (that is, í(x) = x), one Ínay easily check that (i2 + j2)(x)4 O for each x E R #. Hence, the function í2 + f2 is an invertible element of R. Since R (M, j) 2 (í, 1), with ¡2 + j2 E (i, f), this implies that (M. R; in consequence, M is a rnaximal ideal of R. (Here (i, f) denotes the ideal generated by iandf;thatis,(i,j) = tri + sflr,SER}.) .

n

Our immediate goal is to obtain a general resulf assuríng the existen ce of suitably many maximal ideals. As will be seen presently, the crucial step in the proof dependson Zorn's Lemma (see Appendix B), an exceedingly powerful tool which is almost indispensable in modern mathematics. Zom's Lemma (traditionally calIed a lemma, but in fact an equivalent form of the Axiom of Choice) asserts :

72

.,

FIRST COURSE IN RINGS AND IDEALS •

Example 5-1. We propose to show that in. the ring Z of integers the maximal ideáIs correspond to the prime numbers; more precisely: the principal ideal (n), n > 1, is maximal if and only- if n is a prime. Suppose tbat (n) is a maximal ideal of Z. If the integer n is not prime, then n = n l l1 2, where 1 < nl < n2 < n. This impliés that the ideals (nJ ~~~~~' (n) c: (n l ) c:; Z,

..- ' .

MAXIMAL, PRIME, AND PRIMAR Y IDEALS

73

'!I

(n) c: (n 2 ) c: Z,

,;

contrary to the maximality of (n). . ,'. For the opposite direction, ássume now that the integer n is priní~~ If tbe principal ideal (n) i8 not maximal in Z; then either (n) = Z or else~here exists sOD1e proper ideal (m) satisfying (n) c: (m) c:; Z. The first cáse is immediately ruled out. by the fact that 1 is not a multlple of any:¡:iiime number. The aIternativ~ possibility, (n) c: (m), means that n = km foi¡~9'me integer k > 1; tbis is equally untenable, since n is· prime, not comp~&~te. At any rate, we conclude that (n) must be a maximal ideal.~;.)," <:.

simply be their set-theoretic union. For this particular setting, Zorn 's Lemma may be formulated as follows: Let.91 be a nonempty family of subsets of some fixed nonempty' set with' the property tbat for each chain C(j in .91, the union u C(j also belongs . to.9l. Then.91 contains a set which is maximal in the sense that it is not properIy contained in any member of .91. i Because this may be the reader's first contact with Zorn's Lemma, we proceed'in somewhat leisurely fashion to establish ,'o,' Theorem 5-2. lf the ring R is finitely of R is contained in a maximal ideal.

generJi~d, then each proper ideal Y:\ ::;-,';":"

Proof Let 1 b~ any proper ideal of R, a fipüely gene¡:;ated ring; say, R ( a 1 , a2, ... ,an ). We define a family of ideal§ of R by taking d = {Jll ¡:; J; J Isa

proper;W~~l of R}.", : ····1:"(

Zorn's lemma. If (S, ~) is a partially ordered set with the property .. that every chain in S has an upper bound in S, then S possesses at least one maximal element

This family 18 obviously nonempty, for 1 itself belongs todo Now, consider an arbitrary chain {Ji} ofidealsln.9l. Our aim, of course, ís to establish that u li ís again a member of.9l. To this purpose, let the elements a, b E U li and rE R. Then there exist índices i and) for which a E l i , bE lJ' As the colIection {Ji} forms a chain, either li ¡:; lJ or eIse 1) ¡:; li' For definiteness, suppose that li ¡:; l j , so that both a, bE 1,. But 1j ís an ideal of R; hence, the difference a - b E J j ¡:; u 1 i' AI~o, the . products ar and ra E li ¡:; U li' AlI of this shows u li to be an ideal of R. Next, we must verify that u li i8 a pioper ideal of R. Suppose, to tbe contrary, that u li = R = (al' a2, ... ,an ). Then, each generator ak would belong to some ideal l/k of the chain {lJ. There being only finitely many l/k' one contains aIl others, calI it li" . Thus, al' a2' ... , an al11ie in this one.-....J li" In consequence, li' = R,which, is cIearIy impossible. ,FinalIy~ noti{e that 1 ¡:; u 1" whence the union u l¡ E .91. .\ Therefore, on the basis of Zorn's Lemma, the family .91 contains a maximal eIement M. It follows directIy from the definition of.91 that M is aproper ideal of the ring R with 1 ¡:; M. We a8ser.t that M is in fact a maximill ideal. To 8ee this, suppose that J is any ideal of R for which M c: J ¡:; R. . ,Since M is a maximal element of the family .91, J cannot belong to.9l. Accordingly, the ideal J must be improper, which is to say that J = R. We thus concIude that M is a maximal ideal of R, completing the proor.

ClearIy, some partial orderings are more useful than others in applications of Zorn's Lemma. In our later investigations we shall frequently take S to be a fami1y of subsets of a given set and tbe partíal ordering to be the usual inclusion relation; an upper bound of any chain of elements would

The significant point, of course, is that this theorem asserts the existence of certain maximal ideals, but gives no clue as to how actualIyto find them. The chiefvirtue ofTheorem 5-2 is that it Ieads immediately to the following " celebrated result. . '

Example 5-2. For an illustration of the practicality of Theorem 5-1,' we take R = mapR"', a commutative ring with identity (Example 4, Chapter 1). Consider the set M of alI functions which vanish at O:

M = {fERlf(O)

= O}.

EvidentIy, M forms an ideal ofthe ring R; we contend that it is actualIy a maximal ideal. Indeed, iff ~ M and i is the identity map on R"' (that is, í(x) = x), one Ínay easily check that (i2 + j2)(x)4 O for each x E R #. Hence, the function í2 + f2 is an invertible element of R. Since R (M, j) 2 (í, 1), with ¡2 + j2 E (i, f), this implies that (M. R; in consequence, M is a rnaximal ideal of R. (Here (i, f) denotes the ideal generated by iandf;thatis,(i,j) = tri + sflr,SER}.) .

n

Our immediate goal is to obtain a general resulf assuríng the existen ce of suitably many maximal ideals. As will be seen presently, the crucial step in the proof dependson Zorn's Lemma (see Appendix B), an exceedingly powerful tool which is almost indispensable in modern mathematics. Zom's Lemma (traditionally calIed a lemma, but in fact an equivalent form of the Axiom of Choice) asserts :

74

FIRST COURSE IN RINGS AND IDEALS

Theorem 5-3. (Krull-Zorn)o In a ring R with identity each proper ideal is contained in a maximal ideal. Proof. An appeal to Theorem 5-2 is legitima te, since R = (1)0

Corollary. An element of a cornmutative ring R with identity is invertible if and only if it belongs to no maximal ideal of R. o Although maximal ideals were defined for arbitrary rings, we shall abandon a degree of generality and for the time being limit our discussion almost exclusively to commutative rings with identityo A ring of this kind is, of course, much easier to handle than one which is not commutativeo Another advantage stems from the fact that each ideal, other than the ring itself, will be contained in a maximal ideal. Thus, until further notice, we shall assume that al! g(ven rings are commutative with identity,eve~ when this is not explicitly men,tionedo To be sure, a good deal of the subsequent material could be presented without this additional restriotiono o The Krull-Zorn Theorem has many important applications throughout Ideal theoryo For the moment, we content ourselves with giving an ele, mentary proof of a somewhat special result; although the fact involved is rather interesting, there will be no occasion to make use of it. Theorem 5-4. In a ring R having exactly one maximal ideal M, the only idempotents are O and 1. Proof. Assume that the theorem is false; tha t is, su ppose tha t there exists an idempotent a E R with a =1= O, 1. The relation a 2 = a implies a(l - a) = O, so that a and 1 - a are both zero divisors. Hence, by Problem 4(d), Chapter 1, neither the element a nor 1 - a is invertible üilE. But this means that the principal ideals (a) and (1 - a) are both prop~¡:~i~eals of the ring R. As such, they must be contained in M, the sole nj~;xi~al of R. Accordingly, the elements a and 1 - a lie in M, whence 1= a

+

(1 - a)

E

Mo" .

This leads at once to the contradiction that M

'~::R.

Although more elementary proofs are possible, Theorem 5-4 can be used to show that a field has no idempotents, except O and 1. A full justification ofthis statement consists offirst establishing that the zero ideal is the only maximal ideal in a field. We now come to a characterization of maximal ideals in terms of their quotient rings. Theor~m 5-5.

LetI be a proper ideal ofthe ring R. Then l is a maximal ideal if and only if the quotient ring Rjl is a field. Proof. To'begin, let l be a maximal ideal of R. Since R is a commutative rin~ ,:",ith identity, the quotient ring Rjl also has these properties. Thus,

MAXIMAL, PRIME, AND PRIMARY IDEALS

75

to pro ve that Rjl is a field, it suffices to show that each nonzero element of Rjl has a multiplicative inverse. Now, if the coset a + l =1= l, then a f# l. By virtue of the fact that l is a maxirnal ideal, the ideal (l, a) generated by l and a must be the whole ring R: R

(1, a)

=

=

+ rali E l, rE R}o

{i

That is to say, every element of R is expressible in the form i + ra, where i E l and rE R. The identity element 1, in particular, rnay be written as 1 = r + ra for suitable choice of rE 1, r E R. But then, the difference 1 - ra E lo This obviously implies that 1+ l

=

ra

+ l

=

(r

+ lHa + l),

whichassertsthatr + l = (a + 1)-1. Hen<;;e,Rjlisafield. For the opposite direction, we suppose that Rjl is a field and J is any ideal of R for which l e J f; R. The argurnent consists of showing that J = R, for theil l will be a rnaximal ideal. Since l is a proper subset of J, there exists an elernent a E J with a f# l. Consequently, the coset a + l =1= l, the zero elernent of Rjl. Since Rjl is assumed to be a field, a + l must have an in verse under multiplication, (a

+ lHb + l) = ab + l = 1 + l,

for sorne coset b + lE Rjl. It then follows that 1 - ab E l e J. 'But the product ab also lies in J (recall that a is an element of the ideal J), irnplying that the identity 1 = (1 - ab) + ab EJ. This in turn yields J = R, as desired. o Example 5-3. Consider the ring Ze of ~ven integers, a commutative ring without identity. In this I:ilJ,g, the principal ideal (4) generated by the integer 4 is a maxirnal ideal, when:, (4) = {4(2j) + 4klj, k E z} = 4Z. The argument might be expressed as followso If n is any element not in (4), then n is an even integer ~9tdivisible by 4; consequently, n can be expressed in the form n = 4m + 2'for sorne integer m. We then have 2 '= 4(-m)

+ nE((4),n),

so that Ze = (2) = ((4), n)o By virtue of Theorem 5-1, this is sufficient to demonstrate the maxirnality ofthe ideal (4). Now, note that in the quotient ring ZJ(4), (2 + (4))(2 + (4))

=

4 + (4)

=

(4).

74

FIRST COURSE IN RINGS AND IDEALS

Theorem 5-3. (Krull-Zorn)o In a ring R with identity each proper ideal is contained in a maximal ideal. Proof. An appeal to Theorem 5-2 is legitima te, since R = (1)0

Corollary. An element of a cornmutative ring R with identity is invertible if and only if it belongs to no maximal ideal of R. o Although maximal ideals were defined for arbitrary rings, we shall abandon a degree of generality and for the time being limit our discussion almost exclusively to commutative rings with identityo A ring of this kind is, of course, much easier to handle than one which is not commutativeo Another advantage stems from the fact that each ideal, other than the ring itself, will be contained in a maximal ideal. Thus, until further notice, we shall assume that al! g(ven rings are commutative with identity,eve~ when this is not explicitly men,tionedo To be sure, a good deal of the subsequent material could be presented without this additional restriotiono o The Krull-Zorn Theorem has many important applications throughout Ideal theoryo For the moment, we content ourselves with giving an ele, mentary proof of a somewhat special result; although the fact involved is rather interesting, there will be no occasion to make use of it. Theorem 5-4. In a ring R having exactly one maximal ideal M, the only idempotents are O and 1. Proof. Assume that the theorem is false; tha t is, su ppose tha t there exists an idempotent a E R with a =1= O, 1. The relation a 2 = a implies a(l - a) = O, so that a and 1 - a are both zero divisors. Hence, by Problem 4(d), Chapter 1, neither the element a nor 1 - a is invertible üilE. But this means that the principal ideals (a) and (1 - a) are both prop~¡:~i~eals of the ring R. As such, they must be contained in M, the sole nj~;xi~al of R. Accordingly, the elements a and 1 - a lie in M, whence 1= a

+

(1 - a)

E

Mo" .

This leads at once to the contradiction that M

'~::R.

Although more elementary proofs are possible, Theorem 5-4 can be used to show that a field has no idempotents, except O and 1. A full justification ofthis statement consists offirst establishing that the zero ideal is the only maximal ideal in a field. We now come to a characterization of maximal ideals in terms of their quotient rings. Theor~m 5-5.

LetI be a proper ideal ofthe ring R. Then l is a maximal ideal if and only if the quotient ring Rjl is a field. Proof. To'begin, let l be a maximal ideal of R. Since R is a commutative rin~ ,:",ith identity, the quotient ring Rjl also has these properties. Thus,

MAXIMAL, PRIME, AND PRIMARY IDEALS

75

to pro ve that Rjl is a field, it suffices to show that each nonzero element of Rjl has a multiplicative inverse. Now, if the coset a + l =1= l, then a f# l. By virtue of the fact that l is a maxirnal ideal, the ideal (l, a) generated by l and a must be the whole ring R: R

(1, a)

=

=

+ rali E l, rE R}o

{i

That is to say, every element of R is expressible in the form i + ra, where i E l and rE R. The identity element 1, in particular, rnay be written as 1 = r + ra for suitable choice of rE 1, r E R. But then, the difference 1 - ra E lo This obviously implies that 1+ l

=

ra

+ l

=

(r

+ lHa + l),

whichassertsthatr + l = (a + 1)-1. Hen<;;e,Rjlisafield. For the opposite direction, we suppose that Rjl is a field and J is any ideal of R for which l e J f; R. The argurnent consists of showing that J = R, for theil l will be a rnaximal ideal. Since l is a proper subset of J, there exists an elernent a E J with a f# l. Consequently, the coset a + l =1= l, the zero elernent of Rjl. Since Rjl is assumed to be a field, a + l must have an in verse under multiplication, (a

+ lHb + l) = ab + l = 1 + l,

for sorne coset b + lE Rjl. It then follows that 1 - ab E l e J. 'But the product ab also lies in J (recall that a is an element of the ideal J), irnplying that the identity 1 = (1 - ab) + ab EJ. This in turn yields J = R, as desired. o Example 5-3. Consider the ring Ze of ~ven integers, a commutative ring without identity. In this I:ilJ,g, the principal ideal (4) generated by the integer 4 is a maxirnal ideal, when:, (4) = {4(2j) + 4klj, k E z} = 4Z. The argument might be expressed as followso If n is any element not in (4), then n is an even integer ~9tdivisible by 4; consequently, n can be expressed in the form n = 4m + 2'for sorne integer m. We then have 2 '= 4(-m)

+ nE((4),n),

so that Ze = (2) = ((4), n)o By virtue of Theorem 5-1, this is sufficient to demonstrate the maxirnality ofthe ideal (4). Now, note that in the quotient ring ZJ(4), (2 + (4))(2 + (4))

=

4 + (4)

=

(4).

76

.MAXIMAL; PRIME, AND PRIMARY IDEALS

FIRST COURSE IN RINGS AND IDEALS

We no.w shift o.ur attenti~n [ro.m maximal ideals to. prime ideals. Befo.re fo.rmally defining this no.tio.n, let us turn to. the ring Z o.f integers fo.r mo.tivatio.n. Specifically, co.nsider 'the principal' ideal (p) generated by a prime number p. If the pro.duct ab E (p), where a, bE Z, then p divides ab. But if a prime divides a pro.duct, it necessarily divides o.ne o.f the facto.rs. This being the case, either aE (p) o.r bE (p). The ideal (p) thus has the interesting property that, whenever (p) co.ntains a pro.duct, at least o.ne o.f the facto.rs must belo.ng to. (P). This o.bservatio.n serves to. suggest and partly to. illustrate the next defJ.nitio.n. .,\

Definition 5-2. An ideal Tbf the ring R is a prime ide~l if, fo.r all a, b in R, ab E 1 implies that eit?e~ a E 1 o.r b E l. By inductio.n, Defir;¡itio.rii :S-2 can easi1y be extended to. finitely many elements: an ideal 1 o.íR i~,iprime if, whenever a pro.duct a 1a 2 ••• an o.f e1ements o.f R belo.ngs to. 1, then'at least o.ne o.f the a; E l. In this co.nnectio.n, we should cautio.n the reader::üiat many autho.rs insist that the ter~ "prime ideal" always ineans a proper ideal. ' Example 5-4. A co.mmutative ring R with identity is an integral do.main if and o.nly if the zero. ideal {O} is a prime ideal o.f R. Example5-5. The prime ideals of the ring Z are precisely the ideals (n),. where n is a prime number, to.gether with the two. trivial ideals {O} and Z. Fro.m abo.ve, we already kno.w that if n is a prime, then the principal ideal (n) is a prime ideal o.f Z. On the o.ther hand, co.nsider any ideal (n) withn co.mpo.site (n =f. O, 1); say, n = n 1n2 , where 1 < n 1 , n 2 < n. Certainly the product n 1 n2 = n E (n). Ho.wever, sin ce neither n 1 no.r n 2 is an integral multiple o.f n, n 1 ~ (n) and n2 ~ (n). Hence,when n is co.mpo.site, the ideal (n) canno.t be prime. No.tice also that altho.ugh {O} is prime, it is no.t a maximal ideal o.f Z. Example 5-6. Fo.r an illustration o.f a ring po.ssessing allo.ntrivial prime ideal which is no.t maximal, take R = Z x Z, where the o.peratio.ns are perfo.rmed co.mpo.nentwise. One may readily verify that Z x {O} is a prime ideal o.f R. Since

Zx{O} e ZxZe e R, with Z x Ze an ideal o.f R, Z x {O} fails to. be maximal. By analo.gy with Theo.rem 5-5, the prime ideals o.f a ring may be characterizedi,n the fo.llo.wing manner. Theorem 5-6. Let 1 be a pro.per ideal o.f the ring R. Then 1 is a prime ideal if ~nd o.nly if the quotient ring Rjl is an integral do.main.

Proo! First, take 1 to. be a prime ideal o.f R. Since R is a co.mmutative

77

,ring with identity, so. is the quo.tient ring Rji. It remains therefo.re o.nly to. verify that Rj1 is free o.f zero. diviso.rs. Fo.r this, assume that \

(a

+ l)(b +

1) = l.

In o.ther wo.rds, the pro.duct o.f these two. co.sets is the zero. element o.f the ring RjI: The fo.rego.ing equatio.n is plainly equivalent to. requiring that ab + lj= l;o.r what amo.unts to. the same thing, ab E l. Since 1 is assumed to. be a',prime ideal, o.ne o.f the facto.rs a o.r b must be in l. But this means that ei~~rr the. co.set a + 1 = i o.r else b + 1 = 1; hence, Rjl is witho.ut ' zero. diviso.rs. To.:p:~o.ve the co.nverse, we simply reverse the argumento 'Acco.rdingly, suppo.s'e:that Rjl is an integral do.main and the pro.duct ab E 1. In terms o.f co.sets;~J4is mean s that '

h!¡,

(a

+ I)(b +

1) = ab

+

1 = l.

By h;;;rhesis Rjl co.ntains no. diviso.rs o.f zero., so. that a + 1 = 1 o.r b + 1 =;"1. In any event, o.ne o.f a o.rb belo.ngs to. 1, fo.rcing"t.to. be a prime ideal o.f R. There is an impo.rtant c1ass o.f ideals which are always prime, namely, the maximal ideals. Fro.m the several ways o.fpro.ving this result, we cho.o.se the argument given belo.w; anothei' appro.ach invo.lves the use o.f Theo.rems • 5-5 and 5-6~ Theorem 5-7. In a co.mmutative ring with identity, every maximal ideal is a prime ideal.

Proo! Assume that 1 is a maximal ideal o.f the ring R, a co.mmutative ring with identity, and the pro.duct ab El with a ~ l. We pro.po.se to. sho.w·that b E l. The maximality o.f 1 implies that the ideal generated by 1 and a must be the who.le ring: R = (1, a). Hence, there exist elements i E 1, r E R such that 1 = i + ra. Since bo.th ab and i belong to. 1, we co.nc1ude that

b = lb = (i

+ ra)b = ib + r(ab) El,

fro.m which it fo.llo.ws that 1 is a prime ideal o.f R. We sho.uld po.int o.ut that witho.ut the assumptio.n o.f an identity element this last result do.es no.t remain valid; a specific illustratio.n is the ring Z o.f even integers, where (4) fo.rms a maximal ideal which is no.t prime. Mo.r~ generally, o.ne can pr,o.ve the fo.llo.wing: if R is a co.mmutative ring witho.ut identity, bJlt having a single generato.r, then R co.ntains a no.nprime maximal ideal. To. establish this, suppo.se that R = (a). First o.bserve that the principal ideal (a 2 ) is a pro.per ideal o.f R, since the generato.r a ~ (a 2 ). Indeed, were a in (a 2 ), we co.uld write a = ra 2 + na 2 fo.r so.me rE R and n E Z; it is a simple matter to. check that the e1ement e = ra + na wo.uld

76

.MAXIMAL; PRIME, AND PRIMARY IDEALS

FIRST COURSE IN RINGS AND IDEALS

We no.w shift o.ur attenti~n [ro.m maximal ideals to. prime ideals. Befo.re fo.rmally defining this no.tio.n, let us turn to. the ring Z o.f integers fo.r mo.tivatio.n. Specifically, co.nsider 'the principal' ideal (p) generated by a prime number p. If the pro.duct ab E (p), where a, bE Z, then p divides ab. But if a prime divides a pro.duct, it necessarily divides o.ne o.f the facto.rs. This being the case, either aE (p) o.r bE (p). The ideal (p) thus has the interesting property that, whenever (p) co.ntains a pro.duct, at least o.ne o.f the facto.rs must belo.ng to. (P). This o.bservatio.n serves to. suggest and partly to. illustrate the next defJ.nitio.n. .,\

Definition 5-2. An ideal Tbf the ring R is a prime ide~l if, fo.r all a, b in R, ab E 1 implies that eit?e~ a E 1 o.r b E l. By inductio.n, Defir;¡itio.rii :S-2 can easi1y be extended to. finitely many elements: an ideal 1 o.íR i~,iprime if, whenever a pro.duct a 1a 2 ••• an o.f e1ements o.f R belo.ngs to. 1, then'at least o.ne o.f the a; E l. In this co.nnectio.n, we should cautio.n the reader::üiat many autho.rs insist that the ter~ "prime ideal" always ineans a proper ideal. ' Example 5-4. A co.mmutative ring R with identity is an integral do.main if and o.nly if the zero. ideal {O} is a prime ideal o.f R. Example5-5. The prime ideals of the ring Z are precisely the ideals (n),. where n is a prime number, to.gether with the two. trivial ideals {O} and Z. Fro.m abo.ve, we already kno.w that if n is a prime, then the principal ideal (n) is a prime ideal o.f Z. On the o.ther hand, co.nsider any ideal (n) withn co.mpo.site (n =f. O, 1); say, n = n 1n2 , where 1 < n 1 , n 2 < n. Certainly the product n 1 n2 = n E (n). Ho.wever, sin ce neither n 1 no.r n 2 is an integral multiple o.f n, n 1 ~ (n) and n2 ~ (n). Hence,when n is co.mpo.site, the ideal (n) canno.t be prime. No.tice also that altho.ugh {O} is prime, it is no.t a maximal ideal o.f Z. Example 5-6. Fo.r an illustration o.f a ring po.ssessing allo.ntrivial prime ideal which is no.t maximal, take R = Z x Z, where the o.peratio.ns are perfo.rmed co.mpo.nentwise. One may readily verify that Z x {O} is a prime ideal o.f R. Since

Zx{O} e ZxZe e R, with Z x Ze an ideal o.f R, Z x {O} fails to. be maximal. By analo.gy with Theo.rem 5-5, the prime ideals o.f a ring may be characterizedi,n the fo.llo.wing manner. Theorem 5-6. Let 1 be a pro.per ideal o.f the ring R. Then 1 is a prime ideal if ~nd o.nly if the quotient ring Rjl is an integral do.main.

Proo! First, take 1 to. be a prime ideal o.f R. Since R is a co.mmutative

77

,ring with identity, so. is the quo.tient ring Rji. It remains therefo.re o.nly to. verify that Rj1 is free o.f zero. diviso.rs. Fo.r this, assume that \

(a

+ l)(b +

1) = l.

In o.ther wo.rds, the pro.duct o.f these two. co.sets is the zero. element o.f the ring RjI: The fo.rego.ing equatio.n is plainly equivalent to. requiring that ab + lj= l;o.r what amo.unts to. the same thing, ab E l. Since 1 is assumed to. be a',prime ideal, o.ne o.f the facto.rs a o.r b must be in l. But this means that ei~~rr the. co.set a + 1 = i o.r else b + 1 = 1; hence, Rjl is witho.ut ' zero. diviso.rs. To.:p:~o.ve the co.nverse, we simply reverse the argumento 'Acco.rdingly, suppo.s'e:that Rjl is an integral do.main and the pro.duct ab E 1. In terms o.f co.sets;~J4is mean s that '

h!¡,

(a

+ I)(b +

1) = ab

+

1 = l.

By h;;;rhesis Rjl co.ntains no. diviso.rs o.f zero., so. that a + 1 = 1 o.r b + 1 =;"1. In any event, o.ne o.f a o.rb belo.ngs to. 1, fo.rcing"t.to. be a prime ideal o.f R. There is an impo.rtant c1ass o.f ideals which are always prime, namely, the maximal ideals. Fro.m the several ways o.fpro.ving this result, we cho.o.se the argument given belo.w; anothei' appro.ach invo.lves the use o.f Theo.rems • 5-5 and 5-6~ Theorem 5-7. In a co.mmutative ring with identity, every maximal ideal is a prime ideal.

Proo! Assume that 1 is a maximal ideal o.f the ring R, a co.mmutative ring with identity, and the pro.duct ab El with a ~ l. We pro.po.se to. sho.w·that b E l. The maximality o.f 1 implies that the ideal generated by 1 and a must be the who.le ring: R = (1, a). Hence, there exist elements i E 1, r E R such that 1 = i + ra. Since bo.th ab and i belong to. 1, we co.nc1ude that

b = lb = (i

+ ra)b = ib + r(ab) El,

fro.m which it fo.llo.ws that 1 is a prime ideal o.f R. We sho.uld po.int o.ut that witho.ut the assumptio.n o.f an identity element this last result do.es no.t remain valid; a specific illustratio.n is the ring Z o.f even integers, where (4) fo.rms a maximal ideal which is no.t prime. Mo.r~ generally, o.ne can pr,o.ve the fo.llo.wing: if R is a co.mmutative ring witho.ut identity, bJlt having a single generato.r, then R co.ntains a no.nprime maximal ideal. To. establish this, suppo.se that R = (a). First o.bserve that the principal ideal (a 2 ) is a pro.per ideal o.f R, since the generato.r a ~ (a 2 ). Indeed, were a in (a 2 ), we co.uld write a = ra 2 + na 2 fo.r so.me rE R and n E Z; it is a simple matter to. check that the e1ement e = ra + na wo.uld

78

FIRST COURSE IN RINGS AND IDEALS

then serve as a multiplicative identity for Ro' violating our hypothesis. Since (a 2 ) =f R, Theorem 5-2 guarantees the existen ce of a maximal ideal M of R with (a 2 ) S;;; M. However, M is not a prime ideal, as can be seen by considering the product of .elements in the complement of M (given r, s r$ M, the product rs E (a 2 ) S;;; M). Of course, the converse of Theorem 5-7 does not hold; Example 5-6 shows that there exist nontrivial prime ideals which faíl to be maximal ideals. ,The special properties ofBoolean rings aIld principal ideal domains guarantee that the notions of primeness ~d maximality are equivalent for these important classes of rings. Let us look at the details. Theorem 5-8. Let R be a Boolean ringo A nontrivial ideal 1 of R is prime if and only if it is a maximal ideal. , Proo! It is sufficient to show that if the ideal 1 is prime, then 1 is also maximal. To see this, suppose that J is an ideal of R with the property that 1 c J S;;; R; what we must prove is that J R. Ir a is any e1ement of J not in 1, then a(1 - a) = OE l. Using the faet that 1 is a prime ideal with a r$ 1, we infer that 1 - a E 1 c J. As both the elements a and 1 - a He in J, it follows that 1

=

a

+

(1 - a) EJ.

The ideal J thus contains the identíty and, consequentIy, J = R. Since no proper ideal Hes between 1 and the whole ring R, we conclude that 1 is a maximal ideal. Remark. Since every integral domain contains the two trivial prime ideals,

the:4Sf1 of the term "prime ideal" in a principal ideal domain customari1y exc~Ílc:lés these from consideration. Th~orem 5-9. Let R be a principal ideal domain. A non trivial ideal (aj'of R is prime if and only if it is a maximal ideal.

Proo! Assume that (a) is a prime ideal and let 1 be any ideal of R satisfying (a) ¿f '1 s;;; R. Because R is a principal ideal ring, there exists an element bE R,for whichl = (b). Now, a E (a) e (b); bence, a = rb for sorne choice ofr iii R. By supposition, (al is a prime ideal, so that either r E (a) or b E (a). The possibility that b E (a) leads immediately to the contradiction (b) s;;; (a). Therefore, the element r E (a), which implies that r = sa for suitable choice of s in R, or a = rb = (sa)b. Since a =f Oand R is an integral domain, we must have 1 = sb. Thi8, of course, means that the identity element 1 E (b) = 1, whencé 1 = R, making (a) a maximal ideal of R. Theorem 5-7 takes care of the converse.

CoroUary. A nontrivial ideal of the ring Z is prime if and only if it is maximal.

MAXIMAL, PRIME, AND PRIMARY IDEALS

79

Before taking up the matter of primary ideals, let us detour briefiy to introduce a concept which plays an important role inmany aspects of ideal theory. Definition 5-3. Let 1 be an ideal of the ring R. The ni! radical oi 1, designated by .jT,is the set

JI= {r E Rlr" El for sorne n E Z + (n varíes with r)}. We observe that the nil radical of 1 may equally well be characterized as the set of elements rE R whose image r + 1 in the quotient ring Rll is nilpotent. The nil radical of the zero ideal 'is sometimes referred 10 as the nil radical of the ring R; this set consists of all nilpotent elements of R and accounts for the use of the termo Example 5-7. In the ring Z, let us show that if n p~l p~2 ..• P:" is a factorization of the positive integer n =f 1 into distinct primes Pj' then

J(jij = (PIP2 ... Pr)' Indeed, if the integer a PIP2 ... Pr and k = max {kl> k 2, ... , f<;.}, then we have ak E (n); this makes it clear that (PIP2 ... Pr) S;;; J(jij. On the other hand, ir sorne positive integral power of the integer m is divisible by n (that is, if m E J(jij), then m itself must be divisible by each of the primes PI' pz, ... , Pr, and, henee, a member of the ideal (PI) (") (P2) n ... n (Pr) = (PIP2 ••. Pr)' As concrete i11 ustrations ofthis situation, observe that.J{12} (6) and .J(8) = (2).

=

.J{2 2 3} =

AIthough it is not obvious from the definition, .Ji is a,ctually an ideal ofthe ring R which contains l. In the first place, if a and b~a.re elements of .JI, then there exist suitably chosen integers n, m E Z+ sucIl tnat a" E 1 and bm E 1. Now, every term in the binomial expansíon ,of (a "":'b)n+m contains b)"+m lies i1I1and therefore either a" or bmas a factor. This implies that (a b E.Ji. Next,ifrisany element of R, then'(fa)"_= r"d' E 1, theditTerence a so that ra E .JI; thus, .JI is indeed an ideal of R. That 1 $.J1 should be clear from the definition. " Sorne of the basic properties of the nil radical of an ideal are assembled in the theorem below. Theorem 5-10. Ir 1 and J are two ideals of the ring R, then 1) 2)

J

.JIJ = .JI J = .JI n ji, .JI + J = .JI + .JJ 2 .JI + .JY,

3} l k 4)

J for sorne k E Z+ implies that.JI

¡:¡¡ = .Ji.

S;;;

ji, and

78

FIRST COURSE IN RINGS AND IDEALS

then serve as a multiplicative identity for Ro' violating our hypothesis. Since (a 2 ) =f R, Theorem 5-2 guarantees the existen ce of a maximal ideal M of R with (a 2 ) S;;; M. However, M is not a prime ideal, as can be seen by considering the product of .elements in the complement of M (given r, s r$ M, the product rs E (a 2 ) S;;; M). Of course, the converse of Theorem 5-7 does not hold; Example 5-6 shows that there exist nontrivial prime ideals which faíl to be maximal ideals. ,The special properties ofBoolean rings aIld principal ideal domains guarantee that the notions of primeness ~d maximality are equivalent for these important classes of rings. Let us look at the details. Theorem 5-8. Let R be a Boolean ringo A nontrivial ideal 1 of R is prime if and only if it is a maximal ideal. , Proo! It is sufficient to show that if the ideal 1 is prime, then 1 is also maximal. To see this, suppose that J is an ideal of R with the property that 1 c J S;;; R; what we must prove is that J R. Ir a is any e1ement of J not in 1, then a(1 - a) = OE l. Using the faet that 1 is a prime ideal with a r$ 1, we infer that 1 - a E 1 c J. As both the elements a and 1 - a He in J, it follows that 1

=

a

+

(1 - a) EJ.

The ideal J thus contains the identíty and, consequentIy, J = R. Since no proper ideal Hes between 1 and the whole ring R, we conclude that 1 is a maximal ideal. Remark. Since every integral domain contains the two trivial prime ideals,

the:4Sf1 of the term "prime ideal" in a principal ideal domain customari1y exc~Ílc:lés these from consideration. Th~orem 5-9. Let R be a principal ideal domain. A non trivial ideal (aj'of R is prime if and only if it is a maximal ideal.

Proo! Assume that (a) is a prime ideal and let 1 be any ideal of R satisfying (a) ¿f '1 s;;; R. Because R is a principal ideal ring, there exists an element bE R,for whichl = (b). Now, a E (a) e (b); bence, a = rb for sorne choice ofr iii R. By supposition, (al is a prime ideal, so that either r E (a) or b E (a). The possibility that b E (a) leads immediately to the contradiction (b) s;;; (a). Therefore, the element r E (a), which implies that r = sa for suitable choice of s in R, or a = rb = (sa)b. Since a =f Oand R is an integral domain, we must have 1 = sb. Thi8, of course, means that the identity element 1 E (b) = 1, whencé 1 = R, making (a) a maximal ideal of R. Theorem 5-7 takes care of the converse.

CoroUary. A nontrivial ideal of the ring Z is prime if and only if it is maximal.

MAXIMAL, PRIME, AND PRIMARY IDEALS

79

Before taking up the matter of primary ideals, let us detour briefiy to introduce a concept which plays an important role inmany aspects of ideal theory. Definition 5-3. Let 1 be an ideal of the ring R. The ni! radical oi 1, designated by .jT,is the set

JI= {r E Rlr" El for sorne n E Z + (n varíes with r)}. We observe that the nil radical of 1 may equally well be characterized as the set of elements rE R whose image r + 1 in the quotient ring Rll is nilpotent. The nil radical of the zero ideal 'is sometimes referred 10 as the nil radical of the ring R; this set consists of all nilpotent elements of R and accounts for the use of the termo Example 5-7. In the ring Z, let us show that if n p~l p~2 ..• P:" is a factorization of the positive integer n =f 1 into distinct primes Pj' then

J(jij = (PIP2 ... Pr)' Indeed, if the integer a PIP2 ... Pr and k = max {kl> k 2, ... , f<;.}, then we have ak E (n); this makes it clear that (PIP2 ... Pr) S;;; J(jij. On the other hand, ir sorne positive integral power of the integer m is divisible by n (that is, if m E J(jij), then m itself must be divisible by each of the primes PI' pz, ... , Pr, and, henee, a member of the ideal (PI) (") (P2) n ... n (Pr) = (PIP2 ••. Pr)' As concrete i11 ustrations ofthis situation, observe that.J{12} (6) and .J(8) = (2).

=

.J{2 2 3} =

AIthough it is not obvious from the definition, .Ji is a,ctually an ideal ofthe ring R which contains l. In the first place, if a and b~a.re elements of .JI, then there exist suitably chosen integers n, m E Z+ sucIl tnat a" E 1 and bm E 1. Now, every term in the binomial expansíon ,of (a "":'b)n+m contains b)"+m lies i1I1and therefore either a" or bmas a factor. This implies that (a b E.Ji. Next,ifrisany element of R, then'(fa)"_= r"d' E 1, theditTerence a so that ra E .JI; thus, .JI is indeed an ideal of R. That 1 $.J1 should be clear from the definition. " Sorne of the basic properties of the nil radical of an ideal are assembled in the theorem below. Theorem 5-10. Ir 1 and J are two ideals of the ring R, then 1) 2)

J

.JIJ = .JI J = .JI n ji, .JI + J = .JI + .JJ 2 .JI + .JY,

3} l k 4)

J for sorne k E Z+ implies that.JI

¡:¡¡ = .Ji.

S;;;

ji, and

80

FIRST COURSE IN RINGS AND IDEALS

MAXIMAL, PRIME, AND

Proof. Sinc~ .property (1) is the only fact that will be explicitly required ' in the body of the text, we shall content ourselves with its derivation; the proofs of the remaining assertions are quite elementary and are left as an exercise. Now, if an E 1J, then anE 1 n J, and so anEl, anEJ. We thus concIude s.JI n Ji. On the other hand, if it happens that that jTJ S a E -Ji n ,.¡J, there must exist positive integers n, m, for which anE 1 and am E J . . This implies that the element an+ m = anam E IJ; hence, a E Accordingly, -Ji n S .jYJ and the desired equality follows. .:i.

J"Tn7

., '..

JI1.

Ji

In passing, we might point out that although property (1) easíly generalizes to finite intersections, it is false if infinite intersections·'ate allowed. This is best brought out by once again considering the ring Z~aríd the collection of principal ideals (l), where p is a fixed prime and k ·'~·1 ; it foIlows readily that . ' ..:•. : .

nM

=

k

n(p) = (p)

=1=

{O}

=

k'

Jn (pk).

é"1"

k

A. problem of .central interest is that of determining conditions under which a given ideal coincides with its nil radical; in this connection, the foIlowing definition wi1l be useful (the reason for.our choice of terminology appears shortly). . Definition 5-4. An ideal 1 of the ring R is said to be a semiprime ideal if ~nd only if 1 =

Ji

In effect, Definition 5--4 states that an ideal 1 is semiprime if and only if an E 1 for sorne n E Z+ implies that a itself lies in l. bur next result characterizes semiprime ideals by the quotient rings which they determine. Theorem 5-11. An ideal 1 of the ring R is a semiprime ideal if and only if the quotient ring Rjl has no nonzero nilpotent e1ements.

Proof. Suppose that a + JJ is a nilpotent element of Ri-Ji. Then there exists sorne n E Z+ such that (a + .jI)" = an + .JI = .JI; that is to say, the element a" E -Ji. Hence, (a n)'" = anm E 1 for sorne positive integer m. This implies that a E.JI and, consequentIy, that a + -Ji = -Ji, the zero element of Rj.JI . . As regards the converse, as sume that Rj1 has no nonzero nilpotent elements and let a E.JI Then, for sorne positive integer n, an El. Passing to the quotient ring Rjl, this simply means that (a + 1)" = 1; in other words, a + 1 is nilpotent in Rj1. By supposition, we must have a + 1 = 1 and, in con sequen ce, the element a E l. Our argument shows that -Ji S 1; since' the reversed inc]usion always holds, 1 = -Ji, so that 1 is semiprime. This being proved, it is not hard to establish

PRI~ARY

IDEALS

81

CorolIary. If Pis a·prime ideal of the ring R, then Pis semiprime.

Proof. Because P is prime, RjP possesses no zero divisors and, in" particular, no nonzerb nilpotent elements. . . This corollary pro vides another good reason why a semi prime ideal was termed as it was; being a semiprime ideal in a ring is a bit weaker than being prime. There is much more that could bé.t said about semi prime ideals, and more will be said later in the text, but let us now turn our attention to primary ideals.· ., In Chapter 11 we shall show that the ideal s o( a rather wide class of rings (to. be quite explicit, the Noetherian rings),:9¡bey factorization laws which are roughly similar to the prime factorizati9"P- laws for the positive integers. It wiI1 turn out that the primary ideaIs;,:;YIhich we are about to introduce, playa role analogous to the powers ofp¡:üp.e numbers in ordinary .arithmetic. . ,

Definition 5-5. An ideal 1 ofthe ring R is called'jirimary ifthe conditions ab E 1 and a ~ 1 together imply bn E 1 for somepositivdnteger n. Clearly, any prime ideal satisfies this definition with. n = 1, and thus, the concept of a primary ideal may be viewed as a natural generalization of that of a prime ideal. Lest the reader jump to false concIusions, we hasten to point out that a primary ideal is not necessarily a power of a prime ideal (see Example 8~ Chapter 7).' Notice too that Defihition 5-5 may be stated in ·another way: an ideal 1 is primary if ab E 1 and a ~ l' imply b E this formulation in terms of the nil radical is frequently useful. In the ring Z, the primary ideal s are precisely the ideals (pn), w here p is a prime number and n 2 1, together with the two trivial ideals. Our first theorem on primary ideals is simple enough; it shows that to every primary ideal there corresponds a specifié prime ideal.

.JI;

,

.h.'

,.

rheorem 5-12. If Q is a primary ideal of the ring R, then itsnil radical .jQ is a prime ideal, known as the associated prime ideal of Q.

Proof. Suppose that ab E.jQ, with a ~ .JQ. Then (ab)n = a"b" E Q for sorne positive integer n. But a" ~ Q, for otherwise a would lie in .jQ. Since Q is assumed to be primary, we must therefore have (bn)m E Q for suitable . choice of m E Z;.;.and so bE.jQ. This is simply the statement that .JQ is a prime ideal of R. It may very well happen that different primary ideals will have the same associated prime ideal. This is demonstrated rather strikingly in the ring ofintegers where, for any.n E Z +, (p) is the prime ideal associated with each of the primary ideals (p"). It might also be of interest to mention that the nil radical .JQ is the smallest prime ideal to contain a given primary ideal Q. For, suppose that

80

FIRST COURSE IN RINGS AND IDEALS

MAXIMAL, PRIME, AND

Proof. Sinc~ .property (1) is the only fact that will be explicitly required ' in the body of the text, we shall content ourselves with its derivation; the proofs of the remaining assertions are quite elementary and are left as an exercise. Now, if an E 1J, then anE 1 n J, and so anEl, anEJ. We thus concIude s.JI n Ji. On the other hand, if it happens that that jTJ S a E -Ji n ,.¡J, there must exist positive integers n, m, for which anE 1 and am E J . . This implies that the element an+ m = anam E IJ; hence, a E Accordingly, -Ji n S .jYJ and the desired equality follows. .:i.

J"Tn7

., '..

JI1.

Ji

In passing, we might point out that although property (1) easíly generalizes to finite intersections, it is false if infinite intersections·'ate allowed. This is best brought out by once again considering the ring Z~aríd the collection of principal ideals (l), where p is a fixed prime and k ·'~·1 ; it foIlows readily that . ' ..:•. : .

nM

=

k

n(p) = (p)

=1=

{O}

=

k'

Jn (pk).

é"1"

k

A. problem of .central interest is that of determining conditions under which a given ideal coincides with its nil radical; in this connection, the foIlowing definition wi1l be useful (the reason for.our choice of terminology appears shortly). . Definition 5-4. An ideal 1 of the ring R is said to be a semiprime ideal if ~nd only if 1 =

Ji

In effect, Definition 5--4 states that an ideal 1 is semiprime if and only if an E 1 for sorne n E Z+ implies that a itself lies in l. bur next result characterizes semiprime ideals by the quotient rings which they determine. Theorem 5-11. An ideal 1 of the ring R is a semiprime ideal if and only if the quotient ring Rjl has no nonzero nilpotent e1ements.

Proof. Suppose that a + JJ is a nilpotent element of Ri-Ji. Then there exists sorne n E Z+ such that (a + .jI)" = an + .JI = .JI; that is to say, the element a" E -Ji. Hence, (a n)'" = anm E 1 for sorne positive integer m. This implies that a E.JI and, consequentIy, that a + -Ji = -Ji, the zero element of Rj.JI . . As regards the converse, as sume that Rj1 has no nonzero nilpotent elements and let a E.JI Then, for sorne positive integer n, an El. Passing to the quotient ring Rjl, this simply means that (a + 1)" = 1; in other words, a + 1 is nilpotent in Rj1. By supposition, we must have a + 1 = 1 and, in con sequen ce, the element a E l. Our argument shows that -Ji S 1; since' the reversed inc]usion always holds, 1 = -Ji, so that 1 is semiprime. This being proved, it is not hard to establish

PRI~ARY

IDEALS

81

CorolIary. If Pis a·prime ideal of the ring R, then Pis semiprime.

Proof. Because P is prime, RjP possesses no zero divisors and, in" particular, no nonzerb nilpotent elements. . . This corollary pro vides another good reason why a semi prime ideal was termed as it was; being a semiprime ideal in a ring is a bit weaker than being prime. There is much more that could bé.t said about semi prime ideals, and more will be said later in the text, but let us now turn our attention to primary ideals.· ., In Chapter 11 we shall show that the ideal s o( a rather wide class of rings (to. be quite explicit, the Noetherian rings),:9¡bey factorization laws which are roughly similar to the prime factorizati9"P- laws for the positive integers. It wiI1 turn out that the primary ideaIs;,:;YIhich we are about to introduce, playa role analogous to the powers ofp¡:üp.e numbers in ordinary .arithmetic. . ,

Definition 5-5. An ideal 1 ofthe ring R is called'jirimary ifthe conditions ab E 1 and a ~ 1 together imply bn E 1 for somepositivdnteger n. Clearly, any prime ideal satisfies this definition with. n = 1, and thus, the concept of a primary ideal may be viewed as a natural generalization of that of a prime ideal. Lest the reader jump to false concIusions, we hasten to point out that a primary ideal is not necessarily a power of a prime ideal (see Example 8~ Chapter 7).' Notice too that Defihition 5-5 may be stated in ·another way: an ideal 1 is primary if ab E 1 and a ~ l' imply b E this formulation in terms of the nil radical is frequently useful. In the ring Z, the primary ideal s are precisely the ideals (pn), w here p is a prime number and n 2 1, together with the two trivial ideals. Our first theorem on primary ideals is simple enough; it shows that to every primary ideal there corresponds a specifié prime ideal.

.JI;

,

.h.'

,.

rheorem 5-12. If Q is a primary ideal of the ring R, then itsnil radical .jQ is a prime ideal, known as the associated prime ideal of Q.

Proof. Suppose that ab E.jQ, with a ~ .JQ. Then (ab)n = a"b" E Q for sorne positive integer n. But a" ~ Q, for otherwise a would lie in .jQ. Since Q is assumed to be primary, we must therefore have (bn)m E Q for suitable . choice of m E Z;.;.and so bE.jQ. This is simply the statement that .JQ is a prime ideal of R. It may very well happen that different primary ideals will have the same associated prime ideal. This is demonstrated rather strikingly in the ring ofintegers where, for any.n E Z +, (p) is the prime ideal associated with each of the primary ideals (p"). It might also be of interest to mention that the nil radical .JQ is the smallest prime ideal to contain a given primary ideal Q. For, suppose that

82

FIRST COURSE IN 'RINGS AND IDEALS

MAXIMAL, PRIME, AND PRIMAR Y IDEALS

Pis any prime ideal containing Qand let a E JQ. Then there exists a suitable positive integer n such that a" E Q S;; P. Being prime, the ideal P must contain the element a itself, which yields the inclusion .JQ S;; P. The primary ideals of R may be characterized in the folIowing way. Theorem 5-13. Let 1 be an ideal of the ring R. Then 1 is a primary ideal ifand only if every zero divisor ofthe quotient ring RIl is nilpotent. Proa.! . First, suppose that 1 is a primary ideal of R and take a + 1 to be a zero divisor of RIl. Then there exists .sorne coset b + 1 =1= l, the zero element of RIl, for which (a + l)(b + l) = l; that is, ab + 1 = l. Therefore ab El and, since b + 1 =1= l, we al so have b rt l. Now, 1 is assumed to be primary, so that a" E 1 for some positive integer n. This being the case,

+ 1)" = a" + 1 = which shows that the coset a + 1 is nilpotent.

l,

(a

Going in the other direction, we assume that any zero divisor of RI1 is nilpotent and let ab El, with b rt l. It then folIows that (a + I)(b + 1) = l, while b + 1 =1= l; if a + 1 =1= l, this amounts to saying that a + 1 is a zero divisor in RIl. By hypothesis, there must exist some n E Z+ such that (a + l)" = l, which forces the element a" to be in l. Thus, 1 is a primary ideal of R. Theorem 5-13 serves to emphasize the' point that primary ideal s are a modification of the notion of a prime ideal; for, in the quotient ring of a prime ideal, there are no zero divisors (hence, in a vacuous sense, every zero divisor is nilpotent). The folIowing somewhat special result..will be needed later, so we pause ,.,. \ to establish it before proceeding. Corollary. If Ql' Q2' ... , Q" are a frn:ite set of primary ideals of the ring R, alI of them having the sam&associated prime ideal P, then Q = ni=1 Qi is also primary, with JQ; = P. Proo.! Before we del ve into the detail:of the proof, observe that, by Theorem 5-10,

JQ = .J n

Q¡ = n

~

= () p

= P.

Now, suppose that a + Q is a zero divisor of the quotient ring RIQ. In this event, we can find a coset b + Q =1= Q such that ab

+

Q

=

(a

+ Q)(b + Q)

= Q.

Since b rt Q = h Q¡, there exists some index i for which b rt Q¡. Furthermore, ab E Q¡ with Q¡ primary, so that the elernent a E.JQ¡ = P = JQ. This implies a" E Q for sorne integer n; in consequence, (a

+

Q)" = a"

+

Q

= Q,

83

which is to say that a + Q is nilpotent. As every zero divisor ofthe quotient ring RIQ is nilpotent, an appeal to Theorem 5-13 is in order and we may conclude that Q is a primary ideal of R. There is another, frequentIy Useful, criterion for deciding whether a given ideal is actualIy primary. Theorem 5-14. Let P and Q be ideals of the ring R such that 1) Q s;; P s;; JQ, 2) if ab E Q with a rt P, then b E Q. Under these conditions, Q is a primary ideal of R with P = ~ Proa! To see that Q is a primary ideal, 'suppose that the product ab E Q but b rt Q. Using (2), we may conclude that a E P S;; JQ, whence a" E Q for some positive integer n; this shows that Q is primary. In order to prove that P = JQ, we need only establish the inclusion JQ S;; P, since equality would then folIow from (1~. For this, let the. elemen~ b E JQ, so thát there exists sorne n E Z + for WhlCh b" E Q; assurne that n is the smalIest positive integer with this property. If n' = 1, we w~uld h~~e bE Q S;; P, from condition (1). If n' > 1, it folIows that b" = b" 1 b E Q, with b"' -1 rt Q; hence, bE P by (2). In any event, we have shown that bE JQ implies bE P, as required.

A relationship between maximal ideals and primary ideals is brought out in the foIlowing corollary to Theorem 5-14. Corollary. If M is a maximal ideal of the ring R, then all its powers M" (n ;::: 1) areJ?rimary ideals. Proa! Since M" s;;).1 = .JM", we need only verify condition (2) of t~e foregoing theorem.'81lPpose, then, that ab E M" with a rt M. Becaus~ M IS maximal, the ideal (M; a), generated by M and a must be the whole nng R. Hence, the identity"e\ement 1 E (M, a), so, for some m E M and rE R, .we must have 1 = m+ra. Now, m" lies in M". Raising the equatlOn 1 = m + ra to the nth;power and using the binomial theorem. it follows that 1 = m" + r'a, wherer' ER. But then b = bm"

+

r'(ab)

is an element of M", and M" is primary. Another result which has this same general fiavor, but which we leave of an ideal 1 is a rnaximal as an exercise, is the following: If the ni! radical ideal, then 1 itselfis primary. Before closing this chapter, we present two additional theorems regarding prime ideals. The first of these involves the notion of a minimal prime ideal of an ideal.

JI

82

FIRST COURSE IN 'RINGS AND IDEALS

MAXIMAL, PRIME, AND PRIMAR Y IDEALS

Pis any prime ideal containing Qand let a E JQ. Then there exists a suitable positive integer n such that a" E Q S;; P. Being prime, the ideal P must contain the element a itself, which yields the inclusion .JQ S;; P. The primary ideals of R may be characterized in the folIowing way. Theorem 5-13. Let 1 be an ideal of the ring R. Then 1 is a primary ideal ifand only if every zero divisor ofthe quotient ring RIl is nilpotent. Proa.! . First, suppose that 1 is a primary ideal of R and take a + 1 to be a zero divisor of RIl. Then there exists .sorne coset b + 1 =1= l, the zero element of RIl, for which (a + l)(b + l) = l; that is, ab + 1 = l. Therefore ab El and, since b + 1 =1= l, we al so have b rt l. Now, 1 is assumed to be primary, so that a" E 1 for some positive integer n. This being the case,

+ 1)" = a" + 1 = which shows that the coset a + 1 is nilpotent.

l,

(a

Going in the other direction, we assume that any zero divisor of RI1 is nilpotent and let ab El, with b rt l. It then folIows that (a + I)(b + 1) = l, while b + 1 =1= l; if a + 1 =1= l, this amounts to saying that a + 1 is a zero divisor in RIl. By hypothesis, there must exist some n E Z+ such that (a + l)" = l, which forces the element a" to be in l. Thus, 1 is a primary ideal of R. Theorem 5-13 serves to emphasize the' point that primary ideal s are a modification of the notion of a prime ideal; for, in the quotient ring of a prime ideal, there are no zero divisors (hence, in a vacuous sense, every zero divisor is nilpotent). The folIowing somewhat special result..will be needed later, so we pause ,.,. \ to establish it before proceeding. Corollary. If Ql' Q2' ... , Q" are a frn:ite set of primary ideals of the ring R, alI of them having the sam&associated prime ideal P, then Q = ni=1 Qi is also primary, with JQ; = P. Proo.! Before we del ve into the detail:of the proof, observe that, by Theorem 5-10,

JQ = .J n

Q¡ = n

~

= () p

= P.

Now, suppose that a + Q is a zero divisor of the quotient ring RIQ. In this event, we can find a coset b + Q =1= Q such that ab

+

Q

=

(a

+ Q)(b + Q)

= Q.

Since b rt Q = h Q¡, there exists some index i for which b rt Q¡. Furthermore, ab E Q¡ with Q¡ primary, so that the elernent a E.JQ¡ = P = JQ. This implies a" E Q for sorne integer n; in consequence, (a

+

Q)" = a"

+

Q

= Q,

83

which is to say that a + Q is nilpotent. As every zero divisor ofthe quotient ring RIQ is nilpotent, an appeal to Theorem 5-13 is in order and we may conclude that Q is a primary ideal of R. There is another, frequentIy Useful, criterion for deciding whether a given ideal is actualIy primary. Theorem 5-14. Let P and Q be ideals of the ring R such that 1) Q s;; P s;; JQ, 2) if ab E Q with a rt P, then b E Q. Under these conditions, Q is a primary ideal of R with P = ~ Proa! To see that Q is a primary ideal, 'suppose that the product ab E Q but b rt Q. Using (2), we may conclude that a E P S;; JQ, whence a" E Q for some positive integer n; this shows that Q is primary. In order to prove that P = JQ, we need only establish the inclusion JQ S;; P, since equality would then folIow from (1~. For this, let the. elemen~ b E JQ, so thát there exists sorne n E Z + for WhlCh b" E Q; assurne that n is the smalIest positive integer with this property. If n' = 1, we w~uld h~~e bE Q S;; P, from condition (1). If n' > 1, it folIows that b" = b" 1 b E Q, with b"' -1 rt Q; hence, bE P by (2). In any event, we have shown that bE JQ implies bE P, as required.

A relationship between maximal ideals and primary ideals is brought out in the foIlowing corollary to Theorem 5-14. Corollary. If M is a maximal ideal of the ring R, then all its powers M" (n ;::: 1) areJ?rimary ideals. Proa! Since M" s;;).1 = .JM", we need only verify condition (2) of t~e foregoing theorem.'81lPpose, then, that ab E M" with a rt M. Becaus~ M IS maximal, the ideal (M; a), generated by M and a must be the whole nng R. Hence, the identity"e\ement 1 E (M, a), so, for some m E M and rE R, .we must have 1 = m+ra. Now, m" lies in M". Raising the equatlOn 1 = m + ra to the nth;power and using the binomial theorem. it follows that 1 = m" + r'a, wherer' ER. But then b = bm"

+

r'(ab)

is an element of M", and M" is primary. Another result which has this same general fiavor, but which we leave of an ideal 1 is a rnaximal as an exercise, is the following: If the ni! radical ideal, then 1 itselfis primary. Before closing this chapter, we present two additional theorems regarding prime ideals. The first of these involves the notion of a minimal prime ideal of an ideal.

JI

84

Definition 5-6. Let 1 be an ideal 'of the ring R. A prime ideal P of R is ,said to 'be a minimal prime ideal of 1 (sometimes, an isolated prime ideal of 1) if 1 S; P and there exists no prime ideal pi of R such that ls;Pc~ . .

The existence of minimal prime ideals (ofR) is assured by taking 1 in the statement ofTheorem 5-15. .

. Proa! Since any proper ideal of R is contained in a maximal (hence, prime) i<;leal of R, fue theorem can be applied.

{O}

Our final theorem concerns the (set-theoretic) union of a finite number of prime ideals. In thi~ connection, we first observe that if an ideal 1 of the . ring R iscontained in the union J u K of two arbitrary ideal s of R, then 1 must be contained :in one <;lf them. For, suppose that 1 S; J u K with 1 si; J. It is therefor~; possible to choose an element a El n K such tlÍat a rt J. If b E 1 n J,..\~~n the sum a + b rt J (otherwise, a= (a + b) - b isinJ)andsoa + b·'é7(,whencebEK. TheimplicationisthatI n J S; K; consequently,~\:··r . . 1 == f~(J u K) = (1 n J) u (1 n K)

S;

K.

The next pOint\6'::~hich attention should be drawn is that the aboye fact about the·unioil"Ó'r two ideals is no longer true when . we pass to the union of three or more ideals. For a simple example, letR = Z2 x 2 2 : .

Theorem 5-15. Let 1 and P be ideals of the ring R, with P pdme. If 1 S; P, then Pcontains a minimal prime ideal of 1.

Corollary 1. Every proper ideal of the ring R possesses at least one . minimal prime ideal.

=

Corollary 2. If Pis a prime ideal of the ring R, then P contains a mini mal prime ideal of R. '

By abuse of language, we shall refer t~ the minimal prime ideals of the zero ideal {O} as the mini mal prime ideals of the ring R; that is, a prime ideal is a minimal prime ideal (of R) if it does not properly contain any other . . prime ideal. . Let us observ~ that, in the ring Z of Íljtegers, the minimal prime ideals of a nonzero ideal (n) are precisely the piií,iie ideal s (p), where p is a prime dividing n. In particular, one infers tha(eyery ideal in Z possesses only a finite number ofminimal prime ideal s (thigesult is generalized in TheoreÍn 12-3).' ;.: . It is not immediateln:lear that any (pr~per) idea1.admits minimal prime ideals, although this is indeed the case. To dispose of the question requires an appeal to Zorn'~ Lernma; the details
Proa! Denote by ff the family of aH prime ideals of R which contain 1 andare contained in P: ff = {pllp l is a prime ideal of R; 1 S; pi S; ..P}. We, point out that ff is not empty, sin ce P itself belongs to ff. Next, introduce a partial order ~ in ff which is opposite to the usual inc1usion relation; that is to say, if P', P" Eff, interpret pi ~ pI! to mean pI! S; Pi. Consider any nonempty subset {Pi} of ff which is totaHy ordered by ~ (more simply, {P¡} is a chain in 9"). Put P = nP¡. Then P is a prime ideal oi R (Problem 11, Chapter 5) containing 1 and contained in P; hence, PE ff. Sin ce P S; p¡ for every value of i, it foHows that p¡ ~ P, making Pan upper bound for {Pi}' AH the hypotheses ofZorn's Lemma, as applied' .to (ff, ~) are satisfied, so that ff" has a maximal element, say p* (this means that if pi E ff and p* ~ P', then P* = Pi). Inasmuch as P* E ff, it is a prime ideal of R with 1 S; P* S; P. There remains the task of showing that p* is necessarily a m.inimal prime ideal of 1.' Forthis, we suppose that pi is any prime ideal of R satisfying 1 S; pi S; P*. Then pi E ff and p* ~ Pi. By the maximal nature of P*, we thus have P* = P', signifying that P* is a minimal prime ideal of 1.

85

MAXIMAL, PRIME', AND PRIMARY IDEALS

FIRST COURSE IN RINGS AND IDEAL S

R

=

{(O, O), (O, 1), (1, O), (1, In.

We turn R into a ring by taking the addition to be componentwise addition modulo 2 and defining all products to be zero. Then, 1 1 = {(O, O), (O, In,

12 = {(O, 0), (1, 0n,

13

=

{(O, O), (1,

in

are all ideals in R and R = 11 U 12 U 13' It is c1ear, however, ihat R (regarded as an ideal) is not contained in any one of the I¡. . The situation just described can be countered by imposing the demand that each of the ideals irivolved in the union be prime. The theorem we have in mind asserts that if an ideal is contained in a finite union of prime ideals,then it is entirely contained in one of them. ActuaHy, it is easier to prove the contrapositive' ofthis statement, v i z . : ' Theorem 5-16. LetI be an arbitraryideal ofthe ring R and PI' P 2' ... , Pn be prime ideals of R. IfI si; p¡ for aH i, then there exists an element a E 1 such that a rt u Pi; hence, 1 si; u Pi' . i

Proa! The argument will be by induction on the number n ofprime ideals. Assume that the theorem has already been established when there are only n - 1 ideals (when n = 1, the result is trivial). Then, for each i (1 ~ i ~ n), there exists an element r¡ E 1 with r¡ rt Un¡ P j • If, for sorne value of i, it happens that r¡ rt Pi' then r¡ rt u P j and there is nothing to be proved. Thus, we may restrict our attention to the case where r¡ E,P¡ for aH i . In what follows, let a¡ =r 1 .. ··r¡-lr¡+l "·r n • We assert that.a¡rtP¡. Since Pi is prime, the contrary assumption a¡ E p¡ would imply that r j E p¡

84

Definition 5-6. Let 1 be an ideal 'of the ring R. A prime ideal P of R is ,said to 'be a minimal prime ideal of 1 (sometimes, an isolated prime ideal of 1) if 1 S; P and there exists no prime ideal pi of R such that ls;Pc~ . .

The existence of minimal prime ideals (ofR) is assured by taking 1 in the statement ofTheorem 5-15. .

. Proa! Since any proper ideal of R is contained in a maximal (hence, prime) i<;leal of R, fue theorem can be applied.

{O}

Our final theorem concerns the (set-theoretic) union of a finite number of prime ideals. In thi~ connection, we first observe that if an ideal 1 of the . ring R iscontained in the union J u K of two arbitrary ideal s of R, then 1 must be contained :in one <;lf them. For, suppose that 1 S; J u K with 1 si; J. It is therefor~; possible to choose an element a El n K such tlÍat a rt J. If b E 1 n J,..\~~n the sum a + b rt J (otherwise, a= (a + b) - b isinJ)andsoa + b·'é7(,whencebEK. TheimplicationisthatI n J S; K; consequently,~\:··r . . 1 == f~(J u K) = (1 n J) u (1 n K)

S;

K.

The next pOint\6'::~hich attention should be drawn is that the aboye fact about the·unioil"Ó'r two ideals is no longer true when . we pass to the union of three or more ideals. For a simple example, letR = Z2 x 2 2 : .

Theorem 5-15. Let 1 and P be ideals of the ring R, with P pdme. If 1 S; P, then Pcontains a minimal prime ideal of 1.

Corollary 1. Every proper ideal of the ring R possesses at least one . minimal prime ideal.

=

Corollary 2. If Pis a prime ideal of the ring R, then P contains a mini mal prime ideal of R. '

By abuse of language, we shall refer t~ the minimal prime ideals of the zero ideal {O} as the mini mal prime ideals of the ring R; that is, a prime ideal is a minimal prime ideal (of R) if it does not properly contain any other . . prime ideal. . Let us observ~ that, in the ring Z of Íljtegers, the minimal prime ideals of a nonzero ideal (n) are precisely the piií,iie ideal s (p), where p is a prime dividing n. In particular, one infers tha(eyery ideal in Z possesses only a finite number ofminimal prime ideal s (thigesult is generalized in TheoreÍn 12-3).' ;.: . It is not immediateln:lear that any (pr~per) idea1.admits minimal prime ideals, although this is indeed the case. To dispose of the question requires an appeal to Zorn'~ Lernma; the details
Proa! Denote by ff the family of aH prime ideals of R which contain 1 andare contained in P: ff = {pllp l is a prime ideal of R; 1 S; pi S; ..P}. We, point out that ff is not empty, sin ce P itself belongs to ff. Next, introduce a partial order ~ in ff which is opposite to the usual inc1usion relation; that is to say, if P', P" Eff, interpret pi ~ pI! to mean pI! S; Pi. Consider any nonempty subset {Pi} of ff which is totaHy ordered by ~ (more simply, {P¡} is a chain in 9"). Put P = nP¡. Then P is a prime ideal oi R (Problem 11, Chapter 5) containing 1 and contained in P; hence, PE ff. Sin ce P S; p¡ for every value of i, it foHows that p¡ ~ P, making Pan upper bound for {Pi}' AH the hypotheses ofZorn's Lemma, as applied' .to (ff, ~) are satisfied, so that ff" has a maximal element, say p* (this means that if pi E ff and p* ~ P', then P* = Pi). Inasmuch as P* E ff, it is a prime ideal of R with 1 S; P* S; P. There remains the task of showing that p* is necessarily a m.inimal prime ideal of 1.' Forthis, we suppose that pi is any prime ideal of R satisfying 1 S; pi S; P*. Then pi E ff and p* ~ Pi. By the maximal nature of P*, we thus have P* = P', signifying that P* is a minimal prime ideal of 1.

85

MAXIMAL, PRIME', AND PRIMARY IDEALS

FIRST COURSE IN RINGS AND IDEAL S

R

=

{(O, O), (O, 1), (1, O), (1, In.

We turn R into a ring by taking the addition to be componentwise addition modulo 2 and defining all products to be zero. Then, 1 1 = {(O, O), (O, In,

12 = {(O, 0), (1, 0n,

13

=

{(O, O), (1,

in

are all ideals in R and R = 11 U 12 U 13' It is c1ear, however, ihat R (regarded as an ideal) is not contained in any one of the I¡. . The situation just described can be countered by imposing the demand that each of the ideals irivolved in the union be prime. The theorem we have in mind asserts that if an ideal is contained in a finite union of prime ideals,then it is entirely contained in one of them. ActuaHy, it is easier to prove the contrapositive' ofthis statement, v i z . : ' Theorem 5-16. LetI be an arbitraryideal ofthe ring R and PI' P 2' ... , Pn be prime ideals of R. IfI si; p¡ for aH i, then there exists an element a E 1 such that a rt u Pi; hence, 1 si; u Pi' . i

Proa! The argument will be by induction on the number n ofprime ideals. Assume that the theorem has already been established when there are only n - 1 ideals (when n = 1, the result is trivial). Then, for each i (1 ~ i ~ n), there exists an element r¡ E 1 with r¡ rt Un¡ P j • If, for sorne value of i, it happens that r¡ rt Pi' then r¡ rt u P j and there is nothing to be proved. Thus, we may restrict our attention to the case where r¡ E,P¡ for aH i . In what follows, let a¡ =r 1 .. ··r¡-lr¡+l "·r n • We assert that.a¡rtP¡. Since Pi is prime, the contrary assumption a¡ E p¡ would imply that r j E p¡

86

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

for sorne j =1= i, which is impossible by our original choice of r.. On the other hand, if j =1= i, the element aj necessarily lies in Pi (r i being la factor of aj). For the final stage of the proof, put a = L ajo We first note that, because each of al' a2 , ••• , an is in l, the element a E 1. From the relation ai = a - L!'Fiaj, with ¿j'fi aj E P¡, it follows that a ~ Pi; otherwise, we would obtam ai E Pi' an obvious .contradiction. Our construction thus ensures the existence of an element a = L aj which belongs to the ideal l and not to any Pi' thereby proving the theorern. CoroUary. Let l be an arbitrary ideal ofthe ring R and PI' P 2 , be prime ideals of R. If l S;; U Pi' then l S;; Pi for some i.

••• ,

87

9. a) With the aid oC Theorem 5-5 and Example 5-1, obtain another prooC oC the Caet that Zp is a field iC and only iC p is a prime number. b) Prove that in Zn the maximal ideals are the principal ideals (p) p is a prime dividing n.

= pZ",

where

10. Given thatfis a homomorphism Crom the ring R onto the ring R', veriCy that a) R' is a field iC and only iC ker f is a maximal ideal oC R, b) R' is an integral domain iC and only iC ker f is a prime ideal oC R.

'*

11. a) Show that iC PI and P 2 are two ideals oC the ring R sueh that PI P 2 and P 2 $ PI' then the ideal PI n P 2 is not pririle. b) Let {PJ be a ehain oC prime ideals oCthe ring R. Prove that u Pi and n Pi are both prime ideals oC R. '

Pn

12. Prove that iC 1 is an ideal oC the ring R and P is a prime ideal oC 1, then P is an ideal oC the. whore ring R.

PROBLEMS

13. Let R denote the set oC all infinite sequenees {a n} oC rational numbers (that is, an E Q Cor every n). R becomes a eornmutative ring with identity iC the ring operations are defined termwise:

In the Collowing set oC problems, all rings are assumed to be eornmutative with identity.

1. a) Prove that Z Etl Z. is a maximal ideal oC the external direet sum Z Etl Z. b) Show that the ring R is a field iC and only iC {O} is a maximal ideal

oC R.

{a n}

2. Prove that a proper ideal M oC the ring R is maximal.iC and only iC, Cor every element r ~ M, there exists some a E R sueh that 1 + ra E M.

+

{b n}

=

{a n + bn},

{an}'{bn}

=

{anb n}·

re MI and M 2 are distinet maximal ideals oC the ring R, estabJish the equ~Út~ M I M2 = MI nM2 • . ; ';

VerifY eaeh oC the Collowing statements: a) the set B oC bounded sequences is a subring (with identity) oC R, b) the set C oC eonvergent sequenees is a subring (with identity) oC B, e) the set C o oC sequenees whieh converge to zero is a subring oC C, d) C o is an ideal' oC B, but not a prime ideal, e) C o is a maximal ideal oC C, l) the set D oC Cauehy sequences is a subring (with ideotity) oC B, g) C o is a maximal ideal oC D. .,1: Remark. Sinee the field D/C o is isomorphie to R#, thi,~ pfovides an alternative procedure Cor eonstrueting the real numbers [16J.

5. Let M be a proper ideal of the ring R. Prove that M is a maximal ideal if ¡;~d only iC, Cor eaeh ideal 1 oC R, either 1 ~ Mor else 1 + M = R. 't',

14. Assume that P is a proper prime ideal oC the ring R ~i.t~< the property that the quotient ring R/P is finite. Show that P must be a ma~iinal ideal oC R.

3. Letfbe a homomorphism Crom the ring R onto the ring R'. Pro ve that a) iC M is a maximal ideal oC R with M ;2 ker J, thenf(M) is a maximal ideal oC R', b) iCM' is a fl1aKimal ideal oC R', thenf-~(M') is a maximal ideal oC R, e) the mapping M --+ f(M) defines a one-to-one eorrespondenee between the set oC maximal ideals oC R whieh contain ker f and the set oC all maximal ideals oCR'. 4.

6. An ideal 1 oC the ring R is said to be minimal iC 1 =1= {O} and there exists no id~l:l) J oC R sueh that {O} e J e 1. t' a) Prove that a nonzero ideal 1 oC R is a minima1 ideal iC and only iC (a) = 1 for eaeh nomero element a e 1. b) VeriCy that the ring Z oCintegers has no minimal ideals.

'~l'

I 1,

=

1 = RI EB ••• Etl R i _¡ Etl Mi Etl Ri+ ¡ Etl·,· Etl R n, ,

where Mi is a maximal ideal oC R¡. [Hint: Problem 26, Chapter 2.J

7. ~et 1 be a proper ideal oC the ring R. Show that 1 is a prime ideal iC and oIily IC the eomplement (lC 1 is a multiplieatively closed subset oC R. 8. In the ring R

15. Let R = R¡ Etl R 2 Etl ... Etl Rn be the direet sum oC a finite number oC rings Ri' Establish that a proper ideal 1 oC R is a maximal idealjf and only iC, Cor some i (1 :5: i :5: n); 1 is oC the Corm

16. Let P and 1 be idealspCthe ring R, with P prime. ideal P:I = P.

map R"', define the set 1 by

re 1 $

P, prove that the quotient

17. Letfbe a homomorphism Crom the ring R onto the ring R'. Prove that

1 = {fe Rlf(l) = f( -1) = O}.

a) iC P is a prime (primary) ideal oC R with P ideal oC R';

Establish that 1 is an ideal oC R, but not a prime ideal.

I \

I

;2

ker J, thenf(P) is a prime (primary)

86

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

for sorne j =1= i, which is impossible by our original choice of r.. On the other hand, if j =1= i, the element aj necessarily lies in Pi (r i being la factor of aj). For the final stage of the proof, put a = L ajo We first note that, because each of al' a2 , ••• , an is in l, the element a E 1. From the relation ai = a - L!'Fiaj, with ¿j'fi aj E P¡, it follows that a ~ Pi; otherwise, we would obtam ai E Pi' an obvious .contradiction. Our construction thus ensures the existence of an element a = L aj which belongs to the ideal l and not to any Pi' thereby proving the theorern. CoroUary. Let l be an arbitrary ideal ofthe ring R and PI' P 2 , be prime ideals of R. If l S;; U Pi' then l S;; Pi for some i.

••• ,

87

9. a) With the aid oC Theorem 5-5 and Example 5-1, obtain another prooC oC the Caet that Zp is a field iC and only iC p is a prime number. b) Prove that in Zn the maximal ideals are the principal ideals (p) p is a prime dividing n.

= pZ",

where

10. Given thatfis a homomorphism Crom the ring R onto the ring R', veriCy that a) R' is a field iC and only iC ker f is a maximal ideal oC R, b) R' is an integral domain iC and only iC ker f is a prime ideal oC R.

'*

11. a) Show that iC PI and P 2 are two ideals oC the ring R sueh that PI P 2 and P 2 $ PI' then the ideal PI n P 2 is not pririle. b) Let {PJ be a ehain oC prime ideals oCthe ring R. Prove that u Pi and n Pi are both prime ideals oC R. '

Pn

12. Prove that iC 1 is an ideal oC the ring R and P is a prime ideal oC 1, then P is an ideal oC the. whore ring R.

PROBLEMS

13. Let R denote the set oC all infinite sequenees {a n} oC rational numbers (that is, an E Q Cor every n). R becomes a eornmutative ring with identity iC the ring operations are defined termwise:

In the Collowing set oC problems, all rings are assumed to be eornmutative with identity.

1. a) Prove that Z Etl Z. is a maximal ideal oC the external direet sum Z Etl Z. b) Show that the ring R is a field iC and only iC {O} is a maximal ideal

oC R.

{a n}

2. Prove that a proper ideal M oC the ring R is maximal.iC and only iC, Cor every element r ~ M, there exists some a E R sueh that 1 + ra E M.

+

{b n}

=

{a n + bn},

{an}'{bn}

=

{anb n}·

re MI and M 2 are distinet maximal ideals oC the ring R, estabJish the equ~Út~ M I M2 = MI nM2 • . ; ';

VerifY eaeh oC the Collowing statements: a) the set B oC bounded sequences is a subring (with identity) oC R, b) the set C oC eonvergent sequenees is a subring (with identity) oC B, e) the set C o oC sequenees whieh converge to zero is a subring oC C, d) C o is an ideal' oC B, but not a prime ideal, e) C o is a maximal ideal oC C, l) the set D oC Cauehy sequences is a subring (with ideotity) oC B, g) C o is a maximal ideal oC D. .,1: Remark. Sinee the field D/C o is isomorphie to R#, thi,~ pfovides an alternative procedure Cor eonstrueting the real numbers [16J.

5. Let M be a proper ideal of the ring R. Prove that M is a maximal ideal if ¡;~d only iC, Cor eaeh ideal 1 oC R, either 1 ~ Mor else 1 + M = R. 't',

14. Assume that P is a proper prime ideal oC the ring R ~i.t~< the property that the quotient ring R/P is finite. Show that P must be a ma~iinal ideal oC R.

3. Letfbe a homomorphism Crom the ring R onto the ring R'. Pro ve that a) iC M is a maximal ideal oC R with M ;2 ker J, thenf(M) is a maximal ideal oC R', b) iCM' is a fl1aKimal ideal oC R', thenf-~(M') is a maximal ideal oC R, e) the mapping M --+ f(M) defines a one-to-one eorrespondenee between the set oC maximal ideals oC R whieh contain ker f and the set oC all maximal ideals oCR'. 4.

6. An ideal 1 oC the ring R is said to be minimal iC 1 =1= {O} and there exists no id~l:l) J oC R sueh that {O} e J e 1. t' a) Prove that a nonzero ideal 1 oC R is a minima1 ideal iC and only iC (a) = 1 for eaeh nomero element a e 1. b) VeriCy that the ring Z oCintegers has no minimal ideals.

'~l'

I 1,

=

1 = RI EB ••• Etl R i _¡ Etl Mi Etl Ri+ ¡ Etl·,· Etl R n, ,

where Mi is a maximal ideal oC R¡. [Hint: Problem 26, Chapter 2.J

7. ~et 1 be a proper ideal oC the ring R. Show that 1 is a prime ideal iC and oIily IC the eomplement (lC 1 is a multiplieatively closed subset oC R. 8. In the ring R

15. Let R = R¡ Etl R 2 Etl ... Etl Rn be the direet sum oC a finite number oC rings Ri' Establish that a proper ideal 1 oC R is a maximal idealjf and only iC, Cor some i (1 :5: i :5: n); 1 is oC the Corm

16. Let P and 1 be idealspCthe ring R, with P prime. ideal P:I = P.

map R"', define the set 1 by

re 1 $

P, prove that the quotient

17. Letfbe a homomorphism Crom the ring R onto the ring R'. Prove that

1 = {fe Rlf(l) = f( -1) = O}.

a) iC P is a prime (primary) ideal oC R with P ideal oC R';

Establish that 1 is an ideal oC R, but not a prime ideal.

I \

I

;2

ker J, thenf(P) is a prime (primary)

88

FIRST COURSE IN RINGS AND IDEALS

b) if P' is a prime (primary) ideal of R', thenf-1(p') is a prime (prirnary) ideal of R; e) the mapping P -+ f(P) defines a one-to-one corresponden ce between the set of prime (prirnary) ideals of R whieh eontain ker f and the set of aH prime (primary) ideals of R'. 18. If M is a maxirnal ideal ofthe ring R and n E Z +, show that the quotient ring R/Mn has eX/1-etly one proper prime ideal. [Hint: Problem 17(e).] .¡

.JI

19. a) For. any ideal 1 of R, prove that 1 and are eontained in preeisely the same 'prime ideals of R. b) U~i~g part (a), deduoe that whenever 1 is a prime ideal of R, then 1 =

.JI.

20. Letfb,~ a homomorphism from the ring R onto the ri¡;¡g R'. Prove that , a) ifJ¡js an ideal of R with 1 ;2 ker f, then = f(.JI), b) ift';is an ideal of R', theri 1(1') = \Jf)..

.Jr

r

.JJ[i)

21. Verlt'y;that the intersection of semiprime ideals of the ring R is again a semiprime ide~CÓf R. '. . . .

22. If l:í~é:~ ideal ofthe ring R, pro ve that .JYis the smallest (in the set-theoretie sen se) semipdme ideal of R whieh eontains l. . 23. Establish that every divisor of zero in the ring Z p" (p a prime, n > O) is nilpotent. ~, J, and Q be ideals of the ring R, with Q primary. Prove the foHowing statements: a) if 1 $ JQ, then the quotient Q:l ,;. Q; b) if IJ s;: Q and 1 $ JQ, thenJ s:; Q; el if IJ s;: Q and the ideal J is finitely generated, then either 1 s:; Q orelse 1" s;: Q for sorne n E Z+. [Hínt: If 1 $ Q, eaeh generator of J is in .JQ.J

24. Let

If.JI

26. Let R be an integral domain and P be a prime ideal of R. Consider Rp, the ring of quotients of R relative to the corriplement of P: E

d) If 1 is a proper ideal ,of R and bE R - 1, then 1 s:;bR. 29. For a fixed prime p, eonsider the subset of rational numbers defined by

Vp = {a/b

E

Qlp .f' b}.

Show that . a) Vp is a valuation ring of Q; b) the tinique rnaidmal ideal of v" is Mp = {a/bE Qlp1b, butpla}; . e) the field VJMp ~ Zp. [Hint: Let the homomorphismf: Vp -> Zpq~ defined

byf(a/b)

[a] [b]-l.] .

=

"

30. a) Let 11 , 12 , ,In be arbitrary ideals of the ring R and P be a primeicl~al of R. If 1112 In' s:; P, establish that li s:; P for at least one value of i:, :;[Hint: If li $ P for aH i, ehoose a i E li - P and eonsider the element a = ai;a~ an .] b) Assume that M is a maximal ideal of R. Prove that, for eaeh integer n E Z+, the only prime idea,l eontaining M n is M. oo.

oo.

.oo

;:~t;

.';'.

31. Let R be an integral domain with the property that every proper ideal ist}:t~.:produet of maxirnalideals. Prove that . ' . ~:.: ..: a) If M is a rnaximal ideal of R, then there exists an element ,a E.R' iílid ideal K =1= {O} suehthat MK = (a). [Hint: If M =1= {O}, piekO'=I= aEM. Then M 1M 2 oo· M n = (a) s:; M for suitable rnaximal ideals Mi; henee, M = Mi for sorne i.] b) If 1, J, M are ideals of R, with M maximal, then 1M = JM implies 1 = J. 32. a) Ir 1 is an ideal of the ring R sueh that 1 s:; (a), show that there exists an jdeal J of R for which aJ = l. [Hint: Take J = (1: (a)).] b) Prove that if a principal ideal (a) of the ring R properly eontains a prime ideal P, then P s:; (a").

n

n=1

25. Assume that 1 is an ideal of the ring R. is a maximal ideal of R, show that lis primary. [Hint: Mimie the argument of the eorollaryto Theorem 5-14.]

Rp .= {ab- 1

89

PROBLEMS

Qcl(R)laER; b'P}.

Prove that the ring Rp (whieh is known as the localizatíon of R at the prime ideal P) has exaetly one maximal ideal, namely, 1 = {ab- 1 E QclR)la E P; b, P}.

a,

27. A ring R is said to be a local ring if it has a unique maximal ideal. If R is a local rlng with M as its maximal ideal, show that any element M is in vertible in R.,

28. A subring R of a field F is said to be a valuation ring of F if for eaeh nonzero . element a E F at least one of a or a- 1 belongs to R. Assuming that R is a valuation ring of F, prove the following: a) R eontains a1l the idempotent elements of F. b) R is a local ring,with tiniquemaximal ideal M = {a ERla-1 'R}. [Remarlc. iI- 1 denotes the inverse of a in F.] e) For any two elements a, b ER, either aR S;; bR or bR ;2 aRo [Hint: Either ab- 1 ER or ba- 1 ER.]

33. Let 1 be a primary ideal of the ring R. Prove that 1 has exaetly one minimal prime ideal, namely, [Hint: Problem 19.]

F

88

FIRST COURSE IN RINGS AND IDEALS

b) if P' is a prime (primary) ideal of R', thenf-1(p') is a prime (prirnary) ideal of R; e) the mapping P -+ f(P) defines a one-to-one corresponden ce between the set of prime (prirnary) ideals of R whieh eontain ker f and the set of aH prime (primary) ideals of R'. 18. If M is a maxirnal ideal ofthe ring R and n E Z +, show that the quotient ring R/Mn has eX/1-etly one proper prime ideal. [Hint: Problem 17(e).] .¡

.JI

19. a) For. any ideal 1 of R, prove that 1 and are eontained in preeisely the same 'prime ideals of R. b) U~i~g part (a), deduoe that whenever 1 is a prime ideal of R, then 1 =

.JI.

20. Letfb,~ a homomorphism from the ring R onto the ri¡;¡g R'. Prove that , a) ifJ¡js an ideal of R with 1 ;2 ker f, then = f(.JI), b) ift';is an ideal of R', theri 1(1') = \Jf)..

.Jr

r

.JJ[i)

21. Verlt'y;that the intersection of semiprime ideals of the ring R is again a semiprime ide~CÓf R. '. . . .

22. If l:í~é:~ ideal ofthe ring R, pro ve that .JYis the smallest (in the set-theoretie sen se) semipdme ideal of R whieh eontains l. . 23. Establish that every divisor of zero in the ring Z p" (p a prime, n > O) is nilpotent. ~, J, and Q be ideals of the ring R, with Q primary. Prove the foHowing statements: a) if 1 $ JQ, then the quotient Q:l ,;. Q; b) if IJ s;: Q and 1 $ JQ, thenJ s:; Q; el if IJ s;: Q and the ideal J is finitely generated, then either 1 s:; Q orelse 1" s;: Q for sorne n E Z+. [Hínt: If 1 $ Q, eaeh generator of J is in .JQ.J

24. Let

If.JI

26. Let R be an integral domain and P be a prime ideal of R. Consider Rp, the ring of quotients of R relative to the corriplement of P: E

d) If 1 is a proper ideal ,of R and bE R - 1, then 1 s:;bR. 29. For a fixed prime p, eonsider the subset of rational numbers defined by

Vp = {a/b

E

Qlp .f' b}.

Show that . a) Vp is a valuation ring of Q; b) the tinique rnaidmal ideal of v" is Mp = {a/bE Qlp1b, butpla}; . e) the field VJMp ~ Zp. [Hint: Let the homomorphismf: Vp -> Zpq~ defined

byf(a/b)

[a] [b]-l.] .

=

"

30. a) Let 11 , 12 , ,In be arbitrary ideals of the ring R and P be a primeicl~al of R. If 1112 In' s:; P, establish that li s:; P for at least one value of i:, :;[Hint: If li $ P for aH i, ehoose a i E li - P and eonsider the element a = ai;a~ an .] b) Assume that M is a maximal ideal of R. Prove that, for eaeh integer n E Z+, the only prime idea,l eontaining M n is M. oo.

oo.

.oo

;:~t;

.';'.

31. Let R be an integral domain with the property that every proper ideal ist}:t~.:produet of maxirnalideals. Prove that . ' . ~:.: ..: a) If M is a rnaximal ideal of R, then there exists an element ,a E.R' iílid ideal K =1= {O} suehthat MK = (a). [Hint: If M =1= {O}, piekO'=I= aEM. Then M 1M 2 oo· M n = (a) s:; M for suitable rnaximal ideals Mi; henee, M = Mi for sorne i.] b) If 1, J, M are ideals of R, with M maximal, then 1M = JM implies 1 = J. 32. a) Ir 1 is an ideal of the ring R sueh that 1 s:; (a), show that there exists an jdeal J of R for which aJ = l. [Hint: Take J = (1: (a)).] b) Prove that if a principal ideal (a) of the ring R properly eontains a prime ideal P, then P s:; (a").

n

n=1

25. Assume that 1 is an ideal of the ring R. is a maximal ideal of R, show that lis primary. [Hint: Mimie the argument of the eorollaryto Theorem 5-14.]

Rp .= {ab- 1

89

PROBLEMS

Qcl(R)laER; b'P}.

Prove that the ring Rp (whieh is known as the localizatíon of R at the prime ideal P) has exaetly one maximal ideal, namely, 1 = {ab- 1 E QclR)la E P; b, P}.

a,

27. A ring R is said to be a local ring if it has a unique maximal ideal. If R is a local rlng with M as its maximal ideal, show that any element M is in vertible in R.,

28. A subring R of a field F is said to be a valuation ring of F if for eaeh nonzero . element a E F at least one of a or a- 1 belongs to R. Assuming that R is a valuation ring of F, prove the following: a) R eontains a1l the idempotent elements of F. b) R is a local ring,with tiniquemaximal ideal M = {a ERla-1 'R}. [Remarlc. iI- 1 denotes the inverse of a in F.] e) For any two elements a, b ER, either aR S;; bR or bR ;2 aRo [Hint: Either ab- 1 ER or ba- 1 ER.]

33. Let 1 be a primary ideal of the ring R. Prove that 1 has exaetly one minimal prime ideal, namely, [Hint: Problem 19.]

F

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

SIX

91

2) al1 if and only if a is invertible;

3) ifalb,thenaelbe; 4) if alb and ble, then ale; 5) if ela and elb, then el(ax

DIVISIBILITY THEORY IN INTEGRAL DOMAINS As the title sugg~sts, this chapter is concerned with the problem of factoring elements of an 111tegral domain as productsof irreducible elements. The particular impetus is furnished by th~ ring of integers, where the Funda~ental Theorem of Arithmetic states that every integer n' > 1 can be written 111 a~ essentia11y unique way, as a product of prime numbers; for example: the 111teg~r 360 = 2·2· 2·3·3· 5. We are interested here in the possibility of extendl~g the factorization theory of the ring Z and, in particular, the afo~ementloned Fundamental Theorem of Arithmetic to a more general settmg.. Needless to say, any reasonable abstraction of these numberth~or~tic ideas depends on a suitable interpretation of prime elements (the bUlldl?g blocks for the study of divisibility questions in Z) in integral doma111s. Except for certain definitions, which we prefer to have available ~or arbitrary ~ii:J.gs, our hypothesis will, for the most part, restrict us to mtegral. ~o.m.a~ns. !he plan .¡s to proceed from the most general results ab.out dlVlSlblh~y, prIme elemerit~, and uniqueness offactorization to stronger results concern111g specific c1a~ses of integral domains. Throughout this chapter, the. rings considered are assumed to be commuta ti ve; and it is supposed that each possesses an identity element. .

.j •.•.,

D~finition 6-1. If a =1= O !:al¡1~ b are elements of the ring R, then a is sald to divide b, in symbg.1s· alb, provided that there exists sorne e e' R such that b == ae. In caseq does not divide b, we sha11 write a ! b. Oth~r l~~g~age for the divisibility property alb is that a is afaetor of b, th~t b IS dw!s!ble by a, and that b is a multiple of a. Whenever the notation alb IS employed, it is to be uríderstood (even if not explicitIy mentioned) that ~he element a =1= O; on the other hand, not only may b = O, but in such mstances we always have divisibility. Sorne immediate consequences of this definition are listed below' the reader should convince himself of each of them. ' Theorem 6-1. Let the elements a, b, e E R. Then, 1) alO, lla, ala;

90

+ by) for every x, y E R.

Division. of elements in a ring R is c10sely related to ideal inc1usion:

alb if and only if (b)

f;

(a).

Indeed, al b means that b = ae for sorne e E R; thus, b E (a), so that (b) f; (a). Conversely, if(b) f; (a), then there exists an e1ement e in R for which b = ae, implying that alb. Questions concerning divisibility are complicated somewhat by the presence of in vertible elements. For, if u has a multiplicative inverse, any element of a E R can be expressed in the form a = a(uu- 1), so that both ula and u - 11 a. An extreme situation occurs in the case of fields, where every nonzero element divides every other element. On the other hand, in the ring Ze of even integers, the element 2 has no divisors at a11. In order to overcome the difficulty that is produced by invertible elements, we introduce the following definition. Definition 6-2. Two elements a, bE R are said to be associated elements or simply assoeiates if a = bu, where u is an in vertible element of R. A simple argument shows that the relation "', defined on R by taking a '" b if and only if a is an associate of b, is an equivalence relation with equivalence c1asses which are sets of associated elements. The associates of the identity are just the invertible elements of R. Example .6·,.,1. In the case of the ring Z, the only associates of an integer n E Z are;'±n, since ± 1 are the only invertible elements. Example 6-;2. Consider the domain Z(i) of Gaussian integers, a subdomain of the complex number field, whose elements form the set

Z(i) = {a'

+

bila, b E Z; i2 = -1}.

Here, the Qnly invertible elements are ± 1 and ± i. For, suppose a hasamultiplicativeinversee + di. Then,wemusthave(a + bi)(e so that (a '- bi)(e - di) = 1. Therefore,

+ bi E Z(i) + di) = 1,

+ bi)(e + di)(a - bi)(e - di) (a 2 + b2 )(e 2 + d2 ).

1 = (a =

From the fact that a, b, e, d are a11 integers, it follows that a2 + b 2 = 1. The only solutions of this last equation are a = ± 1, b = O or a = O, b = ± 1. This leads to the four invertible elements ± 1, ± i. In con-

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

SIX

91

2) al1 if and only if a is invertible;

3) ifalb,thenaelbe; 4) if alb and ble, then ale; 5) if ela and elb, then el(ax

DIVISIBILITY THEORY IN INTEGRAL DOMAINS As the title sugg~sts, this chapter is concerned with the problem of factoring elements of an 111tegral domain as productsof irreducible elements. The particular impetus is furnished by th~ ring of integers, where the Funda~ental Theorem of Arithmetic states that every integer n' > 1 can be written 111 a~ essentia11y unique way, as a product of prime numbers; for example: the 111teg~r 360 = 2·2· 2·3·3· 5. We are interested here in the possibility of extendl~g the factorization theory of the ring Z and, in particular, the afo~ementloned Fundamental Theorem of Arithmetic to a more general settmg.. Needless to say, any reasonable abstraction of these numberth~or~tic ideas depends on a suitable interpretation of prime elements (the bUlldl?g blocks for the study of divisibility questions in Z) in integral doma111s. Except for certain definitions, which we prefer to have available ~or arbitrary ~ii:J.gs, our hypothesis will, for the most part, restrict us to mtegral. ~o.m.a~ns. !he plan .¡s to proceed from the most general results ab.out dlVlSlblh~y, prIme elemerit~, and uniqueness offactorization to stronger results concern111g specific c1a~ses of integral domains. Throughout this chapter, the. rings considered are assumed to be commuta ti ve; and it is supposed that each possesses an identity element. .

.j •.•.,

D~finition 6-1. If a =1= O !:al¡1~ b are elements of the ring R, then a is sald to divide b, in symbg.1s· alb, provided that there exists sorne e e' R such that b == ae. In caseq does not divide b, we sha11 write a ! b. Oth~r l~~g~age for the divisibility property alb is that a is afaetor of b, th~t b IS dw!s!ble by a, and that b is a multiple of a. Whenever the notation alb IS employed, it is to be uríderstood (even if not explicitIy mentioned) that ~he element a =1= O; on the other hand, not only may b = O, but in such mstances we always have divisibility. Sorne immediate consequences of this definition are listed below' the reader should convince himself of each of them. ' Theorem 6-1. Let the elements a, b, e E R. Then, 1) alO, lla, ala;

90

+ by) for every x, y E R.

Division. of elements in a ring R is c10sely related to ideal inc1usion:

alb if and only if (b)

f;

(a).

Indeed, al b means that b = ae for sorne e E R; thus, b E (a), so that (b) f; (a). Conversely, if(b) f; (a), then there exists an e1ement e in R for which b = ae, implying that alb. Questions concerning divisibility are complicated somewhat by the presence of in vertible elements. For, if u has a multiplicative inverse, any element of a E R can be expressed in the form a = a(uu- 1), so that both ula and u - 11 a. An extreme situation occurs in the case of fields, where every nonzero element divides every other element. On the other hand, in the ring Ze of even integers, the element 2 has no divisors at a11. In order to overcome the difficulty that is produced by invertible elements, we introduce the following definition. Definition 6-2. Two elements a, bE R are said to be associated elements or simply assoeiates if a = bu, where u is an in vertible element of R. A simple argument shows that the relation "', defined on R by taking a '" b if and only if a is an associate of b, is an equivalence relation with equivalence c1asses which are sets of associated elements. The associates of the identity are just the invertible elements of R. Example .6·,.,1. In the case of the ring Z, the only associates of an integer n E Z are;'±n, since ± 1 are the only invertible elements. Example 6-;2. Consider the domain Z(i) of Gaussian integers, a subdomain of the complex number field, whose elements form the set

Z(i) = {a'

+

bila, b E Z; i2 = -1}.

Here, the Qnly invertible elements are ± 1 and ± i. For, suppose a hasamultiplicativeinversee + di. Then,wemusthave(a + bi)(e so that (a '- bi)(e - di) = 1. Therefore,

+ bi E Z(i) + di) = 1,

+ bi)(e + di)(a - bi)(e - di) (a 2 + b2 )(e 2 + d2 ).

1 = (a =

From the fact that a, b, e, d are a11 integers, it follows that a2 + b 2 = 1. The only solutions of this last equation are a = ± 1, b = O or a = O, b = ± 1. This leads to the four invertible elements ± 1, ± i. In con-

92

FIRST COURSE IN RINGS AND IDEALS

sequence, the c1ass of associates determined by any Gaussian integer a consists of exactly four members:

a

+ bi, -a - bi, -b + ai, b -

+ bi

Theorem 6-3. Let ,al' a 2 , ,.;, an be nonzero elements qf the ring R. Then al' a2' ... , an bave a greatest common divisor d, expressible in the form (r¡ ER),

aL

Since associated elements are rather c10sely related, it is not surprising that they have similar properties; for instance: Theo'rem 6-2. Let a, b be nonzero elements Then the Iollowing statements are eq~ivalent:

ifand only iftbe ideal (al' a 2, ''', an) is principal.

Proof Suppose that d gcd (al' a2 , '" , an) exists and can be written in • the form d = rla l + r 2 a 2 + ..,~ + rnan' with r¡ E R. Tben tbe element d líes in the ideal (al' a2, ... , a.)i 'which implies tbat (d) s;:; (al' a 2, .. " an)· To obtain the reverse inc1usion;:'observe that sin ce d = gcd (al' a2' ... , a.), each a¡ is a multiple of d; say, ';-= x¡d, wbere X¡ E R. Thus, for an arbitrary member yla l + yza z + .. , + jy~.an ofthe ideal (al' a2, ... , an), we must have

integral domain R.

1) a and b are associates,

a;

2) both alb and bla,

3).(a)

=

(b).

yla l

Proof To prove the ~\livalence of (1) anó (2), suppose that a = bu, where u is an invertible element; then, also, b. au-l,s~rtbat both alb and bla. Going in the opposite direction, ir al b, we can writeq;~= ax for sorne x E R; while, from bla, ít follows that a by wíth y E R. 1h'erefore, b :::. (by)x· b(yx). Since b =1= ,O, the cancellation law implies that 1 = yx. Hence, y is. an invertible element of R, with a = by, proving that a and b must be associates. The equivalence oI (2) and (3) stems from our earlier remarks

~:"::·:~l.,

(ai' al' ... ;'(sn) = (d)

(d E R).

Qur aim, of course, is to prové tbat d gcd (al' al' ... ,an ). Since each a, E (d), there exist elements b¡ in R for which al = bid, whence dja; for i '= 1,2, ... , n. It remains only to establish that any common divisor e of tbe a· also divides d. Now, a¡ SiC for suitable S¡ E R. As an elementof (al' a~, ... , an), d must have tbe form d = r1a 1 + r2 a2 + ... + rnan' witb r¡ in R. This mean s tbat d (rls l + r2sl + ... + rnsn)e,

We next examine the notion of a greatest common divisor. Definition 6':'3. Let al' a2 , ••• , an be nonzero elements of the ring R. An element dE R is a greatest eommon divisor oI al' a2• ... ,an if it possesses the properties .

which is to say tbat cid. Thus, dis a greatest common divisor of al' a 2 , and has tbe desired representation.

1) dja¡for i = 1,2, ... , n (d is acornmon divisor),

... ,

an

Corollary. Any finite set ofnonzero dementsa l , a2 , ••• , a. ofa principal ideal ring R bas a greatest common divisor; in fact,

= 1,2, ... , n imp,lies that cid.

The use of tbe superlative adjective "greatest" in this definitiol,1 does not imply that d has greater magnitude than any other common divisor e, but only that d is a multiple of any such e. A natural question to ask is wbetber the elements al' a2, ... , an E R can possess two different greatest common dívisors. For an answer, suppose tbat there are two elements d and d' in R satisfying the conditions of Definition 6-3. Then, by (2), we must have dld' as well as d'jd; according to Theorem 6-2, this implies that d and d' are associates. Thus, the greatest common divisor of al' a 2, .. ' , an is unique, wbenever it exists, up to arbitrary invertible faétors. We shall find. it convenient to denote any greatest common divisor of al' a2, ... , all by gcd (al' a 2, ... , an ). The next tbeorem wiI1 prove, at least for principal ideal ríngs, tbat any finite set ofnonzero elements actualIy does have a greatest common divisor.

+ Y2al + ... + Yna~i~ (YlXl + Y2X2 + ... + y.x.)d E (d).

Tbis fact shows that (al' a 2 , ",:¡,!:.'qn) s;:; (d), and equality follows. For the converse, let (al' 1l~?. , a.) be a principal ideal of R:

. relating division of ring elements to ideal inclusion.

2) ela¡ for i

93

DIVISIBILITY THBORY 'IN INTEGRAL DOMAINS

gcd (al' a2 ,

... ,

an)

for suitable choice of r l , r2 ,

I ,1

1

=

... ,

r1a l

+ r2a2 + ... + rnan

rn E R.

When. (al' a 2, ... , an ) = R, the elements al' a2, ... ,an must have a common divisor which is an invertible element of R; in this case, we say tbat al' al' ... , are relatively prime and sball wríte gcd (al' a2, ... ,an ) = 1. Ir al' a2 , '" , a" are nonzero elements of a principal ideal ring R, then the corollary to Theorem 6-3 tells us that al' a2 , '" , an are re1atively prime ¡fand only iftbere exist r l , r2 , .;" r" E R such tbat

a.

rla l

+ r2a2 + ... + rilan

= 1

(Bezout's Identity).

Qne of the most useful applications of Bezout's Identity is the following (it also serves to motívate our coming definition of a prime element).

92

FIRST COURSE IN RINGS AND IDEALS

sequence, the c1ass of associates determined by any Gaussian integer a consists of exactly four members:

a

+ bi, -a - bi, -b + ai, b -

+ bi

Theorem 6-3. Let ,al' a 2 , ,.;, an be nonzero elements qf the ring R. Then al' a2' ... , an bave a greatest common divisor d, expressible in the form (r¡ ER),

aL

Since associated elements are rather c10sely related, it is not surprising that they have similar properties; for instance: Theo'rem 6-2. Let a, b be nonzero elements Then the Iollowing statements are eq~ivalent:

ifand only iftbe ideal (al' a 2, ''', an) is principal.

Proof Suppose that d gcd (al' a2 , '" , an) exists and can be written in • the form d = rla l + r 2 a 2 + ..,~ + rnan' with r¡ E R. Tben tbe element d líes in the ideal (al' a2, ... , a.)i 'which implies tbat (d) s;:; (al' a 2, .. " an)· To obtain the reverse inc1usion;:'observe that sin ce d = gcd (al' a2' ... , a.), each a¡ is a multiple of d; say, ';-= x¡d, wbere X¡ E R. Thus, for an arbitrary member yla l + yza z + .. , + jy~.an ofthe ideal (al' a2, ... , an), we must have

integral domain R.

1) a and b are associates,

a;

2) both alb and bla,

3).(a)

=

(b).

yla l

Proof To prove the ~\livalence of (1) anó (2), suppose that a = bu, where u is an invertible element; then, also, b. au-l,s~rtbat both alb and bla. Going in the opposite direction, ir al b, we can writeq;~= ax for sorne x E R; while, from bla, ít follows that a by wíth y E R. 1h'erefore, b :::. (by)x· b(yx). Since b =1= ,O, the cancellation law implies that 1 = yx. Hence, y is. an invertible element of R, with a = by, proving that a and b must be associates. The equivalence oI (2) and (3) stems from our earlier remarks

~:"::·:~l.,

(ai' al' ... ;'(sn) = (d)

(d E R).

Qur aim, of course, is to prové tbat d gcd (al' al' ... ,an ). Since each a, E (d), there exist elements b¡ in R for which al = bid, whence dja; for i '= 1,2, ... , n. It remains only to establish that any common divisor e of tbe a· also divides d. Now, a¡ SiC for suitable S¡ E R. As an elementof (al' a~, ... , an), d must have tbe form d = r1a 1 + r2 a2 + ... + rnan' witb r¡ in R. This mean s tbat d (rls l + r2sl + ... + rnsn)e,

We next examine the notion of a greatest common divisor. Definition 6':'3. Let al' a2 , ••• , an be nonzero elements of the ring R. An element dE R is a greatest eommon divisor oI al' a2• ... ,an if it possesses the properties .

which is to say tbat cid. Thus, dis a greatest common divisor of al' a 2 , and has tbe desired representation.

1) dja¡for i = 1,2, ... , n (d is acornmon divisor),

... ,

an

Corollary. Any finite set ofnonzero dementsa l , a2 , ••• , a. ofa principal ideal ring R bas a greatest common divisor; in fact,

= 1,2, ... , n imp,lies that cid.

The use of tbe superlative adjective "greatest" in this definitiol,1 does not imply that d has greater magnitude than any other common divisor e, but only that d is a multiple of any such e. A natural question to ask is wbetber the elements al' a2, ... , an E R can possess two different greatest common dívisors. For an answer, suppose tbat there are two elements d and d' in R satisfying the conditions of Definition 6-3. Then, by (2), we must have dld' as well as d'jd; according to Theorem 6-2, this implies that d and d' are associates. Thus, the greatest common divisor of al' a 2, .. ' , an is unique, wbenever it exists, up to arbitrary invertible faétors. We shall find. it convenient to denote any greatest common divisor of al' a2, ... , all by gcd (al' a 2, ... , an ). The next tbeorem wiI1 prove, at least for principal ideal ríngs, tbat any finite set ofnonzero elements actualIy does have a greatest common divisor.

+ Y2al + ... + Yna~i~ (YlXl + Y2X2 + ... + y.x.)d E (d).

Tbis fact shows that (al' a 2 , ",:¡,!:.'qn) s;:; (d), and equality follows. For the converse, let (al' 1l~?. , a.) be a principal ideal of R:

. relating division of ring elements to ideal inclusion.

2) ela¡ for i

93

DIVISIBILITY THBORY 'IN INTEGRAL DOMAINS

gcd (al' a2 ,

... ,

an)

for suitable choice of r l , r2 ,

I ,1

1

=

... ,

r1a l

+ r2a2 + ... + rnan

rn E R.

When. (al' a 2, ... , an ) = R, the elements al' a2, ... ,an must have a common divisor which is an invertible element of R; in this case, we say tbat al' al' ... , are relatively prime and sball wríte gcd (al' a2, ... ,an ) = 1. Ir al' a2 , '" , a" are nonzero elements of a principal ideal ring R, then the corollary to Theorem 6-3 tells us that al' a2 , '" , an are re1atively prime ¡fand only iftbere exist r l , r2 , .;" r" E R such tbat

a.

rla l

+ r2a2 + ... + rilan

= 1

(Bezout's Identity).

Qne of the most useful applications of Bezout's Identity is the following (it also serves to motívate our coming definition of a prime element).

94

FIRST COURSE IN RINGS AND IDEALS

Th~orem 6-4. , Let a, b, e be elements ofthe principal ideal ring R. If elab, wIth a and e relatively prime, then elb. Proof. Since a and e are relatively prime, so that gcd (a, e) = 1, there exist elements r, s E R satisfying 1 = ra + se; hence, b = lb = rab

+ sebo

~~.elab ánd ele, Theorem 6-1(5) guarantees that el(rab + seb), or rather, Dual to the notion of greatest common divisor there is the idea of a least cornmon multiple, defined below. Definition 6-4.. Let al' a2 , ... , an be nonzero elements of a ring R. An element dE R IS a least eommon multiple of al' a2 , ... , an if

1) a¡ld for i

=

1,2, ... ,n (d is a common multiple),

2) ade for i = 1,2, ... , n implies dic.

. . !n brief, an eleme~t dE R is a least cornmon multiple of al' a2 , ... ,a n If It I~ a common multIple of al' a2 , ••• , an which divides any other common ~ult.lple.. Th~ reader should note that a least common multiple, in case It eXIsts, IS ulllque apart from the distinction between associates' indeed if d and d' are both least cornmon multiples of al' a2', ... , an, the~ dld' a~d d'ld; hence, d and d' are associates. We hereafter adopt the standard notation lcm (al' a2 , ... , iln) t~ represent an~ least common multiple of al' a2 , ... , an0 The next result IS a useful companion to Theorem 6-3. Theorem 6-5. Let al' a2 , ... ,all be nonzero elements ott:h~ ring R. Then al' a2 , ... , an have a least common multiple if and only'ifthe ideal n (a¡) is principal. .

Proof. We begin by assuming that d = lcm (al' a2 , ••• ,an) exi~is. Then the ele~ent d ~ies in each of the 'principal ideals (a¡), for i =1 1,2, ... , n, whence rn the rntersection n (a¡). This means that (d) !;; n (di). On the other ~and, any element rE n (a¡) is a common multiple of eadí of the a¡. But d IS a least common multiple, so that dlr, or, equivalently, r~ (d). This leads t~ th~ inc1usion n (a¡) !;; (ti) and the subsequent equality... ?o~g I~ the opposite direction, suppose that the intersection n (a¡) is a prIncIpal Ideal of R, say n (a¡) = (a). Since (a) !;; (a¡), it follows that ada for every i, ma~ng a a common multiple of al' a2 , ... ,an • Given any other common multIple b of al' a2 , ... , an , the condition a;/b implies that (b) !;; (a¡) for each val.ue of i. As a result, (b) !;; n (a¡) = (a) and so alb. Our argument estabhshes that a = lcm (al' a2 , ... ,an ), completing the proof.

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

95

At this point, we introduce two additional definitions. These will help to describe, in a fairly concise manner, certain situations wbich will occur in the sequel. Definition 6-5. A ring R is said to have the ged-property (lem-property) provided that any finite number of nonzero elements of R admit a greatest common divisor (least common multiple). The content of Theorem 6-3 is that a ring R ha.s the gcd-property if and only if every finitely generated ideal of R is principal. Likewise, Theorem 6-5 tells us that R possesses the lcm-property if and only if the intersection of any finite number of principal ideals of R is again principal. Suffice it to say, every principal ideal ring satisfies both these properties. The immediate task is to prove that any integral domain has the gcdproperty if and only if it has the lcm-property. In the pro ces s, we shall acquire certain other facts which have significance for our subsequent investigation. So as to avoid becoming submerged in minor details at a critical stage of the discussion, let us first establish a lemma. Lemma. Let al' a2 , domain R. 1) Iflcm (al' a2 ,

... ,

... ,

an and rbe nonzero elements of an integral

an ) exists, then lcm (ra l , ra 2 ,

lcm (ra l , ra2 , 2) If gcd (ra l , ra 2 ,

... ,

... ,

ra n)

=

r lcm (al' a2 ,

ran ) exists, then gcd (al' a2 ,

gcd (ra l , ra 2 ,

...

,ran )

... ,

= ~

ran) also exists and

... ,

... ,

gcd (al' a2 ,

a,,).

an ) also exists and

... ,

an ).

Proof. First, as sume that d = lcm (al'~ 'q.ú ... ,a,,) exists. Then a¡jd for each value of i, whence ra;jrd. Now, let d' be any common multiple of ra 2 , ra 2 , ... ,ran • Then rld', say d' = rs, w~ere s E R. It follows that a¡ls for every i and so dls. As a result, rdk.if:or rdld'. But this means that lcm (ra l , ra 2 , ... , ran ) exists and equals r{'';' r lcm (al' a2 , ... , an ). As regards the second assertion, suPP
...

,ran)

=

e

=

rt

=

r gcd (al' a2 ,

... ,

an ).

Remark. It is entirely possible for gcd (al' a2 , ... , an ) to exist without the existence of gcd (ra l , ra 2 , ... , ran ); tbis accounts for the lack of symmetry in the statement of the aboye lemma. (See Example 6-4.)

94

FIRST COURSE IN RINGS AND IDEALS

Th~orem 6-4. , Let a, b, e be elements ofthe principal ideal ring R. If elab, wIth a and e relatively prime, then elb. Proof. Since a and e are relatively prime, so that gcd (a, e) = 1, there exist elements r, s E R satisfying 1 = ra + se; hence, b = lb = rab

+ sebo

~~.elab ánd ele, Theorem 6-1(5) guarantees that el(rab + seb), or rather, Dual to the notion of greatest common divisor there is the idea of a least cornmon multiple, defined below. Definition 6-4.. Let al' a2 , ... , an be nonzero elements of a ring R. An element dE R IS a least eommon multiple of al' a2 , ... , an if

1) a¡ld for i

=

1,2, ... ,n (d is a common multiple),

2) ade for i = 1,2, ... , n implies dic.

. . !n brief, an eleme~t dE R is a least cornmon multiple of al' a2 , ... ,a n If It I~ a common multIple of al' a2 , ••• , an which divides any other common ~ult.lple.. Th~ reader should note that a least common multiple, in case It eXIsts, IS ulllque apart from the distinction between associates' indeed if d and d' are both least cornmon multiples of al' a2', ... , an, the~ dld' a~d d'ld; hence, d and d' are associates. We hereafter adopt the standard notation lcm (al' a2 , ... , iln) t~ represent an~ least common multiple of al' a2 , ... , an0 The next result IS a useful companion to Theorem 6-3. Theorem 6-5. Let al' a2 , ... ,all be nonzero elements ott:h~ ring R. Then al' a2 , ... , an have a least common multiple if and only'ifthe ideal n (a¡) is principal. .

Proof. We begin by assuming that d = lcm (al' a2 , ••• ,an) exi~is. Then the ele~ent d ~ies in each of the 'principal ideals (a¡), for i =1 1,2, ... , n, whence rn the rntersection n (a¡). This means that (d) !;; n (di). On the other ~and, any element rE n (a¡) is a common multiple of eadí of the a¡. But d IS a least common multiple, so that dlr, or, equivalently, r~ (d). This leads t~ th~ inc1usion n (a¡) !;; (ti) and the subsequent equality... ?o~g I~ the opposite direction, suppose that the intersection n (a¡) is a prIncIpal Ideal of R, say n (a¡) = (a). Since (a) !;; (a¡), it follows that ada for every i, ma~ng a a common multiple of al' a2 , ... ,an • Given any other common multIple b of al' a2 , ... , an , the condition a;/b implies that (b) !;; (a¡) for each val.ue of i. As a result, (b) !;; n (a¡) = (a) and so alb. Our argument estabhshes that a = lcm (al' a2 , ... ,an ), completing the proof.

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

95

At this point, we introduce two additional definitions. These will help to describe, in a fairly concise manner, certain situations wbich will occur in the sequel. Definition 6-5. A ring R is said to have the ged-property (lem-property) provided that any finite number of nonzero elements of R admit a greatest common divisor (least common multiple). The content of Theorem 6-3 is that a ring R ha.s the gcd-property if and only if every finitely generated ideal of R is principal. Likewise, Theorem 6-5 tells us that R possesses the lcm-property if and only if the intersection of any finite number of principal ideals of R is again principal. Suffice it to say, every principal ideal ring satisfies both these properties. The immediate task is to prove that any integral domain has the gcdproperty if and only if it has the lcm-property. In the pro ces s, we shall acquire certain other facts which have significance for our subsequent investigation. So as to avoid becoming submerged in minor details at a critical stage of the discussion, let us first establish a lemma. Lemma. Let al' a2 , domain R. 1) Iflcm (al' a2 ,

... ,

... ,

an and rbe nonzero elements of an integral

an ) exists, then lcm (ra l , ra 2 ,

lcm (ra l , ra2 , 2) If gcd (ra l , ra 2 ,

... ,

... ,

ra n)

=

r lcm (al' a2 ,

ran ) exists, then gcd (al' a2 ,

gcd (ra l , ra 2 ,

...

,ran )

... ,

= ~

ran) also exists and

... ,

... ,

gcd (al' a2 ,

a,,).

an ) also exists and

... ,

an ).

Proof. First, as sume that d = lcm (al'~ 'q.ú ... ,a,,) exists. Then a¡jd for each value of i, whence ra;jrd. Now, let d' be any common multiple of ra 2 , ra 2 , ... ,ran • Then rld', say d' = rs, w~ere s E R. It follows that a¡ls for every i and so dls. As a result, rdk.if:or rdld'. But this means that lcm (ra l , ra 2 , ... , ran ) exists and equals r{'';' r lcm (al' a2 , ... , an ). As regards the second assertion, suPP
...

,ran)

=

e

=

rt

=

r gcd (al' a2 ,

... ,

an ).

Remark. It is entirely possible for gcd (al' a2 , ... , an ) to exist without the existence of gcd (ra l , ra 2 , ... , ran ); tbis accounts for the lack of symmetry in the statement of the aboye lemma. (See Example 6-4.)

j,l

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96

i

FIRST COURSB IN RINGS AND IDBALS

Although the corning theorern is sornewhat specialized in character, the inforrnation it contains is frequent1y useful.

,

Theorem 6-6. Let al' al• ... , a,. and b l , b2 , ••• , b. be nonzero elernents , of an integral dornain R such that al b 1 = a2 b2 = ... = a.b. = x. 1) Ir !cm (al' a2 , ••• , a.) exists, then gcd (b l , b2 , •• , , b.) also exists and satisfies ' lcm (al' a2 , ••• , a,.) gcd (b l , bl , ... , b.) = x. 2) Ir gcd (rdÚ ra2 , ••• , ra.) exists for all o =1= b.) also ,eiists and satisfies

\';; gcd (al' a2 , ,"',

•••

rE

,a.) lcm (b l , b 2 :

R, then lcm (b l' b2 ,

'" ,

b.)

= gcd (ab 1, ab2 , ... , abn) = a gcd (b 1, bz, ... , bn ) = lcm (al' a2 ,

... ,

... ,

possess a factorization· theory in which the analog of the Fundamental Theorem of Arithmetic holds? To this end, let us introduce two new classes of elements, prime and irreducible elernents; when the ring is specialized to the ring of integers, these concepts are equivalent and yield the usual notion of a prime number. Definition 6-6. 1) A nonzero element p E R is called a prime if and only if p is not invertible and plab implies that either pla or else plb. 2) A nonzero element q E R is said to be irreducible (or nonfaetorizgble) if and only if q is not invertible and every factorization q bc:cw.ith b, e E R, either b or e is invertible. .'

... ,

ro

x.

a.) gcd (b l , b2 ,

i'

:: i

Briefly, an irreducible eleme~t q is an element which cannot be fa~'t~~ed in R in a nontrivial way; the only factors of q are its associates an(í'the inverÍible elements of R. In such rings as division rings and fields;;~,here each nonzero element possesses a rnultiplicative inverse, the concepi:~or:an irreducible element is Mno significance. Observe idso that every element which is an associate oran irreducible (prime) element is itself irreducible (prime). 1t foUows by an easy in'duction argument that ifa product a 1a2 .. : a. is divisible by a prime p, then p must divide at least one ofthe factors a¡ (i = 1, 2, ... , n).

Proo! 'For a":p;oof of statément (1), set a = lcm (al' a2 , ••• , a.). Then ada for i = 1/2,~ ... , n,'say a = ría¡. Frorn the relation x a;b¡, we see tbat ab í = (r¡J:iiJli¡ = riJa and so xlab;. On the other band, consider any divisor y of th.~''izb¡. Then ya¡l(ah¡)a¡, or ya¡lxa, making xa a commOn rnultiple of yal;Y9~, ... , ya•. According to the lemrna, lcm (ya l' ya 2 , '" , ya.) exists and equals ya. Tbus, by tbe definition of least COn;lmon multiple, we conclude tbat yalxa, whence ylx. To recapitulate, we have shown that x/ab¡ for each i and wbenever y/ah¡, then ylx. This simply asserts that 'x

97

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

;:<

b.),

Lemma. In an mtegral domain R, any prime element p is irreducible.

Proo! Suppose that p = ab for SOlDe a, bE R. Since p is prime, either pla or plb; say p divides b, so that there exists some element e in R for which, b =. pe. We then have abe pc = b. It follows from the cancellation law that ae 1; hence, a is invertible. This allows us to conclude that p must be an irreducible'element of R.

where, once again, tbe lernrna has been invoked. We omit the proof of the other half of the tbeorem, wbich follows by mucb the sarne reasoning. In order to apply the lernma, it is now necessary to assurne n .... only that gcd (al' a2 , ••• , a.) exists but, more generally, the existenceof' gcc (raí> ra2 , ... , ra.) for all r f O.

Although prime elements are irreducible in integral domains, the converse is not always true, as we shall see later on: In thecontext of principal ideal domains (our primary interest in this chapter), the no'tions of an irreducible element and a prime element coincide. This is brought out in the theorem below. .

Dur next result is rather striking in that it tells us tbat, at leas't.ror integral domains, tbe gcd-property implies tbe lcm-property, and conversely. Theorem 6-7. An integral dornain R has tbe gcd-property if and only if R has tbe lcrn-property. '

Proo! Let bl , b2 , '" , b.' be nonzero elements of R and suppose that R possesses the lcm-property. Taking x b 1 b2 .. , b. and' ak = bl ... bk - 1 bk+ 1 ... b. for k = 1, 2, ... , n; we may appeal to the first part of Theorem 6-6 to conclude that gcd (b 1 , b2 , ••• ,b.) exists; hence, the gcd-property holds in R. Conversely, if it is hypothesized tbat any finité number of nonzero elements of R adrnit a greatest common divisor, then the existen ce of lcm (Él l , b2 , ... , b,.) can be inferred in the same way.

Theorern 6-8. Let R be a principal ideal domain. A nonzero element PE R is irreducible if and only if it is prime.

Proo! By what we have just proved, p prime always implies p irreducible. So; assurne that p is an irreducible elernent and that p divides the product ab, say pe = ab, with e E R. As R is a principal ideal ring, the ideal generated by p and a, (p, a) = (d)

We now have quite a bit of information about divisibility in integral domains, but the basic question rernains unanswered: when does a ring

for some choice of d in R; hence, p = rd, for suitable rE R. Bllt p is irreducible by hypothesis, so that either r or d must be an ipvertible element.

.1

j,l

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96

i

FIRST COURSB IN RINGS AND IDBALS

Although the corning theorern is sornewhat specialized in character, the inforrnation it contains is frequent1y useful.

,

Theorem 6-6. Let al' al• ... , a,. and b l , b2 , ••• , b. be nonzero elernents , of an integral dornain R such that al b 1 = a2 b2 = ... = a.b. = x. 1) Ir !cm (al' a2 , ••• , a.) exists, then gcd (b l , b2 , •• , , b.) also exists and satisfies ' lcm (al' a2 , ••• , a,.) gcd (b l , bl , ... , b.) = x. 2) Ir gcd (rdÚ ra2 , ••• , ra.) exists for all o =1= b.) also ,eiists and satisfies

\';; gcd (al' a2 , ,"',

•••

rE

,a.) lcm (b l , b 2 :

R, then lcm (b l' b2 ,

'" ,

b.)

= gcd (ab 1, ab2 , ... , abn) = a gcd (b 1, bz, ... , bn ) = lcm (al' a2 ,

... ,

... ,

possess a factorization· theory in which the analog of the Fundamental Theorem of Arithmetic holds? To this end, let us introduce two new classes of elements, prime and irreducible elernents; when the ring is specialized to the ring of integers, these concepts are equivalent and yield the usual notion of a prime number. Definition 6-6. 1) A nonzero element p E R is called a prime if and only if p is not invertible and plab implies that either pla or else plb. 2) A nonzero element q E R is said to be irreducible (or nonfaetorizgble) if and only if q is not invertible and every factorization q bc:cw.ith b, e E R, either b or e is invertible. .'

... ,

ro

x.

a.) gcd (b l , b2 ,

i'

:: i

Briefly, an irreducible eleme~t q is an element which cannot be fa~'t~~ed in R in a nontrivial way; the only factors of q are its associates an(í'the inverÍible elements of R. In such rings as division rings and fields;;~,here each nonzero element possesses a rnultiplicative inverse, the concepi:~or:an irreducible element is Mno significance. Observe idso that every element which is an associate oran irreducible (prime) element is itself irreducible (prime). 1t foUows by an easy in'duction argument that ifa product a 1a2 .. : a. is divisible by a prime p, then p must divide at least one ofthe factors a¡ (i = 1, 2, ... , n).

Proo! 'For a":p;oof of statément (1), set a = lcm (al' a2 , ••• , a.). Then ada for i = 1/2,~ ... , n,'say a = ría¡. Frorn the relation x a;b¡, we see tbat ab í = (r¡J:iiJli¡ = riJa and so xlab;. On the other band, consider any divisor y of th.~''izb¡. Then ya¡l(ah¡)a¡, or ya¡lxa, making xa a commOn rnultiple of yal;Y9~, ... , ya•. According to the lemrna, lcm (ya l' ya 2 , '" , ya.) exists and equals ya. Tbus, by tbe definition of least COn;lmon multiple, we conclude tbat yalxa, whence ylx. To recapitulate, we have shown that x/ab¡ for each i and wbenever y/ah¡, then ylx. This simply asserts that 'x

97

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

;:<

b.),

Lemma. In an mtegral domain R, any prime element p is irreducible.

Proo! Suppose that p = ab for SOlDe a, bE R. Since p is prime, either pla or plb; say p divides b, so that there exists some element e in R for which, b =. pe. We then have abe pc = b. It follows from the cancellation law that ae 1; hence, a is invertible. This allows us to conclude that p must be an irreducible'element of R.

where, once again, tbe lernrna has been invoked. We omit the proof of the other half of the tbeorem, wbich follows by mucb the sarne reasoning. In order to apply the lernma, it is now necessary to assurne n .... only that gcd (al' a2 , ••• , a.) exists but, more generally, the existenceof' gcc (raí> ra2 , ... , ra.) for all r f O.

Although prime elements are irreducible in integral domains, the converse is not always true, as we shall see later on: In thecontext of principal ideal domains (our primary interest in this chapter), the no'tions of an irreducible element and a prime element coincide. This is brought out in the theorem below. .

Dur next result is rather striking in that it tells us tbat, at leas't.ror integral domains, tbe gcd-property implies tbe lcm-property, and conversely. Theorem 6-7. An integral dornain R has tbe gcd-property if and only if R has tbe lcrn-property. '

Proo! Let bl , b2 , '" , b.' be nonzero elements of R and suppose that R possesses the lcm-property. Taking x b 1 b2 .. , b. and' ak = bl ... bk - 1 bk+ 1 ... b. for k = 1, 2, ... , n; we may appeal to the first part of Theorem 6-6 to conclude that gcd (b 1 , b2 , ••• ,b.) exists; hence, the gcd-property holds in R. Conversely, if it is hypothesized tbat any finité number of nonzero elements of R adrnit a greatest common divisor, then the existen ce of lcm (Él l , b2 , ... , b,.) can be inferred in the same way.

Theorern 6-8. Let R be a principal ideal domain. A nonzero element PE R is irreducible if and only if it is prime.

Proo! By what we have just proved, p prime always implies p irreducible. So; assurne that p is an irreducible elernent and that p divides the product ab, say pe = ab, with e E R. As R is a principal ideal ring, the ideal generated by p and a, (p, a) = (d)

We now have quite a bit of information about divisibility in integral domains, but the basic question rernains unanswered: when does a ring

for some choice of d in R; hence, p = rd, for suitable rE R. Bllt p is irreducible by hypothesis, so that either r or d must be an ipvertible element.

.1

98

FIRST COURSE IN RINGS AND IDEALS

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

If d happened to possess an inverse, we would háve (p, a) = R. Thus, there would exist elements s, tER for which 1 = sp + tao Then, b = bl = bsp

+

bta = bsp

+ pct

= p(bs

+

ct),

which implies that plb. On the other hand, if r is invertible in R, then d = r-1p E (p), whence (d) s: (p) .. It follows that the element a E (p) and, in con sequen ce, pla. At any rate, if plab, then p must divide one of the factors, making p a prime ' element of R. We next take up two theorems having to do with the ideal structure of a principal ideal domain; the first result has considerable theoretical importance and will, in particular, serve as our basic tool for tbis section. Theorem 6-9. Let.k be a principal ideal domain. If {In}, n E Z+, is any infinite sequence of ideals of R satisfying

then there exists an integer m such that In = 1m for all n > m. Proa! It is an easy matter to verify that 1 = u In is an ideal of R (see the argument of Theorem 5-2). Being an ideal of a principal ideal ring, 1 = (a) for suitable choice of a E R. Now, the element a must He in one of the ideals of the union, say the ideal 1m. For n > m, it then follows that 1 = (a) s: 1m s: In s: 1; hence, In = 1m , as ass~rted. In asserting the equivalence of maximal and prime ideals in principal ideal domains, Theoreri1h5-9 failed to identify these ideals; this situation is taken care of by our next theorem. First, let us define a principal ideal of the ring R to be a mqfimal principal ideal if it is maximal (with respect to inc1usion) in the set of;proper principal ideals of R. Lemma. Let R be an integral domain. For a nonzero element p E R, the following hold~ a) p is an irreducible element of R if and only if (p) is a maximal principal ideal; b) p is a prime element of R if and only if the principal ideal (P) =F R is prime. Proa! To begin, we suppose that p is an irreducible element of R and that s: R. As p E (a), we must have p = ra for sorne rE R. The fact that p is an irreducible element implies that either r or a is invertible. Were r allowed to possess a multiplicative (a) is any principal ideal for which (p) e (a)

99

inverse, then a = r-1p E (P), from wbich it follows that (a) s: (p), an obvious contradiction. Accordingly, the element a is invertible, whence (a) = R. This argument shows that no principal ideallies between (p) and the whole ring R, so that (p) is a maximal principal ideal. On the other hand, let (p) be a maximal principal ideal of R. For a proof by contradiction, assume that p is not an irreducible elemento Then p admits a factorization p = ab where a, b E R and neither a nor b is invertible (the alternative possibility that p has an inverse implies (p) = R, so may be ruled out). Now, if the element a were in (p), then a = rp ~or sorne choice of r E R; hence, p = ab = (rp)b. Using the cancellation law, we could deduce that 1 = rb; but' this results in the contradiction that b is invertible. Therefore, a ~ (p), yielding the proper inclusion (p) e (a). Next, observe that if (a) = R, then a will possess an inverse, contrary to assumption. We thus conclude that (p) e (a) e R, which denies that (p) is a maximal principal ideal. Our original supposition is false and a must be an irreducible element of R. With regard to the second assertion ofthe lemma, suppose that p is any prime element of R. To see that the principal ideal (p) is in fact a prime ideal, we let the product ab E (p). Then thére exists an element rE R for wbich ab = rp; hence, plab. By hypothesis, p is a prime element, so th~t either pi a or pi b. Translating this into ideals, either a E (p) or b E {p); In consequence, (p) is a prime ideal of R. The converse is proved in much the same way. Let (p) be a prime ideal and plab. Then ab E (p). Using the factthat (p) is a prime ideal, it follows that one of a or b lies in (p). This means that either pla or else plb, and makes 'p a prime element of R. " For principal ide~l domains, aH of this may be summarized by the;;' , following theorem. ,' Theorem 6-10. Let R be a principal ideal domain. The non trivial ideal (p) is a maximal {prime) ideal of R if and only if pis an irreducible (prime), element of R. An immediate consequence of this theorem is that every nonzero noninvertible element of R is divisible by sorne prime. CoroUary. Let a =F O be a noninvertible element of the principal ideal domain R. Then there exists a prime p E R such that pla. Proa! Since a is not invertible, the principal ideal (a) =F R. Thus, by Theorem 5-2, there exists a maximal ideal M of R such that (a) s: M. But the preceding result teHs us that every maximal ideal is ofthe form M = (p), where p is a prime element of R (in this setting, there is no distinction between prime and irreducible elements). Thus, (a) s: (p), wbich is to say that pla. '

98

FIRST COURSE IN RINGS AND IDEALS

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

If d happened to possess an inverse, we would háve (p, a) = R. Thus, there would exist elements s, tER for which 1 = sp + tao Then, b = bl = bsp

+

bta = bsp

+ pct

= p(bs

+

ct),

which implies that plb. On the other hand, if r is invertible in R, then d = r-1p E (p), whence (d) s: (p) .. It follows that the element a E (p) and, in con sequen ce, pla. At any rate, if plab, then p must divide one of the factors, making p a prime ' element of R. We next take up two theorems having to do with the ideal structure of a principal ideal domain; the first result has considerable theoretical importance and will, in particular, serve as our basic tool for tbis section. Theorem 6-9. Let.k be a principal ideal domain. If {In}, n E Z+, is any infinite sequence of ideals of R satisfying

then there exists an integer m such that In = 1m for all n > m. Proa! It is an easy matter to verify that 1 = u In is an ideal of R (see the argument of Theorem 5-2). Being an ideal of a principal ideal ring, 1 = (a) for suitable choice of a E R. Now, the element a must He in one of the ideals of the union, say the ideal 1m. For n > m, it then follows that 1 = (a) s: 1m s: In s: 1; hence, In = 1m , as ass~rted. In asserting the equivalence of maximal and prime ideals in principal ideal domains, Theoreri1h5-9 failed to identify these ideals; this situation is taken care of by our next theorem. First, let us define a principal ideal of the ring R to be a mqfimal principal ideal if it is maximal (with respect to inc1usion) in the set of;proper principal ideals of R. Lemma. Let R be an integral domain. For a nonzero element p E R, the following hold~ a) p is an irreducible element of R if and only if (p) is a maximal principal ideal; b) p is a prime element of R if and only if the principal ideal (P) =F R is prime. Proa! To begin, we suppose that p is an irreducible element of R and that s: R. As p E (a), we must have p = ra for sorne rE R. The fact that p is an irreducible element implies that either r or a is invertible. Were r allowed to possess a multiplicative (a) is any principal ideal for which (p) e (a)

99

inverse, then a = r-1p E (P), from wbich it follows that (a) s: (p), an obvious contradiction. Accordingly, the element a is invertible, whence (a) = R. This argument shows that no principal ideallies between (p) and the whole ring R, so that (p) is a maximal principal ideal. On the other hand, let (p) be a maximal principal ideal of R. For a proof by contradiction, assume that p is not an irreducible elemento Then p admits a factorization p = ab where a, b E R and neither a nor b is invertible (the alternative possibility that p has an inverse implies (p) = R, so may be ruled out). Now, if the element a were in (p), then a = rp ~or sorne choice of r E R; hence, p = ab = (rp)b. Using the cancellation law, we could deduce that 1 = rb; but' this results in the contradiction that b is invertible. Therefore, a ~ (p), yielding the proper inclusion (p) e (a). Next, observe that if (a) = R, then a will possess an inverse, contrary to assumption. We thus conclude that (p) e (a) e R, which denies that (p) is a maximal principal ideal. Our original supposition is false and a must be an irreducible element of R. With regard to the second assertion ofthe lemma, suppose that p is any prime element of R. To see that the principal ideal (p) is in fact a prime ideal, we let the product ab E (p). Then thére exists an element rE R for wbich ab = rp; hence, plab. By hypothesis, p is a prime element, so th~t either pi a or pi b. Translating this into ideals, either a E (p) or b E {p); In consequence, (p) is a prime ideal of R. The converse is proved in much the same way. Let (p) be a prime ideal and plab. Then ab E (p). Using the factthat (p) is a prime ideal, it follows that one of a or b lies in (p). This means that either pla or else plb, and makes 'p a prime element of R. " For principal ide~l domains, aH of this may be summarized by the;;' , following theorem. ,' Theorem 6-10. Let R be a principal ideal domain. The non trivial ideal (p) is a maximal {prime) ideal of R if and only if pis an irreducible (prime), element of R. An immediate consequence of this theorem is that every nonzero noninvertible element of R is divisible by sorne prime. CoroUary. Let a =F O be a noninvertible element of the principal ideal domain R. Then there exists a prime p E R such that pla. Proa! Since a is not invertible, the principal ideal (a) =F R. Thus, by Theorem 5-2, there exists a maximal ideal M of R such that (a) s: M. But the preceding result teHs us that every maximal ideal is ofthe form M = (p), where p is a prime element of R (in this setting, there is no distinction between prime and irreducible elements). Thus, (a) s: (p), wbich is to say that pla. '

100

FIRST COURSE IN RINGS AND IDEALS

Many authors do not insist that an integral domain possess an identity element; for this reason, let us sketch a second proof of the foregoing corollary which avoids the use of Theorem 5-2, First, put (a) I l' If I 1 is not already a maximal ideal, then there exists an ideal 12 of R such that I 1 e 12 , . By tbe same reasoning, if 12 is not maximal, tben 11 e 12 e 13 for sorne ideal 13 , Appealing to Tbeorem 6-9, this process must termínate' after a finite number of steps; in other words, we can eventual1y find a maximal' ideal M oC R containing I 1 = (a), As before, M = (p),witb P a prime In R; tbe remainder ofthe proof is like that abo ve, " ; ; :

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

displayed cbain of ideals eventually terminates; in other words, a" must possess an inverse for sorne n, anQ: (a) e (al) e (a 2) e , .. e (an)

"''.,

·1

"

where p~

Pnan • being an associate of a pri I1l e, is itself prime.

product of a finite number of prime (rna){i~al) ideals. Proo! . Iftbe element O +- a e R is not inverÜ~'Ie, then a bas a representation as a finite product of primes; saya = PlP;',-":~ Pn , wbere each Pi is a prime ff{é';", element of R. It then follows tbat .,"

with (p;) a prime ideal of R.

2) i(a = PlP2 '" Pn = qlq2 ... qm .are two factorizations of a into irreducible elements, tben n = mand tbere is a permutation 1& of tbe indices sucb tbat Pi and q"(i) are associates (i = 1, 2, ... , n).

Specializing to tbe ring of integers, we obtain a celebrated result. TheorellÍ 6-12. (Euclid). Tbere are an ínfinite number oC primes in Z. Proo! Assume that tbe assertion i8 false;' tbat is, suppose tbat there are only a finite number of primes, say PI' P2' ... ,p~. Consider the positive integer.

In sbort, an integral domain is a unique factorization domain if it possesses a factorization theory in wbicb the analog oC the Fundamental Theorem oC Aritbmetic bolds. We intend to sboW tbat any principal ideal domain ls a unique factorization domain; towards tbis goal, let us first prove:

a

(PlP2 ... Pn)'

+

1.

None of tbe listed primes p¡ divides a. If a were divisible by Pi' for instance, we would tben have p¡l(a - PlP2 ... Pm), by Tbeorem 6-1 (S), or p¡J1; but this is impossible by part (2) of the same tbeorem. Since a > 1, Theorem 6-11' asserts that it must llave a prime factor. Accordingly, a isdivisible by a prime whicb is not among those enumerated. Tbis arguments shows tbat tbere is no finite listing of tbe prime numbers.

Tbeorem 6-11. If R is a principal ideal domairi, tben every element oC R wbicb is neitber zero nor invertible bas a factorization into a finite product of primes.

with all-l = p"a" for sorne prime Pn E R. This process goes on as long as all is not an invertible element of R.. But Theorem 6-9 asserts that the

a = PIP2'" Pn-ll!n' ':'

CorolJary. In a principal ideal domain .R.;é·every nontrivial ideal is the

Definition 6-7. Anintegral dornain R is a unique factorizati~i(domain .in case tbe fol1owing two conditions bold: F/.· '." O everyelement a e R, wbicb is neither zero nor invertible, be factored into a finite product of irreducible elements; .

(a) e (al) e (a2 ) e , .. e (a,,) e ... ,

R.

We conclude from tbis that the element a is expressible as the finite product oC primes

If R is an integral domain with tbe property that every noniliVertible element of R can be eJ!:pressed uniquely (up to invertible elementsa:S)actors and the order of factors) as a product of irreducible elements, tbéli'we say tbat R is a unique factorization domain; for a more formal defini~ion:

Proo! Consider any nonzero noninvertible element a e R. By tbe last corollary, tbere exists a prime PI in R witb p1la. Tben a = PI al for sorne (nonzero) ai e R, wbence (a) S (al)' Were (a) = (al), we would have al = raforsuitablereR;itwouldfollowtbata = Pla l = Plra,orl PIr, resulting in tbe contradiction that PI i8 invertible. Consequently, we have the proper inclusion (a) e (al)' Repeat tbe' procedure, now starting witb al> to obtain an increasing cb!!in of principal ideals

101

"

r

!

Having proved tbe existence of a prime factorization in .principal ideal dorna¡ns, one is naturally led to tbe question of uniqueness. Our next theorem is tbe focal point of this cbapter. . Theoiem 6-13. Every principal ideal domain R is a unique factorízation domain. Proo! Tbeorem 6-11 shows that each noninvertible element O +- a E R, bas a prime factorization. To establish uniqueness, let us suppose tbat a can be representeq as a product ofprimes in two ways, say a = PlP2 , .. PII = qlq2 ... qm

(n 5: m)"

100

FIRST COURSE IN RINGS AND IDEALS

Many authors do not insist that an integral domain possess an identity element; for this reason, let us sketch a second proof of the foregoing corollary which avoids the use of Theorem 5-2, First, put (a) I l' If I 1 is not already a maximal ideal, then there exists an ideal 12 of R such that I 1 e 12 , . By tbe same reasoning, if 12 is not maximal, tben 11 e 12 e 13 for sorne ideal 13 , Appealing to Tbeorem 6-9, this process must termínate' after a finite number of steps; in other words, we can eventual1y find a maximal' ideal M oC R containing I 1 = (a), As before, M = (p),witb P a prime In R; tbe remainder ofthe proof is like that abo ve, " ; ; :

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

displayed cbain of ideals eventually terminates; in other words, a" must possess an inverse for sorne n, anQ: (a) e (al) e (a 2) e , .. e (an)

"''.,

·1

"

where p~

Pnan • being an associate of a pri I1l e, is itself prime.

product of a finite number of prime (rna){i~al) ideals. Proo! . Iftbe element O +- a e R is not inverÜ~'Ie, then a bas a representation as a finite product of primes; saya = PlP;',-":~ Pn , wbere each Pi is a prime ff{é';", element of R. It then follows tbat .,"

with (p;) a prime ideal of R.

2) i(a = PlP2 '" Pn = qlq2 ... qm .are two factorizations of a into irreducible elements, tben n = mand tbere is a permutation 1& of tbe indices sucb tbat Pi and q"(i) are associates (i = 1, 2, ... , n).

Specializing to tbe ring of integers, we obtain a celebrated result. TheorellÍ 6-12. (Euclid). Tbere are an ínfinite number oC primes in Z. Proo! Assume that tbe assertion i8 false;' tbat is, suppose tbat there are only a finite number of primes, say PI' P2' ... ,p~. Consider the positive integer.

In sbort, an integral domain is a unique factorization domain if it possesses a factorization theory in wbicb the analog oC the Fundamental Theorem oC Aritbmetic bolds. We intend to sboW tbat any principal ideal domain ls a unique factorization domain; towards tbis goal, let us first prove:

a

(PlP2 ... Pn)'

+

1.

None of tbe listed primes p¡ divides a. If a were divisible by Pi' for instance, we would tben have p¡l(a - PlP2 ... Pm), by Tbeorem 6-1 (S), or p¡J1; but this is impossible by part (2) of the same tbeorem. Since a > 1, Theorem 6-11' asserts that it must llave a prime factor. Accordingly, a isdivisible by a prime whicb is not among those enumerated. Tbis arguments shows tbat tbere is no finite listing of tbe prime numbers.

Tbeorem 6-11. If R is a principal ideal domairi, tben every element oC R wbicb is neitber zero nor invertible bas a factorization into a finite product of primes.

with all-l = p"a" for sorne prime Pn E R. This process goes on as long as all is not an invertible element of R.. But Theorem 6-9 asserts that the

a = PIP2'" Pn-ll!n' ':'

CorolJary. In a principal ideal domain .R.;é·every nontrivial ideal is the

Definition 6-7. Anintegral dornain R is a unique factorizati~i(domain .in case tbe fol1owing two conditions bold: F/.· '." O everyelement a e R, wbicb is neither zero nor invertible, be factored into a finite product of irreducible elements; .

(a) e (al) e (a2 ) e , .. e (a,,) e ... ,

R.

We conclude from tbis that the element a is expressible as the finite product oC primes

If R is an integral domain with tbe property that every noniliVertible element of R can be eJ!:pressed uniquely (up to invertible elementsa:S)actors and the order of factors) as a product of irreducible elements, tbéli'we say tbat R is a unique factorization domain; for a more formal defini~ion:

Proo! Consider any nonzero noninvertible element a e R. By tbe last corollary, tbere exists a prime PI in R witb p1la. Tben a = PI al for sorne (nonzero) ai e R, wbence (a) S (al)' Were (a) = (al), we would have al = raforsuitablereR;itwouldfollowtbata = Pla l = Plra,orl PIr, resulting in tbe contradiction that PI i8 invertible. Consequently, we have the proper inclusion (a) e (al)' Repeat tbe' procedure, now starting witb al> to obtain an increasing cb!!in of principal ideals

101

"

r

!

Having proved tbe existence of a prime factorization in .principal ideal dorna¡ns, one is naturally led to tbe question of uniqueness. Our next theorem is tbe focal point of this cbapter. . Theoiem 6-13. Every principal ideal domain R is a unique factorízation domain. Proo! Tbeorem 6-11 shows that each noninvertible element O +- a E R, bas a prime factorization. To establish uniqueness, let us suppose tbat a can be representeq as a product ofprimes in two ways, say a = PlP2 , .. PII = qlq2 ... qm

(n 5: m)"

102

FI.RST COURSE IN RINGS ANO IDEALS

whe~e. the p¡and q¡ are aH primes. Since Pll(qlq2 '" qm), it foHows that PI dIVIdes sorne q¡ (1 :::;; i :::;; m); renumbering, if necessary, we may suppose that Pllql' Now, P~ and qf are both prime elements of R, with pllql' so they m~st be assoClates: ql = PIU 1 for sorne invertible element U E R. l Cancelhng the cornmon factor PI' we are left with

Although property'(2) seems unsyrnmetric, R is a commutative ring; hence, (2) al so asserts that r5(ab) ;;::: r5(b) as weH as r5(ab) ;;::: r5(a). As simple examples of Euc1idean domains, we may take 1) any p.eld F, with valuation defined by r5(a) = 1 for a11 nonzero a E F; 2) the ring Z, with valuation defined by r5(a) = lal" for a11 nonzero a E Z (fixed n E Z +) ; 3) ,the Gaussian integers Z(i), with valuation defined by r5(a + bi)= a2 + b2

P2"'Pn = ul q2"'qm' ContinuiÍlg this argument, we arrive (after n steps) at

1=

U I U2

103

OIVISIBILITY THEORY IN INTEGRAL OOMAINS

for all nonzero a

'" unq,,+'l '" qm'

+

bi

E

Z(i) (see Theorem 6-17).

Since the q¡ are not invertible, tbis forces m = n. It has also been shown th~t every p! has sorne qj as an associate and conversely. Thus, the two pnme factonzations are identical, apart from the order in which the factors appear and from replac~ment of factors by associates.

Several rudimentary properties of Euclidean domains appear in the lernma below.

Attention is called to' the fact that the converse of Theorem 6-13 is not true; ~n the next chapter, we shall give an exarnple of a unique factorization domalll which is not a principal ideal domain. A useful fact to bear in mind is that in a unique factorization domain ~ ~y irreducible element P E R is necessarily prime. For, suppose that P dIvIdes the prodúct ab, say pe = abo Let

-2) iftwo ÍlOnzero elements a, bE R are associates, then r5(a) = r5(b);

Lernma. Let R be a Euc1idean domain with vaJu,ation r5. Then, 1) for each nonzero a E R, r5(a) ;;::: r5(1); 3) an element O =1= a E R is invertible if and only if r5(a) = r5(1). Proo! Assertion (1) follows from the fact that a = al, whence r5(a) = r5(a1) ;;::: r5(1).

If a and b are associates, then a = bu, with u an invertible element of R ; then, also, b = au - 1. This means that r5(a) = r5(bu) ;;::: r5(b), r5(b) = r5(au- l ) ;;::: r5(a),

a = PI ... Pn , b = ql ... qm' and e = tI ... ts be the unique factorizations of a, b, and e into irreducible factors. We then have

which says that r5(a) = r5(b). To prove (3), suppose that a =1= O has an inverse in R, so that ab = 1 for sorne choice 'of bE R. Then, using (1),

#i.e

Since the factorization of ah into irreducibles is unique~ element P must be an associate of one of the p¡ or q¡, and, consequeti'tly, P divides either

am~

j ".,

r5(a) :::;; r5(ab) = r5(1) :::;; r5(a),

~t

or r5(a) = 1. Conversely, suppose. that the element O =1= a E R is such that r5(a) = 1. Applying the division álgorithm to 1 and a, there exist q, rE R for which 1 = qa + r, where r = O or r5(r) < r5(a). Th~ latter alternative implies that r5(r) < 1, wbich is impossible, so that r = Oi'in other words, 1 = qa and a is invertible in R.

Another interesting c1ass of integral domains, whichiUe propose to look at now, is provided by the so-called Euc1idean domains; these arose out of attempts to generalize the familiar Division AIgorithm feir ordinary integers to arbitrary rings. The precise definition follows. . l.!~ Definition 6-8. An integral domain R is said to be Euclidean if there exists a function r5 (the Euclidean valuation) such' that the following conditions are satisfied: .

Theorem 6-14. The quotient and remainder in condition (3) of the definition of a Euc1idean domain are unique if and only if

1) r5 (a) is a nonnegative integer for every O =1= a E R;

2) for any a, b E R, both nonzero, r5(ab) ;;::: r5(a);

r5(a

3) for any a, bE R, with b =1= O, there exist elements q, rE R (the quotient and remainder) such that a = qb + r, where either r = O or el se r5(r) < r5(b). '

+

b) :::;; max {r5(a), r5(b)}.

Proo! Suppose that there exist nonzero a, bE R such that r5(a max {r5(a), r5(b)}; then,

I

b

=

O(a

+

b)

+

b

=

l(a

+

b) - a,

+

b) >

102

FI.RST COURSE IN RINGS ANO IDEALS

whe~e. the p¡and q¡ are aH primes. Since Pll(qlq2 '" qm), it foHows that PI dIVIdes sorne q¡ (1 :::;; i :::;; m); renumbering, if necessary, we may suppose that Pllql' Now, P~ and qf are both prime elements of R, with pllql' so they m~st be assoClates: ql = PIU 1 for sorne invertible element U E R. l Cancelhng the cornmon factor PI' we are left with

Although property'(2) seems unsyrnmetric, R is a commutative ring; hence, (2) al so asserts that r5(ab) ;;::: r5(b) as weH as r5(ab) ;;::: r5(a). As simple examples of Euc1idean domains, we may take 1) any p.eld F, with valuation defined by r5(a) = 1 for a11 nonzero a E F; 2) the ring Z, with valuation defined by r5(a) = lal" for a11 nonzero a E Z (fixed n E Z +) ; 3) ,the Gaussian integers Z(i), with valuation defined by r5(a + bi)= a2 + b2

P2"'Pn = ul q2"'qm' ContinuiÍlg this argument, we arrive (after n steps) at

1=

U I U2

103

OIVISIBILITY THEORY IN INTEGRAL OOMAINS

for all nonzero a

'" unq,,+'l '" qm'

+

bi

E

Z(i) (see Theorem 6-17).

Since the q¡ are not invertible, tbis forces m = n. It has also been shown th~t every p! has sorne qj as an associate and conversely. Thus, the two pnme factonzations are identical, apart from the order in which the factors appear and from replac~ment of factors by associates.

Several rudimentary properties of Euclidean domains appear in the lernma below.

Attention is called to' the fact that the converse of Theorem 6-13 is not true; ~n the next chapter, we shall give an exarnple of a unique factorization domalll which is not a principal ideal domain. A useful fact to bear in mind is that in a unique factorization domain ~ ~y irreducible element P E R is necessarily prime. For, suppose that P dIvIdes the prodúct ab, say pe = abo Let

-2) iftwo ÍlOnzero elements a, bE R are associates, then r5(a) = r5(b);

Lernma. Let R be a Euc1idean domain with vaJu,ation r5. Then, 1) for each nonzero a E R, r5(a) ;;::: r5(1); 3) an element O =1= a E R is invertible if and only if r5(a) = r5(1). Proo! Assertion (1) follows from the fact that a = al, whence r5(a) = r5(a1) ;;::: r5(1).

If a and b are associates, then a = bu, with u an invertible element of R ; then, also, b = au - 1. This means that r5(a) = r5(bu) ;;::: r5(b), r5(b) = r5(au- l ) ;;::: r5(a),

a = PI ... Pn , b = ql ... qm' and e = tI ... ts be the unique factorizations of a, b, and e into irreducible factors. We then have

which says that r5(a) = r5(b). To prove (3), suppose that a =1= O has an inverse in R, so that ab = 1 for sorne choice 'of bE R. Then, using (1),

#i.e

Since the factorization of ah into irreducibles is unique~ element P must be an associate of one of the p¡ or q¡, and, consequeti'tly, P divides either

am~

j ".,

r5(a) :::;; r5(ab) = r5(1) :::;; r5(a),

~t

or r5(a) = 1. Conversely, suppose. that the element O =1= a E R is such that r5(a) = 1. Applying the division álgorithm to 1 and a, there exist q, rE R for which 1 = qa + r, where r = O or r5(r) < r5(a). Th~ latter alternative implies that r5(r) < 1, wbich is impossible, so that r = Oi'in other words, 1 = qa and a is invertible in R.

Another interesting c1ass of integral domains, whichiUe propose to look at now, is provided by the so-called Euc1idean domains; these arose out of attempts to generalize the familiar Division AIgorithm feir ordinary integers to arbitrary rings. The precise definition follows. . l.!~ Definition 6-8. An integral domain R is said to be Euclidean if there exists a function r5 (the Euclidean valuation) such' that the following conditions are satisfied: .

Theorem 6-14. The quotient and remainder in condition (3) of the definition of a Euc1idean domain are unique if and only if

1) r5 (a) is a nonnegative integer for every O =1= a E R;

2) for any a, b E R, both nonzero, r5(ab) ;;::: r5(a);

r5(a

3) for any a, bE R, with b =1= O, there exist elements q, rE R (the quotient and remainder) such that a = qb + r, where either r = O or el se r5(r) < r5(b). '

+

b) :::;; max {r5(a), r5(b)}.

Proo! Suppose that there exist nonzero a, bE R such that r5(a max {r5(a), r5(b)}; then,

I

b

=

O(a

+

b)

+

b

=

l(a

+

b) - a,

+

b) >

104

FIRST COURSE IN RINGS AND IDEAL S

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

with both ó( -a) = ó(a) < ó(a + b) and ó(b) < ó(a + b). This exhibits the lack of uniqueness of quotient and remainder in condition (3). Conversely, assume that the indicated inequality holds and that the element a E R has two representations; a

=

a

=

with r =1: ;.' and q

qb + r q'b + r'

=1=. q'.

(r (T'

= =

It would seem inappropriate to concIude this chapter without sorne mention of the quadratic number fields; the elements of these domains form the sets Q(Jñ) = {a + bJñla, b E Q},

O or ó(r) < ó(b) ), O or ó(r') < ó(b))

with n =1= 1 a square-free integer (that is, an integer not divisible by the square of any positive integer > 1). When n < O, we may view Q(.Jñ) as a subdomairt ofthe complex number system e and represent its elements in the standar4form a + b.Jñ i. It is not difficult to show that if nI' n 2 are square-freeintegers, then Q(.Jn l ) = Q(Jn 2 ) if and only if nI = n2 • Each ele~ent ex ~ a + bJñ E Q(Jñ) gives rise to, another element .a = a- bjJn of Q(.Jn), which we shall call the conjugate of ex (for n < O, ais the usual: complex conjugate of ex). A simple argument establishes that the mappi~g;j: Q("¡-n) 4 Q(Jñ) defined by f(ex) = a is anisomorphism. To studydivisibility properties of Q(Jñ), it is convenient to make use of the concept of the norm of an element (an analog of the absolute value notion inZY;:. . '.'

Then we:have

ó(b) ~ ó((q - q')b) = ó(r·t. r') < max {ó(r), ó( -r')} < ó(b).

This is only possible if one of r Tr' or q - q' is zero. Since each of these conditions implies the other, uniq'iieness follows .. .,:' CoroJlary. (Division'AlgoritJii¡ffor Z). If a, b E Z, with b exist unique integers q and ~}~uch that

a = qb¡;;; Proof Utilize the valuation ó

=1=

O, then there

O~ r <

la¡. la¡, for all nonzero a E Z.

gi~:~ by ó(a) =

Definition 6-9. For each element ex = a + b.Jñ in Q(.Jñ), the norm N(ex) of ex is simply the product of ex and its conjugate a:

Unique factorization in Z folJows ultimately from the Division Algorithm. It is not surprising that in rings where there is an analog of division with remainder, we can also prove uniqueness of factórization. The rnain line of argument consists of showing that every EucIidean domain is a principal ideal domain. (One need only consider the ring Ze to see that the converse of tbis does not hold.)

N(ex) = exa = (a

Proof Let R be a EucIidean dornain with valuation Ó and 1 be an ideal of 1 =1= {O}, Consider the set Sdefined by =1=

Corollary. Every EucIidean domáin is a unique factorization domain.

= a2

-

b 2 n.

Proof Given ex = a + b.Jñ in Q(Jñ), N(ex) ~ a 2 - b 2 n = O if and only if both a = b = O (that is, ex = O); otherwise, we would contradict the choice of n as a square-free integer. Since the mapping f(ex) = IX is an isomorphism, N is a multiplicative function in the sense that N(ex{J) = exf3exf3 = exf3IXlJ = exaf3p = N(ex)N({J)

O},

Since S ii; a nonempty subset of nonnegative integers, it has a least element by the Well-Ordering Principal. Pick bE 1, so that ó(b) is mini mal in S. Our contention is that 1 = (b). Let a be an arbitrary element of l. By the definition of EucIidean domain, there exist elements q, rE R for which a = qb + r, where either r =/ O or ó(r) < ó(b). Now, r = a - qb El, since 1 is an ideal containing both a and b. The alternative ó(r) < ó(b) would therefore contradict the minimality of ó(b). Consequently, we must have r = 0, and a = qb E (b); this implies that 1 ~ (b). The reverse incIusion cIearIy holds, since bE 1, thereby completing the proof.

b.Jñ) (a - bJñ)

Lemma. For all ex, f3 E Q(Jñ), the following hold: 1) N(ex) = Oif and only if ex = O; 2) N(ex{J) = N(ex)N({J); 3) N(l) = 1;

R; ignoring trivial cases, we may suppose that

{ó(a)laE1; a

+

Sorne properties of the norm function which follow easily from the definition are listed below.

Theo.rem 6-15. Every EucIidean domain is a principal ideal domain.

s=

105

for all ex, f3 E Q(.Jñ). . " The proof of assertion (3) follows from the fact that N(l) .~

= N(1 2 ) = N(l)N(l)

=

N(W,

whence N(l) = L AJthough Q(.jn) has been labeled as a field, we actua1Jy have not proved this to be the case; it is high time to remedy this situation.

104

FIRST COURSE IN RINGS AND IDEAL S

DIVISIBILITY THEORY IN INTEGRAL DOMAINS

with both ó( -a) = ó(a) < ó(a + b) and ó(b) < ó(a + b). This exhibits the lack of uniqueness of quotient and remainder in condition (3). Conversely, assume that the indicated inequality holds and that the element a E R has two representations; a

=

a

=

with r =1: ;.' and q

qb + r q'b + r'

=1=. q'.

(r (T'

= =

It would seem inappropriate to concIude this chapter without sorne mention of the quadratic number fields; the elements of these domains form the sets Q(Jñ) = {a + bJñla, b E Q},

O or ó(r) < ó(b) ), O or ó(r') < ó(b))

with n =1= 1 a square-free integer (that is, an integer not divisible by the square of any positive integer > 1). When n < O, we may view Q(.Jñ) as a subdomairt ofthe complex number system e and represent its elements in the standar4form a + b.Jñ i. It is not difficult to show that if nI' n 2 are square-freeintegers, then Q(.Jn l ) = Q(Jn 2 ) if and only if nI = n2 • Each ele~ent ex ~ a + bJñ E Q(Jñ) gives rise to, another element .a = a- bjJn of Q(.Jn), which we shall call the conjugate of ex (for n < O, ais the usual: complex conjugate of ex). A simple argument establishes that the mappi~g;j: Q("¡-n) 4 Q(Jñ) defined by f(ex) = a is anisomorphism. To studydivisibility properties of Q(Jñ), it is convenient to make use of the concept of the norm of an element (an analog of the absolute value notion inZY;:. . '.'

Then we:have

ó(b) ~ ó((q - q')b) = ó(r·t. r') < max {ó(r), ó( -r')} < ó(b).

This is only possible if one of r Tr' or q - q' is zero. Since each of these conditions implies the other, uniq'iieness follows .. .,:' CoroJlary. (Division'AlgoritJii¡ffor Z). If a, b E Z, with b exist unique integers q and ~}~uch that

a = qb¡;;; Proof Utilize the valuation ó

=1=

O, then there

O~ r <

la¡. la¡, for all nonzero a E Z.

gi~:~ by ó(a) =

Definition 6-9. For each element ex = a + b.Jñ in Q(.Jñ), the norm N(ex) of ex is simply the product of ex and its conjugate a:

Unique factorization in Z folJows ultimately from the Division Algorithm. It is not surprising that in rings where there is an analog of division with remainder, we can also prove uniqueness of factórization. The rnain line of argument consists of showing that every EucIidean domain is a principal ideal domain. (One need only consider the ring Ze to see that the converse of tbis does not hold.)

N(ex) = exa = (a

Proof Let R be a EucIidean dornain with valuation Ó and 1 be an ideal of 1 =1= {O}, Consider the set Sdefined by =1=

Corollary. Every EucIidean domáin is a unique factorization domain.

= a2

-

b 2 n.

Proof Given ex = a + b.Jñ in Q(Jñ), N(ex) ~ a 2 - b 2 n = O if and only if both a = b = O (that is, ex = O); otherwise, we would contradict the choice of n as a square-free integer. Since the mapping f(ex) = IX is an isomorphism, N is a multiplicative function in the sense that N(ex{J) = exf3exf3 = exf3IXlJ = exaf3p = N(ex)N({J)

O},

Since S ii; a nonempty subset of nonnegative integers, it has a least element by the Well-Ordering Principal. Pick bE 1, so that ó(b) is mini mal in S. Our contention is that 1 = (b). Let a be an arbitrary element of l. By the definition of EucIidean domain, there exist elements q, rE R for which a = qb + r, where either r =/ O or ó(r) < ó(b). Now, r = a - qb El, since 1 is an ideal containing both a and b. The alternative ó(r) < ó(b) would therefore contradict the minimality of ó(b). Consequently, we must have r = 0, and a = qb E (b); this implies that 1 ~ (b). The reverse incIusion cIearIy holds, since bE 1, thereby completing the proof.

b.Jñ) (a - bJñ)

Lemma. For all ex, f3 E Q(Jñ), the following hold: 1) N(ex) = Oif and only if ex = O; 2) N(ex{J) = N(ex)N({J); 3) N(l) = 1;

R; ignoring trivial cases, we may suppose that

{ó(a)laE1; a

+

Sorne properties of the norm function which follow easily from the definition are listed below.

Theo.rem 6-15. Every EucIidean domain is a principal ideal domain.

s=

105

for all ex, f3 E Q(.Jñ). . " The proof of assertion (3) follows from the fact that N(l) .~

= N(1 2 ) = N(l)N(l)

=

N(W,

whence N(l) = L AJthough Q(.jn) has been labeled as a field, we actua1Jy have not proved this to be the case; it is high time to remedy this situation.

106

equation is possible only if a = ± 1 and b = O, or a O and b = Rence, the only choice for invertible elernents in Z(i) are ±1 and ± i.

Theorem 6-16. For each square-free integer n, the system Q(.jñ) forms , a field; in fact, Q(.jñ) is a subfield of C.

Proo! The reader may easíly verify that Q(.jñ) is a cornmutative ring with identity. It remains only to establish that each nonzero element of Q(Jñ) has a multiplicative inverse in Q(Jñ). Now, if O =1= a E Q(Jñ), then the element fJ = a/N(a) evidently líes in Q(.jñ); furthermore, the product afJ

= a (fi./N(a) ) = N(a)/N(a) =

I

I

1,

Contained in each quadratic field Q(.jñ) is the integral domain

Theorem 6-17. Each ofthe domains Z(Jñ), where n Euclidean; hence, is a unique factorization dornain.

{a + b.fñla, b E Z}.

whenever IX, fJ =1= O are in Z(Jñ). Since fJ =1= O, the product ap-l E Q(Jñ) and so rnay be written in the form ap-l = a + b.jñ, with a, b E Q. Select integers x and y (the nearest integers to a and b) such that

Proo! As regards (1), observe that if N(a) = ±1, then aa = ±1; thus alI, which is to say that a is in vertible. To prove the converse, let a be an' in vertible ele~ent of Z(~ñ), so that afJ = 1 for sorne fJ in Z(~ñ). Then, N(afJ) = N(l)

=

la Now, set

1.

O'

= x

+ y~ñ.

xl

S;

Then

jb -

1/2, O' E

yl

S;

1/2

Z(~ñ) and norm formula (valilalso in

Q(Jñ») shows that

Since N(a) an,d'N(fJ) are both integers, this implies that N(a) =' ± 1. Next, suppóse that a has the property that N(a) = ±p, where p is a prime num~~r, As N(a) =1= O, 1, the element a is ileither Onor invertible in Z(~ñ). If a '? fJy is a factorization of a in Z(Jñ), then N(fJ)N(y) = N(a) =

-1, -2,2,3, is

c5(afJ) = c5(a)c5(fJ) ~ c5(a)'l = c5(a)

Lemma. For any IX E Z(.jñ), the following hold: 1) N(a) = ±1 if and only if IX is invertible in Z(.jñ); 2) ir N(a) = ±p, where p is a prime number, then a is an irreducible element of Z(.jñ).

=

=

Proo! The strategy ernployed in the proof is to show that the function c5 defined on Z(.fñ) by c5(a) = IN(a)1 is a Euclidean valuation for n = 1, 2, 3. We c1early ha ve c5(a) = Oif and only ir a = O, so tbat c5(a) ;;::: 1 ror all a =1= O. Since both the norm and its absolute value are multiplicative, condition (2) of Definition 6-8 is always satisfied:

Since Z(Jñ) is closed under conjugation, the norm function enables us to get a: clear view of the sets of invertible and irreducible elements in these domains. The multiplicative property of the norm, for instan ce, transfers any factorization a = fJ'Y of an element a E Z(.jñ) into a factorization N(a) = N(fJ)N(y) of the integer N(a). This is particularly helpful in proving

N(a)N(fJ)

± 1.

Perhaps the most obvious approach to the question of unique factorization in the quadratic dornains zCJñ) is to try to show that they are Euclidean domains (a natural candidate for the Euclidean valuation is c5(a) = IN~). In the coming theorern, we shall do precisely this for the dornains Z(~ -1), Z(F'2), Z(J'i), and Z(J3). Although there are other Euclidean quadratic domains, our attention is restricted to these few for which the division algorithrn is easi1y established.

so that fJ serves as the inverse of a. Z(.fñ) =

1'07

DIVISIBILITY THEOR y IN INTEGRAL DOMAINS

FIRST COURSE IN RINGS AND IDEALS

IN(ap-l -

0')1

= =

jN(a - 2X ) + (b ,- y1Jñ) I (a - x) - n(b- y)!I·

...

But the manner in wbich x and y were chosen irnply that

±p,

-n/4 S; (a - X)2 - n(b - y)2 S; 1/4, if n > O, O S; (a - X)2 - n(b - y? S; 1/4 + (-n)1/4, ifn < O.

from which it follows that' one of N(fJ) or N(y) must have the value ± 1. From the first part of the lemma, we may thus conclude that either fJ or y is invertible in Z(Jñ), whlle the other is an associate of IX. Accordingly, a is an irreducible element of Z(Jñ).

In terrns of the function c5 tbis means c5(ap-l - 0')

=

I(a

x)2 - n(b - y)21 < 1

for n = -1, - 2, 2, 3. Putting p = fJ(ap-l 0'), we ha ve a = O'fJ + p. Since a and O'fJ are in Z(~ñ), this equation irnplies that p is in Z(Jñ) also. Moreover, for the previously indicated values of n,

Example 6-3. Let us find all invertible elements in Z(Z) = Z(R), the domain of Gausian integers, by finding those members a of Z(i) for which N(a) = 1 (in this setting, the norm assumes nonnegative values). If a = a + bi E Z(i) and N(a) = 1, then a2 + b2 = 1, with a, bE Z. This

c5(P)

)

,

.1

c5(fJ(ap-l - q)) = c5(fJ)c5(ap-l - 0') < c5(fJ).

106

equation is possible only if a = ± 1 and b = O, or a O and b = Rence, the only choice for invertible elernents in Z(i) are ±1 and ± i.

Theorem 6-16. For each square-free integer n, the system Q(.jñ) forms , a field; in fact, Q(.jñ) is a subfield of C.

Proo! The reader may easíly verify that Q(.jñ) is a cornmutative ring with identity. It remains only to establish that each nonzero element of Q(Jñ) has a multiplicative inverse in Q(Jñ). Now, if O =1= a E Q(Jñ), then the element fJ = a/N(a) evidently líes in Q(.jñ); furthermore, the product afJ

= a (fi./N(a) ) = N(a)/N(a) =

I

I

1,

Contained in each quadratic field Q(.jñ) is the integral domain

Theorem 6-17. Each ofthe domains Z(Jñ), where n Euclidean; hence, is a unique factorization dornain.

{a + b.fñla, b E Z}.

whenever IX, fJ =1= O are in Z(Jñ). Since fJ =1= O, the product ap-l E Q(Jñ) and so rnay be written in the form ap-l = a + b.jñ, with a, b E Q. Select integers x and y (the nearest integers to a and b) such that

Proo! As regards (1), observe that if N(a) = ±1, then aa = ±1; thus alI, which is to say that a is in vertible. To prove the converse, let a be an' in vertible ele~ent of Z(~ñ), so that afJ = 1 for sorne fJ in Z(~ñ). Then, N(afJ) = N(l)

=

la Now, set

1.

O'

= x

+ y~ñ.

xl

S;

Then

jb -

1/2, O' E

yl

S;

1/2

Z(~ñ) and norm formula (valilalso in

Q(Jñ») shows that

Since N(a) an,d'N(fJ) are both integers, this implies that N(a) =' ± 1. Next, suppóse that a has the property that N(a) = ±p, where p is a prime num~~r, As N(a) =1= O, 1, the element a is ileither Onor invertible in Z(~ñ). If a '? fJy is a factorization of a in Z(Jñ), then N(fJ)N(y) = N(a) =

-1, -2,2,3, is

c5(afJ) = c5(a)c5(fJ) ~ c5(a)'l = c5(a)

Lemma. For any IX E Z(.jñ), the following hold: 1) N(a) = ±1 if and only if IX is invertible in Z(.jñ); 2) ir N(a) = ±p, where p is a prime number, then a is an irreducible element of Z(.jñ).

=

=

Proo! The strategy ernployed in the proof is to show that the function c5 defined on Z(.fñ) by c5(a) = IN(a)1 is a Euclidean valuation for n = 1, 2, 3. We c1early ha ve c5(a) = Oif and only ir a = O, so tbat c5(a) ;;::: 1 ror all a =1= O. Since both the norm and its absolute value are multiplicative, condition (2) of Definition 6-8 is always satisfied:

Since Z(Jñ) is closed under conjugation, the norm function enables us to get a: clear view of the sets of invertible and irreducible elements in these domains. The multiplicative property of the norm, for instan ce, transfers any factorization a = fJ'Y of an element a E Z(.jñ) into a factorization N(a) = N(fJ)N(y) of the integer N(a). This is particularly helpful in proving

N(a)N(fJ)

± 1.

Perhaps the most obvious approach to the question of unique factorization in the quadratic dornains zCJñ) is to try to show that they are Euclidean domains (a natural candidate for the Euclidean valuation is c5(a) = IN~). In the coming theorern, we shall do precisely this for the dornains Z(~ -1), Z(F'2), Z(J'i), and Z(J3). Although there are other Euclidean quadratic domains, our attention is restricted to these few for which the division algorithrn is easi1y established.

so that fJ serves as the inverse of a. Z(.fñ) =

1'07

DIVISIBILITY THEOR y IN INTEGRAL DOMAINS

FIRST COURSE IN RINGS AND IDEALS

IN(ap-l -

0')1

= =

jN(a - 2X ) + (b ,- y1Jñ) I (a - x) - n(b- y)!I·

...

But the manner in wbich x and y were chosen irnply that

±p,

-n/4 S; (a - X)2 - n(b - y)2 S; 1/4, if n > O, O S; (a - X)2 - n(b - y? S; 1/4 + (-n)1/4, ifn < O.

from which it follows that' one of N(fJ) or N(y) must have the value ± 1. From the first part of the lemma, we may thus conclude that either fJ or y is invertible in Z(Jñ), whlle the other is an associate of IX. Accordingly, a is an irreducible element of Z(Jñ).

In terrns of the function c5 tbis means c5(ap-l - 0')

=

I(a

x)2 - n(b - y)21 < 1

for n = -1, - 2, 2, 3. Putting p = fJ(ap-l 0'), we ha ve a = O'fJ + p. Since a and O'fJ are in Z(~ñ), this equation irnplies that p is in Z(Jñ) also. Moreover, for the previously indicated values of n,

Example 6-3. Let us find all invertible elements in Z(Z) = Z(R), the domain of Gausian integers, by finding those members a of Z(i) for which N(a) = 1 (in this setting, the norm assumes nonnegative values). If a = a + bi E Z(i) and N(a) = 1, then a2 + b2 = 1, with a, bE Z. This

c5(P)

)

,

.1

c5(fJ(ap-l - q)) = c5(fJ)c5(ap-l - 0') < c5(fJ).

l 108

FIRST COURSE IN RINGS AND lDEALS

PROBLEMS

Definition 6-8 is therefore satisfied in its entirety when n = the corresponding quadratic domains Z(.J'ñ) are Euc1idean.

t,' -

an

of irreducible " de . . . elem¡;nt and of a nonzero .prime do not al ways COInCl In an arbItrary Integral domain .. Specifically, we have (2' + .J -5)1 3 ' 3, but (2 + F)t 3, so that 2 + F5 cannot be a prime element of Z(~.

2, 2, 3 and

Corollary. The domain Z(i) óf Gaussian Íntegers is a unique factorization domain.

PROBLEMS

Remark. By investigating further the divisibility properties of Z(i), one ca!1 prove the c1assic "two squares theorem" of Fermat: every prime number ofthe form 4n + 1 is the sum oftwo squares; the interested reader is advis~d ;.. ,: to consult [13J for the details. .

1. Let R be a commutative ring with identity and the ele~~nts a e E R with Z = Prove that ,:, ' , e e. a) If (a) = (e), then.a arid e are ~ssociates. [Hint: a J, (I - e + a)e.] b) If for. sorner n EllZ+ the elements a" and e are ass6ciates . , then am an d e are aSSOCIates lor a m;?; n. .t;r

Example 6-'4. The various integral domains studied in this chapter might suggest that unique factorization of elements always holds. To round o~t the picture with an example of the failure of unique factorization, let:us consider the quadratic domain Z(.'¡=5); Observe that the element 9 'has two factorizations in Z(F"S), namely,' . . 9

= 3·3 = (2 + .J -5)(2 -:-

2.

R).

c) 1

3,

a2

= 2 + .J - 5,

a3

reducible, we could write a¡ = Pr, where neither ofthe elements yE Z(rs)is invertible. Taking norms, it follows that 9

=

N(a¡}

=

N(fJ)N(y),

N(fJ), N(y)

E

.

p,

Z+,

brs;

which in turn yields N(fJ) = N(y) = 3. Bence, if p = a + we find that we must solve the equation a2 + 5b 2 = 3 for integers and b; but this equation obviously has no solutions in Z (b i= impliesthat a2 + 5b 2 ;;::: 5, and if b = 0, then a2 = 3). Tp.us, we have exhibited two genuinely different factorizations of the e1ement 9 into irreducibles, so that unique factorization does not hold in Z(R). . Notice f~rther that the common. divisors of 9 and 3(2 + F) are . 1,3, and 2. + None ofthese latter elements is divisible by the others, so that gcd (9, 3(2 + .J - 5)) fails '10 exist (in particular, Z(R) does not have the gcd-property). On the other hand, the greatest common divisor does exist and, in fact, gcd (3, 2 + -J=s) = 1. It follows of 3 and 2 + that only the right-hand side of the formula

a

°

3 gcd (3, 2

+

.. ;

= R

for all nonzero' 1; J ..

4. Let R be ~n .integral. domain having the gcd-property. Assuming that equality holds to wlthln assoclates, prove that, for nonzero a, b, e' E R, a) gcd (a, gcd (b, e)) = gcd (gcd (a, b), e) .. b) gcd (a, 1) =:: 1. c) gcd (ea, eb) = e gcd (a, b); in particular, gcd (e, eb) = 1. d) ~f gcd (a, b) ;= 1 and gcd (a, e) = 1, then gcd (a, be) = 1. e) If gcd (a, b) = 1, ale and b/e, then ab/e. ' f) gcd (a, b) 1cm (a, b) =;' abo [Hint: Theorem 6"':6.J

6. !n a princi~al ideal dornain R, establish that the primary ideals are the two trivial Ideals and Ideals of the forrn (p"), where p is a prime element of R d Z . h n an n E +. [H'Int : If JIS'pnrnary, t en -VI = (ji) for sorne prime elernent p. Choose 11 E Z such thatI ~ '(pO), but 1 $ (p"+l), and show that 1 = (p").] . +

7. If R is a pr~nci?a.l idea! doman, we define the Length A(a) for each nonzero a E R a IS. Invertlble, . . (as follows: If. , . then A(a) = O·" otherwise A(a) is the numb er of pnmes not necessanly dls.tlnct) 1ll any factorization of a. Prove the following assertions' a) the length of a IS well-defined' ' . b) if alb, then A(a) 5: A(b); , . c) ~fa/b al1d,A(a) = A(b), then bla; '. d) If a { b and b { a, then there exist nonzero p, q E R such that

.¡=s

+ .J-:::S)) ".;

.1

5. If R i.s an.inte~ral domain having ~he gcd~property, show that a nonzero element of R IS pnme If and only if it is irreducible.

R.

gcd (3'3, 3(2

.'

+

+

K) = (1 J) n (l K).:" . K) = (l n J) + (l n K).

3. Sup?ose, that ~ = Rl EB Rz EB ... EB R", where each R¡ is a principal ideal ringo Venfy tnat R IS also a principal ideal ringo .

F5

= 2-

+ (J n

d) 1 n (J t e) lJ = 1 n J if and only if 1 + J

F

=

Give~ that 1, J, and K are ideals of a principal ideal doá1in R, derive the fOllowing relatIons: .' .!, a) If 1 = (a) and J = (b), then jJ-= (ab); in particuí~r1" = (a").

b) l(J n K) = IJ n lK.

ClearIy, no two ofthe factors 3, 2 + ~, and 2 are assodates of eachother (the only invertible elements in Z(.J - 5) are ± 1), and it can easily be shown that they are all irreducible. Indeed, were any one of al

109

A)

is defined in Z(.J - 5), thereby illustrating the remark 011 page 95. This example has the additional feature of showing that the concepts

A(pa

+

qb) 5: rnin {A(a), A(b)}.

I !

..

__. _ - - - - - - - - - - - - - - -

------------------_._----~-.- ~---_.------

l 108

FIRST COURSE IN RINGS AND lDEALS

PROBLEMS

Definition 6-8 is therefore satisfied in its entirety when n = the corresponding quadratic domains Z(.J'ñ) are Euc1idean.

t,' -

an

of irreducible " de . . . elem¡;nt and of a nonzero .prime do not al ways COInCl In an arbItrary Integral domain .. Specifically, we have (2' + .J -5)1 3 ' 3, but (2 + F)t 3, so that 2 + F5 cannot be a prime element of Z(~.

2, 2, 3 and

Corollary. The domain Z(i) óf Gaussian Íntegers is a unique factorization domain.

PROBLEMS

Remark. By investigating further the divisibility properties of Z(i), one ca!1 prove the c1assic "two squares theorem" of Fermat: every prime number ofthe form 4n + 1 is the sum oftwo squares; the interested reader is advis~d ;.. ,: to consult [13J for the details. .

1. Let R be a commutative ring with identity and the ele~~nts a e E R with Z = Prove that ,:, ' , e e. a) If (a) = (e), then.a arid e are ~ssociates. [Hint: a J, (I - e + a)e.] b) If for. sorner n EllZ+ the elements a" and e are ass6ciates . , then am an d e are aSSOCIates lor a m;?; n. .t;r

Example 6-'4. The various integral domains studied in this chapter might suggest that unique factorization of elements always holds. To round o~t the picture with an example of the failure of unique factorization, let:us consider the quadratic domain Z(.'¡=5); Observe that the element 9 'has two factorizations in Z(F"S), namely,' . . 9

= 3·3 = (2 + .J -5)(2 -:-

2.

R).

c) 1

3,

a2

= 2 + .J - 5,

a3

reducible, we could write a¡ = Pr, where neither ofthe elements yE Z(rs)is invertible. Taking norms, it follows that 9

=

N(a¡}

=

N(fJ)N(y),

N(fJ), N(y)

E

.

p,

Z+,

brs;

which in turn yields N(fJ) = N(y) = 3. Bence, if p = a + we find that we must solve the equation a2 + 5b 2 = 3 for integers and b; but this equation obviously has no solutions in Z (b i= impliesthat a2 + 5b 2 ;;::: 5, and if b = 0, then a2 = 3). Tp.us, we have exhibited two genuinely different factorizations of the e1ement 9 into irreducibles, so that unique factorization does not hold in Z(R). . Notice f~rther that the common. divisors of 9 and 3(2 + F) are . 1,3, and 2. + None ofthese latter elements is divisible by the others, so that gcd (9, 3(2 + .J - 5)) fails '10 exist (in particular, Z(R) does not have the gcd-property). On the other hand, the greatest common divisor does exist and, in fact, gcd (3, 2 + -J=s) = 1. It follows of 3 and 2 + that only the right-hand side of the formula

a

°

3 gcd (3, 2

+

.. ;

= R

for all nonzero' 1; J ..

4. Let R be ~n .integral. domain having the gcd-property. Assuming that equality holds to wlthln assoclates, prove that, for nonzero a, b, e' E R, a) gcd (a, gcd (b, e)) = gcd (gcd (a, b), e) .. b) gcd (a, 1) =:: 1. c) gcd (ea, eb) = e gcd (a, b); in particular, gcd (e, eb) = 1. d) ~f gcd (a, b) ;= 1 and gcd (a, e) = 1, then gcd (a, be) = 1. e) If gcd (a, b) = 1, ale and b/e, then ab/e. ' f) gcd (a, b) 1cm (a, b) =;' abo [Hint: Theorem 6"':6.J

6. !n a princi~al ideal dornain R, establish that the primary ideals are the two trivial Ideals and Ideals of the forrn (p"), where p is a prime element of R d Z . h n an n E +. [H'Int : If JIS'pnrnary, t en -VI = (ji) for sorne prime elernent p. Choose 11 E Z such thatI ~ '(pO), but 1 $ (p"+l), and show that 1 = (p").] . +

7. If R is a pr~nci?a.l idea! doman, we define the Length A(a) for each nonzero a E R a IS. Invertlble, . . (as follows: If. , . then A(a) = O·" otherwise A(a) is the numb er of pnmes not necessanly dls.tlnct) 1ll any factorization of a. Prove the following assertions' a) the length of a IS well-defined' ' . b) if alb, then A(a) 5: A(b); , . c) ~fa/b al1d,A(a) = A(b), then bla; '. d) If a { b and b { a, then there exist nonzero p, q E R such that

.¡=s

+ .J-:::S)) ".;

.1

5. If R i.s an.inte~ral domain having ~he gcd~property, show that a nonzero element of R IS pnme If and only if it is irreducible.

R.

gcd (3'3, 3(2

.'

+

+

K) = (1 J) n (l K).:" . K) = (l n J) + (l n K).

3. Sup?ose, that ~ = Rl EB Rz EB ... EB R", where each R¡ is a principal ideal ringo Venfy tnat R IS also a principal ideal ringo .

F5

= 2-

+ (J n

d) 1 n (J t e) lJ = 1 n J if and only if 1 + J

F

=

Give~ that 1, J, and K are ideals of a principal ideal doá1in R, derive the fOllowing relatIons: .' .!, a) If 1 = (a) and J = (b), then jJ-= (ab); in particuí~r1" = (a").

b) l(J n K) = IJ n lK.

ClearIy, no two ofthe factors 3, 2 + ~, and 2 are assodates of eachother (the only invertible elements in Z(.J - 5) are ± 1), and it can easily be shown that they are all irreducible. Indeed, were any one of al

109

A)

is defined in Z(.J - 5), thereby illustrating the remark 011 page 95. This example has the additional feature of showing that the concepts

A(pa

+

qb) 5: rnin {A(a), A(b)}.

I !

..

__. _ - - - - - - - - - - - - - - -

------------------_._----~-.- ~---_.------

110

FIRST COURSE IN RINGS AND IDEALS PROBLEMS

8. Let a be a nonzero element of the principal ideal domain R. Ir a has length n, prove that there are at most 2n ideal s containing a.

b) Show that in the quadratic domain Z(.J6), the relation 6 not violate unique factorization.

9. Verify that any two nonzero elements of a unique factorization domain possess a greatest cornmon divisor. [Hint: Ir a = P~'P~' ... P~' and b = pl(p~ ... p~ (Pi irreducible), then gcd (a, b) = P{'rl' ... dé, whereji = min (k i , IJ]

19. Show that the quadratic domain Z(.J - 5) is not a principal ideal domain. [Hint: Consider the ideal (3,2

lal is not a Euclidean valuation on the domain Q.

free integer.

below: . a) For nonzero a, bE R, if alb and t5(a) = t5(b), then a and b are associates. [Hi~t: Show that bla.] b) Fór nonzero a, bE R,.t5(ab) > t5(a) if and only if b is not an invertible element. [Hint: Use the division algorithm to wriÚ: a = q(ab) + r.] c) Ir n is any integer such that 15(1) + n ~ O, then the function 15': R - {O} -+ Z defined by t5'(a) = oCa) + n is also a Euclidean valuation on R. 13. For each ideal 1 in Z(i), the domain of Gaussian integers, establish that the quotient ring Z(i)fI is finite. [Hint: Write 1 = (a) and use the division algorithm on ex and any PE Z(i).] 14. Let R be a Euclidean domain with valuation o. a) Determine whether the set 1 = {a E Rlo(a) > o(l)} u {O} is an ideal of R. b) Assuming that the set F = {a ~ Rlo(a) = 1} u {O} is c10sed under addition, verify that F forms a field. 15. a) Prove that if ni and n 2 are square-free integers !!-nd ni -1' n 2 , then the quadratic field Q(.jn;,) is not isomorphic to Q(.jn;). b) For each square-free integer n, determine all~he subfields of the quadratic field Q(.Jñ).

16. Establish the following assertions (where n is a squ~ré-free integer): 'a) For n < -1, the only in vertible elements of thequadratic clomam Z(.Jñ) are

±1.

1 ::;; u(l

1

+

.J2.

+ .J2)-n = a + b.J2, show that a = 1, b = O.] Factor each of the following into primes: 11 + 7i in Z(i); 4 + 7.J2 in Z(.J2);

Assuming that u(l 17. a)

+ .J2)-n <

4- RinZ(R).

+ .j=5).]

20. Describe the field of quotients of the quadratic domain Z(.Jñ) where n'is a square-

12. Assuming that R is a Euclidean domain with valuation 15, prove the statements

b) For n > 1, Z(.Jñ) has infinitely many invertible' elements. [Hint: Ir al' b l is a solution of the equation a2 - nb2 = ±1, conc1ude that ak , bk is also a solution, where ak + bJn = (al + bl.Jñ)k, k r~Z+,] c) The invertible elements of Z(.J2) are precisely the elements of the form ±(1 + .J2)n, n E Z+. [Hint: Ir u is any positive invertible element of Z(.J2), then (1 + J2)n ::;; U < (1 + .J2)n+ 1 for sorne n E Z + ; hence,

3·2 does

18. Prove that the domain Z(R) is not a unique factorization domain by discovering two distinct factorizatioris of the element 10. Do the same for element 9 in the domain Z(J7f).

10. Let R be an integral domain. Prove that R is a unique factorization domain if and only if ,every nontrivial principal ideal óf R is the product of a finite number of maximal principal ideals and these ideals are uniquc up to a permutation of order. 11. Show that t5(a) =

= (.J6)2 =

111

I

I I

I

110

FIRST COURSE IN RINGS AND IDEALS PROBLEMS

8. Let a be a nonzero element of the principal ideal domain R. Ir a has length n, prove that there are at most 2n ideal s containing a.

b) Show that in the quadratic domain Z(.J6), the relation 6 not violate unique factorization.

9. Verify that any two nonzero elements of a unique factorization domain possess a greatest cornmon divisor. [Hint: Ir a = P~'P~' ... P~' and b = pl(p~ ... p~ (Pi irreducible), then gcd (a, b) = P{'rl' ... dé, whereji = min (k i , IJ]

19. Show that the quadratic domain Z(.J - 5) is not a principal ideal domain. [Hint: Consider the ideal (3,2

lal is not a Euclidean valuation on the domain Q.

free integer.

below: . a) For nonzero a, bE R, if alb and t5(a) = t5(b), then a and b are associates. [Hi~t: Show that bla.] b) Fór nonzero a, bE R,.t5(ab) > t5(a) if and only if b is not an invertible element. [Hint: Use the division algorithm to wriÚ: a = q(ab) + r.] c) Ir n is any integer such that 15(1) + n ~ O, then the function 15': R - {O} -+ Z defined by t5'(a) = oCa) + n is also a Euclidean valuation on R. 13. For each ideal 1 in Z(i), the domain of Gaussian integers, establish that the quotient ring Z(i)fI is finite. [Hint: Write 1 = (a) and use the division algorithm on ex and any PE Z(i).] 14. Let R be a Euclidean domain with valuation o. a) Determine whether the set 1 = {a E Rlo(a) > o(l)} u {O} is an ideal of R. b) Assuming that the set F = {a ~ Rlo(a) = 1} u {O} is c10sed under addition, verify that F forms a field. 15. a) Prove that if ni and n 2 are square-free integers !!-nd ni -1' n 2 , then the quadratic field Q(.jn;,) is not isomorphic to Q(.jn;). b) For each square-free integer n, determine all~he subfields of the quadratic field Q(.Jñ).

16. Establish the following assertions (where n is a squ~ré-free integer): 'a) For n < -1, the only in vertible elements of thequadratic clomam Z(.Jñ) are

±1.

1 ::;; u(l

1

+

.J2.

+ .J2)-n = a + b.J2, show that a = 1, b = O.] Factor each of the following into primes: 11 + 7i in Z(i); 4 + 7.J2 in Z(.J2);

Assuming that u(l 17. a)

+ .J2)-n <

4- RinZ(R).

+ .j=5).]

20. Describe the field of quotients of the quadratic domain Z(.Jñ) where n'is a square-

12. Assuming that R is a Euclidean domain with valuation 15, prove the statements

b) For n > 1, Z(.Jñ) has infinitely many invertible' elements. [Hint: Ir al' b l is a solution of the equation a2 - nb2 = ±1, conc1ude that ak , bk is also a solution, where ak + bJn = (al + bl.Jñ)k, k r~Z+,] c) The invertible elements of Z(.J2) are precisely the elements of the form ±(1 + .J2)n, n E Z+. [Hint: Ir u is any positive invertible element of Z(.J2), then (1 + J2)n ::;; U < (1 + .J2)n+ 1 for sorne n E Z + ; hence,

3·2 does

18. Prove that the domain Z(R) is not a unique factorization domain by discovering two distinct factorizatioris of the element 10. Do the same for element 9 in the domain Z(J7f).

10. Let R be an integral domain. Prove that R is a unique factorization domain if and only if ,every nontrivial principal ideal óf R is the product of a finite number of maximal principal ideals and these ideals are uniquc up to a permutation of order. 11. Show that t5(a) =

= (.J6)2 =

111

I

I I

I

POLYNOMIAL RINGS

SE VEN

where, for each k ~ 0,

k=

C

.

POLYNOMIAL RINGS

I

i+ j=k

aibj

0k

is given by

=

aObk +alb k - l

+ ... +

ak-.lb l

+

113

akb o'

°

(It i.s understood t?a~ the aboye s'ummation runs over alI integers i, j ~ subJect to the restnctlOn that i + j = k.) . A routine ~heck e.sta?lis~es that with these two definitions seq R becomes a nng. To.venfy a dlstnbutlve law, for instance, take ., f= (a o, al' ... ), g = (b o, b l , .•. ), h = (c o, C l •••• ).

One findsquickly that

"""," The next step in our program is: to apply some of the previously developed

f.(g:+ h)

";",i_

theory to a particular dass offings, the so-caBed polynomial rings. For the moment, we shaIl nierelY'''i"emark that these are rings whose elements . consist of "polynomials" with coefficients from a fixed, but otherwise arbitrary, ringo (The most intbresting results occur when the coefficients are specialized to a field.) As a first ordér of business, we seek to formalize the intuitive idea of what is meant by a poiynomial. This involves an excursion around the fringes of the more general question of rings of formal power series. Out of the veritable multitude of results concerning polynomials, we have attempted to assemble those facets of the theory w hose discussion reinforces the concepts and theorems expounded earlier; it is hoped thereby to convey a rough idea of how the c1assical arithmetic of polynomials fits into ideal theory. Our investigation conc1udes with a brü;f survey of some of the rudimentary facts relating roots ofpolynomials to field extensions. . To begin with simpler things, given an arbitrary ring R, Íet seq R denote the totality of aB infinite sequences

f of elements a

=

(a o' al' a2 ,

••• ,

ak, ... )

k R. Such sequences are called formal power series, or merely power series,over R. (Our choice of terminology will be justified shortIy.) We intend to introduce suitable operations in the set seq R so that the resulting system forms a ring containing R as a subring. At the outset, it should be made perfectIy c1ear that two power series E

f =

(a o, al' a 2 ,

.,.)

and

g = (b o, b l , b2 ,

••• )

are considered to be equal if and only if they are equal term by. term:

f =

g if and onlyif ak . = bk for aIl k ~ O..

Now, power series may themselves be added and r1mltiplied as follows:

f

+ g = (a o + bo, al + b l , ...), fg

= (co, Cl' C2' ••• ), 112

where

=

(a o, al' ... )(b o

+

ca' bl

+

cl '

.1:,'

... )

=

(do. d l , ... ),

.

dk =

.,4:

.+ J=k

ai(bj

+ c)

=

I

i+ j=k

(aibj

+

aic.) J

=t aibj + I ai c " i+ j=k i+ j=k J

A similar ca1culation of fg + fh leads to the same general termo so that f(g -: h) = f~ + fh. The rest ·of the details are left to the reader's careo We s~m~ly pomt out that the sequence (O, 0, 0, ... ) serves as the zero element of thlS rmg, while the additive inverse of an arbitrary member (a a a' .) fR' f 0' l' 2"" o seq IS, o course, (-aO,-a l , -a 2 • ••• ). To summarize what we know so far: Th~system seq R forms a ring, known as the ring of p.ower. ser~es over R. Furthermore, the ring seq R is commuta-

Theorem 7-1. (form~l)

tlve wlth ldentlty lf and only if the given ring R has these properties .

°

If S represents the subset of all sequences having for every term beyond the first, that is, the set . . S = {(a, 0, 0, ... )Ia E R}, then it is .not. p~ticularl~ difficult to show that S constitutes a subring of' seq R WhlCh IS Isomorphlc to R; one need only consider the mapping that . sends ~~e seq~ence (a, 0, 0, ... ) to the element a. In this sense, seq R contains the ongmal rmg R as a subring. Having reached this stage, we shaIl no longer distinguish bet'ween an element a E R and the special sequence (a, 0, 0, ... ) of seq R: The elements of R, regarded as power series, are hereafter called constant series, or just

constants. . With the aid of some additional notation, it is possible to represen t power senes the way we would like them to look. As a first step in this direction we let ax designate the sequence ' (O, a, 0, 0, ... ).

POLYNOMIAL RINGS

SE VEN

where, for each k ~ 0,

k=

C

.

POLYNOMIAL RINGS

I

i+ j=k

aibj

0k

is given by

=

aObk +alb k - l

+ ... +

ak-.lb l

+

113

akb o'

°

(It i.s understood t?a~ the aboye s'ummation runs over alI integers i, j ~ subJect to the restnctlOn that i + j = k.) . A routine ~heck e.sta?lis~es that with these two definitions seq R becomes a nng. To.venfy a dlstnbutlve law, for instance, take ., f= (a o, al' ... ), g = (b o, b l , .•. ), h = (c o, C l •••• ).

One findsquickly that

"""," The next step in our program is: to apply some of the previously developed

f.(g:+ h)

";",i_

theory to a particular dass offings, the so-caBed polynomial rings. For the moment, we shaIl nierelY'''i"emark that these are rings whose elements . consist of "polynomials" with coefficients from a fixed, but otherwise arbitrary, ringo (The most intbresting results occur when the coefficients are specialized to a field.) As a first ordér of business, we seek to formalize the intuitive idea of what is meant by a poiynomial. This involves an excursion around the fringes of the more general question of rings of formal power series. Out of the veritable multitude of results concerning polynomials, we have attempted to assemble those facets of the theory w hose discussion reinforces the concepts and theorems expounded earlier; it is hoped thereby to convey a rough idea of how the c1assical arithmetic of polynomials fits into ideal theory. Our investigation conc1udes with a brü;f survey of some of the rudimentary facts relating roots ofpolynomials to field extensions. . To begin with simpler things, given an arbitrary ring R, Íet seq R denote the totality of aB infinite sequences

f of elements a

=

(a o' al' a2 ,

••• ,

ak, ... )

k R. Such sequences are called formal power series, or merely power series,over R. (Our choice of terminology will be justified shortIy.) We intend to introduce suitable operations in the set seq R so that the resulting system forms a ring containing R as a subring. At the outset, it should be made perfectIy c1ear that two power series E

f =

(a o, al' a 2 ,

.,.)

and

g = (b o, b l , b2 ,

••• )

are considered to be equal if and only if they are equal term by. term:

f =

g if and onlyif ak . = bk for aIl k ~ O..

Now, power series may themselves be added and r1mltiplied as follows:

f

+ g = (a o + bo, al + b l , ...), fg

= (co, Cl' C2' ••• ), 112

where

=

(a o, al' ... )(b o

+

ca' bl

+

cl '

.1:,'

... )

=

(do. d l , ... ),

.

dk =

.,4:

.+ J=k

ai(bj

+ c)

=

I

i+ j=k

(aibj

+

aic.) J

=t aibj + I ai c " i+ j=k i+ j=k J

A similar ca1culation of fg + fh leads to the same general termo so that f(g -: h) = f~ + fh. The rest ·of the details are left to the reader's careo We s~m~ly pomt out that the sequence (O, 0, 0, ... ) serves as the zero element of thlS rmg, while the additive inverse of an arbitrary member (a a a' .) fR' f 0' l' 2"" o seq IS, o course, (-aO,-a l , -a 2 • ••• ). To summarize what we know so far: Th~system seq R forms a ring, known as the ring of p.ower. ser~es over R. Furthermore, the ring seq R is commuta-

Theorem 7-1. (form~l)

tlve wlth ldentlty lf and only if the given ring R has these properties .

°

If S represents the subset of all sequences having for every term beyond the first, that is, the set . . S = {(a, 0, 0, ... )Ia E R}, then it is .not. p~ticularl~ difficult to show that S constitutes a subring of' seq R WhlCh IS Isomorphlc to R; one need only consider the mapping that . sends ~~e seq~ence (a, 0, 0, ... ) to the element a. In this sense, seq R contains the ongmal rmg R as a subring. Having reached this stage, we shaIl no longer distinguish bet'ween an element a E R and the special sequence (a, 0, 0, ... ) of seq R: The elements of R, regarded as power series, are hereafter called constant series, or just

constants. . With the aid of some additional notation, it is possible to represen t power senes the way we would like them to look. As a first step in this direction we let ax designate the sequence ' (O, a, 0, 0, ... ).

r 114

)

FIRST COURSE IN RINGS AND IDBALS

That is, ax is the specific member of seq R which has the element ti for its second term and for all other terms. More generally, the symbol ax", n ~ 1, will denote the sequence

i

°

x = (0,1,0,0, ...). From this view, ax becomes an actual product of members oC R[[x]]:

ax

(O, ... , 0, a, 0, ... ),

+

where the element a appears as the (n example" we have

ax 2 and

ax 3

(0,0, a, 0, ... ) (0,0,0, a, 0, ...),

(ao, al' a2 ,

...

1+

= (a o, 0,0, ... )

+

(O, al' 0, ...)

+ ... + (O, ... ,0, a., 0, ... ) + ...

x4

+ ... +

x 2•

+ ... E Z[[xJ]

°

with tbe obvious identification oC a o witb the sequence (ao, O, O, ... ). Thus, tbere is no loss in regarding tbe power series ring seq R as consisting of all formal expressions

Suppose f(x), g(x) E R[[x]], with ordf(x) =

n and ord g(x)

= m, so

that f(x) = q.x" + an + lX'+ 1 + ... g(x) = bmx"' + bm+ lX m+ 1 + ...

+ alx + a2 x 2 + ,.. + a.x· + "',

where tbe elements ao, al' ... , an , .. , (tbe coefficients of 1) líe in R. As a notational device, We sball ofien write tbis asf = L akx k (tbe surnmation symbol is not an actual sum and convergence is not at issue bere). Using sigma notation, tbe definitions of addition and multiplication of power series assume the form L ak xk + L bkxk = L (ak + bk)Xk,

(a. =1= O), (b", =1= O).

From the ~efinition ofmultiplication in R[[x]], the reader may easily check that all.coefficients oC f(x)g(x) up to the (n + m)th are zero, whence f(x)g(x) = anbmx·+ m + (a + b + a.b + )x"+m+l + .... n

1

1ll

m

1

If we assume that one of a. or bm is not a divisor of zero in R, then an bm and ., <.,

wbere

+

not all the ak = O) in R[[x]], then the smallest integer n such that a. =1= is called the arder off(x) and denoted by ordf(x).

= ao + alx + a2 x 2 + ... + a.x· + ...

f= a o

X2

as representing the sequence (1,0,1, O, ... ). An important definition in connection with power series is that of order, given below. . k Definition 7-1. If f(x) = L akx is á nonzero power series (that is, if

,a., ...)

may be uniquely expressed in the form

f

(a, 0,.0, ... )(0, 1,0,0, ... ).

Concerning the notation oC power series, it is customary to omit terms with zero coefficients and to replace (-ak)x k by -akx k. Although x is not to be considered as án element oC R[[x]], we shall nonetheless take the liberty of writing the term 1xk as Xk (k ~ 1), With these conventions, one should view, for example, the power series

1)8t term in this sequence; for

By use of these definitions, each power series f

115

POL YNOMIAL RINGS

'ord (¡(x)g(x»)

n

+

m

=1=

°

= ordf(x) + ord g(:*) •. "

1 ; ~

This certainly holds if R is taken to be an integral doma in, qt.~gain if R has an identíty and one of a. or bm is the identity element. The foregoing argument serves to establish the first part of the next theor.em; the prooC oC the second assertion is left as anexercise. <

We sbould empbasize tbat, according to our definition, x ís simply a neW symbol, or indeterminant, totalIy unrelated to the ring R and in no sense represents an element of R. To indicate the indeterminant x, it lS common practice to write R[[ x]] Cor the set seq R, and f(x) for any member oC tbe same. From now on, we sball make exclusive use of tbis notation. Remark. If the ring R bappens to bave a multiplicative ídentity 1, many autbors will identiCy tbe power series + Ix + OX2 + OX3 + ... with x, tbereby treating x itself as a special member oC R[[x]] ; namely, the sequence

°

Theorem 7-2. Iff(x) and g(x) are nonzero power series in R[[x]], then 1) either f(x)g(x) = or ord (J(x)g(x») ~ ord f(x) + ord g(x), with equality if R is an integral domain; 2) either ¡(x) + g(x) = or

°

ord (¡(x)

°

+ g(x»)

~ min {ord f(x), ord.g(x)}.

The notation of order can be used to prove the Collowing corollary,

r 114

)

FIRST COURSE IN RINGS AND IDBALS

That is, ax is the specific member of seq R which has the element ti for its second term and for all other terms. More generally, the symbol ax", n ~ 1, will denote the sequence

i

°

x = (0,1,0,0, ...). From this view, ax becomes an actual product of members oC R[[x]]:

ax

(O, ... , 0, a, 0, ... ),

+

where the element a appears as the (n example" we have

ax 2 and

ax 3

(0,0, a, 0, ... ) (0,0,0, a, 0, ...),

(ao, al' a2 ,

...

1+

= (a o, 0,0, ... )

+

(O, al' 0, ...)

+ ... + (O, ... ,0, a., 0, ... ) + ...

x4

+ ... +

x 2•

+ ... E Z[[xJ]

°

with tbe obvious identification oC a o witb the sequence (ao, O, O, ... ). Thus, tbere is no loss in regarding tbe power series ring seq R as consisting of all formal expressions

Suppose f(x), g(x) E R[[x]], with ordf(x) =

n and ord g(x)

= m, so

that f(x) = q.x" + an + lX'+ 1 + ... g(x) = bmx"' + bm+ lX m+ 1 + ...

+ alx + a2 x 2 + ,.. + a.x· + "',

where tbe elements ao, al' ... , an , .. , (tbe coefficients of 1) líe in R. As a notational device, We sball ofien write tbis asf = L akx k (tbe surnmation symbol is not an actual sum and convergence is not at issue bere). Using sigma notation, tbe definitions of addition and multiplication of power series assume the form L ak xk + L bkxk = L (ak + bk)Xk,

(a. =1= O), (b", =1= O).

From the ~efinition ofmultiplication in R[[x]], the reader may easily check that all.coefficients oC f(x)g(x) up to the (n + m)th are zero, whence f(x)g(x) = anbmx·+ m + (a + b + a.b + )x"+m+l + .... n

1

1ll

m

1

If we assume that one of a. or bm is not a divisor of zero in R, then an bm and ., <.,

wbere

+

not all the ak = O) in R[[x]], then the smallest integer n such that a. =1= is called the arder off(x) and denoted by ordf(x).

= ao + alx + a2 x 2 + ... + a.x· + ...

f= a o

X2

as representing the sequence (1,0,1, O, ... ). An important definition in connection with power series is that of order, given below. . k Definition 7-1. If f(x) = L akx is á nonzero power series (that is, if

,a., ...)

may be uniquely expressed in the form

f

(a, 0,.0, ... )(0, 1,0,0, ... ).

Concerning the notation oC power series, it is customary to omit terms with zero coefficients and to replace (-ak)x k by -akx k. Although x is not to be considered as án element oC R[[x]], we shall nonetheless take the liberty of writing the term 1xk as Xk (k ~ 1), With these conventions, one should view, for example, the power series

1)8t term in this sequence; for

By use of these definitions, each power series f

115

POL YNOMIAL RINGS

'ord (¡(x)g(x»)

n

+

m

=1=

°

= ordf(x) + ord g(:*) •. "

1 ; ~

This certainly holds if R is taken to be an integral doma in, qt.~gain if R has an identíty and one of a. or bm is the identity element. The foregoing argument serves to establish the first part of the next theor.em; the prooC oC the second assertion is left as anexercise. <

We sbould empbasize tbat, according to our definition, x ís simply a neW symbol, or indeterminant, totalIy unrelated to the ring R and in no sense represents an element of R. To indicate the indeterminant x, it lS common practice to write R[[ x]] Cor the set seq R, and f(x) for any member oC tbe same. From now on, we sball make exclusive use of tbis notation. Remark. If the ring R bappens to bave a multiplicative ídentity 1, many autbors will identiCy tbe power series + Ix + OX2 + OX3 + ... with x, tbereby treating x itself as a special member oC R[[x]] ; namely, the sequence

°

Theorem 7-2. Iff(x) and g(x) are nonzero power series in R[[x]], then 1) either f(x)g(x) = or ord (J(x)g(x») ~ ord f(x) + ord g(x), with equality if R is an integral domain; 2) either ¡(x) + g(x) = or

°

ord (¡(x)

°

+ g(x»)

~ min {ord f(x), ord.g(x)}.

The notation of order can be used to prove the Collowing corollary,

POLYNOMIAL RINGS

116

117

FIRST COURSE IN RINGS AND IDEALS

Corollary., If the ring R is an integral domain, then so also is its power series ring R[[ x]]. Proa! We observed earlier that whenever R is a commutative ring with identity, these properties carry over to R[[ x]]. To see that R[[ x]] has no .r ~~ro divisors, selectf(x) =1= O, g(x) =1= O in R[[ x]]. Then, ' .' ord (f(x)g(x))

=

ordf(x)

+

ord g(x) > O ; "

:p:ence, fue productf(x)g(x) cannot be the z~ro series.

J,!;

(/'¡: Although arbitrary pow~r series rings are of sorne interest, the most

, important consequences arise on specializing the discussion to power series 'whose coefficients are taken from a field. These wiII be seen to form principal :~.: ,'ideal domains and, in ¡:onsequence, unique factorizati'on domains. The:" ;i1)f()1l0wing intermediate re~ult is directed towards establishing this fact. :'~ir~: -, Lemma. Let R be a commutative ring with identity. A formal powep,;; "', series f(x) = I akxk is invertible in R[[x]] if and only if the constant '-" term ao has an inverse in R. Proa! If f(x)g(x) = 1, where g(x) = I bkx\ then the definition of multiplication in R[[ x]] shows that aob o = 1; hence, a o is invertible as an element of R. For the converse, suppose that the element ao has an inverse in R. kWe proceed inductively to define the coefficients of a power series I bkx in R[[x]] which is the inverse off(x). To do this, simply take bo = a I and, assurning b l , b2, ... , bk- I have already been defined, let bk = -aOI(aIb k- 1 + a 2bk- 2 + ... -: akbo)'

latter event, choose a nonzero pow~r series f(x) E l of minimal order. Suppose thatf(x) is of order k, so that f(x) = akx k + ak+Ix k+1 + ... = xk(a k + ak+I x + ... ). Sin,ce the coeffieient ak =f= O, ~he p~evious lemma insures that the power senes~k ak+ IX + ... IS an mv~rtlble ~lement of F[[xJ]; in other words, f(x) - x g(x), where g(x) has an mverse m F[[ x]]. But,-then,

t

Xk

has an inverse in F[[x]] if and only if its constant term ao

=1=

O.

Having dealt with these preliminaries, we are now ready to proceed to describe the ideal structure of F[[ xJ]. Theorem 7-3. For any field F, the power series ring F[[x]] is a principal ideal domain; in fact, the nontrivial ideals of F[[x]] are of the form (x k ), where k E Z+. Proa! ' Let 1 be any proper ideal of F[[x]]. Either l = {O}, in which case 1 is just the principal ideal (O), or else 1 contains nonzero elements. In the

f(x)g(X)-1 El,

which leads to the inc1usion (x ) ~ l. On the. other ha.nd, take h(x) to be any nonzero po~gfseries in l, say of ?r?er n. Smcef(x) IS assumed to have least order amon'g: all members of l lt IS c1ear that k :;:; n; thus, h(x) can pe written in the fQrin ' h(x)

~

xk(bnxn-k

+ bn+Ixn-k+1 + ... )~.(~k).

This implies that l ~ (xk), and the equality l

=

(Xk) r611;ws.

~orollary l. The ring F[[ x]] is a local ring with (:i)~s its maximal Ideal. Proa! Inasmuch as the ideal s of F[[ x J] form a chain F[[xJ]

~

(x)

~

(X2)

'~

...

~

{O},

the conc1usion is obvious.

o

Then aob o = 1, while, for k :;::: 1, Ck = I a¡b j = aOb k + aIb k- 1 + ... + akbo =,0.. , ¡+j=k k k By our choice of the bk's, we evidently must have (I akx ) (I bkx ) = 1, and so I akxk possesses aninverse in R[[x]]' Corollary. A power seriesf(x) = I akx k E F[[xJ], where F is a field,

= k

Corollary 2. Any nonzero element f(x) E F[[x J] can be written in the formf(x) = g(x)xk, where g(x) is invertible and k :;::: O. ' To t~is w~add, for future reference, the folIówing as~ertion regarding Ideal s of a power series ring over a commutative ring with ldentlty. ~he ~axlmal

Theorem 7-4 Let R be a commutative ring with identity. There is a one-to-one correspondence between the maximal ideals M of the . R and.the maximal ideal s M' of R[[x]] in such a way that M' to M lf and only if M' is generated by M and x; that is, M' = (M, x). s

corresp;~~g

Proa! Assu~e th.at M is a maximal ideal of R. To see that M' = (M, x) forms a ~axlmalldeal of t~e ring R[[xJ], we need only show that for any , pow~r senes f(x) = I akx ~ M', the element 1 + g(x)f(x) E M' for sorne I g(~).m R[[x]] (Problem 2, Chapter 5). Since the series f(x) does not lie in M, ltS constant term ao ~ M; hence, there exists an element r E R such that 1 + rao E M. This implies that 1

+ rf(x)

=

(1

+ rao) + r(a l + a2x + ... + anx,,-I + .. ·)x E (M, x),

and so M' is a maximal ideal, as required.

POLYNOMIAL RINGS

116

117

FIRST COURSE IN RINGS AND IDEALS

Corollary., If the ring R is an integral domain, then so also is its power series ring R[[ x]]. Proa! We observed earlier that whenever R is a commutative ring with identity, these properties carry over to R[[ x]]. To see that R[[ x]] has no .r ~~ro divisors, selectf(x) =1= O, g(x) =1= O in R[[ x]]. Then, ' .' ord (f(x)g(x))

=

ordf(x)

+

ord g(x) > O ; "

:p:ence, fue productf(x)g(x) cannot be the z~ro series.

J,!;

(/'¡: Although arbitrary pow~r series rings are of sorne interest, the most

, important consequences arise on specializing the discussion to power series 'whose coefficients are taken from a field. These wiII be seen to form principal :~.: ,'ideal domains and, in ¡:onsequence, unique factorizati'on domains. The:" ;i1)f()1l0wing intermediate re~ult is directed towards establishing this fact. :'~ir~: -, Lemma. Let R be a commutative ring with identity. A formal powep,;; "', series f(x) = I akxk is invertible in R[[x]] if and only if the constant '-" term ao has an inverse in R. Proa! If f(x)g(x) = 1, where g(x) = I bkx\ then the definition of multiplication in R[[ x]] shows that aob o = 1; hence, a o is invertible as an element of R. For the converse, suppose that the element ao has an inverse in R. kWe proceed inductively to define the coefficients of a power series I bkx in R[[x]] which is the inverse off(x). To do this, simply take bo = a I and, assurning b l , b2, ... , bk- I have already been defined, let bk = -aOI(aIb k- 1 + a 2bk- 2 + ... -: akbo)'

latter event, choose a nonzero pow~r series f(x) E l of minimal order. Suppose thatf(x) is of order k, so that f(x) = akx k + ak+Ix k+1 + ... = xk(a k + ak+I x + ... ). Sin,ce the coeffieient ak =f= O, ~he p~evious lemma insures that the power senes~k ak+ IX + ... IS an mv~rtlble ~lement of F[[xJ]; in other words, f(x) - x g(x), where g(x) has an mverse m F[[ x]]. But,-then,

t

Xk

has an inverse in F[[x]] if and only if its constant term ao

=1=

O.

Having dealt with these preliminaries, we are now ready to proceed to describe the ideal structure of F[[ xJ]. Theorem 7-3. For any field F, the power series ring F[[x]] is a principal ideal domain; in fact, the nontrivial ideals of F[[x]] are of the form (x k ), where k E Z+. Proa! ' Let 1 be any proper ideal of F[[x]]. Either l = {O}, in which case 1 is just the principal ideal (O), or else 1 contains nonzero elements. In the

f(x)g(X)-1 El,

which leads to the inc1usion (x ) ~ l. On the. other ha.nd, take h(x) to be any nonzero po~gfseries in l, say of ?r?er n. Smcef(x) IS assumed to have least order amon'g: all members of l lt IS c1ear that k :;:; n; thus, h(x) can pe written in the fQrin ' h(x)

~

xk(bnxn-k

+ bn+Ixn-k+1 + ... )~.(~k).

This implies that l ~ (xk), and the equality l

=

(Xk) r611;ws.

~orollary l. The ring F[[ x]] is a local ring with (:i)~s its maximal Ideal. Proa! Inasmuch as the ideal s of F[[ x J] form a chain F[[xJ]

~

(x)

~

(X2)

'~

...

~

{O},

the conc1usion is obvious.

o

Then aob o = 1, while, for k :;::: 1, Ck = I a¡b j = aOb k + aIb k- 1 + ... + akbo =,0.. , ¡+j=k k k By our choice of the bk's, we evidently must have (I akx ) (I bkx ) = 1, and so I akxk possesses aninverse in R[[x]]' Corollary. A power seriesf(x) = I akx k E F[[xJ], where F is a field,

= k

Corollary 2. Any nonzero element f(x) E F[[x J] can be written in the formf(x) = g(x)xk, where g(x) is invertible and k :;::: O. ' To t~is w~add, for future reference, the folIówing as~ertion regarding Ideal s of a power series ring over a commutative ring with ldentlty. ~he ~axlmal

Theorem 7-4 Let R be a commutative ring with identity. There is a one-to-one correspondence between the maximal ideals M of the . R and.the maximal ideal s M' of R[[x]] in such a way that M' to M lf and only if M' is generated by M and x; that is, M' = (M, x). s

corresp;~~g

Proa! Assu~e th.at M is a maximal ideal of R. To see that M' = (M, x) forms a ~axlmalldeal of t~e ring R[[xJ], we need only show that for any , pow~r senes f(x) = I akx ~ M', the element 1 + g(x)f(x) E M' for sorne I g(~).m R[[x]] (Problem 2, Chapter 5). Since the series f(x) does not lie in M, ltS constant term ao ~ M; hence, there exists an element r E R such that 1 + rao E M. This implies that 1

+ rf(x)

=

(1

+ rao) + r(a l + a2x + ... + anx,,-I + .. ·)x E (M, x),

and so M' is a maximal ideal, as required.

118

POLYNOMIAL RINGS

FIRST COURSE IN RINGS AND IDEALS

so that both the'sumf(x) + g(x) and productf(x)g(x) belong to R[xJ. Running parallel to the idea ofthe order of a power series is that ofthe degree of a polynomial, which we introduce at this time.

Next, take MI to be any maximal ideal of R[[x]] and define theset M to consist of the constant terms of power series in MI : M = {ao E RI ¿akxk eMl

Definition 7-3. Given a nonzero polynomial

The reader can painlessly supply a proof that M forms a maximal ideal of the ring R. Notice incidentally that M must be a proper ideal. Were M R, then the~e would exist a power series ¿ bn:x" in MI with constant term bo = t By the last lemma, ¿bnxn would then be an invertible element, so that MI = R[[x]], whichis impossible. Owingto the inc1usion M' ~ (M,x) and the fact that M' is maximal in R[[x]], it now follows that MI = (M, x). To verify that the corresponden ce in question is indeed one-to-one, suppose that (M, x) = (M, x), where M, M are both maximal ideals of the ring R; what we want to prove is that M = M. Let r E M be arbitrary. Givenf(x)E R[[x]], the. sum r.+ f(x)x E (M,x) = (M, x), so that

+ f(x)x

r

°=

r} = ord (g(x)

ord(r

f(x)

f(x»)

+

r

(g(x)

f(x) )x.

ord x 2: 1,

°

alx

An element of R[xJ is called a

°

deg(J(x)

x) over the ring R.

ak

+

¿

f(x) = 1 we obtainf(x)g(x) = 4

a¡b j

=

O fork

~ m

+...n, .

+

2.x,

g(x)

=

4

+ x + 4x 2 ,

+ x + 6x 2 , so that

deg (J(x)g(x»)

2 < 1

+

2

=

degf(x)

+ deg g(x).

Although many properties of the ring R carry over to the associated polynomial ring R[x], it should be pointed out !hat for no ring R does R[x] form a field. In fact, when R is a field (or, for that matter, an integral doma in), no eIement of R[x] which has positive degree can possess a

bk = O fork ~ max{m,n},

I+J=k

max{degf(x), deg g(x)}.

Example 7-1. Aslan illustration ofwhat might happen if R has zero divisors, consider Zg, the ring of integers modulo 8. Taking

°

¿

~

Corollary. Ifthe ring R is an integral domain, then so is its polynomial ring R[xJ. .'

In essence, we are defining a polynbprial to be a finitely nonzero sequence of elements of R. Thus, the sequence (1, 1, 1,0,0, ... ) would be a polynomial over Z2' but (1, 0, 1,0, ... , 1,0, ... ) would noto It is easily verified that R[x] constitutes a subring of R[[x]], the socalled ring of polynomials over R (in an indeterminant x); indeed, if f(x) = akxk, g(x) = bkxk are in R[xJ, with ak for all k 2: n and bk = for all k ~ m, then

°¿

+ g(x»)

Knowing this,:cme could proceed along the lines of the corollary to Theorem 7-2 to 'estáblish

+ '::~ + anx"lak E R; n 2: O}. ~¡¿!ynomial (in

O)

Theorem 7-5. If f(x) and g(x) are nonzero poIynomiaIs in R[x], then 1) either f(x)g(x) = or deg (J(x)g(x») ~ degf(x) + deg g(x), with equality whenever R is an integral domain; 2) either f(x) + g(x) = Oor

Definition 7-2. Let R[x] denote, t~e set of all power series in R[[xJ] whose coefficients are zero from s,o,!llF index onward (the particular index varies from series to series): ..

+

=1=

°

Power series have so far received all the attention, but our primary concern is with polynomials.

{ao

(a"

The degree ofany nonzero polynomial is therefore a nonnegative integer; no degree is assigned to the zero polynomial. Notice that the polynomials of degree are precisely the nonzero constant polynomials. If R is a ring with identity, a polynomial whose leading coefficient is 1 is said to be a monic polynomial. As a matter of notation, we shall hereafter write degf(x) for the degree of any nonzero polynomialf(x) E R[xJ. The result"below is similar to that given for power series and its proof i8 left for the reader to províde; the only change of consequence ís that we now use the notíon of degree rather than order.

an absurdity. In consequence, we must have g(x) - f(x) = which, in its turn, forces r = rE M. The implication is that M ~ M and, since M is maximaI, we end up with M = M. This completes the proof ofthe theorem.

R[xJ

= ao + a1x + ... + a"x"

in R[x], we call a" the leading coefficient of f(x); and the integer n, the degree of the polynomial. .

r + g(x)x

for appropriate r E M and g(x) E R[[x]]' Hence, r If g(x) - f(x) =1= 0, then, upon taking orders,

119

,

:1:

118

POLYNOMIAL RINGS

FIRST COURSE IN RINGS AND IDEALS

so that both the'sumf(x) + g(x) and productf(x)g(x) belong to R[xJ. Running parallel to the idea ofthe order of a power series is that ofthe degree of a polynomial, which we introduce at this time.

Next, take MI to be any maximal ideal of R[[x]] and define theset M to consist of the constant terms of power series in MI : M = {ao E RI ¿akxk eMl

Definition 7-3. Given a nonzero polynomial

The reader can painlessly supply a proof that M forms a maximal ideal of the ring R. Notice incidentally that M must be a proper ideal. Were M R, then the~e would exist a power series ¿ bn:x" in MI with constant term bo = t By the last lemma, ¿bnxn would then be an invertible element, so that MI = R[[x]], whichis impossible. Owingto the inc1usion M' ~ (M,x) and the fact that M' is maximal in R[[x]], it now follows that MI = (M, x). To verify that the corresponden ce in question is indeed one-to-one, suppose that (M, x) = (M, x), where M, M are both maximal ideals of the ring R; what we want to prove is that M = M. Let r E M be arbitrary. Givenf(x)E R[[x]], the. sum r.+ f(x)x E (M,x) = (M, x), so that

+ f(x)x

r

°=

r} = ord (g(x)

ord(r

f(x)

f(x»)

+

r

(g(x)

f(x) )x.

ord x 2: 1,

°

alx

An element of R[xJ is called a

°

deg(J(x)

x) over the ring R.

ak

+

¿

f(x) = 1 we obtainf(x)g(x) = 4

a¡b j

=

O fork

~ m

+...n, .

+

2.x,

g(x)

=

4

+ x + 4x 2 ,

+ x + 6x 2 , so that

deg (J(x)g(x»)

2 < 1

+

2

=

degf(x)

+ deg g(x).

Although many properties of the ring R carry over to the associated polynomial ring R[x], it should be pointed out !hat for no ring R does R[x] form a field. In fact, when R is a field (or, for that matter, an integral doma in), no eIement of R[x] which has positive degree can possess a

bk = O fork ~ max{m,n},

I+J=k

max{degf(x), deg g(x)}.

Example 7-1. Aslan illustration ofwhat might happen if R has zero divisors, consider Zg, the ring of integers modulo 8. Taking

°

¿

~

Corollary. Ifthe ring R is an integral domain, then so is its polynomial ring R[xJ. .'

In essence, we are defining a polynbprial to be a finitely nonzero sequence of elements of R. Thus, the sequence (1, 1, 1,0,0, ... ) would be a polynomial over Z2' but (1, 0, 1,0, ... , 1,0, ... ) would noto It is easily verified that R[x] constitutes a subring of R[[x]], the socalled ring of polynomials over R (in an indeterminant x); indeed, if f(x) = akxk, g(x) = bkxk are in R[xJ, with ak for all k 2: n and bk = for all k ~ m, then

°¿

+ g(x»)

Knowing this,:cme could proceed along the lines of the corollary to Theorem 7-2 to 'estáblish

+ '::~ + anx"lak E R; n 2: O}. ~¡¿!ynomial (in

O)

Theorem 7-5. If f(x) and g(x) are nonzero poIynomiaIs in R[x], then 1) either f(x)g(x) = or deg (J(x)g(x») ~ degf(x) + deg g(x), with equality whenever R is an integral domain; 2) either f(x) + g(x) = Oor

Definition 7-2. Let R[x] denote, t~e set of all power series in R[[xJ] whose coefficients are zero from s,o,!llF index onward (the particular index varies from series to series): ..

+

=1=

°

Power series have so far received all the attention, but our primary concern is with polynomials.

{ao

(a"

The degree ofany nonzero polynomial is therefore a nonnegative integer; no degree is assigned to the zero polynomial. Notice that the polynomials of degree are precisely the nonzero constant polynomials. If R is a ring with identity, a polynomial whose leading coefficient is 1 is said to be a monic polynomial. As a matter of notation, we shall hereafter write degf(x) for the degree of any nonzero polynomialf(x) E R[xJ. The result"below is similar to that given for power series and its proof i8 left for the reader to províde; the only change of consequence ís that we now use the notíon of degree rather than order.

an absurdity. In consequence, we must have g(x) - f(x) = which, in its turn, forces r = rE M. The implication is that M ~ M and, since M is maximaI, we end up with M = M. This completes the proof ofthe theorem.

R[xJ

= ao + a1x + ... + a"x"

in R[x], we call a" the leading coefficient of f(x); and the integer n, the degree of the polynomial. .

r + g(x)x

for appropriate r E M and g(x) E R[[x]]' Hence, r If g(x) - f(x) =1= 0, then, upon taking orders,

119

,

:1:

120

FIRST COVRSE IN. RINGS AND IDEALS . ",.

POLYNOMIAL RINGS

rnultiplicative in verse. For, suppose that f(x) E R[x J, with degf(x) > O; ifj(x)g(x) = 1 for sorne g(x) in R[xJ, we could obtain the contradiction ' O = deg 1 = deg (J(x)g(x») = degf(x)

+ deg g(x)

=/=.

f(x) = ao + alx in R[xJ, we may define f(r)

E

+ g(x),

+ ... + a.r·.

k(x) = f(x)g(x), k(r) = f(r)g(r).

This being so, it may be conc1uded that the mapping rpr: R[xJ --+ R' which sendsf(x) tof(r) is a homomorphism of R[ x Jinto R'. Such a homomorphism , will be called the substítution homomorphísm determined by r and its range denoted by the symbol R[r]: R[r]

=

{J(r)/f(x) E R[xJ}

+ ... + a r"la k ER; n 2: O}. It is a simple matter to show that R[rJ constitutes a subring of R'; in fact, R[rJ is the subring of R' generated by the set R u {r}. (Si~ce R has an =

{ao +

al/'

,f, r(al)r(x) + ... ':

~'.

= ao +,~ ~J r

+ ...

1l

identity element 1, Ix = xER[xJ, and so rER[r].) NotIce also that R[r] = R ifancl only ifr ER. ' The foregoing remarks justify part of the next theorem.

+ r(a.)r(x)· n + anr = f(r) = rpr (J (x) ).

This pro ves that r = rp" yi~!PÍIig the uniqueness conc1usion. ::-~' ':?

Without sorne comm~,tativity assumption, the aboye rernarks need not hold; For, if we let h(x)

=,

(x - il)(x - b) = h(r) = r2

then, h(r) = f(r) +g(r),

r(J(x») = r(a o)

then

The elementf(r) is said to be the result of substituting r for x inf(x). Suffice it to say, the 'addition and multiplication used in defining f(r) are those of the ring R', not those of R[xJ. Now, suppose that f(x), g(x) are polynornials in R[xJ and rE cent R'. We lea ve the reader to pro ve that if h(x) = f(x)

Proof We need only verify that 4>r is uníque. Suppose, then, thát there is another horrtomorphism r: R[xJ --+ R satisfy in the indicated conditions and con~ider any polynornialf(x) = ao..+ alx + '" + a.x·ER[xJ. By ,assu.mptlOn, r(ak) = ak fon each coefficlent ak , while r(x k) = r(x)k = rk. Takmg stock of the fact that r is a homomorphism,

+ '" .f.' a.x·

R' by taking

f(r) = ao + alr

Theorem 7-6. Let R be a ring with identity, R' an extension ring of R, and the element rE cent R'. ,Then there is a unique homomorphism rpr: R[x] --+ R' such that rpAx) = r, rpAa) = a for all a E R.

O. '

The degree of a polynomial is used in the factorization theory of R[xJ in much the same way as the absolute value is employed in Z. For, it is through the degree concept that induction can be utilized in R[ x J to develop a polynomial counterpart of the familiar diyision algorithm. One can subsequent1y establish that the ring F[x}with'Coefficients in a field forms a EucJidean domain in which the degree functio.n}s taken to be the Euc1idean v~luatiort. ' ", ' . " . Before embarking on this program, we Msh to introduce several new ideas. To this purpose, let R be a ring with)dentity; assume further that R' is any ring contain\ng R as a subring (tb,at)s, R' is an extension of R) and let r be an arbitrary element of R'. For;.?,~~h polynornial

121

-

(a

X2 -

(a

+ b)x

;'h

ab;

+ b)r + ab.

Lacking the hypothesis that rE cení R', it cannot be conc1uded that (r - aHr - b) = r2

-

ar - rb

+ ab

will equal h(r); in other words, h(x) = f(x)g(x) does not always imply h(r) = f(r)g(r). Wheneverf(r) = O, we call the element r a root or zero ofthe polynomial f(x). Of cou!se, a given polynomial f(x) E R[x J may not ha ve a root in R; we shall see later that when R is a field, there alwaysexists an extension field R' of R in whichf(x) possesses a root. It is perhaps appropriate to point out at t~is time that the problem of obtaining all roots of a polynornial f(x) E R[ x] is equivalent to that of finding all elements r E R' for which f(x) E ker rpr' ' After this brief digression, let us now state and prove the division algorithm for polynomials.

Theorem 7-7. (Division Algorithm). Let R be a c~mmutative ring with identity and f(x), g(x) =1= O be polynomials in R[xJ, with the leading coefficientof g(x) an invertible element. Then there exist unique polynomials q(x), r(x) E R[xJ such that f(x) where either r(x)

= O or

=

q(x)g(x)

+ r(x),

deg r(x) < deg g(x).

Proof The proof is by induction on the degree off(x)~ First, notice that if f(x) = Oorf(x) =1= Oand degf(x) < deg g(x), a representation meeting the

120

FIRST COVRSE IN. RINGS AND IDEALS . ",.

POLYNOMIAL RINGS

rnultiplicative in verse. For, suppose that f(x) E R[x J, with degf(x) > O; ifj(x)g(x) = 1 for sorne g(x) in R[xJ, we could obtain the contradiction ' O = deg 1 = deg (J(x)g(x») = degf(x)

+ deg g(x)

=/=.

f(x) = ao + alx in R[xJ, we may define f(r)

E

+ g(x),

+ ... + a.r·.

k(x) = f(x)g(x), k(r) = f(r)g(r).

This being so, it may be conc1uded that the mapping rpr: R[xJ --+ R' which sendsf(x) tof(r) is a homomorphism of R[ x Jinto R'. Such a homomorphism , will be called the substítution homomorphísm determined by r and its range denoted by the symbol R[r]: R[r]

=

{J(r)/f(x) E R[xJ}

+ ... + a r"la k ER; n 2: O}. It is a simple matter to show that R[rJ constitutes a subring of R'; in fact, R[rJ is the subring of R' generated by the set R u {r}. (Si~ce R has an =

{ao +

al/'

,f, r(al)r(x) + ... ':

~'.

= ao +,~ ~J r

+ ...

1l

identity element 1, Ix = xER[xJ, and so rER[r].) NotIce also that R[r] = R ifancl only ifr ER. ' The foregoing remarks justify part of the next theorem.

+ r(a.)r(x)· n + anr = f(r) = rpr (J (x) ).

This pro ves that r = rp" yi~!PÍIig the uniqueness conc1usion. ::-~' ':?

Without sorne comm~,tativity assumption, the aboye rernarks need not hold; For, if we let h(x)

=,

(x - il)(x - b) = h(r) = r2

then, h(r) = f(r) +g(r),

r(J(x») = r(a o)

then

The elementf(r) is said to be the result of substituting r for x inf(x). Suffice it to say, the 'addition and multiplication used in defining f(r) are those of the ring R', not those of R[xJ. Now, suppose that f(x), g(x) are polynornials in R[xJ and rE cent R'. We lea ve the reader to pro ve that if h(x) = f(x)

Proof We need only verify that 4>r is uníque. Suppose, then, thát there is another horrtomorphism r: R[xJ --+ R satisfy in the indicated conditions and con~ider any polynornialf(x) = ao..+ alx + '" + a.x·ER[xJ. By ,assu.mptlOn, r(ak) = ak fon each coefficlent ak , while r(x k) = r(x)k = rk. Takmg stock of the fact that r is a homomorphism,

+ '" .f.' a.x·

R' by taking

f(r) = ao + alr

Theorem 7-6. Let R be a ring with identity, R' an extension ring of R, and the element rE cent R'. ,Then there is a unique homomorphism rpr: R[x] --+ R' such that rpAx) = r, rpAa) = a for all a E R.

O. '

The degree of a polynomial is used in the factorization theory of R[xJ in much the same way as the absolute value is employed in Z. For, it is through the degree concept that induction can be utilized in R[ x J to develop a polynomial counterpart of the familiar diyision algorithm. One can subsequent1y establish that the ring F[x}with'Coefficients in a field forms a EucJidean domain in which the degree functio.n}s taken to be the Euc1idean v~luatiort. ' ", ' . " . Before embarking on this program, we Msh to introduce several new ideas. To this purpose, let R be a ring with)dentity; assume further that R' is any ring contain\ng R as a subring (tb,at)s, R' is an extension of R) and let r be an arbitrary element of R'. For;.?,~~h polynornial

121

-

(a

X2 -

(a

+ b)x

;'h

ab;

+ b)r + ab.

Lacking the hypothesis that rE cení R', it cannot be conc1uded that (r - aHr - b) = r2

-

ar - rb

+ ab

will equal h(r); in other words, h(x) = f(x)g(x) does not always imply h(r) = f(r)g(r). Wheneverf(r) = O, we call the element r a root or zero ofthe polynomial f(x). Of cou!se, a given polynomial f(x) E R[x J may not ha ve a root in R; we shall see later that when R is a field, there alwaysexists an extension field R' of R in whichf(x) possesses a root. It is perhaps appropriate to point out at t~is time that the problem of obtaining all roots of a polynornial f(x) E R[ x] is equivalent to that of finding all elements r E R' for which f(x) E ker rpr' ' After this brief digression, let us now state and prove the division algorithm for polynomials.

Theorem 7-7. (Division Algorithm). Let R be a c~mmutative ring with identity and f(x), g(x) =1= O be polynomials in R[xJ, with the leading coefficientof g(x) an invertible element. Then there exist unique polynomials q(x), r(x) E R[xJ such that f(x) where either r(x)

= O or

=

q(x)g(x)

+ r(x),

deg r(x) < deg g(x).

Proof The proof is by induction on the degree off(x)~ First, notice that if f(x) = Oorf(x) =1= Oand degf(x) < deg g(x), a representation meeting the

122

POLYNOMIAL RINGS

FIRST COURSE IN RINGS AND IDIlALS

requirements oC the theorem exists on taking q(x) = O, r(x) f(x). Furthermore, iC degf(x) = deg g(x) = O, f(x) and g(x) are both elements oC the ring R, and it suffices to let q(x) = f(x)g(x)-l, r(x) = O. This being so, aS8ume that the theorem i8 true Cor polynomials of degree ' les8 than n (the induction hypothesis) and let degf(x) = n, deg g(x) m, where n ~ m ~ 1; that is,

123

We now come to a series of theorems concerning the factonzation properties of R[ xJ. Theorem 7-8. (Remainder Theorem). Let R be a commutative ring with identity. If f(x) E R[x] and a E R, then there exists a unique polynomial q(x) in R[x] such thatf(x) = (x - a)q(x) + fea). Proa!. AH this is scarcely more than an application ofthe division algorithm to the polynomials f(x) and x - ,a. We then obtain

g(x) = bo

+ blx + ... +

bmx

nt ,

bm =f= O

(n

~

m).

f(x) = (x - a)q(x)

Now, the polynomial fl(X)

= f(x)

where r(x) = O or deg r(x) < deg (x - a) that r(x) is a constant polynomial, say r(x) x leads to

- (anb~l):x:,-mg(x)

líes in R[x] and, since the coefficient of x· is a. (anb~ l)b m = O, has degree less than n. By supposition, there are polynomials ql(X), r(x) E R[x] such that j¡(x) = ql(X)g(X)

where r(x)

+

fea)

+

=

(ql(X)

r(x),

=

q(x)g(x)

(anb;;;-l)xn-"')g(x)

+

+

CoroUary., The polynomial f(x) only if a Is a.root ofj(x).

f(x)

= q(x)g(x) +

r(x),

r(x)

=

q'(x)g(x)

+

r'(x),

E R[x]

is divisible by x - a if and

Proo!. The argument proceeds by induction on the degree off(x). When degf(x) = O, the result is trivial, since f(x) cannot have any roots. Ir degf(x) = 1, for instance, f(x) = ax + b (a =f= O); 1hen f(x) has at most one root; indeed, if a is an invertible element, it follows that -a-lb is the only root off(x). Now, suppose that the theoremi'~ {rue for all polynomials of degree n - 1 and let degf(x) = n. Irf(x) h~;'a root a, the preceding corollary gives f(x) = (x - a)q(x), where the polynomial q(x) has degree n - 1. Any root a' of f(x) distinct from a must",necessarily be a root of q(x) for, by substitution,

r(x) - r'(x) = (q'{x) - q(x»)g(x).

Sincetheleadingcoefficientofg(x) is in vertible, itfollowsthat q'(x) - q(~J.~ O if and only if r(x) - r'(x) = O. With this in mind, letq'(x) - q(x{~f. O. Knowing that b;,. is not a zero divisor of R, .... ,

=

real = r,

Theorem 7-9. Let R be an integral domain andf(x) E R[x] be a nonzero polynomial of degree n. Then f(x) can have at most n distinct roots in R.

n.

where r(x) and r'(x) satisf)' the requirements of ,the theorem. Subtn:~pting, we obtain , ,

deg (q'(x) - q(x»g(x)

+

It follows in either case rE R. Substitution of a for

Let us next show that a polynomial cannot have more roots in an integral domain than its degree.

r(x)

which shows that the desired representation also exists when degf(x) As for uniqueness, suppose that

(a - a)q(a)

= 1.

as desired.

Oor deg r(x) < deg g(x). Substituting, we obtain theequation f(x)

=

+ r(x),

deg (ql(X) - q(x)) + deg g(x) . ~ deg g(x) > deg (r(x) - rl(x»)¡

a contradiction; the last inequality relies on the fact that the degr~es of r(x) and r'(x) are both less than the degree of g(x). Thus, q'(x) = q(x), .which in turn implies that r'(x) = r(x). . The polynomials q(x) and r(x)appearing in the division algorithm are called, respectively, the quotient and remainder on dividing f(x) by g(x). In this connection, it is important to observe that if g(x) is a monic poly- . nomial, or if R is taken to be a field, one need not assume that the leading coefficient of g(x) is invertible.

O = fea') = (a - al)q(a) and, since R has no zero divisors, q(a') I

=

O. From our induction hypothesis,

q(x) has at most n - 1 distinct roots. As the oniy roots of f(x) are a and those of q(x),J(x) cannot possess more than n distinct roots in R.

With this step forward we can establish CoroUary 1. Letf(x) and g(x) be two n.onzero polynomials of degree n over the integral domain R. If there exist n + 1 distínct elements ak e R (k = 1, 2, ... ,n + 1) such that f(a k ) = g(ak ), then f(x) = g(x).

122

POLYNOMIAL RINGS

FIRST COURSE IN RINGS AND IDIlALS

requirements oC the theorem exists on taking q(x) = O, r(x) f(x). Furthermore, iC degf(x) = deg g(x) = O, f(x) and g(x) are both elements oC the ring R, and it suffices to let q(x) = f(x)g(x)-l, r(x) = O. This being so, aS8ume that the theorem i8 true Cor polynomials of degree ' les8 than n (the induction hypothesis) and let degf(x) = n, deg g(x) m, where n ~ m ~ 1; that is,

123

We now come to a series of theorems concerning the factonzation properties of R[ xJ. Theorem 7-8. (Remainder Theorem). Let R be a commutative ring with identity. If f(x) E R[x] and a E R, then there exists a unique polynomial q(x) in R[x] such thatf(x) = (x - a)q(x) + fea). Proa!. AH this is scarcely more than an application ofthe division algorithm to the polynomials f(x) and x - ,a. We then obtain

g(x) = bo

+ blx + ... +

bmx

nt ,

bm =f= O

(n

~

m).

f(x) = (x - a)q(x)

Now, the polynomial fl(X)

= f(x)

where r(x) = O or deg r(x) < deg (x - a) that r(x) is a constant polynomial, say r(x) x leads to

- (anb~l):x:,-mg(x)

líes in R[x] and, since the coefficient of x· is a. (anb~ l)b m = O, has degree less than n. By supposition, there are polynomials ql(X), r(x) E R[x] such that j¡(x) = ql(X)g(X)

where r(x)

+

fea)

+

=

(ql(X)

r(x),

=

q(x)g(x)

(anb;;;-l)xn-"')g(x)

+

+

CoroUary., The polynomial f(x) only if a Is a.root ofj(x).

f(x)

= q(x)g(x) +

r(x),

r(x)

=

q'(x)g(x)

+

r'(x),

E R[x]

is divisible by x - a if and

Proo!. The argument proceeds by induction on the degree off(x). When degf(x) = O, the result is trivial, since f(x) cannot have any roots. Ir degf(x) = 1, for instance, f(x) = ax + b (a =f= O); 1hen f(x) has at most one root; indeed, if a is an invertible element, it follows that -a-lb is the only root off(x). Now, suppose that the theoremi'~ {rue for all polynomials of degree n - 1 and let degf(x) = n. Irf(x) h~;'a root a, the preceding corollary gives f(x) = (x - a)q(x), where the polynomial q(x) has degree n - 1. Any root a' of f(x) distinct from a must",necessarily be a root of q(x) for, by substitution,

r(x) - r'(x) = (q'{x) - q(x»)g(x).

Sincetheleadingcoefficientofg(x) is in vertible, itfollowsthat q'(x) - q(~J.~ O if and only if r(x) - r'(x) = O. With this in mind, letq'(x) - q(x{~f. O. Knowing that b;,. is not a zero divisor of R, .... ,

=

real = r,

Theorem 7-9. Let R be an integral domain andf(x) E R[x] be a nonzero polynomial of degree n. Then f(x) can have at most n distinct roots in R.

n.

where r(x) and r'(x) satisf)' the requirements of ,the theorem. Subtn:~pting, we obtain , ,

deg (q'(x) - q(x»g(x)

+

It follows in either case rE R. Substitution of a for

Let us next show that a polynomial cannot have more roots in an integral domain than its degree.

r(x)

which shows that the desired representation also exists when degf(x) As for uniqueness, suppose that

(a - a)q(a)

= 1.

as desired.

Oor deg r(x) < deg g(x). Substituting, we obtain theequation f(x)

=

+ r(x),

deg (ql(X) - q(x)) + deg g(x) . ~ deg g(x) > deg (r(x) - rl(x»)¡

a contradiction; the last inequality relies on the fact that the degr~es of r(x) and r'(x) are both less than the degree of g(x). Thus, q'(x) = q(x), .which in turn implies that r'(x) = r(x). . The polynomials q(x) and r(x)appearing in the division algorithm are called, respectively, the quotient and remainder on dividing f(x) by g(x). In this connection, it is important to observe that if g(x) is a monic poly- . nomial, or if R is taken to be a field, one need not assume that the leading coefficient of g(x) is invertible.

O = fea') = (a - al)q(a) and, since R has no zero divisors, q(a') I

=

O. From our induction hypothesis,

q(x) has at most n - 1 distinct roots. As the oniy roots of f(x) are a and those of q(x),J(x) cannot possess more than n distinct roots in R.

With this step forward we can establish CoroUary 1. Letf(x) and g(x) be two n.onzero polynomials of degree n over the integral domain R. If there exist n + 1 distínct elements ak e R (k = 1, 2, ... ,n + 1) such that f(a k ) = g(ak ), then f(x) = g(x).

.

~

.,'

124

FIRST COURSE IN RINGS AND IDEALS

POL YNOMIAL RINGS

p,roof. The p,olynomial h(x) = f(x) - g(x) is such ,that deg h(x) ::;; n and, by supposition, has at least n + 1 distinct roots in R., This is ,impossible unless h(x) = 0, whencef(x) = g(x). Corollary 2. Let f(x) E R[x], where R is an integral domain, and 'let S ,be any infinite subset of R. Uf(a) = O for al1 a E S, thenf(x) is the zero poly:ÍJ.omial. .. ¡. ~

Example.7-2. Consider the polynomial x P - x E Z P [x], where P is a prime number.',Now, the nonzero elements of Z p form a commutative group under muItipliéation of order P - 1. Hence, we have aP - 1 = 1, or aP = a for This is equal1y true if a = O. Our ex.ample shows that it may every a very we'Ifhappen that every element of the underlying ring is a root of a polynOIi~¡~I, yet the polynomiaJ is noi zero. "

Proof. Suppose that R[x] is not a unique factorization dornain and let S be the set of al1 nonconstant polynornials in R[x] which do not have a unique factorization into irreducible elements. Select f(x) E S to be of minimal degree. We may assurne that ' f(x)

J:

The~r~m 7-10. The polynomial ring F[x], where F is a field, forms a Euc1idean dornain.

ó(J(x)g(x») = deg (J(x)g(x») = degf(x)

•...::'

n.= deg ql(X) ~ deg q2(X) ~ ... ~ deg q.(x), ,

,

with n ~, m > O; it is further evident that ~OPi(X) = uq,(~) for any in vertible element u (otherwise, the polynomial obtained on di~iding f(x) by qJx) will have unique factorization; tbis implies that f(x) can also be factored uniquely). Let a, b be the leading coefficients of PI (x),qi:(X), respectively, and d e f i n e : > : , : g(x) = af(x) -, bpI(X)Xn- mq2(X)'" q.(x). "

~:.

On one hand, wehave g(x) = aPI(x)P2(x) ... pAx) - bpI(X)Xn- mq2(X) ... q.(x) . = pI(x)(apix) ... Pr(x) - bxn-mqix) ... q.(x»),

and, on the other hand, g(x) = aql(x)q2(X) ... q.(x) - bpI(X)Xn- m q2(X) ... q.(x) ,

+ deg g(x)

since deg g(x) ~ O. Thus, the function a Euc1idean valuation.

PI(X)P2(X) '" Pr(x) = ql(X)q2(X) '" q.(x),

m = deg PI (x) ~ deg pix) ~ ... ~ deg Pr(x),

'

Proqf. As has been noted, F[x] is an integral domain. Moreover, the function Ó defined by ó(!(x») = degf(x) for any nonzero f(x) E F[x] is a suitable Euclidean valuation. Only condition (2) of Definition 6-6 fails to be immediate. But if f(x) and g(x) are two' nonzero polynomials in F[x], Theorem 7-5 implies that

=;=

where the p¡(x) and qj(x) are a11 irreducible and

'lo.

wih;::ihe Division Algorithm at out disposal, we can prove that the ' ring Fi}: is rich in' s t r u c t u r e . ,

125

~ (aql(x) - bpI(X)xn-m)q2(X) ... qs(x):

~ degf(x) = ó(!(x») ,

Ó satisfies

the requisiteproperties of

The reader is no doubt anticipating the corol1ary below.

Now, either g(x) = 0, which forces aql(x) = bPI(X)Xn- m, or else deg g(x) < degf(x). In the latter event, g(x) must possess a unique factorization into irreducibles, sorne of which are qix), ... , q.(x) and PI (x). The net resuIt of this is that PI(X)/g(x), but PI(X) f q¡(x) for i > 1, so that

Corollary. F[ x] is a principal ideal domain; hence, a unique factorization domain. For a less existential proof of the fact that F[x] is a principal ideal domain and a considerably more precise descripÍion of its ideals, one can repeat (with appropriate modifications) the pedestrian argument used in Theorem 2-3. It will appear that any non trivial ideal l of F[ x] is of the form l = (!(x»), wheref(x) is a nonzero polynomial ofminimal degree in l. Sin ce a field is trivial1y a unique factorization domain, part of the last corolJary couid be regarded as a special case of the coming theorem.

PI(x)/(aql(x)- bpI(X)Xn- m ),

and therefore p¡(x)/aql(x). In either of the two cases considered we are able to conc1ude that PI(X) divides the product aql(x); this b~irig so, aql(x) = PI(x)h(x) for so me polynomial h(x) E R[xJ. Since R is taken to be a unique factorization domain, a has a unique factoriiation as a product of irreducible elements of R - hence, of R[ x] - say, a = C C '" C ,w here each . . d 'b' [] I 2 k C¡ 1S me UC1 le m R x. (The only factorizations of a as an element of R[x] are those it had as an element of R.) Argtiing from the representatlon

Theorem 7-11. If R is a unique factorization domain,'then so is R[x].

C I C 2 ...

-----_._----

ckql(X) = PI (x)h(x),

.

~

.,'

124

FIRST COURSE IN RINGS AND IDEALS

POL YNOMIAL RINGS

p,roof. The p,olynomial h(x) = f(x) - g(x) is such ,that deg h(x) ::;; n and, by supposition, has at least n + 1 distinct roots in R., This is ,impossible unless h(x) = 0, whencef(x) = g(x). Corollary 2. Let f(x) E R[x], where R is an integral domain, and 'let S ,be any infinite subset of R. Uf(a) = O for al1 a E S, thenf(x) is the zero poly:ÍJ.omial. .. ¡. ~

Example.7-2. Consider the polynomial x P - x E Z P [x], where P is a prime number.',Now, the nonzero elements of Z p form a commutative group under muItipliéation of order P - 1. Hence, we have aP - 1 = 1, or aP = a for This is equal1y true if a = O. Our ex.ample shows that it may every a very we'Ifhappen that every element of the underlying ring is a root of a polynOIi~¡~I, yet the polynomiaJ is noi zero. "

Proof. Suppose that R[x] is not a unique factorization dornain and let S be the set of al1 nonconstant polynornials in R[x] which do not have a unique factorization into irreducible elements. Select f(x) E S to be of minimal degree. We may assurne that ' f(x)

J:

The~r~m 7-10. The polynomial ring F[x], where F is a field, forms a Euc1idean dornain.

ó(J(x)g(x») = deg (J(x)g(x») = degf(x)

•...::'

n.= deg ql(X) ~ deg q2(X) ~ ... ~ deg q.(x), ,

,

with n ~, m > O; it is further evident that ~OPi(X) = uq,(~) for any in vertible element u (otherwise, the polynomial obtained on di~iding f(x) by qJx) will have unique factorization; tbis implies that f(x) can also be factored uniquely). Let a, b be the leading coefficients of PI (x),qi:(X), respectively, and d e f i n e : > : , : g(x) = af(x) -, bpI(X)Xn- mq2(X)'" q.(x). "

~:.

On one hand, wehave g(x) = aPI(x)P2(x) ... pAx) - bpI(X)Xn- mq2(X) ... q.(x) . = pI(x)(apix) ... Pr(x) - bxn-mqix) ... q.(x»),

and, on the other hand, g(x) = aql(x)q2(X) ... q.(x) - bpI(X)Xn- m q2(X) ... q.(x) ,

+ deg g(x)

since deg g(x) ~ O. Thus, the function a Euc1idean valuation.

PI(X)P2(X) '" Pr(x) = ql(X)q2(X) '" q.(x),

m = deg PI (x) ~ deg pix) ~ ... ~ deg Pr(x),

'

Proqf. As has been noted, F[x] is an integral domain. Moreover, the function Ó defined by ó(!(x») = degf(x) for any nonzero f(x) E F[x] is a suitable Euclidean valuation. Only condition (2) of Definition 6-6 fails to be immediate. But if f(x) and g(x) are two' nonzero polynomials in F[x], Theorem 7-5 implies that

=;=

where the p¡(x) and qj(x) are a11 irreducible and

'lo.

wih;::ihe Division Algorithm at out disposal, we can prove that the ' ring Fi}: is rich in' s t r u c t u r e . ,

125

~ (aql(x) - bpI(X)xn-m)q2(X) ... qs(x):

~ degf(x) = ó(!(x») ,

Ó satisfies

the requisiteproperties of

The reader is no doubt anticipating the corol1ary below.

Now, either g(x) = 0, which forces aql(x) = bPI(X)Xn- m, or else deg g(x) < degf(x). In the latter event, g(x) must possess a unique factorization into irreducibles, sorne of which are qix), ... , q.(x) and PI (x). The net resuIt of this is that PI(X)/g(x), but PI(X) f q¡(x) for i > 1, so that

Corollary. F[ x] is a principal ideal domain; hence, a unique factorization domain. For a less existential proof of the fact that F[x] is a principal ideal domain and a considerably more precise descripÍion of its ideals, one can repeat (with appropriate modifications) the pedestrian argument used in Theorem 2-3. It will appear that any non trivial ideal l of F[ x] is of the form l = (!(x»), wheref(x) is a nonzero polynomial ofminimal degree in l. Sin ce a field is trivial1y a unique factorization domain, part of the last corolJary couid be regarded as a special case of the coming theorem.

PI(x)/(aql(x)- bpI(X)Xn- m ),

and therefore p¡(x)/aql(x). In either of the two cases considered we are able to conc1ude that PI(X) divides the product aql(x); this b~irig so, aql(x) = PI(x)h(x) for so me polynomial h(x) E R[xJ. Since R is taken to be a unique factorization domain, a has a unique factoriiation as a product of irreducible elements of R - hence, of R[ x] - say, a = C C '" C ,w here each . . d 'b' [] I 2 k C¡ 1S me UC1 le m R x. (The only factorizations of a as an element of R[x] are those it had as an element of R.) Argtiing from the representatlon

Theorem 7-11. If R is a unique factorization domain,'then so is R[x].

C I C 2 ...

-----_._----

ckql(X) = PI (x)h(x),

126

FIRST COURSE IN RINGS AND IDEALS

with p¡(X) an irreducible, it fo11ows that each element a divides h(x). But, then,

C¡

and, in consequence, the

aq¡(x) = p¡(x)ah¡(x)

for sorne h¡(x) in R[x] or, upon canceling, q¡(x) = p¡(x)h¡(x); in other words, p¡(x)lq¡(x). Using the irreducibility of q¡(x) as a member of R[x], ' p¡(x) must be an associate of q¡(x). However, this conflicts with our original assumptions. Thus, we see that R[x] is indeed a unique factorization domain. Remark. For many years, it was an open question as to whether a power series ring over a unique factorization domain is again a unique factorization domain; a negative answer was established not long ago by Samuel [55]. To this we might add, Qn the positive side, that one can prove that the ring of formal power series oyer a principal ideal domain does in fact comprise a unique factorization domain (a not altogether trivial task).

--fo

Coming back to the coro11ary to Theorem 7-10, there is an interesting converse which deserves mention: namely, if R is an integral domain such that the polynomial ring R[xJ forms a principal ideal domain, then R is necessarily a freId. In verifying this, the main point to be proved is that any nonzero element a E R is invertible in R. By virtue of our hypothesis, the ideal generated by x and a must be principal; for instance, (x, a)

=

o =1= f(x) E R[x].

(j(x)),

Since both x, a E (j(x)), it fo11ows that a = g(x)f(x),

and

x

= h(x)f(x)

for suitable g(x), h(x) in R[~J The first of these relations signifies that degf(x) = O, say f(x) = ao;'ilrid as a result deg h(x) = 1, say h(x) = bo + b¡x (b¡ =1= O). We thus obt~in' x = ao(b o + b¡x). But this means that the product aob¡ = 1, therebY,making ao (or, equivalently, f(x») an invertible element of R. The implicatión is that the ideal (x, a) is the entire ring R[x]. It is therefore possible to write the identity element in the form 1:b: xk¡(x)

+

ak 2 (x),

with the two polynomials k¡(x), kix) E R[x]. This can only happen if Co =1= O is thé constant term of kix). In consequence, the element a has a multiplicative inverse in R, which settles the whole affair. At the heart of a11 the interesting questions on factorization in R[x] lies the idea of an irreducible polynomial, which we formulate in a rather general way as fo11ows: ac o = 1, where

Definition 7-4. Let R be an integral domain. A nonconstant polynomialf(x) E R[x] is said to be irreducible over R, or is an irreducible

POLYNOMIAL RINGS

127

polynomial in R[ x], if f(x) cannot be expressed as the product of two polynomials (in R[x]) of positive degree. Otherwise, f(x) is termed reducible in R[x].

In the case of the principal ideal domain F[x], where F is a freId, the irreducible polynomials are precisely the irreducible elements of F[ x] (reca11 that the in vertible elements of the polynomial ring F[x] are just the nonzero constant polynomials); by Theorem 5-9, these coincide with the prime elements of F[x]. Of the equivalent notions, irreducible polynomial, irreducible element, and prime element, the term "irreducible polynomial" is the one customarily preferred for F[x]' Perhaps we should emphasize that Definition 7--4 applies only to polynomials of positive degree; the constant polynomials are neither reducible nor irreducible. Thus, the factorization theory of F[x] concerns only polynomials of degree ~ 1. The dependence of Definition 7-4 upon the polynomial domain R[x] is essential. It may very we11 happen that a given polynomial is irreducible when viewedas an element of one domain, yet reducible in another. One such example is the polynomial X2 + 1; it is irreducible in R #[x], but ¡'~ducible in both C[x], where X2 + 1 = (x + i)(x and Z2[X], where x + 1 = (x + l)(x + 1). Thus, to ask merely whether a polynomial is irreducible, without specifying the coefficient ring involved, is incomplete and meaningless. More often than not, it is a formidable task to decide when a given polynomial is irreducible over a specific ringo If F is a finite field, say one of the fields of integers modulo a prime, we may actually examine a11 of the possible roots. To cite a simple illustration, the polynomial f(x) = x 3 + X + 1 is'irreducible in Z2[X]. Ifthere are any factors ofthis polynomial, at least:one must be linear. But the only possible roots for f(x) are O and 1, yetf(O) .~ f(l) = 1 =1= O, showing that no roots exist in Z2'

n,

+ b, a =1= O, is irreducible in R[x], where R ís an integral domain. Indeed, since the degree of a product of two polynomials is the sum of the degree of the factors, it fo11ows that a representation Exampl~ 7-3. Any linear polynomial ax

ax

+ b = g(x)h(x), g(x), h(x) E R[x], 1 s deg h(x) (S impbssible. This signifies

with 1 s deg g(x), reducible polynomial has degree at least 2.

that every

Example 7-4. The polynomial X2 - 2 is irreducible in Q[x], where Q as usual is the field of rational numbers. Otherwise, we would have X2 - 2 = (ax + b) (cx + d) = (ac)x 2

+

(ad

+

bc)x

+

bd,

,

.;

:~:

126

FIRST COURSE IN RINGS AND IDEALS

with p¡(X) an irreducible, it fo11ows that each element a divides h(x). But, then,

C¡

and, in consequence, the

aq¡(x) = p¡(x)ah¡(x)

for sorne h¡(x) in R[x] or, upon canceling, q¡(x) = p¡(x)h¡(x); in other words, p¡(x)lq¡(x). Using the irreducibility of q¡(x) as a member of R[x], ' p¡(x) must be an associate of q¡(x). However, this conflicts with our original assumptions. Thus, we see that R[x] is indeed a unique factorization domain. Remark. For many years, it was an open question as to whether a power series ring over a unique factorization domain is again a unique factorization domain; a negative answer was established not long ago by Samuel [55]. To this we might add, Qn the positive side, that one can prove that the ring of formal power series oyer a principal ideal domain does in fact comprise a unique factorization domain (a not altogether trivial task).

--fo

Coming back to the coro11ary to Theorem 7-10, there is an interesting converse which deserves mention: namely, if R is an integral domain such that the polynomial ring R[xJ forms a principal ideal domain, then R is necessarily a freId. In verifying this, the main point to be proved is that any nonzero element a E R is invertible in R. By virtue of our hypothesis, the ideal generated by x and a must be principal; for instance, (x, a)

=

o =1= f(x) E R[x].

(j(x)),

Since both x, a E (j(x)), it fo11ows that a = g(x)f(x),

and

x

= h(x)f(x)

for suitable g(x), h(x) in R[~J The first of these relations signifies that degf(x) = O, say f(x) = ao;'ilrid as a result deg h(x) = 1, say h(x) = bo + b¡x (b¡ =1= O). We thus obt~in' x = ao(b o + b¡x). But this means that the product aob¡ = 1, therebY,making ao (or, equivalently, f(x») an invertible element of R. The implicatión is that the ideal (x, a) is the entire ring R[x]. It is therefore possible to write the identity element in the form 1:b: xk¡(x)

+

ak 2 (x),

with the two polynomials k¡(x), kix) E R[x]. This can only happen if Co =1= O is thé constant term of kix). In consequence, the element a has a multiplicative inverse in R, which settles the whole affair. At the heart of a11 the interesting questions on factorization in R[x] lies the idea of an irreducible polynomial, which we formulate in a rather general way as fo11ows: ac o = 1, where

Definition 7-4. Let R be an integral domain. A nonconstant polynomialf(x) E R[x] is said to be irreducible over R, or is an irreducible

POLYNOMIAL RINGS

127

polynomial in R[ x], if f(x) cannot be expressed as the product of two polynomials (in R[x]) of positive degree. Otherwise, f(x) is termed reducible in R[x].

In the case of the principal ideal domain F[x], where F is a freId, the irreducible polynomials are precisely the irreducible elements of F[ x] (reca11 that the in vertible elements of the polynomial ring F[x] are just the nonzero constant polynomials); by Theorem 5-9, these coincide with the prime elements of F[x]. Of the equivalent notions, irreducible polynomial, irreducible element, and prime element, the term "irreducible polynomial" is the one customarily preferred for F[x]' Perhaps we should emphasize that Definition 7--4 applies only to polynomials of positive degree; the constant polynomials are neither reducible nor irreducible. Thus, the factorization theory of F[x] concerns only polynomials of degree ~ 1. The dependence of Definition 7-4 upon the polynomial domain R[x] is essential. It may very we11 happen that a given polynomial is irreducible when viewedas an element of one domain, yet reducible in another. One such example is the polynomial X2 + 1; it is irreducible in R #[x], but ¡'~ducible in both C[x], where X2 + 1 = (x + i)(x and Z2[X], where x + 1 = (x + l)(x + 1). Thus, to ask merely whether a polynomial is irreducible, without specifying the coefficient ring involved, is incomplete and meaningless. More often than not, it is a formidable task to decide when a given polynomial is irreducible over a specific ringo If F is a finite field, say one of the fields of integers modulo a prime, we may actually examine a11 of the possible roots. To cite a simple illustration, the polynomial f(x) = x 3 + X + 1 is'irreducible in Z2[X]. Ifthere are any factors ofthis polynomial, at least:one must be linear. But the only possible roots for f(x) are O and 1, yetf(O) .~ f(l) = 1 =1= O, showing that no roots exist in Z2'

n,

+ b, a =1= O, is irreducible in R[x], where R ís an integral domain. Indeed, since the degree of a product of two polynomials is the sum of the degree of the factors, it fo11ows that a representation Exampl~ 7-3. Any linear polynomial ax

ax

+ b = g(x)h(x), g(x), h(x) E R[x], 1 s deg h(x) (S impbssible. This signifies

with 1 s deg g(x), reducible polynomial has degree at least 2.

that every

Example 7-4. The polynomial X2 - 2 is irreducible in Q[x], where Q as usual is the field of rational numbers. Otherwise, we would have X2 - 2 = (ax + b) (cx + d) = (ac)x 2

+

(ad

+

bc)x

+

bd,

,

.;

:~:

128

POL'¡NOMIA~ RINGS

FIRST COURSE IN RINOS AND IDEALS

with the coefflcients a, b, c,d E Q. Accordingly,

ac whence c = l/a, d obtain

ad + bc

1,

=-

o=

-

now to the real field, we can obtain the form of the prime factorization in

bd

0,

- 2,

2/b. Substituting in the relation ad 2a/b

+ b/a

=

( - 2a

2

+

R#[x] (bear in mind that polynomials with coeflicients from R# are polynomials in C[x] and therefore have..rootsin C).

'1

+

bc

=

0, we

bt)/ab.

Thus -2.a + b = 0, or (b/a)2 = 2, which is im,~ossible beca use ')2 is not a'rational number. Although irreducible in Q[xlthe polynomial X2 2 isnonethelessreducibleinR#[x];inthiscase,x2 -2.,.= (x - JfJ(x + and both fáctors are in R#[x]' , 2

129

2

I

,1

Corollary 2. If f(x) E R#[x] is of positive degree, then f(x) can be factored into linearand irreducible quadratic factors.

I

Proo! Sincef(x) also belongÚo C[x],f(x) factors in C[x] into a product of linear polynomiaIs x - c~, CkE C. If CkER#, then x Ck ER#[x]. Otherwise, Ck = + bi, whe~~ bE R# and b O. But the.complex roots of real polynomials Occur in conjugate pairs (Problem 7-11), so that ck = a - bi is also a root offCx). Thus,

a

J'2)

For ease of reference more than to present rie!y concepts, let us summarize in the next theorem some of the results of prévious chapters (specifically, Theorems 5-5 and ~7) as applied to the pr~P,9pal ideal domain F[x]'

a;

(x -ck ) (x - Ck)";~

X2 -

qg:~ratic

+:

2ax

+ (a 2 +

b 2 )ER!I![x]

is a factor of f(x). The polynomi(!.l X2 - 2ax + (a 2 + b2) i8 irreducible in R#[x], since 'Ji¡Y'factorization in R#[x] is al so valid in C[x] anrl (x - ck ) (x - ck ) Is its"itnlque factorization in C[xJ.

Theorem 7-12. If F is a fiéld, the Collowing s~~tements are equiva~ent: 1) f(x) is ari irreducible polynomial in F[x];J

·',.r

An interesting remark, tobe recorded without prooÚis that if F is a finite field, the ·polynomial ririg F[x] contains irreducible polynomials of every degree (see Theorem 9-10). This may be a convenient place to introduce the notion of a primitive polynomial.

2) the principal ideal (f(x») is a maximal (prime) ideal oC F[ x]; 3) the quotient ring F[ ~]/(f(x») forms a field. The theorem on prime Cactorization of polynomials is stated now. (Unique Factorization in F[x]). Each polynomial f(x) E F[x] ,oC positive degree is the product of a nonzero element of F and irreducible monic polynomials oC F[x]. Apart from the order

Theorem 7-13.

Definition 7-5. Let R be a unique factorization domain. The content· of a nonconstant polynomial f(x) a o + a1x + ... + a" X" E R[x], denoted by the syinbol contf(x), is defined too be the greatest common divisor of Its :coefficients:

of the factors, this factorization is unique. Sufflce it to say, Theorem 7-13 can be marle more explicit for particular polynomialdomains. When we deal with polynomials over the complex Theorem of Algebra. numbers, the crucial tool is the Fundamental .

contf(x) = gcd (ao, al' '" , an ).

We cal1.f(x;)

. ' "

Theorem 7-14. (The Fundamental Theorem of AIgebra). Let C be. the fieId of complex numbers. If f(x) E C[x] ls a polynomial oC positive degree, thenf(x) has at least one root in C. Although many proofs ofthe result are available, none is strictly algebraic in nature' thus we shall assume the vatidity ofTheorem 7-14 withoui prooC. The reader win experience tittle difficulty, however, in establishing the following corollary. Corollary 1. If f(x) E C[xJ is a polynomial of degree n: > O, then f(x) can be expressed in C[x] as a product of n (not necessarily distinct) linear factors. Another way of stating the corollary aboye is that the only irreducible polynomials in C[x] are the linear polynomials. Directing our attention

I

ª primitive polynomial jC contf(x) = 1.

Viewed otherwise, Definition 7-5 asserts that a polynomialf(x) E R[x] is primitive if and only ir there is no irreducible element of R which divides all oC its coeflicients. In this connection, it may be noted that in the domain F[x] of polynomials with coefficients from a field F, every nonconstant polynomial ls primitive (indéed, there are no primes in F). The reader should also take care to' remember that the notion of greatest common divisor and, in consequence, the .content of a polynomial is not determined uniquely, but determined only to within associates. , Given a polynomialf(x) E R[x] ofpositive degree, it is possible to wríte f(x) = Cfl(X), where c E R and fl(X) is primitive; simply letc = contf(x). To a certain extent this reduces the question,of factorization in R[x] (at least, when R is a unique factorization domain) to that of primitive polynomials. By way of specific iIIustrations, we observe that f(x) = 3x 3 _

128

POL'¡NOMIA~ RINGS

FIRST COURSE IN RINOS AND IDEALS

with the coefflcients a, b, c,d E Q. Accordingly,

ac whence c = l/a, d obtain

ad + bc

1,

=-

o=

-

now to the real field, we can obtain the form of the prime factorization in

bd

0,

- 2,

2/b. Substituting in the relation ad 2a/b

+ b/a

=

( - 2a

2

+

R#[x] (bear in mind that polynomials with coeflicients from R# are polynomials in C[x] and therefore have..rootsin C).

'1

+

bc

=

0, we

bt)/ab.

Thus -2.a + b = 0, or (b/a)2 = 2, which is im,~ossible beca use ')2 is not a'rational number. Although irreducible in Q[xlthe polynomial X2 2 isnonethelessreducibleinR#[x];inthiscase,x2 -2.,.= (x - JfJ(x + and both fáctors are in R#[x]' , 2

129

2

I

,1

Corollary 2. If f(x) E R#[x] is of positive degree, then f(x) can be factored into linearand irreducible quadratic factors.

I

Proo! Sincef(x) also belongÚo C[x],f(x) factors in C[x] into a product of linear polynomiaIs x - c~, CkE C. If CkER#, then x Ck ER#[x]. Otherwise, Ck = + bi, whe~~ bE R# and b O. But the.complex roots of real polynomials Occur in conjugate pairs (Problem 7-11), so that ck = a - bi is also a root offCx). Thus,

a

J'2)

For ease of reference more than to present rie!y concepts, let us summarize in the next theorem some of the results of prévious chapters (specifically, Theorems 5-5 and ~7) as applied to the pr~P,9pal ideal domain F[x]'

a;

(x -ck ) (x - Ck)";~

X2 -

qg:~ratic

+:

2ax

+ (a 2 +

b 2 )ER!I![x]

is a factor of f(x). The polynomi(!.l X2 - 2ax + (a 2 + b2) i8 irreducible in R#[x], since 'Ji¡Y'factorization in R#[x] is al so valid in C[x] anrl (x - ck ) (x - ck ) Is its"itnlque factorization in C[xJ.

Theorem 7-12. If F is a fiéld, the Collowing s~~tements are equiva~ent: 1) f(x) is ari irreducible polynomial in F[x];J

·',.r

An interesting remark, tobe recorded without prooÚis that if F is a finite field, the ·polynomial ririg F[x] contains irreducible polynomials of every degree (see Theorem 9-10). This may be a convenient place to introduce the notion of a primitive polynomial.

2) the principal ideal (f(x») is a maximal (prime) ideal oC F[ x]; 3) the quotient ring F[ ~]/(f(x») forms a field. The theorem on prime Cactorization of polynomials is stated now. (Unique Factorization in F[x]). Each polynomial f(x) E F[x] ,oC positive degree is the product of a nonzero element of F and irreducible monic polynomials oC F[x]. Apart from the order

Theorem 7-13.

Definition 7-5. Let R be a unique factorization domain. The content· of a nonconstant polynomial f(x) a o + a1x + ... + a" X" E R[x], denoted by the syinbol contf(x), is defined too be the greatest common divisor of Its :coefficients:

of the factors, this factorization is unique. Sufflce it to say, Theorem 7-13 can be marle more explicit for particular polynomialdomains. When we deal with polynomials over the complex Theorem of Algebra. numbers, the crucial tool is the Fundamental .

contf(x) = gcd (ao, al' '" , an ).

We cal1.f(x;)

. ' "

Theorem 7-14. (The Fundamental Theorem of AIgebra). Let C be. the fieId of complex numbers. If f(x) E C[x] ls a polynomial oC positive degree, thenf(x) has at least one root in C. Although many proofs ofthe result are available, none is strictly algebraic in nature' thus we shall assume the vatidity ofTheorem 7-14 withoui prooC. The reader win experience tittle difficulty, however, in establishing the following corollary. Corollary 1. If f(x) E C[xJ is a polynomial of degree n: > O, then f(x) can be expressed in C[x] as a product of n (not necessarily distinct) linear factors. Another way of stating the corollary aboye is that the only irreducible polynomials in C[x] are the linear polynomials. Directing our attention

I

ª primitive polynomial jC contf(x) = 1.

Viewed otherwise, Definition 7-5 asserts that a polynomialf(x) E R[x] is primitive if and only ir there is no irreducible element of R which divides all oC its coeflicients. In this connection, it may be noted that in the domain F[x] of polynomials with coefficients from a field F, every nonconstant polynomial ls primitive (indéed, there are no primes in F). The reader should also take care to' remember that the notion of greatest common divisor and, in consequence, the .content of a polynomial is not determined uniquely, but determined only to within associates. , Given a polynomialf(x) E R[x] ofpositive degree, it is possible to wríte f(x) = Cfl(X), where c E R and fl(X) is primitive; simply letc = contf(x). To a certain extent this reduces the question,of factorization in R[x] (at least, when R is a unique factorization domain) to that of primitive polynomials. By way of specific iIIustrations, we observe that f(x) = 3x 3 _

130

FIRST COURSE IN RINGS AND IDEALS

POLYNOMIAL RINGS

4x + 35 is a primitivepolynomial in Z[x], while g(x) = 12x2 + 6x - 3 = 3(4x 2 + 2x :.... 1) is not a primitive element of the same, since g(x) has content 3. Here is another new concept: Suppose that l is a (proper) ideal of R, a commutative ring with identity. There is an obvious mapping v: R[x] ~ (RIlJ[x]; for any polynomialf(x) E R[x] simply apply nat¡ to the coefficients off(x), s,o that v(J(x)) = (a o + l) + (al + l)x + ... + (a n + l)xn,

131

Corollary. If R is aunique factorization domain andf(x), g(x) E R[x], then cont (J(x)g(x)) = contf(x) cont g(x). Proof. As noted earlier, we can write f(x) = afl(x), g(x) = bgl(x), where a = contf(x), b = cont g(x) and where fl(X), gl(X) are primitive in R[xJ. Therefore,f(x)g(x) = abfl (x)g 1(x). According to the theorem, the product fl(X)gl(X) is a primitive polynomial of R[x]. This entails that the content off(x)g(x) is simply ab, or, what amounts to the samething, contf(x) cont g(x).

or, more briefiy, v(J(x)) = ~::
Any unique factorization domain R, being an integral domain, possesses a field of quotients K = Qcl(R) and we may consider the ring of polynomials R[ x] as imbedded in the polynomial ring K[ x]. The next theorem deals with the relation between the irreducibility of a polynomial in R[x] as compared to its irreducibility when considered as an element of the larger ring K[ x]. ([he c1assic example of this situation is, of course, the polynomial domain Z[x] S; Q[xJ.) Before concentrating our efforts on this relationship, we require a preliminary lemma.

Theorem 7-15. Let R be a unique factorization domain and let f(x) = ao + alx + :.. + a.x n E R[x], with degf(x) > O. Then f(x) is a primitive polynomial in R[x] if and only if, for each prime element pE R, the reduction of f(x) modulo the principal ideal(p) is nonzero.

Lemma. Let R be a unique factorization domain, with field of quotients K. Given a nonconstant polynomialf(x) E K[x], there exist (nonzero) elements a, bE R and a primitive polynomial fl(X) in R[x] such that f(x) = ab-Ifl(X).

Proof. By definition, the reduction ofj(x) modulo (p) is v(J(x))

=

(a o + (p))

+ (al +

(p))x

+ ... +

(a n + (p))xn.

Thus, to say that v(J(x)) = Ofor sorne prime p E R is equivalent't9 asserting that ak E (p), or rather, plak for all k. But the latter condition' signifies that contf(x) =1= 1; hence,f(x) is not primitive. " ¡: One of the most crucial facts conceming primitive p~jynomials is Gauss's Lemma, which we prove n e x t . ' iY _ Theorem 7-16. (Gauss's Lemma). Let R be a unique,' factorization !- domain. Iff(x), g(x) are both primitive polynomials in R[~], then their . productf(x)g(x) is also primitive in R[xJ. Proof. Given a prime element pE R, (p) is a prime ideal of R, whence the quotient ring R' = Rj(p) forms an integral domain. We next consider the reduction homomorphism v modulo the principal ideal (p). Since R'[x] is an integral domain, it follows that the reduction of f(x)g(x) cannot be the zero polynomial : v(J(x)g(x)) = v(J(x))v(g(x))

=1=

O.

. The assertion of the theorem is now a direct consequence of our last resulto

Furthermore,fl(x) is unique up to invertible elements of Ras factors. Proof. Inasmuch.as K is the field of quotients of R,J(x) can be written in the form f(x) = (aob l ) + (alb11)x + ... + (anbn-l)x n,

o

where a¡, b¡ E R and b¡ =1= O. Take b to b;ni~y common multiple of the b,' for instance, b = bob l ... bn. Then b f"O and, since the coefficients ~f bf(x) alllie in R, we have bf(x) = g(x) E ~tx]. Accordingly, f(x) = b-lg(x) = ab-Ifl(X),

where fl(X) E R[x] is a primitive polyn,bmial and a = cont g(x). We emphasize thatfl(x) is of the same degreé as f(x), so cannot be invertible in R[x]. As for uniqueness, suppose thatf(x) = ab-Ifl(X) = ca- If2(X) are two representations that satisfy the conditions of the theorem. Then, adfl(x)

=

bcfz{x).

Since fl(X) and f2(X) are both primitive, the corollary to Gauss's L~mma implies that we must have ad = ubc for sorne invertible element u E R. In consequence, fl(X) = Uf2(X), showing that fl(X) is unique to within invertible factors in R.

130

FIRST COURSE IN RINGS AND IDEALS

POLYNOMIAL RINGS

4x + 35 is a primitivepolynomial in Z[x], while g(x) = 12x2 + 6x - 3 = 3(4x 2 + 2x :.... 1) is not a primitive element of the same, since g(x) has content 3. Here is another new concept: Suppose that l is a (proper) ideal of R, a commutative ring with identity. There is an obvious mapping v: R[x] ~ (RIlJ[x]; for any polynomialf(x) E R[x] simply apply nat¡ to the coefficients off(x), s,o that v(J(x)) = (a o + l) + (al + l)x + ... + (a n + l)xn,

131

Corollary. If R is aunique factorization domain andf(x), g(x) E R[x], then cont (J(x)g(x)) = contf(x) cont g(x). Proof. As noted earlier, we can write f(x) = afl(x), g(x) = bgl(x), where a = contf(x), b = cont g(x) and where fl(X), gl(X) are primitive in R[xJ. Therefore,f(x)g(x) = abfl (x)g 1(x). According to the theorem, the product fl(X)gl(X) is a primitive polynomial of R[x]. This entails that the content off(x)g(x) is simply ab, or, what amounts to the samething, contf(x) cont g(x).

or, more briefiy, v(J(x)) = ~::
Any unique factorization domain R, being an integral domain, possesses a field of quotients K = Qcl(R) and we may consider the ring of polynomials R[ x] as imbedded in the polynomial ring K[ x]. The next theorem deals with the relation between the irreducibility of a polynomial in R[x] as compared to its irreducibility when considered as an element of the larger ring K[ x]. ([he c1assic example of this situation is, of course, the polynomial domain Z[x] S; Q[xJ.) Before concentrating our efforts on this relationship, we require a preliminary lemma.

Theorem 7-15. Let R be a unique factorization domain and let f(x) = ao + alx + :.. + a.x n E R[x], with degf(x) > O. Then f(x) is a primitive polynomial in R[x] if and only if, for each prime element pE R, the reduction of f(x) modulo the principal ideal(p) is nonzero.

Lemma. Let R be a unique factorization domain, with field of quotients K. Given a nonconstant polynomialf(x) E K[x], there exist (nonzero) elements a, bE R and a primitive polynomial fl(X) in R[x] such that f(x) = ab-Ifl(X).

Proof. By definition, the reduction ofj(x) modulo (p) is v(J(x))

=

(a o + (p))

+ (al +

(p))x

+ ... +

(a n + (p))xn.

Thus, to say that v(J(x)) = Ofor sorne prime p E R is equivalent't9 asserting that ak E (p), or rather, plak for all k. But the latter condition' signifies that contf(x) =1= 1; hence,f(x) is not primitive. " ¡: One of the most crucial facts conceming primitive p~jynomials is Gauss's Lemma, which we prove n e x t . ' iY _ Theorem 7-16. (Gauss's Lemma). Let R be a unique,' factorization !- domain. Iff(x), g(x) are both primitive polynomials in R[~], then their . productf(x)g(x) is also primitive in R[xJ. Proof. Given a prime element pE R, (p) is a prime ideal of R, whence the quotient ring R' = Rj(p) forms an integral domain. We next consider the reduction homomorphism v modulo the principal ideal (p). Since R'[x] is an integral domain, it follows that the reduction of f(x)g(x) cannot be the zero polynomial : v(J(x)g(x)) = v(J(x))v(g(x))

=1=

O.

. The assertion of the theorem is now a direct consequence of our last resulto

Furthermore,fl(x) is unique up to invertible elements of Ras factors. Proof. Inasmuch.as K is the field of quotients of R,J(x) can be written in the form f(x) = (aob l ) + (alb11)x + ... + (anbn-l)x n,

o

where a¡, b¡ E R and b¡ =1= O. Take b to b;ni~y common multiple of the b,' for instance, b = bob l ... bn. Then b f"O and, since the coefficients ~f bf(x) alllie in R, we have bf(x) = g(x) E ~tx]. Accordingly, f(x) = b-lg(x) = ab-Ifl(X),

where fl(X) E R[x] is a primitive polyn,bmial and a = cont g(x). We emphasize thatfl(x) is of the same degreé as f(x), so cannot be invertible in R[x]. As for uniqueness, suppose thatf(x) = ab-Ifl(X) = ca- If2(X) are two representations that satisfy the conditions of the theorem. Then, adfl(x)

=

bcfz{x).

Since fl(X) and f2(X) are both primitive, the corollary to Gauss's L~mma implies that we must have ad = ubc for sorne invertible element u E R. In consequence, fl(X) = Uf2(X), showing that fl(X) is unique to within invertible factors in R.

132 .

POLYNOMIAL RINGS

FIRST COURSE IN RIN6S A:ND f·DEALS

133

Theorern 7-17. Let R be a unique factorization domain, with field of quotients K. Iff(x) E R[x] is an irreducible primitive polynomial, then it is also irreducible as an elernent of K[x]. .

Now consider the reduction off(x) modulo the ideal P. Invoking hypothesis (2), it can be inferred that

Proa! Assume to the contrary that f(x) is reducible over K. Then, f(x) g(x)h(x), where the polynomials g(x), h(x) are in K[x] and are of

Since tlfe polynomial ring (R/P)[.ic] comprises an integral domain, the only possible factorizations of (a n + P)x" are into linear factors. This being so, a moment's refiection shows that

positive degree. By virtue of the lemma just pro ven, g(x) ~ ab-1g1(X),

with a, b, e, d E R

ea- 1h 1(x),

h(x)

anQ.~l(x), h1(x) primitivé in R[x]. ',:':; bdf(x)

Thus,

y;.. / . f(x) = ug1(x)h1(x). Since deg gl(X) = deg g(x) > O, deg h}{x) deg h(x) > O, the o¡¡tcome is a nontrivial factorizatíon off(x) in R[x], contrary to hypothesis. There is an obvious converse to Theorem 7-17, viz.: if the primitive polynomiaI f(x) E R[x] is irreducible as an eIement oC K[x]: it is als.o irreducible in R[xJ. This isjustmed by the fact that R[x] (or an lsomorphlc . copy thereof) appears naturally as a subring oC' K [x] ; thus, if f(x) were reducible in R[x], it would obviously be reducible in tbe larger ring K[x]. Our remarks lead to the following conclusion: Given a primitive polynomial f(x) E R[x], R a unique factorization domain, f(x) is irredúcible in R[x] if and only iff(x) is irreducible in K[xJ. '., Our next concern is a generalízation oC a famous theorem of Elsenstem dealíng with the problem oC irreducibility (this result is oL fundamental importance in the c1assical theory of polynomials with integral coefficients). The generalization which wehave in mind is formulated below. Theorem 7-18. Let R be an integral domain and the nonconstant polynomialf(x) ao + a1x + ... + anx n E R[xJ. Suppose that there exists a prime ideal P of R such that 3) ao f/: p2.

2) ak E P for O :;;; k < n,

Thenf(x) Ís irreducible in R[xJ. Proo! Assume that, cQntrary to. assertion, f(x) is reducible in R[x]; say, f(x) = g(x)h(x) ror polynomials g(x), h(x) E R[x], where g(x) h(x)

= bo + b1x + ... + brxr, = eo + e1x + ... + csx'

v(g(x»)

=

(br ,+ P)x',

v(h(x))

=

(e.

+

+ P)x".

f)x·.

TIÍis mearts that the constant terms of these reductions are zero; that is,

= aeg1(x)h1(x).

Now, Gauss's Letnnl~~~sserts that the product gl(x)h1(x) is a primitlve polynomialinR[x], Whbncef(x)andg1(x)h1(x) differ by aninvertibleelement ofR: .:,!. '

1) an f/: P"

v(g(x»)v(h(x») = v(J(x») = (a n

. '-~, I

bo

+P =

eo

+P

= P.

l.\lt9gether we have proved that both bo, eo E P, reveaHng at the s.ame time ,.¡j:;: iháf ao = boco E p2, which is untenable by (3). Accordingly, no such f;].9t~rization oC f(x) can occur, and f(x) is indeed irreduci,ple in R[ x J. Theorem '7-18 leads almost immediately to the Eisenstein test for . irredtÍcibility. CoroUary. (The Eisenstein Criterion).. Let R be a unique factorization domain and K be its field .of quotients. Letf(x) = ao + a1 x + '" + anx" be a nonconstant polynomial in R[xJ. Suppose further ,that for plak Cor O :;;; k < n, and p2 ari. Then,j(x) is sorne prime ¡j E R, p irreducible in K[xJ.

tan'

t

Proa! We already know tbat (p) is a prime ideal of R. Taking stock of the theorem,J(x) is an irreducible polynomiaI of R[x]; hence, of K[x] (at

tbis point a direct appeal is made to Theorem 7-17). This is probably a good time at whichto examine sorne examples. Example 7-5. Consider the monic polynomial f(x) = xn

+

aEZ[x]

(n >'1),

where a f ± 1 is a nonzero square-Cree integer. For any prime p díviding a, p is certainly a factor of a11 the coefficients except the leading one, and our hypothesis ensures that p" f a. Thus, f(x) fulfils Eisensteín 's criterion, and so lS irreducible over Q. Incidental1y, this example shows that there are irreducible'polynomiaIs in Q[x] of every degree. Ontheotherhand,noticethatx4 + 4 = (X2 + 2x + 2)(x 2 - 2x + 2); one should not expect Theorem 7-18 to lead to a decision in this case since ' , of course, 4 fails to be a square-free integer. Example7-6. Eisenstein's test is not directly applicable to the cyclotonic

(r

+s

= n; r, s

> O).

polynomial

132 .

POLYNOMIAL RINGS

FIRST COURSE IN RIN6S A:ND f·DEALS

133

Theorern 7-17. Let R be a unique factorization domain, with field of quotients K. Iff(x) E R[x] is an irreducible primitive polynomial, then it is also irreducible as an elernent of K[x]. .

Now consider the reduction off(x) modulo the ideal P. Invoking hypothesis (2), it can be inferred that

Proa! Assume to the contrary that f(x) is reducible over K. Then, f(x) g(x)h(x), where the polynomials g(x), h(x) are in K[x] and are of

Since tlfe polynomial ring (R/P)[.ic] comprises an integral domain, the only possible factorizations of (a n + P)x" are into linear factors. This being so, a moment's refiection shows that

positive degree. By virtue of the lemma just pro ven, g(x) ~ ab-1g1(X),

with a, b, e, d E R

ea- 1h 1(x),

h(x)

anQ.~l(x), h1(x) primitivé in R[x]. ',:':; bdf(x)

Thus,

y;.. / . f(x) = ug1(x)h1(x). Since deg gl(X) = deg g(x) > O, deg h}{x) deg h(x) > O, the o¡¡tcome is a nontrivial factorizatíon off(x) in R[x], contrary to hypothesis. There is an obvious converse to Theorem 7-17, viz.: if the primitive polynomiaI f(x) E R[x] is irreducible as an eIement oC K[x]: it is als.o irreducible in R[xJ. This isjustmed by the fact that R[x] (or an lsomorphlc . copy thereof) appears naturally as a subring oC' K [x] ; thus, if f(x) were reducible in R[x], it would obviously be reducible in tbe larger ring K[x]. Our remarks lead to the following conclusion: Given a primitive polynomial f(x) E R[x], R a unique factorization domain, f(x) is irredúcible in R[x] if and only iff(x) is irreducible in K[xJ. '., Our next concern is a generalízation oC a famous theorem of Elsenstem dealíng with the problem oC irreducibility (this result is oL fundamental importance in the c1assical theory of polynomials with integral coefficients). The generalization which wehave in mind is formulated below. Theorem 7-18. Let R be an integral domain and the nonconstant polynomialf(x) ao + a1x + ... + anx n E R[xJ. Suppose that there exists a prime ideal P of R such that 3) ao f/: p2.

2) ak E P for O :;;; k < n,

Thenf(x) Ís irreducible in R[xJ. Proo! Assume that, cQntrary to. assertion, f(x) is reducible in R[x]; say, f(x) = g(x)h(x) ror polynomials g(x), h(x) E R[x], where g(x) h(x)

= bo + b1x + ... + brxr, = eo + e1x + ... + csx'

v(g(x»)

=

(br ,+ P)x',

v(h(x))

=

(e.

+

+ P)x".

f)x·.

TIÍis mearts that the constant terms of these reductions are zero; that is,

= aeg1(x)h1(x).

Now, Gauss's Letnnl~~~sserts that the product gl(x)h1(x) is a primitlve polynomialinR[x], Whbncef(x)andg1(x)h1(x) differ by aninvertibleelement ofR: .:,!. '

1) an f/: P"

v(g(x»)v(h(x») = v(J(x») = (a n

. '-~, I

bo

+P =

eo

+P

= P.

l.\lt9gether we have proved that both bo, eo E P, reveaHng at the s.ame time ,.¡j:;: iháf ao = boco E p2, which is untenable by (3). Accordingly, no such f;].9t~rization oC f(x) can occur, and f(x) is indeed irreduci,ple in R[ x J. Theorem '7-18 leads almost immediately to the Eisenstein test for . irredtÍcibility. CoroUary. (The Eisenstein Criterion).. Let R be a unique factorization domain and K be its field .of quotients. Letf(x) = ao + a1 x + '" + anx" be a nonconstant polynomial in R[xJ. Suppose further ,that for plak Cor O :;;; k < n, and p2 ari. Then,j(x) is sorne prime ¡j E R, p irreducible in K[xJ.

tan'

t

Proa! We already know tbat (p) is a prime ideal of R. Taking stock of the theorem,J(x) is an irreducible polynomiaI of R[x]; hence, of K[x] (at

tbis point a direct appeal is made to Theorem 7-17). This is probably a good time at whichto examine sorne examples. Example 7-5. Consider the monic polynomial f(x) = xn

+

aEZ[x]

(n >'1),

where a f ± 1 is a nonzero square-Cree integer. For any prime p díviding a, p is certainly a factor of a11 the coefficients except the leading one, and our hypothesis ensures that p" f a. Thus, f(x) fulfils Eisensteín 's criterion, and so lS irreducible over Q. Incidental1y, this example shows that there are irreducible'polynomiaIs in Q[x] of every degree. Ontheotherhand,noticethatx4 + 4 = (X2 + 2x + 2)(x 2 - 2x + 2); one should not expect Theorem 7-18 to lead to a decision in this case since ' , of course, 4 fails to be a square-free integer. Example7-6. Eisenstein's test is not directly applicable to the cyclotonic

(r

+s

= n; r, s

> O).

polynomial

134
xl' -J =X -

1

=

p X - 1

+

XP-~

+ ... +

X

+ 1 EZ[X]

(p prime),

beca use no suitable prime is available. Trus problem is resolved by the observation that
With these operations, the set R [x, y] beco mes a ríng contaíning R (or rather an isomorphic copy of R) as a subring. The {total) degree of a nonzero polynomial m

f(x, y) =

n

¿ ¿

aljxiyi

;=0 j=O


(x

+ IV - 1 = (xl' + (f)x p - 1 + ... + px)/x + 1) - 1 x p - 1 + (Dx P- z + ... + pE Z[xJ.

If the Eisenstein criterion is now applied, it ls easy to see that all the requirements for the irreducibility of
Starting with a ring R we can first form the polynomíal ring R[x], with indeterminant x,and then the polynomial nng (R[x])[y] in another indeterminant y. As the notation indicates, the elements of (R[x ])[Y] are simply polynomials g = fo(x)

+

deg (j(x, y)g(x, y)) = degftx, y)

+ ... + amk~mk

From this rule, one can subsequentJy establish that whenever R forms an integral domain, then so does the polynomial ring R[x, y]. Rather than get involved in an elaborate discussion of these matters, we content ourselveswith looking at two examples. Example7-7. To illustrate that the ideal structure of the ring F[x, (F a field) is more complicated than that of F[x], let us show that F[x,

] = {f(x, y)x (aij

E

R).

Consequent1y, g c~n~1;le rewritten as a polynomial in x and y, g

g(x, y) =

¿m ¿" aijxiyi,

i=O

+ deg g(x, y).

y] y]

is not a principal ideal domain. This is accomplished by establishing that ' the ideal] = (x, y) is not principal, where

+ fl(X)y + ... + fn(x)y",

aux

°

is the largest of the integers i + j for which the coefficient aij :f= and is depoted, as before, by degf(x, y). Without going into details here, let us simply state that it is possible to obtain inequalities involying degrees analogous to those ofTheorem 7-5; in particular, if R is an integral domaín, we still have .

where each coefficient.fk(x) E R[x], so that .fk(x) ,= a Ok

135

POLYNOMIAL RINOS

FIRST COURSE IN RINOS AND IDEALS

j=O

with m, n nonnegati;~ integers and aij elements of R (one makes the obvíous conventions that aooxOyO = a oo , a;oxiyO = aiox i and aOjx°yi = aOjyi). In accordance with ttadition, we shall hereafter denote (R[x])[y] by R[x, y] and refer to the members oí thís set as polynomials over R in two indeterminants x and y. ,,' Two such polynomials with coefficients ai¡ and bij are by de:finition equal if al} = bij for all i and j. Addit'ion of polynomials is performed termwise, while multiplication is given by the rule:

+

g(x, y)ylf(x, y), g(x, y) E F[x,

y]} .

• Notice tbat the elements of this ideal are just the polynomials in F[x, y] having zero constant t e r m . , Suppose tbat ] was actually principai, say ] = (h(x, y»), where" deg h{x, y) ~ 1. Since both x, y E ], there would exist polynomials f(x, y);, g(x, y) i11 F[x, y] satisfying x = f{x, y)h{x, y),

y = g(x, y)h(x, y).

Now, degx = degy = 1, which impliesthatdegh(x, y) = 1, anddegf(x,y) =, deg g{x, y) = O; what amounts to the same thing, x = ah{x, y),

y = bh(x, y)

(a,

bE F).

Moreover, h{x, y) musí be a linear polynomíal; for instance,

= Co + C1X + CzY (C¡ E F). :f= 0, then x cannot be a multiple of h(x, y),

h(x, y)

where

But if the coefficient C2 and if C I :f= 0, Y is not a multiple of h(x, y). This being the case, we conclude that C l = C2 = 0, a contradiction to the linearity of h(x, y), and so ] does not form a principal ideal.

134
xl' -J =X -

1

=

p X - 1

+

XP-~

+ ... +

X

+ 1 EZ[X]

(p prime),

beca use no suitable prime is available. Trus problem is resolved by the observation that
With these operations, the set R [x, y] beco mes a ríng contaíning R (or rather an isomorphic copy of R) as a subring. The {total) degree of a nonzero polynomial m

f(x, y) =

n

¿ ¿

aljxiyi

;=0 j=O


(x

+ IV - 1 = (xl' + (f)x p - 1 + ... + px)/x + 1) - 1 x p - 1 + (Dx P- z + ... + pE Z[xJ.

If the Eisenstein criterion is now applied, it ls easy to see that all the requirements for the irreducibility of
Starting with a ring R we can first form the polynomíal ring R[x], with indeterminant x,and then the polynomial nng (R[x])[y] in another indeterminant y. As the notation indicates, the elements of (R[x ])[Y] are simply polynomials g = fo(x)

+

deg (j(x, y)g(x, y)) = degftx, y)

+ ... + amk~mk

From this rule, one can subsequentJy establish that whenever R forms an integral domain, then so does the polynomial ring R[x, y]. Rather than get involved in an elaborate discussion of these matters, we content ourselveswith looking at two examples. Example7-7. To illustrate that the ideal structure of the ring F[x, (F a field) is more complicated than that of F[x], let us show that F[x,

] = {f(x, y)x (aij

E

R).

Consequent1y, g c~n~1;le rewritten as a polynomial in x and y, g

g(x, y) =

¿m ¿" aijxiyi,

i=O

+ deg g(x, y).

y] y]

is not a principal ideal domain. This is accomplished by establishing that ' the ideal] = (x, y) is not principal, where

+ fl(X)y + ... + fn(x)y",

aux

°

is the largest of the integers i + j for which the coefficient aij :f= and is depoted, as before, by degf(x, y). Without going into details here, let us simply state that it is possible to obtain inequalities involying degrees analogous to those ofTheorem 7-5; in particular, if R is an integral domaín, we still have .

where each coefficient.fk(x) E R[x], so that .fk(x) ,= a Ok

135

POLYNOMIAL RINOS

FIRST COURSE IN RINOS AND IDEALS

j=O

with m, n nonnegati;~ integers and aij elements of R (one makes the obvíous conventions that aooxOyO = a oo , a;oxiyO = aiox i and aOjx°yi = aOjyi). In accordance with ttadition, we shall hereafter denote (R[x])[y] by R[x, y] and refer to the members oí thís set as polynomials over R in two indeterminants x and y. ,,' Two such polynomials with coefficients ai¡ and bij are by de:finition equal if al} = bij for all i and j. Addit'ion of polynomials is performed termwise, while multiplication is given by the rule:

+

g(x, y)ylf(x, y), g(x, y) E F[x,

y]} .

• Notice tbat the elements of this ideal are just the polynomials in F[x, y] having zero constant t e r m . , Suppose tbat ] was actually principai, say ] = (h(x, y»), where" deg h{x, y) ~ 1. Since both x, y E ], there would exist polynomials f(x, y);, g(x, y) i11 F[x, y] satisfying x = f{x, y)h{x, y),

y = g(x, y)h(x, y).

Now, degx = degy = 1, which impliesthatdegh(x, y) = 1, anddegf(x,y) =, deg g{x, y) = O; what amounts to the same thing, x = ah{x, y),

y = bh(x, y)

(a,

bE F).

Moreover, h{x, y) musí be a linear polynomíal; for instance,

= Co + C1X + CzY (C¡ E F). :f= 0, then x cannot be a multiple of h(x, y),

h(x, y)

where

But if the coefficient C2 and if C I :f= 0, Y is not a multiple of h(x, y). This being the case, we conclude that C l = C2 = 0, a contradiction to the linearity of h(x, y), and so ] does not form a principal ideal.

136

POLYNOMIAL RINGS

FIRST COURSE IN RI-NGS AND IDEALS

that every field F is an extension of a field isomorphic to Q or to Zp' according as the characteristic of F is Oor a prime p. Assume that F is an extension field of a field F and let theelements r1' rz, ... ,rn all lie in P. The subfield of F' generated by the subset F v {r 1, rz, ... , rn} is customarily denoted by F(r 1, rz' ... , rn):

Anotber point worth mentioning i~ that since F[ x] constitutes a unique factorization domain, so does (F[x])[y] = F[x, y] (Theorem 7-11). The present situation thus furnishes us with an illustration of a unique factorization domain which is not a principal ideal domain. Example 7-8. This example is given to substantiate a cIaim made earlier that a primary ideal need not be a power of a prime ideal. Once again, consider the ideal 1 = (x, y) of. the ring F[ x, y],. where F is a field,. If the polynoinial f(x, y) ~ 1, f(x, y) necessarily ~as a nonzen'> constant, term .a o· But ci o lies in the ideal (l,j(x, yJ) and so 1 = a 1a o E (l,j(~, Y»),for~mg this ideal to be the entire ringo In consequence, 1 = (x, y) IS a,'I?axlmal (hence, prime) ideal of the ring F[x, yJ. (The maximality of 1 CQúld otherwise be deduced from the fact that it is the kernel of the substitutibri homomorphismf(x, y) ~ j(O, O).) ),',,: . . Next, let us look at the ideal (X Z, y) of F[x, y]' As the readet;~ay venfy

F(r 1, rz,,,., rn) = n {KIK Isa subfield of.F'; F S K; ri E K}. Thus, F(r 1; rz, ... , rn) is an extension field6f F containing the elements ri (cIearIy it is the smallest such extension) aq.d one speaks of F(r l' r z, ... , r n) asbeing obtained by adjoining the ri to F, ¿k.by adjunction of the elements ri to F. The purpose of the coming theoJ~~ is to determine, up to isomorphism, thestructure of aH simple exten~i9)iftelds, that is, extension fields F(r) arising from a field F by the adjunctioH'bf a single element r. Now, for each element rE F ' , we havi{!J.f.our disposal the substitution homomorphism I/Jr: F[ x] ~ F' ; the reade'i"will recall that' this is defined by taking
o

IZ

= (X Z, xy, yZ)

S

(X Z, y) S l.

Z Inasmuch as -J(X Z, y) = 1, Problem 25, Chapter 5, g~ara~tees tbitt (X , y) is primary. A straightforward argument shows that (x , y) IS not the power bf any prime ideal ofF[x, yJ. For, in the contrary case, (X Z, y) = pn, where Pis a prime ideal and n ~ 1. Since pn S l,with J prime, we may appeal to Problem 30, Chapter 5, to concIúde that P S; l. By the same token, the incIusions IZ S pn S P, coupled with the fact that Pis a prime ideal, yield . 1 S P. Hence, J = P, so that 1" = (x 2 , y). Now, the element x E 1, while x ~ (x 2 , y), implying that n ~ 1. On the other hand~ y E (~Z, y), bu! y ~ ¡z = ' (x 2 , xy, l), which mean s ¡z e (x 2 , y) e l. These mcIuslOn relatIons show that it is impossible to have In == (x 2 , y) for any n ~ L

F[r] = {f(r)lf(x) E F[x]}.

:'.,.

Let us cIose this phase of our investigation by saying that there is no . difficulty in extending the aboye remarks to polynomials in a finite set of indeterminants. For any ring R,just define recursively R[x 1, X2'

...

,x~]

=

(R[x l' XZ' ... , Xn -l])[Xn J.

It would not be out of place to devote the remainder of this chapter to the matter of field extensions (most notably, algebraic extensions) and splitting fields. The concepts are presen ted he re partly for their own interest and partly to laya foundation. for a proof of the celebrated Wedder?urn theorem on finite 'division rings (Theorem 9-11). We shalI have neIther occasion nor space for more than a passing study, and certain topics are touched upon lightly. . . Byan extension F' ofa field F, we simply mean any field WhlCh contams F as a subfield. For instance, the field of real numbers is an extension of Q, the rational number field. In view of Theorem 4-12, it may be remarked

137

l·

¡he set F[r] forms an integral domain (being a subring. of the field P) and therefore has a·field of quotients K = Qc¡(F[r]) in P. It is apparent that F v {r} S F[r] S K. But F(r) is the smallest subfield of F' to contain both F and r, whence F(r) S K. On the other hand, any subring of F' which contains F and r will necessarily contain the elements of F[r]; in . particular, F[r] S F(r). Since F(r) is a subfield of F', it must also contain alI the quotients of elements in F[r J. Thus, K S F(r) and equality follows. This leads to the more constructive description of F(r) as F(r) = Qc¡(F[r]). The key to cIassifying simple extensions is the nature of the kernel of the substitution homomorphisIIl. .. (b,ear in niind that ker
Proof. By the Fundamental Homomorphism Theorem, we know that F[r] ~ F[x]jker
136

POLYNOMIAL RINGS

FIRST COURSE IN RI-NGS AND IDEALS

that every field F is an extension of a field isomorphic to Q or to Zp' according as the characteristic of F is Oor a prime p. Assume that F is an extension field of a field F and let theelements r1' rz, ... ,rn all lie in P. The subfield of F' generated by the subset F v {r 1, rz, ... , rn} is customarily denoted by F(r 1, rz' ... , rn):

Anotber point worth mentioning i~ that since F[ x] constitutes a unique factorization domain, so does (F[x])[y] = F[x, y] (Theorem 7-11). The present situation thus furnishes us with an illustration of a unique factorization domain which is not a principal ideal domain. Example 7-8. This example is given to substantiate a cIaim made earlier that a primary ideal need not be a power of a prime ideal. Once again, consider the ideal 1 = (x, y) of. the ring F[ x, y],. where F is a field,. If the polynoinial f(x, y) ~ 1, f(x, y) necessarily ~as a nonzen'> constant, term .a o· But ci o lies in the ideal (l,j(x, yJ) and so 1 = a 1a o E (l,j(~, Y»),for~mg this ideal to be the entire ringo In consequence, 1 = (x, y) IS a,'I?axlmal (hence, prime) ideal of the ring F[x, yJ. (The maximality of 1 CQúld otherwise be deduced from the fact that it is the kernel of the substitutibri homomorphismf(x, y) ~ j(O, O).) ),',,: . . Next, let us look at the ideal (X Z, y) of F[x, y]' As the readet;~ay venfy

F(r 1, rz,,,., rn) = n {KIK Isa subfield of.F'; F S K; ri E K}. Thus, F(r 1; rz, ... , rn) is an extension field6f F containing the elements ri (cIearIy it is the smallest such extension) aq.d one speaks of F(r l' r z, ... , r n) asbeing obtained by adjoining the ri to F, ¿k.by adjunction of the elements ri to F. The purpose of the coming theoJ~~ is to determine, up to isomorphism, thestructure of aH simple exten~i9)iftelds, that is, extension fields F(r) arising from a field F by the adjunctioH'bf a single element r. Now, for each element rE F ' , we havi{!J.f.our disposal the substitution homomorphism I/Jr: F[ x] ~ F' ; the reade'i"will recall that' this is defined by taking
o

IZ

= (X Z, xy, yZ)

S

(X Z, y) S l.

Z Inasmuch as -J(X Z, y) = 1, Problem 25, Chapter 5, g~ara~tees tbitt (X , y) is primary. A straightforward argument shows that (x , y) IS not the power bf any prime ideal ofF[x, yJ. For, in the contrary case, (X Z, y) = pn, where Pis a prime ideal and n ~ 1. Since pn S l,with J prime, we may appeal to Problem 30, Chapter 5, to concIúde that P S; l. By the same token, the incIusions IZ S pn S P, coupled with the fact that Pis a prime ideal, yield . 1 S P. Hence, J = P, so that 1" = (x 2 , y). Now, the element x E 1, while x ~ (x 2 , y), implying that n ~ 1. On the other hand~ y E (~Z, y), bu! y ~ ¡z = ' (x 2 , xy, l), which mean s ¡z e (x 2 , y) e l. These mcIuslOn relatIons show that it is impossible to have In == (x 2 , y) for any n ~ L

F[r] = {f(r)lf(x) E F[x]}.

:'.,.

Let us cIose this phase of our investigation by saying that there is no . difficulty in extending the aboye remarks to polynomials in a finite set of indeterminants. For any ring R,just define recursively R[x 1, X2'

...

,x~]

=

(R[x l' XZ' ... , Xn -l])[Xn J.

It would not be out of place to devote the remainder of this chapter to the matter of field extensions (most notably, algebraic extensions) and splitting fields. The concepts are presen ted he re partly for their own interest and partly to laya foundation. for a proof of the celebrated Wedder?urn theorem on finite 'division rings (Theorem 9-11). We shalI have neIther occasion nor space for more than a passing study, and certain topics are touched upon lightly. . . Byan extension F' ofa field F, we simply mean any field WhlCh contams F as a subfield. For instance, the field of real numbers is an extension of Q, the rational number field. In view of Theorem 4-12, it may be remarked

137

l·

¡he set F[r] forms an integral domain (being a subring. of the field P) and therefore has a·field of quotients K = Qc¡(F[r]) in P. It is apparent that F v {r} S F[r] S K. But F(r) is the smallest subfield of F' to contain both F and r, whence F(r) S K. On the other hand, any subring of F' which contains F and r will necessarily contain the elements of F[r]; in . particular, F[r] S F(r). Since F(r) is a subfield of F', it must also contain alI the quotients of elements in F[r J. Thus, K S F(r) and equality follows. This leads to the more constructive description of F(r) as F(r) = Qc¡(F[r]). The key to cIassifying simple extensions is the nature of the kernel of the substitution homomorphisIIl. .. (b,ear in niind that ker
Proof. By the Fundamental Homomorphism Theorem, we know that F[r] ~ F[x]jker
138

FIRST COURSE IN RINGS AND IDEALS

POLYNOMIAL RINGS

139

of F[ X] is not mapped onto zero. Now two possibilities arise, eíther ker 4Jr = {O} or ker 4Jr 1-. {O}. Suppose first that ker 4Jr = {O}. Then the homomorphism 4Jr is actualIy an isomorphism and F[r] ~ F[x]. In the situation at hand, F[r] is not a field, since F[x] fails to be one. However, from Theorem 4-11, we know that the quotient fields of F[r] and F[x] are isomorphic under a mapping induced by the substitution homomorphism. Since F(r) is the quotient 'field of F[r], one thus obtains

over F. Thus, a rational function (in an indeterminant x) over F can be written as a quotient f(x)/g(x), where f(x) and g(x) 1- O are both polynomials in F[x].

F(r) ~ QCI(F[x]).

This theorem completely determines the structure of simple transcendental extensions over F', they are all isomorphic . to the field of rational ../2 functions over F, and, hence, to each other. Thus, for lDstance, Q(n) ~ Q(2 ). As regards simple algebraic extensions, we have

If the kernel of 4Jr is nonzero, then ker 4Jr = (J(x)) for sorne irreducible polynomial (prime element) of F[x], where f(x) can be taken to be monic. Because every nonzero prime ideal of F[x] is maximal, F[x]/(J(x)) forms a field and the same will b,e true of its isomorphic image F[r]. But F(r) is the smallest field to contain 60th F and r, from which it folIows that F[r] = F(r); this leads to the isomorphism F(r) ~ F[x ]/(J(x)).

As ker 4Jr = (J(x)), a polynomial g(x) E F[ x] has r as a root if and only if g(x) is divisible by f(x). Accordingly, if g(x) is any monic irreducible polynomial in F[x] having r as a root, each of f(x) and g(x) divides the other; since both of these polynomials are monic, this is possible only if f(x) = g(x). Thus,f(x) is unique, as asserted in the theorem. . Another virtue of the substitution homomorphism is that it permits us to put the elements of an extension field into one of two essentialIy different categories: -Definition 7-6. Let F' be an exte~sion field of a>ifi~ld F. An element rE F' is said to be algebraic over F if ker 4Jr {O}; otherwise, r is termed transcendental over F. 1!-0"

+;

Definition 7-6 in effect asserts that r is algebraic ¿~~r F if there exists a nonzero polynomialf(x) E F[x] such thatf(r) = O; o~:the other hand, r is a transcendental element over F in case f(r) 1- O fOfi¿very nonzero polynomial f(x) in F[x]. As iIlustrations of these notions take F' = R# and F = Q; then, + .J3 is algebraic over Q, being a rqot of the polynomial .0 - lOx2 + 1 E Q[x], while n and 2../2 are both transcendental over Q. Every element of the field F is trivialIy algebraic over F, for the element rE F is a root of the linear polynomial x - rE F[ x]. The proof ofTheorem 7-19 furnishes us with more detailed information concerning simple extension fields, which we now state as two separate theorems. First, however, let us remark that the field of quotients of the polynomial domain F[x] is traditionally called thefield ofrationaljUnctions

J2

Theorem 7-20. (Simple Transcendental Field Extensiol1s). If rE F' ¿ F is transcendental over F, then F(r) is isomorphic to the field of rational functions over F. In fact, there is an isomorphism ar of F(r) into Qtl(F[x]) such that ar(r) = and ar(a) = a for every a E R.

x

Theorem 7-21. (Simple Algebraic Field Extensions). If r E F' ¿ F is algebraic over F, then there exists a unique monic irreducible polynomialf(x) E F[x] such thatf(r) = O. Furthermore, if g(x) is a polynomial in F[x] for which g(r) = O, necessarily f(x) Ig(x). The unique polynomial f(x) of Theorem 7-21 is referred to as the minimum polynomial of (or belonging to) r over F; as the name suggests, the minimum polynomial of r is the monic polynomial in F[ x] of mini mal degree having r as a root. The degree of an algebraic element r E F' ¿ F is just the degree of its minimum polynomial (this degree is 1 if and only if r E F). . In the course of proving Theorem 7-19, we established the surprising fact that, whenev~r r is algebraic over F, the integral domain F[r] becomes a field, so that F(r) = F[rJ. This amounts to saying that every element of ~(r) is ofthe form f(r), where f(x) isa polynomial in F[ x J. Example 7-9. If n+-1 is any squár~~free integer, the element.Jn E R # ¿ Q is a root of the quadratic polynomiai X2 - n E Q[x] and, hence, is algebraic over Q. From the preceding paragúlph we know that Q[~] is a field and so QJJn] = Q(Jn); in other wor~s, every member of Q(Jn) is of the form f(J n), where f(x) is a polynomial,}n Q[x] : Q(Jn.) = {ao

But (Jn.)2 simplyas

=

+

a¡Jn.

+

ai.Jp)2

+ ... +

ak(Jn.)kla¡EQ; k

~ O}.

n, (Jn.)3 = nJn., ... , so that Q(Jn.) can be described more Q(Jn.) = {a

+

bJn.la, bE Q}.

That is to say, the simple field extension Q(Jn.) is what we referred to as a quadratic field in Chapter 6. Notice also that an arbitrary element a + bJn. E Q(Jn.) satisfies the polynomial . X2 - 2ax + (a 2 - b 2n) E Q[x]'

138

FIRST COURSE IN RINGS AND IDEALS

POLYNOMIAL RINGS

139

of F[ X] is not mapped onto zero. Now two possibilities arise, eíther ker 4Jr = {O} or ker 4Jr 1-. {O}. Suppose first that ker 4Jr = {O}. Then the homomorphism 4Jr is actualIy an isomorphism and F[r] ~ F[x]. In the situation at hand, F[r] is not a field, since F[x] fails to be one. However, from Theorem 4-11, we know that the quotient fields of F[r] and F[x] are isomorphic under a mapping induced by the substitution homomorphism. Since F(r) is the quotient 'field of F[r], one thus obtains

over F. Thus, a rational function (in an indeterminant x) over F can be written as a quotient f(x)/g(x), where f(x) and g(x) 1- O are both polynomials in F[x].

F(r) ~ QCI(F[x]).

This theorem completely determines the structure of simple transcendental extensions over F', they are all isomorphic . to the field of rational ../2 functions over F, and, hence, to each other. Thus, for lDstance, Q(n) ~ Q(2 ). As regards simple algebraic extensions, we have

If the kernel of 4Jr is nonzero, then ker 4Jr = (J(x)) for sorne irreducible polynomial (prime element) of F[x], where f(x) can be taken to be monic. Because every nonzero prime ideal of F[x] is maximal, F[x]/(J(x)) forms a field and the same will b,e true of its isomorphic image F[r]. But F(r) is the smallest field to contain 60th F and r, from which it folIows that F[r] = F(r); this leads to the isomorphism F(r) ~ F[x ]/(J(x)).

As ker 4Jr = (J(x)), a polynomial g(x) E F[ x] has r as a root if and only if g(x) is divisible by f(x). Accordingly, if g(x) is any monic irreducible polynomial in F[x] having r as a root, each of f(x) and g(x) divides the other; since both of these polynomials are monic, this is possible only if f(x) = g(x). Thus,f(x) is unique, as asserted in the theorem. . Another virtue of the substitution homomorphism is that it permits us to put the elements of an extension field into one of two essentialIy different categories: -Definition 7-6. Let F' be an exte~sion field of a>ifi~ld F. An element rE F' is said to be algebraic over F if ker 4Jr {O}; otherwise, r is termed transcendental over F. 1!-0"

+;

Definition 7-6 in effect asserts that r is algebraic ¿~~r F if there exists a nonzero polynomialf(x) E F[x] such thatf(r) = O; o~:the other hand, r is a transcendental element over F in case f(r) 1- O fOfi¿very nonzero polynomial f(x) in F[x]. As iIlustrations of these notions take F' = R# and F = Q; then, + .J3 is algebraic over Q, being a rqot of the polynomial .0 - lOx2 + 1 E Q[x], while n and 2../2 are both transcendental over Q. Every element of the field F is trivialIy algebraic over F, for the element rE F is a root of the linear polynomial x - rE F[ x]. The proof ofTheorem 7-19 furnishes us with more detailed information concerning simple extension fields, which we now state as two separate theorems. First, however, let us remark that the field of quotients of the polynomial domain F[x] is traditionally called thefield ofrationaljUnctions

J2

Theorem 7-20. (Simple Transcendental Field Extensiol1s). If rE F' ¿ F is transcendental over F, then F(r) is isomorphic to the field of rational functions over F. In fact, there is an isomorphism ar of F(r) into Qtl(F[x]) such that ar(r) = and ar(a) = a for every a E R.

x

Theorem 7-21. (Simple Algebraic Field Extensions). If r E F' ¿ F is algebraic over F, then there exists a unique monic irreducible polynomialf(x) E F[x] such thatf(r) = O. Furthermore, if g(x) is a polynomial in F[x] for which g(r) = O, necessarily f(x) Ig(x). The unique polynomial f(x) of Theorem 7-21 is referred to as the minimum polynomial of (or belonging to) r over F; as the name suggests, the minimum polynomial of r is the monic polynomial in F[ x] of mini mal degree having r as a root. The degree of an algebraic element r E F' ¿ F is just the degree of its minimum polynomial (this degree is 1 if and only if r E F). . In the course of proving Theorem 7-19, we established the surprising fact that, whenev~r r is algebraic over F, the integral domain F[r] becomes a field, so that F(r) = F[rJ. This amounts to saying that every element of ~(r) is ofthe form f(r), where f(x) isa polynomial in F[ x J. Example 7-9. If n+-1 is any squár~~free integer, the element.Jn E R # ¿ Q is a root of the quadratic polynomiai X2 - n E Q[x] and, hence, is algebraic over Q. From the preceding paragúlph we know that Q[~] is a field and so QJJn] = Q(Jn); in other wor~s, every member of Q(Jn) is of the form f(J n), where f(x) is a polynomial,}n Q[x] : Q(Jn.) = {ao

But (Jn.)2 simplyas

=

+

a¡Jn.

+

ai.Jp)2

+ ... +

ak(Jn.)kla¡EQ; k

~ O}.

n, (Jn.)3 = nJn., ... , so that Q(Jn.) can be described more Q(Jn.) = {a

+

bJn.la, bE Q}.

That is to say, the simple field extension Q(Jn.) is what we referred to as a quadratic field in Chapter 6. Notice also that an arbitrary element a + bJn. E Q(Jn.) satisfies the polynomial . X2 - 2ax + (a 2 - b 2n) E Q[x]'

140

FIRST COURSE IN RINGS AND IDEALS

I I

POLYNOMIAL RINGS

141

The implication is that, not only .Jñ, but every member of Q(Jñ) is algebraic over Q; our next-proved theorem demonstrates that tbis is no accident.

Let us establish a simple, but nonetheless effective, result about successive extensions.

Before pressing on, sorne additional terminology is required. If F' is an extension field of a field F, F' may be regarded as a vector space over F. We shall call F' a flntte extension of F, if the vector space F' is finite dirnen7 sional over F. For exampIe, the compIex field e is a finite extension of R#, with {l, i} serving as a basis for e (over R#), The dímension of F' as a vector space over F is called the degree ofthe extension and written [F' :F]. The reader who is versed in Uhear algebra will have no difficulty with the next few theorems. ;".'c

Tbeorem 7-24. Ir F ' is a finite extension of F and F" is a finite extension of F'; then FU is a finite exteil.sion of F. Furthermore,

fi~~~

Theorem 7-22. Ir Pis a extension of the fieId.F, then everY eIement of F' is aIgebraic ovciti'F. , , ,~

Proof. Let r E: F and co~ider th<::.,~!@ll1ents 1, r, r'-, '" , 1"', where n = [F' :Fl These n + 1 powers of r are all in!.F"and, hence, must be linearIy dependent OVer F (silice F' is a space of diÍrtension n). Thus, there exist eIements bo,b 1, .. : ,bnEF, which arenotallzéro,suchthat bol + b1r + ... + bnr" = O. But then,f(x) = bo + b1x + ;.. + bnxn is a nonzero polynomíaI in F[x] and J(r) 0, implying that r is algebraic over F. I

This leads us to the following concept: An extension field F ' of F is, said to be algebraic if every elementof F ' is algebraic over F. The content of Theorem 7-22 i8 that a finite extension is an algebraic extension. Theorem 7-23. Let F ' be an extensionof the field F. Then F' 1S an algebraic extension of F if and only if every subríng of F' containing F is a field.

Proof. To begin with, suppose that F' i8 an algebraic extension and let R be a subring of F ' wruch contains F; F s;:; R s;:; F I • For any nonzero element rE R, the inclusion F[r] S R certainly holds; Sin ce r is algeb'raic over F, . we know from what has been estabIished earlier that F[r] coincides with thefieId F(r), But then, r- 1 E F[rJ s;:; R, making thering R a field. As regards the converse, assume that every subring of F' which contains F forms a field. Given an eIement O f rE F', F[r] is a subríng of F' containing F and so must be a field ; in particular, r - 1 E F[r]. K.nowing this, we may inferthe existence ofa polynomialJ(x) in F[x]such thatJ(r) = r- 1. The element r thus becomes a root of thepolynomial g(x) = xJ(x) - 1 and, hence, is algebraic over F. We take this opportunity to mention an interesting theorem due to Steinitz which gives a necessary and sufficient condítion for a finite extension to be simple: Ir F I is a finite extension of the field F, then F' 1S á. simple extension if and only if there are only a finite number of subfields of F' containíng F.

'[F":F]

[F":F'][F' :FJ.

Proof. An 'ibbreviated proof runs as, follows. Suppose that [F' :F] = n and [F": F'J;= m. If {al' a 2 , •• :, a n } is a basis ror F' as a vector space over F and {bl,ill~, "', bm } is a basis for F" over FI, then the set of mn elements ofthe formq¡b j constitutes a basis for F" over F. This implies that ·,.tl

..

[F" :F]

)

=

mn = [F":F'][F' :F].

We stiil' have a few locis~ ends to tie together, inc1uding a more precise descriptióí1.~8f F(r), when r is algebraic over F. i

.F~

.

Theor~mij-25. Let F' be an extension field of F and,r.E F' be aIgébraic over F óf degree n. Then the elements 1, r, ... , rn-l. lorm a basis for . F(r) (considered as a vector space over F).

Proof. Let a be any element of F(r) = F[r]' Then there exists a polynomial = g(r). Applying the division algorithm to g(x) and the mínimum polynomial J(x) of r, we can find q(x) and s(x) in F[x]· satisfying . g(x) E F[x] such that a

°

+

g(x) = q(x)J(x)

s(x),

where either s(x). = or deg s(x) < n.· Since J(r) = 0, it follows that ser) ~ g(r) = a. Ir s(x) = 0, necessari1y a = 0, while ir s(x) = bo + b1x + ... + bmx rn is a nonzero polynomial of degree m < n, then a = b a + b1r + ... + brnr"'. Therefore, the eIements 1, r, ... , 1"'-1 generate F(r) as a vector space over F. It remains to show that the set {1, r, rn-l} is linearly independent. .' Pursuing this aim, let us assume that Col + c1r + ... + cn_1r n - 1 = 0, where the Ck E F. Then the polynomial oo. ,

"h(x)

= Co +

clx

+oo. +

Cn_1Xn-1EF[x]

and clearly h(r) = O, so that h(x) E ker rPr = (J(x»). h(x) = J(x)k(x)- for sorne polynomial k(x) in F[x].

This being the case, But if h(x) f Ó, we

obtain

n > deg h(x)

= degJ(x)

+

deg k(x) ;;::: degJ(x)

n,

a manifestly false conclllsion. Thus, the 'polynomial h(x) = 0, which forces the coefficients Ca = C1 = ... = Cn -1 = O. The proof that the n elements 1, r, ... , rn- 1 constitute a basis for F(,.) over F is now complete. The statement of Theorem 7-25 can be rephrased in severa1 ways.

140

FIRST COURSE IN RINGS AND IDEALS

I I

POLYNOMIAL RINGS

141

The implication is that, not only .Jñ, but every member of Q(Jñ) is algebraic over Q; our next-proved theorem demonstrates that tbis is no accident.

Let us establish a simple, but nonetheless effective, result about successive extensions.

Before pressing on, sorne additional terminology is required. If F' is an extension field of a field F, F' may be regarded as a vector space over F. We shall call F' a flntte extension of F, if the vector space F' is finite dirnen7 sional over F. For exampIe, the compIex field e is a finite extension of R#, with {l, i} serving as a basis for e (over R#), The dímension of F' as a vector space over F is called the degree ofthe extension and written [F' :F]. The reader who is versed in Uhear algebra will have no difficulty with the next few theorems. ;".'c

Tbeorem 7-24. Ir F ' is a finite extension of F and F" is a finite extension of F'; then FU is a finite exteil.sion of F. Furthermore,

fi~~~

Theorem 7-22. Ir Pis a extension of the fieId.F, then everY eIement of F' is aIgebraic ovciti'F. , , ,~

Proof. Let r E: F and co~ider th<::.,~!@ll1ents 1, r, r'-, '" , 1"', where n = [F' :Fl These n + 1 powers of r are all in!.F"and, hence, must be linearIy dependent OVer F (silice F' is a space of diÍrtension n). Thus, there exist eIements bo,b 1, .. : ,bnEF, which arenotallzéro,suchthat bol + b1r + ... + bnr" = O. But then,f(x) = bo + b1x + ;.. + bnxn is a nonzero polynomíaI in F[x] and J(r) 0, implying that r is algebraic over F. I

This leads us to the following concept: An extension field F ' of F is, said to be algebraic if every elementof F ' is algebraic over F. The content of Theorem 7-22 i8 that a finite extension is an algebraic extension. Theorem 7-23. Let F ' be an extensionof the field F. Then F' 1S an algebraic extension of F if and only if every subríng of F' containing F is a field.

Proof. To begin with, suppose that F' i8 an algebraic extension and let R be a subring of F ' wruch contains F; F s;:; R s;:; F I • For any nonzero element rE R, the inclusion F[r] S R certainly holds; Sin ce r is algeb'raic over F, . we know from what has been estabIished earlier that F[r] coincides with thefieId F(r), But then, r- 1 E F[rJ s;:; R, making thering R a field. As regards the converse, assume that every subring of F' which contains F forms a field. Given an eIement O f rE F', F[r] is a subríng of F' containing F and so must be a field ; in particular, r - 1 E F[r]. K.nowing this, we may inferthe existence ofa polynomialJ(x) in F[x]such thatJ(r) = r- 1. The element r thus becomes a root of thepolynomial g(x) = xJ(x) - 1 and, hence, is algebraic over F. We take this opportunity to mention an interesting theorem due to Steinitz which gives a necessary and sufficient condítion for a finite extension to be simple: Ir F I is a finite extension of the field F, then F' 1S á. simple extension if and only if there are only a finite number of subfields of F' containíng F.

'[F":F]

[F":F'][F' :FJ.

Proof. An 'ibbreviated proof runs as, follows. Suppose that [F' :F] = n and [F": F'J;= m. If {al' a 2 , •• :, a n } is a basis ror F' as a vector space over F and {bl,ill~, "', bm } is a basis for F" over FI, then the set of mn elements ofthe formq¡b j constitutes a basis for F" over F. This implies that ·,.tl

..

[F" :F]

)

=

mn = [F":F'][F' :F].

We stiil' have a few locis~ ends to tie together, inc1uding a more precise descriptióí1.~8f F(r), when r is algebraic over F. i

.F~

.

Theor~mij-25. Let F' be an extension field of F and,r.E F' be aIgébraic over F óf degree n. Then the elements 1, r, ... , rn-l. lorm a basis for . F(r) (considered as a vector space over F).

Proof. Let a be any element of F(r) = F[r]' Then there exists a polynomial = g(r). Applying the division algorithm to g(x) and the mínimum polynomial J(x) of r, we can find q(x) and s(x) in F[x]· satisfying . g(x) E F[x] such that a

°

+

g(x) = q(x)J(x)

s(x),

where either s(x). = or deg s(x) < n.· Since J(r) = 0, it follows that ser) ~ g(r) = a. Ir s(x) = 0, necessari1y a = 0, while ir s(x) = bo + b1x + ... + bmx rn is a nonzero polynomial of degree m < n, then a = b a + b1r + ... + brnr"'. Therefore, the eIements 1, r, ... , 1"'-1 generate F(r) as a vector space over F. It remains to show that the set {1, r, rn-l} is linearly independent. .' Pursuing this aim, let us assume that Col + c1r + ... + cn_1r n - 1 = 0, where the Ck E F. Then the polynomial oo. ,

"h(x)

= Co +

clx

+oo. +

Cn_1Xn-1EF[x]

and clearly h(r) = O, so that h(x) E ker rPr = (J(x»). h(x) = J(x)k(x)- for sorne polynomial k(x) in F[x].

This being the case, But if h(x) f Ó, we

obtain

n > deg h(x)

= degJ(x)

+

deg k(x) ;;::: degJ(x)

n,

a manifestly false conclllsion. Thus, the 'polynomial h(x) = 0, which forces the coefficients Ca = C1 = ... = Cn -1 = O. The proof that the n elements 1, r, ... , rn- 1 constitute a basis for F(,.) over F is now complete. The statement of Theorem 7-25 can be rephrased in severa1 ways.

142

Corollary 1. If r e F' ;2 F is algebraic of degree n, then every element of the simple extension F(r) is of the forro f(r), where f(x) e F[ x J is a polynomial of degree less than n: F(r)

=

{ao

+

alr

+ ... +

~_lr"-llak e F}.

Corollary 2. If r e F ' ;2 F is aIgebraic of degree n, .then F(r) is a finite algebraic) extension with [F(r): FJ = n.

that the field Q(J2, A) is actually a simple algebraic extension of Q; in fact, Q(J2, J3) Q(..fi + J3)!..with.Ji + J3 algebraic over Q.. Since the element + .J3 belongs to Q(.j2, 13), we certalOly bave Q(J2 + S;; Q(.j2, J3). As regards the reverse inc1usion, a simple computation shows that

We- inc1ude the next theorem for completeness; it is an immediate consequence ofTheorem 7-22 and Corollary 2 above. Theorem 7-26. Let F' be an extension ofthe field F. An element re F' is algebraic over F if and only if F(r) is a finite extension of F. From this, it is a short step to Corollary. Let re é ;2 F be algebraic and [F': FJ finite. F' = F(r) ifand only if[F(r): FJ [F':F].

Then

Proof By the last-written theorem, F(r) has a finite degree [F(r) :F]. Now, F(r) is a subspace (over F) ofthe vector space F'. This corollary is equivalent

to asserting that a subspace is the entire space ir and only if the dimensions of the two are equal. Example 7-10. Consider the eIement r = .Ji + i E e ;;2 Q, e as usual so that (r 2 - 1)2 = being the complex number fieId. Then r 2 = 1 + 4 2 - 8 or r - 2r + 9 = O. Thus, r is a root of the polynomiaI

2J2i,

2:x;2

+

9 e Q[xJ

and, hence, l~ an algebraic eIement over Q. Now, f(x) has the irreducible factorizatión, ;over e, :;" ~

f(x)

=

,e; - .Ji + i)(x -

J2 -

i)(x

+ .Ji +

i)(x

+

.Ji - i),

which indi~~tes that f(x) has no linear or quadratic factors in QIx]. Therefore,J(x) i$)rreducible asa member of Q[ x J and serves as the minimum polynomialJof r over Q; in particular, the element r has degree 4. By Theorem 7:""25, the simple extension Q(r) is a four-dimensional vector space over Q, with basis

1,

r =

J2 + i,

r2 = 1

+

2.Ji¡,

r3 = -

J2 + 5i.

At the same time r is"a root of the polynomiaI Xl - z.Jix + 3 e R"' [xJ, with X2 2J2x + 3 irreducible over R#; thus, r is of degree 2 over R#. Example 7-11. For a second illustration, we turo to the extension field Q(J2, 13)· The eIements .Ji and are c1earIy aIgebraic over Q, being roots ofthe polynomials xl - 2, X2 ~ 3 E Q[xJ, respectively. Our contention is

13

J2:.

13)

2.j2

(hefi(:~,

f(x) = x 4

143

POL YNOMIAL RINGS

FIRST COURSE IN RINGS AND IDEALS

= (.j2 + .J3)3

- 9(.J2

+ .J3)

+ .J3), and therefore so·is .J2. But then, J3 = (.j2 + J3) - .j2 + J3). This leads to the inclusion Q(.j2, J3)

is a rnember of Q(.j2

also Hes in Q(.J2 S;;; + J3) and the asserted equality. To see that r .j2 + is an algebraic element over Q, notice tbat r 2 = 5 + i.j6, (r 2 5)2 = 24, and, hence, the polynomial f(x) x4 2 10x + 1 has r as a root. One may verify that f(x) is irreducible in Q[x J, making it the mínimum polynomial of r. Perhaps the quickest way to ~ this is as follows. Let F' = Q(.j2); then [F':QJ = 2, with basis {l, .J2}, and [F'(J3):F'J 2, with basis {l, J3J From Theorem 7-24, it !9l1ows that[F'(.J3); QJ = 4anda basis for F'{.J3)overQ isgiven by {1,.j2,.,j3,.j6}. But . Q(.J2

13

J3) = Q(.j2 + J3), and we know that the dimension of Q(.j2 + J3) is equal to the degree of the mínimum polynomial of r = .Ji + J3. . Incidentally, there are five fields between Q and Q(.j2, 13), namt:ly, Q, Q(.j2), Q(J3), Q(.j6), and Q(.Ji, J3). Taking stock of Steinitz's theorem (page 140), it should come as no surprise that Q(J2, J3) can be generated F'(J3)

= Q(.j2)(13)

by a single eIernent.

= Q(.j2,

"

-' '), ','

4

Until now, we have always begun by assuming the existence ófan extension field F' of F and then studied the structure of sii:nple exten~!E>ns F(r) within F'. The subject can be approached from a somewhat diffe~eIlt standpoint. Given a field F and an irreducible polynomialf{x) e F[xJ~ pne may ask whether it is possible to construct a simple extension F ' of F;:in which f(x), thought of as a member of F'[xJ, has a root. (If degf(x) = j~; then, in a trivial sense, F is itself the required extension). To answerthis question, we take our cue from Theorem 7-19. For if such an extension of F can be found at aH, it must be of the form F(r), with r algebraic over F. As pointed out in our earlier discussion, r will possess a minimum polynomial g(x) which is irreducible in F[xJ and such that F(r) ~ F[xJ/(g(x»). This suggests that, when starting with a prescribed irreducible polynomial f(x) e F[ x J, the natural object of interest should be the associated quotient ring F[ x J/(J(x»). . After this preamble, let us proceed to some pertinent details.

142

Corollary 1. If r e F' ;2 F is algebraic of degree n, then every element of the simple extension F(r) is of the forro f(r), where f(x) e F[ x J is a polynomial of degree less than n: F(r)

=

{ao

+

alr

+ ... +

~_lr"-llak e F}.

Corollary 2. If r e F ' ;2 F is aIgebraic of degree n, .then F(r) is a finite algebraic) extension with [F(r): FJ = n.

that the field Q(J2, A) is actually a simple algebraic extension of Q; in fact, Q(J2, J3) Q(..fi + J3)!..with.Ji + J3 algebraic over Q.. Since the element + .J3 belongs to Q(.j2, 13), we certalOly bave Q(J2 + S;; Q(.j2, J3). As regards the reverse inc1usion, a simple computation shows that

We- inc1ude the next theorem for completeness; it is an immediate consequence ofTheorem 7-22 and Corollary 2 above. Theorem 7-26. Let F' be an extension ofthe field F. An element re F' is algebraic over F if and only if F(r) is a finite extension of F. From this, it is a short step to Corollary. Let re é ;2 F be algebraic and [F': FJ finite. F' = F(r) ifand only if[F(r): FJ [F':F].

Then

Proof By the last-written theorem, F(r) has a finite degree [F(r) :F]. Now, F(r) is a subspace (over F) ofthe vector space F'. This corollary is equivalent

to asserting that a subspace is the entire space ir and only if the dimensions of the two are equal. Example 7-10. Consider the eIement r = .Ji + i E e ;;2 Q, e as usual so that (r 2 - 1)2 = being the complex number fieId. Then r 2 = 1 + 4 2 - 8 or r - 2r + 9 = O. Thus, r is a root of the polynomiaI

2J2i,

2:x;2

+

9 e Q[xJ

and, hence, l~ an algebraic eIement over Q. Now, f(x) has the irreducible factorizatión, ;over e, :;" ~

f(x)

=

,e; - .Ji + i)(x -

J2 -

i)(x

+ .Ji +

i)(x

+

.Ji - i),

which indi~~tes that f(x) has no linear or quadratic factors in QIx]. Therefore,J(x) i$)rreducible asa member of Q[ x J and serves as the minimum polynomialJof r over Q; in particular, the element r has degree 4. By Theorem 7:""25, the simple extension Q(r) is a four-dimensional vector space over Q, with basis

1,

r =

J2 + i,

r2 = 1

+

2.Ji¡,

r3 = -

J2 + 5i.

At the same time r is"a root of the polynomiaI Xl - z.Jix + 3 e R"' [xJ, with X2 2J2x + 3 irreducible over R#; thus, r is of degree 2 over R#. Example 7-11. For a second illustration, we turo to the extension field Q(J2, 13)· The eIements .Ji and are c1earIy aIgebraic over Q, being roots ofthe polynomials xl - 2, X2 ~ 3 E Q[xJ, respectively. Our contention is

13

J2:.

13)

2.j2

(hefi(:~,

f(x) = x 4

143

POL YNOMIAL RINGS

FIRST COURSE IN RINGS AND IDEALS

= (.j2 + .J3)3

- 9(.J2

+ .J3)

+ .J3), and therefore so·is .J2. But then, J3 = (.j2 + J3) - .j2 + J3). This leads to the inclusion Q(.j2, J3)

is a rnember of Q(.j2

also Hes in Q(.J2 S;;; + J3) and the asserted equality. To see that r .j2 + is an algebraic element over Q, notice tbat r 2 = 5 + i.j6, (r 2 5)2 = 24, and, hence, the polynomial f(x) x4 2 10x + 1 has r as a root. One may verify that f(x) is irreducible in Q[x J, making it the mínimum polynomial of r. Perhaps the quickest way to ~ this is as follows. Let F' = Q(.j2); then [F':QJ = 2, with basis {l, .J2}, and [F'(J3):F'J 2, with basis {l, J3J From Theorem 7-24, it !9l1ows that[F'(.J3); QJ = 4anda basis for F'{.J3)overQ isgiven by {1,.j2,.,j3,.j6}. But . Q(.J2

13

J3) = Q(.j2 + J3), and we know that the dimension of Q(.j2 + J3) is equal to the degree of the mínimum polynomial of r = .Ji + J3. . Incidentally, there are five fields between Q and Q(.j2, 13), namt:ly, Q, Q(.j2), Q(J3), Q(.j6), and Q(.Ji, J3). Taking stock of Steinitz's theorem (page 140), it should come as no surprise that Q(J2, J3) can be generated F'(J3)

= Q(.j2)(13)

by a single eIernent.

= Q(.j2,

"

-' '), ','

4

Until now, we have always begun by assuming the existence ófan extension field F' of F and then studied the structure of sii:nple exten~!E>ns F(r) within F'. The subject can be approached from a somewhat diffe~eIlt standpoint. Given a field F and an irreducible polynomialf{x) e F[xJ~ pne may ask whether it is possible to construct a simple extension F ' of F;:in which f(x), thought of as a member of F'[xJ, has a root. (If degf(x) = j~; then, in a trivial sense, F is itself the required extension). To answerthis question, we take our cue from Theorem 7-19. For if such an extension of F can be found at aH, it must be of the form F(r), with r algebraic over F. As pointed out in our earlier discussion, r will possess a minimum polynomial g(x) which is irreducible in F[xJ and such that F(r) ~ F[xJ/(g(x»). This suggests that, when starting with a prescribed irreducible polynomial f(x) e F[ x J, the natural object of interest should be the associated quotient ring F[ x J/(J(x»). . After this preamble, let us proceed to some pertinent details.

144

FIRST COURSE IN RINGS AND IDEALS

Theorem 7-27. (Kronecker). If f(x) is an irreducible polynomial in F[x], then there is an extension field of Fin whichf(x) has a,root. Proof. For brevity, we shall write l in place of theprincipal ideal of F[x] generated by polynomial f(x); that i8 to say, 1 (J(x»). Since f(x) is F[x]/1 i8 a assumed to be irreducible, the associated quotient ring F' field. To see that P constitutes an exten8ion of F, consider the natural mapping nat¡: F[x] - t P. According to Theorem 4-7, either the restri~tion nat¡IF is the trivial homomorpbism or else natf (F) forms a field isomqrphic to F, where as usual '~: nat¡(F) = {a + lla EF}. The .first possibility is immediately excluded by the fact that , nat¡(l)

= 1 + 1 1= 1,

POLYNOMIAL RINGS

has degree less than that off(x) or el se is the zero polynomiaL In ract, the cosets of l are uniquely determined by remainders on division by f(x) in the sense that g(x) + 1 = h(x)+ l if and only if g(x) and h(x) leave the same remainder when divided by f(x). Thus, ir degf(x) n > 1 (for instance,f(x) = ao + aí.x + '" + anxn), then the extension field F' may be described by

F'

= {b o + b1x + '" + bn-1X'-~ +

Identifying bk + l with the element bk , we see can be uniquely represented in the form

+ (al + 1)(x + l) + ... + (a n + l)(x + l)" = ao + a 1 x + ... + a.xn + 1= f(x) + l O + l. Ifwe now identify an element ak E F with the coset ak + 1 which it determines in F' (the fact that F is isomorphic to nat¡(F) permits this), we obtain an(x

+

F}.

l)

F' = {b o -i:' b 1A + ... + bn_'.~1~~1IbkEF}:. Observe that since A = x + lis a root off(x)l,Q,P, calculations are carried ,

out with the aid ofthe relation a o + alA + :.. '::p. a.A" '70. The last paragraph serves to bring out the point' tba! F' 18 a finite extension of F with basis {1, A, .1. 2 , ... , A·- 1 }; in particular, we infer that

(a o + 1)

+ a 1(x + 1) + ... +

b 1(x

+

E

+ ... + bn - . + 1)'-1; As a final simplification, let us replace x + 1 by:some new symbol A, so that bo

+

llb k

ás before that a typícal coset

the elements of F' become polynomials in A: ,j' ~

which is the zero e1ement,of Fr. Therefore, F is imbeddable in the (qucltíent) field F' and, in this sen se, F' becomes an extension of F. " .'.. It remains to be established that the polynomialf(x) actually ha:(~'root in P. Assuming that f(x) = ao +a 1 x + ... + a.x", then, from' the . definitions of cosetaddition and multiplication,

ao

145

1)" = O,

whichis equivalent to asserting that f(x + l) = O. In other words, the coset x + l = 1x + lis the rootoff(x) sought in P. ' Sine¡:; each polynomial of positíve degree hasan· irreducible factor (Theorem 7-13), we may drop the restriction thatf(x) be irreducible. Corollary. Ifthe polynomialf(x) E F[x] ls ofpositive degree, then there exists an extension field of F containing a root off(x), To go back to Theorem 7-27 for a moment, let us take a cIoser look at ' the nature of tb,e cosets of l = (¡(x») in F[x], with the aim of expressing the extension field F' = F[x]/l in a Illore convenieht way. As usual, these cosets are of the form g(x) + 1, with g(x) E F[x]. Invoking the division algorithm, for 'each such g{x) there is a unique polynomial r(x) in F[x] satisfying g(x) q(x)f(x) + r(x1 where r(x) = O or deg r(x) < degf(x). Now g(x) r(x) = q(x)f(x) E l, so that g(x) and ¡,(x) determine the same r(x) + 1. From tbis, it is possible to draw the following coset; g(x) + l conc1usion: each coset of l in F[ x] contains exactly one polynomial which

. [F':F] = n = degf(x) . . To recapitulate: iff(x) E F[x] is an irreducible polynomial over F, then there exists a finite extension P of F, such that [P: F] = degf(x), in which f(x) has a root. Moreover, F' is a simple algebraic extenSÍon generated by a root of f(x). (Admittedly, sorne work could be .saved by an appeal to Theorems 7-21 and 7;-25, bjlt our object here is to present an alternative approach to the slibject.) We pause now to examine two concrete examples of the ideas just presented. . Example 7-12. Consider Z2: the fiéld of integers modulo 2, and the polynomial ¡(x) = x 3 + X + 1 E Z2[ x]. Since neither of the elements O and 1 is a root of x 3 + x + l,f(x) must be irreducible in Z2[XJ. Theorem 7-27 thus guarantees the existence of an extension of Z2' . specifical1y, the field Z2[X ]/(J(x»), in which the given polynomial has a root. Denoting this root by A, the discussion aboye tells us that

{a + bA + c,.l2/a,b, e E Z2}

Z2[X]/(J(x»)

= {O, 1, A, 1 + A, ll, 1 +' .1.2 , A + 'where, of course, .1. 3 + A + 1 = O. 1

.1.2 , 1

+ A + A2 },

As an example of operating in this' field, let us calculate the inverse of A + .1.2 • Before starting, observe that by using the relations

+

.1. 3

=

-(A

+

1)

= A + 1,

..14

..1 2

+

A

144

FIRST COURSE IN RINGS AND IDEALS

Theorem 7-27. (Kronecker). If f(x) is an irreducible polynomial in F[x], then there is an extension field of Fin whichf(x) has a,root. Proof. For brevity, we shall write l in place of theprincipal ideal of F[x] generated by polynomial f(x); that i8 to say, 1 (J(x»). Since f(x) is F[x]/1 i8 a assumed to be irreducible, the associated quotient ring F' field. To see that P constitutes an exten8ion of F, consider the natural mapping nat¡: F[x] - t P. According to Theorem 4-7, either the restri~tion nat¡IF is the trivial homomorpbism or else natf (F) forms a field isomqrphic to F, where as usual '~: nat¡(F) = {a + lla EF}. The .first possibility is immediately excluded by the fact that , nat¡(l)

= 1 + 1 1= 1,

POLYNOMIAL RINGS

has degree less than that off(x) or el se is the zero polynomiaL In ract, the cosets of l are uniquely determined by remainders on division by f(x) in the sense that g(x) + 1 = h(x)+ l if and only if g(x) and h(x) leave the same remainder when divided by f(x). Thus, ir degf(x) n > 1 (for instance,f(x) = ao + aí.x + '" + anxn), then the extension field F' may be described by

F'

= {b o + b1x + '" + bn-1X'-~ +

Identifying bk + l with the element bk , we see can be uniquely represented in the form

+ (al + 1)(x + l) + ... + (a n + l)(x + l)" = ao + a 1 x + ... + a.xn + 1= f(x) + l O + l. Ifwe now identify an element ak E F with the coset ak + 1 which it determines in F' (the fact that F is isomorphic to nat¡(F) permits this), we obtain an(x

+

F}.

l)

F' = {b o -i:' b 1A + ... + bn_'.~1~~1IbkEF}:. Observe that since A = x + lis a root off(x)l,Q,P, calculations are carried ,

out with the aid ofthe relation a o + alA + :.. '::p. a.A" '70. The last paragraph serves to bring out the point' tba! F' 18 a finite extension of F with basis {1, A, .1. 2 , ... , A·- 1 }; in particular, we infer that

(a o + 1)

+ a 1(x + 1) + ... +

b 1(x

+

E

+ ... + bn - . + 1)'-1; As a final simplification, let us replace x + 1 by:some new symbol A, so that bo

+

llb k

ás before that a typícal coset

the elements of F' become polynomials in A: ,j' ~

which is the zero e1ement,of Fr. Therefore, F is imbeddable in the (qucltíent) field F' and, in this sen se, F' becomes an extension of F. " .'.. It remains to be established that the polynomialf(x) actually ha:(~'root in P. Assuming that f(x) = ao +a 1 x + ... + a.x", then, from' the . definitions of cosetaddition and multiplication,

ao

145

1)" = O,

whichis equivalent to asserting that f(x + l) = O. In other words, the coset x + l = 1x + lis the rootoff(x) sought in P. ' Sine¡:; each polynomial of positíve degree hasan· irreducible factor (Theorem 7-13), we may drop the restriction thatf(x) be irreducible. Corollary. Ifthe polynomialf(x) E F[x] ls ofpositive degree, then there exists an extension field of F containing a root off(x), To go back to Theorem 7-27 for a moment, let us take a cIoser look at ' the nature of tb,e cosets of l = (¡(x») in F[x], with the aim of expressing the extension field F' = F[x]/l in a Illore convenieht way. As usual, these cosets are of the form g(x) + 1, with g(x) E F[x]. Invoking the division algorithm, for 'each such g{x) there is a unique polynomial r(x) in F[x] satisfying g(x) q(x)f(x) + r(x1 where r(x) = O or deg r(x) < degf(x). Now g(x) r(x) = q(x)f(x) E l, so that g(x) and ¡,(x) determine the same r(x) + 1. From tbis, it is possible to draw the following coset; g(x) + l conc1usion: each coset of l in F[ x] contains exactly one polynomial which

. [F':F] = n = degf(x) . . To recapitulate: iff(x) E F[x] is an irreducible polynomial over F, then there exists a finite extension P of F, such that [P: F] = degf(x), in which f(x) has a root. Moreover, F' is a simple algebraic extenSÍon generated by a root of f(x). (Admittedly, sorne work could be .saved by an appeal to Theorems 7-21 and 7;-25, bjlt our object here is to present an alternative approach to the slibject.) We pause now to examine two concrete examples of the ideas just presented. . Example 7-12. Consider Z2: the fiéld of integers modulo 2, and the polynomial ¡(x) = x 3 + X + 1 E Z2[ x]. Since neither of the elements O and 1 is a root of x 3 + x + l,f(x) must be irreducible in Z2[XJ. Theorem 7-27 thus guarantees the existence of an extension of Z2' . specifical1y, the field Z2[X ]/(J(x»), in which the given polynomial has a root. Denoting this root by A, the discussion aboye tells us that

{a + bA + c,.l2/a,b, e E Z2}

Z2[X]/(J(x»)

= {O, 1, A, 1 + A, ll, 1 +' .1.2 , A + 'where, of course, .1. 3 + A + 1 = O. 1

.1.2 , 1

+ A + A2 },

As an example of operating in this' field, let us calculate the inverse of A + .1.2 • Before starting, observe that by using the relations

+

.1. 3

=

-(A

+

1)

= A + 1,

..14

..1 2

+

A

146

FIRST COURSE IN RrNGS AND IDEALS

POLYNOMIAL RINGS

(our coefficients come from Z2' where -1 1), the degree of any product can be kept less than 3. Now, the problemis to determine elements a, b, e E Z2 for which

Carrying out the multiplication and substituting for A3 , A4 in terms of 1, A, and A7, we obtain (a

+b+

e)

+

aA

+

(a

+ b)A 2 =

1.

This yields the system oflinear equations

a+b

+e=

1,

a

=

0,

+ b = 0, + A + A2)-1 a

withsolutiona b. O,e = l;therefore,{l = A2 • 3 It is worth noting that x + x + 1 factors completely into linear factors in Z2[X]/(J(x») and has the three roots A., A2, and A + A2: x 3 + x + 1 = (x - A)(X - A2)(X - (A + A2»). Example 7-13. The quadratic polynomial X2 + 1 is irreducible in R"'[x]. F or, if X2 + 1 were reducible, it would be of the form X2

+ 1 = (ax + b)(ex + d) 2 = aex + (ad + be)x +

bd,

where a, b, e, d E R "'. It follows at once tbat ac = bd = 1 and ad Therefore, be = - (ad), and

+

be

=

O.

1= (ae)(bd) = (ad)(b4;2{;. -(ad)2,

or rather, (ad)2 = '-1, which is impossible';Í;';" In tbis instance, the extension field R"'[xJ7(x 2

+

1) is described by

:1 ·~t'"

R"'[x]/(x2

+

1)

=

{a + bAla. b$R#; A2 + 1 = O}.

(a

+ bA) + (e + dA) =

(a

+

bA)(e

+ dA)

= (ae =

bd)

(ae - bd)

(a

~t~) + (b + d)A

+ (ad + be)A + + (ad + bd)A.

bd(A 2

+

Before proceeding further, two comments are in order. First, Example 7-12 shows that there exist finite fields other than the fields Zp of integers 8 elemodulo a prime p. The fact that the field of this. example has 2 3 ments is typical ofthe general situation : if F is a finite field, then F contains pn elements, where the prime p is the characteristic of F (Theorem 9-7). In the second place, the construction of Theorem 7-27 yields an extension of the field F in which a given(nonconstant) polynomial f(x) E F[x] splits off one linear factor. By:r;epeated application of tbis procedure, we can build up an extension F' of F in whichf(x), thought of as a member of F'[x], factors into a product of linear factors; that is, the field F' is large eoough to contain all the roots off(x) (technically speaking, the polynomial splits eompletely in F'[x]). We present this result in the form of an existence theorem. Tbeorem 7-28. If f(x) E F[x] is a polynomíal of positive degree, then there exists an extension field F' of Fin whichf(x) factors completely into linear polynomials. Proa! The proof is by induction on n = degf{x). If n = 1, then f(x) lS already linear and F itself is the required extension. Therefore, assume that n > 1 and that the theorem is true for all fields and for all polynomials of degree less than n. Now, the polynomial f(x) must have some irreducible factor g(x). By Theorem 7-27, there is an extension field K of Fin which g(x) and, hence, f(x), has a root r1 ; specifically, the field K = F[x]/{g(x»). Thus,f(x) can be written in K[x] asf(x) = (x - r1)h(x), where deg h(x) = n 1. By our induction assumpti9n, there is an extension F' of K in which h(x) splits completely; say h(x) = a(x - r2)(x - r3) .. , {x - r n ), with r¡ E F', a ::fo O. From this, we s~é that f(x) can be factored ioto linear factors in F'[x]'

Corollary•. Let f(x} E;i,F{x],degf(x) = n > '0. Theo there exists an extension of Fin whlchf(x) has n (not necessarily distinct) roots. (.

Performing the usual operations for polynóíÍ;Úals, we see that and

147

1)

The similarity of these formulas to the usual rules for addition and multiplication of complex numbers should be apparent. As a matter of fact, R#[x]/(x2 + 1) is isomorphic to the field e of complex numbers under the a + bi. Tbis mapping 11>: R#[x]/(x2 + 1) ~ e given by q)(a + bA) provides an elegant way of construoting e from R#.

Example 7-14. To il1~strate tbis situation, let us look at the polynomial f(x) (x 2 - 2)(x 2 - -~~ over the field Q of rational numbers. From Example 7-4, X2 - 2 (a:f¡.d by similar reasoning, X2 - 3) is already known to be irreducible in Q[xJ So we begin by extending Q to the field F 1. where F1

= Q[x]/(x 2

-

2)

= {a + bAla, b E Q; A2

and obtain the factorization f(x) = (x - A)(X

= (x

-

+

A)(X 2

.[2) (x +

3)

.[2)(x

(As A2 = 2, one customari1y identifies A with

2

.[2.)

-

3).

-

2 ;;:: O};

146

FIRST COURSE IN RrNGS AND IDEALS

POLYNOMIAL RINGS

(our coefficients come from Z2' where -1 1), the degree of any product can be kept less than 3. Now, the problemis to determine elements a, b, e E Z2 for which

Carrying out the multiplication and substituting for A3 , A4 in terms of 1, A, and A7, we obtain (a

+b+

e)

+

aA

+

(a

+ b)A 2 =

1.

This yields the system oflinear equations

a+b

+e=

1,

a

=

0,

+ b = 0, + A + A2)-1 a

withsolutiona b. O,e = l;therefore,{l = A2 • 3 It is worth noting that x + x + 1 factors completely into linear factors in Z2[X]/(J(x») and has the three roots A., A2, and A + A2: x 3 + x + 1 = (x - A)(X - A2)(X - (A + A2»). Example 7-13. The quadratic polynomial X2 + 1 is irreducible in R"'[x]. F or, if X2 + 1 were reducible, it would be of the form X2

+ 1 = (ax + b)(ex + d) 2 = aex + (ad + be)x +

bd,

where a, b, e, d E R "'. It follows at once tbat ac = bd = 1 and ad Therefore, be = - (ad), and

+

be

=

O.

1= (ae)(bd) = (ad)(b4;2{;. -(ad)2,

or rather, (ad)2 = '-1, which is impossible';Í;';" In tbis instance, the extension field R"'[xJ7(x 2

+

1) is described by

:1 ·~t'"

R"'[x]/(x2

+

1)

=

{a + bAla. b$R#; A2 + 1 = O}.

(a

+ bA) + (e + dA) =

(a

+

bA)(e

+ dA)

= (ae =

bd)

(ae - bd)

(a

~t~) + (b + d)A

+ (ad + be)A + + (ad + bd)A.

bd(A 2

+

Before proceeding further, two comments are in order. First, Example 7-12 shows that there exist finite fields other than the fields Zp of integers 8 elemodulo a prime p. The fact that the field of this. example has 2 3 ments is typical ofthe general situation : if F is a finite field, then F contains pn elements, where the prime p is the characteristic of F (Theorem 9-7). In the second place, the construction of Theorem 7-27 yields an extension of the field F in which a given(nonconstant) polynomial f(x) E F[x] splits off one linear factor. By:r;epeated application of tbis procedure, we can build up an extension F' of F in whichf(x), thought of as a member of F'[x], factors into a product of linear factors; that is, the field F' is large eoough to contain all the roots off(x) (technically speaking, the polynomial splits eompletely in F'[x]). We present this result in the form of an existence theorem. Tbeorem 7-28. If f(x) E F[x] is a polynomíal of positive degree, then there exists an extension field F' of Fin whichf(x) factors completely into linear polynomials. Proa! The proof is by induction on n = degf{x). If n = 1, then f(x) lS already linear and F itself is the required extension. Therefore, assume that n > 1 and that the theorem is true for all fields and for all polynomials of degree less than n. Now, the polynomial f(x) must have some irreducible factor g(x). By Theorem 7-27, there is an extension field K of Fin which g(x) and, hence, f(x), has a root r1 ; specifically, the field K = F[x]/{g(x»). Thus,f(x) can be written in K[x] asf(x) = (x - r1)h(x), where deg h(x) = n 1. By our induction assumpti9n, there is an extension F' of K in which h(x) splits completely; say h(x) = a(x - r2)(x - r3) .. , {x - r n ), with r¡ E F', a ::fo O. From this, we s~é that f(x) can be factored ioto linear factors in F'[x]'

Corollary•. Let f(x} E;i,F{x],degf(x) = n > '0. Theo there exists an extension of Fin whlchf(x) has n (not necessarily distinct) roots. (.

Performing the usual operations for polynóíÍ;Úals, we see that and

147

1)

The similarity of these formulas to the usual rules for addition and multiplication of complex numbers should be apparent. As a matter of fact, R#[x]/(x2 + 1) is isomorphic to the field e of complex numbers under the a + bi. Tbis mapping 11>: R#[x]/(x2 + 1) ~ e given by q)(a + bA) provides an elegant way of construoting e from R#.

Example 7-14. To il1~strate tbis situation, let us look at the polynomial f(x) (x 2 - 2)(x 2 - -~~ over the field Q of rational numbers. From Example 7-4, X2 - 2 (a:f¡.d by similar reasoning, X2 - 3) is already known to be irreducible in Q[xJ So we begin by extending Q to the field F 1. where F1

= Q[x]/(x 2

-

2)

= {a + bAla, b E Q; A2

and obtain the factorization f(x) = (x - A)(X

= (x

-

+

A)(X 2

.[2) (x +

3)

.[2)(x

(As A2 = 2, one customari1y identifies A with

2

.[2.)

-

3).

-

2 ;;:: O};

148

FIRST COURSE IN RINGS AND IDEALS

However, f(x) does not split completely, since the polynomial X2 - 3 . remains irreducible in F 1 !!:]. For, suppose to the contrary that X2 - 3 has a root in F 1; say e + d.J2, with e, d E Q. Substituting, we find that (c 2

+

2d2

3)

+

2cdJ2 = O,

" the same thing, or, what amounts 'to c2

+ 2d2

3 == Oí::

cd = O. .' The lalter equation implies that either c¿k. O or d = O. But neither c nor d can be zero, since thi~ would mean that #?" = 3/2 or c2 = 3, which is clearIy impossiblé. According1y, X2 - 3 does;~'t;t split in F 1 [x]. In order to factorf(x) into linear fagtQrs, it becomes necessary to extend the coefficient field further. We therefqr~' construct a second extension F 2' where ;,. '.'

;2P~la,PEF1;fl2 - 3 = The elementsof F 2 can be expressed ~it~tnatively in the form F2

(a

F 1[x]/(X 2' - 3) =:= {a

+ bJ2)

+(e

+ d.J2).J3

= a

+ b.J2 +

O}.

cJ3 + d.j6,

where, of course, the coefficients a, b, c, d all lie in Q. It follows without difficuIty that the original polynomial n9W factors in F 2 [x] as f(x)

+ 'A)(X fl)(X + fl) .J2)(x + .J2)(x .J3)(x + .J3).

(x - A)(X = (x -

149

POLYNOMIAL RINGS

!-~..

Let a field F be given and consider a nonconstant polynomialf(x) E F[x]. An extension field F' of Fis said to be a splitting field for f(x) over F provided thatf(x) can be factored completely into linear factors in F'[x], but not so factored over any proper subfield of F' containing F (this minimum nature of the splitting field is n.ot required by all authors). LooseIy speaking, a splitting fieId is the smaIlest extension fieId F ' in which the prescribed poIynomial factors linearIy: f(x) = a(x - 'l)(X - r2 ) ••. (x - rn) (r i EF' ). To obtain a splitting field for f(x), we need onIy consíder the family {Fi} of all extension fields Fi in whichf(x) can be decomposed as a product of linear factors (Theorem7-27 guarantees the existence of such extensions); then ( l Fi serves as a splitting field for f(x) over F. . Having thus indicated the existence of a spIítting field for an arbitrary polynomial in F[ x], it is natural to Collow this up with a query as to uniqueness. For a final topic, we shall prove that any two splitting fields oC the same (nonconstant) polynomial are isomorphic; this being so, one is justified in using the definite article and speaking oC the splitting field of a given polynomíal.

BeCore presenting the main theorem, two preparatory results of a somewhat techniéal nature are needed: Lemma. Letf(x) be an irreducible poIynomiaI in F[x] and r be a root oC f(x) in sorne extension field K oC F. Then F(r) ~ F[x]/(f(x») under an isomorphism whereby the element r corresponds to the coset x'+ (f(x»). Proof. Since the eIement r is algebraic over F, it folIows directly. from Theorem 7-19 tha~; F(r) ~ F[xJ/(f(x)) via an isomorphism B with the property that
r .i:!:~r(X) = (B o nat(f(x)})(x) = B(x

+ (f(x))).

The chief valúe:~~í this lemma ls that it leads almost immediately to the Collowing theorem~: .'~ . - ~ •.;>:¡

.

Theorem7-29. -r(Isomoiphism Extension Theorem): Let (J be an isomorphismfrom thefieldF onto thefieldF'. Afilo,letf(x) a o + al x + ... + a"xn be an irreducible polynomial in F[x] and f'(y) = (J(a o) + (J(aJy + ... + (J(anly" he the corresponding polynomial in F'EyJ. Then, f'(y) is likewise irreducible. Furthermore, if r is a root of f(x} in sorne extension ·field oC F and ís a rootof f'(y) in some extensíon field of F', then (J can be extended to an isomorphism (J> of F(r) onto F'(r') with
r

Proof Let us first extend (J to a mapping ii between the polynomial rings F[x] and F'[y] by setting

+ ... + bnx~) = (J(b o) + (J(b1)y + ... + (J(b,,)y" polynomial g(x) = b o + b 1 x + ... + bnr E F[x]' We bequeath

iig(x) = ii(b o

+

b1x

for any to the reader the task of supplying the necessary details that ii is an isomorphism of F[ x ] onto F,[yJ. 1t is important to notice that for any polynomial g(x) in F[x], an element a E F is a root of g(x) iC and only if (J(a) is a rootoC iig(x). Indeed, if, as before, g(x) = bo + b 1 x + ... + bnx", then, upon evaluating iig(x) at (J(a), (¡¡g(x))((J(a))

+ (J(b1)c¡(a) + ... + (J(bn)(J(a)n (J(b o + b1a + ... + bna") oí.b o)

=

= (J(g(a)),

. from which our assertion foIlows. In particular, we infer that the poIynomiaIs g(x) and iig(x) are simultaneously reducible or irreducible in F[ x] and F'[y], respectiveIy. ,This being SO,J'(y) = ¡¡f(x) is irreducible in F'[y J.

148

FIRST COURSE IN RINGS AND IDEALS

However, f(x) does not split completely, since the polynomial X2 - 3 . remains irreducible in F 1 !!:]. For, suppose to the contrary that X2 - 3 has a root in F 1; say e + d.J2, with e, d E Q. Substituting, we find that (c 2

+

2d2

3)

+

2cdJ2 = O,

" the same thing, or, what amounts 'to c2

+ 2d2

3 == Oí::

cd = O. .' The lalter equation implies that either c¿k. O or d = O. But neither c nor d can be zero, since thi~ would mean that #?" = 3/2 or c2 = 3, which is clearIy impossiblé. According1y, X2 - 3 does;~'t;t split in F 1 [x]. In order to factorf(x) into linear fagtQrs, it becomes necessary to extend the coefficient field further. We therefqr~' construct a second extension F 2' where ;,. '.'

;2P~la,PEF1;fl2 - 3 = The elementsof F 2 can be expressed ~it~tnatively in the form F2

(a

F 1[x]/(X 2' - 3) =:= {a

+ bJ2)

+(e

+ d.J2).J3

= a

+ b.J2 +

O}.

cJ3 + d.j6,

where, of course, the coefficients a, b, c, d all lie in Q. It follows without difficuIty that the original polynomial n9W factors in F 2 [x] as f(x)

+ 'A)(X fl)(X + fl) .J2)(x + .J2)(x .J3)(x + .J3).

(x - A)(X = (x -

149

POLYNOMIAL RINGS

!-~..

Let a field F be given and consider a nonconstant polynomialf(x) E F[x]. An extension field F' of Fis said to be a splitting field for f(x) over F provided thatf(x) can be factored completely into linear factors in F'[x], but not so factored over any proper subfield of F' containing F (this minimum nature of the splitting field is n.ot required by all authors). LooseIy speaking, a splitting fieId is the smaIlest extension fieId F ' in which the prescribed poIynomial factors linearIy: f(x) = a(x - 'l)(X - r2 ) ••. (x - rn) (r i EF' ). To obtain a splitting field for f(x), we need onIy consíder the family {Fi} of all extension fields Fi in whichf(x) can be decomposed as a product of linear factors (Theorem7-27 guarantees the existence of such extensions); then ( l Fi serves as a splitting field for f(x) over F. . Having thus indicated the existence of a spIítting field for an arbitrary polynomial in F[ x], it is natural to Collow this up with a query as to uniqueness. For a final topic, we shall prove that any two splitting fields oC the same (nonconstant) polynomial are isomorphic; this being so, one is justified in using the definite article and speaking oC the splitting field of a given polynomíal.

BeCore presenting the main theorem, two preparatory results of a somewhat techniéal nature are needed: Lemma. Letf(x) be an irreducible poIynomiaI in F[x] and r be a root oC f(x) in sorne extension field K oC F. Then F(r) ~ F[x]/(f(x») under an isomorphism whereby the element r corresponds to the coset x'+ (f(x»). Proof. Since the eIement r is algebraic over F, it folIows directly. from Theorem 7-19 tha~; F(r) ~ F[xJ/(f(x)) via an isomorphism B with the property that
r .i:!:~r(X) = (B o nat(f(x)})(x) = B(x

+ (f(x))).

The chief valúe:~~í this lemma ls that it leads almost immediately to the Collowing theorem~: .'~ . - ~ •.;>:¡

.

Theorem7-29. -r(Isomoiphism Extension Theorem): Let (J be an isomorphismfrom thefieldF onto thefieldF'. Afilo,letf(x) a o + al x + ... + a"xn be an irreducible polynomial in F[x] and f'(y) = (J(a o) + (J(aJy + ... + (J(anly" he the corresponding polynomial in F'EyJ. Then, f'(y) is likewise irreducible. Furthermore, if r is a root of f(x} in sorne extension ·field oC F and ís a rootof f'(y) in some extensíon field of F', then (J can be extended to an isomorphism (J> of F(r) onto F'(r') with
r

Proof Let us first extend (J to a mapping ii between the polynomial rings F[x] and F'[y] by setting

+ ... + bnx~) = (J(b o) + (J(b1)y + ... + (J(b,,)y" polynomial g(x) = b o + b 1 x + ... + bnr E F[x]' We bequeath

iig(x) = ii(b o

+

b1x

for any to the reader the task of supplying the necessary details that ii is an isomorphism of F[ x ] onto F,[yJ. 1t is important to notice that for any polynomial g(x) in F[x], an element a E F is a root of g(x) iC and only if (J(a) is a rootoC iig(x). Indeed, if, as before, g(x) = bo + b 1 x + ... + bnx", then, upon evaluating iig(x) at (J(a), (¡¡g(x))((J(a))

+ (J(b1)c¡(a) + ... + (J(bn)(J(a)n (J(b o + b1a + ... + bna") oí.b o)

=

= (J(g(a)),

. from which our assertion foIlows. In particular, we infer that the poIynomiaIs g(x) and iig(x) are simultaneously reducible or irreducible in F[ x] and F'[y], respectiveIy. ,This being SO,J'(y) = ¡¡f(x) is irreducible in F'[y J.

150

FIRST COURSE IN RINGS AND IDEALS

a(a1)y + ... + a(anJY" be the corresponding polynomial in F'ey]' If K is a splitting field of f(x) and K' a splitting fieId of f'(y), then a can be ex~ended to an isomorphisrp.
By the foregoing lemma, we know that there exist isomorphisms a: F(r) --+ F[x]j{J(x)) and p: F'(r') --+ F'ey]/{J'(Y)), with

a(r) = x

+

(J(x)),

(J(r') = y

+

(J'(y)).

Proof. Our argument will be by induction on the number n of roots ofj(x) that lie outside F, but (needless to say) in K. When n = 0, all the roots of

Moreover, it is an easy matter to show that there is also an i,somorphism r of F[x]/{J(x)) onto F'[y]j{J'(y)) defined by ,

r(g(x)

+ (f(x))) =

ag(x)

+

(J'(y))

(g(x)

E

F[x]).

Observe particular1y that r carries the coset x + (J(x)) onto y We contend that F(r) ~ F'(r') under the composition of maps «l> = P- 1 o r o a,

+

(f'(y)).

where
:I

~~,F(l

F[x ]/(J(x))

-----?

F'ey]/{J'(y))

Certainly,

= (fJ-l o r)(a(a)) = (P-l or)(a + p-l(a(a) + (f'(y))) = a(a),

(f(x)))

=

whence
P-l(y + '(f'(y))) = 1", asrequired, and the theorem is pro ved in its entirety. =

.

-..""

,For a simple, but nonetheless satisfYing, illustration of this last result, take both F and P to be the real number field R # ; let f(x) E R # [x] be the . ;. irreducible polynomial f(x) = X2 + 1, so that f'(y) = y2 + 1 (recall that th~ identity map is the only isomorphism of R# ontoitself). Finally, choose .;; l' '= i and 1" = - i. Theorem 7-29 then asserts that R # (i) ~ R # ( - i) under ao.isomorphismwhichcarriesionto-i, InasmuchasR#(i) = R#(-i) = e, the isomorphism in question is jus~ the correspondence between a complex number and its conjugate. We now have the mathematical machinery to show the uniqueness (to within isomorphism) of splitting fields. Actua1ly, we shall prove a somewhat more general result. ' Theorem 7-30. Let a be an isomorphism of the field F onto the field P. Let f(x) = ao + a1x + anxn EF[x] and f'(y) = a(ao) +

+ ...

151

PROBLEMS

f(x) belong to F and F is itself the splitting field of f(x); that is, K = F. This in turn induces a splitting of the polynomial f'(y) into a product of linear factors in F'[y], so that K', = F'. Thus, when it happens that n = 0, the isomorphism a is, in a trivial sense, the desired extension to the splitting fields. Let us next assume, inductively, that the theorem holds true for any pair of corresponding polynomials f(x) and f'(y) over isomorphíc fields E and E', provided that the number of roots of roots of f(x) outside of E is less than n (n ~ 1). If f(x) E F[x] is a polynomial having n roots outside of F, then not all of the irreducible factors of f(x) can be linear in F[x] ; for, otherwise,f(x) would split cbmpletely in F, contrary to assumption. Accordingly, f(x) must have sorne factor g(x) of degree m > 1 which is irreducible in F[x]. Let g'(y) denote the corresponding irreducible factor of f'(y). Since K is a splitting field of f(x) over F, g(x) in particular must have a root in K; call it r. Similar1y, one of the roots of the polynomialf'(y), say 1", is a root of g'(y) in K'. By Theorem 7-29, a can be extended to an isomorphism a' between the fields F(r) and F'(r'). Now, K is a splitting field off(x), viewed as a polynomial with coefficients from F(r); in a like manner, K' can be regan;led as a splitting field off'(y) over the field F'(r'). Because the number of roots off(x) lying outside of F(r) is les s than n,the induction hypothesis permits us to extend a' (itse1f an extension of a) to an is'Ó'fuorphism such that the restriction
Proof. This is ah immediate consequence of the theorem on taking F and a to be the identity isomorphism iF •

=

P

PROBLEMS 1. If R is a commutative ring with identity, prove that a) The set 1 = {¡(x) e R[[x]]lordf(x) > O} u {O} forros an ideal of the ring R[[x]]; in fact, 1 = (x).

150

FIRST COURSE IN RINGS AND IDEALS

a(a1)y + ... + a(anJY" be the corresponding polynomial in F'ey]' If K is a splitting field of f(x) and K' a splitting fieId of f'(y), then a can be ex~ended to an isomorphisrp.
By the foregoing lemma, we know that there exist isomorphisms a: F(r) --+ F[x]j{J(x)) and p: F'(r') --+ F'ey]/{J'(Y)), with

a(r) = x

+

(J(x)),

(J(r') = y

+

(J'(y)).

Proof. Our argument will be by induction on the number n of roots ofj(x) that lie outside F, but (needless to say) in K. When n = 0, all the roots of

Moreover, it is an easy matter to show that there is also an i,somorphism r of F[x]/{J(x)) onto F'[y]j{J'(y)) defined by ,

r(g(x)

+ (f(x))) =

ag(x)

+

(J'(y))

(g(x)

E

F[x]).

Observe particular1y that r carries the coset x + (J(x)) onto y We contend that F(r) ~ F'(r') under the composition of maps «l> = P- 1 o r o a,

+

(f'(y)).

where
:I

~~,F(l

F[x ]/(J(x))

-----?

F'ey]/{J'(y))

Certainly,

= (fJ-l o r)(a(a)) = (P-l or)(a + p-l(a(a) + (f'(y))) = a(a),

(f(x)))

=

whence
P-l(y + '(f'(y))) = 1", asrequired, and the theorem is pro ved in its entirety. =

.

-..""

,For a simple, but nonetheless satisfYing, illustration of this last result, take both F and P to be the real number field R # ; let f(x) E R # [x] be the . ;. irreducible polynomial f(x) = X2 + 1, so that f'(y) = y2 + 1 (recall that th~ identity map is the only isomorphism of R# ontoitself). Finally, choose .;; l' '= i and 1" = - i. Theorem 7-29 then asserts that R # (i) ~ R # ( - i) under ao.isomorphismwhichcarriesionto-i, InasmuchasR#(i) = R#(-i) = e, the isomorphism in question is jus~ the correspondence between a complex number and its conjugate. We now have the mathematical machinery to show the uniqueness (to within isomorphism) of splitting fields. Actua1ly, we shall prove a somewhat more general result. ' Theorem 7-30. Let a be an isomorphism of the field F onto the field P. Let f(x) = ao + a1x + anxn EF[x] and f'(y) = a(ao) +

+ ...

151

PROBLEMS

f(x) belong to F and F is itself the splitting field of f(x); that is, K = F. This in turn induces a splitting of the polynomial f'(y) into a product of linear factors in F'[y], so that K', = F'. Thus, when it happens that n = 0, the isomorphism a is, in a trivial sense, the desired extension to the splitting fields. Let us next assume, inductively, that the theorem holds true for any pair of corresponding polynomials f(x) and f'(y) over isomorphíc fields E and E', provided that the number of roots of roots of f(x) outside of E is less than n (n ~ 1). If f(x) E F[x] is a polynomial having n roots outside of F, then not all of the irreducible factors of f(x) can be linear in F[x] ; for, otherwise,f(x) would split cbmpletely in F, contrary to assumption. Accordingly, f(x) must have sorne factor g(x) of degree m > 1 which is irreducible in F[x]. Let g'(y) denote the corresponding irreducible factor of f'(y). Since K is a splitting field of f(x) over F, g(x) in particular must have a root in K; call it r. Similar1y, one of the roots of the polynomialf'(y), say 1", is a root of g'(y) in K'. By Theorem 7-29, a can be extended to an isomorphism a' between the fields F(r) and F'(r'). Now, K is a splitting field off(x), viewed as a polynomial with coefficients from F(r); in a like manner, K' can be regan;led as a splitting field off'(y) over the field F'(r'). Because the number of roots off(x) lying outside of F(r) is les s than n,the induction hypothesis permits us to extend a' (itse1f an extension of a) to an is'Ó'fuorphism such that the restriction
Proof. This is ah immediate consequence of the theorem on taking F and a to be the identity isomorphism iF •

=

P

PROBLEMS 1. If R is a commutative ring with identity, prove that a) The set 1 = {¡(x) e R[[x]]lordf(x) > O} u {O} forros an ideal of the ring R[[x]]; in fact, 1 = (x).

152

,l·

FIRST COURSE IN RINGS ANO IOEALS

b) 'The ideal 1" eonsists ofall power series having order e) zJ' = {O}.

n..

~n,

together with O,

9. Consider the polynomial domain F[x], where F is a field, and a fixed element rE F. Show that t,he set of al1 polynomials having r as a' root,

2. For any field F, eonsider the set F(x> eonsisting of all expressions of the form

f

ak xk = IL.x-'

+ a_.+1x-·+l +

oo'

+ a_lx- 1 + ao + alx + a2 x 2 + '"

k=-n

" where aH the ak E F and n ~ Ovaries. I

. ·If addition and multiplieation are defined in the obvious way, F(x> becomes a "i\,ring, known as the ring af extended (formal) .pawer series aver F. Show that F(x> /('is in faet the field of quotients of the domain F[[x]J. [Hint: Given nEZ+, \YQc¡(F[[xJ]) must eontain x-'.J .3/''t,et R be a eornmutative ring with identity. If R is a local ring, prove that the ~':''''power series ring R[[x]] is also local.

~~)íÚiven that R is a eom~utative ring with identity, deduce that

;~:';:~) No monic polynomial is R[x] is a zero divisor. . "'}!:') If the polynomial f(x) = ao + a1x + + a.X' is a zero divisor in R[x], oO.

then there exists ari element O -+ rE R sueh that rf(x) = O. [Hint: Assume that f(x)g(x) = O. Use the polynomials akg(x) to obtain O =1= h(x) E R[x], with . deg h(x) < deg f(x), satisfying h(x)f(x) = O.J

5. If R is a eommutative ring with identity, verify that the polynomial 1 + ax is invertible in R[x J if and only if the element a is nilpotent in R. [Hint: Problem 10, Chapter L]

153

PROBLEMS

M,

= {¡(x) E F[x]lf(r) = O},

forms a maximal ideal of F[xJ,wlth F[x]jMr ~ F. [Hint: Mr -+ F is the substitu tion homomorphism indueed by r.]

= ker

10. Regarding thering ofExample 8, Chapter 1, show thatthe polynom¡al (a, 0)x 2 E R[x] has infinitely many roots in R[x].. . 11. Given f(x)

= ao + a1x +

oO,

+

a.x' E C[x], define the polyn¿inial f(x) by

+ a.x', iik denotes the usual eomplex eonjugate of dk • Verify t,§ii' ](x) = iio

+ ii1x +

'oO

where a) rE C is a root of/(x) if and only if r is a r90t ofJ(x). [Hint':]fr) = 1(r).J b) If f(x) E R#[xJ ~ C[x] and r is a eomplex root of f(x), th,en'r is also a root of f(x),,~.';';c '::::':: . :;.

12. Let R be a eommutative ring with identity and let f(x) E R[~J: The fundion 1: R -+ R defined by taking l(r) = f(r) for every rE R is eal1ea .the palynamial jimetian indueed by f(x). Assuming that P R denotes the set of all polynomial . funetions indueed by elements of R[x], prove that a) PR forms a subring of map R, known as the ring of polynomial funetions on R; b) the mapping a: R[xJ -+ PR given by a(!(x)) = 1 is a homomorphism of R[xJ onto PR ; . e) if the element rE R is fixed and 1, = {lE P Rll
6. For an arbitrary ring R, prove that . a) If 1 is an ideal of R, then I[x] forms an ideal ofthe polynomial ring R[xJ. b) If R and R' are isomorphie rings, then R[x] is isomorphie to R'ex]. e) ehar R = ehar R[x] = ehar R[[x]J. d) If 1 is a ni! ideal of R, then I[x] is a ni! ideal of R[xJ. [Hint: Induet on the degree of polynomials in I[x].J

13. a) When R is an integral domain, show that <;Iistinet polynomials in R[x] induce distinet polynomial funetions (in other words, the mapping a: R[x J -+ P R is one-to-one) if ando only if R has an infinite number of elements. b) Give an example oftwo distinet polynomials whieh induce the same polynomial funetion.

7. Establish the following assertions eoneerning the polynomialring Z[xJ: a) The ideal

14. Let R be a eommutative ring with identity and·define the funetion 15: R[xJ the so-ealled derivative fonctian, as fo1!ows: If f(x) = ao + a1x + + a.X' E R[x],

(x)

=

{a1x

+ a2 r +

oO,

+

a.X'lak E Z; n ~ 1}

is a prime ideal of Z[x], but not a maximal ideaL Ineidentally, (x) is maximal . in F[x], where F is a field.· . . b) Z[x] is not a principal ideal domain. [Hint: Conslder (x, 2), the (maxlmal) ideal of polynomials with even eonstant terros.J e) The primary ideal (x, 4) is not the power of any prime ideal of Z[x]. [Hint: (x, 2) is the only prime ideal eontaining (x,4).J 8. Let P be a prime ideal of R, a eommutative ring with identity. Prove that P[xJ is a prime ideal of the polynomial ring R[x]. If M is a maximal ideal of R, is M[xJ a maximal ideal of R[x]?

-+

R[x],

.oO

then For any f(x), g(x) E R[x] and any rE R, establish that + 15g(x).

a) 15(!(x) + g(x)) = 15f(x) b) 15(rf(x)) = r15f(x).

e) 15(!(x)g(x)) off(x).J

= 15f(x)'g(x) + f(x)·15g(x). [Hint: Induet on the number of terms

15. Suppose that R is a eommutative ring with identity and let r E R be a root of the nonzero polynomial f(x) E R[x]. We eall r a multiple raat of f(x) provided that

f(x)

=

(x - r)'g(x)

(n > 1),

152

,l·

FIRST COURSE IN RINGS ANO IOEALS

b) 'The ideal 1" eonsists ofall power series having order e) zJ' = {O}.

n..

~n,

together with O,

9. Consider the polynomial domain F[x], where F is a field, and a fixed element rE F. Show that t,he set of al1 polynomials having r as a' root,

2. For any field F, eonsider the set F(x> eonsisting of all expressions of the form

f

ak xk = IL.x-'

+ a_.+1x-·+l +

oo'

+ a_lx- 1 + ao + alx + a2 x 2 + '"

k=-n

" where aH the ak E F and n ~ Ovaries. I

. ·If addition and multiplieation are defined in the obvious way, F(x> becomes a "i\,ring, known as the ring af extended (formal) .pawer series aver F. Show that F(x> /('is in faet the field of quotients of the domain F[[x]J. [Hint: Given nEZ+, \YQc¡(F[[xJ]) must eontain x-'.J .3/''t,et R be a eornmutative ring with identity. If R is a local ring, prove that the ~':''''power series ring R[[x]] is also local.

~~)íÚiven that R is a eom~utative ring with identity, deduce that

;~:';:~) No monic polynomial is R[x] is a zero divisor. . "'}!:') If the polynomial f(x) = ao + a1x + + a.X' is a zero divisor in R[x], oO.

then there exists ari element O -+ rE R sueh that rf(x) = O. [Hint: Assume that f(x)g(x) = O. Use the polynomials akg(x) to obtain O =1= h(x) E R[x], with . deg h(x) < deg f(x), satisfying h(x)f(x) = O.J

5. If R is a eommutative ring with identity, verify that the polynomial 1 + ax is invertible in R[x J if and only if the element a is nilpotent in R. [Hint: Problem 10, Chapter L]

153

PROBLEMS

M,

= {¡(x) E F[x]lf(r) = O},

forms a maximal ideal of F[xJ,wlth F[x]jMr ~ F. [Hint: Mr -+ F is the substitu tion homomorphism indueed by r.]

= ker

10. Regarding thering ofExample 8, Chapter 1, show thatthe polynom¡al (a, 0)x 2 E R[x] has infinitely many roots in R[x].. . 11. Given f(x)

= ao + a1x +

oO,

+

a.x' E C[x], define the polyn¿inial f(x) by

+ a.x', iik denotes the usual eomplex eonjugate of dk • Verify t,§ii' ](x) = iio

+ ii1x +

'oO

where a) rE C is a root of/(x) if and only if r is a r90t ofJ(x). [Hint':]fr) = 1(r).J b) If f(x) E R#[xJ ~ C[x] and r is a eomplex root of f(x), th,en'r is also a root of f(x),,~.';';c '::::':: . :;.

12. Let R be a eommutative ring with identity and let f(x) E R[~J: The fundion 1: R -+ R defined by taking l(r) = f(r) for every rE R is eal1ea .the palynamial jimetian indueed by f(x). Assuming that P R denotes the set of all polynomial . funetions indueed by elements of R[x], prove that a) PR forms a subring of map R, known as the ring of polynomial funetions on R; b) the mapping a: R[xJ -+ PR given by a(!(x)) = 1 is a homomorphism of R[xJ onto PR ; . e) if the element rE R is fixed and 1, = {lE P Rll
6. For an arbitrary ring R, prove that . a) If 1 is an ideal of R, then I[x] forms an ideal ofthe polynomial ring R[xJ. b) If R and R' are isomorphie rings, then R[x] is isomorphie to R'ex]. e) ehar R = ehar R[x] = ehar R[[x]J. d) If 1 is a ni! ideal of R, then I[x] is a ni! ideal of R[xJ. [Hint: Induet on the degree of polynomials in I[x].J

13. a) When R is an integral domain, show that <;Iistinet polynomials in R[x] induce distinet polynomial funetions (in other words, the mapping a: R[x J -+ P R is one-to-one) if ando only if R has an infinite number of elements. b) Give an example oftwo distinet polynomials whieh induce the same polynomial funetion.

7. Establish the following assertions eoneerning the polynomialring Z[xJ: a) The ideal

14. Let R be a eommutative ring with identity and·define the funetion 15: R[xJ the so-ealled derivative fonctian, as fo1!ows: If f(x) = ao + a1x + + a.X' E R[x],

(x)

=

{a1x

+ a2 r +

oO,

+

a.X'lak E Z; n ~ 1}

is a prime ideal of Z[x], but not a maximal ideaL Ineidentally, (x) is maximal . in F[x], where F is a field.· . . b) Z[x] is not a principal ideal domain. [Hint: Conslder (x, 2), the (maxlmal) ideal of polynomials with even eonstant terros.J e) The primary ideal (x, 4) is not the power of any prime ideal of Z[x]. [Hint: (x, 2) is the only prime ideal eontaining (x,4).J 8. Let P be a prime ideal of R, a eommutative ring with identity. Prove that P[xJ is a prime ideal of the polynomial ring R[x]. If M is a maximal ideal of R, is M[xJ a maximal ideal of R[x]?

-+

R[x],

.oO

then For any f(x), g(x) E R[x] and any rE R, establish that + 15g(x).

a) 15(!(x) + g(x)) = 15f(x) b) 15(rf(x)) = r15f(x).

e) 15(!(x)g(x)) off(x).J

= 15f(x)'g(x) + f(x)·15g(x). [Hint: Induet on the number of terms

15. Suppose that R is a eommutative ring with identity and let r E R be a root of the nonzero polynomial f(x) E R[x]. We eall r a multiple raat of f(x) provided that

f(x)

=

(x - r)'g(x)

(n > 1),

154

FIRST COURSE IN RINGS ANO IDEALS

where g(x)ER[xJ is a polynomial such that g(r) -+ O. Prove that an element rE R is a multiple root of f(x) if and only if ,. is a root of both f(x) and c5f(x). 16. Let F be a field andf(x) E F[xJ be a polynomial of degree 2 or 3. Deduce that f(x) is irreducible in F[xJ if and only if it has no root in F. Give an example which shows that this result need not hold if degf(x) ~ 4.

17. Prove that if a polynomial f(x) E F[x J (F a field of characteristic O) is irreducible, then a)1 of its roots in any field containing F must be distinct. [Hint: First show tha~ gcd (J(x), c5f(x)) = 1.J 18. Given that l is a proper ideal of R, a cornmutative ring with identity, establish the assertions below: a) If v: R[xJ -> (R/I)[xJ is the reduction homomorphism modulo the ideal l, thenker v = l[xJ;hence,R[xJ/l[xJ ~ (R/l)[xJ. b) If the polynomiaIJ(x)E R[xJ is such that v(J(x)) is irreducible in (R/l)[xJ, thenf(x) is irreducíl;>le in R[xJ. c) The polynomial f(xl = x 3 - X2 + 1 is irreducible in Z[xJ. [Hint: Reduce the coefficients modulo 2.J 19. Let R be a unique factorization domain. Show that any nonconstant divisor of a primitive polynomial in R[xJ is again primitive. 20. Utilize Gauss's Lemma to give an alternative proof of the fact that if R is a unique factorization domain, then so is the polynomial ring R[xJ. [Hint: For f(x) E R[xJ, induct on degf(x); if degf(x) > O, write f(x) = Cfl(X), where e E R and fl(X) is primitive; if fl(X) is reducible, apply induction to its factors.]

0-+

21. Apply the Eisenstein Criterion to establish that the following polynomials are irreducibleinQ[xJ:f(x) = X2 + 1,g(x) = X2 - X + 1,andh(x) = 2x s - 6x 3 +

9X2 - 15. [Hint: Considerf(x

+

1), g(-x).J

22. Let R be a unique factorization doniih.t and K its field of quotients. Assume that ab- l E K (where a and b are relativéi)(p'rime) is a nonzero root of the polynomial f(x) = a o + alx + ... + a.X' E R[x;J" Verify that aJa o and bJa•. 23. Prove ,the following assertions concér~,ng the polynomial ring Z[x, y]: a) The ideals (x), (x, y) and (2, x, y) i§''e aH prime in Z[x, yJ, but only the last is maximal. . :. :' b) (x, y) = .J(x2, y) = .J(x2, xy, y2):;::; c) The ideal (xl', xy, y2) is primary iifZ[x, yJ for any integer k E Z+. d) If l = (x 2, xy), then.JI is a prime i?eal, but lis not primary. [Hint:.JI = (x).J 24. Consider the polynomial domain F[Jé, yJ, where F forms a field. a) Show that (x 2, xy, y2) is not a prin'cipal ideal of F[ x, y J. ' b) Establish the isomorphlsm F[ x, y J/(x + y) ~ F[ xJ. 25. Let the element r be algebraic over the field F and let f(x) E F[ x J be a monic polynomial such thatf(r) = O. Prove thatf(x) is the mínimum polynomial of r over F if and only iff(x) is irreducible in F[xJ. 26. Assuming that F' is a finite extension of the field F, verify each of the statements below:

PROBLEMS

155

a) When [F': F] is prilI!e, F' is a simple extension of F; in fact, F' = F(I') for every element rE F' - F. b) Iff(x) E F[xJ is an irreducible polynomial whose degree is relatively prime to [F':FJ, thenf(x) has no roots in F'. c) If r E F' is algebraic of degree n, then each element of F(r) has as its degree an integer dividing n. d) Given fields Ki (i = 1,2) such that F' 2 Ki 2 F, with [Kl:F] and [K2:FJ relatively prime integers, necessarily Kl n K 2 = F. 27. Show that the following extension fields of Q are simple extensions and determine their respective degrees: Q(.j3, .j7), Q(.j3, 0, Q(.ji, ~), 28. a) Prove that the extension field F' = F(r l , r 2, ... , r.), where each element r i is algebraic over F, forms a finite extension of F. [Hint: If Fi = Fi_l(r;), then F. = F' and [F':FJ = IT¡[Fi+1:FiJ.] b) If F" is an algebraic extension of F' and F' is an algebraic extension of F, show that F" is an algebraic extension of F. [Hint: Each r E FU is a root of sorne polynomial f(x) = a o + alx + ... + a.x' E F'exJ; consider the extension fields K ~ F(a o, al' ... , a.) and K' = K(r); r is algebraic over K'.] 29. Let F' be an extension of the field F. Prove that the set of aH elements in F' which are algebraic over F constitute a s,ubfield of F'; applied to the case where F' = C and F = Q, this yields the field of algebraic numbers. [Hint: If r, s are algebraic over F, [F(r, s):FJ is finite; hence, F(r, s) is an algebraic extension of F.]

JO. a) Granting that f(x) = X2 + X + 2 is an irreducible polynomial in Z3[ x J, construct the multiplication table for the field Z3[XJ/(f(x)). b) Show that the polynomial f(x) = x 3 + X2 + 1 E Z2[XJ factors into linear factors in Z2[XJ/(f(x)) by actually finding the factorization. 31. If n -+ 1 is a (nonzero) square-free integer, verify that Q[xJ/(x 2 - n) forros a field isomorph¡c to, tIle quadratic field

Q(Jñ) = {a

+ bJñJa, b E Q}.

32. Describe the spíitting fields' of the following polynomials: a) x 3 - 3 E Q[;J, b) X2 + X + Zs[xJ, c) x 4 + 2X2 ER#[xJ, d) (xl - 2)(x 2 .,+ l)EQ[xJ.

"í'e +1

33. Let r be a rodi of the polynomial f(x) = x 3 - X

of 1 - 2r

+

+

1 E Q[ x J. Find the inverse

3;'2 in Q(r).

34. Letf(x) E F[xJ be an irreducible polynomial and r, s be two roots off(x) in sorne

splitting field. ' Show that F[rJ ~ F[sJ, by a unique isomorphlsm that leaves every element of F fixed and takes r into s.

35. Suppose that F' is the splitting field for the polynomialf(x) E F[xJ; say

f(x) = a(x - r1)(x - rz} ... (x - r.)

(ri E F', a

-+

O).

154

FIRST COURSE IN RINGS ANO IDEALS

where g(x)ER[xJ is a polynomial such that g(r) -+ O. Prove that an element rE R is a multiple root of f(x) if and only if ,. is a root of both f(x) and c5f(x). 16. Let F be a field andf(x) E F[xJ be a polynomial of degree 2 or 3. Deduce that f(x) is irreducible in F[xJ if and only if it has no root in F. Give an example which shows that this result need not hold if degf(x) ~ 4.

17. Prove that if a polynomial f(x) E F[x J (F a field of characteristic O) is irreducible, then a)1 of its roots in any field containing F must be distinct. [Hint: First show tha~ gcd (J(x), c5f(x)) = 1.J 18. Given that l is a proper ideal of R, a cornmutative ring with identity, establish the assertions below: a) If v: R[xJ -> (R/I)[xJ is the reduction homomorphism modulo the ideal l, thenker v = l[xJ;hence,R[xJ/l[xJ ~ (R/l)[xJ. b) If the polynomiaIJ(x)E R[xJ is such that v(J(x)) is irreducible in (R/l)[xJ, thenf(x) is irreducíl;>le in R[xJ. c) The polynomial f(xl = x 3 - X2 + 1 is irreducible in Z[xJ. [Hint: Reduce the coefficients modulo 2.J 19. Let R be a unique factorization domain. Show that any nonconstant divisor of a primitive polynomial in R[xJ is again primitive. 20. Utilize Gauss's Lemma to give an alternative proof of the fact that if R is a unique factorization domain, then so is the polynomial ring R[xJ. [Hint: For f(x) E R[xJ, induct on degf(x); if degf(x) > O, write f(x) = Cfl(X), where e E R and fl(X) is primitive; if fl(X) is reducible, apply induction to its factors.]

0-+

21. Apply the Eisenstein Criterion to establish that the following polynomials are irreducibleinQ[xJ:f(x) = X2 + 1,g(x) = X2 - X + 1,andh(x) = 2x s - 6x 3 +

9X2 - 15. [Hint: Considerf(x

+

1), g(-x).J

22. Let R be a unique factorization doniih.t and K its field of quotients. Assume that ab- l E K (where a and b are relativéi)(p'rime) is a nonzero root of the polynomial f(x) = a o + alx + ... + a.X' E R[x;J" Verify that aJa o and bJa•. 23. Prove ,the following assertions concér~,ng the polynomial ring Z[x, y]: a) The ideals (x), (x, y) and (2, x, y) i§''e aH prime in Z[x, yJ, but only the last is maximal. . :. :' b) (x, y) = .J(x2, y) = .J(x2, xy, y2):;::; c) The ideal (xl', xy, y2) is primary iifZ[x, yJ for any integer k E Z+. d) If l = (x 2, xy), then.JI is a prime i?eal, but lis not primary. [Hint:.JI = (x).J 24. Consider the polynomial domain F[Jé, yJ, where F forms a field. a) Show that (x 2, xy, y2) is not a prin'cipal ideal of F[ x, y J. ' b) Establish the isomorphlsm F[ x, y J/(x + y) ~ F[ xJ. 25. Let the element r be algebraic over the field F and let f(x) E F[ x J be a monic polynomial such thatf(r) = O. Prove thatf(x) is the mínimum polynomial of r over F if and only iff(x) is irreducible in F[xJ. 26. Assuming that F' is a finite extension of the field F, verify each of the statements below:

PROBLEMS

155

a) When [F': F] is prilI!e, F' is a simple extension of F; in fact, F' = F(I') for every element rE F' - F. b) Iff(x) E F[xJ is an irreducible polynomial whose degree is relatively prime to [F':FJ, thenf(x) has no roots in F'. c) If r E F' is algebraic of degree n, then each element of F(r) has as its degree an integer dividing n. d) Given fields Ki (i = 1,2) such that F' 2 Ki 2 F, with [Kl:F] and [K2:FJ relatively prime integers, necessarily Kl n K 2 = F. 27. Show that the following extension fields of Q are simple extensions and determine their respective degrees: Q(.j3, .j7), Q(.j3, 0, Q(.ji, ~), 28. a) Prove that the extension field F' = F(r l , r 2, ... , r.), where each element r i is algebraic over F, forms a finite extension of F. [Hint: If Fi = Fi_l(r;), then F. = F' and [F':FJ = IT¡[Fi+1:FiJ.] b) If F" is an algebraic extension of F' and F' is an algebraic extension of F, show that F" is an algebraic extension of F. [Hint: Each r E FU is a root of sorne polynomial f(x) = a o + alx + ... + a.x' E F'exJ; consider the extension fields K ~ F(a o, al' ... , a.) and K' = K(r); r is algebraic over K'.] 29. Let F' be an extension of the field F. Prove that the set of aH elements in F' which are algebraic over F constitute a s,ubfield of F'; applied to the case where F' = C and F = Q, this yields the field of algebraic numbers. [Hint: If r, s are algebraic over F, [F(r, s):FJ is finite; hence, F(r, s) is an algebraic extension of F.]

JO. a) Granting that f(x) = X2 + X + 2 is an irreducible polynomial in Z3[ x J, construct the multiplication table for the field Z3[XJ/(f(x)). b) Show that the polynomial f(x) = x 3 + X2 + 1 E Z2[XJ factors into linear factors in Z2[XJ/(f(x)) by actually finding the factorization. 31. If n -+ 1 is a (nonzero) square-free integer, verify that Q[xJ/(x 2 - n) forros a field isomorph¡c to, tIle quadratic field

Q(Jñ) = {a

+ bJñJa, b E Q}.

32. Describe the spíitting fields' of the following polynomials: a) x 3 - 3 E Q[;J, b) X2 + X + Zs[xJ, c) x 4 + 2X2 ER#[xJ, d) (xl - 2)(x 2 .,+ l)EQ[xJ.

"í'e +1

33. Let r be a rodi of the polynomial f(x) = x 3 - X

of 1 - 2r

+

+

1 E Q[ x J. Find the inverse

3;'2 in Q(r).

34. Letf(x) E F[xJ be an irreducible polynomial and r, s be two roots off(x) in sorne

splitting field. ' Show that F[rJ ~ F[sJ, by a unique isomorphlsm that leaves every element of F fixed and takes r into s.

35. Suppose that F' is the splitting field for the polynomialf(x) E F[xJ; say

f(x) = a(x - r1)(x - rz} ... (x - r.)

(ri E F', a

-+

O).

156

FIRST COURSE 'IN RINGS AND IDEALS

EIGHT

Prove that F' = F(r l' r2' ... , r.). As a particular illustration, establish' that Q(J2,J3) is the splitting field of (x 2 - 2)(x 2 - 3) E Q[x].

36. Letf(x) E Zp[x] be an irreducible polynomial of degree n, p a prime. Verify that the field F = Zp[xJ/(J(x)) contains p. elements. 37. If F' is a splitting field of a polynomial of degree n over F, show that [F' :F] ::;; n! 38. A field F' is said to be algebraica/ly e/osed if F' has 'n'p proper algebraic extensions. Assurnlng that F' is an algebraic extension of F, prove the equivalence of the follciwing statements: a) F' is algebraically closed. b) Every irreducible polynomial in F'[x] is linear..... e) Every polynomial in F[x] splits in F'. '. . (For a proof that every field has an algebraic eXf~,nsion which is algebraicalIy i! ,,' . closed, the reader is re(erred to [23].)

'..

,

: . ~. ~~ ~ ...

";j,

I

CERTAIN RADICALS OF A RING '.'

"'.:

We touched earlier on the~rádica1 concept by briefiy considering the notion of the ni! radical of an ¡-¿le·al. There are a number of other radicals in circuhition ;. several of the.' more prominent ones are introduced jn this chapter. These various fofiriulations are not, in general, equivalent to one another and this has given.tise to certain confusion and a111biguity in the use ofthe termo (Indeed, whenever the reader encounters the word "radical" by itself, he should take sorne pains to discover just what ismeant by it.) By way of removing sorne of this confusion, qualifying adjectives are given to the different types ofradicals which appear here. In addition to indicating the importance of these new radical s in the structure theory, we will be . con cerned with the pature of the inclusion relations between them and the circumstances under which various radicalscoincide. The reader is agaÍn reminded that, inthe absence ol any statement to the contrary, ihe term "ring" will always mean a commutative ring with identity. It appears in order to defipe one of the radicals around which our interest centers. Definition 8-1. The Jacobsonradical of a ring R, denoted by rad R, is the set

rad R =

n {MIM is a maximal ideal of R}.

If rad R = {O}, then R is said to be a ring without Jacobson radical or,

more briefiy, R is a semisimple ringo The Jacobson radical always exists, since we knOw by Theorem 5-2 that any commutative ring with identity contains at least one maximal ideal. It is also irnmediately obvious from the definition that rad R forms an ideal of R which is contained in each maximal ideal. To fix ideas, let us determine the Jacobson radical in several concrete rings. Example 8-1. The ring Z of integers is a semi simple ringo For, according 157

156

FIRST COURSE 'IN RINGS AND IDEALS

EIGHT

Prove that F' = F(r l' r2' ... , r.). As a particular illustration, establish' that Q(J2,J3) is the splitting field of (x 2 - 2)(x 2 - 3) E Q[x].

36. Letf(x) E Zp[x] be an irreducible polynomial of degree n, p a prime. Verify that the field F = Zp[xJ/(J(x)) contains p. elements. 37. If F' is a splitting field of a polynomial of degree n over F, show that [F' :F] ::;; n! 38. A field F' is said to be algebraica/ly e/osed if F' has 'n'p proper algebraic extensions. Assurnlng that F' is an algebraic extension of F, prove the equivalence of the follciwing statements: a) F' is algebraically closed. b) Every irreducible polynomial in F'[x] is linear..... e) Every polynomial in F[x] splits in F'. '. . (For a proof that every field has an algebraic eXf~,nsion which is algebraicalIy i! ,,' . closed, the reader is re(erred to [23].)

'..

,

: . ~. ~~ ~ ...

";j,

I

CERTAIN RADICALS OF A RING '.'

"'.:

We touched earlier on the~rádica1 concept by briefiy considering the notion of the ni! radical of an ¡-¿le·al. There are a number of other radicals in circuhition ;. several of the.' more prominent ones are introduced jn this chapter. These various fofiriulations are not, in general, equivalent to one another and this has given.tise to certain confusion and a111biguity in the use ofthe termo (Indeed, whenever the reader encounters the word "radical" by itself, he should take sorne pains to discover just what ismeant by it.) By way of removing sorne of this confusion, qualifying adjectives are given to the different types ofradicals which appear here. In addition to indicating the importance of these new radical s in the structure theory, we will be . con cerned with the pature of the inclusion relations between them and the circumstances under which various radicalscoincide. The reader is agaÍn reminded that, inthe absence ol any statement to the contrary, ihe term "ring" will always mean a commutative ring with identity. It appears in order to defipe one of the radicals around which our interest centers. Definition 8-1. The Jacobsonradical of a ring R, denoted by rad R, is the set

rad R =

n {MIM is a maximal ideal of R}.

If rad R = {O}, then R is said to be a ring without Jacobson radical or,

more briefiy, R is a semisimple ringo The Jacobson radical always exists, since we knOw by Theorem 5-2 that any commutative ring with identity contains at least one maximal ideal. It is also irnmediately obvious from the definition that rad R forms an ideal of R which is contained in each maximal ideal. To fix ideas, let us determine the Jacobson radical in several concrete rings. Example 8-1. The ring Z of integers is a semi simple ringo For, according 157

158

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

to Example 5-1, the maximal ideals of Z are precisely the principal ideals generated by the prime numbers; thus, rad R

=

n {(p)lp is a prime number}.

Since no nonzero integer is divisible by every prime, we see at once that rad R = {O}. . Example'8-2. A more penetrating illustration is furnished by the ring R = map(X, F), where X is an arbitrary set and F a field. For any element

x E X, consider the function <,:/ = f(x) which assigns to each function in R its value at x. It is easily checked that
'Mx = {fE Rlf(x) = O}. Because rad R ~ n Mx = {fE Rlf(x) that R must be a semisimple ringo

=

O for all x E X}

=

{O}, it follows

Example 8-3. For a final example, we turn to the ring R[[x]] of formal power series. Here, there is a one-to-one correspondence between the maximal ideals M of R and maxirnal ideal s M' of R[[xJ] in such a way that M' corresponds to M if and only if M' is generated by M and x (Theorem 7-4). Thus, rad R[[xJ]

=

n {M'IM' is a maximal ideal of R[[xJ]}

= n {(M, x)IM is a maximal ideal of R} =

(n M, x)

=

To pro ve the converse, suppose that each member. of 1 + 1 has a multiplicative inverse in R, but 1 $ rad R. By definition of the Jacobson radical, there will exist a maximal ideal M of R with 1 $ M. Now, if a is any element of 1 which is not in M, the maximality of M implies that the ideal (M, a) = R. Knowing this, the identity element 1 can be expressed in the form 1 = m + ra for suitable choice of m E M and r E R. But then, m = 1 - ra E 1 + 1, so that m possesses an inverse. The conc1usion is untenable, since no proper ideal contains an invertible element. The form which this result takes when 1 is the principal ideal generated by a Erad R furnishes a characterization of the Jacobson radical in terms of elements rather than ideals. Although actually a corollary to the theorem just proved, it is iIilportant enough to be singled out as a theorem. Theorem 8-2. In any ring R, an element a Erad R if and only if 1 - ra is invertible for each r E R. This theorem adapts itself' to many uses. instructive applications are presented below. .

Three fairly short and

Corollary 1. An element a is invertible in the ring R if and only if the coset a + rad R is invertible in the quotient ring Rlrad R. Proof. Assume that the coset a + rad ~ has an inverse in Rlrad R, so that (a + rad R)(b + rad R) = 1 + rad R for sorne bE R. Then 1 - ab lies in rad R. We now appeal to Theorem 8-2, with r = 1, to conc1ude that the product ab = 1 - 1(1 - ab) is invertible; this, in turn, forces the element a to h~ve an inverse in R. The

other direction of the corbllary is all but obvious.";'<·"

(rad R, x).

In particular, if R is taken to be a field F, we have rad F[[xJ] principal ideal generated by X.

159

Corollary 2. The only idempotent element =

(x),;Jhe :

Our first theorem establishes a basic connection between the Jacobson radical and invertibility of ring elements. . Theorem 8-1. Let 1 be an ideal of the ring R. Then 1 ~ r~d R i(and only if each element of the coset 1 + 1 has ·an inverse in R. ; Proof. We begin by assuming that 1 ~ rad R and that there is sorne element a El for which 1 + a is not invertible. Our object, of course, 'is

to derive a contradiction. By the corollary to Theorem 5-3, the element 1 + a must belong to sorne maximal ideal M ofthe ring R. Since a E rao R, a is also contained in M, and therefore 1 = (1 + a) - a líes in M. But this mean s that M = R, which is c1early impossible.

i~;';;~d R is O.

Proof. Let the element a Erad R with a 2 = a. TaÍcing r = 1 in the pre-

ceding theorem, we see that 1 - a has an inverse:iiJ R; say (1 - ",)b where b E R. This 1eads irnmediately to '

=

1,

a = a(l - a)b = (a - a 2 )b'e O,

which completes the proof. Corollary 3. Every nil ideal of R is contained in rad R. Proof. Let N be a nil ideal of R and suppose that a E N. For every rE R, we then have ra E N, so that the product ra is nilpotent. Problem 10, Chapter 1, therefore implíes that 1 - ra is invertible in R. This shows that the element a líes in rad R, from which it follows that N ~ rad R.

Although the Jacobson radical of a ring R is not necessarily a nil ideal; very little restriction on R force s it to be nil. We shall see subsequentIy

158

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

to Example 5-1, the maximal ideals of Z are precisely the principal ideals generated by the prime numbers; thus, rad R

=

n {(p)lp is a prime number}.

Since no nonzero integer is divisible by every prime, we see at once that rad R = {O}. . Example'8-2. A more penetrating illustration is furnished by the ring R = map(X, F), where X is an arbitrary set and F a field. For any element

x E X, consider the function <,:/ = f(x) which assigns to each function in R its value at x. It is easily checked that
'Mx = {fE Rlf(x) = O}. Because rad R ~ n Mx = {fE Rlf(x) that R must be a semisimple ringo

=

O for all x E X}

=

{O}, it follows

Example 8-3. For a final example, we turn to the ring R[[x]] of formal power series. Here, there is a one-to-one correspondence between the maximal ideals M of R and maxirnal ideal s M' of R[[xJ] in such a way that M' corresponds to M if and only if M' is generated by M and x (Theorem 7-4). Thus, rad R[[xJ]

=

n {M'IM' is a maximal ideal of R[[xJ]}

= n {(M, x)IM is a maximal ideal of R} =

(n M, x)

=

To pro ve the converse, suppose that each member. of 1 + 1 has a multiplicative inverse in R, but 1 $ rad R. By definition of the Jacobson radical, there will exist a maximal ideal M of R with 1 $ M. Now, if a is any element of 1 which is not in M, the maximality of M implies that the ideal (M, a) = R. Knowing this, the identity element 1 can be expressed in the form 1 = m + ra for suitable choice of m E M and r E R. But then, m = 1 - ra E 1 + 1, so that m possesses an inverse. The conc1usion is untenable, since no proper ideal contains an invertible element. The form which this result takes when 1 is the principal ideal generated by a Erad R furnishes a characterization of the Jacobson radical in terms of elements rather than ideals. Although actually a corollary to the theorem just proved, it is iIilportant enough to be singled out as a theorem. Theorem 8-2. In any ring R, an element a Erad R if and only if 1 - ra is invertible for each r E R. This theorem adapts itself' to many uses. instructive applications are presented below. .

Three fairly short and

Corollary 1. An element a is invertible in the ring R if and only if the coset a + rad R is invertible in the quotient ring Rlrad R. Proof. Assume that the coset a + rad ~ has an inverse in Rlrad R, so that (a + rad R)(b + rad R) = 1 + rad R for sorne bE R. Then 1 - ab lies in rad R. We now appeal to Theorem 8-2, with r = 1, to conc1ude that the product ab = 1 - 1(1 - ab) is invertible; this, in turn, forces the element a to h~ve an inverse in R. The

other direction of the corbllary is all but obvious.";'<·"

(rad R, x).

In particular, if R is taken to be a field F, we have rad F[[xJ] principal ideal generated by X.

159

Corollary 2. The only idempotent element =

(x),;Jhe :

Our first theorem establishes a basic connection between the Jacobson radical and invertibility of ring elements. . Theorem 8-1. Let 1 be an ideal of the ring R. Then 1 ~ r~d R i(and only if each element of the coset 1 + 1 has ·an inverse in R. ; Proof. We begin by assuming that 1 ~ rad R and that there is sorne element a El for which 1 + a is not invertible. Our object, of course, 'is

to derive a contradiction. By the corollary to Theorem 5-3, the element 1 + a must belong to sorne maximal ideal M ofthe ring R. Since a E rao R, a is also contained in M, and therefore 1 = (1 + a) - a líes in M. But this mean s that M = R, which is c1early impossible.

i~;';;~d R is O.

Proof. Let the element a Erad R with a 2 = a. TaÍcing r = 1 in the pre-

ceding theorem, we see that 1 - a has an inverse:iiJ R; say (1 - ",)b where b E R. This 1eads irnmediately to '

=

1,

a = a(l - a)b = (a - a 2 )b'e O,

which completes the proof. Corollary 3. Every nil ideal of R is contained in rad R. Proof. Let N be a nil ideal of R and suppose that a E N. For every rE R, we then have ra E N, so that the product ra is nilpotent. Problem 10, Chapter 1, therefore implíes that 1 - ra is invertible in R. This shows that the element a líes in rad R, from which it follows that N ~ rad R.

Although the Jacobson radical of a ring R is not necessarily a nil ideal; very little restriction on R force s it to be nil. We shall see subsequentIy

160

Again appealing to Theorem 8-2, we conclude that the element

that, if every ideal of R is finitely generated, then rad R is not only ni! but nilpotent. This is a cpnvenient place to also point out that a homomorphic image of a semisimple ríng need not be semisimpleo An explicit example of tbis situation can easily be obtained from the ring Z of integerso While Z forms a ring without a Jacobson radical, its homomorphic image Zpn (p a prime; n > 1) eqntains the nil ideal (p); appealing to Corollary 3 aboye, we see that Zp. cannot be semisimpleo Example 8-4.;iConsider F[[x ]], the ríng of formal power series over a field F. From th6';lemma 011 page 116, it is known that an element f(x) = ao + a 1x + '.o:.~ + a.x· + .0. has an inverse in F[[x]] if and only if the constant ter~~ap =1= 00 This observatión (in conjunction with Theorem 8-2) implies that:if g(x) = biJ + b1x + o.. + bllx" + ... , then , rad

F[[~:iJ~'~

{j(x)!

i - f(x)g(x) is invertible for aH g(x) E F[[xJ]} F} ,

• ":,;'h {j(x)!1 - aob o =1= O for all bo E

= {j(x)!a o =

O} = (x).

Wenext prove several resuItsbearing on the Jacobson radical of qt¡.otient, ríngs. The first.of these provides a convenient method for manufacturil.lg semisimple rings; its proof utilizes both implications of the iast theorem. Theorem 8-3. For any ring R, the quotient ring Rjrad R is semisimple; that is, rad(Rjrad R) = {O}o Proo/. Before becoming involved in details, let us remark that since rad R ,constitutes an ideal of R, we may certainly form the quotient ring RjradR. To simplify no tati o n somewhat, we will temporarily denote rae!. R by 1. Suppose that the coset a + 1 Erad (Rjl). Our strategy is to show that the element a E 1, for then a + 1 = 1, which would imply that rad (Rjl) eonsists of only the zero element of Rjlo Sinee a + 1 is a member of rad (Rjl), Theorem 8-2 asserts that

+ 1)

- (r+ l)(a

+

1) = 1 - ra

+

1

is invertible in Rjl for each choice of r E Ro Accordingly, there exists a coset b + 1 (depending, of course,.on both r and a) such that (1 - ra

+ l)(b +

b - rab

= 1 -

1(1 - b

+ rab)

has an inve,fse e in R. But then (1 - ra)(bc) = (b - rab)c

=

1,

so that 1 - ra possesses a multiplicative inverse in R. As this argl,lment holds for every r E R, it follows that a Erad R = 1, as desiredo " '-, .

.,

'0'.

Continuing this theme, let us expre~s the Jacobson radical ofthe qiJotient ríng Rj1 as a function of the radical of R. ;':: ,'¡ • ~ •

.

Theorem 8-4. If 1 is an ideal of the ríng R, then rad R 1) rad (Rjl) 2 1 2) whenever 1

5;

+1 "

1) = 1

+

This is plainly equivalent to requiring 1 - (b - rab) E 1 = rad R.

lo

;,';'

:..

,'

and,

rad R, rad (Rjl) = (rad R)/1.

Proo/. Perhapsthe q~ickest way to establish the first assertion is by Íiieans of the Correspondence Theorem; using this, one has <

Thus, we have a second proof ofthe fact that the Jacobson radical of F[[xJ] is the principal ideal generatedby Xo ,

(1

161

CERTAIN RADICALS OF A RIN'G

FIRST COURSE IN RINGS AND IDEALS

rad (R/I)

=

•

n {M'! M' is a maximal ideal of RjI} ,

= n {nat¡M!M is a maximal ideal of <

2 nat¡ (rad R

+ 1), =

R with 1

5;

M}

rad R+ 1 1 '

which is the first part of our theorem (the crucial step requires the inclusion nJ<;MM 2 1 + rad R)o ' ' With an eye to proving (2), notice that whenever 1 5; rad R, then

,

'radR+l rad (Rjl) 2

.

l ' ~ (rad ,~)j!: :

Thus, we need only to show the inclusion (rad R)jl 2 rad (Rjl)o To this purpose, choose the coset a + 1 Erad (Rjl) and let M be an arbitrary maximal ideal of R. Since 1 5; rad R 5; M, the image nat¡M = Mjl must be a maximal ideal of the qUbtient ríng Rjl (Problem 3, Chapter 5)0 But ' then,

a

+ 1 Erad (R/I)

5;

Mjl,

foreing the element ato lie in Mo As this holds for every maxirnal ideal of R, it follows that a Erad R and so a + 1 E (rad R)/lo All in all, we have proved that rad (Rjl) 5; (rad R)jl, which, combined with our earlier inclusion, leads to (2). Armed with Theorem 8-4, we are in a position to establish:

160

Again appealing to Theorem 8-2, we conclude that the element

that, if every ideal of R is finitely generated, then rad R is not only ni! but nilpotent. This is a cpnvenient place to also point out that a homomorphic image of a semisimple ríng need not be semisimpleo An explicit example of tbis situation can easily be obtained from the ring Z of integerso While Z forms a ring without a Jacobson radical, its homomorphic image Zpn (p a prime; n > 1) eqntains the nil ideal (p); appealing to Corollary 3 aboye, we see that Zp. cannot be semisimpleo Example 8-4.;iConsider F[[x ]], the ríng of formal power series over a field F. From th6';lemma 011 page 116, it is known that an element f(x) = ao + a 1x + '.o:.~ + a.x· + .0. has an inverse in F[[x]] if and only if the constant ter~~ap =1= 00 This observatión (in conjunction with Theorem 8-2) implies that:if g(x) = biJ + b1x + o.. + bllx" + ... , then , rad

F[[~:iJ~'~

{j(x)!

i - f(x)g(x) is invertible for aH g(x) E F[[xJ]} F} ,

• ":,;'h {j(x)!1 - aob o =1= O for all bo E

= {j(x)!a o =

O} = (x).

Wenext prove several resuItsbearing on the Jacobson radical of qt¡.otient, ríngs. The first.of these provides a convenient method for manufacturil.lg semisimple rings; its proof utilizes both implications of the iast theorem. Theorem 8-3. For any ring R, the quotient ring Rjrad R is semisimple; that is, rad(Rjrad R) = {O}o Proo/. Before becoming involved in details, let us remark that since rad R ,constitutes an ideal of R, we may certainly form the quotient ring RjradR. To simplify no tati o n somewhat, we will temporarily denote rae!. R by 1. Suppose that the coset a + 1 Erad (Rjl). Our strategy is to show that the element a E 1, for then a + 1 = 1, which would imply that rad (Rjl) eonsists of only the zero element of Rjlo Sinee a + 1 is a member of rad (Rjl), Theorem 8-2 asserts that

+ 1)

- (r+ l)(a

+

1) = 1 - ra

+

1

is invertible in Rjl for each choice of r E Ro Accordingly, there exists a coset b + 1 (depending, of course,.on both r and a) such that (1 - ra

+ l)(b +

b - rab

= 1 -

1(1 - b

+ rab)

has an inve,fse e in R. But then (1 - ra)(bc) = (b - rab)c

=

1,

so that 1 - ra possesses a multiplicative inverse in R. As this argl,lment holds for every r E R, it follows that a Erad R = 1, as desiredo " '-, .

.,

'0'.

Continuing this theme, let us expre~s the Jacobson radical ofthe qiJotient ríng Rj1 as a function of the radical of R. ;':: ,'¡ • ~ •

.

Theorem 8-4. If 1 is an ideal of the ríng R, then rad R 1) rad (Rjl) 2 1 2) whenever 1

5;

+1 "

1) = 1

+

This is plainly equivalent to requiring 1 - (b - rab) E 1 = rad R.

lo

;,';'

:..

,'

and,

rad R, rad (Rjl) = (rad R)/1.

Proo/. Perhapsthe q~ickest way to establish the first assertion is by Íiieans of the Correspondence Theorem; using this, one has <

Thus, we have a second proof ofthe fact that the Jacobson radical of F[[xJ] is the principal ideal generatedby Xo ,

(1

161

CERTAIN RADICALS OF A RIN'G

FIRST COURSE IN RINGS AND IDEALS

rad (R/I)

=

•

n {M'! M' is a maximal ideal of RjI} ,

= n {nat¡M!M is a maximal ideal of <

2 nat¡ (rad R

+ 1), =

R with 1

5;

M}

rad R+ 1 1 '

which is the first part of our theorem (the crucial step requires the inclusion nJ<;MM 2 1 + rad R)o ' ' With an eye to proving (2), notice that whenever 1 5; rad R, then

,

'radR+l rad (Rjl) 2

.

l ' ~ (rad ,~)j!: :

Thus, we need only to show the inclusion (rad R)jl 2 rad (Rjl)o To this purpose, choose the coset a + 1 Erad (Rjl) and let M be an arbitrary maximal ideal of R. Since 1 5; rad R 5; M, the image nat¡M = Mjl must be a maximal ideal of the qUbtient ríng Rjl (Problem 3, Chapter 5)0 But ' then,

a

+ 1 Erad (R/I)

5;

Mjl,

foreing the element ato lie in Mo As this holds for every maxirnal ideal of R, it follows that a Erad R and so a + 1 E (rad R)/lo All in all, we have proved that rad (Rjl) 5; (rad R)jl, which, combined with our earlier inclusion, leads to (2). Armed with Theorem 8-4, we are in a position to establish:

162

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

163

Theorem 8-5. For any ring R, rad R is the smallest ideal l of R such that the quotient ring Rll is semisimple (in other words, if Rll is a semisimple ring, then rad R S;;; l).

yielding the contradiction 1 e Mi' But it is known that every proper ideal of R is contained in a maximal ideal of R (Theorem 5-2). From this contradiction we conclude that J must be finite.

Proof. From Theorem 8-3, it is already known that Rlrad R is without { Jacobson radical. Now, assume that l is any ideal of R fúr which the associated quotient ring Rll is semisimple. Using part (1) of the preceding theorem, we can then deduce the equality (I + rad R)ll = 1. This in turn leads to the inclusion rad R S;;; l, which is what we sought to pro ve.

Let us now turn to a consideration of another radical which plays an essential role in ring theory, to wit, the prime. radical. Its definition may also be framed in terms of the intersection of certain ideals.

This may be a good place to mention two theorems concerning the number of maximal ideals in a ring; these are of a rather special character, but typify the results that can be obtained. Theorem 8-6. Let F- be a principal ideal domain. Then, R is semisimple if and only if R is either a field or has an infinite number of maximal ideals. Proof. Let {p;} be the sét of prime elements of R. According to Theorem 6~·-10, themaximal ideals of R are simply the principal ideals (p¡). It follows that an element a erad R if and only if a is divisible by each prime Pi- If R has an infinite set of maximal ideals, then a = O, since every nonzero noninvertible element of R is uniquely representable as a finite product of

primes. On the other hand, if R contains only a finite number of primes PI' P2' ... , p", we have . rad R =

n (p¡) = (PIP2 ... Pn) +- {O}, n

,

i=1

so that R cannot be semisimple.. Finally, observe that if the set {Pi} is empty, then each nonzero element oJ Ris invertible and R is a field (in which case rad R = { O } ) . ' ( ! CoroUary. The ring Z of,integers is semisimple. Theorem 8-7. Let {M;},i' F J, be the set of maximal ideals of the ring R. If, for each i, there !e?Cists an element ai e Mi such that 1 - a i e rad R - Mi' then {Ma i~ a finite set. Proof. Suppose that the indéx set J is infinite. Then there exists a wellordering ~ of J under which J has no last el~ent. (See Appendix A for terminology.) For each ieJ, we define li ~"I( li
Definition 8-2. The prime radical of a ring R, denoted by Rad R (in contrast with rad R), is the set Rad R

=

n {plp is a prime ideal of R}.

If Rad R = {O}, we say that the ring R is without prime radical or has zero prime radical.

Theorem 5-7, together with Definition 8-1, shows that the prime radical exists, forms an ideal of R, and satisfies the inclusion Rad R S;;; rad R. It is useful to keep in mind that, for any integral domain, the zero ideal is a prime ideal ;for these rings, Rad R = {O}. In particular, the ring F[ [xJ] of formal power series over a field F has zero prime radical but, as we already know, a non trivial Jacobson radical. Perhaps the most striking result of the present chapter is that the prime radical, although seemingly quite different, is actually equal to the nil radical of a ringo The lemma below provides the key to establishing this assertion. Lemma. Let l be an ideal ofthe ring R. Further, assume that the subset . S S;;; R is closed under multiplication and disjoint from 1. Then there exists an ideal P which is maximal in the set of ideals which contain l aná '40 not meet s; any such ideal is necessarily prime. Proof.Consider the family $' of all ideals J of R such that l S;;; J and J n S == ifJ. This family is not empty since l itself satisfies the indicated conditions. Our immediate aim is to show that for any chain of ideals {J;} in $', their union u Ji also belongs to $'. It has already be'en established in Theorem 5-2 that the union of a chain of ideals is again an ideal; moreover, since l S;;; Ji for each i, we certainly have l S;;; u Ji' Finally, observe that ' (u J¡) n S = u (Ji n S) = u ifJ = ifJ.

The crux of the matter is that Zorn 's Lemma can now be applied to inter that $' has a maximal element P; this is the ideal that we want. By definition, P is maximal in the set of ideal s which contain l but do not meet S. To settle the whole affair there remains simply to show that P is a prime ideal. For this purpose, assume that the product ab e P but that a ~ P and b ~ P. Since it is strictIy larger than P, the ideal (P, a) must

.)

.'

162

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

163

Theorem 8-5. For any ring R, rad R is the smallest ideal l of R such that the quotient ring Rll is semisimple (in other words, if Rll is a semisimple ring, then rad R S;;; l).

yielding the contradiction 1 e Mi' But it is known that every proper ideal of R is contained in a maximal ideal of R (Theorem 5-2). From this contradiction we conclude that J must be finite.

Proof. From Theorem 8-3, it is already known that Rlrad R is without { Jacobson radical. Now, assume that l is any ideal of R fúr which the associated quotient ring Rll is semisimple. Using part (1) of the preceding theorem, we can then deduce the equality (I + rad R)ll = 1. This in turn leads to the inclusion rad R S;;; l, which is what we sought to pro ve.

Let us now turn to a consideration of another radical which plays an essential role in ring theory, to wit, the prime. radical. Its definition may also be framed in terms of the intersection of certain ideals.

This may be a good place to mention two theorems concerning the number of maximal ideals in a ring; these are of a rather special character, but typify the results that can be obtained. Theorem 8-6. Let F- be a principal ideal domain. Then, R is semisimple if and only if R is either a field or has an infinite number of maximal ideals. Proof. Let {p;} be the sét of prime elements of R. According to Theorem 6~·-10, themaximal ideals of R are simply the principal ideals (p¡). It follows that an element a erad R if and only if a is divisible by each prime Pi- If R has an infinite set of maximal ideals, then a = O, since every nonzero noninvertible element of R is uniquely representable as a finite product of

primes. On the other hand, if R contains only a finite number of primes PI' P2' ... , p", we have . rad R =

n (p¡) = (PIP2 ... Pn) +- {O}, n

,

i=1

so that R cannot be semisimple.. Finally, observe that if the set {Pi} is empty, then each nonzero element oJ Ris invertible and R is a field (in which case rad R = { O } ) . ' ( ! CoroUary. The ring Z of,integers is semisimple. Theorem 8-7. Let {M;},i' F J, be the set of maximal ideals of the ring R. If, for each i, there !e?Cists an element ai e Mi such that 1 - a i e rad R - Mi' then {Ma i~ a finite set. Proof. Suppose that the indéx set J is infinite. Then there exists a wellordering ~ of J under which J has no last el~ent. (See Appendix A for terminology.) For each ieJ, we define li ~"I( li
Definition 8-2. The prime radical of a ring R, denoted by Rad R (in contrast with rad R), is the set Rad R

=

n {plp is a prime ideal of R}.

If Rad R = {O}, we say that the ring R is without prime radical or has zero prime radical.

Theorem 5-7, together with Definition 8-1, shows that the prime radical exists, forms an ideal of R, and satisfies the inclusion Rad R S;;; rad R. It is useful to keep in mind that, for any integral domain, the zero ideal is a prime ideal ;for these rings, Rad R = {O}. In particular, the ring F[ [xJ] of formal power series over a field F has zero prime radical but, as we already know, a non trivial Jacobson radical. Perhaps the most striking result of the present chapter is that the prime radical, although seemingly quite different, is actually equal to the nil radical of a ringo The lemma below provides the key to establishing this assertion. Lemma. Let l be an ideal ofthe ring R. Further, assume that the subset . S S;;; R is closed under multiplication and disjoint from 1. Then there exists an ideal P which is maximal in the set of ideals which contain l aná '40 not meet s; any such ideal is necessarily prime. Proof.Consider the family $' of all ideals J of R such that l S;;; J and J n S == ifJ. This family is not empty since l itself satisfies the indicated conditions. Our immediate aim is to show that for any chain of ideals {J;} in $', their union u Ji also belongs to $'. It has already be'en established in Theorem 5-2 that the union of a chain of ideals is again an ideal; moreover, since l S;;; Ji for each i, we certainly have l S;;; u Ji' Finally, observe that ' (u J¡) n S = u (Ji n S) = u ifJ = ifJ.

The crux of the matter is that Zorn 's Lemma can now be applied to inter that $' has a maximal element P; this is the ideal that we want. By definition, P is maximal in the set of ideal s which contain l but do not meet S. To settle the whole affair there remains simply to show that P is a prime ideal. For this purpose, assume that the product ab e P but that a ~ P and b ~ P. Since it is strictIy larger than P, the ideal (P, a) must

.)

.'

164

FIRST COURSE IN RINGS ANO IOEALS

CERTAIN RAOICALS OF A RING

contain sorne element r of S; similarly, we can find an element s E S such that s E (P, b). This means that rs E (P, a)(P, b)

~

(P, ab)

~

Po

As S is hypothesized to be Closed under multiplication, the product rs al so lies in So But this obviously contradicts the fact that P n S = 00 Our argumenttherefore shows that either a or bis a member of P, which pro ves that PiSa prime ideal. " Remarko The ideal P need not be a maximal ideal of;R, in the usual meaning of the term, but only maximal with respect to exclusIdn of the set So To put it another way, if .lis any ideal of the ring R whicil properly contains P, fhen J must contain elements of So ,'\':: "

Two special cases of, this general setting are particularly noteworthy: S = {1} and l = {OJo In the event S = {l}, the i~¡é1iI P mentioned in the lemma is actual1y a maximal ideal (in the usuáCl<:1eal-theoretic sense); consequentIy, we have a somewhat different proof onhe facts that (i) every proper ideal is contained in a maximal ideal and (ii) each maximal ideal is primeo The case where l is the zero ideal is the subject ofthe following corollary, a result which will be utilized on several occasions in thesequel. Corollary. Let S be a subset of the ring R which is closed under multiplication and does not contain 00 Then there exists an ideal maximal in the set of ideals disjoint froro S; any such ideal is prime. As it stands, the preceding lemma is just the opening wedge; we can exploit it rather effectively by now proving Theorem 8-8. The intersection of all prime ideals of R which contain a given ideal l is precisely the nilradical of l :

)7 = n {plp

2, l; P

is a prime ideal}o

Ji,

Proa! Ifthe element a ~ then the set S = {a"ln E Z+} does not intersect lo Sin ce S is closed under multiplication, the preceding lemma insures the existence ofsome prime ideal P which contains l,but not a; that is, a does not belong to the intersection of prime ideals containing lo This establishes the incIusion n {plp

2 l; P

is a prime ideal} ~

.JI.

The reverse inclusion folIows readily upon noting that ifthere exists a prime ideal which contains l but not a, then a ~ )7,' since no power of a belongs toPo

165

As with the case of the Jaco bson radical, the prime radical may be characterized by its elements; this is brought out by a result promised earliero ' Corollary. The prime radIcal of a ring R coincides with the nil radical of R; that is, Rad/R is simply the ideal of all nilpotent elements of R. Proa! The assertion is aIl b~t obvious upon taking 8-80 '

r= {O} in Theorem

An immediate conseque~ce of this last corollary is the potentially powerful statement: every·nji~ideal of R is contained in the prime radical, not simply contained in the larger Jacobson radical (Corollary 3 to Theorem 8-2)0 ,/1, ' Example 8-5. For an il1usi~aÚon of Theorem 8-8, let us fall back on the ring Z ofintegerso In this segjí1;g, the nontrivial prime ideals are the principal ideal s (p), where p is a pri~e,fiumbero Given n > 1, the ideal (n) ~ (p) if and only if p divides n; this beihg so,

J[n)

=

n(Pi)o

pil·

Thus, if we assume that n has the prime power factorization ,n = p11p~2 000 p~r (k¡ E Z+), it follows that = (Pl) n (P2) n 000 n (p,) = (P1P2 000 p,)o

M

Let us go back to Theorem 8-8 for a momento Another ofits advantages is that it permits a rather simple characterization of' semiprime idealso (The reader is reminded that we defined an ideal l to be semiprime provided that l = )7)0 Theorem 8-? An ideal l of the ring R is a semiprime ideal if and only if 1 is an intersection of prime ideals of R. Proa! The proof is left to the reader; it should offer no difficultieso Corollary. The prime radical Rad R is a semi prime ideal which is contained in every semiprime ideal of R. ' Before pressing on, we should also prove the prime radical counterpart of Theorem 8-30 Theorem 8-10. For any ring R, the quotient ring R/Rad R is without prime radical. Proa! For clarity of exposition, set l = Rad R. Suppose that a anY nilpotent element of Rilo Then, foi: sorne positive integer n, (a + l)n = an + l = l,

+ l is

164

FIRST COURSE IN RINGS ANO IOEALS

CERTAIN RAOICALS OF A RING

contain sorne element r of S; similarly, we can find an element s E S such that s E (P, b). This means that rs E (P, a)(P, b)

~

(P, ab)

~

Po

As S is hypothesized to be Closed under multiplication, the product rs al so lies in So But this obviously contradicts the fact that P n S = 00 Our argumenttherefore shows that either a or bis a member of P, which pro ves that PiSa prime ideal. " Remarko The ideal P need not be a maximal ideal of;R, in the usual meaning of the term, but only maximal with respect to exclusIdn of the set So To put it another way, if .lis any ideal of the ring R whicil properly contains P, fhen J must contain elements of So ,'\':: "

Two special cases of, this general setting are particularly noteworthy: S = {1} and l = {OJo In the event S = {l}, the i~¡é1iI P mentioned in the lemma is actual1y a maximal ideal (in the usuáCl<:1eal-theoretic sense); consequentIy, we have a somewhat different proof onhe facts that (i) every proper ideal is contained in a maximal ideal and (ii) each maximal ideal is primeo The case where l is the zero ideal is the subject ofthe following corollary, a result which will be utilized on several occasions in thesequel. Corollary. Let S be a subset of the ring R which is closed under multiplication and does not contain 00 Then there exists an ideal maximal in the set of ideals disjoint froro S; any such ideal is prime. As it stands, the preceding lemma is just the opening wedge; we can exploit it rather effectively by now proving Theorem 8-8. The intersection of all prime ideals of R which contain a given ideal l is precisely the nilradical of l :

)7 = n {plp

2, l; P

is a prime ideal}o

Ji,

Proa! Ifthe element a ~ then the set S = {a"ln E Z+} does not intersect lo Sin ce S is closed under multiplication, the preceding lemma insures the existence ofsome prime ideal P which contains l,but not a; that is, a does not belong to the intersection of prime ideals containing lo This establishes the incIusion n {plp

2 l; P

is a prime ideal} ~

.JI.

The reverse inclusion folIows readily upon noting that ifthere exists a prime ideal which contains l but not a, then a ~ )7,' since no power of a belongs toPo

165

As with the case of the Jaco bson radical, the prime radical may be characterized by its elements; this is brought out by a result promised earliero ' Corollary. The prime radIcal of a ring R coincides with the nil radical of R; that is, Rad/R is simply the ideal of all nilpotent elements of R. Proa! The assertion is aIl b~t obvious upon taking 8-80 '

r= {O} in Theorem

An immediate conseque~ce of this last corollary is the potentially powerful statement: every·nji~ideal of R is contained in the prime radical, not simply contained in the larger Jacobson radical (Corollary 3 to Theorem 8-2)0 ,/1, ' Example 8-5. For an il1usi~aÚon of Theorem 8-8, let us fall back on the ring Z ofintegerso In this segjí1;g, the nontrivial prime ideals are the principal ideal s (p), where p is a pri~e,fiumbero Given n > 1, the ideal (n) ~ (p) if and only if p divides n; this beihg so,

J[n)

=

n(Pi)o

pil·

Thus, if we assume that n has the prime power factorization ,n = p11p~2 000 p~r (k¡ E Z+), it follows that = (Pl) n (P2) n 000 n (p,) = (P1P2 000 p,)o

M

Let us go back to Theorem 8-8 for a momento Another ofits advantages is that it permits a rather simple characterization of' semiprime idealso (The reader is reminded that we defined an ideal l to be semiprime provided that l = )7)0 Theorem 8-? An ideal l of the ring R is a semiprime ideal if and only if 1 is an intersection of prime ideals of R. Proa! The proof is left to the reader; it should offer no difficultieso Corollary. The prime radical Rad R is a semi prime ideal which is contained in every semiprime ideal of R. ' Before pressing on, we should also prove the prime radical counterpart of Theorem 8-30 Theorem 8-10. For any ring R, the quotient ring R/Rad R is without prime radical. Proa! For clarity of exposition, set l = Rad R. Suppose that a anY nilpotent element of Rilo Then, foi: sorne positive integer n, (a + l)n = an + l = l,

+ l is

166

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

so that d' E l. But 1 consists of all nilpotent elements of R. Thus, we must have (d,)m = O for suitably chosen m E Z+; this is simply the statement that a El, and, hence, a + 1 is the zero element of R/I. Our argument ~mp1ies that the quotient ring R/I has no nonzero nilpoient elements, which lS to say that Rad (R/I) = {O}.

Having dealt with 'these preliminaries we are now ready to prove Theorem 8-13. For any ring R, rad R[x]

tells us that 1

~

+ 1, and,

Rad R, Rad (R/I)

= (Rad R)/l.

where (Rad R)[x J denotes the ring of polynomials in x with coefficients from Rad R; In fact, the inclusion Rad R[x J ~ (Rad R)[x J is implicit in the foregoing proof; the opposite inclusion requires the corollary to Theorem: 8-8. By virtue of the displayed equation, we have

Lemma. A polynomialf(x) = €lo + alx + ... + a"x" is invertible in R[ x J if and only if a o is invertible in R and aIl the other coefficients al> al' ... , a" are nilpotent elements of R.

Proof. If aohas an inversein R and al' al' ... , a" are all nilpotent, then the polynomial f(x) = a o + alx + ... + anxn is tbe sum of irt ,invertible element ando a nilpotent elemento Hence, f(x) must ítself be a'n"invertible element of R[x J (Problem 5, Chapter 7). , k,' , Going in the other direction, assume that the polyno~~l f(x) = a o + alx + ... + an~ER[xJ possesses an ,nverse. That'áo is then in;ertí?le in R should be obvíous. Forany prime ideal P of R,P[xJ is a pnme Ideal of R[xJ and the quotient ring R[xJ/P[xJ ~ (R/P}[xJ. Thus, the homomorpbic image off(x) in (R/P)[x J,

rad F[xJ = (Rad F)[x]

'1

for any field F. That is to say, the polynomíal ring F[x J constitutes a semi" simple ringo Suppose for the moment that 1 is an ideal of the ring R with 1 ~ Rad R. Given añ idempotent element e =1= O in.,\R, we know that the coset e + 1 will be idempotent in R/I. What is not SQobvious lS that e + 1 =1= 1; tbis follows from the fact that Rad R containsIiÓ'11onzero idempotents (Corollary to Theorem 8-2). We are mainly conceined with the converse here: If u + 1 is a nonzero idempotent of the quotient ring R/I, does there exist an idempotent e E R for which e + 1 = u + I? Before becoming involved in this ,discussion, let us give a general ' "• definition. ' ' ' -,

+ P = al + P = ... = al! + P = P; , I

hence, the elements al' al' ... , an all He in P. As this statement holds for every prime ideal of R, it follows that al' al' ... , an ERad R. By the corollary to The'orem 8-8, the elements al' al' ... , an must therefore be nilpotent.

= {O}

DefinidOR 8-3. Let 1 be an arbitrary ideal of the ring R. We say that the idempotents of R/I can be raised ,or lifted into R in case every idempotent element of R/I is pf the form e + 1, where e is idempotent in R.

P)~

must have an inverse. Since R/P is an integral domain, the invertible elements in (R/P)[xJ are nonzero constant polynomials. This impHes that al

+ aox + ... + allx"+ l

Rad R[x J = (Rad ~)[x J,

A problem exerting a natural appeal is that of describing the prime radical of the polynomial ring R[x J in terms of the prime radical of R. As a starting point, let us first prove a lemma which is of interest for various parts of ring theory.

+ (al + P)x + ... + (a n +

1

The assertion ofTheorem 8-13 can be improved upon. For the reader will have little difficulty in now convincing bimself that

Theorem 8-12. For uny ring R, Rad R is the smallest ideal 1 of R such that the quotient ring R/I is without prime radical.

(a o + P)

+ xf(x) =

must be invertible in R[x J. Hence, by the aboye lemma, the coefficients ao, al' ... , a" are all nilpotent elements of R. For a sufficiently large power, f(x) will then be nilpotent in R[xJ and thus be in Rad R[xJ.

Theorem 8-11. If 1 is an ideal of the ring R, then

2) whenever 1

= Rad R[x].

Proof. It is enough to establish the inc1usion rad R[x J ~ Rad R[xJ. If the polynomial f(x) = ao + alx + ... + an~ Erad R[xJ, Theorem 8-2

To round out the picture, two theorems are stated without proof; it will be o bserved that these take the same form as the corresponding result established for the Jacobson radical (Theorems 8-4 and 8-5). 1) Rad (R/I);:2 Rad ~

167

Definition 8-3 means just this: the idempotents of R/I can be lifted if for each element u E R such that ul - U El there exists some element el e E R with e - u E l. Although ít is surely too much to expect the lifting ofidempotents to take place for every 1, we shalI see that tbis situation does occur whenever 1 is a nil ideal (or, equívalently, whenever 1 ~ Rad R). Let us begin with a lemma, important in itself.

166

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

so that d' E l. But 1 consists of all nilpotent elements of R. Thus, we must have (d,)m = O for suitably chosen m E Z+; this is simply the statement that a El, and, hence, a + 1 is the zero element of R/I. Our argument ~mp1ies that the quotient ring R/I has no nonzero nilpoient elements, which lS to say that Rad (R/I) = {O}.

Having dealt with 'these preliminaries we are now ready to prove Theorem 8-13. For any ring R, rad R[x]

tells us that 1

~

+ 1, and,

Rad R, Rad (R/I)

= (Rad R)/l.

where (Rad R)[x J denotes the ring of polynomials in x with coefficients from Rad R; In fact, the inclusion Rad R[x J ~ (Rad R)[x J is implicit in the foregoing proof; the opposite inclusion requires the corollary to Theorem: 8-8. By virtue of the displayed equation, we have

Lemma. A polynomialf(x) = €lo + alx + ... + a"x" is invertible in R[ x J if and only if a o is invertible in R and aIl the other coefficients al> al' ... , a" are nilpotent elements of R.

Proof. If aohas an inversein R and al' al' ... , a" are all nilpotent, then the polynomial f(x) = a o + alx + ... + anxn is tbe sum of irt ,invertible element ando a nilpotent elemento Hence, f(x) must ítself be a'n"invertible element of R[x J (Problem 5, Chapter 7). , k,' , Going in the other direction, assume that the polyno~~l f(x) = a o + alx + ... + an~ER[xJ possesses an ,nverse. That'áo is then in;ertí?le in R should be obvíous. Forany prime ideal P of R,P[xJ is a pnme Ideal of R[xJ and the quotient ring R[xJ/P[xJ ~ (R/P}[xJ. Thus, the homomorpbic image off(x) in (R/P)[x J,

rad F[xJ = (Rad F)[x]

'1

for any field F. That is to say, the polynomíal ring F[x J constitutes a semi" simple ringo Suppose for the moment that 1 is an ideal of the ring R with 1 ~ Rad R. Given añ idempotent element e =1= O in.,\R, we know that the coset e + 1 will be idempotent in R/I. What is not SQobvious lS that e + 1 =1= 1; tbis follows from the fact that Rad R containsIiÓ'11onzero idempotents (Corollary to Theorem 8-2). We are mainly conceined with the converse here: If u + 1 is a nonzero idempotent of the quotient ring R/I, does there exist an idempotent e E R for which e + 1 = u + I? Before becoming involved in this ,discussion, let us give a general ' "• definition. ' ' ' -,

+ P = al + P = ... = al! + P = P; , I

hence, the elements al' al' ... , an all He in P. As this statement holds for every prime ideal of R, it follows that al' al' ... , an ERad R. By the corollary to The'orem 8-8, the elements al' al' ... , an must therefore be nilpotent.

= {O}

DefinidOR 8-3. Let 1 be an arbitrary ideal of the ring R. We say that the idempotents of R/I can be raised ,or lifted into R in case every idempotent element of R/I is pf the form e + 1, where e is idempotent in R.

P)~

must have an inverse. Since R/P is an integral domain, the invertible elements in (R/P)[xJ are nonzero constant polynomials. This impHes that al

+ aox + ... + allx"+ l

Rad R[x J = (Rad ~)[x J,

A problem exerting a natural appeal is that of describing the prime radical of the polynomial ring R[x J in terms of the prime radical of R. As a starting point, let us first prove a lemma which is of interest for various parts of ring theory.

+ (al + P)x + ... + (a n +

1

The assertion ofTheorem 8-13 can be improved upon. For the reader will have little difficulty in now convincing bimself that

Theorem 8-12. For uny ring R, Rad R is the smallest ideal 1 of R such that the quotient ring R/I is without prime radical.

(a o + P)

+ xf(x) =

must be invertible in R[x J. Hence, by the aboye lemma, the coefficients ao, al' ... , a" are all nilpotent elements of R. For a sufficiently large power, f(x) will then be nilpotent in R[xJ and thus be in Rad R[xJ.

Theorem 8-11. If 1 is an ideal of the ring R, then

2) whenever 1

= Rad R[x].

Proof. It is enough to establish the inc1usion rad R[x J ~ Rad R[xJ. If the polynomial f(x) = ao + alx + ... + an~ Erad R[xJ, Theorem 8-2

To round out the picture, two theorems are stated without proof; it will be o bserved that these take the same form as the corresponding result established for the Jacobson radical (Theorems 8-4 and 8-5). 1) Rad (R/I);:2 Rad ~

167

Definition 8-3 means just this: the idempotents of R/I can be lifted if for each element u E R such that ul - U El there exists some element el e E R with e - u E l. Although ít is surely too much to expect the lifting ofidempotents to take place for every 1, we shalI see that tbis situation does occur whenever 1 is a nil ideal (or, equívalently, whenever 1 ~ Rad R). Let us begin with a lemma, important in itself.

1.68

CERTAIN RADICALS OF A RING

F!RS,T COURSE IN RINGS AND IDEALS

Lemma. If e and e' are two idempotent elements ofthe ring R such that e - el ERad R, then e = el. el){l - (e + el») O, it lS enough to show that 1 - (e + el) is an invertible element of R. Now, one may write 1 - (e + e') in the form

Proof Inasmuch as the product (e

1 - (e',;4- e1 = (1 - 2e)

+ (e

e),

Corollary. LetI, J be:jdeals of the ring R with 1 J S;;; Rad R. If the idempotents of RfJ cá~,be lifted into R, then so can the idempotents of

R/l~

r~t.i:'i

'

Proof Suppose that u ~;l}s any idempotent of RjI. Since 1 S;; J, it follows U E 1 S;; J. By assumpthat u + J is an idempoterit element of RfJ; u2 tion, there must exist somee2 = e in R such that e + J u + J, whence e - u E J. But then (e

+ 1)

- (u

+ 1)=

e - u

+ 1 E J/l

S;;;

(Rad R)/1 = Rad (R/l).

Applying the lemma to the quotient ring R/l, we conclude that the coset u + 1 = e + land so the idempotents of R/l can be lifted. The key to showing that the idempotents of RJRad R are liftable is the circumsta~'ce that certain quadrati~ equations have a solution in the prime radical of R.

Theorem 8-14. For any ring R, the idempotents of RJRad R can be lifted into R. Proof Let u· + Rad R be an idempotent element of RJRad R, so that u2 - u = r ERad R. The problem is to find an idempotent e E R with e - u ERad R or, putting it another way, to obtain a solution a of the equation (u + a)2 = u + a, with a ERad R. Wefirst set a = x(! 2u), where x is yet to be determined. Now, the u + x(1 - 2u) be ídempotent is requirement that the element u + a equivalent to the equation (x 2

x)(l

+

4r)

+

r = O.

By the quadratic formula, tbis has a formal solution

x =

Hl . 1 1/2(2r

@r2

+

(~)r3

... ).

Since r is nilpotent (being a member of Rad R), the displayed series will . terminate in a finite number of steps; the result is a perfectly meaningful polynomial in r with integral coefficients. Thus, the desired idempotent is e = u + x(1 - 2u), where x ERad R.

CoroUary. For any nil ideal 1 of R, the idempotents of R/l can be lífted.

Proof Because 1

where(l' - 2e)2 = 1 - 4e + ~ = 1. Theimplicationisthat 1 -:- (e + e'), being the sum of a nilpote~relement and an ínvertible element, is necessarily invertible in- R (Problem~~;; COOpter 1).

169

S;;

Rad R,an appeal to the last lemma (with J = Rad R)

is legitimate. \f:,et us define a ring R to be primary whenever the zero ideal is a primary ideáljbf R. This readily translates into a state'ment involving the elements of]{ :.R is a primary ring if and only if every zero divisor of R is nilpotent. Int~.g¡.al domains are examples of primary pngs. In general, primary rings caíi'bt:, obtained by constructing quotient rings R/Q, where Q is a primary ideaLof R. - ';-;:'N,s anapplication of the preceding ideas; we shall characterize such rings;in terms ,of minimal prime ideaIs. (A prime ideal is said to be a mini mal prime ideal if it is minimal in the set of prime idel:!.ls; in a cornmutative ring with identity, such ideals are necessarily proper.) .The crucial step in the proof is the corollary on page 164. ~'}-

~

Theorem 8-15. A ring R is a primary ring if and only if R has a minimal prime ideal which contains aH zero divisors.

Proof F or the first half of the proof, let R' be a primary ringo Then the set of zero divisors of R, along with zero, coincides with the ideal N of nilpotent elements, and N will be prime. Being equal. to the prime radical of R, N is necessarily contained in every prime ideal of R; that lS to say, N is a minimal prime ideal. The converse is less obvious; in fact, it is easiest to prove the contraposítive form of the converse. Suppose, then, that·R ha~ a mínimal. prime ideal P which contalns all zero divisors and let a E R be any nonnilpotent element. We define the set S by

S

= {r~nlr~P; n ~

O}.

S is easily seen to be closed under multiplication and 1 E S, Notice, particularly, that the Zero element does not líe in S, for, otherwise, we would have ran = O with an =1= O; this ímplies that r lS a zero divisor and therefore a member'of P. We now appeal to the corollary on page 164 to infer that the complement of S contains a prime ideal Pi. Since pi S;; R - S S;;; P, S = P. But a E S, with P being a mínimal prime ideal, it follows that R whence a ~ P, so that a cannot be a zero divisor of R. In other words, every zero divisor of R is nilpotent, which completes the proof. For the sake of refinement, let us temporarily drop the assumption that all rings must have a multiplicative identity (commutativity could al so be

,

;~

1.68

CERTAIN RADICALS OF A RING

F!RS,T COURSE IN RINGS AND IDEALS

Lemma. If e and e' are two idempotent elements ofthe ring R such that e - el ERad R, then e = el. el){l - (e + el») O, it lS enough to show that 1 - (e + el) is an invertible element of R. Now, one may write 1 - (e + e') in the form

Proof Inasmuch as the product (e

1 - (e',;4- e1 = (1 - 2e)

+ (e

e),

Corollary. LetI, J be:jdeals of the ring R with 1 J S;;; Rad R. If the idempotents of RfJ cá~,be lifted into R, then so can the idempotents of

R/l~

r~t.i:'i

'

Proof Suppose that u ~;l}s any idempotent of RjI. Since 1 S;; J, it follows U E 1 S;; J. By assumpthat u + J is an idempoterit element of RfJ; u2 tion, there must exist somee2 = e in R such that e + J u + J, whence e - u E J. But then (e

+ 1)

- (u

+ 1)=

e - u

+ 1 E J/l

S;;;

(Rad R)/1 = Rad (R/l).

Applying the lemma to the quotient ring R/l, we conclude that the coset u + 1 = e + land so the idempotents of R/l can be lifted. The key to showing that the idempotents of RJRad R are liftable is the circumsta~'ce that certain quadrati~ equations have a solution in the prime radical of R.

Theorem 8-14. For any ring R, the idempotents of RJRad R can be lifted into R. Proof Let u· + Rad R be an idempotent element of RJRad R, so that u2 - u = r ERad R. The problem is to find an idempotent e E R with e - u ERad R or, putting it another way, to obtain a solution a of the equation (u + a)2 = u + a, with a ERad R. Wefirst set a = x(! 2u), where x is yet to be determined. Now, the u + x(1 - 2u) be ídempotent is requirement that the element u + a equivalent to the equation (x 2

x)(l

+

4r)

+

r = O.

By the quadratic formula, tbis has a formal solution

x =

Hl . 1 1/2(2r

@r2

+

(~)r3

... ).

Since r is nilpotent (being a member of Rad R), the displayed series will . terminate in a finite number of steps; the result is a perfectly meaningful polynomial in r with integral coefficients. Thus, the desired idempotent is e = u + x(1 - 2u), where x ERad R.

CoroUary. For any nil ideal 1 of R, the idempotents of R/l can be lífted.

Proof Because 1

where(l' - 2e)2 = 1 - 4e + ~ = 1. Theimplicationisthat 1 -:- (e + e'), being the sum of a nilpote~relement and an ínvertible element, is necessarily invertible in- R (Problem~~;; COOpter 1).

169

S;;

Rad R,an appeal to the last lemma (with J = Rad R)

is legitimate. \f:,et us define a ring R to be primary whenever the zero ideal is a primary ideáljbf R. This readily translates into a state'ment involving the elements of]{ :.R is a primary ring if and only if every zero divisor of R is nilpotent. Int~.g¡.al domains are examples of primary pngs. In general, primary rings caíi'bt:, obtained by constructing quotient rings R/Q, where Q is a primary ideaLof R. - ';-;:'N,s anapplication of the preceding ideas; we shall characterize such rings;in terms ,of minimal prime ideaIs. (A prime ideal is said to be a mini mal prime ideal if it is minimal in the set of prime idel:!.ls; in a cornmutative ring with identity, such ideals are necessarily proper.) .The crucial step in the proof is the corollary on page 164. ~'}-

~

Theorem 8-15. A ring R is a primary ring if and only if R has a minimal prime ideal which contains aH zero divisors.

Proof F or the first half of the proof, let R' be a primary ringo Then the set of zero divisors of R, along with zero, coincides with the ideal N of nilpotent elements, and N will be prime. Being equal. to the prime radical of R, N is necessarily contained in every prime ideal of R; that lS to say, N is a minimal prime ideal. The converse is less obvious; in fact, it is easiest to prove the contraposítive form of the converse. Suppose, then, that·R ha~ a mínimal. prime ideal P which contalns all zero divisors and let a E R be any nonnilpotent element. We define the set S by

S

= {r~nlr~P; n ~

O}.

S is easily seen to be closed under multiplication and 1 E S, Notice, particularly, that the Zero element does not líe in S, for, otherwise, we would have ran = O with an =1= O; this ímplies that r lS a zero divisor and therefore a member'of P. We now appeal to the corollary on page 164 to infer that the complement of S contains a prime ideal Pi. Since pi S;; R - S S;;; P, S = P. But a E S, with P being a mínimal prime ideal, it follows that R whence a ~ P, so that a cannot be a zero divisor of R. In other words, every zero divisor of R is nilpotent, which completes the proof. For the sake of refinement, let us temporarily drop the assumption that all rings must have a multiplicative identity (commutativity could al so be

,

;~

170

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

b = -a - a2 - ••• _. an - 1 is a quasi-inverse of a. Notice also that zero is the only idempotent which is quasi-regular. For, if a2 = a and a + b ab = O for sorne b in R, then we have

abandoned, but this seerns unnecessarily elaborate for the purposes in rnind). To make things more specific, we pose the problem of constructing a radical which will agree with the Jacobson radical when an identity element is available. Of course, it is always possible to imbed a ring in a ring with identity, but the imbedding is often unnatural and distorts essential features of the given ringo The direct approach of considering the intersection of all maximal ideals is not very effective, because one no longer knows that such ideals exist (Theorem 5-2, our basic existence theorem for rnaximal ideals, c1early requires the presence of an identity). A more useful clue is provided by Theorem 8-2, wbich asserts that an element a Erad R if and only if 1 - ra 1S invertible for every choice of r in R, or, to put it somewhat differentIy, the principal ideal (1 ~, ra) = R for all r E R. This latter condition can be \Hitten as {x - raxlx E.,R} = R for each r in R, and is meaningful in the absence of a rnultiplicative identity. It thus would appear that Theorem 8-2 constitutes a hopeful starting point to the solution of the problem before uso Needless to say, it will be necessary to introduce concepts capable of replacing the notions of an invertible element and 'maximal ideal which were so essential to our earlier work: One begins by associating with each elernent a E R the set la

=

a

Example 8-6. Consider the assertion: if every element of a commutati~e ring R is quasi-regular, with exactIy one exception, then R must be a fieI'9" 1;" To see this let us take the element e to be the one exception; certairily'~" ~ ~ ','é =1= O, since O is quasi-regular. .' . 2 Now, e o a = e o (-e o a) =1= O for each a E R, from which we infer t that e2 = e. Observe also that if e o a =1= e, then there would exist so~e ..... elementb E R such that(e o a) o b = O. Associating, weobtaine o (a o b) =~:O ", and so e is a quasi-regular element, a contradiction. Accordingly, we musF '" have e o a = e for every choice of a E R or, upon expanding, ae = a for all in R; this implies that e acts as a rnultiplicative identity for R. Finally, given an element x =1= O, we wrHe x = e - a, with a E R. Then, since a =1= e, ,,¡

,

b - ab = O.

The element b),atisfying this equation is called a quasi-inverse of a. Proof. Suppose tb;~t a is quasi-regular, so that la = R. Since'a E la, we must have a = abj& b for suitable b in R, whence a + b -:- ab = O. On the other hand, if there exists sorne elernent b E R satisfying a + b - ab = O, then a E la. Thus, for any r in R, ar E la. By virtue of the definition of la' we also have ar - r E la' which implies that r = ar - (ar - r) E la. f;

la' or rather R

+ ab - ab = a(a + b - ab) = O.

Accordingly,a is quasi-regular if and only if 1 - a is an invertible element of R. (Iying this idea more c10sely to Theorem 8-2, we see that the product ra is quasi-regular for every rE R if and only if 1 - ra is invertible for all r in R.)

Theorem 8-16:'An element a of the ring R is quasi-regular if and only if there exists ~o~e b E R such that

This means R

a2

With tbis notation, Theorem 8~16 may be rephrased so asto assert that an element a E R is quasi-regular if and only if there exists sorne second element b E R for which a o b = O. It is a simple matter to verify that the pair (R, o) is a semigroup with identity element O; in particular, one infers from tbis that quasi-inverses are unique, whenever they exist. An even stronger result is that the quasiregular elernents of R form a group with respect to the circ1e operation. LastIy, let us 'call attention to the fact that if R possesses a multiplicative identity 1, then ' (1 - a)(1 - b) = 1 - a o b.

{ax - xix E R}.

+

=

One of the most useful tools in handling the concept of quasi-regularity is the so-called "circ1e operation" of Perlis [54]. Given a, bE R, we define a o b by a o b = a + b - ab.

A rnoment's thought shows la to be an ideal of R. Now, it may very well happen that la = R; in this event, we shall say that a is a quasi-regular elernent. There is ~other way of looking at quasi-regularity:

a

171

;

•

,

, 1

x(e - b)

= la' and a is quasi-regular.

=

(e' - a)(e - b)

=e

- aob

=

e

Corollary. An element a E R is quasi-regular ifand only if a E la.

for suitable b E R. In other words, every nonzero elernent of R is in vertible, confirming R to be a field.

Here are sorne consequences: Every nilpotent elernent of R is quasiregular. Indeed, if an = O, a stiaightforward calculation will establish that

As heralded by our earlier remarks, we now define the J-radical (fo,1' Jacobson, naturally enough) of a ring R to consist of those elements a for

1

.t.

170

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

b = -a - a2 - ••• _. an - 1 is a quasi-inverse of a. Notice also that zero is the only idempotent which is quasi-regular. For, if a2 = a and a + b ab = O for sorne b in R, then we have

abandoned, but this seerns unnecessarily elaborate for the purposes in rnind). To make things more specific, we pose the problem of constructing a radical which will agree with the Jacobson radical when an identity element is available. Of course, it is always possible to imbed a ring in a ring with identity, but the imbedding is often unnatural and distorts essential features of the given ringo The direct approach of considering the intersection of all maximal ideals is not very effective, because one no longer knows that such ideals exist (Theorem 5-2, our basic existence theorem for rnaximal ideals, c1early requires the presence of an identity). A more useful clue is provided by Theorem 8-2, wbich asserts that an element a Erad R if and only if 1 - ra 1S invertible for every choice of r in R, or, to put it somewhat differentIy, the principal ideal (1 ~, ra) = R for all r E R. This latter condition can be \Hitten as {x - raxlx E.,R} = R for each r in R, and is meaningful in the absence of a rnultiplicative identity. It thus would appear that Theorem 8-2 constitutes a hopeful starting point to the solution of the problem before uso Needless to say, it will be necessary to introduce concepts capable of replacing the notions of an invertible element and 'maximal ideal which were so essential to our earlier work: One begins by associating with each elernent a E R the set la

=

a

Example 8-6. Consider the assertion: if every element of a commutati~e ring R is quasi-regular, with exactIy one exception, then R must be a fieI'9" 1;" To see this let us take the element e to be the one exception; certairily'~" ~ ~ ','é =1= O, since O is quasi-regular. .' . 2 Now, e o a = e o (-e o a) =1= O for each a E R, from which we infer t that e2 = e. Observe also that if e o a =1= e, then there would exist so~e ..... elementb E R such that(e o a) o b = O. Associating, weobtaine o (a o b) =~:O ", and so e is a quasi-regular element, a contradiction. Accordingly, we musF '" have e o a = e for every choice of a E R or, upon expanding, ae = a for all in R; this implies that e acts as a rnultiplicative identity for R. Finally, given an element x =1= O, we wrHe x = e - a, with a E R. Then, since a =1= e, ,,¡

,

b - ab = O.

The element b),atisfying this equation is called a quasi-inverse of a. Proof. Suppose tb;~t a is quasi-regular, so that la = R. Since'a E la, we must have a = abj& b for suitable b in R, whence a + b -:- ab = O. On the other hand, if there exists sorne elernent b E R satisfying a + b - ab = O, then a E la. Thus, for any r in R, ar E la. By virtue of the definition of la' we also have ar - r E la' which implies that r = ar - (ar - r) E la. f;

la' or rather R

+ ab - ab = a(a + b - ab) = O.

Accordingly,a is quasi-regular if and only if 1 - a is an invertible element of R. (Iying this idea more c10sely to Theorem 8-2, we see that the product ra is quasi-regular for every rE R if and only if 1 - ra is invertible for all r in R.)

Theorem 8-16:'An element a of the ring R is quasi-regular if and only if there exists ~o~e b E R such that

This means R

a2

With tbis notation, Theorem 8~16 may be rephrased so asto assert that an element a E R is quasi-regular if and only if there exists sorne second element b E R for which a o b = O. It is a simple matter to verify that the pair (R, o) is a semigroup with identity element O; in particular, one infers from tbis that quasi-inverses are unique, whenever they exist. An even stronger result is that the quasiregular elernents of R form a group with respect to the circ1e operation. LastIy, let us 'call attention to the fact that if R possesses a multiplicative identity 1, then ' (1 - a)(1 - b) = 1 - a o b.

{ax - xix E R}.

+

=

One of the most useful tools in handling the concept of quasi-regularity is the so-called "circ1e operation" of Perlis [54]. Given a, bE R, we define a o b by a o b = a + b - ab.

A rnoment's thought shows la to be an ideal of R. Now, it may very well happen that la = R; in this event, we shall say that a is a quasi-regular elernent. There is ~other way of looking at quasi-regularity:

a

171

;

•

,

, 1

x(e - b)

= la' and a is quasi-regular.

=

(e' - a)(e - b)

=e

- aob

=

e

Corollary. An element a E R is quasi-regular ifand only if a E la.

for suitable b E R. In other words, every nonzero elernent of R is in vertible, confirming R to be a field.

Here are sorne consequences: Every nilpotent elernent of R is quasiregular. Indeed, if an = O, a stiaightforward calculation will establish that

As heralded by our earlier remarks, we now define the J-radical (fo,1' Jacobson, naturally enough) of a ring R to consist of those elements a for

1

.t.

172

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

which lar = R for every r E R; recasting this in terms of the notion of quasiregularity : . Definition 8-4. The J-radical J(R) of a ríng R, with or without an identity, is the set J(R)

If J(R)

=

{a E Rlar is quasi-regular for all rE R}.

= {O}, then R is said to be a J-semisimple ring·f

To 'reinforce these ideas, let us consid~r several examplé~. .

.'~~'

Example 8-7. The ring Z. of even integers is J-semisimpl~.;¡:For, suppose that the integer n E J(Z.). Then, in particular, n2 is quasi-"r~gular. But the equation.'·; n2 + x - n2 x = O, :~) or, equivalent1y, n2 = (,F - l)x, has no solution among·'~h~:eveIi. integers • unless n = O. This implies thatJ(Z.) = {O}.L~;~::, ~ .<.,.~

Exampl~ 8-8. Any commutative regular ring R is J-semislIi1ple .. Indeed, giveil a E J(R), there is some a' in R such that a2 a' = a. Now, aa' must be quasi-~egular, so we can find an element x E R satisfying aa'

+x -

aa' x

= O.

Multiplying this equation by a and using the fact that a2 a' = O, we deduce tbat a = O, whence J(R) = {O}. Example 8-9. Consider the ring F[[x]] of formal power series over the field F. As we know, F[[x]] is a commutative ríng with identity in which an element f(x) = akX' is invertible if and only if ao =1= o. If f(x) belongs to the principal ideal (x), thenf(x) has zero constant term; hence, (1 - f(x)) - 1 exists in F[[x]]' Takingf(x)' = (1 - f(X)tI, we see thatf(x) o f(x)' = O. Thus,everymemberof(x)isquasi-regular, whichimpliesthat(x) S J(F[[x]]). On the other hand, any element not in (x) is invertible and therefore cannot be in the J-radical. (In general, ir a E J(R) has a multiplicative inverse, then 1 = aa -1 is quasi-regular; but zero is the only quasi-regular idempotent, so that 1 = O, a contradiction.) The implication is that J(F[[ x]]) S (x) and equality follows: J(F[[x]]) = (x) = rad F[[x]]'

L

Turning once again to generalities, let us show that any element a E J(R) is itself quasi-regular. Since ar is quasi-regular for each choice of r in R, it folIows that a2 in particular wiII be quasi-regular. Therefore, we can obtain an element bE R for which a2 o b = O. But a simple computation shows that ao((-a)ob) = (ao(-a))ob = a20 b = O

173

One fincls in .this way that the element a is quasi-regular with quasi-inverse (-a)

o

b.

It is by no means apparent from Definition 8-3 that J(R) forms an ideal of R; our next concern is to establish that this is actually the case.

Theorem 8-17. For any ríng R, the J-radical J(R) is an ideal of R. Proof. Suppose that the element a E JfR), so that for any choice of x E R, ax is quasi-regular. If rE R, then ce~ainly (ar)y = a(ry) must be quasiregular for alI y in R, and therefore aréJ(R).

It remains to show that whenever;~'.b E J(R), then the difference a - b lies in J(R). Given x, u, v E R, a fair1yJroutine ca1culation establishes the identity .,,;, (a - b)x

o

(u

o

v)

=

a(x - ux),o.v:

+

(-bx ou) - (-bx

o

u)v.

Taking stock of the fact that a; b belop,g;to J(R), we can select an element u such that -(bx) o u = b( -x) o u =:0 'ahd a second element v for which a(x- ax) o V.= O; in consequeIi.ce, (é~' b)x o (u o v) =0. This being the case, (a - b)x is qua si-regular for every x E R, whence a - bE J(R). Thus, the J-radical satisfies the defining conditions for an ideal of R. As one would expect, there are many theorems concerning the J-radical which are completely analagous to theorems stated in terms ofthe Jacobson radical. Although it would be tedious to prove aIl of these results, the following deserves to be carried through. Theorem 8-18. For any ríng R, tbe quotient ring RjJ(R) is J-semisimple. Pro~f. Take a

+

J(R) to be an arbitrary element of J(RjJ(R)). Then (a

+

J(R)) (x

+

J(R))

=

ax

+

J(R)

is quasi-regular for each, x E R. . Accordingly, tbere exists some coset y + J(R) in RjJ(R), depending on both a and x, for which (ax

+

J(R))

o

(y

+ J(R)) =

J(R).

But this implies that the element ax o y lies in J(R), and, hence, is quasiregular as a member of R ; say (ax o y) o Z = O, where z, E R. It foIlows from the associativity. of o that ax is itself quasi-regular in R, with quasi-inverse y o z. Since tbis holds for every x E R, the element a belongs to J(R), and we have a + J(R) = J(R), the zero of the quotient ring RjJ(R). It tums out that the c1ass of ideals which must replace the maximal ideals are precisely those ideals whose quotient rings possess a multiplicative identity.

Definition 8-5. An ideal l of the ring R is calIed modular (or regular, in the older terminology) if and' only if there exists an .element e E R such

172

FIRST COURSE IN RINGS AND IDEALS

CERTAIN RADICALS OF A RING

which lar = R for every r E R; recasting this in terms of the notion of quasiregularity : . Definition 8-4. The J-radical J(R) of a ríng R, with or without an identity, is the set J(R)

If J(R)

=

{a E Rlar is quasi-regular for all rE R}.

= {O}, then R is said to be a J-semisimple ring·f

To 'reinforce these ideas, let us consid~r several examplé~. .

.'~~'

Example 8-7. The ring Z. of even integers is J-semisimpl~.;¡:For, suppose that the integer n E J(Z.). Then, in particular, n2 is quasi-"r~gular. But the equation.'·; n2 + x - n2 x = O, :~) or, equivalent1y, n2 = (,F - l)x, has no solution among·'~h~:eveIi. integers • unless n = O. This implies thatJ(Z.) = {O}.L~;~::, ~ .<.,.~

Exampl~ 8-8. Any commutative regular ring R is J-semislIi1ple .. Indeed, giveil a E J(R), there is some a' in R such that a2 a' = a. Now, aa' must be quasi-~egular, so we can find an element x E R satisfying aa'

+x -

aa' x

= O.

Multiplying this equation by a and using the fact that a2 a' = O, we deduce tbat a = O, whence J(R) = {O}. Example 8-9. Consider the ring F[[x]] of formal power series over the field F. As we know, F[[x]] is a commutative ríng with identity in which an element f(x) = akX' is invertible if and only if ao =1= o. If f(x) belongs to the principal ideal (x), thenf(x) has zero constant term; hence, (1 - f(x)) - 1 exists in F[[x]]' Takingf(x)' = (1 - f(X)tI, we see thatf(x) o f(x)' = O. Thus,everymemberof(x)isquasi-regular, whichimpliesthat(x) S J(F[[x]]). On the other hand, any element not in (x) is invertible and therefore cannot be in the J-radical. (In general, ir a E J(R) has a multiplicative inverse, then 1 = aa -1 is quasi-regular; but zero is the only quasi-regular idempotent, so that 1 = O, a contradiction.) The implication is that J(F[[ x]]) S (x) and equality follows: J(F[[x]]) = (x) = rad F[[x]]'

L

Turning once again to generalities, let us show that any element a E J(R) is itself quasi-regular. Since ar is quasi-regular for each choice of r in R, it folIows that a2 in particular wiII be quasi-regular. Therefore, we can obtain an element bE R for which a2 o b = O. But a simple computation shows that ao((-a)ob) = (ao(-a))ob = a20 b = O

173

One fincls in .this way that the element a is quasi-regular with quasi-inverse (-a)

o

b.

It is by no means apparent from Definition 8-3 that J(R) forms an ideal of R; our next concern is to establish that this is actually the case.

Theorem 8-17. For any ríng R, the J-radical J(R) is an ideal of R. Proof. Suppose that the element a E JfR), so that for any choice of x E R, ax is quasi-regular. If rE R, then ce~ainly (ar)y = a(ry) must be quasiregular for alI y in R, and therefore aréJ(R).

It remains to show that whenever;~'.b E J(R), then the difference a - b lies in J(R). Given x, u, v E R, a fair1yJroutine ca1culation establishes the identity .,,;, (a - b)x

o

(u

o

v)

=

a(x - ux),o.v:

+

(-bx ou) - (-bx

o

u)v.

Taking stock of the fact that a; b belop,g;to J(R), we can select an element u such that -(bx) o u = b( -x) o u =:0 'ahd a second element v for which a(x- ax) o V.= O; in consequeIi.ce, (é~' b)x o (u o v) =0. This being the case, (a - b)x is qua si-regular for every x E R, whence a - bE J(R). Thus, the J-radical satisfies the defining conditions for an ideal of R. As one would expect, there are many theorems concerning the J-radical which are completely analagous to theorems stated in terms ofthe Jacobson radical. Although it would be tedious to prove aIl of these results, the following deserves to be carried through. Theorem 8-18. For any ríng R, tbe quotient ring RjJ(R) is J-semisimple. Pro~f. Take a

+

J(R) to be an arbitrary element of J(RjJ(R)). Then (a

+

J(R)) (x

+

J(R))

=

ax

+

J(R)

is quasi-regular for each, x E R. . Accordingly, tbere exists some coset y + J(R) in RjJ(R), depending on both a and x, for which (ax

+

J(R))

o

(y

+ J(R)) =

J(R).

But this implies that the element ax o y lies in J(R), and, hence, is quasiregular as a member of R ; say (ax o y) o Z = O, where z, E R. It foIlows from the associativity. of o that ax is itself quasi-regular in R, with quasi-inverse y o z. Since tbis holds for every x E R, the element a belongs to J(R), and we have a + J(R) = J(R), the zero of the quotient ring RjJ(R). It tums out that the c1ass of ideals which must replace the maximal ideals are precisely those ideals whose quotient rings possess a multiplicative identity.

Definition 8-5. An ideal l of the ring R is calIed modular (or regular, in the older terminology) if and' only if there exists an .element e E R such

174

that ae - a E 1 for every a in R. Such an element e is said to be an . identity for R relative to 1, or modulo 1. In passing, we should remark that whenever R has an identity element 1, then 1 can be taken as the element e of Definition 8-5, and all ideals of R are modular. Notice, too, that if e is an identity for R relative to 1, then the same is true for the elements e + i, where i E 1, and e" (n E Z). By a ¡modular maximal ideal, we shall mean a maximal (hence, proper) ideal which is also modular. Parallelíng the proof of Theorem 5-5, it can be shown without too much difficulty that a'proper ideal M of R is a modular maximal ideal if and only if the quotient ring R/M forms a field. The existence of suitably many modular maximal ideals is assured by the following: Theorem 8-19. Each proper modular ideal of the ring R is contained in a modular maxiIÍial ideal of R. Proof. Let 1 be a proper modular ideal of R and e be an identity element for R relative to 1. We consider the family d ofall proper ideal s of R which contain 1; because 1 itself is such an ideal, d is certainly nonempty. It is important to observe that the element e líes outside each ideal J of d. Indeed, if e did belong to J, we would then have ae E J for all a in R. By virtue ofthe fact that 1 is modular, ae . . . :. a El!:;;; J, from which it follows that a = ae - (ae - a) E J. One finds in this way that J = R, a flat contradiction, inasmuch as J is a proper ideal of R by definition of d. Now, let {l¡} be any chain of ideals from d. When the set-theoretic union u Ji fonns an ideal·of R containing 1. Since"e:rj!uJ¡, this ideal is proper, whence u J¡ E d. Thus, Zom's Lemma asserts:the existence of a maximal ideal M of R with 1 !:;;; M. Any such ideal wiHb.e modular, because ae - aE 1 !:;;; M for each element a E R. . This t~eorem .has a num?er of i~portant. conse~~~;nces (whic,b. we list as corollanes) havmg to do wlth quasl-regulanty. 'y .

175

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

.,.}.

CoroUary 1. If the element a E R is not quasi-regular, then there exists a modular maximal ideal M of R such that a rj 1v[. Proof. Since a is not quasi-regular, la = {ra - rlr ER} forms a proper ideal of R. Moreover, la is modular, with the element a as an identity for R modulo la. Knowing this, it follows from the theorem that there exists a modular maximal ideal M of R containing lo and exc1uding a.

CoroUary 2. If J(R) =f R, then the ring R contains modular maximal ideals; in fact, for any a rj J(R), there exists a modular maximal ideal M with arjM.

Proof. If the element a.rj J(R), thtm there is sorne x E R such that ax is not quasi-regular. Corollary 1 asserts the existence of a modular maximal ideal of R which exc1udes ax and, in consequence, does not contain a.

CoroUary 3. An element a E R is quasi-regular if and only if, for each modular maximal ideal M of R, there exists an element b such that a o b EM. Proof. The indicated condition is c1early necessary, for it suffices to take the quasi-inverse of a as the element b. Suppose now that the condition is satisfied, but that a does not possess a quasi-inverse. Then the modular ideal la = {ar r E R} will be contained in sorne modular maximal ideal M of R. By assumption, we can find an element b in R for which a o b E M. But ah - b E la, whence

rl

a = aob

+

(ab - b) E M.

It follows that ar E M for arbitrary r in R, and, consequentIy, that r = ar -'(aro - r) E M. Therefore, M = R, which is impossible.

In the presence of an identity element, the Jacobson radical rad R is the intersection of all the maximal ideals of a ring R. One would rightIy suspect that there is a similar characterization of the J-radical in terms of modular maximal ideals· (the sole difference being that, in the present setting, we must impose the demand that J(R) does not exhaust the ring R). Theorem 8-20. If R is a ring such that J(R) J(R)

= n

=f

R, then

{MIM is:~ modular maximal ideal of R}.

Proof. As so often happens, one inc1usion will be quite straightforward and easy, and the other will be deeper-and more complícated. In the first place, suppose that the element a lies i~every modular maximal ideal of R, but that a rj J(R). Using Corollary Z;we could then find a modular maximal ideal M for which a rj M, a contra,giction; consequentIy, a E J(R). Going in the other direction,take a in J(R). We wish to show that a E M, where M is any modular maximal ideal of R. Assume for the moment that a rj M. Owing to the fact that M is maximal, the ideal generated by M and a must be the whole ring R; Jherefore (in the absence of an identity), R = {i

+

ra

+

naliEM, rER, nEZ}.

Now, let e be an identity element for R modulo M. Then there exist suitable i E M, rE R and an integer n for which

+ na. an ideal of R, the sum ra + na E J(R), so that e

e=

As J(R) fonns

i

+

ra

- i

E

J(R).

174

that ae - a E 1 for every a in R. Such an element e is said to be an . identity for R relative to 1, or modulo 1. In passing, we should remark that whenever R has an identity element 1, then 1 can be taken as the element e of Definition 8-5, and all ideals of R are modular. Notice, too, that if e is an identity for R relative to 1, then the same is true for the elements e + i, where i E 1, and e" (n E Z). By a ¡modular maximal ideal, we shall mean a maximal (hence, proper) ideal which is also modular. Parallelíng the proof of Theorem 5-5, it can be shown without too much difficulty that a'proper ideal M of R is a modular maximal ideal if and only if the quotient ring R/M forms a field. The existence of suitably many modular maximal ideals is assured by the following: Theorem 8-19. Each proper modular ideal of the ring R is contained in a modular maxiIÍial ideal of R. Proof. Let 1 be a proper modular ideal of R and e be an identity element for R relative to 1. We consider the family d ofall proper ideal s of R which contain 1; because 1 itself is such an ideal, d is certainly nonempty. It is important to observe that the element e líes outside each ideal J of d. Indeed, if e did belong to J, we would then have ae E J for all a in R. By virtue ofthe fact that 1 is modular, ae . . . :. a El!:;;; J, from which it follows that a = ae - (ae - a) E J. One finds in this way that J = R, a flat contradiction, inasmuch as J is a proper ideal of R by definition of d. Now, let {l¡} be any chain of ideals from d. When the set-theoretic union u Ji fonns an ideal·of R containing 1. Since"e:rj!uJ¡, this ideal is proper, whence u J¡ E d. Thus, Zom's Lemma asserts:the existence of a maximal ideal M of R with 1 !:;;; M. Any such ideal wiHb.e modular, because ae - aE 1 !:;;; M for each element a E R. . This t~eorem .has a num?er of i~portant. conse~~~;nces (whic,b. we list as corollanes) havmg to do wlth quasl-regulanty. 'y .

175

CERTAIN RADICALS OF A RING

FIRST COURSE IN RINGS AND IDEALS

.,.}.

CoroUary 1. If the element a E R is not quasi-regular, then there exists a modular maximal ideal M of R such that a rj 1v[. Proof. Since a is not quasi-regular, la = {ra - rlr ER} forms a proper ideal of R. Moreover, la is modular, with the element a as an identity for R modulo la. Knowing this, it follows from the theorem that there exists a modular maximal ideal M of R containing lo and exc1uding a.

CoroUary 2. If J(R) =f R, then the ring R contains modular maximal ideals; in fact, for any a rj J(R), there exists a modular maximal ideal M with arjM.

Proof. If the element a.rj J(R), thtm there is sorne x E R such that ax is not quasi-regular. Corollary 1 asserts the existence of a modular maximal ideal of R which exc1udes ax and, in consequence, does not contain a.

CoroUary 3. An element a E R is quasi-regular if and only if, for each modular maximal ideal M of R, there exists an element b such that a o b EM. Proof. The indicated condition is c1early necessary, for it suffices to take the quasi-inverse of a as the element b. Suppose now that the condition is satisfied, but that a does not possess a quasi-inverse. Then the modular ideal la = {ar r E R} will be contained in sorne modular maximal ideal M of R. By assumption, we can find an element b in R for which a o b E M. But ah - b E la, whence

rl

a = aob

+

(ab - b) E M.

It follows that ar E M for arbitrary r in R, and, consequentIy, that r = ar -'(aro - r) E M. Therefore, M = R, which is impossible.

In the presence of an identity element, the Jacobson radical rad R is the intersection of all the maximal ideals of a ring R. One would rightIy suspect that there is a similar characterization of the J-radical in terms of modular maximal ideals· (the sole difference being that, in the present setting, we must impose the demand that J(R) does not exhaust the ring R). Theorem 8-20. If R is a ring such that J(R) J(R)

= n

=f

R, then

{MIM is:~ modular maximal ideal of R}.

Proof. As so often happens, one inc1usion will be quite straightforward and easy, and the other will be deeper-and more complícated. In the first place, suppose that the element a lies i~every modular maximal ideal of R, but that a rj J(R). Using Corollary Z;we could then find a modular maximal ideal M for which a rj M, a contra,giction; consequentIy, a E J(R). Going in the other direction,take a in J(R). We wish to show that a E M, where M is any modular maximal ideal of R. Assume for the moment that a rj M. Owing to the fact that M is maximal, the ideal generated by M and a must be the whole ring R; Jherefore (in the absence of an identity), R = {i

+

ra

+

naliEM, rER, nEZ}.

Now, let e be an identity element for R modulo M. Then there exist suitable i E M, rE R and an integer n for which

+ na. an ideal of R, the sum ra + na E J(R), so that e

e=

As J(R) fonns

i

+

ra

- i

E

J(R).

176

FIRST COURSE IN RINGS AND IDEALS PROBLEMS

Tbis implies that e - i is quasi-regular, say with quasi-inverse x. From the equation (e ...:. i) o x = 0, together witb. the modularíty of M, we obtain

+ (xe

e = i - ix

x) E M.

In con sequen ce, M = R, and an obvious contradiction ensues. Thus, a E M, from which one concludes that J{R) is contained in theintersection of the modular maximal ideals of ~, completing the proof. I

.

The hypothesis that J{R) =1= R: is certainly fulfilled whenever the ring R possesses a multiplicative identity 1. Specifically, the element 1 itself is not quasi-regular, whence 1 f/: J(~h in fact, if 1 + b - lb = O for some b in R, we would have 1 == O, a cótÍtradictio;o.. When an identity element is available, all ideals of R are automatically' modular. In tbis situation, the J-radical will coincide ?Vith the Ja'cpbson radical of R: J{R) = rad R. If J{R) = R, then th.e ríng R. ¡n!ly contain maximal ideals, but no such ideal can be modular. Índeed,:~~ppose that l is any modular ideal of R, R, with e acting as an identity foi.,R,.modulo 1. By suppósition, the element e E J(R), so that e has a quasi-invérse e'. The modularity of l then yields e == e'e - e' E l, .which implies that l = R. Accordingly, the ring R possesses no proper modular ideals and, in particular, no modular maximal ideals. However, the possibility of the existence of maximal ideal s in R is not excluded. The following theorem provides a convenient result with which to close this chapter. Theorem 8.21. A ring R can be imbedded in a ring R' with identity such that J(R) rad R'. Proof. If R already has an identity, we simply take R' = R. Otherwise, we imbed R in the ring R' = R x Z in the.standard way (see TheoreÍn 2-12 for details). Then R, or more precisely, its isomorphic image R x {O}, is an ideal of R' and R'IR !:::! Z; thus, R'IR is semi simple. This being so, it follows from Theorem 8-5 that rad R' S;;; R. Since R is an ideal of R', we also have J(R) = J(R') n R = rad R' n R (Problem 26). But rad R' 5· R, which implies that J(R) = rad R' .

177

3. Prove that rad R is the largest (in the set-theoretie sense) ideal 1 of R sueh that 1 + a is invertibJe for all a e: l. 4. a) Let the ring R have the property that al! zero divisors líe in rad R. If (a) = (b), show that the elements a and b must be associates. . b) Verify that ifthe element a e: rad R and ax x for some x e: R, then x = O.. 5. ~rovethata.P?werseriesf(x) = ao + a1x+ .,. +a.,x" + ... belongstoradR[[x]] lf and only .,!f lts constant term ao belongs to cad R. 6. Prove the .f,ollowing assertions eonceming semisimple rings: a) A ring !i.tis semisimple if and only if a :f= O implies that there exists some element . r e R fdt'\.vhich 1 - ra is not invertible. b) Every :&:Aunutative regular ring is semisimple. e) Suppo,s~:·that {J,} is a family of ideals of R sueh that R¡1 , is semisimple for eaeh i,a~~ n 1i == {O}. ThenRitselfisasemisimplering. [Hint:Theorem 8-5.J

7. 1: R (1

=

"

,./,

.

= R.~,@J R 2 ® ... ® R. is the direct sum of a finite number of rings R¡ 1, 2¡!:;>i'n), prove that "

,i~r.(

..,

cad R

rad R 1 (!l rad R 2 ® ... ® rad Rn' .

8. Establish that the conditions below are equivalenl: al the ring R has exaetly one maximal ideal (that ís, R is a local ring); b) rad R is a maximal ideal of R; el the set of noninvertible ·elementsof R coincides with rad R' d) the set of noninvertible elements of R form an ideal' ' el lhe sum of two noninvertible elements of R is again'noninvertible; f) for each element r E R, either r or 1 r is in vertible. 9. Let R be a principal ideal domain. Ifthe clement a E R has the prime factorization a == ph~~ ... ¡f-, prove that the Jacobson radical of the guotient ring Rj(a) is (p lP2 ... Pr )¡(a). [Hint: The maximal ideals of R containing (a) are (p 1)' (P2)' ... , (Pr)'] 10.

J..etf be a homomorphism from the ring R onto the ring R'. Show that f(rad R) ¡;; rad R ' and, whenever kerf ¡;; rad R, then rad R = f-1(rad R'); do the same for .RadR..

11. Prove: An ideal 1 of R is semiprime if and only if a2

E

1 implies that a e: 1.

12. Establish that an ideal 1 of R contains a prime ideal ¡f and only if fo! eaeh 11 a1a2 ... a" = Óimplies that ak E 1 for some k. [Hint: The set ' PROBLEMS Unless indicated to thecontrary,allrings are assumed to beeommutativewith identity.

1. Describe the Jacobson radical ofthe ring Zn ofintegers modulo n. [Hint: Consider the prime factorization of I1.J In particular, show that Zn is semisimple if and only if 11 is a square-free integer. 2. Prove that F(xJ, the ring of polynomials in x over a field F, is semisimple.

S= {b1b2· .. b"lbkEi1;n~ 1} is cJosed under multiplication and

a~ S.]

13. Show that the prime radical of a ring R contains the sum of all nilpotent ideals of R. 14. Establish the equivalence ofthe statements below: a) {a} is the only nilpotent ideal of R; b) R is without prime radical; that is, Rad R = {O}; e) for any ideals 1 and J of R, 1J = {O} implies that 1 n J

= {O}.

176

FIRST COURSE IN RINGS AND IDEALS PROBLEMS

Tbis implies that e - i is quasi-regular, say with quasi-inverse x. From the equation (e ...:. i) o x = 0, together witb. the modularíty of M, we obtain

+ (xe

e = i - ix

x) E M.

In con sequen ce, M = R, and an obvious contradiction ensues. Thus, a E M, from which one concludes that J{R) is contained in theintersection of the modular maximal ideals of ~, completing the proof. I

.

The hypothesis that J{R) =1= R: is certainly fulfilled whenever the ring R possesses a multiplicative identity 1. Specifically, the element 1 itself is not quasi-regular, whence 1 f/: J(~h in fact, if 1 + b - lb = O for some b in R, we would have 1 == O, a cótÍtradictio;o.. When an identity element is available, all ideals of R are automatically' modular. In tbis situation, the J-radical will coincide ?Vith the Ja'cpbson radical of R: J{R) = rad R. If J{R) = R, then th.e ríng R. ¡n!ly contain maximal ideals, but no such ideal can be modular. Índeed,:~~ppose that l is any modular ideal of R, R, with e acting as an identity foi.,R,.modulo 1. By suppósition, the element e E J(R), so that e has a quasi-invérse e'. The modularity of l then yields e == e'e - e' E l, .which implies that l = R. Accordingly, the ring R possesses no proper modular ideals and, in particular, no modular maximal ideals. However, the possibility of the existence of maximal ideal s in R is not excluded. The following theorem provides a convenient result with which to close this chapter. Theorem 8.21. A ring R can be imbedded in a ring R' with identity such that J(R) rad R'. Proof. If R already has an identity, we simply take R' = R. Otherwise, we imbed R in the ring R' = R x Z in the.standard way (see TheoreÍn 2-12 for details). Then R, or more precisely, its isomorphic image R x {O}, is an ideal of R' and R'IR !:::! Z; thus, R'IR is semi simple. This being so, it follows from Theorem 8-5 that rad R' S;;; R. Since R is an ideal of R', we also have J(R) = J(R') n R = rad R' n R (Problem 26). But rad R' 5· R, which implies that J(R) = rad R' .

177

3. Prove that rad R is the largest (in the set-theoretie sense) ideal 1 of R sueh that 1 + a is invertibJe for all a e: l. 4. a) Let the ring R have the property that al! zero divisors líe in rad R. If (a) = (b), show that the elements a and b must be associates. . b) Verify that ifthe element a e: rad R and ax x for some x e: R, then x = O.. 5. ~rovethata.P?werseriesf(x) = ao + a1x+ .,. +a.,x" + ... belongstoradR[[x]] lf and only .,!f lts constant term ao belongs to cad R. 6. Prove the .f,ollowing assertions eonceming semisimple rings: a) A ring !i.tis semisimple if and only if a :f= O implies that there exists some element . r e R fdt'\.vhich 1 - ra is not invertible. b) Every :&:Aunutative regular ring is semisimple. e) Suppo,s~:·that {J,} is a family of ideals of R sueh that R¡1 , is semisimple for eaeh i,a~~ n 1i == {O}. ThenRitselfisasemisimplering. [Hint:Theorem 8-5.J

7. 1: R (1

=

"

,./,

.

= R.~,@J R 2 ® ... ® R. is the direct sum of a finite number of rings R¡ 1, 2¡!:;>i'n), prove that "

,i~r.(

..,

cad R

rad R 1 (!l rad R 2 ® ... ® rad Rn' .

8. Establish that the conditions below are equivalenl: al the ring R has exaetly one maximal ideal (that ís, R is a local ring); b) rad R is a maximal ideal of R; el the set of noninvertible ·elementsof R coincides with rad R' d) the set of noninvertible elements of R form an ideal' ' el lhe sum of two noninvertible elements of R is again'noninvertible; f) for each element r E R, either r or 1 r is in vertible. 9. Let R be a principal ideal domain. Ifthe clement a E R has the prime factorization a == ph~~ ... ¡f-, prove that the Jacobson radical of the guotient ring Rj(a) is (p lP2 ... Pr )¡(a). [Hint: The maximal ideals of R containing (a) are (p 1)' (P2)' ... , (Pr)'] 10.

J..etf be a homomorphism from the ring R onto the ring R'. Show that f(rad R) ¡;; rad R ' and, whenever kerf ¡;; rad R, then rad R = f-1(rad R'); do the same for .RadR..

11. Prove: An ideal 1 of R is semiprime if and only if a2

E

1 implies that a e: 1.

12. Establish that an ideal 1 of R contains a prime ideal ¡f and only if fo! eaeh 11 a1a2 ... a" = Óimplies that ak E 1 for some k. [Hint: The set ' PROBLEMS Unless indicated to thecontrary,allrings are assumed to beeommutativewith identity.

1. Describe the Jacobson radical ofthe ring Zn ofintegers modulo n. [Hint: Consider the prime factorization of I1.J In particular, show that Zn is semisimple if and only if 11 is a square-free integer. 2. Prove that F(xJ, the ring of polynomials in x over a field F, is semisimple.

S= {b1b2· .. b"lbkEi1;n~ 1} is cJosed under multiplication and

a~ S.]

13. Show that the prime radical of a ring R contains the sum of all nilpotent ideals of R. 14. Establish the equivalence ofthe statements below: a) {a} is the only nilpotent ideal of R; b) R is without prime radical; that is, Rad R = {O}; e) for any ideals 1 and J of R, 1J = {O} implies that 1 n J

= {O}.

178

PROBLEMS

FIRST COURSE IN RINGS· AND IDEALS

b) Verify that the annihilator of a J-semisimple ring R is zero; in other words, ann R = {O}o

15. A multiplieatively elosed subset S of the ring R is said to be saturated if ab E S implies thatboth a E S and bES. Prove that a) S is a saturated multiplieatively elosed subset of R if and only if its eomplement R - S is a union ofprime ideals; b) the set of non-zero-divisors of R is a saturated multiplieatively elosed subset (hence, the set of zero divisors of R, along with zero, is a union of prime ideals)o

23. If f is a homomorphism froID the ring R onto the ring R', establish the inelusion f(J(R») S;;; J(R'); also show that ir kerf S;;; J(R), then J(R) = rl(J(R'»o 24. Prove eaeh of the statements below: a) J(R) eontains every nil ideal of the ring R. b) J(R) is a semiprime ideal of R. [Hint: Theorem 5-11.] e) For any ring R, Rad R S;;; J(R)o

16. a) Prove that Rad R is the maximal nil ideal of R (maximal among the set of nil ideals); this property is often taken as the definition of the prime radical of R. b) If R has no nonzero nil ideals, deduce that the polynomial ring R[x] is semisimpleo e) Let ehar R = n > 00 Prove that if R is without prime radical, then n is a square-free integero [Hint: Assume that n = p2q for some prime p; then there exists an element a E R sueh that pqa =1 O, but(pqaV = 00]

25. Ifwe define R = {2n/(2m + 1)ln, m E Z}, then R forms a eommutative ring under ordinary addition and multiplieationo Show that J(R) = R, whíle Rad R = {O}o

[Hint:(~)o( 2m + 1 2( -

b) Any1:lomomorphie image of a Hilbert ring is again a Hilbert ringo e) Ifthe'p'olynomial ring R[x] is a Hilbert ring, then R is one alsoo [Hint: Utilize (b) an'eHhe faet that R[xJ/(x) ~ Ro]

In .Problems 21-30, the ring R need not possess an identity elemento 21. Consider the ring P(X) of subsets of some (nonempty) set Xo Show that in this setting the eircle operation reduces to the union operation and determine the' quasi-regular elementso 22. a) Prove that if the element a E R has the property that a" is quasi-regular for so me n E Z+, then a itselfmust be quasi-regularo [Hint: a" = a o( - L:;;:l ak)o]

)= O]

1

o

í'I

lo

29. a) Prove that an element a E R fails to be quasi-regular if and only if a is an identity for R relative to some proper modular ideal of R. b) If lis an ideal ofthe ring R and K is a modular ideal of 1, show that K is also an ideal of R. .

rad (R/I) = Rad (R/I)o

Ji is the only minimal prime ideal of lo

+

28. Assume that the ideal I of R consists of elements whieh are quasi-regular modulo J(R)o (J-Ie say that a is quasi-regular modulo J(R) provided that there exists an e!ement b E R sueh that a o b E J(R»)o Establish that I S;;; J(R)o

19. A ring R is termed a Hilbert ring if eaeh proper prime ideal of R is an interseetion of maximal idealso Prove that: a) R is a Hilbert ring if and op,ly if for every proper ideal I of R,

d) If I is a primary ideal of R,' then [Hint: Problem 19, Chapter 50]

-2n + m)

27_ Prove that ifthe element a fE J(R) is idempotent modulo the ideal I (in other words, (a + I? = a + 1), then a E 1.

18. Supply a proof ofTheorems 8-11 and 8-120

S = {r E Rlra ERad R, for sorne a rt Rad R}o

n

26. LetI be an ideal of the ring R. Regarding I as a ring, deduce thatJ(l) = J(R)

17. Prove that the following.statements are equivalent: a) R has a uniq ue prope~' prime ideal; . b) R is a local ring with rad R = Rad R; e) every noninvertible element of R is nilpotent; d) R is a primary ring and every noninvertible element of R is either a zerb divisor or zeroo

20. If I is anjdeal of the ring R, prove eaeh of the following statements: a) The nílradieal of I is the interseetion of all the minimal prime ideals of lo b) Rad R is the interseetion of all the minimal prime ideals of R. e) The uníon of aIl theminimal prime ideals of R is the set

179

j'

,

;

30. We shall eall an ide'al I of R regular if the quotient ring R/I is a regular ring; in other words, if for eaeh a E R, there exists an element b E R sueh that a2 b - a E lo Prove that o '~,' :; a) Every modular maximal ideal of R is regular. ."-,:;¡~ b) If I I and 12 are both regular ideals of R, then so also is I I í'l1 2 o [Hint: Given a E R, there exist b, e E R sueh that a2 b - a E I I and (a 2 b - a)2c -'''(a 2 b - a) belongs to 12 ; rewrite the last expressiono] ,', e) J(R) = í'I {III is a regular ideal of R}o ""5 [Hint: Assume that a E J(R), but not the right-hand side, so that a rtUor some regular ideal lo If S = I í'I J(R), then a1b - a E S for some b in',Ro Take e = ab E J(R)o Then, el - e E So Show that e E S, whieh leads to the eontradietion that a E lo]

178

PROBLEMS

FIRST COURSE IN RINGS· AND IDEALS

b) Verify that the annihilator of a J-semisimple ring R is zero; in other words, ann R = {O}o

15. A multiplieatively elosed subset S of the ring R is said to be saturated if ab E S implies thatboth a E S and bES. Prove that a) S is a saturated multiplieatively elosed subset of R if and only if its eomplement R - S is a union ofprime ideals; b) the set of non-zero-divisors of R is a saturated multiplieatively elosed subset (hence, the set of zero divisors of R, along with zero, is a union of prime ideals)o

23. If f is a homomorphism froID the ring R onto the ring R', establish the inelusion f(J(R») S;;; J(R'); also show that ir kerf S;;; J(R), then J(R) = rl(J(R'»o 24. Prove eaeh of the statements below: a) J(R) eontains every nil ideal of the ring R. b) J(R) is a semiprime ideal of R. [Hint: Theorem 5-11.] e) For any ring R, Rad R S;;; J(R)o

16. a) Prove that Rad R is the maximal nil ideal of R (maximal among the set of nil ideals); this property is often taken as the definition of the prime radical of R. b) If R has no nonzero nil ideals, deduce that the polynomial ring R[x] is semisimpleo e) Let ehar R = n > 00 Prove that if R is without prime radical, then n is a square-free integero [Hint: Assume that n = p2q for some prime p; then there exists an element a E R sueh that pqa =1 O, but(pqaV = 00]

25. Ifwe define R = {2n/(2m + 1)ln, m E Z}, then R forms a eommutative ring under ordinary addition and multiplieationo Show that J(R) = R, whíle Rad R = {O}o

[Hint:(~)o( 2m + 1 2( -

b) Any1:lomomorphie image of a Hilbert ring is again a Hilbert ringo e) Ifthe'p'olynomial ring R[x] is a Hilbert ring, then R is one alsoo [Hint: Utilize (b) an'eHhe faet that R[xJ/(x) ~ Ro]

In .Problems 21-30, the ring R need not possess an identity elemento 21. Consider the ring P(X) of subsets of some (nonempty) set Xo Show that in this setting the eircle operation reduces to the union operation and determine the' quasi-regular elementso 22. a) Prove that if the element a E R has the property that a" is quasi-regular for so me n E Z+, then a itselfmust be quasi-regularo [Hint: a" = a o( - L:;;:l ak)o]

)= O]

1

o

í'I

lo

29. a) Prove that an element a E R fails to be quasi-regular if and only if a is an identity for R relative to some proper modular ideal of R. b) If lis an ideal ofthe ring R and K is a modular ideal of 1, show that K is also an ideal of R. .

rad (R/I) = Rad (R/I)o

Ji is the only minimal prime ideal of lo

+

28. Assume that the ideal I of R consists of elements whieh are quasi-regular modulo J(R)o (J-Ie say that a is quasi-regular modulo J(R) provided that there exists an e!ement b E R sueh that a o b E J(R»)o Establish that I S;;; J(R)o

19. A ring R is termed a Hilbert ring if eaeh proper prime ideal of R is an interseetion of maximal idealso Prove that: a) R is a Hilbert ring if and op,ly if for every proper ideal I of R,

d) If I is a primary ideal of R,' then [Hint: Problem 19, Chapter 50]

-2n + m)

27_ Prove that ifthe element a fE J(R) is idempotent modulo the ideal I (in other words, (a + I? = a + 1), then a E 1.

18. Supply a proof ofTheorems 8-11 and 8-120

S = {r E Rlra ERad R, for sorne a rt Rad R}o

n

26. LetI be an ideal of the ring R. Regarding I as a ring, deduce thatJ(l) = J(R)

17. Prove that the following.statements are equivalent: a) R has a uniq ue prope~' prime ideal; . b) R is a local ring with rad R = Rad R; e) every noninvertible element of R is nilpotent; d) R is a primary ring and every noninvertible element of R is either a zerb divisor or zeroo

20. If I is anjdeal of the ring R, prove eaeh of the following statements: a) The nílradieal of I is the interseetion of all the minimal prime ideals of lo b) Rad R is the interseetion of all the minimal prime ideals of R. e) The uníon of aIl theminimal prime ideals of R is the set

179

j'

,

;

30. We shall eall an ide'al I of R regular if the quotient ring R/I is a regular ring; in other words, if for eaeh a E R, there exists an element b E R sueh that a2 b - a E lo Prove that o '~,' :; a) Every modular maximal ideal of R is regular. ."-,:;¡~ b) If I I and 12 are both regular ideals of R, then so also is I I í'l1 2 o [Hint: Given a E R, there exist b, e E R sueh that a2 b - a E I I and (a 2 b - a)2c -'''(a 2 b - a) belongs to 12 ; rewrite the last expressiono] ,', e) J(R) = í'I {III is a regular ideal of R}o ""5 [Hint: Assume that a E J(R), but not the right-hand side, so that a rtUor some regular ideal lo If S = I í'I J(R), then a1b - a E S for some b in',Ro Take e = ab E J(R)o Then, el - e E So Show that e E S, whieh leads to the eontradietion that a E lo]

TWO CLASSIC THEOREMS

NINE.

a Boole~n ring, f(x) O,then ,

181

w~ proceed as follows: If the funetion f E R is sueh that .

.

(j2)(x) = f(x) '2f(x) = Ó '2 O = 0,

1, then

whereas if f(x)

1.

TWO CLASSIC THEOREMS .-:,.•,... ¡" ;:" ..

.

; ~:~

The two theorems whichwe shall be primarily eoneemed with are the Stone Representation Theorem for Boolean rings and Wedderburn's Theorembn finite division rings. O[ the two, the proof of Wedderburn's Theorem..is more troublesome 'and occupies the major portion of our effort in '"üti~ chapter. Before embarkiríg on this latter task, it is necessary to assemble'~ number of results pertinent to the strueture of finite, fields. OUT aim haír been to spell out all the important details and the reacier may find some of the arguments ra thei complex.' . We first take up the celebrated result of Stone, which asserts that eaeh .Boolean ring can be represented by a ring of sets. At the outset, let us recall that by a Boolean ring is meant a ring with identity every element of which is idempotent. ~t may be well to emphasize that the existen ce of an identity is frequently omitted in the definition of a Boolean ring; for the applications which we have in mind, the presence of such an e1ement will be convenient. (One can show that if the number of elements of a Boolean ring is finite, then a multiplicative identity always exists.) Let us pause long enough to indicate several standard examples of Boolean rings. ' Example 9-1. The ring Z2 of integers moduló 2. Example 9-2. The ring (P(X), A, n) of subsets of a nonempty set X, with the usual interpretation of A and n. Example 9-3. For a less obvious illustration, let R = map(X, Z2), where . X is an arbitrary nonempty .set. ·As is eustomary in this setting, the ring operations are define4 pointwise; that 1S, ifj and g ,are in R, then (f + g)(x) = f(x) (fg) (x)

f(x)

'2

g(x)

+2 g(x), (x E X).

It is already kllown that R forms a commutative ring with identity (Example

1-4). To establish the idempotency condition and thereby show that R is 180

In any event, (j2)(X)

f(x) for all x

E

X; hence,j2;

!

Th~ idempotency proviso in the definition or:;~ Boolean ring has a strong mfluence on the structure of such rings (in f~Gt, Boolean rings have an almost embarrassingly rieh strueture). Twoot the most important eonsequences are that (1) a Boolean ring is of el,iaiacteristic 2and (2) a Boolean ring is commutative. Although these facts',have appeared in the exercises, they have never been formally proved; laii~ly to assure completeness, let us indicate the argument here. . . :.~Tr . Given arbitrary elements 'eJ, b of a Boolean ring~, it folIows that a + b=:,' (a + b)2 a2 + ab + ba + b2 = a'+ ab + ba + b and, henee, ab

+

ba == O. Setting a 2a

a

= b in this last equation, we obtain

+ 'a =

a2

a2

°

=. , The relation ab + ba

: whieh shows that ehar R= 2.. commutativity of R; indeed, since a. ab

+

-ba

= O now yields the

-a for any a in R, ba.

The maximal ideals of a Boolean ring are characterized by some interesting algebraic properties; to elaborate more fully on this, If J is a proper ideal of the Boolean ring R then the Theorem 9-1. fOllowing conditions are equivalent: ' 1) J is a maximal ideal; 2) 1 is a prime ideal; 3) for every element a E R, either a or 1 - a (but not both) belongs to J. Proo! . The equivalen ce of assertions (1) and (2) 1S the content of Theorem 5-8. We now assume (2). Siliee a(l a) OE J for aH a in R the fact t~at J is prime implies that either a or 1 a must lie in J (but ~ot both, ~lllce 1~ a + (1 - a)). The proof ofthe theorem is completed by showing that (1) lS a consequence pf (3). For this, suppose that J is an ideal of R with the prope~ty t.hat J c: J S;; R; what needs to be proved is the equality J = R. Now, lf a lS any element of J whieh is not in J, then by supposition 1 - a E J. But this means 1 - a E J, so that 1 a + (1 a) E J and, as a result, J = R.

TWO CLASSIC THEOREMS

NINE.

a Boole~n ring, f(x) O,then ,

181

w~ proceed as follows: If the funetion f E R is sueh that .

.

(j2)(x) = f(x) '2f(x) = Ó '2 O = 0,

1, then

whereas if f(x)

1.

TWO CLASSIC THEOREMS .-:,.•,... ¡" ;:" ..

.

; ~:~

The two theorems whichwe shall be primarily eoneemed with are the Stone Representation Theorem for Boolean rings and Wedderburn's Theorembn finite division rings. O[ the two, the proof of Wedderburn's Theorem..is more troublesome 'and occupies the major portion of our effort in '"üti~ chapter. Before embarkiríg on this latter task, it is necessary to assemble'~ number of results pertinent to the strueture of finite, fields. OUT aim haír been to spell out all the important details and the reacier may find some of the arguments ra thei complex.' . We first take up the celebrated result of Stone, which asserts that eaeh .Boolean ring can be represented by a ring of sets. At the outset, let us recall that by a Boolean ring is meant a ring with identity every element of which is idempotent. ~t may be well to emphasize that the existen ce of an identity is frequently omitted in the definition of a Boolean ring; for the applications which we have in mind, the presence of such an e1ement will be convenient. (One can show that if the number of elements of a Boolean ring is finite, then a multiplicative identity always exists.) Let us pause long enough to indicate several standard examples of Boolean rings. ' Example 9-1. The ring Z2 of integers moduló 2. Example 9-2. The ring (P(X), A, n) of subsets of a nonempty set X, with the usual interpretation of A and n. Example 9-3. For a less obvious illustration, let R = map(X, Z2), where . X is an arbitrary nonempty .set. ·As is eustomary in this setting, the ring operations are define4 pointwise; that 1S, ifj and g ,are in R, then (f + g)(x) = f(x) (fg) (x)

f(x)

'2

g(x)

+2 g(x), (x E X).

It is already kllown that R forms a commutative ring with identity (Example

1-4). To establish the idempotency condition and thereby show that R is 180

In any event, (j2)(X)

f(x) for all x

E

X; hence,j2;

!

Th~ idempotency proviso in the definition or:;~ Boolean ring has a strong mfluence on the structure of such rings (in f~Gt, Boolean rings have an almost embarrassingly rieh strueture). Twoot the most important eonsequences are that (1) a Boolean ring is of el,iaiacteristic 2and (2) a Boolean ring is commutative. Although these facts',have appeared in the exercises, they have never been formally proved; laii~ly to assure completeness, let us indicate the argument here. . . :.~Tr . Given arbitrary elements 'eJ, b of a Boolean ring~, it folIows that a + b=:,' (a + b)2 a2 + ab + ba + b2 = a'+ ab + ba + b and, henee, ab

+

ba == O. Setting a 2a

a

= b in this last equation, we obtain

+ 'a =

a2

a2

°

=. , The relation ab + ba

: whieh shows that ehar R= 2.. commutativity of R; indeed, since a. ab

+

-ba

= O now yields the

-a for any a in R, ba.

The maximal ideals of a Boolean ring are characterized by some interesting algebraic properties; to elaborate more fully on this, If J is a proper ideal of the Boolean ring R then the Theorem 9-1. fOllowing conditions are equivalent: ' 1) J is a maximal ideal; 2) 1 is a prime ideal; 3) for every element a E R, either a or 1 - a (but not both) belongs to J. Proo! . The equivalen ce of assertions (1) and (2) 1S the content of Theorem 5-8. We now assume (2). Siliee a(l a) OE J for aH a in R the fact t~at J is prime implies that either a or 1 a must lie in J (but ~ot both, ~lllce 1~ a + (1 - a)). The proof ofthe theorem is completed by showing that (1) lS a consequence pf (3). For this, suppose that J is an ideal of R with the prope~ty t.hat J c: J S;; R; what needs to be proved is the equality J = R. Now, lf a lS any element of J whieh is not in J, then by supposition 1 - a E J. But this means 1 - a E J, so that 1 a + (1 a) E J and, as a result, J = R.

182

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

A natural undertaking is to determine which Boolean rings are also fields. We may dispose of tliis question rather quickly: up to isomorphism, the only BooIean field is the ring of integers modulo 2. Theorem 9-2., A Boolean ring R is a field if and only if R Proof. Let R be a Boolean field. For any nonzero

el~ment

~

1 +2 f(a)

= f(l)

+ a El

+zf(a)

2 2,

a E R, we then

have a = al = a(aa- 1 ) = a2a- 1 = aa- 1 = 1.

This reasoning shows that the only nonzero element of R is the multiplicative identity; in other words, R = {O, 1}. But any two-element field is isomorphic to 2 2 , The Converse of the theorem is fairly obvious. This gives rise to another characterization of maximal (equivalently, prime) ideals in Boolean"rings. CoroUary. A proper ideal l of a Boolean ring R is a maximal ideal if and onIy ifR.jl ~ 2 2 , Proof. Since the quotient ring Rjl inherits the idempotency condition, Rjl is itself a Boolean ringo By Theorem 5-5, l is a maximal ideal if and only if Rjl is a (Boo.1ean) field. An appeal to the aboye theorem now completes the p~oof.

Tlie next theorem is a major one and requires a preliminary lemma of sorne intrinsic interest. Lemma. Let R be a Boolean ringo For, each n~nzero element a E R, there exists a homomorphism f from R;~'~pto the field 2 2 such that f(a) = 1 . ) / . ,

f

= (1 q) generated by the element 1 + a. Now,l =1= R,sincetheidentityelemeni,ls.notamemberofI. Indeed, "1"'11" if 1 E l, then 1 = r(1 + a) for sorne choice o'fr in R; this means . . /. Proof. Consider the principal ideal l

such functions. Inasmuqh as 1 so that

= r(l + a)2 = (r(l + a})(l +::'a) = 1(1 + a), "H from which it follows that a = O, contrary to 'assumption. 1

Because lis a proper ideal, Theorem 5-2guarantees theexistence of a maximal ideal M of R with l s; M. In the \ight of the result just proven, the corresponding quotient ring RjM will be isomorphic to 2 2 via sorne homomorphism g. We may therefore define a function f: R -+ 2 2 by taking f = g o nat M , where natM is the usual natural mapping ofR onto RjM.

The remainder of the proof amounts to showing that the function J, so defined, has the properties asserted in the statement ofthe theotem. Plainly, fis both an onto rnap and a homornorphisrn, being the cornposition oftwo

S;

M, the coset 1

= f(l + = g(l +

+

a

+

183

M = M,

a) a

+ M) =

g(M) = O.

But 1 +2 f(a) = O if and only iff(a) = 1, which finishes the proof. That any Boolean ring is semisimple is an immediate consequence of ' the foregoing result. CoroUary. Every Boolean ring R is a sernisimple ring; that is, rad R = {O}. Proof. In order to arrive at a contradiction, we assume that a Erad R with a =1= O. Then there'exist~ an onto homomorphismf: R -+ 2 2 for which f(a) = 1. It follows that kerfmust be a proper ideal ofthe ring R. Hence, there exists sorne maximal ideal M of R with ker f S; M. In particular, the element 1 - ti E ker f S; M. But also, a Erad R S; M, which implies that 1 = a + (1 I a) E M. This at once leads' to M = R, the desired contradiction.

We now set our~elves to the prinCipal task, that of showing that each Boolean ring is essentially a ring of sets. Theorem 9-3. (Stone Representation Theorem). Any Boolean ring R is isornorphic to a ring of subsets of sorne fixed seto Proof. To begin the attac~, let H denote the collection of all hornornorphisrns of R onto the field 2 2 , ~ext, define a function h: R -+ P(H) by assigning to each element a E R thQsemembers of H which as sume the value 1 at a; in other words, .. "

':

h(ii) = {fE Hlf(a) = 1}. .-,:.

Although the notation is perfectly c1ear, let us stress that h is a set-valued function in the sense that:#~ functional values are certain subsets of H. By means of this function, we'shall establish the isornorphism mentioned in the theorern. ' [ Let us now give the details. For any fE H, the productf(a) '2f(b) = 1 if and only if both f(a) = 1 and f(b) = 1. This being so, one conc1udes that h(ab) = {fE Hlf(ab) = 1}

= {fE Hlf(a) '2f(b) =

1}

= {fE Hlf(a) = 1} n {fE Hlf(b) = 1} = h(a) n h(b),

182

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

A natural undertaking is to determine which Boolean rings are also fields. We may dispose of tliis question rather quickly: up to isomorphism, the only BooIean field is the ring of integers modulo 2. Theorem 9-2., A Boolean ring R is a field if and only if R Proof. Let R be a Boolean field. For any nonzero

el~ment

~

1 +2 f(a)

= f(l)

+ a El

+zf(a)

2 2,

a E R, we then

have a = al = a(aa- 1 ) = a2a- 1 = aa- 1 = 1.

This reasoning shows that the only nonzero element of R is the multiplicative identity; in other words, R = {O, 1}. But any two-element field is isomorphic to 2 2 , The Converse of the theorem is fairly obvious. This gives rise to another characterization of maximal (equivalently, prime) ideals in Boolean"rings. CoroUary. A proper ideal l of a Boolean ring R is a maximal ideal if and onIy ifR.jl ~ 2 2 , Proof. Since the quotient ring Rjl inherits the idempotency condition, Rjl is itself a Boolean ringo By Theorem 5-5, l is a maximal ideal if and only if Rjl is a (Boo.1ean) field. An appeal to the aboye theorem now completes the p~oof.

Tlie next theorem is a major one and requires a preliminary lemma of sorne intrinsic interest. Lemma. Let R be a Boolean ringo For, each n~nzero element a E R, there exists a homomorphism f from R;~'~pto the field 2 2 such that f(a) = 1 . ) / . ,

f

= (1 q) generated by the element 1 + a. Now,l =1= R,sincetheidentityelemeni,ls.notamemberofI. Indeed, "1"'11" if 1 E l, then 1 = r(1 + a) for sorne choice o'fr in R; this means . . /. Proof. Consider the principal ideal l

such functions. Inasmuqh as 1 so that

= r(l + a)2 = (r(l + a})(l +::'a) = 1(1 + a), "H from which it follows that a = O, contrary to 'assumption. 1

Because lis a proper ideal, Theorem 5-2guarantees theexistence of a maximal ideal M of R with l s; M. In the \ight of the result just proven, the corresponding quotient ring RjM will be isomorphic to 2 2 via sorne homomorphism g. We may therefore define a function f: R -+ 2 2 by taking f = g o nat M , where natM is the usual natural mapping ofR onto RjM.

The remainder of the proof amounts to showing that the function J, so defined, has the properties asserted in the statement ofthe theotem. Plainly, fis both an onto rnap and a homornorphisrn, being the cornposition oftwo

S;

M, the coset 1

= f(l + = g(l +

+

a

+

183

M = M,

a) a

+ M) =

g(M) = O.

But 1 +2 f(a) = O if and only iff(a) = 1, which finishes the proof. That any Boolean ring is semisimple is an immediate consequence of ' the foregoing result. CoroUary. Every Boolean ring R is a sernisimple ring; that is, rad R = {O}. Proof. In order to arrive at a contradiction, we assume that a Erad R with a =1= O. Then there'exist~ an onto homomorphismf: R -+ 2 2 for which f(a) = 1. It follows that kerfmust be a proper ideal ofthe ring R. Hence, there exists sorne maximal ideal M of R with ker f S; M. In particular, the element 1 - ti E ker f S; M. But also, a Erad R S; M, which implies that 1 = a + (1 I a) E M. This at once leads' to M = R, the desired contradiction.

We now set our~elves to the prinCipal task, that of showing that each Boolean ring is essentially a ring of sets. Theorem 9-3. (Stone Representation Theorem). Any Boolean ring R is isornorphic to a ring of subsets of sorne fixed seto Proof. To begin the attac~, let H denote the collection of all hornornorphisrns of R onto the field 2 2 , ~ext, define a function h: R -+ P(H) by assigning to each element a E R thQsemembers of H which as sume the value 1 at a; in other words, .. "

':

h(ii) = {fE Hlf(a) = 1}. .-,:.

Although the notation is perfectly c1ear, let us stress that h is a set-valued function in the sense that:#~ functional values are certain subsets of H. By means of this function, we'shall establish the isornorphism mentioned in the theorern. ' [ Let us now give the details. For any fE H, the productf(a) '2f(b) = 1 if and only if both f(a) = 1 and f(b) = 1. This being so, one conc1udes that h(ab) = {fE Hlf(ab) = 1}

= {fE Hlf(a) '2f(b) =

1}

= {fE Hlf(a) = 1} n {fE Hlf(b) = 1} = h(a) n h(b),

184

FIRST COURSE IN RINGS AND IDEALS

185

TWO CLASSIC THEOREMS

showing that the function h preserves muItiplication. The verification that h(a + b) = h(a) 8 h(b) is equally straightforward, depending chiefiy on the observation that the sumf(a) +d(b) = 1 if and only if one off(a) or f(b) is 1, while the other is O; the reader may easily fill in the steps for himselL These remarks serve to demonstrate the fact that h is a homomorphism from' R into the ring of sets (P(H), 8, ()). . , AH that is needed to complete the proof is to show that h is a one-to-one function'or:, equivalentIy, that ker h = {O}. But this foHows immediately from the preceding lemma, which asserts that h(a) is nonempty if and only if a +- O; w h e n c e ' · . . kerh

= {aERlh(~);:~::

0} = {O}.

AH the pieces faH into place and we s.~:~that the ring R i8 isomorphic to a subring (namely, the sll,bring h(R») oqI:~JH), 8, tl). •

"/-;·.'l

We shall return to Boolean rings:frqm time to time in the sequel, but for the moment let us raise the foIlowirig:question (which may have already occurred to some readers): whatis the effect.of requiring that each ideal of a ring R be indempotent? In the theorem below, we characterize the regular ring8 by a condition on one-sided ideals which in the commutative case i8 equivalent to the idempotency of the ideals of R. To recall the earlier definition, a regular ring is a ring R with the property that every element a has a pseudo-inver8e a' E R satisIYing aa' a = a; it is worth emphasizing that regular ri~gs need not be commutative or póssess an identity. Tbeorem 9-4. A ring R is regular if and only if IJ=1tlJ

Proof. Let 1 be a rigbt ideal and J a left ideal of the regular ring R. Since the inclusion IJ S;; 1 () J always obtains, we have only ·to show that any element a of 1 tl J is in U. By the regularity of R, aa' a = a for some a' E R. Rere a'a is a me~ber ofthe left ideal J, so that the product a(a'a) E IJ, from which one infers that a E U. For the converse, we assume that the indicated condition holds and ]2roceed to establish that R is regular. Let a be an arbitrary elemenf of R. Then (in the absence of an identity) the right ideal generate9 by a is the set aR + Za {ar + nalr E R, n E Z}. By virtue oC our hypothesis,

+ Za

;: (aR

+ Za)

() R = (aR

+ Za)R

and so a E aR. Analogously, a ERa, which leads to

a E aR

tl

Ra = aR2a.

.

I

!

'

CorolJary 1. A commutative ring R is regular if and only if evei'y ideal Qf R is idempotept. ..

Pro?f. If R is regular, .ve may take 1 = J in the theorem to conclude that the ld~ls o~ R are idempotent. On the other hand, let 1 and J be arbitrary (two-slded) ldeals R. Then the idempotency of 1 tl J yields :\'1 tl J = (1 tl J)2 S;; 1J S;; 1 tl J,

0l

and so 1J = 1 ()~!/ Appealing to the theorem once more it follows that R must be a reguli;lt ringo . , ..... .,.." . CorolIary commutative ring R with identity, the condition

2.fídl.

. "c:;', (ab) = (a) • ,-? '. lS eqUlvalent t.q,,tegularity. ~

tl

(b),

a, bE R

:,~~,!

.

.l::,

Proof. If the indicated condition hólds, then in particular a E R implies 2 (a ) = (a) () (a). Rence, for each a in R, there exists some a' E R such that 2 a = a' a = 00' a and R is regular. Conversely, if R is regular, it follows from the theorem that (a) tl (b) = (a)(b) = (ah) for alI a, b E R. . . F~o~ no,: on, we shall assume that any regular ring possesses a multíphc~t1ve. l~entIty 1; this requirement is not essential, buthas the advantage of slmphfymg our arguments. To continue with the analysis of regular rings,

:Lem~a. A ring R ls regular if and only if ev~ry principal right (left) ldealls generated by an idempotent.

holds for every right ideal 1 and left ideal J of R;

aR

~hus; there exists an element a' E R such that a = aa'a, making R a regular ~S·.

= aR,

Proof.. S~ppos~ that R is regular and aR ii> a principal right ideal.. If the element a E R lS such that aa'a = a, then (aa')2 = (aa'a)a'

= aa',

= 00' is !dempot~nt ..

so that e From a .= ea E eR, we deduce that aR S;; eR. Rowever~ e ;: aa E aR, Yleldtng eR S;; aR and the subsequent equ:ality. ~ Turmng matte~s around, let us now assume that every principal. right Ideal of R has an ldempotent generator: Given an element a E R 'ch 'd ' oose ~ I empotent e such that ~R = eR: Then, for suitable r, s in R, the equatlons a = er and e = as wIIl be sattsfied. But this implies that

asa'= ea = e(er) = er = a, whence R forms a regular ringo !he elementary lemma just proved enables us to throw new light on theldeaI structure of regular rings. .

184

FIRST COURSE IN RINGS AND IDEALS

185

TWO CLASSIC THEOREMS

showing that the function h preserves muItiplication. The verification that h(a + b) = h(a) 8 h(b) is equally straightforward, depending chiefiy on the observation that the sumf(a) +d(b) = 1 if and only if one off(a) or f(b) is 1, while the other is O; the reader may easily fill in the steps for himselL These remarks serve to demonstrate the fact that h is a homomorphism from' R into the ring of sets (P(H), 8, ()). . , AH that is needed to complete the proof is to show that h is a one-to-one function'or:, equivalentIy, that ker h = {O}. But this foHows immediately from the preceding lemma, which asserts that h(a) is nonempty if and only if a +- O; w h e n c e ' · . . kerh

= {aERlh(~);:~::

0} = {O}.

AH the pieces faH into place and we s.~:~that the ring R i8 isomorphic to a subring (namely, the sll,bring h(R») oqI:~JH), 8, tl). •

"/-;·.'l

We shall return to Boolean rings:frqm time to time in the sequel, but for the moment let us raise the foIlowirig:question (which may have already occurred to some readers): whatis the effect.of requiring that each ideal of a ring R be indempotent? In the theorem below, we characterize the regular ring8 by a condition on one-sided ideals which in the commutative case i8 equivalent to the idempotency of the ideals of R. To recall the earlier definition, a regular ring is a ring R with the property that every element a has a pseudo-inver8e a' E R satisIYing aa' a = a; it is worth emphasizing that regular ri~gs need not be commutative or póssess an identity. Tbeorem 9-4. A ring R is regular if and only if IJ=1tlJ

Proof. Let 1 be a rigbt ideal and J a left ideal of the regular ring R. Since the inclusion IJ S;; 1 () J always obtains, we have only ·to show that any element a of 1 tl J is in U. By the regularity of R, aa' a = a for some a' E R. Rere a'a is a me~ber ofthe left ideal J, so that the product a(a'a) E IJ, from which one infers that a E U. For the converse, we assume that the indicated condition holds and ]2roceed to establish that R is regular. Let a be an arbitrary elemenf of R. Then (in the absence of an identity) the right ideal generate9 by a is the set aR + Za {ar + nalr E R, n E Z}. By virtue oC our hypothesis,

+ Za

;: (aR

+ Za)

() R = (aR

+ Za)R

and so a E aR. Analogously, a ERa, which leads to

a E aR

tl

Ra = aR2a.

.

I

!

'

CorolJary 1. A commutative ring R is regular if and only if evei'y ideal Qf R is idempotept. ..

Pro?f. If R is regular, .ve may take 1 = J in the theorem to conclude that the ld~ls o~ R are idempotent. On the other hand, let 1 and J be arbitrary (two-slded) ldeals R. Then the idempotency of 1 tl J yields :\'1 tl J = (1 tl J)2 S;; 1J S;; 1 tl J,

0l

and so 1J = 1 ()~!/ Appealing to the theorem once more it follows that R must be a reguli;lt ringo . , ..... .,.." . CorolIary commutative ring R with identity, the condition

2.fídl.

. "c:;', (ab) = (a) • ,-? '. lS eqUlvalent t.q,,tegularity. ~

tl

(b),

a, bE R

:,~~,!

.

.l::,

Proof. If the indicated condition hólds, then in particular a E R implies 2 (a ) = (a) () (a). Rence, for each a in R, there exists some a' E R such that 2 a = a' a = 00' a and R is regular. Conversely, if R is regular, it follows from the theorem that (a) tl (b) = (a)(b) = (ah) for alI a, b E R. . . F~o~ no,: on, we shall assume that any regular ring possesses a multíphc~t1ve. l~entIty 1; this requirement is not essential, buthas the advantage of slmphfymg our arguments. To continue with the analysis of regular rings,

:Lem~a. A ring R ls regular if and only if ev~ry principal right (left) ldealls generated by an idempotent.

holds for every right ideal 1 and left ideal J of R;

aR

~hus; there exists an element a' E R such that a = aa'a, making R a regular ~S·.

= aR,

Proof.. S~ppos~ that R is regular and aR ii> a principal right ideal.. If the element a E R lS such that aa'a = a, then (aa')2 = (aa'a)a'

= aa',

= 00' is !dempot~nt ..

so that e From a .= ea E eR, we deduce that aR S;; eR. Rowever~ e ;: aa E aR, Yleldtng eR S;; aR and the subsequent equ:ality. ~ Turmng matte~s around, let us now assume that every principal. right Ideal of R has an ldempotent generator: Given an element a E R 'ch 'd ' oose ~ I empotent e such that ~R = eR: Then, for suitable r, s in R, the equatlons a = er and e = as wIIl be sattsfied. But this implies that

asa'= ea = e(er) = er = a, whence R forms a regular ringo !he elementary lemma just proved enables us to throw new light on theldeaI structure of regular rings. .

186

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

Proo! First, take R to be a regular ring and J and arbitrary ideal of R. Since we always ha ve J S;; it is enough to establish the reverse inclusion here. For this, let a be any element of so that el' E J for sorne positive integer n. By the regularity oC R, there exists an a' E R with a = a2 a'. Repeated multiplication of this relation by aa' leads to a = a/l(a,)n-l E J, S;; 1. whence Going in the other direction, assume that R is a ring in which every ideal coincides with its nil radical. Given an element a E R, we consider the principal ideal (a 2). Because a 3 'E (a 2), it follows that a E -J(a2) and so, by our hypothesis, that a E (a 2). But then, a = a2a' for a suitable choice of a' in R, making R aregular ringo

Theorem 9,·5. The sum of two principal right (left) ideal s of a regular ring R is itself a principal right (left) ideal.

.JI,

Proo! Consider the right ideal aR + bR. As reference to the lemma will confirm, aR is of the form aR = eR, with e an idempotent. We also ha ve eR + bR = {ex + bylx,YER} = =

+ eby + (1 {e(x + by) + (1 {ex

= {ez

+

- e)bylx, y

E

R}

= eR + (1 - e)bR.

Put e = (1 - e)b, so that aR + bR = eR + eR, where the element e has thepropertyee.= e(1 - e)b = (e - e2)b = O. Weagaininvokethelemma to come up with eR ~ fR, f being an idempotent of R; since f is of the formf = ex, the product ef = O. As a final preliminary, let g = f(1 - e). To see that the element g is idempotertt, observe that

As a prelude to the second of the two theorems on our program, it is essential to exaffiine the structure of fields with a finite number of elements. In this connection, the most reasonable question is whether there exist finite fields other than the fields Zp (p a prime) and, if so, whether they can be completely described. We launch our inquiry by proving that every finite field is of prime-power order.

gf = f(1 - e)f = fU - ef) = f2 = j, which subsequently yields g2 = gf(1 - e) very important deduction is that

eg ge

= f(1 - e) = g. A further and

Theorem 9-7. If F is a finite field, then F has exactly p" elements for sorne prime p and n E Z+.

°

= ef(1 - e) = 0, = f(1 - e)e = O.

As bothf = gfe gR and g = f(1 - e) EfR, the ideal fR = gR. One finds in this way that the sum aR + bR = eR + gR. . , ,Our remaining objective is to demonstrate that eR + gR = (e + g)R. Tlíe reasoning proceeds as follows. Since e + g e eR + gR, the inclusion . ..1 (e. g)R S;; eR + gR evidentIy holds. On the other hand, the element é.~ e2 + ge = (e + g)e E (e + g)R,and,atthesametime,g = eg + g2 = (~.;.+ g)g E (e + g)R, which forces eR + gR S;; (e + g)R. Thisleads to the Üi. t:gp.ality eR + gR = (e + g)R and in turn to aR + bR = (e + g)R, t concluding the proof.

+

.: By an easy induction, we arrive naturally at Corollary. A right (left) ideal of a regular ring is finitely generated if and only if it is principal.

.JI,

.JI

- e)bylx, y E R}

(1 - e)bylz, y E R}

187

'~'.

Proo! Since the prime subfield of a field of characteristic has infinitely many elements, F must necessarily be of characteristic p, where p is a prime. Nothing will be lost by assuming, as we shall henceforth, that the prime subfield of F is actually the field Zp (Theorem 4-12). In the light of the finitenessofF,itmayberegardedasafiniteextensionofZp , say [F: Zp] = n. Suppose that the n elements r 1, r2 , ... , rn. constitute:aJJasis for F as a vector space over Zp. Then every element r in F can De ',üniquely written "~~ ,;': in the form r = a1r 1 + a2r 2 + ... + anrn , . ·t , ~

~

where the a¡ E Zp. Now, each coefficient a¡ is capable of assuming p values, so that the total number of such linear combinations is gn; It follows that there are pn distinct members of F . . ' " Corollary. The number of .elements in a finite field i is pn, where the prime p = char F and n = [F: Zp]' .

We conclude this phase of our investigation with a rather surprising characterization of commutative regular rings in terms of semi prime ideaÍs. (Lest one forget, an ideal J of a ring R is said to be semiprime if and only if J =

The implications of Theorem 9-7 is that one cannot construct finite fields with q elements unless q is a power of a prime. This immediately raises the question: Given q, a prescribed power of a prime, do there actually exist fields with q elements? In obtaining an answer, it is crucial to know the following. .

Theorem 9-6. A commutative ring R is regular if and only if every ideal of R is semiprime.

Lemma. Every element of a finite field F with pn elements isa root of the polynomialf(x) = xpn - x e F[x].

.JI.)

186

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

Proo! First, take R to be a regular ring and J and arbitrary ideal of R. Since we always ha ve J S;; it is enough to establish the reverse inclusion here. For this, let a be any element of so that el' E J for sorne positive integer n. By the regularity oC R, there exists an a' E R with a = a2 a'. Repeated multiplication of this relation by aa' leads to a = a/l(a,)n-l E J, S;; 1. whence Going in the other direction, assume that R is a ring in which every ideal coincides with its nil radical. Given an element a E R, we consider the principal ideal (a 2). Because a 3 'E (a 2), it follows that a E -J(a2) and so, by our hypothesis, that a E (a 2). But then, a = a2a' for a suitable choice of a' in R, making R aregular ringo

Theorem 9,·5. The sum of two principal right (left) ideal s of a regular ring R is itself a principal right (left) ideal.

.JI,

Proo! Consider the right ideal aR + bR. As reference to the lemma will confirm, aR is of the form aR = eR, with e an idempotent. We also ha ve eR + bR = {ex + bylx,YER} = =

+ eby + (1 {e(x + by) + (1 {ex

= {ez

+

- e)bylx, y

E

R}

= eR + (1 - e)bR.

Put e = (1 - e)b, so that aR + bR = eR + eR, where the element e has thepropertyee.= e(1 - e)b = (e - e2)b = O. Weagaininvokethelemma to come up with eR ~ fR, f being an idempotent of R; since f is of the formf = ex, the product ef = O. As a final preliminary, let g = f(1 - e). To see that the element g is idempotertt, observe that

As a prelude to the second of the two theorems on our program, it is essential to exaffiine the structure of fields with a finite number of elements. In this connection, the most reasonable question is whether there exist finite fields other than the fields Zp (p a prime) and, if so, whether they can be completely described. We launch our inquiry by proving that every finite field is of prime-power order.

gf = f(1 - e)f = fU - ef) = f2 = j, which subsequently yields g2 = gf(1 - e) very important deduction is that

eg ge

= f(1 - e) = g. A further and

Theorem 9-7. If F is a finite field, then F has exactly p" elements for sorne prime p and n E Z+.

°

= ef(1 - e) = 0, = f(1 - e)e = O.

As bothf = gfe gR and g = f(1 - e) EfR, the ideal fR = gR. One finds in this way that the sum aR + bR = eR + gR. . , ,Our remaining objective is to demonstrate that eR + gR = (e + g)R. Tlíe reasoning proceeds as follows. Since e + g e eR + gR, the inclusion . ..1 (e. g)R S;; eR + gR evidentIy holds. On the other hand, the element é.~ e2 + ge = (e + g)e E (e + g)R,and,atthesametime,g = eg + g2 = (~.;.+ g)g E (e + g)R, which forces eR + gR S;; (e + g)R. Thisleads to the Üi. t:gp.ality eR + gR = (e + g)R and in turn to aR + bR = (e + g)R, t concluding the proof.

+

.: By an easy induction, we arrive naturally at Corollary. A right (left) ideal of a regular ring is finitely generated if and only if it is principal.

.JI,

.JI

- e)bylx, y E R}

(1 - e)bylz, y E R}

187

'~'.

Proo! Since the prime subfield of a field of characteristic has infinitely many elements, F must necessarily be of characteristic p, where p is a prime. Nothing will be lost by assuming, as we shall henceforth, that the prime subfield of F is actually the field Zp (Theorem 4-12). In the light of the finitenessofF,itmayberegardedasafiniteextensionofZp , say [F: Zp] = n. Suppose that the n elements r 1, r2 , ... , rn. constitute:aJJasis for F as a vector space over Zp. Then every element r in F can De ',üniquely written "~~ ,;': in the form r = a1r 1 + a2r 2 + ... + anrn , . ·t , ~

~

where the a¡ E Zp. Now, each coefficient a¡ is capable of assuming p values, so that the total number of such linear combinations is gn; It follows that there are pn distinct members of F . . ' " Corollary. The number of .elements in a finite field i is pn, where the prime p = char F and n = [F: Zp]' .

We conclude this phase of our investigation with a rather surprising characterization of commutative regular rings in terms of semi prime ideaÍs. (Lest one forget, an ideal J of a ring R is said to be semiprime if and only if J =

The implications of Theorem 9-7 is that one cannot construct finite fields with q elements unless q is a power of a prime. This immediately raises the question: Given q, a prescribed power of a prime, do there actually exist fields with q elements? In obtaining an answer, it is crucial to know the following. .

Theorem 9-6. A commutative ring R is regular if and only if every ideal of R is semiprime.

Lemma. Every element of a finite field F with pn elements isa root of the polynomialf(x) = xpn - x e F[x].

.JI.)

188

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

Proo! The n.onzer.o elements .of F f.orm a muItiplicative gr.oup .of .order p" .....: 1, s.o thát f.or each such a E F, we must have a P'- l = 1. Since

aP'

-

a ==

a(aP"-l -

Proo! Consider the splitting fieId F' of tbe p.oIynomial f(x) ,= xpn - x in Zp [x]. Since F' contains Zp' it has finite cbaracteristic p. Now, the derivative .off(x) is

1),

of(x) = p"xP''-¡

it f.olJ.ows tbat every eIement a E F (zer.o .or n.onzer.o) satisfies t):¡e equati.on aP" - a = O. '

.

189

-

1.

1 =

By virtue of ProbIem 15, Cbapter 7, this mean s that th~ poIynomiaI has n.o repeated r.o.ots in F ' . Let the subset F'!;;; F' c.onsist orihe p' distinct roots off(x)inF': - ,

CoroUary 1. Any finite field F with p. elements is the splitting field .oC , p = x ' - XE Zp[X].i, the'p.olyn.omiaIf(x) . ~

,~_Proo! We begin by recalIing that f(x) can have at m.ost aegf(x) = p~,:;;! .~:.:.distinct r.oóts in F. But p' r.o.ots are already kn.own, namely, the elements;¡\ }\;of F; whence, the p.olyn.omiaI f(x} splits c.ompIetely int.o linear fact.ors in,,':;} \:;'F[x]. Needless t.o say, it cann.ot split in any pr.oper subfield .of F, f.or n?".; ;li,,~"pr.oper subfieId c.ontairi~ p' eIements. Thus, we c.onclude that F lS th~ 'l. ;:;t'~plitting field .ofj(x) .over Zp and '. ',; ~I,·l\.;,',r 'Ü,,' ¡~:<"'~~~ p '!:5¡Z x ' - x == TI (x - r¡), \~;i~'

, F = {a

E

CIearly, the elements O and 1 He in F. Ir a, b E F, (a -

risF

aP"(bp"r = ab-

'.

{.S;:b,

From this, it is a short step to

Corollary2. Any two finite fields having the: same number of elements

Corollary. F.or any finite field Fand positive integer n, Jbere exists an extension fieId .of F of degree n.

are is.om.orphic. Before going on, let us digress to p.oint .out that this Iemma has an interesting appHcati.on in the the.ory .of numbers. We wish to esta.bli~h w?at, in the líterature, goes by the name of Wilson's Theorem; t.o Wlt, If P IS a prime number, then

Proo! Supp.ose that F has, q ,pm eIements, where p is the characteristic .of F and m [F: Zpl By tbe the.oremjust pr.oved, there exists a field F ' with pmn qn eIements, nameIy, the splitting fieId of xP"'" x over Zp. We contend that F' i8 actuaIly an extension of F. Indeed, every eIernent of F is a rO.ot of tbe p.olyn.omiaI' xP'" - x; this fact, t.ogether with tbe relati.on pmk = pmpm(k-.I), implies that for any a e F

(p - 1)! == -1 (modp).

Since this holds trivialIy f.or p = 2, .one may assume that p > 2 ;thus, p is an .odd prime. We talce f.or F the field Zp' Then, from ab.ove, every nonzero 1 e Z p [x] : ' eIement of Z p is a ro.ot of the p.olynomial xP- 1

Thus, each eIement .of F is also a r.o.ot .of the polynomial x p"'· t.o say that F !;;; F'. FinalIy, .observe that

1 == (x - 1)(x - 2) ... (x - (p - 1)) (m.od p). O (m.od p) in the just-written equation, it f.olI.ows that

mn

(-1)( -2) ." ('-(p - 1)) = (_1)P-l(P - 1)! (m.od p).

=

x, which is

[F':Zp] = [F':F][F:Z p] = [F':F]m,

whence [P: F] = n, completing the pr.o.or.

As p - 1 is even, this leads directly t.o Wilsón's Theorem. We now settle the questi.on of the existence of finite fields.

Theorem 9-8. F.or any prime number p and positive integer n, there exists a fieId (unique up t.o isomorphism) with exactly p" eIements.

1

(again using the fact char F' p), so that b.oth a ab- 1 E F. Consequently, the set F c.onstitutes a subfield .of F' and, lhus, a field with p" elements. Fr.om the corolIary of tbe Iast theorem, we ¡nfer that F, = F ' and F'.is the desired field. Uniqueness folIows from the result that any two frel7s having p" elements are isom.orphic (Cor.olIary 2).

'

Since any tw.o splitting fields of a given nonconstant polyn.omial over a given field are isomorphic (The.orem 7-30), Cor.olJary 1 leads to:

Putting x

'Si t;'

(ab-IV"

where the r¡ are the distinct elements .of F,

XP-l

wit~<Ú: 1= 0, then we bave

bV" = a P"

.'f

":"

"! .

F'la P' = a}.

I

l

I

We next examine the multiplicative structure of a finite freId. The finiteness assumpti.on Ieads to a particuIarly simple description : the nonzero elements <;.omprise a cyclic gr.oup under multiplication.

Theorllm 9-9. The muItiplicative gr.oup of a finite field is cycIic.

188

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

Proo! The n.onzer.o elements .of F f.orm a muItiplicative gr.oup .of .order p" .....: 1, s.o thát f.or each such a E F, we must have a P'- l = 1. Since

aP'

-

a ==

a(aP"-l -

Proo! Consider the splitting fieId F' of tbe p.oIynomial f(x) ,= xpn - x in Zp [x]. Since F' contains Zp' it has finite cbaracteristic p. Now, the derivative .off(x) is

1),

of(x) = p"xP''-¡

it f.olJ.ows tbat every eIement a E F (zer.o .or n.onzer.o) satisfies t):¡e equati.on aP" - a = O. '

.

189

-

1.

1 =

By virtue of ProbIem 15, Cbapter 7, this mean s that th~ poIynomiaI has n.o repeated r.o.ots in F ' . Let the subset F'!;;; F' c.onsist orihe p' distinct roots off(x)inF': - ,

CoroUary 1. Any finite field F with p. elements is the splitting field .oC , p = x ' - XE Zp[X].i, the'p.olyn.omiaIf(x) . ~

,~_Proo! We begin by recalIing that f(x) can have at m.ost aegf(x) = p~,:;;! .~:.:.distinct r.oóts in F. But p' r.o.ots are already kn.own, namely, the elements;¡\ }\;of F; whence, the p.olyn.omiaI f(x} splits c.ompIetely int.o linear fact.ors in,,':;} \:;'F[x]. Needless t.o say, it cann.ot split in any pr.oper subfield .of F, f.or n?".; ;li,,~"pr.oper subfieId c.ontairi~ p' eIements. Thus, we c.onclude that F lS th~ 'l. ;:;t'~plitting field .ofj(x) .over Zp and '. ',; ~I,·l\.;,',r 'Ü,,' ¡~:<"'~~~ p '!:5¡Z x ' - x == TI (x - r¡), \~;i~'

, F = {a

E

CIearly, the elements O and 1 He in F. Ir a, b E F, (a -

risF

aP"(bp"r = ab-

'.

{.S;:b,

From this, it is a short step to

Corollary2. Any two finite fields having the: same number of elements

Corollary. F.or any finite field Fand positive integer n, Jbere exists an extension fieId .of F of degree n.

are is.om.orphic. Before going on, let us digress to p.oint .out that this Iemma has an interesting appHcati.on in the the.ory .of numbers. We wish to esta.bli~h w?at, in the líterature, goes by the name of Wilson's Theorem; t.o Wlt, If P IS a prime number, then

Proo! Supp.ose that F has, q ,pm eIements, where p is the characteristic .of F and m [F: Zpl By tbe the.oremjust pr.oved, there exists a field F ' with pmn qn eIements, nameIy, the splitting fieId of xP"'" x over Zp. We contend that F' i8 actuaIly an extension of F. Indeed, every eIernent of F is a rO.ot of tbe p.olyn.omiaI' xP'" - x; this fact, t.ogether with tbe relati.on pmk = pmpm(k-.I), implies that for any a e F

(p - 1)! == -1 (modp).

Since this holds trivialIy f.or p = 2, .one may assume that p > 2 ;thus, p is an .odd prime. We talce f.or F the field Zp' Then, from ab.ove, every nonzero 1 e Z p [x] : ' eIement of Z p is a ro.ot of the p.olynomial xP- 1

Thus, each eIement .of F is also a r.o.ot .of the polynomial x p"'· t.o say that F !;;; F'. FinalIy, .observe that

1 == (x - 1)(x - 2) ... (x - (p - 1)) (m.od p). O (m.od p) in the just-written equation, it f.olI.ows that

mn

(-1)( -2) ." ('-(p - 1)) = (_1)P-l(P - 1)! (m.od p).

=

x, which is

[F':Zp] = [F':F][F:Z p] = [F':F]m,

whence [P: F] = n, completing the pr.o.or.

As p - 1 is even, this leads directly t.o Wilsón's Theorem. We now settle the questi.on of the existence of finite fields.

Theorem 9-8. F.or any prime number p and positive integer n, there exists a fieId (unique up t.o isomorphism) with exactly p" eIements.

1

(again using the fact char F' p), so that b.oth a ab- 1 E F. Consequently, the set F c.onstitutes a subfield .of F' and, lhus, a field with p" elements. Fr.om the corolIary of tbe Iast theorem, we ¡nfer that F, = F ' and F'.is the desired field. Uniqueness folIows from the result that any two frel7s having p" elements are isom.orphic (Cor.olIary 2).

'

Since any tw.o splitting fields of a given nonconstant polyn.omial over a given field are isomorphic (The.orem 7-30), Cor.olJary 1 leads to:

Putting x

'Si t;'

(ab-IV"

where the r¡ are the distinct elements .of F,

XP-l

wit~<Ú: 1= 0, then we bave

bV" = a P"

.'f

":"

"! .

F'la P' = a}.

I

l

I

We next examine the multiplicative structure of a finite freId. The finiteness assumpti.on Ieads to a particuIarly simple description : the nonzero elements <;.omprise a cyclic gr.oup under multiplication.

Theorllm 9-9. The muItiplicative gr.oup of a finite field is cycIic.

190

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS ANO IDEALS

Proof. Let E be a finite field with p" elements and E* be its multiplicative group of nonzero elements; this group has order p" - 1. The argument about to be presented hinges on finding an element in E* of order h = p" - 1. To this end, we first consider the prime factorization of h: h =

qí'q~'

.,. if",m,

where th~ q¡ are distinct primes and r¡ E Z+. For i = 1,2, ... , m, set h¡ = h/q¡. Now, there exists a nonzero element a¡ E E which is not a root of the polynomial X" - 1 E E[x] ; for this polynomial has at most h¡ distinct roots in E and h¡ < h, the number of nonzero elements of E. Next, take

b¡ -and define b = b1 b2

a"I., ¡,

wh ere

.. ·b m •

-- q¡r, (¡. := .'" 1 2' .... m)

S¡

We certainly have

bf'

=

a7

= 1,

so that the order b¡ must divide S¡ = qf'. On the other hand, if then contrary to our original choice of the element a¡. The implication is that has order q~'. To settle the whole affair, we will show that the element b is of order h. In the contrary case, the order of b must be a proper divisor of h (since b" = 1, the order of b certainly divide.~. h) and therefore divides at least one ofthe integers h¡ (i = 1,2, '" , m), s~y:ihl' We then have b¡

1

=

bh1

=;,:.bi'b~' "

~

... b~l.

,

If 2 ~ i ~ m, then qi'lh 1 , which im~~í.~s that b~' = 1 and so b~' = 1. This means that q~' (the order of b1 ) divid1eis h1 , which is impossible. Thus, the element b has order h and, in corisequence, the cyclic subgroup of E* generated by b will also be of orderJ; since E* contains only p" - 1 = h elements, this cyclic group must be al! of E*. It is not surprising and is quite easy to prove:

Corollary. Any finite field E with p" elements is a simple algebraic extension of the field Zp.

Proof. We already know that E is an algebraic extension of degree n of its prime subfield Zp. The theorem aboye indicates that the p" elements of E can be written as 0, 1, b, b2 , ••• ,bP" - 2 for sorne b E E* ; in other words, the field E = Zp(b).

191

As an application of these ideas, let us prove a statement' made earlier to the effect that, for any finite field E, the polynomial domain E[x] contains irreducible polynomials of arbitrary ordjr. Theorem 9-10. Let E be afinite field. For each positive integer n, there exists an irreducible polynomialf(x) E E[x] with degf(x) = n.

Proof. Suppose that E' is an extension of E with [E': E] = n. As was just seen, there exists an element b in E' such that E' = E(b). If f(x) is the minimum polynomial of b oyerE, then (invoking Corollary 2 of Theorem 7-25) degf(x)

= [E':

E]

=

n.

Therefore, f(x) E E[x] is the required irreducible polynomial of degree n and the theorem follows. Finite fields are called Galois fields after the French mathematician Evariste GalQis, who first discovered the existence offinite fields aside from those of the form Zp. The (essentially unique) field with p" elements is commonly denoted by the symbol GF(p"). To construct GF(p"), we need only determine an irreducible polynomial f(x) of degree n in Zp[x]; then Zp [x]/(J(x)) is the required Galois field with p" elements. It is now time to redeem a promise made earlier to provide a proof that every finite division ring is a field (Wedderburn's Theorem). Our approach is founded on a treatment by Herstein [43]. Althóugh tbis is perhaps the most elementary, other proofs of Wedderburn's Theorem are common; an 'el;t~jrely different one requiring the concept of cyclotonic polynomials appears in [5]. . The argumeptwhich we are about to give is lengthy and will be prefaced by two simplify!ng lemmas (the student who is pressed for time may wish to omit all tl¡.is·on a first reading). Much of our success, both with Wedderburn's:Theorem and its applications, inevitably fiows from the result below. :' Lemma 1. Let R be a division ring of characteristic p > 0, p a prime. pm Suppose that the element a E R, a ~ cent R, is such that a = a for some m >. Q. Then there exists an x E R for which 1) xax- 1 =1= a, 2) xax- 1 E Zp(a), the extension field obtained by adjoining a to Zp. Proof. Let Zp be the prime subfield of R. Since apm - a = 0, a is algebraic over Zp. By Theorem 7-25, we know that the extension Zp(a) is a finite field and therefore must have p" elements for some.n E Z+. Furthermore, each rE Zp(a) satisfies r P" = r.

190

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS ANO IDEALS

Proof. Let E be a finite field with p" elements and E* be its multiplicative group of nonzero elements; this group has order p" - 1. The argument about to be presented hinges on finding an element in E* of order h = p" - 1. To this end, we first consider the prime factorization of h: h =

qí'q~'

.,. if",m,

where th~ q¡ are distinct primes and r¡ E Z+. For i = 1,2, ... , m, set h¡ = h/q¡. Now, there exists a nonzero element a¡ E E which is not a root of the polynomial X" - 1 E E[x] ; for this polynomial has at most h¡ distinct roots in E and h¡ < h, the number of nonzero elements of E. Next, take

b¡ -and define b = b1 b2

a"I., ¡,

wh ere

.. ·b m •

-- q¡r, (¡. := .'" 1 2' .... m)

S¡

We certainly have

bf'

=

a7

= 1,

so that the order b¡ must divide S¡ = qf'. On the other hand, if then contrary to our original choice of the element a¡. The implication is that has order q~'. To settle the whole affair, we will show that the element b is of order h. In the contrary case, the order of b must be a proper divisor of h (since b" = 1, the order of b certainly divide.~. h) and therefore divides at least one ofthe integers h¡ (i = 1,2, '" , m), s~y:ihl' We then have b¡

1

=

bh1

=;,:.bi'b~' "

~

... b~l.

,

If 2 ~ i ~ m, then qi'lh 1 , which im~~í.~s that b~' = 1 and so b~' = 1. This means that q~' (the order of b1 ) divid1eis h1 , which is impossible. Thus, the element b has order h and, in corisequence, the cyclic subgroup of E* generated by b will also be of orderJ; since E* contains only p" - 1 = h elements, this cyclic group must be al! of E*. It is not surprising and is quite easy to prove:

Corollary. Any finite field E with p" elements is a simple algebraic extension of the field Zp.

Proof. We already know that E is an algebraic extension of degree n of its prime subfield Zp. The theorem aboye indicates that the p" elements of E can be written as 0, 1, b, b2 , ••• ,bP" - 2 for sorne b E E* ; in other words, the field E = Zp(b).

191

As an application of these ideas, let us prove a statement' made earlier to the effect that, for any finite field E, the polynomial domain E[x] contains irreducible polynomials of arbitrary ordjr. Theorem 9-10. Let E be afinite field. For each positive integer n, there exists an irreducible polynomialf(x) E E[x] with degf(x) = n.

Proof. Suppose that E' is an extension of E with [E': E] = n. As was just seen, there exists an element b in E' such that E' = E(b). If f(x) is the minimum polynomial of b oyerE, then (invoking Corollary 2 of Theorem 7-25) degf(x)

= [E':

E]

=

n.

Therefore, f(x) E E[x] is the required irreducible polynomial of degree n and the theorem follows. Finite fields are called Galois fields after the French mathematician Evariste GalQis, who first discovered the existence offinite fields aside from those of the form Zp. The (essentially unique) field with p" elements is commonly denoted by the symbol GF(p"). To construct GF(p"), we need only determine an irreducible polynomial f(x) of degree n in Zp[x]; then Zp [x]/(J(x)) is the required Galois field with p" elements. It is now time to redeem a promise made earlier to provide a proof that every finite division ring is a field (Wedderburn's Theorem). Our approach is founded on a treatment by Herstein [43]. Althóugh tbis is perhaps the most elementary, other proofs of Wedderburn's Theorem are common; an 'el;t~jrely different one requiring the concept of cyclotonic polynomials appears in [5]. . The argumeptwhich we are about to give is lengthy and will be prefaced by two simplify!ng lemmas (the student who is pressed for time may wish to omit all tl¡.is·on a first reading). Much of our success, both with Wedderburn's:Theorem and its applications, inevitably fiows from the result below. :' Lemma 1. Let R be a division ring of characteristic p > 0, p a prime. pm Suppose that the element a E R, a ~ cent R, is such that a = a for some m >. Q. Then there exists an x E R for which 1) xax- 1 =1= a, 2) xax- 1 E Zp(a), the extension field obtained by adjoining a to Zp. Proof. Let Zp be the prime subfield of R. Since apm - a = 0, a is algebraic over Zp. By Theorem 7-25, we know that the extension Zp(a) is a finite field and therefore must have p" elements for some.n E Z+. Furthermore, each rE Zp(a) satisfies r P" = r.

q

I I

i

192

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

Now, define the function f: R ~ R by setting f(x) = xa - ax for all x in R. Using induction, it is not difficult to show that the composite k

I

fk(x) =

(-lima i xa k- i

(k

.~

CorolJary. In the lemma, xax - 1

i=O

To cope with the' problem at hand, we shall also need the following: Lemma 2. If P:'i~ finite field and O 9= a E F, then there exist elements a, b E F such tpi..at a = a2 + b2 •

a

yp" - y

I1

=

'

(y - r),

reZp(a)

or, what amou~ts to the same thing, yp n

_

y = y

I1

This formal identity requires only that y cornmute with all elements rE Zp(a). Taking stock of the fact thatf o T,. = T,. o J, as well as the relation fP" = J, we thereby obtain o = fP - f = foIl (f - 'T,.). O'i"reZp(a)

epi

(In essence, one applies the substitution homomorphism to the ring of polynomials whose coefficients are homomorphisms on (R, +).) If, for every r 9= O in Zp(a), it happens that (f - T,.)(x) = O implies x = O thetl the last-written equation would necessarily lead to f = O. This ~ould mean that xa - ax = O for all x E R, forcing a to He in the center of R, contrary to hypothesis. Consequently, there must exist sorne O 9= rE Zp(a) and sorne element x 9= O in R for which (f - T,.)(x) = O; that is to say, xa - ax = rx and so xax- 1 = r + a EZp(a).

9= O, certainly the product xax- 1 9= a.

{

~'

The iemma is thereby established on taking a = a2"-1 and b = O. Now, if the characteristic of F is an odd prime p, then F will contain pn elements. Let f be the mapping of F* into itself defined by f(x) = X2 (as usual, F* denotes the multiplicative group of F). Then f is a group homomorphism, with

kerf= {x

(y - r).

o 'i"reZp(a)

Since r

~

Proo! We first di#;énse with the case where char F = 2. In this special situation, F. has 2~Jiements and any element of F satisfies the equation , x 2 " = X. Thus, every nonzero member a of F is a square and, in particular, a = a 2" = (a 2"-1)2.

r(xa ...,:.' ax) = (T,. o f)(x). Z P [y]

sorne integer k E Z +.

k r.:, . = a for sorne k, with 2 :::;; k :::;; s,- 1. '

r.

E

9= a for

In consequence, xai- 1

But aP" = a, whence fP"(x) = xa - ax= f(x) for all x E R, which is equivalent to asserting thatf P" = f. :'/ For each element rE Zp(a), consider':the functionT,. on R defined by T,.(x) = rx. Our cont~ntion is that commutes with all such T,.. The reasoning proceeds asJollows: Being afi,eld, Zp(a) is commutative, so that, ifxER, ' ,', (f o T,.)(x) = f(rx) ~ (rx)~:,t a(rx) = rxa - rax

This in short, means thatf o T,. = T,. o ffor every r in Zp(a). , From the corollary on page 188, the polynomial yp" - y factors completely in Zp(a); in other words, we ha ve

ak

Proo! Since aP"-l = 1, the element a has finite order as a member of the multiplicative group R*. Let sbe the/order of a. Then, in the field Zp(a), each of the s elements 1, a; a2, .:. , a -1 is a root of the polynomial yS y E Zp[xJ. This polynomial can possess at most s I"oots in Zp(a) and 1, a, ... , as - 1 are all'distinct. But xax- 1 E Zia) and c1early : (xax- 1)S = .xas x- 1 = xx- 1 = 1.

1).

When k = p, the foregoing equation reduces simply to f P(x) == xa P - aPx, because plm for O < i < p (recall also that char R = p). Another routine induction argument extends this to " ' fP"(x) = xa P", '-:- aP"x.

=

=

193

E

F*lx2 = 1} = {1, -1}.

Since char F 9= 2, 1 and -1 are necessarily distinct., This implies that, for eachp Ef(F*), thereexistexactlytwoelementsa1,a 2 in F* withai = a~ = P; in fact, a 2 = -al' To put it another way,for each pair ofelements al and -':a¡'in F*, we get one element which is a square. Hence, halfthe elements of F* will be squares, call these P1' P2' ... ,Pk' where the integer k = (p" - 1)/2. Given O 9= a E F, assume that a is not a square and consider the set s = {a - p;ji = 1,2, ... , k}. If it turns out that a - Pi is not a square for any value of i, the set S (which contains k distinct elements) mustcoincide with the k nonsquares of F*. But then a willlie in S, yielding a = a - Pi for sorne choice of i; whence Pi = O, an obvious conttadiction. This being the case, we conc1ude that a - Pi = Pj for suitable integers i and j, or a = Pi + pj ' Thus, a is the sum oftwo squares in R and the requisite equation holds. Corollary. If F is a finite field and O 9= a E F, then there exist elements a, b in F such that 1 + a2 - ab 2 = O.

q

I I

i

192

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

Now, define the function f: R ~ R by setting f(x) = xa - ax for all x in R. Using induction, it is not difficult to show that the composite k

I

fk(x) =

(-lima i xa k- i

(k

.~

CorolJary. In the lemma, xax - 1

i=O

To cope with the' problem at hand, we shall also need the following: Lemma 2. If P:'i~ finite field and O 9= a E F, then there exist elements a, b E F such tpi..at a = a2 + b2 •

a

yp" - y

I1

=

'

(y - r),

reZp(a)

or, what amou~ts to the same thing, yp n

_

y = y

I1

This formal identity requires only that y cornmute with all elements rE Zp(a). Taking stock of the fact thatf o T,. = T,. o J, as well as the relation fP" = J, we thereby obtain o = fP - f = foIl (f - 'T,.). O'i"reZp(a)

epi

(In essence, one applies the substitution homomorphism to the ring of polynomials whose coefficients are homomorphisms on (R, +).) If, for every r 9= O in Zp(a), it happens that (f - T,.)(x) = O implies x = O thetl the last-written equation would necessarily lead to f = O. This ~ould mean that xa - ax = O for all x E R, forcing a to He in the center of R, contrary to hypothesis. Consequently, there must exist sorne O 9= rE Zp(a) and sorne element x 9= O in R for which (f - T,.)(x) = O; that is to say, xa - ax = rx and so xax- 1 = r + a EZp(a).

9= O, certainly the product xax- 1 9= a.

{

~'

The iemma is thereby established on taking a = a2"-1 and b = O. Now, if the characteristic of F is an odd prime p, then F will contain pn elements. Let f be the mapping of F* into itself defined by f(x) = X2 (as usual, F* denotes the multiplicative group of F). Then f is a group homomorphism, with

kerf= {x

(y - r).

o 'i"reZp(a)

Since r

~

Proo! We first di#;énse with the case where char F = 2. In this special situation, F. has 2~Jiements and any element of F satisfies the equation , x 2 " = X. Thus, every nonzero member a of F is a square and, in particular, a = a 2" = (a 2"-1)2.

r(xa ...,:.' ax) = (T,. o f)(x). Z P [y]

sorne integer k E Z +.

k r.:, . = a for sorne k, with 2 :::;; k :::;; s,- 1. '

r.

E

9= a for

In consequence, xai- 1

But aP" = a, whence fP"(x) = xa - ax= f(x) for all x E R, which is equivalent to asserting thatf P" = f. :'/ For each element rE Zp(a), consider':the functionT,. on R defined by T,.(x) = rx. Our cont~ntion is that commutes with all such T,.. The reasoning proceeds asJollows: Being afi,eld, Zp(a) is commutative, so that, ifxER, ' ,', (f o T,.)(x) = f(rx) ~ (rx)~:,t a(rx) = rxa - rax

This in short, means thatf o T,. = T,. o ffor every r in Zp(a). , From the corollary on page 188, the polynomial yp" - y factors completely in Zp(a); in other words, we ha ve

ak

Proo! Since aP"-l = 1, the element a has finite order as a member of the multiplicative group R*. Let sbe the/order of a. Then, in the field Zp(a), each of the s elements 1, a; a2, .:. , a -1 is a root of the polynomial yS y E Zp[xJ. This polynomial can possess at most s I"oots in Zp(a) and 1, a, ... , as - 1 are all'distinct. But xax- 1 E Zia) and c1early : (xax- 1)S = .xas x- 1 = xx- 1 = 1.

1).

When k = p, the foregoing equation reduces simply to f P(x) == xa P - aPx, because plm for O < i < p (recall also that char R = p). Another routine induction argument extends this to " ' fP"(x) = xa P", '-:- aP"x.

=

=

193

E

F*lx2 = 1} = {1, -1}.

Since char F 9= 2, 1 and -1 are necessarily distinct., This implies that, for eachp Ef(F*), thereexistexactlytwoelementsa1,a 2 in F* withai = a~ = P; in fact, a 2 = -al' To put it another way,for each pair ofelements al and -':a¡'in F*, we get one element which is a square. Hence, halfthe elements of F* will be squares, call these P1' P2' ... ,Pk' where the integer k = (p" - 1)/2. Given O 9= a E F, assume that a is not a square and consider the set s = {a - p;ji = 1,2, ... , k}. If it turns out that a - Pi is not a square for any value of i, the set S (which contains k distinct elements) mustcoincide with the k nonsquares of F*. But then a willlie in S, yielding a = a - Pi for sorne choice of i; whence Pi = O, an obvious conttadiction. This being the case, we conc1ude that a - Pi = Pj for suitable integers i and j, or a = Pi + pj ' Thus, a is the sum oftwo squares in R and the requisite equation holds. Corollary. If F is a finite field and O 9= a E F, then there exist elements a, b in F such that 1 + a2 - ab 2 = O.

194

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

195

After this preparation, we now undertake the task of proving the theorem which serves as the focal point ofthe present chapter.

whence r" = 1. Because n is a prime; the order of r (as a member of R*) must be n. Finally, it is worth noticing that

Theorem 9-11. (Wedderburn). Every finite division ring is a field.

b" ".; rObo = (rb)" = (a- 1 ba)" = a- 1 b"a,

Proa! Suppose, for purposes of contradiction, that the theorem is not true for all finite division rings. Let R have minimal order among the set of noncomrp.utative division rings, so that any division ring with fewer elements tban R will be commutative. Before becoming involved in the technical argument of the proof, let us note that if the elements a, b E R satisfy abk = bka, but ab =1= ba, then bk E cent R. For, consider the centralizer of bk in R: CW) = {x E Rlxb k = bkx}.

from which we derive abo = bOa. Again in the light of our opening remarks, since a commutes with b" but not with b, necessarily b" E cent R. We now assert that whenever an element y of R satisfies y" = 1, then it must be of the form y = rí , where O ~ i ~ n - 1. Indeed, the extension field (cent R)(y) (although awkward, the notation conveys the point) contains at most n roots of the polynomial z" - 1. But, since r is of prime order n, the elements 1, r, r2 , ••• , r"-l comprise n distinct roots of z" - 1 in tbis field. These remarks should make it plain that y = rí for some i. In passing, we might also observe that, because y E cent R, (cent R)(y) = centRo With reference to Theorem 9-9, inasmuch as cent R constitutes a finite field, its multiplicative group of nonzero elements must be cyc1ic; say with generator S. Accordingly, a" = si, b" = Si for suitable integers j and l. Furthermore,. n divides neither j nor l. To see this, suppose that j = nk; then, a" = si = s"k, whence a"(s-~" = 1. As the element s lies in cent R, we would have (as-k)" = 1. But the preceding paragraph then yields as- k = rí for some integer i, or a = rísk E cent R, which is impossible. In a similar fashion, one is able to estabHsh tbat n does not divide l. We now set e = d,.d = bi . Then,

It follows without difficulty that CW) comprises a division rlng (a division subring of R). If C(b~ =F R, then by our hypothesis C(b k) would necessarily be commutative. But a, b both lie in C(b~ and these elements c1early do not commute. This entails that C(b~ = R, which is scarcely more tban a restatement that bk E cent R. Now to the proof proper. Since the multiplicative group R* is finite, every nonzero element of R must have finite order and, as a result, some power of it belongs to the center of R. By virtue of this circumstance, the set S

= {m E Z+ Ifor some e fj cent R,

cm

E

cent R}

is not empty. Pick the integer n to be minimal in S. Then there exists an element a fj cent R such tbat a" E cent R. We assert that n is a prime number. Indeed, weren = n1 n2 , with1 < n 1, n2 < n,itwould followthata"1 fj ceut R, yet (a"1)"' = "a" E cent R. In other words, the integer n2 is a membe¡¡,o.f,S, a contradiction to the minimal nature of n . ¡ N~xt apply Lemma 1 to obtain an element x E R and an integer k súch tbat xax- 1 = ak =1= a. At the outset, observe that

This relation, in conjunction with ba = rab, leads to ed = tde, where the element t = r- ji E cent R. A fact wbich will not d~tain us long is that t =1= 1. In the contrary case, r- jl = 1, wbich implies thkt.h~/; since n is a prime number, either n\j or nl/, resulting in a contradictjon. One can deduce a .Httle more, namely, that t"

so, by induction, x"-l ax -(n-1) = al
rOan = (ra)" = (bab- 1 )" = ba"b- 1 = a",

=

(r- j/)"

=

(r")-jl

=

e

Pausing for a moment to tidy up, let us point-out tbat the reasoning so far has succeeded in producing two elements e, dE R with the following properties : 1) e" = d" = ex E cent R, 2) ed = tde, with t E cent R, 3) t =1= 1, but t" = 1. From these relations, we may easily compute (e- 1 d)". In this connection, notice that

194

TWO CLASSIC THEOREMS

FIRST COURSE IN RINGS AND IDEALS

195

After this preparation, we now undertake the task of proving the theorem which serves as the focal point ofthe present chapter.

whence r" = 1. Because n is a prime; the order of r (as a member of R*) must be n. Finally, it is worth noticing that

Theorem 9-11. (Wedderburn). Every finite division ring is a field.

b" ".; rObo = (rb)" = (a- 1 ba)" = a- 1 b"a,

Proa! Suppose, for purposes of contradiction, that the theorem is not true for all finite division rings. Let R have minimal order among the set of noncomrp.utative division rings, so that any division ring with fewer elements tban R will be commutative. Before becoming involved in the technical argument of the proof, let us note that if the elements a, b E R satisfy abk = bka, but ab =1= ba, then bk E cent R. For, consider the centralizer of bk in R: CW) = {x E Rlxb k = bkx}.

from which we derive abo = bOa. Again in the light of our opening remarks, since a commutes with b" but not with b, necessarily b" E cent R. We now assert that whenever an element y of R satisfies y" = 1, then it must be of the form y = rí , where O ~ i ~ n - 1. Indeed, the extension field (cent R)(y) (although awkward, the notation conveys the point) contains at most n roots of the polynomial z" - 1. But, since r is of prime order n, the elements 1, r, r2 , ••• , r"-l comprise n distinct roots of z" - 1 in tbis field. These remarks should make it plain that y = rí for some i. In passing, we might also observe that, because y E cent R, (cent R)(y) = centRo With reference to Theorem 9-9, inasmuch as cent R constitutes a finite field, its multiplicative group of nonzero elements must be cyc1ic; say with generator S. Accordingly, a" = si, b" = Si for suitable integers j and l. Furthermore,. n divides neither j nor l. To see this, suppose that j = nk; then, a" = si = s"k, whence a"(s-~" = 1. As the element s lies in cent R, we would have (as-k)" = 1. But the preceding paragraph then yields as- k = rí for some integer i, or a = rísk E cent R, which is impossible. In a similar fashion, one is able to estabHsh tbat n does not divide l. We now set e = d,.d = bi . Then,

It follows without difficulty that CW) comprises a division rlng (a division subring of R). If C(b~ =F R, then by our hypothesis C(b k) would necessarily be commutative. But a, b both lie in C(b~ and these elements c1early do not commute. This entails that C(b~ = R, which is scarcely more tban a restatement that bk E cent R. Now to the proof proper. Since the multiplicative group R* is finite, every nonzero element of R must have finite order and, as a result, some power of it belongs to the center of R. By virtue of this circumstance, the set S

= {m E Z+ Ifor some e fj cent R,

cm

E

cent R}

is not empty. Pick the integer n to be minimal in S. Then there exists an element a fj cent R such tbat a" E cent R. We assert that n is a prime number. Indeed, weren = n1 n2 , with1 < n 1, n2 < n,itwould followthata"1 fj ceut R, yet (a"1)"' = "a" E cent R. In other words, the integer n2 is a membe¡¡,o.f,S, a contradiction to the minimal nature of n . ¡ N~xt apply Lemma 1 to obtain an element x E R and an integer k súch tbat xax- 1 = ak =1= a. At the outset, observe that

This relation, in conjunction with ba = rab, leads to ed = tde, where the element t = r- ji E cent R. A fact wbich will not d~tain us long is that t =1= 1. In the contrary case, r- jl = 1, wbich implies thkt.h~/; since n is a prime number, either n\j or nl/, resulting in a contradictjon. One can deduce a .Httle more, namely, that t"

so, by induction, x"-l ax -(n-1) = al
rOan = (ra)" = (bab- 1 )" = ba"b- 1 = a",

=

(r- j/)"

=

(r")-jl

=

e

Pausing for a moment to tidy up, let us point-out tbat the reasoning so far has succeeded in producing two elements e, dE R with the following properties : 1) e" = d" = ex E cent R, 2) ed = tde, with t E cent R, 3) t =1= 1, but t" = 1. From these relations, we may easily compute (e- 1 d)". In this connection, notice that

196

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

while a similar approach leads to (e= A straightforward induction argument, which we bequeath to the reader, extends this to (e- 1d)n = t1+2+"'+(n-1)e-ndn = t1t2+"'+(n-l) = ~(n-1){2. 1d)3

t 1 + 2 e- 3 d3 •

The final stage ofthe proof achieves the long-sought contradiction. We consider two cases in tum: n is an 0dd prime and n = 2. Ifn>. 2, then (n - 1)/2 is an integer and so ,. t"(n-I){2= w)(n-I)/2 = 1, "

"'....:

wbich i}:Íiplies that (e-l~ = 1. Being a solution of the equation y" = 1, it follow§1 from earlier reasoning that, e-Id = r! E cent R for sorne choice ofi. B~t then r le = (e-Id) -1 E cent R and so (using (2) abo ve ), !

.,.,

~-Ite

,.:,.]

= (dc- 1d- 1 )e =

d(d- 1 e}e- 1

an ob~ii~,s contradictiolI. Thus, the theorem is proved, at least when n is an odd.ipilme. Turníj:lg to the more troublesome possibility, we now suppose that n = 2. In tbis event, t 2 = 1 and, of course, t =1= 1, whence t = -l. Then, ed = -de =1= de; consequently, the characteristic of R is differeÍlt from 2. Applying Lemma 2 to the field cent R, we can find elements Xi (i = 1,2) in cent R satisfying

Armed with this, a direct computation shows that (i::

+ dX 1 +

cdx::Y

e2(1

+ xi -

which, because R is a division ring, leads to e clinch matters, since char R =1= 2,

O =1= 2e 2 = e(e

+ dx 1

O,

!Xxi)

+

dX 1

+

edx2 =

+edx:z} + (e + dX1 + edx )e 2

O. To

O,

an absurdity. This contradiction finally completes the proof ofWedderburn's Theorem. ' We next proceed to take up a c1ass of rings introduced by Jacobson. Definition 9-1. A ring R with identity"is called a J-ring if, for each x E R, there exists an integer n(x) > 1 (depending on x) such that x"(x) = x.

Our immediate goal is to prove that every J-ríng is commutatíve. (In a very natural way, this can be regarded as a generalization ofWedderburn's Theorem). Before establishing the quoted result in fuIl generality, we first settle the question for the special case of division rings; the argument relies heavily on the Wedderburn Theorem.

197

Theorem 9-12. Let R be a J-ring. If R forms a division ring, then R is commutative (hence, a field) .. ,

Pr.oof. As a first step, let us show tbat R is of characteristic p > O, p'a pnme. If char R = 2, then thére is nothing to prove; thus, it rnay be assurned ~hat. char R =1= 2. Consider any elernent a =1= Oin R. By hypothesis, Setting there eXIst mtegers h, k > 1 for which ah = a, (2a)k = 2a. q = (h - l)(k - 1) + 1 > 1, it follows that both aq =' a and (2a)q = 2a. From tbis, we obtain (2q ...:.. 2)a' = O, with 2q - 2 =1= O. Therefore, there' exists a least positive integer p such that pa = O, which implies tbat char R = p, p a prime [Theorem 1 - 6 ) . " Let Zp be the prime sub:fleld of R. Since ah ::, a, the ~í6~ent a is algeb~aic over Zp and, hence, the extension Zp(a) constitutes aflnite field; say wlth p" elements. In particular, a itself líes in Zp(a), so Ü:í'át a pn = a. Ir we now assume that a ~ cent R, then all fue hypothesis of tém:ma 1 will be satisfied ; thus, there exists an element b E R and integer k ~':t:~atisfying bab- 1 ak =F a. SiÍllilar reasoning applied to the extensioií.'~field Z (b) indicates that bpm = b for sorne integer m > 1. :' p At this point we tum our attention to the set of finÍte sums

It should be apparent that W is a finite set which is closed under addition. Since the relation akb = ba allows us to bring the a's and b's together in a product, W is also c10sed under multiplication. Whatever further it may be, W has at least been shown to be a ring.· As a Bnite ~ubring of a division rirtg, W is more than just a ring; it lS, in fact, a Bnite division rirtg (Problem 32). Hence, by Wedderburn's Theorem, we know that W lS necessarily commutative. In particular, a and b are both members of W; so that ab = ba, contradicting the relation bab -1 =' ak '=1= a. Having arrived at a suitable contradiction, we infer that the choice of a ~ cent R is impossible and R must be cornmutative. The transition of Theorem 9-12 from the division rings case arbitrary rings is accomplished by two lemillas. Lemma 1. Let R be a J-ring. Then every right ideal 1 of R is a twosided ideal of R.

Proo! To begin with, we assert that R can possess no, nonzero nilpotent elements. Indeed, if x =1= O, the condition xn(x) = x necessarily implies that xm =1= O for all m ~ 1. Now, suppose that e is any idempotent elemeI1t of R; then, for any x E R, , , . (xe - exe)2

=:

(ex

exe)2

=

O,

196

FIRST COURSE IN RINGS AND IDEALS

TWO CLASSIC THEOREMS

while a similar approach leads to (e= A straightforward induction argument, which we bequeath to the reader, extends this to (e- 1d)n = t1+2+"'+(n-1)e-ndn = t1t2+"'+(n-l) = ~(n-1){2. 1d)3

t 1 + 2 e- 3 d3 •

The final stage ofthe proof achieves the long-sought contradiction. We consider two cases in tum: n is an 0dd prime and n = 2. Ifn>. 2, then (n - 1)/2 is an integer and so ,. t"(n-I){2= w)(n-I)/2 = 1, "

"'....:

wbich i}:Íiplies that (e-l~ = 1. Being a solution of the equation y" = 1, it follow§1 from earlier reasoning that, e-Id = r! E cent R for sorne choice ofi. B~t then r le = (e-Id) -1 E cent R and so (using (2) abo ve ), !

.,.,

~-Ite

,.:,.]

= (dc- 1d- 1 )e =

d(d- 1 e}e- 1

an ob~ii~,s contradictiolI. Thus, the theorem is proved, at least when n is an odd.ipilme. Turníj:lg to the more troublesome possibility, we now suppose that n = 2. In tbis event, t 2 = 1 and, of course, t =1= 1, whence t = -l. Then, ed = -de =1= de; consequently, the characteristic of R is differeÍlt from 2. Applying Lemma 2 to the field cent R, we can find elements Xi (i = 1,2) in cent R satisfying

Armed with this, a direct computation shows that (i::

+ dX 1 +

cdx::Y

e2(1

+ xi -

which, because R is a division ring, leads to e clinch matters, since char R =1= 2,

O =1= 2e 2 = e(e

+ dx 1

O,

!Xxi)

+

dX 1

+

edx2 =

+edx:z} + (e + dX1 + edx )e 2

O. To

O,

an absurdity. This contradiction finally completes the proof ofWedderburn's Theorem. ' We next proceed to take up a c1ass of rings introduced by Jacobson. Definition 9-1. A ring R with identity"is called a J-ring if, for each x E R, there exists an integer n(x) > 1 (depending on x) such that x"(x) = x.

Our immediate goal is to prove that every J-ríng is commutatíve. (In a very natural way, this can be regarded as a generalization ofWedderburn's Theorem). Before establishing the quoted result in fuIl generality, we first settle the question for the special case of division rings; the argument relies heavily on the Wedderburn Theorem.

197

Theorem 9-12. Let R be a J-ring. If R forms a division ring, then R is commutative (hence, a field) .. ,

Pr.oof. As a first step, let us show tbat R is of characteristic p > O, p'a pnme. If char R = 2, then thére is nothing to prove; thus, it rnay be assurned ~hat. char R =1= 2. Consider any elernent a =1= Oin R. By hypothesis, Setting there eXIst mtegers h, k > 1 for which ah = a, (2a)k = 2a. q = (h - l)(k - 1) + 1 > 1, it follows that both aq =' a and (2a)q = 2a. From tbis, we obtain (2q ...:.. 2)a' = O, with 2q - 2 =1= O. Therefore, there' exists a least positive integer p such that pa = O, which implies tbat char R = p, p a prime [Theorem 1 - 6 ) . " Let Zp be the prime sub:fleld of R. Since ah ::, a, the ~í6~ent a is algeb~aic over Zp and, hence, the extension Zp(a) constitutes aflnite field; say wlth p" elements. In particular, a itself líes in Zp(a), so Ü:í'át a pn = a. Ir we now assume that a ~ cent R, then all fue hypothesis of tém:ma 1 will be satisfied ; thus, there exists an element b E R and integer k ~':t:~atisfying bab- 1 ak =F a. SiÍllilar reasoning applied to the extensioií.'~field Z (b) indicates that bpm = b for sorne integer m > 1. :' p At this point we tum our attention to the set of finÍte sums

It should be apparent that W is a finite set which is closed under addition. Since the relation akb = ba allows us to bring the a's and b's together in a product, W is also c10sed under multiplication. Whatever further it may be, W has at least been shown to be a ring.· As a Bnite ~ubring of a division rirtg, W is more than just a ring; it lS, in fact, a Bnite division rirtg (Problem 32). Hence, by Wedderburn's Theorem, we know that W lS necessarily commutative. In particular, a and b are both members of W; so that ab = ba, contradicting the relation bab -1 =' ak '=1= a. Having arrived at a suitable contradiction, we infer that the choice of a ~ cent R is impossible and R must be cornmutative. The transition of Theorem 9-12 from the division rings case arbitrary rings is accomplished by two lemillas. Lemma 1. Let R be a J-ring. Then every right ideal 1 of R is a twosided ideal of R.

Proo! To begin with, we assert that R can possess no, nonzero nilpotent elements. Indeed, if x =1= O, the condition xn(x) = x necessarily implies that xm =1= O for all m ~ 1. Now, suppose that e is any idempotent elemeI1t of R; then, for any x E R, , , . (xe - exe)2

=:

(ex

exe)2

=

O,

198

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

so that xe - exe = O = ex- exe. Therefore, ex = exe = xe, in consequence of which e E cent R. It follows that every idempotent of R must be in the center. Given that a E 1, with an = a (n > 1), it is easy to show that e = an - 1 is an idempotent element of R:

Hence, a" - 1 E cent R and so, for any r in R, .

where r' = an - 2 ra. Since ar' e 1, this shows that ra el also, making 1 a two-sided ideal of R. Lemma 2. Let R be,a J-ring. For all a, b e R, the element ab in radR.

ba líes

Proo! A standard argument, using Zom's Lemma, shows that R is endowed with maximal right ideals M, which are two-sided from Lemma 1 (the presence oC an identity element in R enters here). By virtue of the fact that R/M has no nontrivial ideals, the quotient ring R/M becomes a division ringo Being a homomorphic image of R, R/M inherits the property that x"cx) = x. Thus, we are thrown back to a situation where Theorem 9-12 can act, and the quotient ring R/M is thereby rendered commutative. In other words, (a + M),(b + M) = (b + M)(a + M)

for all a, b in R, or, equivlÍleritIy, ab - ba e M. As this last relation holds for every maximal ideal dfR,. it follows that ab - ba Erad R. ~r".

With these preliminariesestablished, we now have the constituent pieces to prove ',: Theorem 9-13. (JacoosO'n). If R is a J-ring, then R is commutative. Proo! Suppose that the elj~hlent x Erad R. As. in the proof oC Lemma 1, some power of x is an id~mpotent; to be quite explicit, if xn x, then e = x n - 1 turns out to be idempotent. Since rad R forms an ideal of R, the element e willlie in rad R.'But, according to the corollary ofTheorem 8-2, Oistheonlyidempotentbelongingtorad R;hence, theelement e xn- 1 = O n 1 and so x = x" = xx - = O. The implication of this is that R comprises ba erad R = {O} for all a semisimple ringo Lemma 2 tells us that ah a, b in R. The net result is that any two elements of R commute, thereby completing the proof. '

As an interesting application of Jacobson's Theorem, we cite

199

Corollary. Let R be a ring with the pr.operty that every nonzero subring of R forms a division ringo Then R is a field. Proo! Observe first that the ring R has prime characteristic. Indeed, iC R were of characteristic zero, it would contain a proper subfield isomorphic to Q and, hence, a proper subring isomorphic to Z. Since the ring Z of integers is not a division ring, we obtain a contradiction. Now, let S be the subring of R generated by any nonzero element a e R. Then S oonsists of all polynomials in a over the prime subfield of R; that is to say, S = Zp[aJ, for some prime p. Since the e1ement a- 1 e S, a- 1 must be a polynomial in a, which implies that a is a root oC some polynomial with coefficients from Zp. In consequence, S forms a simple algebraic extension (field) of Zp. By Theorem 7-26, we also know that S is a finite field. This being the case, anCa) = a, where n(a) is the number oC elements in S. From Jacobson's result, it follows that R is necessarily commutative; hence, a field. There are a number of other fairIy general assumptions which at a glance seem quite far removed from commutativity, bui when imposed on a given ring render it commutative. In this connection, we might mention without proof Theorem 9-14. (Herstein). Let R be a ring with the property that, for each x E R, there exists an integer n(x) > 1 dep.ending on x such that xn(x) x E cent R. Then R is commutative. We have noted that in a J-ring some positive power of every element lies in the center. This provides another path along which to proceed to comntutativity. Theo,rem 9-15. (Herstein). Let R be a ring with the property that for each x E R there exists a positive integer n(x) depending on x such that xn(x) E cent R. If R contains no nonzeronil ideals, then R is commutative.

PROBLEMS

1. a) If the BooIean ring R has at least three elements, prove that every nonzero eIement except the identity is a zero divisor oC R. b) Verify that the idempotent eIements oC any commutative rlng with identity of characteristic 2 Corm a BooIean subring. 2. Show that any ring R (not necessarily with identity) in which each element is idempotent can be imbedded in a BooIean ringo [Hint: Let R' R x Zz and mimic the argument oC Theorem 2-U.] 3. a) Let R be a commutative ring with identity. and S the set oC all idempotent elements of R. Define a new sum of a and b in S by taking

a

+' b = a + b -

2ab.

198

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

so that xe - exe = O = ex- exe. Therefore, ex = exe = xe, in consequence of which e E cent R. It follows that every idempotent of R must be in the center. Given that a E 1, with an = a (n > 1), it is easy to show that e = an - 1 is an idempotent element of R:

Hence, a" - 1 E cent R and so, for any r in R, .

where r' = an - 2 ra. Since ar' e 1, this shows that ra el also, making 1 a two-sided ideal of R. Lemma 2. Let R be,a J-ring. For all a, b e R, the element ab in radR.

ba líes

Proo! A standard argument, using Zom's Lemma, shows that R is endowed with maximal right ideals M, which are two-sided from Lemma 1 (the presence oC an identity element in R enters here). By virtue of the fact that R/M has no nontrivial ideals, the quotient ring R/M becomes a division ringo Being a homomorphic image of R, R/M inherits the property that x"cx) = x. Thus, we are thrown back to a situation where Theorem 9-12 can act, and the quotient ring R/M is thereby rendered commutative. In other words, (a + M),(b + M) = (b + M)(a + M)

for all a, b in R, or, equivlÍleritIy, ab - ba e M. As this last relation holds for every maximal ideal dfR,. it follows that ab - ba Erad R. ~r".

With these preliminariesestablished, we now have the constituent pieces to prove ',: Theorem 9-13. (JacoosO'n). If R is a J-ring, then R is commutative. Proo! Suppose that the elj~hlent x Erad R. As. in the proof oC Lemma 1, some power of x is an id~mpotent; to be quite explicit, if xn x, then e = x n - 1 turns out to be idempotent. Since rad R forms an ideal of R, the element e willlie in rad R.'But, according to the corollary ofTheorem 8-2, Oistheonlyidempotentbelongingtorad R;hence, theelement e xn- 1 = O n 1 and so x = x" = xx - = O. The implication of this is that R comprises ba erad R = {O} for all a semisimple ringo Lemma 2 tells us that ah a, b in R. The net result is that any two elements of R commute, thereby completing the proof. '

As an interesting application of Jacobson's Theorem, we cite

199

Corollary. Let R be a ring with the pr.operty that every nonzero subring of R forms a division ringo Then R is a field. Proo! Observe first that the ring R has prime characteristic. Indeed, iC R were of characteristic zero, it would contain a proper subfield isomorphic to Q and, hence, a proper subring isomorphic to Z. Since the ring Z of integers is not a division ring, we obtain a contradiction. Now, let S be the subring of R generated by any nonzero element a e R. Then S oonsists of all polynomials in a over the prime subfield of R; that is to say, S = Zp[aJ, for some prime p. Since the e1ement a- 1 e S, a- 1 must be a polynomial in a, which implies that a is a root oC some polynomial with coefficients from Zp. In consequence, S forms a simple algebraic extension (field) of Zp. By Theorem 7-26, we also know that S is a finite field. This being the case, anCa) = a, where n(a) is the number oC elements in S. From Jacobson's result, it follows that R is necessarily commutative; hence, a field. There are a number of other fairIy general assumptions which at a glance seem quite far removed from commutativity, bui when imposed on a given ring render it commutative. In this connection, we might mention without proof Theorem 9-14. (Herstein). Let R be a ring with the property that, for each x E R, there exists an integer n(x) > 1 dep.ending on x such that xn(x) x E cent R. Then R is commutative. We have noted that in a J-ring some positive power of every element lies in the center. This provides another path along which to proceed to comntutativity. Theo,rem 9-15. (Herstein). Let R be a ring with the property that for each x E R there exists a positive integer n(x) depending on x such that xn(x) E cent R. If R contains no nonzeronil ideals, then R is commutative.

PROBLEMS

1. a) If the BooIean ring R has at least three elements, prove that every nonzero eIement except the identity is a zero divisor oC R. b) Verify that the idempotent eIements oC any commutative rlng with identity of characteristic 2 Corm a BooIean subring. 2. Show that any ring R (not necessarily with identity) in which each element is idempotent can be imbedded in a BooIean ringo [Hint: Let R' R x Zz and mimic the argument oC Theorem 2-U.] 3. a) Let R be a commutative ring with identity. and S the set oC all idempotent elements of R. Define a new sum of a and b in S by taking

a

+' b = a + b -

2ab.

200

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

+

With as the addition, pro ve that S eonstitutes a Boolean ring, known as the idempotent Boolean ring of Ro b) In particular, obtain the idempotent Boolean ring of Z120 , e) Show that the idempotent Boolean ring of any integral domain is isomorphie to Z20

201

then the ring R is isom~rphie to a subring of the ring (P(X), d, n)o [Hint: Consider the mappingf: R -+ P(X) defined by fea) = Sao]

11. Assume that a and b are elements of the Boolean ring R with a i bo Deduce the existenee of a maximal ideal M qf R sueh that a ~ M, but bE Mo 12. Pro ve that any proper ideal of a Boolean ring R is a semiprime ideal o

4. EstabHsh the statements below: a) Up to isomorphism, the only simple Boolean ring is Z2":' b) If R is a Boolean ring, then R = Qcl(R)o . e) A ring R is a Boolean ring if and only if R is eornmutative with identity and . ab(a + b) = O for all a, bE R. o 5. In any BooleaIl ring R, an order relation ::; may be intro,d.uced by taking a::; b if and only if ab = ao Given that the elements a, b, e, d :,~U: He in R, eonfirm the following order -properties : '-. a) a ::; a, O ::; a ::; 1 for every aE R; b) a ::; b and b ::; e imply a ::; e; . > e) a ::; b and b ::; a imply a = b; d)a ::; b and e ::; d imply ac ::; bd; e) be = O implies ac= O (e =f O) if and only if a :S bo ':!"

6. a) Let 1 be a nortempty subset of the Boolean ring Ro Show that 1 is an ideal of R if and only if i) a, bE 1 imply a + bE 1, and' ii) a El and r E R with r ::; a imply r E lo b) Verify that the setI a = {r E Rlr ::; a} forms an ideal of Ro

8. Let Rbe a Boolean ringo For any nonzero element a E R, show that there exists a maximal ideal M of R such that a ~ M from this, deduce that R is semisimpleo [Hint: Apply Zorn's Lemrna to the family of all ideals of R which eontain 1 - a,

.

9. For any nonzero element a E R, R a Boolean ring, define the set Sa by

Sa = {MIM is a. maximal ideal of R; a ~ M}o . Establish the following properties of the sets Sa: a) Sa =1= 0 whenever a =f 00 b) Sa+b = Sa d Sb e) Sab = Sa n SbO d) Sa :::: Sb ir and only if a = bo [Hint: a ~ M if and only if 1 - a E Mo] o

10. With referenee to Problem 9, prove that if X = {MIM is a maximal ideal of R},

l'.i, {1 -

alaEI}o

Show that 1 u ]' is the smallest s~~ring of R' in which 1 is a maximal ideal.

14. Silppose that S is a subring of theBoolean ring R. Prove that any homomorphism f from S onto the field Z2 can be:e~tended to all of R. [Hint: Use Theorern 2-6; ker f is contained üi a maximal ideal M, where R/M ~ Z2o] .

~dregular if and only if every principal ideal of

15. Prove that a commutative ring R is a direet summand of R.":,:'

R

16. Let R be a regular ringo EstaWish,that has no nonzero nilpotent elements if and only i'f for eaeh a E R there exists an element a' in R sueh that a = a2 a'0 [Hint: If R has no nonzero nilpotent elements, then aa' being idempotent Hes in ~~

.

17. Show that if R is a regular ring, then eent R is also regular. [Hint: Given that a E centR, then aa' a = a for sorne a' E R; show that axa :::: a, where x = 'a(a')2 belongs to eent Ro]

7. Byan atom of a Boolean ring R is meant an element a =1= Osueh that r ::; a implies either r = a or r = 00 Prove that a) the ideal la ¡s' maximal if and only if 1 - a is an atom of R (see Problem 6 for the definition of la); b) any maximal ideal eontains all the atoms of R, except at most oneo [Hint: Use Theorem 9-1.]

W~~

13. Let 1 be a proper ideal of a Boolean ring R and define the set l' by

18. Assuming that R is a regular ring with identity, prove the statements below: a) if O and 1 are the only idempotent elements of R, then R is a division ring (this holds, in particular, if R has no divisors of zero); . b) ir R is of positive eharaeteristic n, then n is a square-free integer; e) R has nO'nonzero nilpotent (two-sided) ideals; d) for every right ideal 1 of R, 1 = ann,(ann¡l)o [Hint: If 1 = eR, where e is idempotent, thenann¡l = R(l - e).} e) If R has no nonzero nilpotent elements, then aR = Ra for every a E Ro [Hint: Choose a' E R such that aa'a = a; sinee thiddempotent a'a E eentR, ar = ar(a'a) = sao for any r E R.] 19. If R = R 1 Ei3l R 2 Ei3l ooo Ei3l Rn is the direet sum of a finite number of regular rings R¡ (i = 1, 2, o, o, n), show that R is also regularo 20;' Verify that the ring L(V) of linear mappings of an n-dimensional vector spaee V into itself forms a regular ring; in this setting, ring multiplication is taken to be funetional compositiono [Hint: Starting with O =1= fE L(V), a basis {Xl' 000, x k } fOI kerfand a basis {Xl> 000, Xk> 000' xn} for V, extend the linearly independent set {j(Xk+ ¡), 000 J(xn)} to a basis {y¡, ooo, Yk,!(Xk+ ¡), ooo,f(Xn )} for Vo Given any k elements Z1, 000, Zk E V, define j' E L(V) by j'(y¡) = Z¡ for 1 ::; i ::; k,!'{J(x¡») = X¡ for k + 1 :S i ::; n,] 21. Prove that an integer n > 1 is prime if and only if (n - 1)!

+

1 is divisible by no

200

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

+

With as the addition, pro ve that S eonstitutes a Boolean ring, known as the idempotent Boolean ring of Ro b) In particular, obtain the idempotent Boolean ring of Z120 , e) Show that the idempotent Boolean ring of any integral domain is isomorphie to Z20

201

then the ring R is isom~rphie to a subring of the ring (P(X), d, n)o [Hint: Consider the mappingf: R -+ P(X) defined by fea) = Sao]

11. Assume that a and b are elements of the Boolean ring R with a i bo Deduce the existenee of a maximal ideal M qf R sueh that a ~ M, but bE Mo 12. Pro ve that any proper ideal of a Boolean ring R is a semiprime ideal o

4. EstabHsh the statements below: a) Up to isomorphism, the only simple Boolean ring is Z2":' b) If R is a Boolean ring, then R = Qcl(R)o . e) A ring R is a Boolean ring if and only if R is eornmutative with identity and . ab(a + b) = O for all a, bE R. o 5. In any BooleaIl ring R, an order relation ::; may be intro,d.uced by taking a::; b if and only if ab = ao Given that the elements a, b, e, d :,~U: He in R, eonfirm the following order -properties : '-. a) a ::; a, O ::; a ::; 1 for every aE R; b) a ::; b and b ::; e imply a ::; e; . > e) a ::; b and b ::; a imply a = b; d)a ::; b and e ::; d imply ac ::; bd; e) be = O implies ac= O (e =f O) if and only if a :S bo ':!"

6. a) Let 1 be a nortempty subset of the Boolean ring Ro Show that 1 is an ideal of R if and only if i) a, bE 1 imply a + bE 1, and' ii) a El and r E R with r ::; a imply r E lo b) Verify that the setI a = {r E Rlr ::; a} forms an ideal of Ro

8. Let Rbe a Boolean ringo For any nonzero element a E R, show that there exists a maximal ideal M of R such that a ~ M from this, deduce that R is semisimpleo [Hint: Apply Zorn's Lemrna to the family of all ideals of R which eontain 1 - a,

.

9. For any nonzero element a E R, R a Boolean ring, define the set Sa by

Sa = {MIM is a. maximal ideal of R; a ~ M}o . Establish the following properties of the sets Sa: a) Sa =1= 0 whenever a =f 00 b) Sa+b = Sa d Sb e) Sab = Sa n SbO d) Sa :::: Sb ir and only if a = bo [Hint: a ~ M if and only if 1 - a E Mo] o

10. With referenee to Problem 9, prove that if X = {MIM is a maximal ideal of R},

l'.i, {1 -

alaEI}o

Show that 1 u ]' is the smallest s~~ring of R' in which 1 is a maximal ideal.

14. Silppose that S is a subring of theBoolean ring R. Prove that any homomorphism f from S onto the field Z2 can be:e~tended to all of R. [Hint: Use Theorern 2-6; ker f is contained üi a maximal ideal M, where R/M ~ Z2o] .

~dregular if and only if every principal ideal of

15. Prove that a commutative ring R is a direet summand of R.":,:'

R

16. Let R be a regular ringo EstaWish,that has no nonzero nilpotent elements if and only i'f for eaeh a E R there exists an element a' in R sueh that a = a2 a'0 [Hint: If R has no nonzero nilpotent elements, then aa' being idempotent Hes in ~~

.

17. Show that if R is a regular ring, then eent R is also regular. [Hint: Given that a E centR, then aa' a = a for sorne a' E R; show that axa :::: a, where x = 'a(a')2 belongs to eent Ro]

7. Byan atom of a Boolean ring R is meant an element a =1= Osueh that r ::; a implies either r = a or r = 00 Prove that a) the ideal la ¡s' maximal if and only if 1 - a is an atom of R (see Problem 6 for the definition of la); b) any maximal ideal eontains all the atoms of R, except at most oneo [Hint: Use Theorem 9-1.]

W~~

13. Let 1 be a proper ideal of a Boolean ring R and define the set l' by

18. Assuming that R is a regular ring with identity, prove the statements below: a) if O and 1 are the only idempotent elements of R, then R is a division ring (this holds, in particular, if R has no divisors of zero); . b) ir R is of positive eharaeteristic n, then n is a square-free integer; e) R has nO'nonzero nilpotent (two-sided) ideals; d) for every right ideal 1 of R, 1 = ann,(ann¡l)o [Hint: If 1 = eR, where e is idempotent, thenann¡l = R(l - e).} e) If R has no nonzero nilpotent elements, then aR = Ra for every a E Ro [Hint: Choose a' E R such that aa'a = a; sinee thiddempotent a'a E eentR, ar = ar(a'a) = sao for any r E R.] 19. If R = R 1 Ei3l R 2 Ei3l ooo Ei3l Rn is the direet sum of a finite number of regular rings R¡ (i = 1, 2, o, o, n), show that R is also regularo 20;' Verify that the ring L(V) of linear mappings of an n-dimensional vector spaee V into itself forms a regular ring; in this setting, ring multiplication is taken to be funetional compositiono [Hint: Starting with O =1= fE L(V), a basis {Xl' 000, x k } fOI kerfand a basis {Xl> 000, Xk> 000' xn} for V, extend the linearly independent set {j(Xk+ ¡), 000 J(xn)} to a basis {y¡, ooo, Yk,!(Xk+ ¡), ooo,f(Xn )} for Vo Given any k elements Z1, 000, Zk E V, define j' E L(V) by j'(y¡) = Z¡ for 1 ::; i ::; k,!'{J(x¡») = X¡ for k + 1 :S i ::; n,] 21. Prove that an integer n > 1 is prime if and only if (n - 1)!

+

1 is divisible by no

202

FIRST COURSE IN RINGS ANO IDEALS

PROBLEMS

22. Show that GF(p"') is (isomorphie to) a subfield of GF(p") if and only if mln; in faet, ifmln,then there is exaetly one subfield with p" elements. [Hint: In case mln, use the faet that a - llak - 1 for k > 1 t6 eonclude that xP'" - xlx P" - x.] 23. Establish the following assertions: a) given that an irreducible polynomial f(x) E Zp[X], then f(x)lx P " - x if and . only if degf(x)ln; bl if an irreducible polynomialf(x) E Zp[X] has a root in GF(p"), thenf(x) splits eompletely in GF(p"); e) x}'" - x is the product of all the irreducible monic polynomialsf(x) E Zp[x] sueh tbat degf(x)ln. 24. If pisan odd prime, prove that the Galois field GF(p") eontains an élement which is not a square.

25. Let P be a prime ideal of R, a commutative ring with identity. If the quotient ring ,R./P has only a finite number of eIements, verify that R/P is a Galois field. 26. Prove tbat if F is a finite field and K is a subfield of F, then F forms a simple extension field of K. [Hint: Any generator of F* will generate F as a vector space over K.]

34. Show that the assumption of an identity element ls unneeessary in proving that J-rings are eommutative; in other words, if R is a ring with the property that for every a E R there is an integer n(a) > 1 for whleh an(a) = a, then R must be commutative. [Hint: Sineeany idempotenteisin cent R, thesubringS ... eR ... Re has e for an identity and, hence, is eommutative by Theorem 9-13; then "-1 • (xy - yx)e = O. for all x, y E R; now use the faet that e ( xy - yx) lS idempotent.] 35. A ring R is ealled an H-ring if for every x E R there exists an integer n(x) ~ 1 such that x"(x) x E cent R. Assuming that R ls an H-ring, prove the followmg assertions: a) Any homomorphic image of R is againan H-ring. b) For eaeh x E R, there exist arbitrarily large n for which x" x E eent R. e) Al! the idempotentand nilpotent elements of R lie in eent R. d) If a E R is a zero divisor, then there exists sorne nonzero e E cent R sueh that ac= O. [Hint: If ab = O, with b O, then e b" bE cent R and ac O; if e = O, look at the idempotent d = b"-l.]

+

27. Let F be a finite field with p" elements. Prove that the mapping O' p: F -+ F defined by taking O'p(a) = aP is an automorphism, the so-eaUed Frobenius automorphism of F; furthermore, ~ = ir 28. al Suppose that R is a ring with identity (not necessarily commutative). If R has

no nontrivial ideals, establish that R is a division ringo b) Show that jf f is a homomorphism from a ring R onto a division ring, then ker f forms a maximal ideal of R. 29. Prove that any finite subring of a division rÍJlg is again a division'ring.

.. ..

JO. For any element a E R, a division ring, define C(a) by C(a)

{r E Rlra = arlo

l'·

a) Show that C(a) is a division subring of R eontaining cent R. ¡' . b) If R is finite and there are q elements in cent R, prove that thereare q" elements in C(a) for sorne n E Z+. [Hint: C(a) may be regarded as a ve~~or space over . the finite field eent R.] , 31. If R is a division ring, show that its dimension as a vector spaee over cent R eannot equa12. 32. lf an integral domain R is finite dimensional as a vector spaee over its center, prove that R forrns a division ringo [Hint: For fixed a O, the linear mapping T.,x = ax is one-to-one; henee, onto R.]

+

33. a) Prove that every finite field is a J -ringo b) More generally, establish that a field F is a J-ring if and only if F is of prime eharacteristie and is an algebraic extension of its prime subfield.

203

.

202

FIRST COURSE IN RINGS ANO IDEALS

PROBLEMS

22. Show that GF(p"') is (isomorphie to) a subfield of GF(p") if and only if mln; in faet, ifmln,then there is exaetly one subfield with p" elements. [Hint: In case mln, use the faet that a - llak - 1 for k > 1 t6 eonclude that xP'" - xlx P" - x.] 23. Establish the following assertions: a) given that an irreducible polynomial f(x) E Zp[X], then f(x)lx P " - x if and . only if degf(x)ln; bl if an irreducible polynomialf(x) E Zp[X] has a root in GF(p"), thenf(x) splits eompletely in GF(p"); e) x}'" - x is the product of all the irreducible monic polynomialsf(x) E Zp[x] sueh tbat degf(x)ln. 24. If pisan odd prime, prove that the Galois field GF(p") eontains an élement which is not a square.

25. Let P be a prime ideal of R, a commutative ring with identity. If the quotient ring ,R./P has only a finite number of eIements, verify that R/P is a Galois field. 26. Prove tbat if F is a finite field and K is a subfield of F, then F forms a simple extension field of K. [Hint: Any generator of F* will generate F as a vector space over K.]

34. Show that the assumption of an identity element ls unneeessary in proving that J-rings are eommutative; in other words, if R is a ring with the property that for every a E R there is an integer n(a) > 1 for whleh an(a) = a, then R must be commutative. [Hint: Sineeany idempotenteisin cent R, thesubringS ... eR ... Re has e for an identity and, hence, is eommutative by Theorem 9-13; then "-1 • (xy - yx)e = O. for all x, y E R; now use the faet that e ( xy - yx) lS idempotent.] 35. A ring R is ealled an H-ring if for every x E R there exists an integer n(x) ~ 1 such that x"(x) x E cent R. Assuming that R ls an H-ring, prove the followmg assertions: a) Any homomorphic image of R is againan H-ring. b) For eaeh x E R, there exist arbitrarily large n for which x" x E eent R. e) Al! the idempotentand nilpotent elements of R lie in eent R. d) If a E R is a zero divisor, then there exists sorne nonzero e E cent R sueh that ac= O. [Hint: If ab = O, with b O, then e b" bE cent R and ac O; if e = O, look at the idempotent d = b"-l.]

+

27. Let F be a finite field with p" elements. Prove that the mapping O' p: F -+ F defined by taking O'p(a) = aP is an automorphism, the so-eaUed Frobenius automorphism of F; furthermore, ~ = ir 28. al Suppose that R is a ring with identity (not necessarily commutative). If R has

no nontrivial ideals, establish that R is a division ringo b) Show that jf f is a homomorphism from a ring R onto a division ring, then ker f forms a maximal ideal of R. 29. Prove that any finite subring of a division rÍJlg is again a division'ring.

.. ..

JO. For any element a E R, a division ring, define C(a) by C(a)

{r E Rlra = arlo

l'·

a) Show that C(a) is a division subring of R eontaining cent R. ¡' . b) If R is finite and there are q elements in cent R, prove that thereare q" elements in C(a) for sorne n E Z+. [Hint: C(a) may be regarded as a ve~~or space over . the finite field eent R.] , 31. If R is a division ring, show that its dimension as a vector spaee over cent R eannot equa12. 32. lf an integral domain R is finite dimensional as a vector spaee over its center, prove that R forrns a division ringo [Hint: For fixed a O, the linear mapping T.,x = ax is one-to-one; henee, onto R.]

+

33. a) Prove that every finite field is a J -ringo b) More generally, establish that a field F is a J-ring if and only if F is of prime eharacteristie and is an algebraic extension of its prime subfield.

203

.

r

function O:.!I -> v R¡ .defined by taking O(i) = OE R¡ for every index i; similarIy, the negative -a of a function a E ¿ EB R¡ is given by the rule

TEN

(- a)(í)

.

•

If {R¡} ís an indexed famil),:'of rings (not necessarily distinct), it is reasonable to ask whether there is'.so~·· promising way too use the rings R¡ to build up new rings. Towards thi '" we now introduce the notion of the complete direct sum of a set of ririgs ',the term "direct product" is also employe
(d¡~:a2' ... ,an ) (al' a2 ,

¿

+ b)(i) = a(i) +

b(z),

(ab)(i)

= a(i)b(¡}

204

a.)(b l , b 2 ,

.:.,

b~)

(al

+

bl , a2

(a1b 1, a2 b2 ,

+

...

b 2, ... ,an

+

bn ),

,anbn ).

(a¡ E R¡).

Furthermore, e.veryelement of R has a uÍüque representation in the form (al' az, ... , an ) = (al' O, ... ,O)

EB R¡ and,

for aIl í E .!l.

It follows without difficulty from the ring axioms in each component that the résulting system comprises a ringo The zero element of ¿ EB R¡ is the

••• ,

+ (b l , b2, ... , bn )

a¡ -> (O, ... , O, al' O, ... , O)

Addition and multiplication may be introduced in the set ¿ EB R¡ by means of the corresponding operations in the individual components; writing this as a formula, we have (a

.

Now, let us define 11 to be the set of all n-tuples (al' a~, ... , a.) E R with the property that ak = O for k =f i. It is easily 9hecked that 11 constitutes an 'ideal of R, which is isomorphic to the ring R¡ under the assignment

RJ

The rings R¡ are called the component ríngs of the sum moré specifically, we say that R¡ is the ith component.

a(i).

·.·.:,·:;;1

Definition 10-1. Let {R¡} be a family of rings indexed by some set .!l. The complete direct sum of the rings R í , denoted by ¿ EB R¡, consists of all functions a defined on the index set .!I subject to the condition tha!. for each element ¡E .!I the functional value a(i) lies inRi; {ala:.!I -> v R¡; a(i) E

=

At tbis point, we should m,ake several remarks. For one thing, in defining the complete direct sum ¿ EB R¡, the component rings were not required to be distinct; some, or even aH, ·of these rings may coincide. A case of particular interest occurs when R¡ = R for every value of i E.!I. Under these circumstances, ¿ EB R¡ becomes the set of all functions defined on .fand ha ving values in the given ring R; in short, ¿ EB R¡ = map (.!l., R). ~econdly, if i runs J)ver a finite index set .!I (there is no loss in assuming that¿pí = {l,2, ... , n}), then the situation is even simpler than it first app~~s. When this happens, the ringR = ¿ EB R¡ may be interpreted as consisting of all ordered n-tuples (al' a 2 , ... , a.), where the element a¡ E R¡. Addition and multiplication are still to be carried out componentwise; that lS tá's:tY, .

DIRECT SUMS OF RINGS . ,1"

205

DlRECT SUMS OF RINGS

l.

\

I¡;

+ (O, az, ... , O) + .. ~ + (O, ... , 0, an ).

Tbis feature throws us back into the situation described in Chapter 2 (see page 21). If we invoke Theorem 2-4, it follows that the. ring R is the direct sum (in the sen se ofOefinitlon 2-4) ofthe ideals li' The point which we wish to make is that the concept of complete direct sum extends our previously defined direct sum; in the finite case, the two notions coincide up to isomorphism of components. The particular ring $0 obtained is customarily denoted by either ¿7= 1 EB R¡ or Rl EB Rz EB ... EB R". We might also mention in passing that if.!l is the positive integers, then ¿ EB R¡ may be viewedas the set of alI infinite sequences (al' az, ... , an , .•• ) such that a¡ E R¡ for each i E .!l. Since the generality of the com plete direct sum confronts the imagination with sucn a hurdle, we shall seldom have occasion to use it. Certain subrings ofthe complete direct sum are more manageable and more interesting, 'For instance, the discrete direct sum of the rings R¡ is the subring of ¿ EB R¡ consisting of those functions which are zero for almost all i; here the phrase "for almost all i" is short for "for all i with at most a finite number of

r

function O:.!I -> v R¡ .defined by taking O(i) = OE R¡ for every index i; similarIy, the negative -a of a function a E ¿ EB R¡ is given by the rule

TEN

(- a)(í)

.

•

If {R¡} ís an indexed famil),:'of rings (not necessarily distinct), it is reasonable to ask whether there is'.so~·· promising way too use the rings R¡ to build up new rings. Towards thi '" we now introduce the notion of the complete direct sum of a set of ririgs ',the term "direct product" is also employe
(d¡~:a2' ... ,an ) (al' a2 ,

¿

+ b)(i) = a(i) +

b(z),

(ab)(i)

= a(i)b(¡}

204

a.)(b l , b 2 ,

.:.,

b~)

(al

+

bl , a2

(a1b 1, a2 b2 ,

+

...

b 2, ... ,an

+

bn ),

,anbn ).

(a¡ E R¡).

Furthermore, e.veryelement of R has a uÍüque representation in the form (al' az, ... , an ) = (al' O, ... ,O)

EB R¡ and,

for aIl í E .!l.

It follows without difficulty from the ring axioms in each component that the résulting system comprises a ringo The zero element of ¿ EB R¡ is the

••• ,

+ (b l , b2, ... , bn )

a¡ -> (O, ... , O, al' O, ... , O)

Addition and multiplication may be introduced in the set ¿ EB R¡ by means of the corresponding operations in the individual components; writing this as a formula, we have (a

.

Now, let us define 11 to be the set of all n-tuples (al' a~, ... , a.) E R with the property that ak = O for k =f i. It is easily 9hecked that 11 constitutes an 'ideal of R, which is isomorphic to the ring R¡ under the assignment

RJ

The rings R¡ are called the component ríngs of the sum moré specifically, we say that R¡ is the ith component.

a(i).

·.·.:,·:;;1

Definition 10-1. Let {R¡} be a family of rings indexed by some set .!l. The complete direct sum of the rings R í , denoted by ¿ EB R¡, consists of all functions a defined on the index set .!I subject to the condition tha!. for each element ¡E .!I the functional value a(i) lies inRi; {ala:.!I -> v R¡; a(i) E

=

At tbis point, we should m,ake several remarks. For one thing, in defining the complete direct sum ¿ EB R¡, the component rings were not required to be distinct; some, or even aH, ·of these rings may coincide. A case of particular interest occurs when R¡ = R for every value of i E.!I. Under these circumstances, ¿ EB R¡ becomes the set of all functions defined on .fand ha ving values in the given ring R; in short, ¿ EB R¡ = map (.!l., R). ~econdly, if i runs J)ver a finite index set .!I (there is no loss in assuming that¿pí = {l,2, ... , n}), then the situation is even simpler than it first app~~s. When this happens, the ringR = ¿ EB R¡ may be interpreted as consisting of all ordered n-tuples (al' a 2 , ... , a.), where the element a¡ E R¡. Addition and multiplication are still to be carried out componentwise; that lS tá's:tY, .

DIRECT SUMS OF RINGS . ,1"

205

DlRECT SUMS OF RINGS

l.

\

I¡;

+ (O, az, ... , O) + .. ~ + (O, ... , 0, an ).

Tbis feature throws us back into the situation described in Chapter 2 (see page 21). If we invoke Theorem 2-4, it follows that the. ring R is the direct sum (in the sen se ofOefinitlon 2-4) ofthe ideals li' The point which we wish to make is that the concept of complete direct sum extends our previously defined direct sum; in the finite case, the two notions coincide up to isomorphism of components. The particular ring $0 obtained is customarily denoted by either ¿7= 1 EB R¡ or Rl EB Rz EB ... EB R". We might also mention in passing that if.!l is the positive integers, then ¿ EB R¡ may be viewedas the set of alI infinite sequences (al' az, ... , an , .•• ) such that a¡ E R¡ for each i E .!l. Since the generality of the com plete direct sum confronts the imagination with sucn a hurdle, we shall seldom have occasion to use it. Certain subrings ofthe complete direct sum are more manageable and more interesting, 'For instance, the discrete direct sum of the rings R¡ is the subring of ¿ EB R¡ consisting of those functions which are zero for almost all i; here the phrase "for almost all i" is short for "for all i with at most a finite number of

206

FIRST COURSE IN RINGS AND IDEALS

DIRECT 8UM8 OF RlNG8

exceptions." It would not be too far removed from traditional connotations to represent the discrete direct sum of the rings RI by ¿d $ R I :

¿d $

R¡

=

{a E ¿ $ R;\a(¡)

Ofor all buta finite number of i}.

Again, if the index set J is taken to be finite, say Jí = {l, 2, ... , n}, then the stipulation "for almost all i" is redundant and may be dropped from the descriptión of ¿d (fl R¡; in this latter setting,

¿d $

R¡

Rl (fl R 2$ ... (fl R,..

Another special subring of the complete direct sum ¿ $ RI which is worthy of consideration is the so~called subdirect sumo Let us proceed to examine tbis particular concept in sorne detaiL First, observe that for a fixed index i, we may defiÚe a function 11:1: ¿ $ R¡ --? RI by the equation

1I:¡(a) = a(i). One can verify that 11:1 is a homomorphism of ¿ EB RI onto the ring R¡, called the ith component projection. Ir S is any subring of ¿ $ R¡, the restriction 1I:/IS defines a homomorpbism of S into R¡ and, hence, onto a subring 11:1(S) of Rí' The case ofprincipal interest is that in wbich 1I:¡(S) = R¡ for each index i; in this event, we call S a subdirect sum of the rings RI' Let us record these remarks as a formal definition. Definition 10-2. A subring S of the complete direct sum ¿ EB R¡ is said to be a subdirect sum of the rings R í , written S = ¿s $ R¡, if the induced projectiol'l 1I:¡ls: S --? R¡ is an onto mapping for each i. The subdirect sUrii;:1S nontrivial if none of the mappings 1I:¡j$ is one-to-one (hence, S is nqt isomorphic to any R i ). In effect, a suprlng S 5; ¿ $ Ri is a subdirect sum of tl1e rings R¡ if and only ir, for ea.~h index i, every element of RI appears as the functional ' ' value at i of sorne ;function ln S. Definition 10~2,raises a rather significant question: What necessary and sufficient conditiQI),S upon a ring R will enable us to write it (up to isomorpbism) as the'subdirect sum of more tractable rings R¡? Up to this point, everything' has been a matter oC definition and observation; with the needed preliminar,ies finally compiled, let us make a start at providing an answer to the a:bove problem. Lemma. A ring R is isomorphic to a subdirect sum of the rings R¡ if and only if there exists an isomorphism f: R --? ¿ $ R¡ such that, for , each i, 11:1o f is a homomorphism of R onto R¡.

Proo! Given an isomorphism f of R onto a subdirect sum L $ R¡ of the rings Ro the composition 11:1o f: R --? R¡ defines a homomorphism oC R into S

207

R¡ (11:; itself being a homori:lOrphism). Since ¿' $ R¡ is a subdirect sum, 11:¡ o f actually carries R onto R¡. On the other. hand, if there happens to exist an isomorphism f satisfying the indicated conditions, then '\Ve certainly have R :::=: f(R)

=

¿' (fl R¡.

lt is helpfuL to translate the foregoing lemma into a condition on the ideals of a given ring; in what fúllows we describe just such a condition. Theorem 10-1. A ring R is isomorpbic to a subdirect sum of rings R¡ ir and only if R contains a collection of ideals {1¡} such that RJl¡ :::=: RI and n 11 {O}.

Proo! To start, we assume that R :::=: ¿B $ R¡. Then there exists an isomorphism f: R --? ¿s EB R¡ such that the "natuml" homomorphisms 11:1 0 f: R --? R¡ are all onto mappings. Using the Fundamental Homomorphism T.heorem, this implies that RJl¡ :::=: R¡, where 11 = ker(1I:¡ 01). Note Curther that ker f = {r E Rl!(r)

= O}

= {r E RI(1I:¡ o f)(r) = Ofor all i}

=

nI¡.

Since f is a one-to-one function, kerf= {O}, from wmch it Collows that n

11 = {O}.

Going in the other direction,suppose that we are given a set of ideals = {O}. Define a functionf: R - L (fl R/ ., by requiringf(a) to be such that its ith projection 1I:i( ({(a») = a + 1/. ('fP.e essentíal point here is that any element of¿ $ R¡ is completely determined by its projections.) Then R is isomorphic by mean s offto a subring oC the : '. direct sum L EB R¡. To see thatfis one-to-one, Cor instance, simply obsex:ye: that .

{11} of R with RJl¡ ~ R¡ and n li

l¡ for all i}

ker f = {a E RI(1I:1 o fHa) =

{a E Rla + li =

JI

)'.

for all i}

= nI, = {O}. We leave the checking oCthe remaining details as an exercise. Most applications depend more directly on the following version of Theorem 10-L CoroDary. A ring R is isomorphic to Si subdirect sum of the quotient rings RJ1i if and only if R contains a collection oC ideals {11} such that n li = {O}. Furthermore, the subdirect sum is nontrivial if and only ir 11 {O} for all i.

+

206

FIRST COURSE IN RINGS AND IDEALS

DIRECT 8UM8 OF RlNG8

exceptions." It would not be too far removed from traditional connotations to represent the discrete direct sum of the rings RI by ¿d $ R I :

¿d $

R¡

=

{a E ¿ $ R;\a(¡)

Ofor all buta finite number of i}.

Again, if the index set J is taken to be finite, say Jí = {l, 2, ... , n}, then the stipulation "for almost all i" is redundant and may be dropped from the descriptión of ¿d (fl R¡; in this latter setting,

¿d $

R¡

Rl (fl R 2$ ... (fl R,..

Another special subring of the complete direct sum ¿ $ RI which is worthy of consideration is the so~called subdirect sumo Let us proceed to examine tbis particular concept in sorne detaiL First, observe that for a fixed index i, we may defiÚe a function 11:1: ¿ $ R¡ --? RI by the equation

1I:¡(a) = a(i). One can verify that 11:1 is a homomorphism of ¿ EB RI onto the ring R¡, called the ith component projection. Ir S is any subring of ¿ $ R¡, the restriction 1I:/IS defines a homomorpbism of S into R¡ and, hence, onto a subring 11:1(S) of Rí' The case ofprincipal interest is that in wbich 1I:¡(S) = R¡ for each index i; in this event, we call S a subdirect sum of the rings RI' Let us record these remarks as a formal definition. Definition 10-2. A subring S of the complete direct sum ¿ EB R¡ is said to be a subdirect sum of the rings R í , written S = ¿s $ R¡, if the induced projectiol'l 1I:¡ls: S --? R¡ is an onto mapping for each i. The subdirect sUrii;:1S nontrivial if none of the mappings 1I:¡j$ is one-to-one (hence, S is nqt isomorphic to any R i ). In effect, a suprlng S 5; ¿ $ Ri is a subdirect sum of tl1e rings R¡ if and only ir, for ea.~h index i, every element of RI appears as the functional ' ' value at i of sorne ;function ln S. Definition 10~2,raises a rather significant question: What necessary and sufficient conditiQI),S upon a ring R will enable us to write it (up to isomorpbism) as the'subdirect sum of more tractable rings R¡? Up to this point, everything' has been a matter oC definition and observation; with the needed preliminar,ies finally compiled, let us make a start at providing an answer to the a:bove problem. Lemma. A ring R is isomorphic to a subdirect sum of the rings R¡ if and only if there exists an isomorphism f: R --? ¿ $ R¡ such that, for , each i, 11:1o f is a homomorphism of R onto R¡.

Proo! Given an isomorphism f of R onto a subdirect sum L $ R¡ of the rings Ro the composition 11:1o f: R --? R¡ defines a homomorphism oC R into S

207

R¡ (11:; itself being a homori:lOrphism). Since ¿' $ R¡ is a subdirect sum, 11:¡ o f actually carries R onto R¡. On the other. hand, if there happens to exist an isomorphism f satisfying the indicated conditions, then '\Ve certainly have R :::=: f(R)

=

¿' (fl R¡.

lt is helpfuL to translate the foregoing lemma into a condition on the ideals of a given ring; in what fúllows we describe just such a condition. Theorem 10-1. A ring R is isomorpbic to a subdirect sum of rings R¡ ir and only if R contains a collection of ideals {1¡} such that RJl¡ :::=: RI and n 11 {O}.

Proo! To start, we assume that R :::=: ¿B $ R¡. Then there exists an isomorphism f: R --? ¿s EB R¡ such that the "natuml" homomorphisms 11:1 0 f: R --? R¡ are all onto mappings. Using the Fundamental Homomorphism T.heorem, this implies that RJl¡ :::=: R¡, where 11 = ker(1I:¡ 01). Note Curther that ker f = {r E Rl!(r)

= O}

= {r E RI(1I:¡ o f)(r) = Ofor all i}

=

nI¡.

Since f is a one-to-one function, kerf= {O}, from wmch it Collows that n

11 = {O}.

Going in the other direction,suppose that we are given a set of ideals = {O}. Define a functionf: R - L (fl R/ ., by requiringf(a) to be such that its ith projection 1I:i( ({(a») = a + 1/. ('fP.e essentíal point here is that any element of¿ $ R¡ is completely determined by its projections.) Then R is isomorphic by mean s offto a subring oC the : '. direct sum L EB R¡. To see thatfis one-to-one, Cor instance, simply obsex:ye: that .

{11} of R with RJl¡ ~ R¡ and n li

l¡ for all i}

ker f = {a E RI(1I:1 o fHa) =

{a E Rla + li =

JI

)'.

for all i}

= nI, = {O}. We leave the checking oCthe remaining details as an exercise. Most applications depend more directly on the following version of Theorem 10-L CoroDary. A ring R is isomorphic to Si subdirect sum of the quotient rings RJ1i if and only if R contains a collection oC ideals {11} such that n li = {O}. Furthermore, the subdirect sum is nontrivial if and only ir 11 {O} for all i.

+

20a

DIRECT SUMS OF RINGS

FIRST COURSE IN RINGS AN'D IDEA.LS

¿'

If a ring R is isomorphic to a subdirect sum $ R¡ oC rings R¡. it is convenient to speak oC $ Ri as being a representatíon oC R ,(as a subdirect sum oC the rings R¡). The last corollary, although satisCying in the sense that it reduces the problem oC finding such representations to that oC establishing the existence oC certain ideals, is actually a stepping stone to the more CruitCul results below. These theorems tellus under what conditions a ring R lis isomorphic to a subdirect sum oC rings whose struCture is welI . known.

¿'

Theorem 10-2. A ring R is isomorphic to a sqbdirect sifm oC fields ir. and only jC .R is semisimple. . Proof. A ring R is semisimple iCand only iCthe intersection ofallits maximal ideals Mi is the zero id~. By the previous corollary, this lali~r condition is a necessary and suffiCient condition that R be isomorphici~to, a subdirect sum oCthe quotient ring~ RIMi , each oCwhich is a field.>':¡'\' Corollary. For any ring R, R/rad R is isomorphic to a'>Úibdirect sum oCfields. . 'o

. Going one móre step in this direction, we also have Theorem 10-3. A ring R is isomorphic to a subdirect sum oC integral domains iC and only iC Ris without prime radical. Corollary. For any ring R, RIRad R is isomorphic to a subdirect sum oC integral domains. Since any integral dornain can be imbedded in a field, Theorem 10-3 implies the ColIowing: a (commutative) ring R with no nonzero nilpotent elements is isomorphic to a subdirect sum oC fields. Example 10-1. The ring Z oC integers fumishes a simple illustration oC the lack oC any kind oC uniqueness in the representation oC a ring as a subdirect sumo Since Z is semi simple, Theorem 10-1 ensures that it is isomorphic to a subdirect sum oC the rings Z/(p) Zp, wher.e pis a prime number: Z~

¿s

it being understood that the summation runs over al! primes. At the same time, Z can be represented as a subdirect sum of the rings ZpI, since the intersection of the ideals (p2) isalso the zero ideal:

¿s

Example 10-2. For anotber application oC Theorem 10-1, consider the ring map R # oC real-valued Cunctions on R #. As we know, each oC the ideals Mx

I

.!

{fe mapR# If(x) = O},

xER#

is maximal in mapR#. Sinée (\.eRMx {O}, it Collows that mapR# is the subdirect sum oC uncountably many copies oC the real field-one for each point oC R #. (This should come as ,no surprise, being essen tially the d6finition oC map R#. ) : Simply as an application of the Coregoing ideas (Cor we shalI make no su bsequent use oC the result), let us estabIl'sh ' Theorem 10-4. A ring R is isomorphic to a subdirect sum oC fields iC . and only iC Cor each nonzero ideal J{q,f R, there exists an ideal J =1= R such that 1 + J = R. ' ,." . Proof. Let 1 =1= {O} bean ideal oC R, where R is isomorphicto a subdirect sum oC fields. Then R conrains a coUé~~i¡on {Mi} oC maximal ideals with n Mi {O}. ' Since 1 is nonzero, this entails that 1 $ Mi Cor sorne value oC i; Cor any such i, we necessarily have 1 + Mi = R . Conversely, as sume that the indicated condition holds. We shall argue that each nonzero element is excIuded by sorne maximal ideal oC R, whence rad R = {O}. Pursuing this end, let Ó =1= a E R, sothat the principal ideal (a) =1= {O} (there is no loss in supposing also that (a) =1= R). By our hypo" thesis, (a) + J = R Cor sorne proper ideal J oC R. Now, Zorn's Lernma implies the existence oC an ideal M which is chosen maximal in the set of ideal s satisCying (í) J S; M and (ii) a fj M. To see thaf Mis actualIy a maximal ideal oC R, consider any ideal K with M e K S; R. Then, by the maximal nature of M, the element a E K; hence, R = (a) + J S; (a) + K S; K, or R = K. The outcome is that the intersection oC alI tI1e maxirnaI ideals oC R is zero. This being so, Theorem 10-2 aIlows us to concIude that R is isomorphic to á subdirect sum of ftelds. One direction oC Theorem 10-3 can be sharpened conslderably, as the next result shows. .

$ Zp.

p prime

Z ~

209

Theorem 10-5. Let R be a ring containing no nonzero nH ideals. Then R is isomorphic to a subdirect sum oC integral domains. Proof. For each nonnilpotent element a e R, the set

Sa

$ ZpI'

pprime

. AH the component rings in the ftrst representation are fields, while none is a field in the second. This shows that a given ring may be representable as a subdirect sum oC rings having quite differerit properties.

=

{a,a 2 ,

... ,

a", ... }

is cIosed under multiplication and does not contain O. Thus, there exists a . prime ideal Po oC R, with Po n So = 0 (corollary on page 164). We assert that R ~ $ (RIPa ), where the summation ranges over aH the nonnilpotent elements oC R.

¿$

20a

DIRECT SUMS OF RINGS

FIRST COURSE IN RINGS AN'D IDEA.LS

¿'

If a ring R is isomorphic to a subdirect sum $ R¡ oC rings R¡. it is convenient to speak oC $ Ri as being a representatíon oC R ,(as a subdirect sum oC the rings R¡). The last corollary, although satisCying in the sense that it reduces the problem oC finding such representations to that oC establishing the existence oC certain ideals, is actually a stepping stone to the more CruitCul results below. These theorems tellus under what conditions a ring R lis isomorphic to a subdirect sum oC rings whose struCture is welI . known.

¿'

Theorem 10-2. A ring R is isomorphic to a sqbdirect sifm oC fields ir. and only jC .R is semisimple. . Proof. A ring R is semisimple iCand only iCthe intersection ofallits maximal ideals Mi is the zero id~. By the previous corollary, this lali~r condition is a necessary and suffiCient condition that R be isomorphici~to, a subdirect sum oCthe quotient ring~ RIMi , each oCwhich is a field.>':¡'\' Corollary. For any ring R, R/rad R is isomorphic to a'>Úibdirect sum oCfields. . 'o

. Going one móre step in this direction, we also have Theorem 10-3. A ring R is isomorphic to a subdirect sum oC integral domains iC and only iC Ris without prime radical. Corollary. For any ring R, RIRad R is isomorphic to a subdirect sum oC integral domains. Since any integral dornain can be imbedded in a field, Theorem 10-3 implies the ColIowing: a (commutative) ring R with no nonzero nilpotent elements is isomorphic to a subdirect sum oC fields. Example 10-1. The ring Z oC integers fumishes a simple illustration oC the lack oC any kind oC uniqueness in the representation oC a ring as a subdirect sumo Since Z is semi simple, Theorem 10-1 ensures that it is isomorphic to a subdirect sum oC the rings Z/(p) Zp, wher.e pis a prime number: Z~

¿s

it being understood that the summation runs over al! primes. At the same time, Z can be represented as a subdirect sum of the rings ZpI, since the intersection of the ideals (p2) isalso the zero ideal:

¿s

Example 10-2. For anotber application oC Theorem 10-1, consider the ring map R # oC real-valued Cunctions on R #. As we know, each oC the ideals Mx

I

.!

{fe mapR# If(x) = O},

xER#

is maximal in mapR#. Sinée (\.eRMx {O}, it Collows that mapR# is the subdirect sum oC uncountably many copies oC the real field-one for each point oC R #. (This should come as ,no surprise, being essen tially the d6finition oC map R#. ) : Simply as an application of the Coregoing ideas (Cor we shalI make no su bsequent use oC the result), let us estabIl'sh ' Theorem 10-4. A ring R is isomorphic to a subdirect sum oC fields iC . and only iC Cor each nonzero ideal J{q,f R, there exists an ideal J =1= R such that 1 + J = R. ' ,." . Proof. Let 1 =1= {O} bean ideal oC R, where R is isomorphicto a subdirect sum oC fields. Then R conrains a coUé~~i¡on {Mi} oC maximal ideals with n Mi {O}. ' Since 1 is nonzero, this entails that 1 $ Mi Cor sorne value oC i; Cor any such i, we necessarily have 1 + Mi = R . Conversely, as sume that the indicated condition holds. We shall argue that each nonzero element is excIuded by sorne maximal ideal oC R, whence rad R = {O}. Pursuing this end, let Ó =1= a E R, sothat the principal ideal (a) =1= {O} (there is no loss in supposing also that (a) =1= R). By our hypo" thesis, (a) + J = R Cor sorne proper ideal J oC R. Now, Zorn's Lernma implies the existence oC an ideal M which is chosen maximal in the set of ideal s satisCying (í) J S; M and (ii) a fj M. To see thaf Mis actualIy a maximal ideal oC R, consider any ideal K with M e K S; R. Then, by the maximal nature of M, the element a E K; hence, R = (a) + J S; (a) + K S; K, or R = K. The outcome is that the intersection oC alI tI1e maxirnaI ideals oC R is zero. This being so, Theorem 10-2 aIlows us to concIude that R is isomorphic to á subdirect sum of ftelds. One direction oC Theorem 10-3 can be sharpened conslderably, as the next result shows. .

$ Zp.

p prime

Z ~

209

Theorem 10-5. Let R be a ring containing no nonzero nH ideals. Then R is isomorphic to a subdirect sum oC integral domains. Proof. For each nonnilpotent element a e R, the set

Sa

$ ZpI'

pprime

. AH the component rings in the ftrst representation are fields, while none is a field in the second. This shows that a given ring may be representable as a subdirect sum oC rings having quite differerit properties.

=

{a,a 2 ,

... ,

a", ... }

is cIosed under multiplication and does not contain O. Thus, there exists a . prime ideal Po oC R, with Po n So = 0 (corollary on page 164). We assert that R ~ $ (RIPa ), where the summation ranges over aH the nonnilpotent elements oC R.

¿$

210

FIRST COURSE IN RINGS ANO IDEALS

DIRECT SUMS OF RINGS

ClearIy, 1 = n Pa comprises an ideal of R and is not nil by hypothesis. If 1 1= {O}, we can select sorne nonnilpotent element b E I. But then 1 S;;; Pb , while b ~ Pb , an obvious contradiction. This being the case, we must have 1 = n Pa = {O}. It follows from Theorem 10-1 that R is isomorphic to a subdirect sum ofthe quotient rings (actually integral domains) R/Pa. Before pressing forward with the main line of investigation, let us look at a special case which will prove useful when, at a later stage, we study Artinian rings. Theorem 10-6. Let 11 , 12 , ... ,In be afinite set of (nontrivial) ideals of the ring R. If li + lj = R whenever i 1= j, then R/n 1; ~ ¿ Et> (R/IJ . Proo! To start, we define a mappingf: R ~ f(x)' = (x + 11 , X + 12 ,

¿

$ (R/l;) by x+

... ,

U.

The reader can painlessIy supply a ptoof that f is a homomorphism with ker f = nI;. Our problem is to show that, under tbis homomorphism, any element (Xl + 1 1 , x 2 + 12 " " , x n + In) of the complete diI:ect sum ¿ Et> (R/l;) appears as the image of sorne element in R; the stated result then hinges upon an application of the Fundamental Homomorphism Theorem. Fix the indexj for the moment Using the fact that 1; + lj = R whenever i 1= j, there exist e1ements a; El;, b; E lj with a; + b; = 1. This ensures that the product r¡

= a1a 2

...

aj - 1a j + 1 ... an

E

nI;. ; 'fj '"

Furthermore, since 1 - a; E l j , the coset a; + lj ={l:'+ lj for all i 1= j, whence rj + lj = 1 + l j ' ( " Now, pick arbitrary elements x; E R (i = 1,2, ... ·in); our.contention is that :'i'f· . f(x)

where x

= ¿ r;x;. x + lj

= {Xl + 11 , x 2 + 12 ,

... ,

xn

+':ln ), j

To see tbis, observe that we may write x

=

¿

+

lj as

(r; + 1}(x; + lj) + (r j + 1)(xj + lj)'

;'fj

But r; E lj for i 1= j, while r j + lj = 1 + l j , so the displayed equation reduces to x + lj = x j + lj U = 1,2, ... ,n). This substantiates the elaim that f is actually an onto mapping, leading to the isomorphism R/n 1; ~

¿

Et> (R/IJ

Careful scrutiny of the aboye argument shows that we have proved a subresult of independent interest ; name1y,

I I

211

Corollary. Let 1 1,1 i, ... , 1" be a finite set of ideals of the ring R with the property that 1; + lj = R whenever i 1= j. Given any n elements Xl' x 2 , ••• , Xn E R, there exists sorne x E R such that x - x; E 1; for i = 1,2, ... , n. This corollary may be applied to the ring Z of integers and to the principal ideals (m 1), (m 2 ), ' •• , (~), where the integers m; are relatively prime in pairs. One then obtains an old and famous theorem about congruences which goes by the name of the Chinese Remainder Theorem (the result being known to Chinese mathematicians as early as A.D. 250): Theorem 10-7. (Chinese Remainder Theorem). Let m1' m 2 , .... , m n be positive integers such that gcd (m;, mj) = 1 for i 1= j. If al' a 2 , ... , an are any n integers, then the system of congruences

x == al (mod m 1 ), admits a simultaneous solution. Furthermore, tbis solution is unique modulo m' = m 1m 2 ... mn , The hypothesis in Theorem 10-6 is conveniently expressed in terms of the following: a finite set of ideals 1 1,12 , ... , In of a ring R is said to be pairwise comaximal (or pairwise relative1y prime, in the older terminology) if 1; 1= R and 1; + lj = R for i 1= j; when n = 2, we simply term 1 1 and 12 comaximal. Thus, the condition on the ideal s in Theorem 10-6 is that they be pairwise comaximal. EvidentIy, the definition ofpairwise comaximal implies that 1; 1= lj for i 1= j, as well as 1; 1= {O} for all i. If, in the representation of a ring R as a subdirect sum of the rings R¡, the "natural" homomorphism ofi:onto R; happens to be an isomorphism for sorne i, then the representatio~ ls, termed trivial; in the contrary case it is non trivial. (A nontrivial repres~ntation does not rule out the possibility that R ~ R; by way of sorne mapping other than the "natural" homomorphism of R onto R;.) A ring.,R is called subdirectly irreducible if there is no nontrivial representation of R as a subdirect sumo Let us summarize these remarks in a definition. Definition 10-3. A ring R is said to be subdirectly irreducible if, in any representation of R as a subdirect sum ofthe rings R;, at least one ofthe associated homomorphisms of R onto R; is actual1y an isomorphism; otherwise, .R is subdirectly reducible. The corollary to Theorem 10-1 may be taken as asserting that R is subdirectIy reducible if and only if there exists in R a set of nonzero ideal s with zero intersection. An equivalent and often handier formulation is the following: a ring R is subdirectIy irreducible if and only if the intersection of all the nonzero ideals of R is different from the zero ideal.

210

FIRST COURSE IN RINGS ANO IDEALS

DIRECT SUMS OF RINGS

ClearIy, 1 = n Pa comprises an ideal of R and is not nil by hypothesis. If 1 1= {O}, we can select sorne nonnilpotent element b E I. But then 1 S;;; Pb , while b ~ Pb , an obvious contradiction. This being the case, we must have 1 = n Pa = {O}. It follows from Theorem 10-1 that R is isomorphic to a subdirect sum ofthe quotient rings (actually integral domains) R/Pa. Before pressing forward with the main line of investigation, let us look at a special case which will prove useful when, at a later stage, we study Artinian rings. Theorem 10-6. Let 11 , 12 , ... ,In be afinite set of (nontrivial) ideals of the ring R. If li + lj = R whenever i 1= j, then R/n 1; ~ ¿ Et> (R/IJ . Proo! To start, we define a mappingf: R ~ f(x)' = (x + 11 , X + 12 ,

¿

$ (R/l;) by x+

... ,

U.

The reader can painlessIy supply a ptoof that f is a homomorphism with ker f = nI;. Our problem is to show that, under tbis homomorphism, any element (Xl + 1 1 , x 2 + 12 " " , x n + In) of the complete diI:ect sum ¿ Et> (R/l;) appears as the image of sorne element in R; the stated result then hinges upon an application of the Fundamental Homomorphism Theorem. Fix the indexj for the moment Using the fact that 1; + lj = R whenever i 1= j, there exist e1ements a; El;, b; E lj with a; + b; = 1. This ensures that the product r¡

= a1a 2

...

aj - 1a j + 1 ... an

E

nI;. ; 'fj '"

Furthermore, since 1 - a; E l j , the coset a; + lj ={l:'+ lj for all i 1= j, whence rj + lj = 1 + l j ' ( " Now, pick arbitrary elements x; E R (i = 1,2, ... ·in); our.contention is that :'i'f· . f(x)

where x

= ¿ r;x;. x + lj

= {Xl + 11 , x 2 + 12 ,

... ,

xn

+':ln ), j

To see tbis, observe that we may write x

=

¿

+

lj as

(r; + 1}(x; + lj) + (r j + 1)(xj + lj)'

;'fj

But r; E lj for i 1= j, while r j + lj = 1 + l j , so the displayed equation reduces to x + lj = x j + lj U = 1,2, ... ,n). This substantiates the elaim that f is actually an onto mapping, leading to the isomorphism R/n 1; ~

¿

Et> (R/IJ

Careful scrutiny of the aboye argument shows that we have proved a subresult of independent interest ; name1y,

I I

211

Corollary. Let 1 1,1 i, ... , 1" be a finite set of ideals of the ring R with the property that 1; + lj = R whenever i 1= j. Given any n elements Xl' x 2 , ••• , Xn E R, there exists sorne x E R such that x - x; E 1; for i = 1,2, ... , n. This corollary may be applied to the ring Z of integers and to the principal ideals (m 1), (m 2 ), ' •• , (~), where the integers m; are relatively prime in pairs. One then obtains an old and famous theorem about congruences which goes by the name of the Chinese Remainder Theorem (the result being known to Chinese mathematicians as early as A.D. 250): Theorem 10-7. (Chinese Remainder Theorem). Let m1' m 2 , .... , m n be positive integers such that gcd (m;, mj) = 1 for i 1= j. If al' a 2 , ... , an are any n integers, then the system of congruences

x == al (mod m 1 ), admits a simultaneous solution. Furthermore, tbis solution is unique modulo m' = m 1m 2 ... mn , The hypothesis in Theorem 10-6 is conveniently expressed in terms of the following: a finite set of ideals 1 1,12 , ... , In of a ring R is said to be pairwise comaximal (or pairwise relative1y prime, in the older terminology) if 1; 1= R and 1; + lj = R for i 1= j; when n = 2, we simply term 1 1 and 12 comaximal. Thus, the condition on the ideal s in Theorem 10-6 is that they be pairwise comaximal. EvidentIy, the definition ofpairwise comaximal implies that 1; 1= lj for i 1= j, as well as 1; 1= {O} for all i. If, in the representation of a ring R as a subdirect sum of the rings R¡, the "natural" homomorphism ofi:onto R; happens to be an isomorphism for sorne i, then the representatio~ ls, termed trivial; in the contrary case it is non trivial. (A nontrivial repres~ntation does not rule out the possibility that R ~ R; by way of sorne mapping other than the "natural" homomorphism of R onto R;.) A ring.,R is called subdirectly irreducible if there is no nontrivial representation of R as a subdirect sumo Let us summarize these remarks in a definition. Definition 10-3. A ring R is said to be subdirectly irreducible if, in any representation of R as a subdirect sum ofthe rings R;, at least one ofthe associated homomorphisms of R onto R; is actual1y an isomorphism; otherwise, .R is subdirectly reducible. The corollary to Theorem 10-1 may be taken as asserting that R is subdirectIy reducible if and only if there exists in R a set of nonzero ideal s with zero intersection. An equivalent and often handier formulation is the following: a ring R is subdirectIy irreducible if and only if the intersection of all the nonzero ideals of R is different from the zero ideal.

212

DIRECT SUMS OF RINGS

FIRST COURSE IN RINGS AND IDEALS

Proo! First, suppose that the element r
The importance of subdirectly irreducible rings is demonstrated by the following representation theorem due to Birkhoff. Theorem 10-8. (Birkhoff). Every ring R is isomorphic to a subdirect surn of subdirectly irreducible rings.

5).

Proo! For each element a =f O of R, Zorn's Lemrna can be used to seleá an ideal 1 which is maximal in the family of all ideals of R contained in R - {~}I; this family is evidently nonempty, since the zero ideal belongs to i1. Our definition of l. implies that if 1 i,s any ideal of R with the property that 1 e 1, then a E l. We~hould also point out that the intersection of the id~alsla (where a runs ovet:'all nonzero elements of R) is zero. Indeed, if b E na'/' o l. with b =f O, th~n b must, in particular, He in the ideal J b;

Regard,ing the second assertion of the theorem, choose r to be any zero divisor of,¡R. Then ann (r) =f {O} and, since R v is contained inevery nonzero i4:~al of R, ann (r) ;2 R v. This last inclusion simply asserts that ' r E ann (ai1p. (r)) S;; ann R v, so that ann R v consists of all zero divisors, together wjth O. We n;óvi pass to a proof of (3). According to the hypothesis, the ideal (R vf =f::{P} (by Problem 14, Chapter 8, {O} is the only nilpotent ideal of R). Thus, tht:Ty,;exists sorne element tE R v for which rR v =f {O}. The implication of thi,~dact is that R v $ ann R v. Inasmuch as R v is minimal in the set of nom#ro ideals of R, we conclude at once that ann R v = {O}. The rest follow(Irom (1): {O} is a maximal ideal of R and so R forms l;l field.

this contradicts the fact thatdvwas originally chosen so as to exclude the element b; hence, n.tO l.~:{a}. It now follows from the corollary to Theorem 10-1 that R 'ís isómorphic to a subdirect sum of the quotient rings RIl.. ir·,) The proofis completed up'ón showing thateach ring Rila is itself subdirectly irreducible or, more to the point, that the intersection 'of all th~ nonzero ideals oY Rila is nonzero. By the Corre~pondence Theorem, it suffices to establish that the intersection of all the ideals of R properly containing l. again contains l. as a proper su bse1. In Iight of the maximality of 1 the element ti must belong to all such ideal s ; therefore, their intersecti¿n contains a and, hence, contains l. properly. The implication is that the coset a + l. is nonzero and lies in every nonzero ideal of RI1•. Thus, OUf goal is achieved. Before announcing the next result, let us introduce sorne convenient notation. Definition 10-4. For any ring R, the heart of R is the ideal RV =

í'I

As a special case of part (3) aboye, we might point out that any subdirectly irreducible Boolean ring must be a field, which is c1early isomorphic to Z2 (Theorem 9-2). There is a corollary to Theorem 10-9 that wíIl be usefullater on. Corollary. If R V =f {a}, then the annihilator ofthe set ofzero divisors of R is precisely R v. Proo! With reference to the theorem, it is enough to prove that ann (ann R V) = R v • Since one always has R v S;; ann (ann R V), let us concentrate on the reverse inclusion. If a is any nonzero element of ann (ann R V), then R v 5; (a) and, hence, O =f 'ar E R v for sorne choice of , r ~ ann R v (in other words, r is not a zero divisor of R). As in the proof of Theorem 10-9, we can find an element s E R for which 1 - rs E ann R v • This means that a(1 - rs) = O and so a = (ar)s E R v. It follows that aun (ann R V) 5; R v, which completes the argumen1. '

I

{Jll is a nonzero ideal of R}.

We observe that R v is a minimal ideal of R which is contained in each nonzero ideal of R; for this reason, R v is frequently called the minimal ideal of R. When R v =f {O}, it is not hard to see that R v constitutes a principal ideal with any ofits nonzero elements serving as a generator. The relation of this notion to the concept of a subdirect sum should be fairly obvious: a ring R is subdirectly irreducible if and only if R v =f {O}. A definition deserves a theorem, so we oblige with the following: v

Theorem 10-9. (McCoy). If R is a ring for which'R =f {a}, then 1) ann R v is a maximal ideal of R; 2) ann R v consists of all zero divisors of R, plus zero; 3) whenever Ris 'without prime radical, R forms a field.

213

i

There are a number of situaiions where the hypothesis of Theorem 10-9 occurs quite naturally. By way ofexample, the hypothesis ís certainly fulfilled in any field. A more interesting illustration is provided by the ring R = Zpn of integers modulo a power of a prime; in this setting, one has R V = (pn-l) and annR v = (p). Although no further attempt is made to discuss the subject of subdirect sums systematically, we shall continue to throw sidelong glances in this direction (for a more thoroughtreatment, the reader is invited to consult [49J). Sorne of these ideas will be put to work in the next section when rings with chain conditions are discussed..

212

DIRECT SUMS OF RINGS

FIRST COURSE IN RINGS AND IDEALS

Proo! First, suppose that the element r
The importance of subdirectly irreducible rings is demonstrated by the following representation theorem due to Birkhoff. Theorem 10-8. (Birkhoff). Every ring R is isomorphic to a subdirect surn of subdirectly irreducible rings.

5).

Proo! For each element a =f O of R, Zorn's Lemrna can be used to seleá an ideal 1 which is maximal in the family of all ideals of R contained in R - {~}I; this family is evidently nonempty, since the zero ideal belongs to i1. Our definition of l. implies that if 1 i,s any ideal of R with the property that 1 e 1, then a E l. We~hould also point out that the intersection of the id~alsla (where a runs ovet:'all nonzero elements of R) is zero. Indeed, if b E na'/' o l. with b =f O, th~n b must, in particular, He in the ideal J b;

Regard,ing the second assertion of the theorem, choose r to be any zero divisor of,¡R. Then ann (r) =f {O} and, since R v is contained inevery nonzero i4:~al of R, ann (r) ;2 R v. This last inclusion simply asserts that ' r E ann (ai1p. (r)) S;; ann R v, so that ann R v consists of all zero divisors, together wjth O. We n;óvi pass to a proof of (3). According to the hypothesis, the ideal (R vf =f::{P} (by Problem 14, Chapter 8, {O} is the only nilpotent ideal of R). Thus, tht:Ty,;exists sorne element tE R v for which rR v =f {O}. The implication of thi,~dact is that R v $ ann R v. Inasmuch as R v is minimal in the set of nom#ro ideals of R, we conclude at once that ann R v = {O}. The rest follow(Irom (1): {O} is a maximal ideal of R and so R forms l;l field.

this contradicts the fact thatdvwas originally chosen so as to exclude the element b; hence, n.tO l.~:{a}. It now follows from the corollary to Theorem 10-1 that R 'ís isómorphic to a subdirect sum of the quotient rings RIl.. ir·,) The proofis completed up'ón showing thateach ring Rila is itself subdirectly irreducible or, more to the point, that the intersection 'of all th~ nonzero ideals oY Rila is nonzero. By the Corre~pondence Theorem, it suffices to establish that the intersection of all the ideals of R properly containing l. again contains l. as a proper su bse1. In Iight of the maximality of 1 the element ti must belong to all such ideal s ; therefore, their intersecti¿n contains a and, hence, contains l. properly. The implication is that the coset a + l. is nonzero and lies in every nonzero ideal of RI1•. Thus, OUf goal is achieved. Before announcing the next result, let us introduce sorne convenient notation. Definition 10-4. For any ring R, the heart of R is the ideal RV =

í'I

As a special case of part (3) aboye, we might point out that any subdirectly irreducible Boolean ring must be a field, which is c1early isomorphic to Z2 (Theorem 9-2). There is a corollary to Theorem 10-9 that wíIl be usefullater on. Corollary. If R V =f {a}, then the annihilator ofthe set ofzero divisors of R is precisely R v. Proo! With reference to the theorem, it is enough to prove that ann (ann R V) = R v • Since one always has R v S;; ann (ann R V), let us concentrate on the reverse inclusion. If a is any nonzero element of ann (ann R V), then R v 5; (a) and, hence, O =f 'ar E R v for sorne choice of , r ~ ann R v (in other words, r is not a zero divisor of R). As in the proof of Theorem 10-9, we can find an element s E R for which 1 - rs E ann R v • This means that a(1 - rs) = O and so a = (ar)s E R v. It follows that aun (ann R V) 5; R v, which completes the argumen1. '

I

{Jll is a nonzero ideal of R}.

We observe that R v is a minimal ideal of R which is contained in each nonzero ideal of R; for this reason, R v is frequently called the minimal ideal of R. When R v =f {O}, it is not hard to see that R v constitutes a principal ideal with any ofits nonzero elements serving as a generator. The relation of this notion to the concept of a subdirect sum should be fairly obvious: a ring R is subdirectly irreducible if and only if R v =f {O}. A definition deserves a theorem, so we oblige with the following: v

Theorem 10-9. (McCoy). If R is a ring for which'R =f {a}, then 1) ann R v is a maximal ideal of R; 2) ann R v consists of all zero divisors of R, plus zero; 3) whenever Ris 'without prime radical, R forms a field.

213

i

There are a number of situaiions where the hypothesis of Theorem 10-9 occurs quite naturally. By way ofexample, the hypothesis ís certainly fulfilled in any field. A more interesting illustration is provided by the ring R = Zpn of integers modulo a power of a prime; in this setting, one has R V = (pn-l) and annR v = (p). Although no further attempt is made to discuss the subject of subdirect sums systematically, we shall continue to throw sidelong glances in this direction (for a more thoroughtreatment, the reader is invited to consult [49J). Sorne of these ideas will be put to work in the next section when rings with chain conditions are discussed..

214

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

8. Pro ve that an irredundant subdireet sum of a finite number of simple rings is their direet Sunl.

In the set ofproblems below, aH rings are assumed to be eonmlUtative with identity.

1. Prove each ofthe foIJowing assertions regarding the complete direet sum ¿ El:> R¡: a) If a and b ar~ elements of ¿ El:> R¡ sueh that n¡(a) = n¡(b) for eaeh index i, then a = b. b) If an element r¡ E R¡ is given for eaeh i, then there exists a unique a E ¿ El:> R¡ satisfying ¡¡Aa) = r¡. e) 'If R¡ =1= 0 for aIJ i, then the ith projection n¡ maps ¿ El:> R¡ onto R¡.

9. If R is a (eonmlUtative) regular ring, verify that R is isomorphie to a subdireet Sunl of fields.

I

.[

2. Prove that an arbitrary funetion f from a ring R into the complete direet Sunl ¿ El:> R¡ of the rings R¡ is a homomorphism if and only if the eomposition n¡ o f: R -+ R¡ is itself a homomorphism for each value of i. 3. Consider the complete direct sum

1;

=

¿

= Oforj

=1=

Pi is the ith prime.]

12. a) Let al> a2' ... ,anbe a finite set ofnonzero elements ofthe principal ideal domain R sueh that a¡ and aj are relatively prime for i =1= j. If a = lem (al> a2, ... ,' an), show that RI(a) ~ ¿ EB (RI(a¡». b) Prove that if the integer n > 1 has the prime faetorization n = p~'~2 ... p~',

i},

J¡ = {a E ¿ El:> R;ja(i) = O}. Verify that li and J¡ are both ideals of the ring ¿ El:> R¡ and that

¿ El:> R¡ =

then Z" ~

1¡ El:> J¡.

1112

R

n

7. GiventhattheringR = {(a,b)la,bEZ; a - bEZ.},showthatRisanirredundant subdirect sum (hut not the direct sum) of two copies of Z.

In = 11

Il

12

Il ... Il

JI¡ are

In'

= Rn = n(In + l i) S; In + (Il l i) S;

Il .:;; Il

l n-l>

',,1',0',,'

R

for 1 ::;; i ::;; n - 1.] 14. Assume that the ring R is subdireetly irreducible. Establish that thereit::~ists an element O =1= rE R with f(r) = O for every homomorphism f on R whieh is not one-to·one.

Z;i".R¡

ni"')

",

[Hint: Use induetion on n. Notiee that In is eomaximal with 11 sinee

5. Establish that each of the given rings has a representation as a subdireet sum of the ri~gs R¡{i = 1,2, 3, ... ): a) zi R¡ = Zpl, where P is a fixed prime. b) Z; R¡ = Zp, where PI is an odd prime. , e) Z.;){¡ = Z~/(2¡). (In the situation eonsidered, (2¡) = {2¡r + 21nlr E Z.; n E Z} = 2¡Z.) d) = Z.I(Pi)' where p¡ is an odd prime. Suppós~ that R is isomorphie to a subdireet sum of the rings R¡ under the homomorphismf We say that the subdireet sum is irredundant ir, for eaeh indexj, the mapping h): R -+ ¿¡"'j EB Ri defined by hk) = f(r)I¿¡'fj El:> R¡ is not one-to-one (that is, ker h) =1= {O}. Prove the equivalenee ofthe foIlowing statements: a) the subdirect sum ¿' EB R¡ is irredundant; b) ker (nj o f) ;j2 ni"') ker (n¡ e f) for eaeh indexj; e) thereexists a eollection ofideals {I¡} of R sueh that (1) RI ~ Rll¡, (2) 1¡ = {O}, and (3) 1¡ =1= {O} for eaeh index j.

¿ EB Zp,"

13. Let 1 l> 12 , ... , In be a finite set of ideals of the ring R. Prove that a) the ideals 1¡ are pairwise eomaximal if and only if their ni! radical s pairwise eomaximal; b) if the ideals 11 are pairwise eomaximal, then their produet

4. Prove that a ring R is isomorphic to a subdireet sum of rings R¡ if and only if, for eaeh i" there exists a homomorphism gl of R onto R¡ sueh that if Ú =1= rE R, then gl(r) =1= Ofor at least one value ofi. [Hint: Assume that the stated eondition holds. For fixed rE R, define f,. E L El:> R¡ by f,.(i) = g¡(r). Now, eonsider the mapping f: R -+ L El:> 'R¡ in whieh f(r) = f,..]

6.

10. a) Prove that a ring Ris isomorphie to the complete direet Sunl of a finite number of fields if and only if (i) R eontains only a finite number of ideals and (ii) rad R = {O}. b) Prove that a finite ring R is a direet sum offields if and only ifit has no nonzero nilpotent elements.

11. Demonstrate that the conc1usion ofTheorem 10-6 is false if an infinite numbér of ideals li are allowed. [Hint: Consider the ring Z and the ideals li = (p¡), where

El:> R¡. For a fixed index i, define the sets

{aE¿ El:> R;ja(j)

215

PROBLEMS

I

1

15. Prove that any subdireetly irreducible ring has eharacteristie zero or a pdwer of a prime. In partir;ular, eonclude that Zn is subdirectly irreducible if and 'only if n is a power of a prime. , 16. If R is a subdireetly irreducible ring, show that O and 1 are the only idempotents of R. [Hint: For an idempotent e E R, consider the principal ideals (e) and (1 - e).] 17. a) Verify that any subdirectly irreducible Boolean ring is a field. b) Prove that a ring R is a Boolean ring if and only if R is isomorphie to a subdireet sum offields Z2' [Hint: Theorenl 10-9 and part (a).]

214

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

8. Pro ve that an irredundant subdireet sum of a finite number of simple rings is their direet Sunl.

In the set ofproblems below, aH rings are assumed to be eonmlUtative with identity.

1. Prove each ofthe foIJowing assertions regarding the complete direet sum ¿ El:> R¡: a) If a and b ar~ elements of ¿ El:> R¡ sueh that n¡(a) = n¡(b) for eaeh index i, then a = b. b) If an element r¡ E R¡ is given for eaeh i, then there exists a unique a E ¿ El:> R¡ satisfying ¡¡Aa) = r¡. e) 'If R¡ =1= 0 for aIJ i, then the ith projection n¡ maps ¿ El:> R¡ onto R¡.

9. If R is a (eonmlUtative) regular ring, verify that R is isomorphie to a subdireet Sunl of fields.

I

.[

2. Prove that an arbitrary funetion f from a ring R into the complete direet Sunl ¿ El:> R¡ of the rings R¡ is a homomorphism if and only if the eomposition n¡ o f: R -+ R¡ is itself a homomorphism for each value of i. 3. Consider the complete direct sum

1;

=

¿

= Oforj

=1=

Pi is the ith prime.]

12. a) Let al> a2' ... ,anbe a finite set ofnonzero elements ofthe principal ideal domain R sueh that a¡ and aj are relatively prime for i =1= j. If a = lem (al> a2, ... ,' an), show that RI(a) ~ ¿ EB (RI(a¡». b) Prove that if the integer n > 1 has the prime faetorization n = p~'~2 ... p~',

i},

J¡ = {a E ¿ El:> R;ja(i) = O}. Verify that li and J¡ are both ideals of the ring ¿ El:> R¡ and that

¿ El:> R¡ =

then Z" ~

1¡ El:> J¡.

1112

R

n

7. GiventhattheringR = {(a,b)la,bEZ; a - bEZ.},showthatRisanirredundant subdirect sum (hut not the direct sum) of two copies of Z.

In = 11

Il

12

Il ... Il

JI¡ are

In'

= Rn = n(In + l i) S; In + (Il l i) S;

Il .:;; Il

l n-l>

',,1',0',,'

R

for 1 ::;; i ::;; n - 1.] 14. Assume that the ring R is subdireetly irreducible. Establish that thereit::~ists an element O =1= rE R with f(r) = O for every homomorphism f on R whieh is not one-to·one.

Z;i".R¡

ni"')

",

[Hint: Use induetion on n. Notiee that In is eomaximal with 11 sinee

5. Establish that each of the given rings has a representation as a subdireet sum of the ri~gs R¡{i = 1,2, 3, ... ): a) zi R¡ = Zpl, where P is a fixed prime. b) Z; R¡ = Zp, where PI is an odd prime. , e) Z.;){¡ = Z~/(2¡). (In the situation eonsidered, (2¡) = {2¡r + 21nlr E Z.; n E Z} = 2¡Z.) d) = Z.I(Pi)' where p¡ is an odd prime. Suppós~ that R is isomorphie to a subdireet sum of the rings R¡ under the homomorphismf We say that the subdireet sum is irredundant ir, for eaeh indexj, the mapping h): R -+ ¿¡"'j EB Ri defined by hk) = f(r)I¿¡'fj El:> R¡ is not one-to-one (that is, ker h) =1= {O}. Prove the equivalenee ofthe foIlowing statements: a) the subdirect sum ¿' EB R¡ is irredundant; b) ker (nj o f) ;j2 ni"') ker (n¡ e f) for eaeh indexj; e) thereexists a eollection ofideals {I¡} of R sueh that (1) RI ~ Rll¡, (2) 1¡ = {O}, and (3) 1¡ =1= {O} for eaeh index j.

¿ EB Zp,"

13. Let 1 l> 12 , ... , In be a finite set of ideals of the ring R. Prove that a) the ideals 1¡ are pairwise eomaximal if and only if their ni! radical s pairwise eomaximal; b) if the ideals 11 are pairwise eomaximal, then their produet

4. Prove that a ring R is isomorphic to a subdireet sum of rings R¡ if and only if, for eaeh i" there exists a homomorphism gl of R onto R¡ sueh that if Ú =1= rE R, then gl(r) =1= Ofor at least one value ofi. [Hint: Assume that the stated eondition holds. For fixed rE R, define f,. E L El:> R¡ by f,.(i) = g¡(r). Now, eonsider the mapping f: R -+ L El:> 'R¡ in whieh f(r) = f,..]

6.

10. a) Prove that a ring Ris isomorphie to the complete direet Sunl of a finite number of fields if and only if (i) R eontains only a finite number of ideals and (ii) rad R = {O}. b) Prove that a finite ring R is a direet sum offields if and only ifit has no nonzero nilpotent elements.

11. Demonstrate that the conc1usion ofTheorem 10-6 is false if an infinite numbér of ideals li are allowed. [Hint: Consider the ring Z and the ideals li = (p¡), where

El:> R¡. For a fixed index i, define the sets

{aE¿ El:> R;ja(j)

215

PROBLEMS

I

1

15. Prove that any subdireetly irreducible ring has eharacteristie zero or a pdwer of a prime. In partir;ular, eonclude that Zn is subdirectly irreducible if and 'only if n is a power of a prime. , 16. If R is a subdireetly irreducible ring, show that O and 1 are the only idempotents of R. [Hint: For an idempotent e E R, consider the principal ideals (e) and (1 - e).] 17. a) Verify that any subdirectly irreducible Boolean ring is a field. b) Prove that a ring R is a Boolean ring if and only if R is isomorphie to a subdireet sum offields Z2' [Hint: Theorenl 10-9 and part (a).]

T 216

FIRST COURSE IN RINGS AND IDEALS

ELEVEN

18. Prove that a ring R is subdirectIy irreducible ir and only ir R contains an element r V with the Collowing two properties: i) the principal ideal (r V ) has nonzero intersection with every nopzero ideal oC R; ii) ann (r V) is a maximal ideal of R. . [Hint: Assume the conditions and let a 1= O; from (a) () (r V ) 1= {O}; deduce that . (r V ) !;; (a).] 19. Prove that the idempotent Boolean ring oC Z. is isomorphic to the Boolean ring of 2/1 elements, where k is the number of distinct prime divísors oC n. [Hint: Shpw . tllat Z. has exactly 2k idempotents or that X2 == x (med n) has 2& solutions;Jor k > 1 use the Chin~se Remainder TheoreÍll.] ,~.l í ,

RINGS WITH CHAIN !(ONDITIONS

,ir..

(~:, . ~;:':i ~:, '(:1 :

.

..... ',.

In pursuit of the deeper results of ideal theory, it wiÚ;:Qe necessary to limit ourselves somewhat and hereafter study special classe.s' ófrings. Noetherian rings, which we are about to introduce, are particY:la:tly versatile. These satisfya certain finiteness condition, namely, that e'Jlry ideal of the ring should be finitely generated. As will be seen pre~~ntly, an equivalent formulation oí the Noetherian requirement is that the ideals of the ring satisfy the so-called ascending chain condition. From this idea, we are led in a natural way to consider a number of results relevant to rings with descending chain condition for ideals. Our investigation culminates in a structure theorem for semisimple Artinian' rings which dates back to . Wedderburn. (By a ring, we shall continue to mean a commutative ring with identity.) The foIlowing definition serves as a convenient starting point. Definition 11-1. A ring R satisfieS;.. the ascending chaín condition for ideals ir, given any sequence of ideáIs 1 i, 12 , ... of R with

there exists an integer n (deperÍding on the sequence) such that 1m for all m ~ n.

=

In

Definition 11-1 amounts to saying that every infinite ascending chain of ideals of R must "break off" at sorne point; that is, equality must hold beyond sorne inde.x. In the case 01 noncommutative rings, it should be apparent how to define the ascending chain condition for Ieft ideal s or for right ideaIs. We illustrate this idea with several examples. Example 11-1. In a trivial sense (being simple rings), every fieId and the ring M,,(F) of matrices over a field F satisfy the ascending chain condition. So also do the rings Zn, for they have only a finite number of ideals. 217

T 216

FIRST COURSE IN RINGS AND IDEALS

ELEVEN

18. Prove that a ring R is subdirectIy irreducible ir and only ir R contains an element r V with the Collowing two properties: i) the principal ideal (r V ) has nonzero intersection with every nopzero ideal oC R; ii) ann (r V) is a maximal ideal of R. . [Hint: Assume the conditions and let a 1= O; from (a) () (r V ) 1= {O}; deduce that . (r V ) !;; (a).] 19. Prove that the idempotent Boolean ring oC Z. is isomorphic to the Boolean ring of 2/1 elements, where k is the number of distinct prime divísors oC n. [Hint: Shpw . tllat Z. has exactly 2k idempotents or that X2 == x (med n) has 2& solutions;Jor k > 1 use the Chin~se Remainder TheoreÍll.] ,~.l í ,

RINGS WITH CHAIN !(ONDITIONS

,ir..

(~:, . ~;:':i ~:, '(:1 :

.

..... ',.

In pursuit of the deeper results of ideal theory, it wiÚ;:Qe necessary to limit ourselves somewhat and hereafter study special classe.s' ófrings. Noetherian rings, which we are about to introduce, are particY:la:tly versatile. These satisfya certain finiteness condition, namely, that e'Jlry ideal of the ring should be finitely generated. As will be seen pre~~ntly, an equivalent formulation oí the Noetherian requirement is that the ideals of the ring satisfy the so-called ascending chain condition. From this idea, we are led in a natural way to consider a number of results relevant to rings with descending chain condition for ideals. Our investigation culminates in a structure theorem for semisimple Artinian' rings which dates back to . Wedderburn. (By a ring, we shall continue to mean a commutative ring with identity.) The foIlowing definition serves as a convenient starting point. Definition 11-1. A ring R satisfieS;.. the ascending chaín condition for ideals ir, given any sequence of ideáIs 1 i, 12 , ... of R with

there exists an integer n (deperÍding on the sequence) such that 1m for all m ~ n.

=

In

Definition 11-1 amounts to saying that every infinite ascending chain of ideals of R must "break off" at sorne point; that is, equality must hold beyond sorne inde.x. In the case 01 noncommutative rings, it should be apparent how to define the ascending chain condition for Ieft ideal s or for right ideaIs. We illustrate this idea with several examples. Example 11-1. In a trivial sense (being simple rings), every fieId and the ring M,,(F) of matrices over a field F satisfy the ascending chain condition. So also do the rings Zn, for they have only a finite number of ideals. 217

218

FIRST COURS.E IN RINGS AND IDEALS

Example 11-2. In the ring of integers, the inclusion (n) S;; (m) implies that m divides n. Since a nonzero integer can have only a finite number of distinct divisors, the ring Z evidently satisfies Definition 11-l. Example 11-3. As a more interesting example, let us show that the ascending chain condition is satisfied by any principal ideal ring R. For this purpose, consider an increasing sequence of ideal s of R,

11

S;;

12

S;; ••• S;;

In

S;; ••••.

It is easily checked that the set theoretic union 1 = U In is also an ideal of R. Moreover, since R is principal, we must have 1 = (a) for suitable a E R. Now, the element a lies in one of the ideals of the union, say the ideal In' For m ~ n, it then fo11ows that 1 = (a)

whence 1m

=

S;;

In

S;;

1m

S;;

.

,

219

in Y; hence, 1 1 is properly contained in sorne ideal 12 E Y. Likewise, 12 is not maximal, so there exists an ideal 13 in Y wiÍh 12 e 13' Continuing in this fashion, we obtain an infinite ascending chain of ideals of R, 11 e 12 e 13 c···, all ofwhose inclusions are proper; this violates the ascending chain condition. We now assume that the maximum condition holds and let 1 beany ideal of R. If 1 = {O}, then 1 is generated by one element, namely, O. Otherwise, choose a' nonzero element al E l. Either the principal ideal (al) = 1 and we are through, or else there is an element a2 E 1 which does not lie in (al); then, (al) e (al' a 2) S;; l. Again, if (a 1 ,a2)=/= 1, there exists sorne a3 in 1 such that (al' a 2) e (al' a2, a3)' This reasoning leads to an ascending chain of ideals of R:

1,

In' as desired.

Example 11-4. To provide an illustration of a ring in which the ascending chain condition fails to hold, let R denote the co11ection of a11 firiite subsets of Z+. Then (R,.1, n).is a commutative ring without identity (in fact, R . is an ideal of the ring of sets P(Z +)). lf In = {l, 2, ... , n}, then the reader may verify that P(1 1 ) e P(I2) e P(1 3) c··· forms an increasing chain of ideals of R which terminates at no point. Our first theorem establishes several equivalent formuhttions of the ascending chain condition. Before presenÚ'ilg this, we make one definition. i

RINGS WITH CHAIN CONDITIONS

/L'" ~~ \

Definition 11-2. The maximum conditiQ1J..(for ideal s) is said to hold in a ring R if every nonempty set of ic!eals of R, partiaUy ordered by inclusion, has at least one maximal elenient (that is, an ideal which is not properly contained in any other i4~~ of the set).. We make immediate use of this idea i1l::~~roving .~;(,¡.

Theorem 11-1. The fo11owing statem~p.ts concerning the ideals of a ring R are equivalent: 1) R satisfies the ascending chain condition for ideals. 2) The maximum condition holds in R. 3) Every ideal of R is finitely generated. Proof. With an eye to proving that the ascending chain condition implies statement (2), let Y be a nonempty co11ection of ideals of R. We sha11 suppose that Y has no maximal element and derive a contradiction. Since y is not empty, pick an ideal 11 E Y. By assumption, 1 1 cannot be maximal

(al) e (al' a 2) e (al' a 2, a 3) e .... The maximum condition assures uS that the aboye set of ideals possesses a maximal element, say the ideal (al' a 2, ... ,an). Were 1 =/= (al' a 2, ... , an), we could then find sorne a E 1 with a fj (al' a 2, ... , an); accordingly, the ideal (a, al' ... , an) would properly contain (al' a 2, ... , an), which is impossible. Thus, 1 is generated by the n elements al' a2, ... , an0 The proof of the theorem is completed by showing that (1) is a consequence of (3). For this, as sume that we ha"e an ascending chain of ideals of R,

and let 1 = U li' TheJi) is an ideal of R which, by hypothesis, must be finitely generated; suppb"se, for instance, that 1 = (al' a 2, ... ,ar ). Now, each generator ak is an elénient of sorne ideal!¡k ofthe given chain. Choosing n to be the largest of the indices ik , it fo11ows that a11 the ak lie in the ideal In' But then, for m ~ ~>

.

1 = (~i, a2 ,

.•• ,

ar )

S;;

In

S;;

1m

S;;

1;

hence,lm = In' Our argurnent shows that every ascending chain of ideals R terminates at sorne point. Rings satisfying any one of the three equivalent conditions of Theorern 11-1 (hence, a11 three cohditions) are ca11ed Noetherian rings, in honor of Ernrny Noether, who first initiated their study. The fact that, when dealing with Noetherian rings, we can restrict our attention to finitely generated ideals is of great advantage; the next two results should arnply illustrate this. Theorem 11-2. If 1 is an ideal of the Noetherian ring R, then 1 contains sorne power of its nil radical; that is, (Ji)n S;; 1 for sorne n E Z +.

218

FIRST COURS.E IN RINGS AND IDEALS

Example 11-2. In the ring of integers, the inclusion (n) S;; (m) implies that m divides n. Since a nonzero integer can have only a finite number of distinct divisors, the ring Z evidently satisfies Definition 11-l. Example 11-3. As a more interesting example, let us show that the ascending chain condition is satisfied by any principal ideal ring R. For this purpose, consider an increasing sequence of ideal s of R,

11

S;;

12

S;; ••• S;;

In

S;; ••••.

It is easily checked that the set theoretic union 1 = U In is also an ideal of R. Moreover, since R is principal, we must have 1 = (a) for suitable a E R. Now, the element a lies in one of the ideals of the union, say the ideal In' For m ~ n, it then fo11ows that 1 = (a)

whence 1m

=

S;;

In

S;;

1m

S;;

.

,

219

in Y; hence, 1 1 is properly contained in sorne ideal 12 E Y. Likewise, 12 is not maximal, so there exists an ideal 13 in Y wiÍh 12 e 13' Continuing in this fashion, we obtain an infinite ascending chain of ideals of R, 11 e 12 e 13 c···, all ofwhose inclusions are proper; this violates the ascending chain condition. We now assume that the maximum condition holds and let 1 beany ideal of R. If 1 = {O}, then 1 is generated by one element, namely, O. Otherwise, choose a' nonzero element al E l. Either the principal ideal (al) = 1 and we are through, or else there is an element a2 E 1 which does not lie in (al); then, (al) e (al' a 2) S;; l. Again, if (a 1 ,a2)=/= 1, there exists sorne a3 in 1 such that (al' a 2) e (al' a2, a3)' This reasoning leads to an ascending chain of ideals of R:

1,

In' as desired.

Example 11-4. To provide an illustration of a ring in which the ascending chain condition fails to hold, let R denote the co11ection of a11 firiite subsets of Z+. Then (R,.1, n).is a commutative ring without identity (in fact, R . is an ideal of the ring of sets P(Z +)). lf In = {l, 2, ... , n}, then the reader may verify that P(1 1 ) e P(I2) e P(1 3) c··· forms an increasing chain of ideals of R which terminates at no point. Our first theorem establishes several equivalent formuhttions of the ascending chain condition. Before presenÚ'ilg this, we make one definition. i

RINGS WITH CHAIN CONDITIONS

/L'" ~~ \

Definition 11-2. The maximum conditiQ1J..(for ideal s) is said to hold in a ring R if every nonempty set of ic!eals of R, partiaUy ordered by inclusion, has at least one maximal elenient (that is, an ideal which is not properly contained in any other i4~~ of the set).. We make immediate use of this idea i1l::~~roving .~;(,¡.

Theorem 11-1. The fo11owing statem~p.ts concerning the ideals of a ring R are equivalent: 1) R satisfies the ascending chain condition for ideals. 2) The maximum condition holds in R. 3) Every ideal of R is finitely generated. Proof. With an eye to proving that the ascending chain condition implies statement (2), let Y be a nonempty co11ection of ideals of R. We sha11 suppose that Y has no maximal element and derive a contradiction. Since y is not empty, pick an ideal 11 E Y. By assumption, 1 1 cannot be maximal

(al) e (al' a 2) e (al' a 2, a 3) e .... The maximum condition assures uS that the aboye set of ideals possesses a maximal element, say the ideal (al' a 2, ... ,an). Were 1 =/= (al' a 2, ... , an), we could then find sorne a E 1 with a fj (al' a 2, ... , an); accordingly, the ideal (a, al' ... , an) would properly contain (al' a 2, ... , an), which is impossible. Thus, 1 is generated by the n elements al' a2, ... , an0 The proof of the theorem is completed by showing that (1) is a consequence of (3). For this, as sume that we ha"e an ascending chain of ideals of R,

and let 1 = U li' TheJi) is an ideal of R which, by hypothesis, must be finitely generated; suppb"se, for instance, that 1 = (al' a 2, ... ,ar ). Now, each generator ak is an elénient of sorne ideal!¡k ofthe given chain. Choosing n to be the largest of the indices ik , it fo11ows that a11 the ak lie in the ideal In' But then, for m ~ ~>

.

1 = (~i, a2 ,

.•• ,

ar )

S;;

In

S;;

1m

S;;

1;

hence,lm = In' Our argurnent shows that every ascending chain of ideals R terminates at sorne point. Rings satisfying any one of the three equivalent conditions of Theorern 11-1 (hence, a11 three cohditions) are ca11ed Noetherian rings, in honor of Ernrny Noether, who first initiated their study. The fact that, when dealing with Noetherian rings, we can restrict our attention to finitely generated ideals is of great advantage; the next two results should arnply illustrate this. Theorem 11-2. If 1 is an ideal of the Noetherian ring R, then 1 contains sorne power of its nil radical; that is, (Ji)n S;; 1 for sorne n E Z +.

220

r I

FIRST COURSE IN RINGS AND IDEALS

Proa! In view of Theorem 11-1, JI is a finitely generated ideal, say .JI = (al' a~, ... "a m )· Since each a¡E.JI, there exist positive integers n¡ for which aí' E l. Take n = n1 + n2 + ... + nm· Now, a generating system for (.JI)n is provided by the products a1 1 a~2 ... a~,,; where k¡ E Z ' and n = k 1 + k 2 + ... + km' But, if k1 + k2 +

... + km

=

n1

+

n2

+ ... +

nm,

then we must have k¡ ~ n¡ for sorne index, i (i = 1, 2, ... ,m). This implies that a~' El, hence that the dement a1 1 a~2 :/:, a~m E l. Since aH the generators of (.JI)n lie in 1, it foHows that (.JI)n S; l.":' . ~

'.-

Corollary. Let Q be a primary ideah.of the Noetherian ring R and 1 and J be ideals with 1J S; Q. Then,:.~ither 1 S; Q or else (JJ)n S; Q for sorne n E Z+. ' .,','

Proa! Taking stock of Problem 24(c),¿~apter 5, the condition 1J S; Q implies that either 1 S; Q or there existS:;,a positive integer m for which ¡m ~ Q. Since R is Noetherian, we also liiÍve (.jJ)k S; J fol' sorne k E Z+. This being so,

RINGS WITH CHA1N CONDJTIONS

condition holds in R, there exists an integer n such that 1k = In for aH k ~ n. Moreover, each of the ideal s 1¡ (i = 1,2, ... , n) has a finite basis, say (i = 0, 1.... , n),

where a¡j is the leading coefficient of !;j(x), a polynomial of degree i in l. We now set ourse1ves to the prime task of proving that the mo + ... + mn polynomials !;)x) generate l. The ideal J = (foü .... ,JOmo, : .. ,f,,1' ... ,JnmJ is finitely generated and, by our choice of the j';j(x), must be contained in l. To obtain the reverse inc1usiori and thereby complete the proof, consider an arbitrary polynomial f(x) E 1, say, of degren.: f(x) ~'.

Theorem 11-3. (Hilbert Basis Theorern). If R is a Noetherian ring, then the polynomial ring R[x] is al so Noetherian.

Proo! Let 1 be an arbitrary nonzero ideal of R[x]' To prove that R[x] is Noetherian, it is enough to show that 1, is finite1y generated. For each integer k ~ 0, we first consider the set· 1k consisting of zero and those element~ rE R which appear as the leading (nonzero) coefficient of sorne polynomial of degree k lying in 1 :

= {r E Rla o + a1x + ... +

b.o +

b 1x

+' ... +

br _ 1x r - 1

+

bx'.

The argument procee~rby induction on r. If r = 0, then we have f(x) = bo E los; J and nothin~ needs to be prQven. Next, assume inductively that any polynomial of degree r - 1lying in 1 also belongs to the ideal generated

,

~~k~ l'

> n, the leading coefficient b E Ir = In and one may write

for suitable choice of CíE R. Then the polynomial

The Hilbert Basis Theorem asserts that if R is a Noetherian ring (comrnutative with identity), then the polynomial ring R[ x] inherits this property. Since any principal ideal domain and, in particular, any fie1d, is Noetherian, Hilbert's Theorem pro vides us with a rather extensive c1a.ss of Noetherian rings. The proof is somewhat demanding, but the result so elegant, that we hope aH readers will work through the details.

1~

.

~,r

:, :

When

which is what had to be proven.

221

k I'X E

I}

U

f1(X) = f(x) - x r - n(C 1f,,1(X)

+

c2fn2(X)

+ ... +

cmnf"m.(x))

belongs to 1 and has degree les S than r; indeed, the coefficient of x' in this pplynomial is . . mn b - L c¡a., = O. ¡=1

(Notice particular1y thatf1(x) differs fromf(x) by an element of J.) At this point, theinductive assumption can be applied to f1(X) to conc1ude that f1(X) and, in turn,f(x) lie in the ideal J. If r :s; n, a similar line of reasoning can be employed. Indeed, since bE 1" we can always find elements d 1 , d2 , ... , dm• in R such that the polynomial

{O}.

I t is easily checked that 1k forms an ideal of the ring R with 1k S; 1k+ l' (The second assertion follows from the fact that if r E 1k' then r occurs as the leading coefficient of ~+ 1 when the corresponding polynomial is multiplied by x; hence, r E 1k+1') Since we are assuming that the ascending chain

is an e1ement of 1 with degree r - 1 oro less. In either case, our argument leads to the inclusion 1 S; J and the subsequént equality 1 = J. By induction, Hilbert's Theorem can be extended to polynomials in several indeterminants.

'

220

r I

FIRST COURSE IN RINGS AND IDEALS

Proa! In view of Theorem 11-1, JI is a finitely generated ideal, say .JI = (al' a~, ... "a m )· Since each a¡E.JI, there exist positive integers n¡ for which aí' E l. Take n = n1 + n2 + ... + nm· Now, a generating system for (.JI)n is provided by the products a1 1 a~2 ... a~,,; where k¡ E Z ' and n = k 1 + k 2 + ... + km' But, if k1 + k2 +

... + km

=

n1

+

n2

+ ... +

nm,

then we must have k¡ ~ n¡ for sorne index, i (i = 1, 2, ... ,m). This implies that a~' El, hence that the dement a1 1 a~2 :/:, a~m E l. Since aH the generators of (.JI)n lie in 1, it foHows that (.JI)n S; l.":' . ~

'.-

Corollary. Let Q be a primary ideah.of the Noetherian ring R and 1 and J be ideals with 1J S; Q. Then,:.~ither 1 S; Q or else (JJ)n S; Q for sorne n E Z+. ' .,','

Proa! Taking stock of Problem 24(c),¿~apter 5, the condition 1J S; Q implies that either 1 S; Q or there existS:;,a positive integer m for which ¡m ~ Q. Since R is Noetherian, we also liiÍve (.jJ)k S; J fol' sorne k E Z+. This being so,

RINGS WITH CHA1N CONDJTIONS

condition holds in R, there exists an integer n such that 1k = In for aH k ~ n. Moreover, each of the ideal s 1¡ (i = 1,2, ... , n) has a finite basis, say (i = 0, 1.... , n),

where a¡j is the leading coefficient of !;j(x), a polynomial of degree i in l. We now set ourse1ves to the prime task of proving that the mo + ... + mn polynomials !;)x) generate l. The ideal J = (foü .... ,JOmo, : .. ,f,,1' ... ,JnmJ is finitely generated and, by our choice of the j';j(x), must be contained in l. To obtain the reverse inc1usiori and thereby complete the proof, consider an arbitrary polynomial f(x) E 1, say, of degren.: f(x) ~'.

Theorem 11-3. (Hilbert Basis Theorern). If R is a Noetherian ring, then the polynomial ring R[x] is al so Noetherian.

Proo! Let 1 be an arbitrary nonzero ideal of R[x]' To prove that R[x] is Noetherian, it is enough to show that 1, is finite1y generated. For each integer k ~ 0, we first consider the set· 1k consisting of zero and those element~ rE R which appear as the leading (nonzero) coefficient of sorne polynomial of degree k lying in 1 :

= {r E Rla o + a1x + ... +

b.o +

b 1x

+' ... +

br _ 1x r - 1

+

bx'.

The argument procee~rby induction on r. If r = 0, then we have f(x) = bo E los; J and nothin~ needs to be prQven. Next, assume inductively that any polynomial of degree r - 1lying in 1 also belongs to the ideal generated

,

~~k~ l'

> n, the leading coefficient b E Ir = In and one may write

for suitable choice of CíE R. Then the polynomial

The Hilbert Basis Theorem asserts that if R is a Noetherian ring (comrnutative with identity), then the polynomial ring R[ x] inherits this property. Since any principal ideal domain and, in particular, any fie1d, is Noetherian, Hilbert's Theorem pro vides us with a rather extensive c1a.ss of Noetherian rings. The proof is somewhat demanding, but the result so elegant, that we hope aH readers will work through the details.

1~

.

~,r

:, :

When

which is what had to be proven.

221

k I'X E

I}

U

f1(X) = f(x) - x r - n(C 1f,,1(X)

+

c2fn2(X)

+ ... +

cmnf"m.(x))

belongs to 1 and has degree les S than r; indeed, the coefficient of x' in this pplynomial is . . mn b - L c¡a., = O. ¡=1

(Notice particular1y thatf1(x) differs fromf(x) by an element of J.) At this point, theinductive assumption can be applied to f1(X) to conc1ude that f1(X) and, in turn,f(x) lie in the ideal J. If r :s; n, a similar line of reasoning can be employed. Indeed, since bE 1" we can always find elements d 1 , d2 , ... , dm• in R such that the polynomial

{O}.

I t is easily checked that 1k forms an ideal of the ring R with 1k S; 1k+ l' (The second assertion follows from the fact that if r E 1k' then r occurs as the leading coefficient of ~+ 1 when the corresponding polynomial is multiplied by x; hence, r E 1k+1') Since we are assuming that the ascending chain

is an e1ement of 1 with degree r - 1 oro less. In either case, our argument leads to the inclusion 1 S; J and the subsequént equality 1 = J. By induction, Hilbert's Theorem can be extended to polynomials in several indeterminants.

'

222

RINGS WITH CHAIN CONDITIONS

FIRST COURSE IN RINGS AND IDEALS

223

will serve as a simple illustration: in a power series ring F[[x]] over a field F, rad F[[x]] = (x), but

Corollary. If R is a Noetherían ring, then so is the polynomial ring in a finit~ number of indeterminants Xl' Xl' ••• , X •.

R[Xl' Xl' ... , X.]

(X)" = (x")

We recall that an ideal 1 is nilpotent provided that there exists an integer n for which 111 = {O}, whereas 1 is said to be a ni! ideal if every element of 1 is nilpotent. It is not hard to see that any nilpotent ideal is a nil ideal.

:/= {O}

Levitsky proved that for Noetherian rings the converse also holds: nil ideals are nilpótent. This fact is brought out as a corolIary to our next theorem.

ror all n E Z + (Problem 1, Chapter 7). Let us now broaden the outlook by considering rings with the descending chain condition.

Theorem 11-4. (Levitsky). In a Noetherian ring R, the prime radical Rad R is the largest nilpotent ideal of R.

Definition 11-3. A ring R is said to satisry the descending chain condition ror ideals ir, given any descending chain of ideals of R,

Proof. .At the outset, observe that since R is Noetherian, we can use the maximum condition to select an ideal N of R which is maximal with respect to being nilpotent Our contention is that N is the largest nilpotent ideal of R (in the sense of containing a11 other nilpotent ideals). To set this in evidence, let N 1 be an arbitrary nilpotent ideal of R, say N~ = {O}; assume further that Ni = {O}. Tben (N + N ly+k {O}, so that the ideal N + N 1 is nilpotent. From the inclusion N S;;; N + N 1 and the maximal property of N, it follows that N = N + NI' One is then left with Ni S;;; N, which settles the point. Now every nilpotent ideal must also be nil and thus N S;;; Rad R by the corollary to Theorem 8-8. To derive the reverse inc1usion, assume that a + N is any nilpotent element of the quotient ring RfN. Then a" + N = (a + N)" = N for some n E Z +, implying that a" E N. Becau$e N is a nil ideal, there exists a positive integer m for which (a"'j" ,,;. 0, and so a is nilpotent as an element of R. This being the case, we conc1ude that the principal ideal (a) is nilpotent; hence, (a) S;;; N, by the maximality of N. Tbe rest should be c1ear: since a E (a) S;;; N, the coset

d':+-

N

= N.

" Our reasoning shows that the quotient ring R/N contains no nonzero nilpotent elements, which is to say that R/N has zero prime radical. But it is aIready known that Rad R is the smallest ideal of R possessing a quotient ring without prime radical (Theorem 8-12). Tberefore, Rad R S;;; N, which yields the desired equality N = Rad R; the theorem is now established. As corollaries we have Corollary 1. In a Noetherian ring, any nil ideal is nilpotent.

Proo! The proof amounts to the observation that any nil ideal is contained in the prime radical of a ringo Corollary 2. A semisimple N oetherian ring contains no nonzero nilpotent ideals. The breakdown of Levitsky's Theorem is rather dramatic when one replaces the prime radical by the Jacobson radical. The foIlowing example

there exists an integer n such that 1"

111+1

= 1.+2 =

As in Theorem 11-1, this definition leads to The following statements concerning the ideals of a Theorem 11-5. ring R are equivalent: 1) R satisfies the descending chain condition for ideals. 2) Every nonempty set of ideals of R, partial1y ordered by inclusion, contains a mínimal element (the mínimum condítion holds). A ring satisfying either bf these conditions is said to be Artinian (after Emíl Artin).

.

It would be repetitious to prove this modified version ofTheorem 11-1 and we shall refrain from doing so. However, lest some re.ader try to obtain the exact analog for Artinian ríngs ofTheorem 11-1, we hasten to point out that every ideal in the ring Z of integers is finitely gener.~t~d, but Z is not Artinian. Indeed, if(n) is any nonzero ideal of Z, then (2n)~ls'~ nonzero ideal proper1y contained in (n); thus, the set of all nonzero i4e~ls of Z has no minimal element. . In the light of the equivalence of the ascending (descending) chain condition with the maximum (minimum) condition, the two will be used interchangeably. Certain results are more easily proved 'in terms of one than the other, and convenience will be our guidé. '

Example 11-5. Tbe statement of the Hilbert Basis Tbeorem is no longer true ir Artinian is subtituted for Noetherian. For example, if F is any field, then (x) ~ {X2) ~ (x 3) ~ •.. is a strictly descending chain of principal ideals oC F[x]. descending chain condition fails to be satisfied in F[ X J.

Thus, the

222

RINGS WITH CHAIN CONDITIONS

FIRST COURSE IN RINGS AND IDEALS

223

will serve as a simple illustration: in a power series ring F[[x]] over a field F, rad F[[x]] = (x), but

Corollary. If R is a Noetherían ring, then so is the polynomial ring in a finit~ number of indeterminants Xl' Xl' ••• , X •.

R[Xl' Xl' ... , X.]

(X)" = (x")

We recall that an ideal 1 is nilpotent provided that there exists an integer n for which 111 = {O}, whereas 1 is said to be a ni! ideal if every element of 1 is nilpotent. It is not hard to see that any nilpotent ideal is a nil ideal.

:/= {O}

Levitsky proved that for Noetherian rings the converse also holds: nil ideals are nilpótent. This fact is brought out as a corolIary to our next theorem.

ror all n E Z + (Problem 1, Chapter 7). Let us now broaden the outlook by considering rings with the descending chain condition.

Theorem 11-4. (Levitsky). In a Noetherian ring R, the prime radical Rad R is the largest nilpotent ideal of R.

Definition 11-3. A ring R is said to satisry the descending chain condition ror ideals ir, given any descending chain of ideals of R,

Proof. .At the outset, observe that since R is Noetherian, we can use the maximum condition to select an ideal N of R which is maximal with respect to being nilpotent Our contention is that N is the largest nilpotent ideal of R (in the sense of containing a11 other nilpotent ideals). To set this in evidence, let N 1 be an arbitrary nilpotent ideal of R, say N~ = {O}; assume further that Ni = {O}. Tben (N + N ly+k {O}, so that the ideal N + N 1 is nilpotent. From the inclusion N S;;; N + N 1 and the maximal property of N, it follows that N = N + NI' One is then left with Ni S;;; N, which settles the point. Now every nilpotent ideal must also be nil and thus N S;;; Rad R by the corollary to Theorem 8-8. To derive the reverse inc1usion, assume that a + N is any nilpotent element of the quotient ring RfN. Then a" + N = (a + N)" = N for some n E Z +, implying that a" E N. Becau$e N is a nil ideal, there exists a positive integer m for which (a"'j" ,,;. 0, and so a is nilpotent as an element of R. This being the case, we conc1ude that the principal ideal (a) is nilpotent; hence, (a) S;;; N, by the maximality of N. Tbe rest should be c1ear: since a E (a) S;;; N, the coset

d':+-

N

= N.

" Our reasoning shows that the quotient ring R/N contains no nonzero nilpotent elements, which is to say that R/N has zero prime radical. But it is aIready known that Rad R is the smallest ideal of R possessing a quotient ring without prime radical (Theorem 8-12). Tberefore, Rad R S;;; N, which yields the desired equality N = Rad R; the theorem is now established. As corollaries we have Corollary 1. In a Noetherian ring, any nil ideal is nilpotent.

Proo! The proof amounts to the observation that any nil ideal is contained in the prime radical of a ringo Corollary 2. A semisimple N oetherian ring contains no nonzero nilpotent ideals. The breakdown of Levitsky's Theorem is rather dramatic when one replaces the prime radical by the Jacobson radical. The foIlowing example

there exists an integer n such that 1"

111+1

= 1.+2 =

As in Theorem 11-1, this definition leads to The following statements concerning the ideals of a Theorem 11-5. ring R are equivalent: 1) R satisfies the descending chain condition for ideals. 2) Every nonempty set of ideals of R, partial1y ordered by inclusion, contains a mínimal element (the mínimum condítion holds). A ring satisfying either bf these conditions is said to be Artinian (after Emíl Artin).

.

It would be repetitious to prove this modified version ofTheorem 11-1 and we shall refrain from doing so. However, lest some re.ader try to obtain the exact analog for Artinian ríngs ofTheorem 11-1, we hasten to point out that every ideal in the ring Z of integers is finitely gener.~t~d, but Z is not Artinian. Indeed, if(n) is any nonzero ideal of Z, then (2n)~ls'~ nonzero ideal proper1y contained in (n); thus, the set of all nonzero i4e~ls of Z has no minimal element. . In the light of the equivalence of the ascending (descending) chain condition with the maximum (minimum) condition, the two will be used interchangeably. Certain results are more easily proved 'in terms of one than the other, and convenience will be our guidé. '

Example 11-5. Tbe statement of the Hilbert Basis Tbeorem is no longer true ir Artinian is subtituted for Noetherian. For example, if F is any field, then (x) ~ {X2) ~ (x 3) ~ •.. is a strictly descending chain of principal ideals oC F[x]. descending chain condition fails to be satisfied in F[ X J.

Thus, the

224

RINGS WITH CHAIN CONDITlONS

FIRST COURSE IN RINGS AND IDEALS

Example 11-6. Consider the ring R = mapR# ofreal-valued funetions on R #. Given im arbitrary real number r > O, we define Ir

= {fE RI!(x)

O for -r

:5; X :5;

r}.

Then Ir is an ideal of R and it is not difficult to seethat , ... c:

13 c:: 12

c:

I1

C;

225

In the sequel, ther~ occur eertain results which hold for both Noetherian and Artinian rings. Where the proofs are virtually the same, our policy will be to establish the theoremin question only in the Noetherian case. Let us first show that the chain conditions are not deStroyed by homomorphisms. ' Theorem 11-6. If R is a Noetherian (Artinian) ring, then any homomorphic image of R is also Noetherian (Artinian) .. ,!:

I 1/ 2 c: I 1 / 3 c: ....

)rhe iQlplication lS that R eontains ascending and descending chains that do' :)ot becomestationary, whenee R is neither Artinian nor Noetherian. It~'i¡ 'ís perhaps appropriate to call attention to the ,fact that eaeh of the ideals";: 'Jr is properly contained in the maximal ideal M = {fE Rlf(O) = O}. ""

not':¡

: "Example 11-7. We next give' an example oC an Artinian ring which is ,Noetherian. For this"purpose, let p be a fixed prime. Consider the group~"A p::;Z(pOO) of all rational 1'l,umbers r between O and 1 of the form r = m/P~;,:} (y..here m is an arbitrary integer and n runs through the nonnegative integers':,~:\ '<':'"~nder the operation of addhion modulo 1: : : ¡ \ ¡ ¡

Z(pOO) = {mjpnlO :s; m < pn; mEZ; n = 0,1,2, ... }. ' We make Z(pOO) into a ring (without identity) by defining the produet ab to be zero forall a, bE Z(p""). It is important to observe that the ideals of the resulting ring are simply the subgroups of the additive group of Z(pOO). Now, let I be any nontrivial ideal of Z(pOO) and choose kto be the smaIlest positive integer such that for some a, ajpk 1 I; we implicitly assume that a and p are relatively prime. Then I must contain aH the elements 0, 1jl~ 1, 2jpk-1, ... ,(pk-l l)jpk~l. Our contention is that these are theonly members of l. To support tbis, suppose to the contrary that bjpi E I, where i ¿ k and, of course, b and p are relatively prime. One can then find integers r, s for which rb + sp = 1. Sinceboth the rational numbers (reduced modulo 1)

ProoJ. Let f be a homomorphism of the Noetherían ti,p.g R onto the ring , R' and consider any aseending chain 1'1 s; 1; s; ... s;:J~ s; ... of ideals of R '. Put Ik = f-l(Ik), for, k = 1,2,.... Then 1 1 s;t;i. s; ... s; In S; ...

forms an ascending ehain of ideals of R which, accordil1 gto our hypothesis, must eventually be constant; that is, there is sorne index:f¡ such tha t I m = I n for all m ~ n. Takíng stock of the fact that fis an ootó mapping, we have , f(I k ) = I~. Hence, l~ = I~, whenever m ~ n, so tbat\Jhe original chain ';'. ::: also stabilizes at some point. . J

Lettingfbe the natural mapping, we have as a

co~61Iary:

Corollary. If lis an ideal of the Noetherian (Artinian) ring R, then the quotient ring Rj1 is Noetherian (Artiniílll). Further progress will be facilitated by the technical lemma below. Lernma. Ir 1, 1, and K are ideal s of a ring R such that (1) 1 ~ K, (2) 1 n I = K nI, and then 1 K. .

(3) Jj1 = Kj1,

ProoJ. EvidentIy, we need only establish the inclusion K S; 1. To this purpose, seleet any, member k of K. On the basís of (3), thereexists an elementj El for whichj + 1 = k + 1, which signifies that k - j = i for some choice of i in l. But, since 1 S; K, the difference k - j also líes in the ideal K. Using eondition (2), we thus find that

and He in I, it follows that (rb + Sp)jpk 1jpk also belongs to 1, contradicting therninimaJity of k. Thus, the ideal 1 is finite and is given by

•

i = k - j

and, in consequence, k

=

i

+j

E

E

1n K

=

J n I,

1.

This faet ü¡ enough to enableus to prove a partíal converse of the last coróllary. ' . Representing 1 by the symbol1 k _ 1, we eonclude that the only ideals of Z(pOO) are those which appear in the chain

{O} e 11 c: 12

c: ... e

1k c: ... e Z(paJ).

Therefore, Z(p"") possesses an infinite (strict1y) ascending chain of ideals, but any descending chain is of finite length.

Theorem 11-7. Let 1 be an ideal of the ring R. If 1 and R/ I are both Noetherian (Artinian) rings, then R is also Noetherian (Artinian).

ProoJ. To begin, let 1 1

~ 1 2 s;;; .. , S;;; 1 n S;;; ... be any ascending chain of ideals of R. Fromthis, we may eonstruct a chain ofideals of 1,

11 n 1

~

12 n 1

~

'" s;;; ln nI

S;;; ... ,

224

RINGS WITH CHAIN CONDITlONS

FIRST COURSE IN RINGS AND IDEALS

Example 11-6. Consider the ring R = mapR# ofreal-valued funetions on R #. Given im arbitrary real number r > O, we define Ir

= {fE RI!(x)

O for -r

:5; X :5;

r}.

Then Ir is an ideal of R and it is not difficult to seethat , ... c:

13 c:: 12

c:

I1

C;

225

In the sequel, ther~ occur eertain results which hold for both Noetherian and Artinian rings. Where the proofs are virtually the same, our policy will be to establish the theoremin question only in the Noetherian case. Let us first show that the chain conditions are not deStroyed by homomorphisms. ' Theorem 11-6. If R is a Noetherian (Artinian) ring, then any homomorphic image of R is also Noetherian (Artinian) .. ,!:

I 1/ 2 c: I 1 / 3 c: ....

)rhe iQlplication lS that R eontains ascending and descending chains that do' :)ot becomestationary, whenee R is neither Artinian nor Noetherian. It~'i¡ 'ís perhaps appropriate to call attention to the ,fact that eaeh of the ideals";: 'Jr is properly contained in the maximal ideal M = {fE Rlf(O) = O}. ""

not':¡

: "Example 11-7. We next give' an example oC an Artinian ring which is ,Noetherian. For this"purpose, let p be a fixed prime. Consider the group~"A p::;Z(pOO) of all rational 1'l,umbers r between O and 1 of the form r = m/P~;,:} (y..here m is an arbitrary integer and n runs through the nonnegative integers':,~:\ '<':'"~nder the operation of addhion modulo 1: : : ¡ \ ¡ ¡

Z(pOO) = {mjpnlO :s; m < pn; mEZ; n = 0,1,2, ... }. ' We make Z(pOO) into a ring (without identity) by defining the produet ab to be zero forall a, bE Z(p""). It is important to observe that the ideals of the resulting ring are simply the subgroups of the additive group of Z(pOO). Now, let I be any nontrivial ideal of Z(pOO) and choose kto be the smaIlest positive integer such that for some a, ajpk 1 I; we implicitly assume that a and p are relatively prime. Then I must contain aH the elements 0, 1jl~ 1, 2jpk-1, ... ,(pk-l l)jpk~l. Our contention is that these are theonly members of l. To support tbis, suppose to the contrary that bjpi E I, where i ¿ k and, of course, b and p are relatively prime. One can then find integers r, s for which rb + sp = 1. Sinceboth the rational numbers (reduced modulo 1)

ProoJ. Let f be a homomorphism of the Noetherían ti,p.g R onto the ring , R' and consider any aseending chain 1'1 s; 1; s; ... s;:J~ s; ... of ideals of R '. Put Ik = f-l(Ik), for, k = 1,2,.... Then 1 1 s;t;i. s; ... s; In S; ...

forms an ascending ehain of ideals of R which, accordil1 gto our hypothesis, must eventually be constant; that is, there is sorne index:f¡ such tha t I m = I n for all m ~ n. Takíng stock of the fact that fis an ootó mapping, we have , f(I k ) = I~. Hence, l~ = I~, whenever m ~ n, so tbat\Jhe original chain ';'. ::: also stabilizes at some point. . J

Lettingfbe the natural mapping, we have as a

co~61Iary:

Corollary. If lis an ideal of the Noetherian (Artinian) ring R, then the quotient ring Rj1 is Noetherian (Artiniílll). Further progress will be facilitated by the technical lemma below. Lernma. Ir 1, 1, and K are ideal s of a ring R such that (1) 1 ~ K, (2) 1 n I = K nI, and then 1 K. .

(3) Jj1 = Kj1,

ProoJ. EvidentIy, we need only establish the inclusion K S; 1. To this purpose, seleet any, member k of K. On the basís of (3), thereexists an elementj El for whichj + 1 = k + 1, which signifies that k - j = i for some choice of i in l. But, since 1 S; K, the difference k - j also líes in the ideal K. Using eondition (2), we thus find that

and He in I, it follows that (rb + Sp)jpk 1jpk also belongs to 1, contradicting therninimaJity of k. Thus, the ideal 1 is finite and is given by

•

i = k - j

and, in consequence, k

=

i

+j

E

E

1n K

=

J n I,

1.

This faet ü¡ enough to enableus to prove a partíal converse of the last coróllary. ' . Representing 1 by the symbol1 k _ 1, we eonclude that the only ideals of Z(pOO) are those which appear in the chain

{O} e 11 c: 12

c: ... e

1k c: ... e Z(paJ).

Therefore, Z(p"") possesses an infinite (strict1y) ascending chain of ideals, but any descending chain is of finite length.

Theorem 11-7. Let 1 be an ideal of the ring R. If 1 and R/ I are both Noetherian (Artinian) rings, then R is also Noetherian (Artinian).

ProoJ. To begin, let 1 1

~ 1 2 s;;; .. , S;;; 1 n S;;; ... be any ascending chain of ideals of R. Fromthis, we may eonstruct a chain ofideals of 1,

11 n 1

~

12 n 1

~

'" s;;; ln nI

S;;; ... ,

226

FIRST COURSE IN RINGS AND IDEALS

RINGS WITH CHAIN CONDITIONS

as well as a chain of ideals of the quotient ring Rll, where Since we are told that 1 and Rll are both Noetherian, each of these chains becomes stationary from sorne point on; say after r and s steps, respectively. Now, take the integer n to be the larger of r and s, so that

227

One can say considerably more about the ideal structure of an Artinian ring R: R has only a finite number of prime (hence, .maximal) ideals. For, suppose that there exists an infinite sequence {P¡} of distinct proper prime ideals of R. We would then be able to form a descending chain of ideals PI

~

P IP 2

~

P I P 2P 3

~ ....

Since R is Artinian, there exists a positive integer n for which J", nI = J n n 1

and

for alI m ~ n. This being the case, an invocation ofthe lemma is permissible; it folIows that J m = J n whenever m ~ n, whence R comprises a Noetherian ringo Artinian rings are generalIy more restrictive than Noetherian rings; for instance, the only integral domains which satisfy the descending chain condition are fields (this 1S not to suggest, however, that Artinian rings are without interest). Theorem 11-8. Any Artinian domain R (integral domain and Artinian

ring) is a field.

PIP2

Pn = P I P 2 ... PnP n+ l • It folIows from this that P I P 2 '" Pn S; Pn+ l , whence P k S; P n+ 1 for sorne k ::; n. But P k is a maximal ideal of R, so that we must have P k = Pn+l' ...

contrary to the fact that the Pi are distinct. These observations are summarized as Theorem 11-10. Every Artinian ring has only a finite number of proper prime ideals, each of which is maximal.

We now come to the interesting part ofthe theory; namely, the extension of Levitsky's' Theorem to Artinian rings. Theorem 11-11. If R is an Artinian ring, then rad R forms a nilpotent

Proof. It obviously suffices to show that each nonzero element of R has a multiplicative in verse. Thus, suppose that a =1= O in R and consider the descending chain of ideals (a) ~ (a 2) ~ {a 3 ) ~

:~

rad R

=

(a n + 2)

= ....

.~'

Then there exists an element r E Rs~ch that an = ra n + l ; using the cancelIation law, it folIows that 1 = ra, whjch, pro ves our assertion.

Corollary. An integral domain 'with only a finite number of ideal s is a field. With the aid of this result, we cannow prove that in the presence of the descending chain condition the Jacobson radical and prime radical coincide. Theorem 11-9. If R is an Artinian ring, then every proper prime ideal of R is a maximal ideal. .

Proof. Suppose that 1 is a proper prime ideal of R. Then, the quotient ring Rll forms an integral domain which satisfies the descending chain condition because R does. It folIows from Theorem 11-8 that Rll must be a field, whence 1 is a maximal ideal of R.

CorolJary. In any Artinian ring R, rad R

Proof. The descending chain condition applied to the chain

....

By the descending chain condition, this chain must be of finite length, say (a) ~ (a 2) ~ ... ~ (anfe:: (a n + l )

ideal.

=

Rad R.

~

(rad R)2

~

(rad R)3

~

...

shows that there exists an integer n for which (rad R)" = (rad R)n+r" If we put 1 = (rad R)", then 1 S; rad R and 12 = 1. Our contention is that

1

=

{O}.

.

Assume-{qr. the moment that 1 =1= {O} and consider the family of alI ideal s J of R such that (i) J S; 1 and (ii) JI =f {O}. This colIection is not empty since jt "contains 1 and, hence, it admits a minimal member K. By (ii), Kl =f {O},~so that al =f {O} for sorne nonzeroelement a E K. Thus, (al)l = aJ2 = al

=f {O},

with al S; K 's; 1; hence, al = K by the minimality of K. This being the case, there exists an element bE 1 such that ab = a. But bE 1 S; rad R, which implies that 1 - b must be an invertible element of R (Theorem 8-2); in other words, (1 - b)c = 1 for suitable e E R. We then have

= a(1 - b)c = (a - ab)c = O, contradicting the fact that al =f {O}. This contradiction signifies that 1 = (rad R)n = {O}, as asserted. a

With littIe additional effort we can learn a good de al more about nilpotent ideals in rings with the descending chain condition.

226

FIRST COURSE IN RINGS AND IDEALS

RINGS WITH CHAIN CONDITIONS

as well as a chain of ideals of the quotient ring Rll, where Since we are told that 1 and Rll are both Noetherian, each of these chains becomes stationary from sorne point on; say after r and s steps, respectively. Now, take the integer n to be the larger of r and s, so that

227

One can say considerably more about the ideal structure of an Artinian ring R: R has only a finite number of prime (hence, .maximal) ideals. For, suppose that there exists an infinite sequence {P¡} of distinct proper prime ideals of R. We would then be able to form a descending chain of ideals PI

~

P IP 2

~

P I P 2P 3

~ ....

Since R is Artinian, there exists a positive integer n for which J", nI = J n n 1

and

for alI m ~ n. This being the case, an invocation ofthe lemma is permissible; it folIows that J m = J n whenever m ~ n, whence R comprises a Noetherian ringo Artinian rings are generalIy more restrictive than Noetherian rings; for instance, the only integral domains which satisfy the descending chain condition are fields (this 1S not to suggest, however, that Artinian rings are without interest). Theorem 11-8. Any Artinian domain R (integral domain and Artinian

ring) is a field.

PIP2

Pn = P I P 2 ... PnP n+ l • It folIows from this that P I P 2 '" Pn S; Pn+ l , whence P k S; P n+ 1 for sorne k ::; n. But P k is a maximal ideal of R, so that we must have P k = Pn+l' ...

contrary to the fact that the Pi are distinct. These observations are summarized as Theorem 11-10. Every Artinian ring has only a finite number of proper prime ideals, each of which is maximal.

We now come to the interesting part ofthe theory; namely, the extension of Levitsky's' Theorem to Artinian rings. Theorem 11-11. If R is an Artinian ring, then rad R forms a nilpotent

Proof. It obviously suffices to show that each nonzero element of R has a multiplicative in verse. Thus, suppose that a =1= O in R and consider the descending chain of ideals (a) ~ (a 2) ~ {a 3 ) ~

:~

rad R

=

(a n + 2)

= ....

.~'

Then there exists an element r E Rs~ch that an = ra n + l ; using the cancelIation law, it folIows that 1 = ra, whjch, pro ves our assertion.

Corollary. An integral domain 'with only a finite number of ideal s is a field. With the aid of this result, we cannow prove that in the presence of the descending chain condition the Jacobson radical and prime radical coincide. Theorem 11-9. If R is an Artinian ring, then every proper prime ideal of R is a maximal ideal. .

Proof. Suppose that 1 is a proper prime ideal of R. Then, the quotient ring Rll forms an integral domain which satisfies the descending chain condition because R does. It folIows from Theorem 11-8 that Rll must be a field, whence 1 is a maximal ideal of R.

CorolJary. In any Artinian ring R, rad R

Proof. The descending chain condition applied to the chain

....

By the descending chain condition, this chain must be of finite length, say (a) ~ (a 2) ~ ... ~ (anfe:: (a n + l )

ideal.

=

Rad R.

~

(rad R)2

~

(rad R)3

~

...

shows that there exists an integer n for which (rad R)" = (rad R)n+r" If we put 1 = (rad R)", then 1 S; rad R and 12 = 1. Our contention is that

1

=

{O}.

.

Assume-{qr. the moment that 1 =1= {O} and consider the family of alI ideal s J of R such that (i) J S; 1 and (ii) JI =f {O}. This colIection is not empty since jt "contains 1 and, hence, it admits a minimal member K. By (ii), Kl =f {O},~so that al =f {O} for sorne nonzeroelement a E K. Thus, (al)l = aJ2 = al

=f {O},

with al S; K 's; 1; hence, al = K by the minimality of K. This being the case, there exists an element bE 1 such that ab = a. But bE 1 S; rad R, which implies that 1 - b must be an invertible element of R (Theorem 8-2); in other words, (1 - b)c = 1 for suitable e E R. We then have

= a(1 - b)c = (a - ab)c = O, contradicting the fact that al =f {O}. This contradiction signifies that 1 = (rad R)n = {O}, as asserted. a

With littIe additional effort we can learn a good de al more about nilpotent ideals in rings with the descending chain condition.

T 228

FIRST COURSE IN RINGS ANO IOEALS,

RINGS WITH CHAIN CONOlTlONS '

Corollary. In any Artinian ring R, the following hold:

229

Consider the ideals lk = ann (a k ); c1early, we have

1) rad R Is the largest nilpotent ideal of R. 2) Every nil ideal of R is nilpotent. 3) Rlrad R contains no nonzero nilpotent ideals. Proa! By Theorem 8-8, any nilpotent ideal of R is contained in the prime radical a.pd this coincides with rad R. Conceming (2), each nil ideal is contained in Rad R = rad R, which is a nilpotent idea!. The final assertion follows from the fact that Rlrad R is a semisimple~rtinian ringo ,

",

Now, x" is not in In' but a(q"x n}' = ar v = 0, since a Ís a zero divisor of R. Therefore, x n lies in ln+1 and the lk form a properly ascending ch,ain. This contradicts the ascending ch~n condition and no such element a exists. , I

I

Remark. Over the course oC th,e next several pages, we shall often simply say "the set ofzero divisors':of R form anideal" when what is really meant is "the set of zero divisors,í()gether with zero, form an ideal".

The next theorem is perhaps of secondary intei~st, but it affords us an opportunity to discuss subdirectly irreducible rings'again. The reader wjll recall that these are rings R possessing a smallest nónzero ideal R v (the heart of R). Clearly, R v is a principal ideal generate_d byany of its elements, other than zero. We shaIl, in the proofbeJow, let rY',designate a fixed nonzero element of R v, so that R v = (r v ) " , ; ¡ Observe also that for any nonzero element a~;R, (a) IS a nonzero ideal of R, and, hence, must contain R v; thus, there exists an element x in R such that ax = rV:. The only other fact which we wiII require is tha! the annihilator of the set of zerodivisors of Ris precisely the ideal R v = (r

Proa! Suppose that a andp~~e both zero divisors ofR. Then ax = O = by for sorne nonzero x, y in R:!t:Inasmuch as the principal ideals (x)and' (y) have nonzero intersectioú;_ th~re also exist elements u, v E R such that xu = yv =F O. But then,¡;

Theor:em 11-12. If R is a subdirectly irreducible ring satisfying either chain condition, then every zero divisor of R is nilpotent (that is, R is a primary rin~).

in consequence of which a - b is a divisor of zero. Certainly, the product ra \viII b'e a zero divisor for any choice of r E R. The implícation is that the set of aIl zero divisors of R constitute an ideal (indeed, this is true in any subdirectly irreducible ring); by the theorem, such -an ideal must be ni!.

Corollary. If R is a subdirectIy irreducible ring satisfying either chain condition, then the set,Rl,zero divisors of R form a nil ideal:

(a - b)xu

= -bxu = -byv ,,;, O,

V

).

Proa! In the first place, we take R to be Artinian. Suppose further that, R is a zero divisor which is not nilpotent and consider the descending chain of principal ideals aE

,'.,

.

........

.

.

By assumption, none of these is the zero ideal and, because of the minimum cOridition, we must have (a n) = (a n + 1 ) for sorne n E Z+. This being the case, a n = ra n+ 1 or a"(1 - ra) = O, with rE R. Inasmuch as a" =F O, the expression in parentheses is a zero divisor of R and, hence, lies in ann R v by Theoreml0-9, Thus,foranynonzero elementx E R V, wehavex(l - ra) = O. But xa =- O, since a also belongs to ann R v , and SO x = O. This contradiction forces the element a to be nilpotent, as desired. We next extend the stated result to rings with the ascending chain condition. As in the previous paragraph, suppose that the element a is a zero divisor of R which is oot nilpotent. Then all the powers a 2 , a 3 , .. , , an, .,. are zero divisors and, of course, none is zero. Thus, for every power a", there exists an element XII such that aX 1

=

a2x 2

= ... =

anx"

= .,. =

r V =F O.

'In the líght ofthe corolIary aboye, it would appear natural to study rings whose zero divisors form an ideal which is contained in the Jacobson radical (we point out that this condition holds trivially in any integral domain). Our next two results pre,sent criteria for these rings to become local rings, Theorem 11-13. Let 1 be an ideal of the ring R with 1 R is a local ring if and only if RIl is a local ringo

S;

rad R.

Then-

Proa! One direction is fairly obvious, since the homomorphic image of a local ring is necessarily local. Going the other way, suppose that RII is local and let a + 1 be any invertible element of RIl. Then ax + 1 = 1 + 1 for sorne x in R, or, equivalently, ax = 1 + r with r E 1, Since the ideal 1 S; rad R, Theorem 8-2 tells us ihat ax is an in vertible element of R., But then a itself will possess an inverse in R Now, let a and b be two non-in vertible elements of R. The reasoning of the previous paragraph shows that the cosets a + J and b + 1 lack inverses in RIl. Since the quotient ring RIl constitutes a local ring, their sum (a + 1) + (b + 1) = a + b + 1 is again a non-in vertible element' (Problem 8, Chapter 8). This means that a + b fails to have an in verse in R, forcing'R to be a localring.

T 228

FIRST COURSE IN RINGS ANO IOEALS,

RINGS WITH CHAIN CONOlTlONS '

Corollary. In any Artinian ring R, the following hold:

229

Consider the ideals lk = ann (a k ); c1early, we have

1) rad R Is the largest nilpotent ideal of R. 2) Every nil ideal of R is nilpotent. 3) Rlrad R contains no nonzero nilpotent ideals. Proa! By Theorem 8-8, any nilpotent ideal of R is contained in the prime radical a.pd this coincides with rad R. Conceming (2), each nil ideal is contained in Rad R = rad R, which is a nilpotent idea!. The final assertion follows from the fact that Rlrad R is a semisimple~rtinian ringo ,

",

Now, x" is not in In' but a(q"x n}' = ar v = 0, since a Ís a zero divisor of R. Therefore, x n lies in ln+1 and the lk form a properly ascending ch,ain. This contradicts the ascending ch~n condition and no such element a exists. , I

I

Remark. Over the course oC th,e next several pages, we shall often simply say "the set ofzero divisors':of R form anideal" when what is really meant is "the set of zero divisors,í()gether with zero, form an ideal".

The next theorem is perhaps of secondary intei~st, but it affords us an opportunity to discuss subdirectly irreducible rings'again. The reader wjll recall that these are rings R possessing a smallest nónzero ideal R v (the heart of R). Clearly, R v is a principal ideal generate_d byany of its elements, other than zero. We shaIl, in the proofbeJow, let rY',designate a fixed nonzero element of R v, so that R v = (r v ) " , ; ¡ Observe also that for any nonzero element a~;R, (a) IS a nonzero ideal of R, and, hence, must contain R v; thus, there exists an element x in R such that ax = rV:. The only other fact which we wiII require is tha! the annihilator of the set of zerodivisors of Ris precisely the ideal R v = (r

Proa! Suppose that a andp~~e both zero divisors ofR. Then ax = O = by for sorne nonzero x, y in R:!t:Inasmuch as the principal ideals (x)and' (y) have nonzero intersectioú;_ th~re also exist elements u, v E R such that xu = yv =F O. But then,¡;

Theor:em 11-12. If R is a subdirectly irreducible ring satisfying either chain condition, then every zero divisor of R is nilpotent (that is, R is a primary rin~).

in consequence of which a - b is a divisor of zero. Certainly, the product ra \viII b'e a zero divisor for any choice of r E R. The implícation is that the set of aIl zero divisors of R constitute an ideal (indeed, this is true in any subdirectly irreducible ring); by the theorem, such -an ideal must be ni!.

Corollary. If R is a subdirectIy irreducible ring satisfying either chain condition, then the set,Rl,zero divisors of R form a nil ideal:

(a - b)xu

= -bxu = -byv ,,;, O,

V

).

Proa! In the first place, we take R to be Artinian. Suppose further that, R is a zero divisor which is not nilpotent and consider the descending chain of principal ideals aE

,'.,

.

........

.

.

By assumption, none of these is the zero ideal and, because of the minimum cOridition, we must have (a n) = (a n + 1 ) for sorne n E Z+. This being the case, a n = ra n+ 1 or a"(1 - ra) = O, with rE R. Inasmuch as a" =F O, the expression in parentheses is a zero divisor of R and, hence, lies in ann R v by Theoreml0-9, Thus,foranynonzero elementx E R V, wehavex(l - ra) = O. But xa =- O, since a also belongs to ann R v , and SO x = O. This contradiction forces the element a to be nilpotent, as desired. We next extend the stated result to rings with the ascending chain condition. As in the previous paragraph, suppose that the element a is a zero divisor of R which is oot nilpotent. Then all the powers a 2 , a 3 , .. , , an, .,. are zero divisors and, of course, none is zero. Thus, for every power a", there exists an element XII such that aX 1

=

a2x 2

= ... =

anx"

= .,. =

r V =F O.

'In the líght ofthe corolIary aboye, it would appear natural to study rings whose zero divisors form an ideal which is contained in the Jacobson radical (we point out that this condition holds trivially in any integral domain). Our next two results pre,sent criteria for these rings to become local rings, Theorem 11-13. Let 1 be an ideal of the ring R with 1 R is a local ring if and only if RIl is a local ringo

S;

rad R.

Then-

Proa! One direction is fairly obvious, since the homomorphic image of a local ring is necessarily local. Going the other way, suppose that RII is local and let a + 1 be any invertible element of RIl. Then ax + 1 = 1 + 1 for sorne x in R, or, equivalently, ax = 1 + r with r E 1, Since the ideal 1 S; rad R, Theorem 8-2 tells us ihat ax is an in vertible element of R., But then a itself will possess an inverse in R Now, let a and b be two non-in vertible elements of R. The reasoning of the previous paragraph shows that the cosets a + J and b + 1 lack inverses in RIl. Since the quotient ring RIl constitutes a local ring, their sum (a + 1) + (b + 1) = a + b + 1 is again a non-in vertible element' (Problem 8, Chapter 8). This means that a + b fails to have an in verse in R, forcing'R to be a localring.

230

FIRST COURSE IN RINGS AND IDEALS

RINGS WITH CHAl N CONDITIONS

Coro]]ary. Let R be a ring in which the set of zero divisors of R forms an ideal D, with D S; rad R, Then R is a local ring if and only if R/D is a local ringo Theorem 11-14. Let R be a ring which satisfiesthe following conditions: 1) rad R is a nonzero prime ideal of R; 2) al~ the ideals containing rad R are principal; 3) the set of zero divisors D S; rad R. Then R forms a local ringo

natural mapping nat Ai,: R ..... R/Mi induces an isomorphism 1i ~ R/Mi' making 1i a fieldo o o By virtue ofthe relation R = 1i + Mi' one can wnte, for each element aER:

a

=

Xi

+

Xi E1 i , YiEMi

Yi'

(i

=

1, 2, ooo, n)o

Given any integer k between 1 and n, a -.:.

Proof With reference to Problem 8, Chapter 8, it suffices to show that rad R coincides with the set of all noninvertible elements of R. For this purpose, let us suppose that rad R = (x) and choose an arbitrary a rj rad R; the strategy is to show tbat a has an inversé. Now, our hypothesis signifies that the ideal (rad R, a)' must be principal; say (rad R, a) = (b), where b rj rad R. Thus, x = by for sorne choice of y E R. Since rad R is a prime ideal, a further deduction is that y Erad R. Knowing this, we can write y = ex with e E R. But then x = by = bex, or x(1 - be) = O, which implies that 1 - be lies in D S; rad Ro Falling back on Theorem 8-2, the product be and, in turn, the element b are necessarily invertible in R. Accordingly, the ideal (rad R, a) = R. It follows that 1 - ra Erad R for suitable r E R, making ra an invertible element. From tbis we may conclude that a itself possesses an inverse in R, as desired. Coro]]ary. Let R be a principal ideal ring with D S; rad R. is a local ring whenever rad R is á nonzero prime ideal.

231

Then R .~~ ! •.

Although the hypotheses of Theorem 11-14 appear somewhát formidable, it is worth remarking that the power series ring F[[xJ] (F a fi~l~) satisfies the requisite conditionso ' . \¡Our next goal is to describe semisimple Artinian rings; crucialtd tli'e discussion is the fact that such rings have only a finite number of max~i:nal ideal s with zero intersectiono The theorembelow is the cornmutative veision of Wedderburn's fundamental result (Theorem 13-3)0

Th~orem 11-15. (Wedderburn). Any semisimple Artinian ring R is;íhe direct sum of a finite number of fieldso Proof Since R has only a finite number of maximal ideals, we may assume that if any one of these ideals is omitted the intersection of the others is different from zero. (If this is not the case, a set with the desired property may be obtained by simply deleting certain idealso) Accordingly, there exist maximal ideals M l' M 2' ooo, M n of R such that n Mi = {O}, but the ideals 1i = nk'f'iMk =1= {O}foreveryio InasmuchasMiisniaximal,R = 1i + Mi; moreover, 1i n Mi = {O}, so that this sum is actually directo Hence, the

since a - x k

=

f

i= 1

Xi = (a - Xk) -

Yk E M k, while Xi E 1i

a -

S;

¿ Xi E n

¿ Xi E M k,

i'f'k

M k for i

=1=

ko Thus,

M k = {O}

and so a = ¿ xio This implies that the ring R. m~y be represented as R = 1 1 + 12 + '0' + 1n ;thattheforegoingsumlsdlrectfollowsfromthe fact that ¿i'f'k 1i ~ M k, whence 1k n

(¿

i'f'k

1i ) ~ 1k n M k = {O}o

It is of interest to compare Theorems 10-2 and 11-150 We have exchanged the subdirect sum part ofTheorem 10-2 for a direct sum (in fact, a finite direct sum) in Theorem 11-15; however, Theorem 11-15 :vas 0 obtained at the cost of.an additional hypothesis: the ring must be Artlm~no o We offer a second proof of Wedderburn's Theorem, the relatlve ments of which can be weighed by the reader; although ,othis second prooof is clearly less complicated than the first, it nonetheless ~elies more heavIly on the '. results of the previous chapter.

Seeond proof ofTheorem 11-150 Let MI' M 2' ooo';!lfn be the ~axin:"al ideals of R (there is no harm in assuming that {O} i~ ~o~ a ~axlmal lde~l, for otherwÍse the theorem follows trivially)o The m~IJ(lmahty of these ldea~s implies that Mi + Mj = R whenever i =1= j. :r;hus, by Theorem 10-5, lt follows that R = R/n Mi ~

¿ EB> (R/M¡)

(complete direct sum),

where each ofthe quotient rings R/Mi is a fiel~ (i = 1,2, o.. ,n): But, in the finite case, the complete direct sum coincides with the usual dlfect sumo Notice that, in carrying out the aboye argument, we have proved a o subresult which is interesting in its own right: If a ring R has a fimte number of maximal ideals M. with zero intersection, then R ~ ¿ (f) (R/M¡). There is a coroll~ry to Theorem 11-15 which is worthy of emphasiso Coro]]ary. Any semisimple Artinian ring is Noetheriano

230

FIRST COURSE IN RINGS AND IDEALS

RINGS WITH CHAl N CONDITIONS

Coro]]ary. Let R be a ring in which the set of zero divisors of R forms an ideal D, with D S; rad R, Then R is a local ring if and only if R/D is a local ringo Theorem 11-14. Let R be a ring which satisfiesthe following conditions: 1) rad R is a nonzero prime ideal of R; 2) al~ the ideals containing rad R are principal; 3) the set of zero divisors D S; rad R. Then R forms a local ringo

natural mapping nat Ai,: R ..... R/Mi induces an isomorphism 1i ~ R/Mi' making 1i a fieldo o o By virtue ofthe relation R = 1i + Mi' one can wnte, for each element aER:

a

=

Xi

+

Xi E1 i , YiEMi

Yi'

(i

=

1, 2, ooo, n)o

Given any integer k between 1 and n, a -.:.

Proof With reference to Problem 8, Chapter 8, it suffices to show that rad R coincides with the set of all noninvertible elements of R. For this purpose, let us suppose that rad R = (x) and choose an arbitrary a rj rad R; the strategy is to show tbat a has an inversé. Now, our hypothesis signifies that the ideal (rad R, a)' must be principal; say (rad R, a) = (b), where b rj rad R. Thus, x = by for sorne choice of y E R. Since rad R is a prime ideal, a further deduction is that y Erad R. Knowing this, we can write y = ex with e E R. But then x = by = bex, or x(1 - be) = O, which implies that 1 - be lies in D S; rad Ro Falling back on Theorem 8-2, the product be and, in turn, the element b are necessarily invertible in R. Accordingly, the ideal (rad R, a) = R. It follows that 1 - ra Erad R for suitable r E R, making ra an invertible element. From tbis we may conclude that a itself possesses an inverse in R, as desired. Coro]]ary. Let R be a principal ideal ring with D S; rad R. is a local ring whenever rad R is á nonzero prime ideal.

231

Then R .~~ ! •.

Although the hypotheses of Theorem 11-14 appear somewhát formidable, it is worth remarking that the power series ring F[[xJ] (F a fi~l~) satisfies the requisite conditionso ' . \¡Our next goal is to describe semisimple Artinian rings; crucialtd tli'e discussion is the fact that such rings have only a finite number of max~i:nal ideal s with zero intersectiono The theorembelow is the cornmutative veision of Wedderburn's fundamental result (Theorem 13-3)0

Th~orem 11-15. (Wedderburn). Any semisimple Artinian ring R is;íhe direct sum of a finite number of fieldso Proof Since R has only a finite number of maximal ideals, we may assume that if any one of these ideals is omitted the intersection of the others is different from zero. (If this is not the case, a set with the desired property may be obtained by simply deleting certain idealso) Accordingly, there exist maximal ideals M l' M 2' ooo, M n of R such that n Mi = {O}, but the ideals 1i = nk'f'iMk =1= {O}foreveryio InasmuchasMiisniaximal,R = 1i + Mi; moreover, 1i n Mi = {O}, so that this sum is actually directo Hence, the

since a - x k

=

f

i= 1

Xi = (a - Xk) -

Yk E M k, while Xi E 1i

a -

S;

¿ Xi E n

¿ Xi E M k,

i'f'k

M k for i

=1=

ko Thus,

M k = {O}

and so a = ¿ xio This implies that the ring R. m~y be represented as R = 1 1 + 12 + '0' + 1n ;thattheforegoingsumlsdlrectfollowsfromthe fact that ¿i'f'k 1i ~ M k, whence 1k n

(¿

i'f'k

1i ) ~ 1k n M k = {O}o

It is of interest to compare Theorems 10-2 and 11-150 We have exchanged the subdirect sum part ofTheorem 10-2 for a direct sum (in fact, a finite direct sum) in Theorem 11-15; however, Theorem 11-15 :vas 0 obtained at the cost of.an additional hypothesis: the ring must be Artlm~no o We offer a second proof of Wedderburn's Theorem, the relatlve ments of which can be weighed by the reader; although ,othis second prooof is clearly less complicated than the first, it nonetheless ~elies more heavIly on the '. results of the previous chapter.

Seeond proof ofTheorem 11-150 Let MI' M 2' ooo';!lfn be the ~axin:"al ideals of R (there is no harm in assuming that {O} i~ ~o~ a ~axlmal lde~l, for otherwÍse the theorem follows trivially)o The m~IJ(lmahty of these ldea~s implies that Mi + Mj = R whenever i =1= j. :r;hus, by Theorem 10-5, lt follows that R = R/n Mi ~

¿ EB> (R/M¡)

(complete direct sum),

where each ofthe quotient rings R/Mi is a fiel~ (i = 1,2, o.. ,n): But, in the finite case, the complete direct sum coincides with the usual dlfect sumo Notice that, in carrying out the aboye argument, we have proved a o subresult which is interesting in its own right: If a ring R has a fimte number of maximal ideals M. with zero intersection, then R ~ ¿ (f) (R/M¡). There is a coroll~ry to Theorem 11-15 which is worthy of emphasiso Coro]]ary. Any semisimple Artinian ring is Noetheriano

232

PROBLEMS

FIRST COURSE IN RINOS AND IDEALS

Proof. In ~onjunction with Theorem

233

9. Le! p be a fixed prime number and put.

11-15, one needs only the faet that a

finite direet suin ofNoetherian rings (in this case, fields) is again Noetherian.

Qp = {mfp~lmE2; n = 0,1,2, ... }.

We shalI see later that the imposed semisiinplicity condition is unnecessaríly stringent; indeed, the foregoing result can be sharpened to read that everycommutative Artinian ring with identity lS Noetherian. The ring of integers shows that the converse need not hold.

In Qp, define addition lo be ordinary addition of ralional numbers and mllltiplication to be the trivial multiplication (i.e. ab = O for alI a, b E Qp). Establish that a) Z forms an ideal ofthe resulting ring Qp; b) the quotient ring Qp/Z is isomorphic to 2(p
¿

. , 10. a) Prove that a finite direct sum ® R¡ js Noetheí:ian (Artinian) if and only' if each of the component rings R¡ is Noetherian (Artinían). [Hi/lt: Ifn = 2, sáy R = Rl E9 R 2 , then R/Rl c:: R 2 .] . b) LetR bearinghavingllfinitenumberofidealsl¡, 12, ... , lnsuch that n li = {Ó}'; If each of the quotient rings R/I ¡ is Noetherian (Artinian), show !hat R ís also' Noetherian (Artinian). ' ,,;;:

PROBLEMS In all problems, R is, ~conm1Utative ring with identity. l. Le! 1 be a nonzéió.ideal of !he principal ideal domaín R Prove that the quotlerit ring R/l satisfie~' both cPain conditions.

" 11. In an Artinían ring R, prove that the zero ideal is a prodllct of maximal ideáIs.: , [Hint: rad R = M 1 n M 2 n ... n'M., where each Mi ís maxhllal; use Theoré~;' 11-11 and Problern 13, Chapter 10, to conclude {O} (rad R)k MfM~ .. , M~"

2. Prove that everyJi,omomorphismfofa Noetberian ring R ontoitselfis necessarily one-to-one. [lfiilfi Consider the ascending chain {O} ~ kerf~ kerf2 ~ ... of ideals of R]" \'

. .

b~~~~

;

.JI/l

3. a) Ir 1 is an ideal of the Noetherian rine; R, show that forms a rulpotent ideal of R/l. ' , b) Let 1 and J be two ideals of tbe Noetherian ring R. Establish that 1" ~ J for ~ .jI . , some integer n E 2 + if and only if

12. Establish that an Artinian ring R is isomorphic to a finite diiect sum of Artinian local rings. [Hint: From Problem 11 and the fact!hat the ideals M} arecomaximal, we have M~ n ... n M~ ... M~ '" M! = {O}. By Theorem 10-1, R c:: ¿ ® RIM~.

4. Prove that every id~al of a NO,etherian ringR contains a product of prinle ideals. [Hint: If not, let S be the set of those ideals of R which do not contain a product of prime idealsand apply ,the maximllm condition.] .

13. Prove that any semisinlple Artinian ring possesses only a finite number of ideals. [Hlnt: Assume that R admits that decomposition R = F1 ® F2 ® ... ® F., F¡ a fie1d; if 1 is an ideiÚ, of R, then 1 = 11 ® ... ® In witb li an ideal of Fd

5. a)Obtain the converse of the Hilbert Basis Theorem: if R[x] is a Noetherian , ring, then, so also ís R ' , b) Verify that the power series ring R[[x]] is Noetherian if and only if R is a Noetherian ringo [Hint: Mimic Hilbert's Theorem, now using elements of lowest order when defining the ideals Id e)Let R' be an extension ring of !he Noetherian ring R. For a fixed elernent r E R', show tliat the ring

14. a) Let 1 be a nontrívial mínimal ideal of the Artinian ring R Show that the annihilator ai:m 1 forms a prime and, hence, maxinlal, ideal of R [Hint: If a ~ ann 1, al ~ l.] b) Assume tbat 1 is a nonzero ideal of the ring R (no chain conditions). If Pis a maxinlal member of the eollection {ann (x)IO -+ x E 1), d~uc~ thatP is a prime ideal. [Hlnt: Let abEP= ann(r), with a~P; then, P 5;; (P,b) ~ ann(ar).]

.JI

Now use Problem 5-18.] .

<;

15. A ring R is temled divisible if every non-zero-divisor of R is invertible.

R[r] == (f(r)jf(x) e R[x]} is Noetberian. 6. Prove that if R is a Noetherian ring, then the matrix ring M.(R) is a!so Noetherian. [Hint " Problem 28, Chapter 2.] 7. Assuming that R is a Noetherian ring, establish that a) Rad R is the sum of al1 the nilpotent ideals of R; b) tbe quotient ring R/Rad R has no nonzero nilpotent ideals. 8. Let R be a ring wíth at least one non-zero-divisor. Prove that ir R is Noetherian, then itsclassical ringofquotientsQ'I(R)is also Noetherian. [Hint: If J 1 ~ J 1 ~ ... is an ascending chain of ideals of Qcl(R), then, by Problem 29, Chapter 4, 11 ~ 12 S .. , forms an ascending ehain of ideals of R, where 1k "" J k n R.]

Assuming ' that R is a divisible ring prove the following: a) R is a local ring if and only jf the set D of a1l zero divisors (together with zero) is included in a proper ideal of R; in this case, D itself becomes an ideal. b) If 11 n 12 {O} for any two nonzero ideals of R, then R ís a local ringo [Hint: Show !hat the sum pf two noninvertible elenlents of R is again noninvertible.]

+.

,

16. Let R be a principal ideal ring which is not an integral domain. If!he set of al! zero divisors D = rad R, verify that R is a local ring, 17. a) If R is a finite Boolean ring, prove that R is isomorphic to the direct sum of a finite number of fields 2 2 , [Hint: See the remark following the second proof ofTheorem 11-15.] b) Prove that a finite Boolean ring has 2" elements for some n E Z+.

232

PROBLEMS

FIRST COURSE IN RINOS AND IDEALS

Proof. In ~onjunction with Theorem

233

9. Le! p be a fixed prime number and put.

11-15, one needs only the faet that a

finite direet suin ofNoetherian rings (in this case, fields) is again Noetherian.

Qp = {mfp~lmE2; n = 0,1,2, ... }.

We shalI see later that the imposed semisiinplicity condition is unnecessaríly stringent; indeed, the foregoing result can be sharpened to read that everycommutative Artinian ring with identity lS Noetherian. The ring of integers shows that the converse need not hold.

In Qp, define addition lo be ordinary addition of ralional numbers and mllltiplication to be the trivial multiplication (i.e. ab = O for alI a, b E Qp). Establish that a) Z forms an ideal ofthe resulting ring Qp; b) the quotient ring Qp/Z is isomorphic to 2(p
¿

. , 10. a) Prove that a finite direct sum ® R¡ js Noetheí:ian (Artinian) if and only' if each of the component rings R¡ is Noetherian (Artinían). [Hi/lt: Ifn = 2, sáy R = Rl E9 R 2 , then R/Rl c:: R 2 .] . b) LetR bearinghavingllfinitenumberofidealsl¡, 12, ... , lnsuch that n li = {Ó}'; If each of the quotient rings R/I ¡ is Noetherian (Artinian), show !hat R ís also' Noetherian (Artinian). ' ,,;;:

PROBLEMS In all problems, R is, ~conm1Utative ring with identity. l. Le! 1 be a nonzéió.ideal of !he principal ideal domaín R Prove that the quotlerit ring R/l satisfie~' both cPain conditions.

" 11. In an Artinían ring R, prove that the zero ideal is a prodllct of maximal ideáIs.: , [Hint: rad R = M 1 n M 2 n ... n'M., where each Mi ís maxhllal; use Theoré~;' 11-11 and Problern 13, Chapter 10, to conclude {O} (rad R)k MfM~ .. , M~"

2. Prove that everyJi,omomorphismfofa Noetberian ring R ontoitselfis necessarily one-to-one. [lfiilfi Consider the ascending chain {O} ~ kerf~ kerf2 ~ ... of ideals of R]" \'

. .

b~~~~

;

.JI/l

3. a) Ir 1 is an ideal of the Noetherian rine; R, show that forms a rulpotent ideal of R/l. ' , b) Let 1 and J be two ideals of tbe Noetherian ring R. Establish that 1" ~ J for ~ .jI . , some integer n E 2 + if and only if

12. Establish that an Artinian ring R is isomorphic to a finite diiect sum of Artinian local rings. [Hint: From Problem 11 and the fact!hat the ideals M} arecomaximal, we have M~ n ... n M~ ... M~ '" M! = {O}. By Theorem 10-1, R c:: ¿ ® RIM~.

4. Prove that every id~al of a NO,etherian ringR contains a product of prinle ideals. [Hint: If not, let S be the set of those ideals of R which do not contain a product of prime idealsand apply ,the maximllm condition.] .

13. Prove that any semisinlple Artinian ring possesses only a finite number of ideals. [Hlnt: Assume that R admits that decomposition R = F1 ® F2 ® ... ® F., F¡ a fie1d; if 1 is an ideiÚ, of R, then 1 = 11 ® ... ® In witb li an ideal of Fd

5. a)Obtain the converse of the Hilbert Basis Theorem: if R[x] is a Noetherian , ring, then, so also ís R ' , b) Verify that the power series ring R[[x]] is Noetherian if and only if R is a Noetherian ringo [Hint: Mimic Hilbert's Theorem, now using elements of lowest order when defining the ideals Id e)Let R' be an extension ring of !he Noetherian ring R. For a fixed elernent r E R', show tliat the ring

14. a) Let 1 be a nontrívial mínimal ideal of the Artinian ring R Show that the annihilator ai:m 1 forms a prime and, hence, maxinlal, ideal of R [Hint: If a ~ ann 1, al ~ l.] b) Assume tbat 1 is a nonzero ideal of the ring R (no chain conditions). If Pis a maxinlal member of the eollection {ann (x)IO -+ x E 1), d~uc~ thatP is a prime ideal. [Hlnt: Let abEP= ann(r), with a~P; then, P 5;; (P,b) ~ ann(ar).]

.JI

Now use Problem 5-18.] .

<;

15. A ring R is temled divisible if every non-zero-divisor of R is invertible.

R[r] == (f(r)jf(x) e R[x]} is Noetberian. 6. Prove that if R is a Noetherian ring, then the matrix ring M.(R) is a!so Noetherian. [Hint " Problem 28, Chapter 2.] 7. Assuming that R is a Noetherian ring, establish that a) Rad R is the sum of al1 the nilpotent ideals of R; b) tbe quotient ring R/Rad R has no nonzero nilpotent ideals. 8. Let R be a ring wíth at least one non-zero-divisor. Prove that ir R is Noetherian, then itsclassical ringofquotientsQ'I(R)is also Noetherian. [Hint: If J 1 ~ J 1 ~ ... is an ascending chain of ideals of Qcl(R), then, by Problem 29, Chapter 4, 11 ~ 12 S .. , forms an ascending ehain of ideals of R, where 1k "" J k n R.]

Assuming ' that R is a divisible ring prove the following: a) R is a local ring if and only jf the set D of a1l zero divisors (together with zero) is included in a proper ideal of R; in this case, D itself becomes an ideal. b) If 11 n 12 {O} for any two nonzero ideals of R, then R ís a local ringo [Hint: Show !hat the sum pf two noninvertible elenlents of R is again noninvertible.]

+.

,

16. Let R be a principal ideal ring which is not an integral domain. If!he set of al! zero divisors D = rad R, verify that R is a local ring, 17. a) If R is a finite Boolean ring, prove that R is isomorphic to the direct sum of a finite number of fields 2 2 , [Hint: See the remark following the second proof ofTheorem 11-15.] b) Prove that a finite Boolean ring has 2" elements for some n E Z+.

FURTHER RESULTS ON NOETHERIAN RINGS

TWELVE

235

arbitrary Noetherian ringo A convenient vehic1e for this discussion is the notion of an irreducible ideal. Definition 12-1. Let] be an ideal of the ring R. Then] is said to be irreducible if it is not a finite intersection of ideals of R properly containing ]; otherwise, ] is termed reducible.

FURTHER RESULTS ON NOETHERIAN RINGS

In the present chapter, emphasis is laid on certain aspects of ideal theory in which the ascending chain condition píays a dominant role. Although our treatment is rather selective, it may fairly c1aim to cover most of the high spots, as we11 as utilize a cross-section of the previously developed material. A special concern will be the proof of a fundamental theorem by Emmy Noether which asserts that ev¡;:ry ideal in a Noetherian ring isthe intersection of primary ideals; to sorne extent, this reduces the study of arbitrary ideals in such rings to that of primary ideals. Particular attention will also be paid to a number of results dealing with the intersection of the powers of an ideal in a Noetherian ringo The latter portio n of this chapter furnishes the reader with a brief introduction to module theory (roughly speaking, a module is a ve.ctor space over a ring rather than a ~eld); the ultima te aim being to prove that every cornmutative Artinian ring with identity is necessarily Noetherian. Underlying a11 our arguments is ·the equivalence of the ascending chain condition for ideals and the maximum condition. Failing any indicatioIltp the contrary, a11 rings considered are'assumed to be commutative and h~l.\ú~ an identity element. Ofien jt is not essential to stipulate both these hYP9theses and this will be revealed from a careful examination of the proof:'ill question. Let us begin our dev,elopment by showing how primary ideals fit into the iheory of Noetherian.rings. One of the basic decomposition theorems concerning the ring of int¡;:gers (itself a Noetherian ring) is that every ideal cl!-n be expressed as the intersection of a finite number of primary ideals.. Indeed, if n = p11p~2 ... p~r is a factorization of the positive integer n into distinct primes Pi' then an integer m is divisible by n if and only if m is divisible by each p~'; in the notation of principal ideals, this amounts to asserting that where each of the (P~') is a primary ideal of Z. Our immediate aim is to prove that a representation of the aboye type (that is, as a finite intersection of primary ideals) is valid for ideals in an 234

As a general cornment, it is worth remarking that any prime ideal P is always irreducible. For, suppose that there exist ideals ] and J of R satisfying P = ] n J, P c: ], P c: J. We can then select elements a E] - P and bE J - P. Now, the product ab lies in both ] and J, whencé it is a member of P. From this it fo11ows that P cannot be a prime ideal. On the other hand, we note that there exist (non-prime) primary ideals which are not irreducible. A simple illustration is furnished by the polynomial ring F[x, y], where F is a field. Here the ideal M = (x, y) is maximal, so that its square M 2 = (x 2, xy, y2) must be priniary (see Exaniple 7-8); however, M 2 has the fo11owing representation as an intersection ofproper ideal s of F[x, y]: M2 = (M 2, x) n (M2, y).

Our program is somewhat lengthy and will be completed in Theorem 12-5; we prepare the way by first establishing two auxiliary results. Lemma 1. Every ideal in a Noetherian ring R is a finite intersection of irreducible ideals. Proof.. Let ~ be the family of a11 ideals of R which are not finite intersections ofirreducible ideals. Hit happens that ~ 1- 0, then Theorem 11-1 asserts the existence of an ideal] which is maximal in the set ~ (this is where the Noetherian hypothesis enters). Then any ideal of R properly containing ] must b.~a finite intersection ofirreducible ideals. Since] E~, ] is not itself irreducible. Thus, we can write ] = J n K, where J and K are ideals of R strictly containing l. The maximal nature of ] implies that J and K both are finite intersections of irreducible ideals; hence, ] is one also. But this c1early contradicts the fact that ] E~. Accordingly, the set ~ is empty, thereby proving the assertion.

To exploit this situation fu11y, we also require: Lemma 2. In a Noetherian ring R, every irreducible ideal is primary. Proof. Our plan is to prove that any ideal] of R which is not primary is necessarily reducible; in other words, we will deduce the contrapositive form of the theorem. Since] is not primary, there exist a pair of elements a, b in R such that ab E], b rf:] a1?-d no power of a belongs to 1. Now, ]: (a) ~ ]: (a 2 ) ~

... ~ ]:

(a") ~ ...

FURTHER RESULTS ON NOETHERIAN RINGS

TWELVE

235

arbitrary Noetherian ringo A convenient vehic1e for this discussion is the notion of an irreducible ideal. Definition 12-1. Let] be an ideal of the ring R. Then] is said to be irreducible if it is not a finite intersection of ideals of R properly containing ]; otherwise, ] is termed reducible.

FURTHER RESULTS ON NOETHERIAN RINGS

In the present chapter, emphasis is laid on certain aspects of ideal theory in which the ascending chain condition píays a dominant role. Although our treatment is rather selective, it may fairly c1aim to cover most of the high spots, as we11 as utilize a cross-section of the previously developed material. A special concern will be the proof of a fundamental theorem by Emmy Noether which asserts that ev¡;:ry ideal in a Noetherian ring isthe intersection of primary ideals; to sorne extent, this reduces the study of arbitrary ideals in such rings to that of primary ideals. Particular attention will also be paid to a number of results dealing with the intersection of the powers of an ideal in a Noetherian ringo The latter portio n of this chapter furnishes the reader with a brief introduction to module theory (roughly speaking, a module is a ve.ctor space over a ring rather than a ~eld); the ultima te aim being to prove that every cornmutative Artinian ring with identity is necessarily Noetherian. Underlying a11 our arguments is ·the equivalence of the ascending chain condition for ideals and the maximum condition. Failing any indicatioIltp the contrary, a11 rings considered are'assumed to be commutative and h~l.\ú~ an identity element. Ofien jt is not essential to stipulate both these hYP9theses and this will be revealed from a careful examination of the proof:'ill question. Let us begin our dev,elopment by showing how primary ideals fit into the iheory of Noetherian.rings. One of the basic decomposition theorems concerning the ring of int¡;:gers (itself a Noetherian ring) is that every ideal cl!-n be expressed as the intersection of a finite number of primary ideals.. Indeed, if n = p11p~2 ... p~r is a factorization of the positive integer n into distinct primes Pi' then an integer m is divisible by n if and only if m is divisible by each p~'; in the notation of principal ideals, this amounts to asserting that where each of the (P~') is a primary ideal of Z. Our immediate aim is to prove that a representation of the aboye type (that is, as a finite intersection of primary ideals) is valid for ideals in an 234

As a general cornment, it is worth remarking that any prime ideal P is always irreducible. For, suppose that there exist ideals ] and J of R satisfying P = ] n J, P c: ], P c: J. We can then select elements a E] - P and bE J - P. Now, the product ab lies in both ] and J, whencé it is a member of P. From this it fo11ows that P cannot be a prime ideal. On the other hand, we note that there exist (non-prime) primary ideals which are not irreducible. A simple illustration is furnished by the polynomial ring F[x, y], where F is a field. Here the ideal M = (x, y) is maximal, so that its square M 2 = (x 2, xy, y2) must be priniary (see Exaniple 7-8); however, M 2 has the fo11owing representation as an intersection ofproper ideal s of F[x, y]: M2 = (M 2, x) n (M2, y).

Our program is somewhat lengthy and will be completed in Theorem 12-5; we prepare the way by first establishing two auxiliary results. Lemma 1. Every ideal in a Noetherian ring R is a finite intersection of irreducible ideals. Proof.. Let ~ be the family of a11 ideals of R which are not finite intersections ofirreducible ideals. Hit happens that ~ 1- 0, then Theorem 11-1 asserts the existence of an ideal] which is maximal in the set ~ (this is where the Noetherian hypothesis enters). Then any ideal of R properly containing ] must b.~a finite intersection ofirreducible ideals. Since] E~, ] is not itself irreducible. Thus, we can write ] = J n K, where J and K are ideals of R strictly containing l. The maximal nature of ] implies that J and K both are finite intersections of irreducible ideals; hence, ] is one also. But this c1early contradicts the fact that ] E~. Accordingly, the set ~ is empty, thereby proving the assertion.

To exploit this situation fu11y, we also require: Lemma 2. In a Noetherian ring R, every irreducible ideal is primary. Proof. Our plan is to prove that any ideal] of R which is not primary is necessarily reducible; in other words, we will deduce the contrapositive form of the theorem. Since] is not primary, there exist a pair of elements a, b in R such that ab E], b rf:] a1?-d no power of a belongs to 1. Now, ]: (a) ~ ]: (a 2 ) ~

... ~ ]:

(a") ~ ...

FURTHER RESULTS ON NOETHERIAN RINGS

236

237

FIRST COURSE IN RINGS AND IDEALS

if .jQ¡, = .JQ¡, = ... ;= .jQ¡ , simply take Qí = Q, n Q, n ... n Q, By ,the corollary to Theore~ 5-13, the ideal Qí is "itself '~rimary witb JQf = . .jQ¡k and, ofcour~e, wehave J = nQí. In this way, the components of a pnmary representabon can be combined so that their ni! radicals are all ~istinct. . Next, strike out one at a time those ideal s Qí which contain the mters~ctlOn ?f the ~emaining ones. The result of removing these superfluous pnmary Ideals IS that conditiqn (1) of Definition 12-2 now holds. In this manner, the given primar y representation can be transformed into an irredundant o n e . : Using the language of irredurid~nce, we can now state our basic representation theorem as

n

forms an ascending chain ofideals of R; indeed, ifran El, certainly ra + 1 E l. Because R is taken to be Noetherian, we can therefore find an integer k for which 1: (a k) = 1: (aH l). The bulk of our argument consists of showing that 1 can be expressed as 1 = (1, ak ) n (1, b).

EvidentIy, each ideal on the rigbt-hand side of this equation contains 1, so that 1 .S;;; (1, ak ) n (1, b). To obtain the opposite inc1usion, select an arbitrary x E (1, a~ n (1, b). Then, (.:

x = i

+ rak =

i'

+

r' b

Theorem 12-2. Every ideal in a ~,betherian ring has a flnite irredundant primary representation. . -

for suitably chosen elements i, i' El and r, r' E R. Consecju~ntIy, the product

raHl = (i'

- i)a

+

r'(ab)

E

J,

We shall have occasion to use theJollowing lemma. '."

which in turn implies that rEl:(a ) = J:(a ). But~this signifies that rak E J and so the element x = i + rak lies in 1, as weWished to show. To complete the proof, observe that 1 e (1, ak ), for our hypothesis as sures us that ak ~ 1; furthenllore, the element b ~ 1, whence 1 e (1, b). Inasmuch as both the ideals in equation (1) properIy contain 1, it follows that 1 must be reducible. Hl

k

.'--.

~emma. ~et ~ be an arbitraryt'iíig and 1 an ideal ofR having a finite medundant pnmary representatioil 1 = n7=l Q¡. Thén, a prime ideal P of R contains 1 if and only if P contains sorne .j.Q¡. Proo! The if part is trivial: .jQ¡ S; P implies that J S; Q¡ ~ P. Conversely, assume that there is no .JQ¡ which is contained in P. For each in~ex, i; .we ca~ then choose an element a¡ E.JQ¡ with a¡ ~ P. There also eXIst sUltable mtegers k¡ such that a~' E Q¡. Setting a = af'a~' ... akn it follows that a E n7='l'Q¡ = 1 S;;; P. Now, the ptoduct n'

These results may now be put into the form of a ~ecomposition theorem, the so-called Primary Decomposition Theorem of Noether. Theorem 12-1. (Noether). Every ideal of a Noetherian ring can be represented as a finite intersection of primary ideals. Let us call a representation of an ideal 1 in the form J- ~ n¡ Q¡, where eacp Q¡ is a primary ideal, a primary representation of J; the individual Q¡ are said to be the primary components of the representation, while JQ¡ are the prime ideals associated witb l. What was just pro~ed is that, in a Noetherian ring, every ideal admits a finite primary representation. Before announcing the next result concerning primary representations, we wish to introduce a new idea. Definition 12-2. A primary representation 1 = n~= 1 Q¡ will be termed irredundant if it satisfies the following two conditions. 1) No Q¡ contains the intersection of the other primary components; that is to say, n¡;=j Q¡ =1= n Q¡ for any j = 1,2, ... , n. 2) .jQ¡

=1=

.jQj for i

=1=

j.

If an ideal 1 admits a finite primary representation, say 1 = n7=l Q¡, then enough of tbe Q¡'s can be omitted to yield an irredundant representation. To make this precise, we first let Qí be the intersection of all those primary' components which have the same associated prime ideal; that is,

a~'(a~' '" a~n)

P

wi,th al ~ P and so, by the definition of prime ideal, (a~' ... a~n) E P. Repeating thlS argument, we finally obtain a~n E P, whence an E P, which is impossible.

\

RecaH that a prime idéal of R is said to be a mini mal prime of the ideal 1 if it.is minimal in t,he set of prime ideals containing l. Keeping the same notatlOn, the foregomg lemma asserts that the minimal prime ideals of J ar~ the ~i~im~l elements ofthe family {.jQ¡}, partiaHy ordered by inc1usion. WIth thls m mmd, we can now formulate . ,

! I

I

Theore~

12-3. Any ideal of a Noetherian ring has a finite number of minimal prime ideals. ,

I

"\'

E

i

• One of ~he tasks which is still ahead of us is the burden of showing umqueness (m sorne sense) ofthe primary representation. Given an irredundant representation 1 = n7=l Q¡ of an ideal 1 as a pnite intersection of primar~ ideals Q¡, we .do not ~laim that these primary ideals are uniquely determmed by 1; the I1lustratlve example below shows that this is not to be expected, AH that can be established is that the associated prime ideals

:/

FURTHER RESULTS ON NOETHERIAN RINGS

236

237

FIRST COURSE IN RINGS AND IDEALS

if .jQ¡, = .JQ¡, = ... ;= .jQ¡ , simply take Qí = Q, n Q, n ... n Q, By ,the corollary to Theore~ 5-13, the ideal Qí is "itself '~rimary witb JQf = . .jQ¡k and, ofcour~e, wehave J = nQí. In this way, the components of a pnmary representabon can be combined so that their ni! radicals are all ~istinct. . Next, strike out one at a time those ideal s Qí which contain the mters~ctlOn ?f the ~emaining ones. The result of removing these superfluous pnmary Ideals IS that conditiqn (1) of Definition 12-2 now holds. In this manner, the given primar y representation can be transformed into an irredundant o n e . : Using the language of irredurid~nce, we can now state our basic representation theorem as

n

forms an ascending chain ofideals of R; indeed, ifran El, certainly ra + 1 E l. Because R is taken to be Noetherian, we can therefore find an integer k for which 1: (a k) = 1: (aH l). The bulk of our argument consists of showing that 1 can be expressed as 1 = (1, ak ) n (1, b).

EvidentIy, each ideal on the rigbt-hand side of this equation contains 1, so that 1 .S;;; (1, ak ) n (1, b). To obtain the opposite inc1usion, select an arbitrary x E (1, a~ n (1, b). Then, (.:

x = i

+ rak =

i'

+

r' b

Theorem 12-2. Every ideal in a ~,betherian ring has a flnite irredundant primary representation. . -

for suitably chosen elements i, i' El and r, r' E R. Consecju~ntIy, the product

raHl = (i'

- i)a

+

r'(ab)

E

J,

We shall have occasion to use theJollowing lemma. '."

which in turn implies that rEl:(a ) = J:(a ). But~this signifies that rak E J and so the element x = i + rak lies in 1, as weWished to show. To complete the proof, observe that 1 e (1, ak ), for our hypothesis as sures us that ak ~ 1; furthenllore, the element b ~ 1, whence 1 e (1, b). Inasmuch as both the ideals in equation (1) properIy contain 1, it follows that 1 must be reducible. Hl

k

.'--.

~emma. ~et ~ be an arbitraryt'iíig and 1 an ideal ofR having a finite medundant pnmary representatioil 1 = n7=l Q¡. Thén, a prime ideal P of R contains 1 if and only if P contains sorne .j.Q¡. Proo! The if part is trivial: .jQ¡ S; P implies that J S; Q¡ ~ P. Conversely, assume that there is no .JQ¡ which is contained in P. For each in~ex, i; .we ca~ then choose an element a¡ E.JQ¡ with a¡ ~ P. There also eXIst sUltable mtegers k¡ such that a~' E Q¡. Setting a = af'a~' ... akn it follows that a E n7='l'Q¡ = 1 S;;; P. Now, the ptoduct n'

These results may now be put into the form of a ~ecomposition theorem, the so-called Primary Decomposition Theorem of Noether. Theorem 12-1. (Noether). Every ideal of a Noetherian ring can be represented as a finite intersection of primary ideals. Let us call a representation of an ideal 1 in the form J- ~ n¡ Q¡, where eacp Q¡ is a primary ideal, a primary representation of J; the individual Q¡ are said to be the primary components of the representation, while JQ¡ are the prime ideals associated witb l. What was just pro~ed is that, in a Noetherian ring, every ideal admits a finite primary representation. Before announcing the next result concerning primary representations, we wish to introduce a new idea. Definition 12-2. A primary representation 1 = n~= 1 Q¡ will be termed irredundant if it satisfies the following two conditions. 1) No Q¡ contains the intersection of the other primary components; that is to say, n¡;=j Q¡ =1= n Q¡ for any j = 1,2, ... , n. 2) .jQ¡

=1=

.jQj for i

=1=

j.

If an ideal 1 admits a finite primary representation, say 1 = n7=l Q¡, then enough of tbe Q¡'s can be omitted to yield an irredundant representation. To make this precise, we first let Qí be the intersection of all those primary' components which have the same associated prime ideal; that is,

a~'(a~' '" a~n)

P

wi,th al ~ P and so, by the definition of prime ideal, (a~' ... a~n) E P. Repeating thlS argument, we finally obtain a~n E P, whence an E P, which is impossible.

\

RecaH that a prime idéal of R is said to be a mini mal prime of the ideal 1 if it.is minimal in t,he set of prime ideals containing l. Keeping the same notatlOn, the foregomg lemma asserts that the minimal prime ideals of J ar~ the ~i~im~l elements ofthe family {.jQ¡}, partiaHy ordered by inc1usion. WIth thls m mmd, we can now formulate . ,

! I

I

Theore~

12-3. Any ideal of a Noetherian ring has a finite number of minimal prime ideals. ,

I

"\'

E

i

• One of ~he tasks which is still ahead of us is the burden of showing umqueness (m sorne sense) ofthe primary representation. Given an irredundant representation 1 = n7=l Q¡ of an ideal 1 as a pnite intersection of primar~ ideals Q¡, we .do not ~laim that these primary ideals are uniquely determmed by 1; the I1lustratlve example below shows that this is not to be expected, AH that can be established is that the associated prime ideals

:/

238

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

JQ¡ are unique and are the same in all irredundant primary representations of I; thus, it is the number of primary components that will be unique. To veriCy this, it is enough to show that the associated prime ideals can be characterized in terrns of the properties oC I alone, independent oC any particular primary representation considered. Before proceeding to the prooC, let us illustrate the fact that an ideal in a Noetherian ring need not have a unique irredundant primary representation. {

Example 12-1. In the polynomial ring F[x, yJ, where F is any field, consider the ideal (x2, xy). It is easy to see that (x2; xy) consists of those polynomials which have x as a factor and which do not possess linear terms. Now, the nonzero elements of(x2, xy, are precisely the polynomials each ofwhose terms are of degree ¿2; hence the intersection (x2, xy, y2) n (x) contains the zero polynomial together with aH polynomials of degree ¿2 which bave x as a factor. Th~s, we have

r)

(x 2, xy)

(x 2, xy, y2) n (x).

Besides this irredundant representation, there is yet another: (x 2, xy) = (x2, y) n (x)

To derive the relation above, notice that for a polynomial to He in (x 2, xy), it is suflicient to require that the polynomial be divisible by x and that the coeflicient of y be zero. As has been seen in Example 7-8, (x 2, xy, y2) and (x 2, y) are both primary ideals, while (x) is prime in F[x, y] and, hence, also primary. We next determine a characteristic ofthe primary representation which , .' is uniquely determined by the ideal in question. Theorem 12-4. Suppose that an ideal I oc" the ring R has a finite ni=l Q¡, arid let P be any irredundant primary representation, say J .JQ.i for sorne i if and only'if there exists prime ideal of R. Then P JI: (a). . an element a ~ I such that P

Proo!. To start with, assume that P .JQI for a given inde,x i. We shall argue that the hypothesis (and, hence, the conc1usion) of Theorem 5-14 holds. Now, the irredundancy of the representation I = Qk implies that there exists an element a E nk'f i Qk' but a ~ 1. For any such element a, we must have

nk

I: (a) :;;;;

.JQ, :; ;

JI: (a).

The first inclusion is justified by the fact that, since a(I: (a») :;;;; I :;;;; Q¡ with a ~ Q¡, necessari1y I: (a) :;;;;.J(j¡. To see 'tb,e second inc1usion, simply note ,that aQ¡ :;;;; I, whence Q¡ :;;;; I: (a).

239

Next, suppose that the product be E I: (a), but b ~.J(j¡. Then a(be) El:;;;; Q¡. Since Q¡ is primary and b ~.J(j¡, it follows that ac E Q¡. Also, ae (a) S;;; nk'l'¡ Qk' which gives ae Qk = I, forcing the element e to líe in I: (a). In other words, be El: (a) with b ~ J(I¡ implíes that e El: (a). Thus, all the conditions of Theorem 5-14 are satisfied and we may conc1ude that P = JQ¡ = JI: (a). Going in the other direction, let the element a ~ I be such that P = .JI: (a). With níerence to Theorem 2-5,

E

Enk

I: (a)

=

(n Q¡): (a) = n (Q¡: (a») i

i

n¡

and, the intersection being finite, P = JQ¡: (a). Observe that if a E Q¡, then JQ¡: (a) = Q¡: (a) = R. On the other band, if a ~ Q¡, reason as in the first part ofthe proofwith I now replaced by the primary ideal Q¡; for these Q¡, we then obtain JQ¡: (a) = J(I¡. In consequence, P is the intersection of sorne of the .J(j¡; let us say that P

= JQ;. n JO:;. n

... n

JQ:..

Knowing this, the proof is easi1y completed; for, by Problern 30(a), Chapter 5, P must contain one of the ideals ..JQ;. and is obviously contained by it, whence P = ..jQik' What we are realIy after is the corollary below. Corollary. Let I be an ideal of the Noetherian ring R. Suppose that I = Ql n ... n Qn = Q~ n ... n Q~ are two finite irredundant primary representations of 1. Then, n = m and the assoclated prime ideals of these two representations are equal (t,bat is, with a sllÍtable renumbering of the indices .JQ. = for 1 .$ ,í :s;¡ n = m).

.Jffi

Proo!. The theorem provides an intrinsic cl'laracterization of the associated prime ideals in terms of I alone. Example 12-2. In the polynomial ring F[x, y], the ideal (Xl, xy) has the irredundant primary representation (Xl, xy) = (x 2, xy;y") n (x) ;

regardless of the choice of n > 1. The corresponding nil radicals (that is, the associated prime ideals) are (x, y) and (x). Thus, in any irredundant represen ta tion (x 2, xy) = Ql n Q2 by primary ideals Ql and Q2' we must have

.JQ;

(x, y),

JQ;

= (x).

Of these, only (x) is a mínimal prime of (x 2, xy). Now, it so happens that

238

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

JQ¡ are unique and are the same in all irredundant primary representations of I; thus, it is the number of primary components that will be unique. To veriCy this, it is enough to show that the associated prime ideals can be characterized in terrns of the properties oC I alone, independent oC any particular primary representation considered. Before proceeding to the prooC, let us illustrate the fact that an ideal in a Noetherian ring need not have a unique irredundant primary representation. {

Example 12-1. In the polynomial ring F[x, yJ, where F is any field, consider the ideal (x2, xy). It is easy to see that (x2; xy) consists of those polynomials which have x as a factor and which do not possess linear terms. Now, the nonzero elements of(x2, xy, are precisely the polynomials each ofwhose terms are of degree ¿2; hence the intersection (x2, xy, y2) n (x) contains the zero polynomial together with aH polynomials of degree ¿2 which bave x as a factor. Th~s, we have

r)

(x 2, xy)

(x 2, xy, y2) n (x).

Besides this irredundant representation, there is yet another: (x 2, xy) = (x2, y) n (x)

To derive the relation above, notice that for a polynomial to He in (x 2, xy), it is suflicient to require that the polynomial be divisible by x and that the coeflicient of y be zero. As has been seen in Example 7-8, (x 2, xy, y2) and (x 2, y) are both primary ideals, while (x) is prime in F[x, y] and, hence, also primary. We next determine a characteristic ofthe primary representation which , .' is uniquely determined by the ideal in question. Theorem 12-4. Suppose that an ideal I oc" the ring R has a finite ni=l Q¡, arid let P be any irredundant primary representation, say J .JQ.i for sorne i if and only'if there exists prime ideal of R. Then P JI: (a). . an element a ~ I such that P

Proo!. To start with, assume that P .JQI for a given inde,x i. We shall argue that the hypothesis (and, hence, the conc1usion) of Theorem 5-14 holds. Now, the irredundancy of the representation I = Qk implies that there exists an element a E nk'f i Qk' but a ~ 1. For any such element a, we must have

nk

I: (a) :;;;;

.JQ, :; ;

JI: (a).

The first inclusion is justified by the fact that, since a(I: (a») :;;;; I :;;;; Q¡ with a ~ Q¡, necessari1y I: (a) :;;;;.J(j¡. To see 'tb,e second inc1usion, simply note ,that aQ¡ :;;;; I, whence Q¡ :;;;; I: (a).

239

Next, suppose that the product be E I: (a), but b ~.J(j¡. Then a(be) El:;;;; Q¡. Since Q¡ is primary and b ~.J(j¡, it follows that ac E Q¡. Also, ae (a) S;;; nk'l'¡ Qk' which gives ae Qk = I, forcing the element e to líe in I: (a). In other words, be El: (a) with b ~ J(I¡ implíes that e El: (a). Thus, all the conditions of Theorem 5-14 are satisfied and we may conc1ude that P = JQ¡ = JI: (a). Going in the other direction, let the element a ~ I be such that P = .JI: (a). With níerence to Theorem 2-5,

E

Enk

I: (a)

=

(n Q¡): (a) = n (Q¡: (a») i

i

n¡

and, the intersection being finite, P = JQ¡: (a). Observe that if a E Q¡, then JQ¡: (a) = Q¡: (a) = R. On the other band, if a ~ Q¡, reason as in the first part ofthe proofwith I now replaced by the primary ideal Q¡; for these Q¡, we then obtain JQ¡: (a) = J(I¡. In consequence, P is the intersection of sorne of the .J(j¡; let us say that P

= JQ;. n JO:;. n

... n

JQ:..

Knowing this, the proof is easi1y completed; for, by Problern 30(a), Chapter 5, P must contain one of the ideals ..JQ;. and is obviously contained by it, whence P = ..jQik' What we are realIy after is the corollary below. Corollary. Let I be an ideal of the Noetherian ring R. Suppose that I = Ql n ... n Qn = Q~ n ... n Q~ are two finite irredundant primary representations of 1. Then, n = m and the assoclated prime ideals of these two representations are equal (t,bat is, with a sllÍtable renumbering of the indices .JQ. = for 1 .$ ,í :s;¡ n = m).

.Jffi

Proo!. The theorem provides an intrinsic cl'laracterization of the associated prime ideals in terms of I alone. Example 12-2. In the polynomial ring F[x, y], the ideal (Xl, xy) has the irredundant primary representation (Xl, xy) = (x 2, xy;y") n (x) ;

regardless of the choice of n > 1. The corresponding nil radicals (that is, the associated prime ideals) are (x, y) and (x). Thus, in any irredundant represen ta tion (x 2, xy) = Ql n Q2 by primary ideals Ql and Q2' we must have

.JQ;

(x, y),

JQ;

= (x).

Of these, only (x) is a mínimal prime of (x 2, xy). Now, it so happens that

240

FURTHER RESULTS ON NOETHERIAN RINGS

FIRST COURSE IN RINGS AND IDEALS

Theorem 12-6. (Cohen). A ring R is Noetherian if and only ifevery prime ideal of R is finitely genera ted. ,

the nil radical-JI of an ideal l in a Noetherian ring is the intersection of the minimal priines of 1; granting this fact, one finds that

-J(x 2 , xy)

=

(x).

Incidentally, our example has the added advantage of showing that -JI can ' be prime without the ideal l being primary.

Proo! The "only W' part'is an immediate consequence of Theorem 11-1. Passing to the less trivial assertion, assume that every prime ideal of R is finitely generated, but that R is not Noetherian. This assures that the collection ff of ideals of R which are not finitely generated is nonempty. Appealing to Zorn's lernma, ff must contain a maximall1leent, call it l. :~,Y v!rtue of our hypothesis, l 'c~nnot be a pr~me ideal of . Consequently, ther!'!. eXlst elements a, b of R WhlCh are not m l such t at their product ab,~ .:J. Now, both the ideals (1, b) and 1: (b) properIy contain 1; in particular, a e:¡.,A; (b). By the maximal nature of l in ff, these ideals are finitely generated. F<:>,~/definiteness, let us suppose that

'}

At this stage, it is reasonable to inquire under what circumstances (if any at ~1I) the ideal s in;a given primary representation will turn out to be prime ideals. The follo~ing theorem supplies an answer. Theorem 12-5. Let'j:be an ideal ofthe ring R with a finite irredundant primary representati<;m l = n7=l Qi' Then lis semiprime (that is, l = -JI) if and onlyif each Qi is a prime ideal of R. Proo! We begin by a~~u'~ng that all the Q¡ in the given primary represe~~a-' tion of l are prime ideaJs.···1f the element a E -JI, then a" E l for sorne poslbve integer n; hence, a" lies:i~ each Qi' As Q¡ is taken to be prime, this implies that a itselfbelongs to QJor every i and so a E l. Our argument shows that S; 1, froID which the desired equality follows. . .... With regard'to the converse, suppose that the Ideal l comcldes wIth ltS ni! radical. Then, usingTheorem 5-10 again,

<::,,':,,1. (1, b) ;-?:.:!::; T~~p; C¡ = a¡

=

+

(el) c2 ,

... ,

... ,

dm ).

L 2, ... , n), so that

a", b) .

Next, consider the ideal J generated by the elements a¡ and bd j ; in other ' " ' words, the ideal J = (al' ... , a", bd l , ... , bdm)·

i

i

Sin ce bd j E l for every j, the inclusion J S; l certainly holds. What is not so o bvious is that l S; J. To see this, let x be an'arbitrary member of 1; because x E (1, b), it may be written in the forO).

It is important to point out that this is actually an irredundant representation of las an intersection of primary (in f~ct, prime) ideals. Suppose not; there would then exist sorne positive integer j such that l = n¡ti J[. But then

n-JQi nQi ;2

i'f'i

1: (b) == (d 1 , d2 ,

and

en)

br¡, where a¡ E l and r¡ E R (i =

=-JI =-Jn Qi = n.JQi·

l =

... ,

(1, b) = (al' a 2 ,

.JI

l

241

;2

(Y¡, Y E R);

1,

iti

which means that l = ni'f'j Qi' This, however, contradicts the hypothesis that the given representation of l is irredundant. , Next, fix the integer j and let a be any element of the ideal~. Since nitj J[ =1= n -JQ¡, we can find sorne b E n¡~j.JQ; with b ~ 'VQj' Then the product ab E J[ = l ~ Qj' whence a lS a member of Qi: .The~e­ fore, .J(E S; Qi' yielding the subsequent equality Qi = .J(E. ThlS lmpbes that Qi is a prime ideal, which was what had to be proved.

As each a¡ lies in the ideal 1, so do es by, which is simply' to assert that Z¡ sR such that

y El: (b). Kn()wing this, we areable to find elements

I

I

y = dlz l

n

Let us change direction now. The reader will no doubt recall that a ring R is Noetherian if and only if every ideal of R is finitely generated (Theorem 11-1). Actually, it is enough to consider just the prime ideals of R, the proof being due to l. S. Cohen.

d2 z 2

+ ... + dmz m,

leading directly to

+ ... + a"y" + b(dlz l + ... + dmz m ) a'lYl + ... + a"y" + (bdl)Zl + .... ~ (bdm)zm EJ.

x = aly =

With this information at our disposal, we can now state Corollary. In a Noetherian ring, any semiprime ideal is a finite intersection of prime ideals.

+

I

The equality l = J now follows and so one concludes that l itself is finitely generated, an impossibility since l E ff. This contradiction completes the proof. Scrutiny of the preceding argument reveals a fact which is important enough to be stated independently: Let l be an ideal of the ring R and b an element of R; if the ideals (1, b) and 1: (b) are both finitely generated,

,/,;(,:

{.;~,f

l· ','.

240

FURTHER RESULTS ON NOETHERIAN RINGS

FIRST COURSE IN RINGS AND IDEALS

Theorem 12-6. (Cohen). A ring R is Noetherian if and only ifevery prime ideal of R is finitely genera ted. ,

the nil radical-JI of an ideal l in a Noetherian ring is the intersection of the minimal priines of 1; granting this fact, one finds that

-J(x 2 , xy)

=

(x).

Incidentally, our example has the added advantage of showing that -JI can ' be prime without the ideal l being primary.

Proo! The "only W' part'is an immediate consequence of Theorem 11-1. Passing to the less trivial assertion, assume that every prime ideal of R is finitely generated, but that R is not Noetherian. This assures that the collection ff of ideals of R which are not finitely generated is nonempty. Appealing to Zorn's lernma, ff must contain a maximall1leent, call it l. :~,Y v!rtue of our hypothesis, l 'c~nnot be a pr~me ideal of . Consequently, ther!'!. eXlst elements a, b of R WhlCh are not m l such t at their product ab,~ .:J. Now, both the ideals (1, b) and 1: (b) properIy contain 1; in particular, a e:¡.,A; (b). By the maximal nature of l in ff, these ideals are finitely generated. F<:>,~/definiteness, let us suppose that

'}

At this stage, it is reasonable to inquire under what circumstances (if any at ~1I) the ideal s in;a given primary representation will turn out to be prime ideals. The follo~ing theorem supplies an answer. Theorem 12-5. Let'j:be an ideal ofthe ring R with a finite irredundant primary representati<;m l = n7=l Qi' Then lis semiprime (that is, l = -JI) if and onlyif each Qi is a prime ideal of R. Proo! We begin by a~~u'~ng that all the Q¡ in the given primary represe~~a-' tion of l are prime ideaJs.···1f the element a E -JI, then a" E l for sorne poslbve integer n; hence, a" lies:i~ each Qi' As Q¡ is taken to be prime, this implies that a itselfbelongs to QJor every i and so a E l. Our argument shows that S; 1, froID which the desired equality follows. . .... With regard'to the converse, suppose that the Ideal l comcldes wIth ltS ni! radical. Then, usingTheorem 5-10 again,

<::,,':,,1. (1, b) ;-?:.:!::; T~~p; C¡ = a¡

=

+

(el) c2 ,

... ,

... ,

dm ).

L 2, ... , n), so that

a", b) .

Next, consider the ideal J generated by the elements a¡ and bd j ; in other ' " ' words, the ideal J = (al' ... , a", bd l , ... , bdm)·

i

i

Sin ce bd j E l for every j, the inclusion J S; l certainly holds. What is not so o bvious is that l S; J. To see this, let x be an'arbitrary member of 1; because x E (1, b), it may be written in the forO).

It is important to point out that this is actually an irredundant representation of las an intersection of primary (in f~ct, prime) ideals. Suppose not; there would then exist sorne positive integer j such that l = n¡ti J[. But then

n-JQi nQi ;2

i'f'i

1: (b) == (d 1 , d2 ,

and

en)

br¡, where a¡ E l and r¡ E R (i =

=-JI =-Jn Qi = n.JQi·

l =

... ,

(1, b) = (al' a 2 ,

.JI

l

241

;2

(Y¡, Y E R);

1,

iti

which means that l = ni'f'j Qi' This, however, contradicts the hypothesis that the given representation of l is irredundant. , Next, fix the integer j and let a be any element of the ideal~. Since nitj J[ =1= n -JQ¡, we can find sorne b E n¡~j.JQ; with b ~ 'VQj' Then the product ab E J[ = l ~ Qj' whence a lS a member of Qi: .The~e­ fore, .J(E S; Qi' yielding the subsequent equality Qi = .J(E. ThlS lmpbes that Qi is a prime ideal, which was what had to be proved.

As each a¡ lies in the ideal 1, so do es by, which is simply' to assert that Z¡ sR such that

y El: (b). Kn()wing this, we areable to find elements

I

I

y = dlz l

n

Let us change direction now. The reader will no doubt recall that a ring R is Noetherian if and only if every ideal of R is finitely generated (Theorem 11-1). Actually, it is enough to consider just the prime ideals of R, the proof being due to l. S. Cohen.

d2 z 2

+ ... + dmz m,

leading directly to

+ ... + a"y" + b(dlz l + ... + dmz m ) a'lYl + ... + a"y" + (bdl)Zl + .... ~ (bdm)zm EJ.

x = aly =

With this information at our disposal, we can now state Corollary. In a Noetherian ring, any semiprime ideal is a finite intersection of prime ideals.

+

I

The equality l = J now follows and so one concludes that l itself is finitely generated, an impossibility since l E ff. This contradiction completes the proof. Scrutiny of the preceding argument reveals a fact which is important enough to be stated independently: Let l be an ideal of the ring R and b an element of R; if the ideals (1, b) and 1: (b) are both finitely generated,

,/,;(,:

{.;~,f

l· ','.

242

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

then 1 is al so finitely generatedo As an application of Cohen's Theorem, we present

It now suffices to take 1 - r¡+ 1 = (1 - r;)(1 - r¡ - b¡¡); clearly, ri+ 1 E J and, as a little computation will show,

Corollary. If R is a ring in which each maximal ideal is generated by an idempotent, then R is Noetheriano Proa! We first prove that every primary ideal of R is maximal. Suppose otherwis~; that is, let 1 be a primary ideal which is not maximal in R. Now, 1 will be contained in sorne maximal ideal Mo By hypothesis, M has an idempotent generator; say M = (e), where.e is an idempotent'different from O or 1 (if e = O, R becomes a field and there is nothing to prove)o Then e(1 - e) = O El and, since 1 is a primary ideal, it follows that

(1 :-- e)n E 1

~

M

for sorne positive integer no This implies that 1 - e E M = (e), so that 1 E M, an obvious contradictiono Because every primary ideal bf R is maximal, the notions of maximal, prime, and primary ideal all agreeo In the light of our hypothesis, every maximal ideal (hence, every prime ideal) is finitely generatedo That R is necessarily Noetherian now follows from Cohen's result. We next propose to take a look at several results concerning the in tersection ofthe powers ofan ideal in a Noetherian ringo Before any deductions can be made, it will be convenient to establish a technicallemmao Lemma. Let 1 and J be two ideals ofthe ring R, with 1 finitely generatedo If IJ = 1, then there exists an element rE J such that (1 - r)l = {O}o Proa! Supposethat 1 is generated by the elements al' a2 , an0 Let 1¡ denote the ideaL(a¡;a¡+l' an) and put ln+l = {O}o By induction on i, we shall pro ve the existence of an element r¡ E J such that (1 - r¡)l !;; 1¡ (i = 1,2, n T ~); in particular, rn+l will be the element mentioned in the statement of t4e theoremo When i = 1, the ideal 11 = 1 and one can simply take r 1 = 00 Using the induction hypothesis that (1 - rJl ~ li for sorne r¡ E J, together with the fact that 1 ~ 1J, we have 000'

000'

000'

(1 - rJl

!;;

(1 - r;)IJ

~

l¡Jo

(1 - r¡+l)l

n

In consequence,

=

I

k=i

b¡kak

(b ik E 1)0

=

(1 - r;)(1 - r¡ - b¡Jl

!;;

(1 - r i

bu )l í

-

This proves the lemmao In a moment, we shall appeal to this lemma to characterize the elements which belong to the intersection of the powers of an ideal o Let us temporarily turn aside from this pursuit, however, to call attention to a noteworthy result of Nakayamao . Theorem 12-7. (Nakayama's Lemma)o Let 1 be a finitely generated ideal ofthe ring R. If l(rad R) = 1, then 1 = {O}o Proa! The foregoing lemma tells us that there exists an element r Erad R for which (1 - r)I = {O}o If 1 - r were not invertible in R, then it would be contained in sorne maximal ideal Mo But rE rad R ~ M, leading to the contradiction that 1 E Mo Accordingly, 1 - r is an invertible element of R, which forces 1 = {O}. Remarko It is possible to prove somewhat more than is asserted aboye, for one may replace rad R by any ideal which is contained in rad R. What is important in the present situation is that Nakayama's Lemma holds iti any Noetherian ringo

We now come to the theorem that was alluded to earliero Theorem 12-8 Let 1 be a proper ideal of the Noetherian ring R. TMil"'

n In n=l IX)

=

{r E RI(1 - a)r = O for sorne a E I}o o

. Proa! For ease ofnotation, let S denote the right-hand side ofthe indicate,d, equationo If the element r E S, so that (1 - a)r = O for suitable a E 1, would necessarily have

we.

r.

Since each generator a¡ lies in 1, it follows that (1 - r;)a¡ E 1;1 and so (1 - r;)a¡

243

= ar = a2 r =

o o o

=

a"r

=

The implication of these relations is that r belongs to In for every integer n and, hence, r E 1" ; in other words, S ~ ln The opposite inclusion is les s obviouso To obtain this, put J = In and consider the irredundant primary representation of the ideal IJ:

nn

nn

IJ =

n Qi' i

o

(Qi primary)o

n"

242

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

then 1 is al so finitely generatedo As an application of Cohen's Theorem, we present

It now suffices to take 1 - r¡+ 1 = (1 - r;)(1 - r¡ - b¡¡); clearly, ri+ 1 E J and, as a little computation will show,

Corollary. If R is a ring in which each maximal ideal is generated by an idempotent, then R is Noetheriano Proa! We first prove that every primary ideal of R is maximal. Suppose otherwis~; that is, let 1 be a primary ideal which is not maximal in R. Now, 1 will be contained in sorne maximal ideal Mo By hypothesis, M has an idempotent generator; say M = (e), where.e is an idempotent'different from O or 1 (if e = O, R becomes a field and there is nothing to prove)o Then e(1 - e) = O El and, since 1 is a primary ideal, it follows that

(1 :-- e)n E 1

~

M

for sorne positive integer no This implies that 1 - e E M = (e), so that 1 E M, an obvious contradictiono Because every primary ideal bf R is maximal, the notions of maximal, prime, and primary ideal all agreeo In the light of our hypothesis, every maximal ideal (hence, every prime ideal) is finitely generatedo That R is necessarily Noetherian now follows from Cohen's result. We next propose to take a look at several results concerning the in tersection ofthe powers ofan ideal in a Noetherian ringo Before any deductions can be made, it will be convenient to establish a technicallemmao Lemma. Let 1 and J be two ideals ofthe ring R, with 1 finitely generatedo If IJ = 1, then there exists an element rE J such that (1 - r)l = {O}o Proa! Supposethat 1 is generated by the elements al' a2 , an0 Let 1¡ denote the ideaL(a¡;a¡+l' an) and put ln+l = {O}o By induction on i, we shall pro ve the existence of an element r¡ E J such that (1 - r¡)l !;; 1¡ (i = 1,2, n T ~); in particular, rn+l will be the element mentioned in the statement of t4e theoremo When i = 1, the ideal 11 = 1 and one can simply take r 1 = 00 Using the induction hypothesis that (1 - rJl ~ li for sorne r¡ E J, together with the fact that 1 ~ 1J, we have 000'

000'

000'

(1 - rJl

!;;

(1 - r;)IJ

~

l¡Jo

(1 - r¡+l)l

n

In consequence,

=

I

k=i

b¡kak

(b ik E 1)0

=

(1 - r;)(1 - r¡ - b¡Jl

!;;

(1 - r i

bu )l í

-

This proves the lemmao In a moment, we shall appeal to this lemma to characterize the elements which belong to the intersection of the powers of an ideal o Let us temporarily turn aside from this pursuit, however, to call attention to a noteworthy result of Nakayamao . Theorem 12-7. (Nakayama's Lemma)o Let 1 be a finitely generated ideal ofthe ring R. If l(rad R) = 1, then 1 = {O}o Proa! The foregoing lemma tells us that there exists an element r Erad R for which (1 - r)I = {O}o If 1 - r were not invertible in R, then it would be contained in sorne maximal ideal Mo But rE rad R ~ M, leading to the contradiction that 1 E Mo Accordingly, 1 - r is an invertible element of R, which forces 1 = {O}. Remarko It is possible to prove somewhat more than is asserted aboye, for one may replace rad R by any ideal which is contained in rad R. What is important in the present situation is that Nakayama's Lemma holds iti any Noetherian ringo

We now come to the theorem that was alluded to earliero Theorem 12-8 Let 1 be a proper ideal of the Noetherian ring R. TMil"'

n In n=l IX)

=

{r E RI(1 - a)r = O for sorne a E I}o o

. Proa! For ease ofnotation, let S denote the right-hand side ofthe indicate,d, equationo If the element r E S, so that (1 - a)r = O for suitable a E 1, would necessarily have

we.

r.

Since each generator a¡ lies in 1, it follows that (1 - r;)a¡ E 1;1 and so (1 - r;)a¡

243

= ar = a2 r =

o o o

=

a"r

=

The implication of these relations is that r belongs to In for every integer n and, hence, r E 1" ; in other words, S ~ ln The opposite inclusion is les s obviouso To obtain this, put J = In and consider the irredundant primary representation of the ideal IJ:

nn

nn

IJ =

n Qi' i

o

(Qi primary)o

n"

244

FURTHER RESULTS ON NOETHERIAN RINGS

FIRST COURSE IN RINGS AND IDEALS

We wish to establish the equality 1J = J (once this has been accomplished . the previous Íemma can be applied). Since 1J f;; J, it will be enough to sh<;>w that J f;; Q¡ for each index i. Now, 1J f;; Q¡, so, by the corollary to Theorem 11-2, either J S; Q¡ or eIse (Ji)" S; Q¡ for some n E Z+. But if (,JJ)v S; Q¡; then

In any 'event, J S; Q¡ for each value of i, whence J S; 1J and eq:u:ality follows. . From the lemma just proved, there exists an elementd 'i¡; 1 such that (1 - a)J = {O}. But this amounts to asserting that J = n~.t~ is contained in the set S and thereby completes the proof. ")

This leads almost immediately to an important theorem. of Krull. Theorem 12-9.. (KrhUIntersection Theorem). Let 1 be á¡j~oper ideal of . the N oetherian ring R. Then ¡n {O} if and on.i:y:'íf no element of the set 1 ~ 1 = {1 - ala E 1} is a zero divisor in R. .: '.

n:= 1

There are, of.course, a number oC interesting consequences of this last result. Some of these are' the content of the corollaries below. Corollary 1. If l ls a proper ideal of the Noetherian domain R (in l" = {O}. particulár, if R is a principal ideal domain), then

n:=1

COfolIary 2~ In any Noetherian ring R,

n:=1 (radR)" = {O}.

Proo! Take l ~ rad R in Krull's Theorem. By Theorem 8-2, every elem.ent of 1 - radR is invertible and thus cannot be a zero divisor. COfolIary 3. In a Noetherian domain R, any prime principal ideal (a) is a minimal prime idealof R.

Proo! Suppose to the contrai:y that there eiists a prime ideal P of R satisfying {O} c: P c: (a). Since the element a rt P, the condition ra E P implies that rE P; hence, P = aP. Utilizing Corollary 1, we therefore conclude that P

aP

a2 p = ...

S;

n(a") = 11

n

(a)" = {O},

"

which is impossible. This line of reasoning makes (a) a minimal .prime ideal of R. Given a zero divisor r = 1 - a, with a E l, the element 1 - r evidently belongs to the ideal l. In the light of this, the Intersection Theorem is often phrased in a slightly different form: Let l be an ideal of the Noetherian In = {O} if and only if there is no zero divisor r of R ring R; then, such that 1 - r E 1. .

n:=l

245

We .conclude the prese~t ph~se of our investigation by showing that, In is equal to the intersectioñ of for any Ideal l of a Noethenan nng, certain primary components of the zero ideal. This result, which may be regarded as a refinement of the Krull Inter8ection Theorem, i8 due to Northcott [53].

n:=l

Theorem 12-10. {O} = Q1 n Q2 of {O}. Assume but not for 1 ~

Let 1 be an ideal of the Noetherian ring R and let n ... n Qn be an irredundant primary representation further that ..JQ¡ n¡(1 - 1) =/:; 0 for m + 1 ~ i ~ n, i ~ m. Then,::': ,'1'\ .

n 1" "=1 co

Proo! According to Theorem 12-8, S ~ {r E RI(l

a)r

<,

;

Ql n.·Q~ n ... n Qm'

..

i\:i~;enough to show th~t the set

O for some¿'~,E l} = Ql n Q2 n '" n Qm' "'¡( '".-:

Suppose that the elernent r E S, SO thát~t O E Q¡ for suitable X E 1 - l. For 1 ~ i ::;;'m, our hypothesis implies' ihat X rt ,JQ¡, whence rE Q.' this " establishes the inclusion S f;; Ql I'l Q2 n ... n Qm' Now, let y be an arbitrary member of the intersection Q1 n. Q2 n ... n Qm ~When the integer i ;;:::: m + 1, it is possible to choose an element a¡ E ,JQ¡ n (1 - 1). For k sufficiently large, we. will then have . y(am+ 1 ... an)k E Ql n ... n Qm n ... n Qn = {O}, that is to say, y(a m+ 1 ... an)k = O. Because 1 - 1 forms a multiplicatively closed 8ubset of R, the product am + 1 ... a" El - 1 and so, by definition, y ES. Our argumentgives Q1 n Q2 n ... n Qm S; S, from which the desired . equality follows. . One.could rephrase Theorem 12-10 to read as follows. Given an ideal 1 of the Noetherian ring R, ():=¿n i8 the intersection of those primary components Q¡ of {O} for which ,JQ¡ n (1 - 1) = 0· We next intend to prove the following result On local rings: if a local Mil = {O}, then R must ring R has principal maximal ideal M, with be Noetherian. Such rings have an extremely simple ideal theory in that every.no~triv~al i~eal is a power of the maximal ideal. The obvious example oC thIS sltuatlOn 18 the power series ring F[[xJ], where F is a field; by Theorem 7-3, we know that F[[xJ] forms a local ring with maximal ideal I (x), in consequence of which

n:=l

n:=l (xt = n:~l (x") = {O}. Before we consider the stated result, a lemma is required. Lemma. Let R be a local ring whose maximal ideal M is principal. Then,

244

FURTHER RESULTS ON NOETHERIAN RINGS

FIRST COURSE IN RINGS AND IDEALS

We wish to establish the equality 1J = J (once this has been accomplished . the previous Íemma can be applied). Since 1J f;; J, it will be enough to sh<;>w that J f;; Q¡ for each index i. Now, 1J f;; Q¡, so, by the corollary to Theorem 11-2, either J S; Q¡ or eIse (Ji)" S; Q¡ for some n E Z+. But if (,JJ)v S; Q¡; then

In any 'event, J S; Q¡ for each value of i, whence J S; 1J and eq:u:ality follows. . From the lemma just proved, there exists an elementd 'i¡; 1 such that (1 - a)J = {O}. But this amounts to asserting that J = n~.t~ is contained in the set S and thereby completes the proof. ")

This leads almost immediately to an important theorem. of Krull. Theorem 12-9.. (KrhUIntersection Theorem). Let 1 be á¡j~oper ideal of . the N oetherian ring R. Then ¡n {O} if and on.i:y:'íf no element of the set 1 ~ 1 = {1 - ala E 1} is a zero divisor in R. .: '.

n:= 1

There are, of.course, a number oC interesting consequences of this last result. Some of these are' the content of the corollaries below. Corollary 1. If l ls a proper ideal of the Noetherian domain R (in l" = {O}. particulár, if R is a principal ideal domain), then

n:=1

COfolIary 2~ In any Noetherian ring R,

n:=1 (radR)" = {O}.

Proo! Take l ~ rad R in Krull's Theorem. By Theorem 8-2, every elem.ent of 1 - radR is invertible and thus cannot be a zero divisor. COfolIary 3. In a Noetherian domain R, any prime principal ideal (a) is a minimal prime idealof R.

Proo! Suppose to the contrai:y that there eiists a prime ideal P of R satisfying {O} c: P c: (a). Since the element a rt P, the condition ra E P implies that rE P; hence, P = aP. Utilizing Corollary 1, we therefore conclude that P

aP

a2 p = ...

S;

n(a") = 11

n

(a)" = {O},

"

which is impossible. This line of reasoning makes (a) a minimal .prime ideal of R. Given a zero divisor r = 1 - a, with a E l, the element 1 - r evidently belongs to the ideal l. In the light of this, the Intersection Theorem is often phrased in a slightly different form: Let l be an ideal of the Noetherian In = {O} if and only if there is no zero divisor r of R ring R; then, such that 1 - r E 1. .

n:=l

245

We .conclude the prese~t ph~se of our investigation by showing that, In is equal to the intersectioñ of for any Ideal l of a Noethenan nng, certain primary components of the zero ideal. This result, which may be regarded as a refinement of the Krull Inter8ection Theorem, i8 due to Northcott [53].

n:=l

Theorem 12-10. {O} = Q1 n Q2 of {O}. Assume but not for 1 ~

Let 1 be an ideal of the Noetherian ring R and let n ... n Qn be an irredundant primary representation further that ..JQ¡ n¡(1 - 1) =/:; 0 for m + 1 ~ i ~ n, i ~ m. Then,::': ,'1'\ .

n 1" "=1 co

Proo! According to Theorem 12-8, S ~ {r E RI(l

a)r

<,

;

Ql n.·Q~ n ... n Qm'

..

i\:i~;enough to show th~t the set

O for some¿'~,E l} = Ql n Q2 n '" n Qm' "'¡( '".-:

Suppose that the elernent r E S, SO thát~t O E Q¡ for suitable X E 1 - l. For 1 ~ i ::;;'m, our hypothesis implies' ihat X rt ,JQ¡, whence rE Q.' this " establishes the inclusion S f;; Ql I'l Q2 n ... n Qm' Now, let y be an arbitrary member of the intersection Q1 n. Q2 n ... n Qm ~When the integer i ;;:::: m + 1, it is possible to choose an element a¡ E ,JQ¡ n (1 - 1). For k sufficiently large, we. will then have . y(am+ 1 ... an)k E Ql n ... n Qm n ... n Qn = {O}, that is to say, y(a m+ 1 ... an)k = O. Because 1 - 1 forms a multiplicatively closed 8ubset of R, the product am + 1 ... a" El - 1 and so, by definition, y ES. Our argumentgives Q1 n Q2 n ... n Qm S; S, from which the desired . equality follows. . One.could rephrase Theorem 12-10 to read as follows. Given an ideal 1 of the Noetherian ring R, ():=¿n i8 the intersection of those primary components Q¡ of {O} for which ,JQ¡ n (1 - 1) = 0· We next intend to prove the following result On local rings: if a local Mil = {O}, then R must ring R has principal maximal ideal M, with be Noetherian. Such rings have an extremely simple ideal theory in that every.no~triv~al i~eal is a power of the maximal ideal. The obvious example oC thIS sltuatlOn 18 the power series ring F[[xJ], where F is a field; by Theorem 7-3, we know that F[[xJ] forms a local ring with maximal ideal I (x), in consequence of which

n:=l

n:=l (xt = n:~l (x") = {O}. Before we consider the stated result, a lemma is required. Lemma. Let R be a local ring whose maximal ideal M is principal. Then,

246

FIRST COURSE IN RINGS ANO IDEALS

1) for nonzero a, b E M, (a) = (b) if and only if a = bu, where u is invertible in R; 2) if M = (p) =1= {O}, then p is an irreducible element of R; 3) if O =1= a = qnu, with q E M and u in vertible, then this factorization is essentially unique (that is, the integer n is uniquely determined by q). Proo! If (a) = (b), then a = br and b = as for suitable r, s E R. Accordingly, a '= asr or a(l - sr) = O. Suppose that one of the elements r or s lies in M, so that the product sr E M. We observe that 1 - sr fj M, for other-· wise 1 E M. This means that 1 - sr is an invertible element of R (recall that M consists of all the noninvertible elements of R). Then the relation a(l - sr) = O yields a = O, a contradiction. Thus, neither r nor s belongs

to M, which signifies that they are both invertible. The conve~se should' be obvious. Concerning (2), let R = ab. If the element b is not in vertible, then part (1) forces M = (p) e (a). The inaximality of M then ensures that (a) = R, whence a has an inverse in R. To see the final assertion, assume that a = qnu = qmv, with m > n. Then qnu(l - qm-nvu- 1 ) = O. By the I:).rgument of the first paragraph, this relation implies that a = O. Theorem 12-11. Let R be a local ring with principal maximal ideal M = (p). Then every element O =1= a E M has a factorization in the form a = pmu, where u is invertible, if and only if n~= 1 (pn) = {O}.

nn

Proo! Assume that the intersection (p") = {O}. If O =1= a E M, then a = pr for sorne r in R. If it happens that r is not an invertible element, then r E M; thus, we can .write r = ps or, upon substituting, a = p2 s. This process must eventually termínate, for otherwisea wou1dlie in nn (pn) = {O}. For the converse, suppose that there exists sorne 'nQi1zero element a in (p"). In particular, a E (p) = M, so that a = pmu for sorne invertible U.

n"

For any integer k > m, we then have (pm)

S;;;

(a)

S;;;

FURTHER RESULTS ON NOETHERIAN RINGS

Corollary 2. If R i8 a Noetherian local ring whose maximal ideal is principal, then R is a principal ideal ringo Our development has now reached a point where, in order to make further progress, we need to bring in certain results that belong primarily to the theory of modules. The concept of a module is the natural generalization of that of a vector space; instead of requiring the "scalars" to be elements of a field, we now allow them to lie in an arbitrary ring with identity. The major theorem to be established is a remarkable result of Hopkins that every commutative Artinian ring with identity is Noetherian. This theorem does not extend to rings lacking an identity;· indeed, Z(pOO) shows that it is possible for the descending chain condition tobe satisfied in a ring without the ascending chain condition also holding. Apart from sorne standard results about ideals, Hopkin's argument requires only the JordanHolder Theorem for modules (including the fact that a composition series exists if and only if both chain conditions on submodules are satisfied). The proof will not be given immediately, but only after we assemble sorne of the module-theoretic prerequisites. Our discussion is not entirely selfcontained in this regard and certain facts will be presented without proof. The reader who is unfamiliar with modules would profit from working out the details. It is time for these somewhat vague preliminaries to give way to a more precise definition. Definition 12-3. Let R be a ring with identity. Bya lejt module over R (or a left R-module), we mean a commutative group M (written additively) together with an operation of IIlultiplication which associa~es with each r E R and a E M a uniqueelement ra E M such that the following conditions are satisfied: 1) (r

+

2) (rs)a (pk)

S;;;

(pm),

whence the equality (pm) = (pk). The lemma now tells us that pm = pkV, where v is an invertible element of R.. This mean s that a = pkVU, with vu invertible, contradicting the last assertion of the lemma. Corollary 1. LetR bea local ring with principal maximal ideal M = (p). Assume further that n:'=l M" = {O}. If 1 is any nontrivial ideal of R, then 1 = M k for sorne k E Z + (hence, R is Noetherian). Proo! Clearly, 1 S;;; M, so that each nonzero element of 1 can be written . as pnu, with u invertible. Take k to be the least integer such that pku E l. It then follows that 1 S;;; (pk). On the other hand, since pku E 1, so does pk = (pkU)U- 1 ; this implies that (pk) S;;; 1 and equality follows.

247

3) r(a

s)a

=

=

r(sa),

+

ra

b) = ra

+

sa,

+

rb,

4) la = a,

for all r, s E R and a, b E M. The parallel notion of a right R-module can be defined symmetrically, Technically speaking, (left) module multiplication is a function M ~ M, where a(r, a) is denoted by ra. The element ra is often called the module product of r and a. In effect, Definition 12-3 states that a left R-module is an ordered pair (M, a); this approach gets a little cumbersorne and so, when there is no possibility of confusion, we shall lapse into saying "the left R-module M". We pause to look at sorne typical examples of modules. a: R x

246

FIRST COURSE IN RINGS ANO IDEALS

1) for nonzero a, b E M, (a) = (b) if and only if a = bu, where u is invertible in R; 2) if M = (p) =1= {O}, then p is an irreducible element of R; 3) if O =1= a = qnu, with q E M and u in vertible, then this factorization is essentially unique (that is, the integer n is uniquely determined by q). Proo! If (a) = (b), then a = br and b = as for suitable r, s E R. Accordingly, a '= asr or a(l - sr) = O. Suppose that one of the elements r or s lies in M, so that the product sr E M. We observe that 1 - sr fj M, for other-· wise 1 E M. This means that 1 - sr is an invertible element of R (recall that M consists of all the noninvertible elements of R). Then the relation a(l - sr) = O yields a = O, a contradiction. Thus, neither r nor s belongs

to M, which signifies that they are both invertible. The conve~se should' be obvious. Concerning (2), let R = ab. If the element b is not in vertible, then part (1) forces M = (p) e (a). The inaximality of M then ensures that (a) = R, whence a has an inverse in R. To see the final assertion, assume that a = qnu = qmv, with m > n. Then qnu(l - qm-nvu- 1 ) = O. By the I:).rgument of the first paragraph, this relation implies that a = O. Theorem 12-11. Let R be a local ring with principal maximal ideal M = (p). Then every element O =1= a E M has a factorization in the form a = pmu, where u is invertible, if and only if n~= 1 (pn) = {O}.

nn

Proo! Assume that the intersection (p") = {O}. If O =1= a E M, then a = pr for sorne r in R. If it happens that r is not an invertible element, then r E M; thus, we can .write r = ps or, upon substituting, a = p2 s. This process must eventually termínate, for otherwisea wou1dlie in nn (pn) = {O}. For the converse, suppose that there exists sorne 'nQi1zero element a in (p"). In particular, a E (p) = M, so that a = pmu for sorne invertible U.

n"

For any integer k > m, we then have (pm)

S;;;

(a)

S;;;

FURTHER RESULTS ON NOETHERIAN RINGS

Corollary 2. If R i8 a Noetherian local ring whose maximal ideal is principal, then R is a principal ideal ringo Our development has now reached a point where, in order to make further progress, we need to bring in certain results that belong primarily to the theory of modules. The concept of a module is the natural generalization of that of a vector space; instead of requiring the "scalars" to be elements of a field, we now allow them to lie in an arbitrary ring with identity. The major theorem to be established is a remarkable result of Hopkins that every commutative Artinian ring with identity is Noetherian. This theorem does not extend to rings lacking an identity;· indeed, Z(pOO) shows that it is possible for the descending chain condition tobe satisfied in a ring without the ascending chain condition also holding. Apart from sorne standard results about ideals, Hopkin's argument requires only the JordanHolder Theorem for modules (including the fact that a composition series exists if and only if both chain conditions on submodules are satisfied). The proof will not be given immediately, but only after we assemble sorne of the module-theoretic prerequisites. Our discussion is not entirely selfcontained in this regard and certain facts will be presented without proof. The reader who is unfamiliar with modules would profit from working out the details. It is time for these somewhat vague preliminaries to give way to a more precise definition. Definition 12-3. Let R be a ring with identity. Bya lejt module over R (or a left R-module), we mean a commutative group M (written additively) together with an operation of IIlultiplication which associa~es with each r E R and a E M a uniqueelement ra E M such that the following conditions are satisfied: 1) (r

+

2) (rs)a (pk)

S;;;

(pm),

whence the equality (pm) = (pk). The lemma now tells us that pm = pkV, where v is an invertible element of R.. This mean s that a = pkVU, with vu invertible, contradicting the last assertion of the lemma. Corollary 1. LetR bea local ring with principal maximal ideal M = (p). Assume further that n:'=l M" = {O}. If 1 is any nontrivial ideal of R, then 1 = M k for sorne k E Z + (hence, R is Noetherian). Proo! Clearly, 1 S;;; M, so that each nonzero element of 1 can be written . as pnu, with u invertible. Take k to be the least integer such that pku E l. It then follows that 1 S;;; (pk). On the other hand, since pku E 1, so does pk = (pkU)U- 1 ; this implies that (pk) S;;; 1 and equality follows.

247

3) r(a

s)a

=

=

r(sa),

+

ra

b) = ra

+

sa,

+

rb,

4) la = a,

for all r, s E R and a, b E M. The parallel notion of a right R-module can be defined symmetrically, Technically speaking, (left) module multiplication is a function M ~ M, where a(r, a) is denoted by ra. The element ra is often called the module product of r and a. In effect, Definition 12-3 states that a left R-module is an ordered pair (M, a); this approach gets a little cumbersorne and so, when there is no possibility of confusion, we shall lapse into saying "the left R-module M". We pause to look at sorne typical examples of modules. a: R x

248

FIRST COURSE IN RINGS AND IDEALS

If R =' F, where F is any field, then a left R-module is simply a veCtor space over F.

Example 12-3.

Example 12-4. Every cornmutative group (G, +) can be considered as a'

left Z-module in a natural way. For, given an integer n and element a E G, na has a well-defined meaning.: (n summands). na = a + +a

a+ ...

Example 12-5. If 1 is a left ideal of a ring R with identity, then the underlying additive group (1, +) ofjforms a left R~module. Indeed, the definition

of left ideal insures that thedng product ra E 1 for every rE R and a El. As a special case, any ring R'with identity I::an be viewed as a left (or right) R-module over itself. :' ' , Example 12-6. Consider th~set hom G of all homomorphisms of a com-

mutative group (G, + )""into 'itsdf (that is, the set of endomorphisms of G). It is aIread y known that (hpIp.~, +, o) constitutes a ring with identity, where o indica tes the operation of ftiri.étional composition. To provide G with the structure of a left module over hom G, we define the module product fa by putting (f E hom G, a E G). fa = f(a) Condition (3) of Definition 12-3 is satisfied by virtue of the fact that f is a homomorphism. . . To avoid a proliferation of symbols, O will be used to designate the additive identity element of the group (M, +) as well as the zero element of R. This convention should lead to no ambiguity if the reader attends . c10sely to the context in which the notation is employed: As with vector spaces, we have the laws (i) Oa = rO = O, (ii) r(-a) = (-r)a = -(ra), for all r E R and a E M. Oile can introduce the notions of submodule, quotient module, and module homomorphisms, all by natural definitions. These are of fundamental importance for our theory and from them our ultimate goal, Hopkin's Theorem, will follow easily. In the remainder of this discussion, we shall drop the prefix "left", so that the ter m "R-module" will always mean "left R-module"; it should be apparent that the entire discussion applies equally well to right R-modules. Of course, when R is acommutative ring, any left R-module can be turned into a right R-module simply by putting ar = ra. Modules over commuta:tive rings are essentially two-sided and all distinction between left and right disappears (it is merely a matter of personal preference whether one writes the ring elements on the left or on the right). A natural starting point is, perhaps, to .examine the concept of a submodule. Suppose then that M is an arbitrary module over the ring R. By

FURTHER RESULTS ON NOETHERIAN RINGS

249

an R-submodule of M we shall mean a nonempty.subset N of M which is . itself a module relative to the addition and module multiplication of M. To make this idea more precise;' Definition 12-4. A nonempty subset N of the R~module M is an Rsubmodule (or simply a submodule) of M provided that ;/ 1) (N, + ) is a subgroup of (M, +); 2) for~~l rE R and a E N, th~ module product ra E N. Needl~ss,to say, the first condition.in Definition'12-4 is equivalent to requiring;that- if a, b E N, then the difference a - bE N. Every R-module M c1early h~s two trivial submodules, namely, {O} and M itself; a submodule distinct from M is termed proper. Paralle~ing our discussion of rings, we shall call,1I:n R-module M simple, if M =1= {O} and the trivial submodules are its only sijbinodules. It is~ell' worth noting that if M is a vector space over a field F, then any F-su?m~du.~e is }ust a vector subspace of M. A further illustration arises by ~011s1dermga rmg R. as a module over itself; when thisjs done the (left) ldeals of R becomes ltS R-submodules. . The concept of a quotient structure carries over to modules as expected. To be more concrete, let N be a submodule of a given R-module M. Since Misa commutative group, N is automatically normal in M and we can form the quotient group M/N. The elements of this group are just the cosets a + N, with a E M; coset addition is given, as usual, by

(a

+

N)

+

(b

+

N) = a

+

b

+

N.

To equip M/N with the structure of a module, anotionof multiplication by elements of R is introduced by writing .

r(a

+

N)

=

ra

+

N.

We must first satisfy ourselves that module multiplication is una~bi­ guou,sly defined, depending only on the coset a + N and element rE R. This amounts to showing that whenever a + N = a' + N, then r(a + N) = r(a' + N), or, rather, ra + N = ra' + N. Our aim would obviously be achieved if we knew that ra - ra'. = r(a - a') E N. But this follows directlyfrom the fact that a - a' E N and that N is assumed to be a submodule over R. Thus, the module product in M/N is independent of coset representatives. One can easily check that M/N, with the abo ve operations, forms an R-module (the so-caBed quotient module of M by its submodule N). When forming quotient rings, it became necessary to introduce a special subsystem (namely, ideal s) in order to ensure that the operations of the

248

FIRST COURSE IN RINGS AND IDEALS

If R =' F, where F is any field, then a left R-module is simply a veCtor space over F.

Example 12-3.

Example 12-4. Every cornmutative group (G, +) can be considered as a'

left Z-module in a natural way. For, given an integer n and element a E G, na has a well-defined meaning.: (n summands). na = a + +a

a+ ...

Example 12-5. If 1 is a left ideal of a ring R with identity, then the underlying additive group (1, +) ofjforms a left R~module. Indeed, the definition

of left ideal insures that thedng product ra E 1 for every rE R and a El. As a special case, any ring R'with identity I::an be viewed as a left (or right) R-module over itself. :' ' , Example 12-6. Consider th~set hom G of all homomorphisms of a com-

mutative group (G, + )""into 'itsdf (that is, the set of endomorphisms of G). It is aIread y known that (hpIp.~, +, o) constitutes a ring with identity, where o indica tes the operation of ftiri.étional composition. To provide G with the structure of a left module over hom G, we define the module product fa by putting (f E hom G, a E G). fa = f(a) Condition (3) of Definition 12-3 is satisfied by virtue of the fact that f is a homomorphism. . . To avoid a proliferation of symbols, O will be used to designate the additive identity element of the group (M, +) as well as the zero element of R. This convention should lead to no ambiguity if the reader attends . c10sely to the context in which the notation is employed: As with vector spaces, we have the laws (i) Oa = rO = O, (ii) r(-a) = (-r)a = -(ra), for all r E R and a E M. Oile can introduce the notions of submodule, quotient module, and module homomorphisms, all by natural definitions. These are of fundamental importance for our theory and from them our ultimate goal, Hopkin's Theorem, will follow easily. In the remainder of this discussion, we shall drop the prefix "left", so that the ter m "R-module" will always mean "left R-module"; it should be apparent that the entire discussion applies equally well to right R-modules. Of course, when R is acommutative ring, any left R-module can be turned into a right R-module simply by putting ar = ra. Modules over commuta:tive rings are essentially two-sided and all distinction between left and right disappears (it is merely a matter of personal preference whether one writes the ring elements on the left or on the right). A natural starting point is, perhaps, to .examine the concept of a submodule. Suppose then that M is an arbitrary module over the ring R. By

FURTHER RESULTS ON NOETHERIAN RINGS

249

an R-submodule of M we shall mean a nonempty.subset N of M which is . itself a module relative to the addition and module multiplication of M. To make this idea more precise;' Definition 12-4. A nonempty subset N of the R~module M is an Rsubmodule (or simply a submodule) of M provided that ;/ 1) (N, + ) is a subgroup of (M, +); 2) for~~l rE R and a E N, th~ module product ra E N. Needl~ss,to say, the first condition.in Definition'12-4 is equivalent to requiring;that- if a, b E N, then the difference a - bE N. Every R-module M c1early h~s two trivial submodules, namely, {O} and M itself; a submodule distinct from M is termed proper. Paralle~ing our discussion of rings, we shall call,1I:n R-module M simple, if M =1= {O} and the trivial submodules are its only sijbinodules. It is~ell' worth noting that if M is a vector space over a field F, then any F-su?m~du.~e is }ust a vector subspace of M. A further illustration arises by ~011s1dermga rmg R. as a module over itself; when thisjs done the (left) ldeals of R becomes ltS R-submodules. . The concept of a quotient structure carries over to modules as expected. To be more concrete, let N be a submodule of a given R-module M. Since Misa commutative group, N is automatically normal in M and we can form the quotient group M/N. The elements of this group are just the cosets a + N, with a E M; coset addition is given, as usual, by

(a

+

N)

+

(b

+

N) = a

+

b

+

N.

To equip M/N with the structure of a module, anotionof multiplication by elements of R is introduced by writing .

r(a

+

N)

=

ra

+

N.

We must first satisfy ourselves that module multiplication is una~bi­ guou,sly defined, depending only on the coset a + N and element rE R. This amounts to showing that whenever a + N = a' + N, then r(a + N) = r(a' + N), or, rather, ra + N = ra' + N. Our aim would obviously be achieved if we knew that ra - ra'. = r(a - a') E N. But this follows directlyfrom the fact that a - a' E N and that N is assumed to be a submodule over R. Thus, the module product in M/N is independent of coset representatives. One can easily check that M/N, with the abo ve operations, forms an R-module (the so-caBed quotient module of M by its submodule N). When forming quotient rings, it became necessary to introduce a special subsystem (namely, ideal s) in order to ensure that the operations of the

(

250

FIRST COURSE IN RINGS AND IDEALS

condition can also be applied to R-modules, the sole difference being that, in our earlier definitions, the term"ideal" must now be replaced by the word "submodule". Adapting the argument of Theorem 11-2, it is a simple matter to show that an R-niodule M satisfies the ascending (descending) chain condition on submodules if and only if M satisfies the maximum (minimum) condition on submodüles; we leave the verification of this to the reader. The corning theorem indicates how the chain conditions on submodules are affected by certain operations.

in?n

Theorem 12-12. Let N be a submodule ofthe R-module M. there is' a one-to-one inclusion preserving correspondence between the submodules of M/N and those submodules of M which contain N. The notions of homomorphism and isomorphism can be defined for modules in the obvious way.

Theorem 12-14. 1) If the R-module M satisfies the ascending (descending) chain condition, then so does every homomorphic image of M. 2) Let N be a submodule of the R-module M. Then M satisfies the ascending (descending) chain condition if and only if N and M/N both satisfy it.

Definition 12-5. Given two R-modules M and N, a mappingf: M -> N is called a module ~omomorphism or merely an (R-) homomorphism if 1) fis a gro~p homomorphism from (M, + ) into (N, +); 2) f(ra) = rf(a) for all r E R and a E M. Whenfis one-to-one and onto N, it is caBed an (R-) isomorphism; one then says that M and N are (R-) isomorphic and writes M ~ N.

For the most part, the stated results are merely a translation ofTheorems 11-6 and 11-7 into the language of modules. What is new in the present situation is that any submodule N of M inherits the ascending (descending) chain condition. This follows from the fact that any submodule of N is itself a submodule of M (a marked contrast to the behavior of ideals). Before the reader collapses under a burden of definitions, let us turn our attention to the matter of normal and composition series. By a normal series for an R-module M is meant a (finite) chain of Rsubmodules running from M to {O}:

Example 12-7. If R = F, where F is an arbitrary field, the R-homomorphisms are just the linear mappings (linear transformations) from M to N. Example 12-8. Let N be a submodule of the R-module M. The function nat N : M -> M/N which assigns to each element a E M its coset a + N is an R-homomorphism; for, by definition, nat N (ra)

=

ra

+N =

r(a

+

N) = r nat N (a).

As in the"ring-theoretic case, we shall call nat N the natural mapping of M onto the ,qüotient module M/N. With ihe aboye definitions in view, the reader will experience no difticulty in proving'the appropriate results. These are set out in the following omnibus theorem.

Theotem 12-13. Let M and N be two R-modules and f: M -> N be an R-homomorphism from M into N. Then, 1) the kernel off, kerf = {a E Mlf(a) = O}, is a submodule of M; 2) the image of M under f,f(M) = {f(a)la E M}, is a submodule of N; 3) fis a one-to-one function if and only ifker f = {O}; 4) Mjkerf

~

\\, )

,----'

FURTHER RESULTS bN NOETHERIAN RINGS ~··i51

quotient structure were well-defined. Let us emphasize that, in the case of modules, no such distinguished subsystem need be defined; for each submodule of an R-module M, we can construct a quotient module of M. The counterpart of the Correspondence Theorem remains valid for modules and will be relevant to our discussion ; we take the opportunity to t• record this as

·~I,'.

¡'"/ ,

f(M).

At this point, it should come as no surprise that such ideas as the ascending (descending) chain condition and the maximum (minimum)

M = Mo

;;2

I

, ... o',

1

M1

;;2 ... ;;2

M n- I

;;2

Mn =

{O}.

A given normal series can be lengthened or refined by the inserdon of new submodules between those already present. In technical t~r~s, a second normal series

J

is .said to be a refinement of M = Mo ::::> M1 -

::::> ... ::::> -

M n-1

::::> -

Mn

= {O}

provided that there exists a one-to-one function f from {O, 1, ... , n} into {O, 1, ... , m} such that Mi = N f(i)' This amounts to saying that every Mi must appear as one of the N j • A refinement of a normal series is termed proper if the refinement contains a submodule not in the original series. A normal series which adrnits no proper refinement is called a composition series. We summarize this in the following definition.

DefinitioR 12-6. A composition series for an R-module M is a normal series (without repetitions) M = Mo

::::>

MI

::::> '"

:;:>

Mn-I

::::>

M n = {O}

./

(

250

FIRST COURSE IN RINGS AND IDEALS

condition can also be applied to R-modules, the sole difference being that, in our earlier definitions, the term"ideal" must now be replaced by the word "submodule". Adapting the argument of Theorem 11-2, it is a simple matter to show that an R-niodule M satisfies the ascending (descending) chain condition on submodules if and only if M satisfies the maximum (minimum) condition on submodüles; we leave the verification of this to the reader. The corning theorem indicates how the chain conditions on submodules are affected by certain operations.

in?n

Theorem 12-12. Let N be a submodule ofthe R-module M. there is' a one-to-one inclusion preserving correspondence between the submodules of M/N and those submodules of M which contain N. The notions of homomorphism and isomorphism can be defined for modules in the obvious way.

Theorem 12-14. 1) If the R-module M satisfies the ascending (descending) chain condition, then so does every homomorphic image of M. 2) Let N be a submodule of the R-module M. Then M satisfies the ascending (descending) chain condition if and only if N and M/N both satisfy it.

Definition 12-5. Given two R-modules M and N, a mappingf: M -> N is called a module ~omomorphism or merely an (R-) homomorphism if 1) fis a gro~p homomorphism from (M, + ) into (N, +); 2) f(ra) = rf(a) for all r E R and a E M. Whenfis one-to-one and onto N, it is caBed an (R-) isomorphism; one then says that M and N are (R-) isomorphic and writes M ~ N.

For the most part, the stated results are merely a translation ofTheorems 11-6 and 11-7 into the language of modules. What is new in the present situation is that any submodule N of M inherits the ascending (descending) chain condition. This follows from the fact that any submodule of N is itself a submodule of M (a marked contrast to the behavior of ideals). Before the reader collapses under a burden of definitions, let us turn our attention to the matter of normal and composition series. By a normal series for an R-module M is meant a (finite) chain of Rsubmodules running from M to {O}:

Example 12-7. If R = F, where F is an arbitrary field, the R-homomorphisms are just the linear mappings (linear transformations) from M to N. Example 12-8. Let N be a submodule of the R-module M. The function nat N : M -> M/N which assigns to each element a E M its coset a + N is an R-homomorphism; for, by definition, nat N (ra)

=

ra

+N =

r(a

+

N) = r nat N (a).

As in the"ring-theoretic case, we shall call nat N the natural mapping of M onto the ,qüotient module M/N. With ihe aboye definitions in view, the reader will experience no difticulty in proving'the appropriate results. These are set out in the following omnibus theorem.

Theotem 12-13. Let M and N be two R-modules and f: M -> N be an R-homomorphism from M into N. Then, 1) the kernel off, kerf = {a E Mlf(a) = O}, is a submodule of M; 2) the image of M under f,f(M) = {f(a)la E M}, is a submodule of N; 3) fis a one-to-one function if and only ifker f = {O}; 4) Mjkerf

~

\\, )

,----'

FURTHER RESULTS bN NOETHERIAN RINGS ~··i51

quotient structure were well-defined. Let us emphasize that, in the case of modules, no such distinguished subsystem need be defined; for each submodule of an R-module M, we can construct a quotient module of M. The counterpart of the Correspondence Theorem remains valid for modules and will be relevant to our discussion ; we take the opportunity to t• record this as

·~I,'.

¡'"/ ,

f(M).

At this point, it should come as no surprise that such ideas as the ascending (descending) chain condition and the maximum (minimum)

M = Mo

;;2

I

, ... o',

1

M1

;;2 ... ;;2

M n- I

;;2

Mn =

{O}.

A given normal series can be lengthened or refined by the inserdon of new submodules between those already present. In technical t~r~s, a second normal series

J

is .said to be a refinement of M = Mo ::::> M1 -

::::> ... ::::> -

M n-1

::::> -

Mn

= {O}

provided that there exists a one-to-one function f from {O, 1, ... , n} into {O, 1, ... , m} such that Mi = N f(i)' This amounts to saying that every Mi must appear as one of the N j • A refinement of a normal series is termed proper if the refinement contains a submodule not in the original series. A normal series which adrnits no proper refinement is called a composition series. We summarize this in the following definition.

DefinitioR 12-6. A composition series for an R-module M is a normal series (without repetitions) M = Mo

::::>

MI

::::> '"

:;:>

Mn-I

::::>

M n = {O}

./

T

FU~ THER RESUL TS ON NOETHERIAN RINGS

I

, FIRST COURSE IN RINGS AND IDEALS

252

Proa! Suppose that b9th chain conditions and, hence, the maximum and minimum conditions, hold in M. Applying the maximum condition to the set of submodules different from M, we can select a maximal submodule MI e M. Now, either MI = {O} and the proof halts, or there exists a submodule M 2 of MI which is maximal with respect to being proper. Continuing in this way, we get a strict1y decreasing chain of R-submodules ofM:

such that the quotient modules M¡/M¡+ I are all simple; in other words, the inclusions M¡ ;2 N ;2 M¡+I' where N is a submodule of M¡,imply that either N = Mi or N = M¡+I' The number. of submodules in a composition series is called the length of the series. Two normal series for the R-module M, M = Mo :::J MI :::J .,. :::J M n - I ,:::J' M n = {O} and' M

= No :::J NI

::::l. •••

:::J N m -

I

:::J N m

= {O}

M

th~';~,

.J.,!,

,

Expressed differentl'y, two normal series are equivalent iftheir associateq quotient modules are pairwise isomorphic in sorne order. ;" : ........ ..,

\.

Using these definitions, the classic Jordan-Holder Theorem assertsthilf anytwo composition series for an R-module M are equivalent and therefore have' the same length; this commón value is called the length of M and ' denoted by I(M). In effect, an R-module has essentially one composition series. We omit the proof, but the details can be found, for example, in the admirable book by Northcott [:28].

1) M

Z24

:::J (3) :::J (6) :::J (12) :::J {O}

To go still further we need a criterion for the existence of ~omposition series, Theorem 12-15. An R-module M has a composition series if and only if M satisfies both chain conditions for submodules.

---------_..

=

Mo:::J MI :::J ... :::J M n

=

=

{O},

{O}

This implies that the module NjN¡+1 is simple and so the chain (2) actualIy comprises a composition series for M/Mn - I. As a result, we are able to conc1ude that I(M/Mn _ l ) = n - 1. By our induction assumption, the quotient module M/Mn - I must satisfy both the ascending ,and descending chain conditions. Since M n - I is a simple R-module, an appeal to Theorem 12-14 is legitimate; we thus deduce that M itself satisfies both chain conditions for submodules.

and both form composition series for Z24' One way to verify this is to check the orders of the subgroups involved. For instance, to inserf a submodule between (2) and (4) there wquld have to exist a subgroup of Z24 of order n, 6 < n < 12, such that n divides 12 and is itself divisible by 6; clearly, no such subgroup exists.

MI :::J M 2 :::J .... ',

NjN i + 1 = (MjMn-I)/(M¡+dMn-l) ~ MjM¡+I'

is not a composition series, since it may be refined by inserting either of the submodules (4) or (6). (In the situation considered, the notation (n) stands for the cyclic subgroup generated by n.) On the other hand" ' :::J (2) :::J (4) :::J (8) :::J {O}

:::::J

will form a normal chain for the quotient module N = M/M n _ l . Using the first isomorphism theorem for modules (Problem 26, Chapter 12),

:::J (2) :::J (12) :::J {O}

Z24

Mo

then,upon setting Ni = MjMn- l , the chain 2) N = No:::J NI :::J ... :::J N n -¡

Example 12-9., In the Z-module Z24; the normal series Z24

=

By virtue of the descending chain condition such aF~ain must eventually terminate; thus, M n = {O} for sorne integer n and a:'ó6mposition series for ' ' M is o btained. As regards the converse, we proceed by'inductiorÍ on the length I(M) of M. If I(M) = 1, then M ::::l. {O} is a composition,~eries; hence, 'M is a simple module and both chain conditions hold t'tivlally. Next, assume inductively that all R-modules of ltmgth n - 1 sati~fy ,the two chain conditions and let I(M) =:= n. Given any composition series,for M, say

are termed equivalent if they have the same length (n = m) and exists a one-to-one correspondence f between their indices such tha~: ;

253

I

It is hardly necéssary to point out that the concepts of normal ;~ries and compositiori series apply equally well to the ideal s (the R-subrriodules) , of a ring R. In what follows, whenever we speak of a composition series for a ring R, we shall mean a composition series for R considered as a module over itself. We are now ina position, having assembled the necessary mathematical machinery, to aWick Hopkin's Theorem. To set the stage for our presentation, part of the argument is separated off as two lemmas. Hereafter, R will , denote a commutative ring with identity. .

_----~--,,-----------------------

---

------~

--

- - -

T

FU~ THER RESUL TS ON NOETHERIAN RINGS

I

, FIRST COURSE IN RINGS AND IDEALS

252

Proa! Suppose that b9th chain conditions and, hence, the maximum and minimum conditions, hold in M. Applying the maximum condition to the set of submodules different from M, we can select a maximal submodule MI e M. Now, either MI = {O} and the proof halts, or there exists a submodule M 2 of MI which is maximal with respect to being proper. Continuing in this way, we get a strict1y decreasing chain of R-submodules ofM:

such that the quotient modules M¡/M¡+ I are all simple; in other words, the inclusions M¡ ;2 N ;2 M¡+I' where N is a submodule of M¡,imply that either N = Mi or N = M¡+I' The number. of submodules in a composition series is called the length of the series. Two normal series for the R-module M, M = Mo :::J MI :::J .,. :::J M n - I ,:::J' M n = {O} and' M

= No :::J NI

::::l. •••

:::J N m -

I

:::J N m

= {O}

M

th~';~,

.J.,!,

,

Expressed differentl'y, two normal series are equivalent iftheir associateq quotient modules are pairwise isomorphic in sorne order. ;" : ........ ..,

\.

Using these definitions, the classic Jordan-Holder Theorem assertsthilf anytwo composition series for an R-module M are equivalent and therefore have' the same length; this commón value is called the length of M and ' denoted by I(M). In effect, an R-module has essentially one composition series. We omit the proof, but the details can be found, for example, in the admirable book by Northcott [:28].

1) M

Z24

:::J (3) :::J (6) :::J (12) :::J {O}

To go still further we need a criterion for the existence of ~omposition series, Theorem 12-15. An R-module M has a composition series if and only if M satisfies both chain conditions for submodules.

---------_..

=

Mo:::J MI :::J ... :::J M n

=

=

{O},

{O}

This implies that the module NjN¡+1 is simple and so the chain (2) actualIy comprises a composition series for M/Mn - I. As a result, we are able to conc1ude that I(M/Mn _ l ) = n - 1. By our induction assumption, the quotient module M/Mn - I must satisfy both the ascending ,and descending chain conditions. Since M n - I is a simple R-module, an appeal to Theorem 12-14 is legitimate; we thus deduce that M itself satisfies both chain conditions for submodules.

and both form composition series for Z24' One way to verify this is to check the orders of the subgroups involved. For instance, to inserf a submodule between (2) and (4) there wquld have to exist a subgroup of Z24 of order n, 6 < n < 12, such that n divides 12 and is itself divisible by 6; clearly, no such subgroup exists.

MI :::J M 2 :::J .... ',

NjN i + 1 = (MjMn-I)/(M¡+dMn-l) ~ MjM¡+I'

is not a composition series, since it may be refined by inserting either of the submodules (4) or (6). (In the situation considered, the notation (n) stands for the cyclic subgroup generated by n.) On the other hand" ' :::J (2) :::J (4) :::J (8) :::J {O}

:::::J

will form a normal chain for the quotient module N = M/M n _ l . Using the first isomorphism theorem for modules (Problem 26, Chapter 12),

:::J (2) :::J (12) :::J {O}

Z24

Mo

then,upon setting Ni = MjMn- l , the chain 2) N = No:::J NI :::J ... :::J N n -¡

Example 12-9., In the Z-module Z24; the normal series Z24

=

By virtue of the descending chain condition such aF~ain must eventually terminate; thus, M n = {O} for sorne integer n and a:'ó6mposition series for ' ' M is o btained. As regards the converse, we proceed by'inductiorÍ on the length I(M) of M. If I(M) = 1, then M ::::l. {O} is a composition,~eries; hence, 'M is a simple module and both chain conditions hold t'tivlally. Next, assume inductively that all R-modules of ltmgth n - 1 sati~fy ,the two chain conditions and let I(M) =:= n. Given any composition series,for M, say

are termed equivalent if they have the same length (n = m) and exists a one-to-one correspondence f between their indices such tha~: ;

253

I

It is hardly necéssary to point out that the concepts of normal ;~ries and compositiori series apply equally well to the ideal s (the R-subrriodules) , of a ring R. In what follows, whenever we speak of a composition series for a ring R, we shall mean a composition series for R considered as a module over itself. We are now ina position, having assembled the necessary mathematical machinery, to aWick Hopkin's Theorem. To set the stage for our presentation, part of the argument is separated off as two lemmas. Hereafter, R will , denote a commutative ring with identity. .

_----~--,,-----------------------

---

------~

--

- - -

r· 254

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

Lemma l.. Suppose that in the ring R the zero ideal is a product of maximal ideals, say {O} = M 1M 2 ••• Mil' If (M 1M 2

•••

M¡-1)/(M 1M 2

•••

M¡)

viewed as a vector space over R/M¡is finite dimensionalfor i = 1, 2, ... , n, then R has a composition series. Proof. We first set Ni = M 1M 2 ••• Mi for i = 1,2, ... , n. Observe that the quotient R-module N¡_ ¡/N i can be regarded as a vector space over the field R/M¡. Its elements are simply the cósets x + Ni' with x E N i - l , and scalar multiplication is defined (on the right) by (x

+

N¡)(r

+

M¡) = xr

+

(r

N¡

E

R).

Since N¡_¡/N¡ is annihilated by Mi' this definition makes sense; in fact, if x - x' E N¡, where x, x.' lie in N i - l , and r - r' E Mi' we necessarily have xr -'- x'r'

=

x(r - r')

+ (x

- x')r' E Ni-1M i

+

Ni S Ni'

Let us now consider the descending chain R

=

No

;;2

Nl

;;2

N2

;;2 ••• ;;2

N.

= {O}

of R-submodules. It is well known that any finite dimensional vector space admits a composition series [33]. This being so, our hypothesis guarantees that N¡_¡/N i has a composition series as an R/Mi-module and, hence, ·as an R-module. (Let us stress that, by virtue of the definition of scalar multiplication, the R/Mesubmodules of Ni-l/N¡ are identical with the R-submodules of Ni_¡/N i .) Using Theorem 12-g a composition series can therefore be inserted between N¡-l and Ni,~'By putting all thesé series together, we obtain a composition series for.R.itself. Lemma 2. If in the ring R, {O} = M1Prf~ ... M., where the Mi are maximal ideals, then either chain condition implies the other. '-1',"".

Proof. Again, let Ni = M 1M 2 ••• Mi for i ·~"1, 2, ... , n and consider the quotient module N¡_l/N¡ as a vector space)wer R/M i . Now, N¡-l/N i fórms an R-submodule of R/N¡ which, in tum; is a homomorphic image of R; it follows that if either chain condition on ideals holds in R, then N i - ¡/N ¡ must satisfy the corresponding chain conditio'n on R-submodules (hence, on R/M¡-subspac~s). But, in a vector space, either chain condition implies that the space is finite dimensional [33]. From Lemma 1, the ring R thereby admits a composition series and so, with the aid of Theorem 12-15, we conc1ude that both chain conditions hold in R.

Here now is the main result of this chapter; our proof follows the lines of [36].

255

Theorem 12-16. (Akizuki-Hopkins). A ring R is Artinian if and only if R is Noetherian and every proper prime "ideal of R is maximal. Proof. We first suppose that R satisfies the ascending chain condition and - that every proper prime ideal of R is maximal. By Problem 4, Chapter 11, every ideal of R contains a product of prime ideals (the Noetherian hypothesis ensures this). In particular, {O} must be a product of prime, and therefore maximal, ideals. That R is Artinian follows immediately from Lemma 2. Conversely, let us now as sume that R satisfies the descending chain condition. If Pis any proper prime ideal of R, the quotient ring R/P also satisfies this chain condition and, of course, is an integral domain. Appealing to Theorem 11-8, we see that R/P is in fact a field, whence P forms a maximal ideal. To prove that R is Noetherian, it is again enough to establish that {O} is a product of prime (hence, maximal) ideals. We assert first that forevery proper ideal l. of R there exists a prime ideal P for which l el: P. To see this, define the family fF by fF

=

{JIJ is an ideal of R; 1: J

=1=

R}.

fF surely is not empty, because R is a member of fF. Use the minimum condition to select an ideal J' which is minimal in this collection. Then, P = 1: J' forms a prime ideal of R. If not, there would exist elements a, b not in P such that their product ab E P. Therefore, P e P: (a) e R.

That is to say, 1: J' e l :,J'a =1= R, whence J'a e J'. Since this contradicts the minimal nature of J':'in $', P must indeed be a prime ideal. Now, the quotient ideal 1: P ;;2 1; inasmuch as 1: P ;;2 J, which is not contained in 1, it follows that l e 1: Pand our assertion is proved. (IncidentaUy, this argument proves the existtmce of prime ideals in R.) For the final stage of,the proof, let K be minimal in the set of those ideal s of R which are products 0fiprime ideals. If K =1= {O}, then the ideal l = O: K is different from R, for 1 fft l. By the last paragraph, there is a prime ideal P of R such that l el: P; in other words, O: K e O: KP. This implies that the ideal KP e K and contradicts the minimality of K. In consequence, K = {O}, so that {O} is a product of prime (maximal) ideals. Lemnia 2 now completes our task. .Corollary. Any commutative Artinian ring with identity is Noetherian. Having come this far, it might be useful to pro ve Fitting's Lemma, a result which requires both chain conditions on submodules. First, we pause to establish a fact of independent interest. .

r· 254

FIRST COURSE IN RINGS AND IDEALS

FURTHER RESULTS ON NOETHERIAN RINGS

Lemma l.. Suppose that in the ring R the zero ideal is a product of maximal ideals, say {O} = M 1M 2 ••• Mil' If (M 1M 2

•••

M¡-1)/(M 1M 2

•••

M¡)

viewed as a vector space over R/M¡is finite dimensionalfor i = 1, 2, ... , n, then R has a composition series. Proof. We first set Ni = M 1M 2 ••• Mi for i = 1,2, ... , n. Observe that the quotient R-module N¡_ ¡/N i can be regarded as a vector space over the field R/M¡. Its elements are simply the cósets x + Ni' with x E N i - l , and scalar multiplication is defined (on the right) by (x

+

N¡)(r

+

M¡) = xr

+

(r

N¡

E

R).

Since N¡_¡/N¡ is annihilated by Mi' this definition makes sense; in fact, if x - x' E N¡, where x, x.' lie in N i - l , and r - r' E Mi' we necessarily have xr -'- x'r'

=

x(r - r')

+ (x

- x')r' E Ni-1M i

+

Ni S Ni'

Let us now consider the descending chain R

=

No

;;2

Nl

;;2

N2

;;2 ••• ;;2

N.

= {O}

of R-submodules. It is well known that any finite dimensional vector space admits a composition series [33]. This being so, our hypothesis guarantees that N¡_¡/N i has a composition series as an R/Mi-module and, hence, ·as an R-module. (Let us stress that, by virtue of the definition of scalar multiplication, the R/Mesubmodules of Ni-l/N¡ are identical with the R-submodules of Ni_¡/N i .) Using Theorem 12-g a composition series can therefore be inserted between N¡-l and Ni,~'By putting all thesé series together, we obtain a composition series for.R.itself. Lemma 2. If in the ring R, {O} = M1Prf~ ... M., where the Mi are maximal ideals, then either chain condition implies the other. '-1',"".

Proof. Again, let Ni = M 1M 2 ••• Mi for i ·~"1, 2, ... , n and consider the quotient module N¡_l/N¡ as a vector space)wer R/M i . Now, N¡-l/N i fórms an R-submodule of R/N¡ which, in tum; is a homomorphic image of R; it follows that if either chain condition on ideals holds in R, then N i - ¡/N ¡ must satisfy the corresponding chain conditio'n on R-submodules (hence, on R/M¡-subspac~s). But, in a vector space, either chain condition implies that the space is finite dimensional [33]. From Lemma 1, the ring R thereby admits a composition series and so, with the aid of Theorem 12-15, we conc1ude that both chain conditions hold in R.

Here now is the main result of this chapter; our proof follows the lines of [36].

255

Theorem 12-16. (Akizuki-Hopkins). A ring R is Artinian if and only if R is Noetherian and every proper prime "ideal of R is maximal. Proof. We first suppose that R satisfies the ascending chain condition and - that every proper prime ideal of R is maximal. By Problem 4, Chapter 11, every ideal of R contains a product of prime ideals (the Noetherian hypothesis ensures this). In particular, {O} must be a product of prime, and therefore maximal, ideals. That R is Artinian follows immediately from Lemma 2. Conversely, let us now as sume that R satisfies the descending chain condition. If Pis any proper prime ideal of R, the quotient ring R/P also satisfies this chain condition and, of course, is an integral domain. Appealing to Theorem 11-8, we see that R/P is in fact a field, whence P forms a maximal ideal. To prove that R is Noetherian, it is again enough to establish that {O} is a product of prime (hence, maximal) ideals. We assert first that forevery proper ideal l. of R there exists a prime ideal P for which l el: P. To see this, define the family fF by fF

=

{JIJ is an ideal of R; 1: J

=1=

R}.

fF surely is not empty, because R is a member of fF. Use the minimum condition to select an ideal J' which is minimal in this collection. Then, P = 1: J' forms a prime ideal of R. If not, there would exist elements a, b not in P such that their product ab E P. Therefore, P e P: (a) e R.

That is to say, 1: J' e l :,J'a =1= R, whence J'a e J'. Since this contradicts the minimal nature of J':'in $', P must indeed be a prime ideal. Now, the quotient ideal 1: P ;;2 1; inasmuch as 1: P ;;2 J, which is not contained in 1, it follows that l e 1: Pand our assertion is proved. (IncidentaUy, this argument proves the existtmce of prime ideals in R.) For the final stage of,the proof, let K be minimal in the set of those ideal s of R which are products 0fiprime ideals. If K =1= {O}, then the ideal l = O: K is different from R, for 1 fft l. By the last paragraph, there is a prime ideal P of R such that l el: P; in other words, O: K e O: KP. This implies that the ideal KP e K and contradicts the minimality of K. In consequence, K = {O}, so that {O} is a product of prime (maximal) ideals. Lemnia 2 now completes our task. .Corollary. Any commutative Artinian ring with identity is Noetherian. Having come this far, it might be useful to pro ve Fitting's Lemma, a result which requires both chain conditions on submodules. First, we pause to establish a fact of independent interest. .

256

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

exists somey E M withjn(x) = pn(y). Thus,f"(X lently, x - f"(y) E ker f". As a result,

Lemma. Let M be an R-module satisfying both chain conditions and let f: M- M be an R-homomorphism. Then f is a one-to-one function if and only iffmaps onto M.

f"(y»)

257

O, or, equiva-

+ (x - f"(y») E f"(M) + ker f", + ker 1".

x = f"(y)

Proa! To start, suppose that f is one-to-one and consider the chain of

which forces R = f"(M)

R-submodules

As we noted eartier, any finite dimens~onal vector space satisfies both chain conditions (on subspaces). This being the case, Fítting's Lemma can be interpreted in v,~ctorspace terminology as asserting

.

M I

2.

f(M)

2.

f2(M)

2. ....

.

Since M satisfies the descending chain:conditiori, this chain will termínate after a finite number of steps, say n steps; then ¡n(M} = ¡n+ ¡(M). Given an arbitrary x E M, ¡n(x) = ¡n+ ¡(y) for' suitable y in M. As f is assumed to be a one-to-one function,¡n also enjoys this property, whence x = f(y}· The implication is that M = f(M) andso fmaps onto M. . Next, letf carry the, set Monto itself: NQtice that we have the following ascending chain of R-sQbmodules:, ..

Corollary. Lta(V be a finite dimensional vector space and let f: V - V be a linear tEtinsformation. Then V = W¡ (El W2,· where, W¡ and Wz are both inv
Proa! Take Wi: =:= f"(V) and Wz

= ker f", . as indícated aboye. By the lemma to Theo~eni~12-17, the restríction flW¡ being an onto mapping is also one-to-one;'hence, a ve~tor space isomorphistn (to put it another way, fl W¡ is a nonsingular transfoÍmation),

{~} ~ kerf ~."t¿erp ~"'. ::'. . ~

By hypothésis, there exists an integet m for which kerfm = kerfm+l. Select any XE M wÍthf(x} = O. Inasmuch asfmaps onto M, so also must fm. Thus, it is possible to choose an element y E M s:uch that fm(y) = x. Butthen¡m+l(y) =f(x} = O,implyingthaty.Ekerfm+l = kerfm, Accordingly,x = fm(y} = O and, hence, ker f 7= {O}. This makes f a one-to-one function and we are done.

PROBLEMS Unles8 indicated to the contrary, all rings considered are assumed to be commutative . with identity,

1. Let 1 be a semiprime ideal of the ring

R; Prove that 1 is a prime ideal if and only . ifit is irreducible. [Hint: If1 is irreducible, but not prime, then there exist elements a, b ~ 1 with ab El; argue tha! (1, a) n (J, b) ~ 1.J

. The result which we have in mind is stated below.

-JT:2

Theorem 12-17. (Fitting's Lemma}.Let the R-module M satisfy both chain conditions. Given an R-homomorphismf: M - M, there exists sorne n. E Z + such that M

= ¡n(M) (El

2. a) In the polynomial ring F[xJ, where F is a field, show thatthe ideal (x 2 , 2x, 4) Is primary, but reducible. [Hint: (X2, 2x, 4) = (X2, 2) n (x, 4).] b) Express the ideal (x 2 , xy, 3) as an intersection ofprimary ideals in F[x, y].

ker ¡no

3. Let R be a Noetherian ring, and 1 and J two ideals of R with J

n7=

submodules at our disposal: .

M2. f(M)

2.

P(M)

2. ... ,

{O} ~ kerf ~ kerf2 ~ ... , Because both chain conditions hold, each of these chains ultimately stops, for instance, after r and s steps, respectively. The theorem now follows on taking n to be the larger of r and S. For, suppose that x E¡n(M) (\ ker!"; then f"(y) = x for sorne y E M, O. Therefóre, pn(y) = f"(x) = O, so that y lies in ker pn = while ¡n{x) ker 1". But this means that x = ¡n(y) = O, . whence the intersectíon ¡n(M) (\ ker ¡n = {O}. Now, pick any element x E M. Since ¡n(x) E ¡n(M) = P"(M), there

= ni=

S;;; 1. If 1 1 Q¡ is an irredundant primary representation of 1, establish that . a) 1/1 = ¡ (QJJ) is an irredundant primary representation of the ideal 1/1 in R/J; b) .J(QJJ) =JQJJ are the associated prime ideals of J/J.

Proa! As observea in the proof of the lemma, we have two chains of R-

4. Find an irredundant primary representation for the ideal (x 2 ,2xy) in F[x, y J, F a field.; detemúne the asso.ciated prime ideals of (x?, 2xy), as well as its mínima! primes. [Hint: (x A, 2xy) (x 2 , xy, yA) n (x) n (x A,2x, 4).J

I~

,'¡

1:

5. Let 1 be an ideal of the Noetherian ring R. Without recourse to Problem 20 Chapter 8, prove the statements below: . . '. al .JI is the intersection of the mínimal prime ideals of 1. [Hint: If 1 has the ni=¡ .JQ¡.] irredundant, primary representation 1 = ni=¡ Q¡, then.JI b) The set of nllpotent elements of R í8 the intersection of the mínimal prime ideals of R.

256

PROBLEMS

FIRST COURSE IN RINGS AND IDEALS

exists somey E M withjn(x) = pn(y). Thus,f"(X lently, x - f"(y) E ker f". As a result,

Lemma. Let M be an R-module satisfying both chain conditions and let f: M- M be an R-homomorphism. Then f is a one-to-one function if and only iffmaps onto M.

f"(y»)

257

O, or, equiva-

+ (x - f"(y») E f"(M) + ker f", + ker 1".

x = f"(y)

Proa! To start, suppose that f is one-to-one and consider the chain of

which forces R = f"(M)

R-submodules

As we noted eartier, any finite dimens~onal vector space satisfies both chain conditions (on subspaces). This being the case, Fítting's Lemma can be interpreted in v,~ctorspace terminology as asserting

.

M I

2.

f(M)

2.

f2(M)

2. ....

.

Since M satisfies the descending chain:conditiori, this chain will termínate after a finite number of steps, say n steps; then ¡n(M} = ¡n+ ¡(M). Given an arbitrary x E M, ¡n(x) = ¡n+ ¡(y) for' suitable y in M. As f is assumed to be a one-to-one function,¡n also enjoys this property, whence x = f(y}· The implication is that M = f(M) andso fmaps onto M. . Next, letf carry the, set Monto itself: NQtice that we have the following ascending chain of R-sQbmodules:, ..

Corollary. Lta(V be a finite dimensional vector space and let f: V - V be a linear tEtinsformation. Then V = W¡ (El W2,· where, W¡ and Wz are both inv
Proa! Take Wi: =:= f"(V) and Wz

= ker f", . as indícated aboye. By the lemma to Theo~eni~12-17, the restríction flW¡ being an onto mapping is also one-to-one;'hence, a ve~tor space isomorphistn (to put it another way, fl W¡ is a nonsingular transfoÍmation),

{~} ~ kerf ~."t¿erp ~"'. ::'. . ~

By hypothésis, there exists an integet m for which kerfm = kerfm+l. Select any XE M wÍthf(x} = O. Inasmuch asfmaps onto M, so also must fm. Thus, it is possible to choose an element y E M s:uch that fm(y) = x. Butthen¡m+l(y) =f(x} = O,implyingthaty.Ekerfm+l = kerfm, Accordingly,x = fm(y} = O and, hence, ker f 7= {O}. This makes f a one-to-one function and we are done.

PROBLEMS Unles8 indicated to the contrary, all rings considered are assumed to be commutative . with identity,

1. Let 1 be a semiprime ideal of the ring

R; Prove that 1 is a prime ideal if and only . ifit is irreducible. [Hint: If1 is irreducible, but not prime, then there exist elements a, b ~ 1 with ab El; argue tha! (1, a) n (J, b) ~ 1.J

. The result which we have in mind is stated below.

-JT:2

Theorem 12-17. (Fitting's Lemma}.Let the R-module M satisfy both chain conditions. Given an R-homomorphismf: M - M, there exists sorne n. E Z + such that M

= ¡n(M) (El

2. a) In the polynomial ring F[xJ, where F is a field, show thatthe ideal (x 2 , 2x, 4) Is primary, but reducible. [Hint: (X2, 2x, 4) = (X2, 2) n (x, 4).] b) Express the ideal (x 2 , xy, 3) as an intersection ofprimary ideals in F[x, y].

ker ¡no

3. Let R be a Noetherian ring, and 1 and J two ideals of R with J

n7=

submodules at our disposal: .

M2. f(M)

2.

P(M)

2. ... ,

{O} ~ kerf ~ kerf2 ~ ... , Because both chain conditions hold, each of these chains ultimately stops, for instance, after r and s steps, respectively. The theorem now follows on taking n to be the larger of r and S. For, suppose that x E¡n(M) (\ ker!"; then f"(y) = x for sorne y E M, O. Therefóre, pn(y) = f"(x) = O, so that y lies in ker pn = while ¡n{x) ker 1". But this means that x = ¡n(y) = O, . whence the intersectíon ¡n(M) (\ ker ¡n = {O}. Now, pick any element x E M. Since ¡n(x) E ¡n(M) = P"(M), there

= ni=

S;;; 1. If 1 1 Q¡ is an irredundant primary representation of 1, establish that . a) 1/1 = ¡ (QJJ) is an irredundant primary representation of the ideal 1/1 in R/J; b) .J(QJJ) =JQJJ are the associated prime ideals of J/J.

Proa! As observea in the proof of the lemma, we have two chains of R-

4. Find an irredundant primary representation for the ideal (x 2 ,2xy) in F[x, y J, F a field.; detemúne the asso.ciated prime ideals of (x?, 2xy), as well as its mínima! primes. [Hint: (x A, 2xy) (x 2 , xy, yA) n (x) n (x A,2x, 4).J

I~

,'¡

1:

5. Let 1 be an ideal of the Noetherian ring R. Without recourse to Problem 20 Chapter 8, prove the statements below: . . '. al .JI is the intersection of the mínimal prime ideals of 1. [Hint: If 1 has the ni=¡ .JQ¡.] irredundant, primary representation 1 = ni=¡ Q¡, then.JI b) The set of nllpotent elements of R í8 the intersection of the mínimal prime ideals of R.

258

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

.JI

e) is a prime ideal if and only if 1 has a single minimal prime. d) If Pis a minimal prinle ideal of 1, then the primary eomponent eorresponding to P is the same for all irredundant primary representations of l. 6. Let 1 be an ideal of a Noetherian ring R in whieh every nontrivial prime ideal is maxima!. Show that 1 is a produet of primary ideals. [Hint: If 1 = ni Qi' then the ideals .jQ¡ are pairwise eomaximal when non trivial ; now use Problem 13, Chapter 1O.J

7. Let R be a Noetherian ring, and 1 and J two ideals of R with 1 f K Prove that l:J = 1 if and only if J is eontained in no assoeiated prime ideal of l. [Hint: Assume that 1 = ni Q¡, Qi prinlary. If J .jQ; for all i, then, by Problem 24(a), Chapter 5, Q¡:J = Q¡. Conversely, let l:J = l. If J s .jQ;., then Jn s (.jQ,Jn S Qk for some n; whenee 1 = 1:J" = ni (Qi :J") = ni'/'k (Qi :J") 2 ni'/'k Q¡ 2 J.J

*

8. Given that R is a Noetherian ring, prove thilt a) An element a belongs to some associated prime ideal of the ideal 1 if and only if there exists some b ~ 1 for whieh ab E l. [Hint: Apply Problem 7 to the ideal J = (a).J b) The set of all zero divisors of R together with zero is the union of the associated prime ideals of {O}. [Hint: Part (a) with 1 = {O}.J e) An ideal 1 of R eonsists entirely of zero divisor s (along with O) if and only if 1 is eontained in some assoeiated prime ideal of {O}. [Hint: Part (b) and Theorem 5-16.J 9. Let 1 be an ideal of the ring KAn element a E R is said to be related to 1 if there exists some r ~ 1 su eh that ar E 1. Prove eaeh of the assertions below: a) An elemerit ti E R is related to 1 if and only if the quotiellt ideal 1 :(a) f l. b) An element a E R is related to 1 if and only if the eoset a + 1 is either zero or a divisor of zero in R/l. e) Every elemen t of the ni! radical .jI is related to l. d) If R is Noetherian and 1.= Q¡ is an irredundant primary representation of 1, then an element a E R is related to 1 if and only if a E Ui .jQ¡. [Hint: Problem 8(a).]

.nr=¡

10. Assume that R is a principal ideal ring with zero prinle radica!. Deduce that the .. zero ideal is the interseetion of a finite number of prinle ideals. 11. Given that 1 is an ideal of the Noetherian ring R, establish the following: a) If 1 S rad R, then 1" = {O}. b) (1 + (rad R)n) = 1. [Hint: Apply part (a) to R/l.J e) If 1 + radR = R, then 1 = K [Hint: R = Rn = (1 + radRf S {l + (rad R)n) = 1.J

n:=l

n:..¡

n:=¡

n:..¡

12. Let R be a Noetherian local ring with maximal ideal M. a) Verify that the interseetion Mn = {O}. b) If 1 is any ideal of R for whieh M = 1 + M 2 , prove that 1 = M. [Hint: From M = 1 + M = 1 + M(I + M 2 ) = 1 + M = "', deduce that M = ¡ (I + Mn) = 1.J

n:..¡

2

3

n:=

259

13. a) Derive the Krull Interseetion Theorem from Theorem 12-10. [Hint: Problem 8(b).J . . b) Show that the set-theoretie eondition .jQ; n (1 - l·) = 0 appearmg In Theorem 12-10 is equivalent to requiring that .jQi + 1 f K 14. Let 1 be a proper ideal of the integral domain K Assume further that, for any ideal J of R, there exists an integer k for whieh lk n J S IJ (when R is a Noetherian domain, every ideal! has this property [33J). Prove that the interseetion 1" = {O}. [Hint: Takea E In and eonsider the prineipalidealJ = (a).J

n:..¡

n:= ¡

15. Suppose that R is a local ring whose maxinlal ideal M is principal, say M = (p). If M is a ni! ideal of R, pro ve that a) M is a nilpotent ideal of K b) For any proper ideal 1 of R, 1 = ann (ann 1). [Hint: By Theorem 12-11, 1 = (p~ for sorne integer k.J 16. Let R be a ring possessing an ideal M whieh is both maximill and ni!. Verify that R is a local ring with unique maximal ideal M. [Hint: Ifthe element a ~ M, show that a is invertible by expanding (ab - l)n.J In Problems 17-26, the term R-module means left R-module.

17. Prove the following statements coneerning submodules of the R-modules M: a) A nonempty subset N !;;; M forms a submodule of M if and only if (i) x, y E N imply x + yE N and (ii) XE N, r E R imply that rx E N .. b) If S is a subring ofthe ring R, then every R-submodule of M can also be regarded as an S-submodule. e) If 1 is an ideal of R and x a fixed element of M, then the set N:x; = {rxlx E l} forms a submodule of M. 18. a) Verify that the submodule [SJ of ~e R-module M genetat~d by a nonempty subset S S;;; M eonsists. of all finite R-linear eombinations''Of elements of S; that is, [SJ = rix;jri E R, Xi E S}.

U::

. .,~. ~

b) Let j; g: M -> N be two R-homomorphisms of the R-IÍl~dule M into the R-module N. Iff(x) = g(x) for every x in a nonempty subs~t $ !;;; M, show that f and 9 agree on the submodule [S]. ., 19. An element x of an R-module Mis said to be a torsion element'if there exists sorne r f O in R for whieh rx = O. Show that the set T of torsion elements of M forms a submoduleof M and that the quotient module M/T is torsion-free (in other words, M/T has no nonzero torsion elements). 20. Let f: M -> N be an R-homomorphism of the sinlple R-module M into the Rmodule N. Establish thatf(M) is a simple submodule of N and thatfis one-to-one whenever f(M) f {O}. 21. Let MI> M 2 , ... , M n be submodules ofthe R-module M. We eall M the (internal) direct sum of MI> M 2 , ... , M n and write M = MI ® M 2 ® ... E9 M n if i) M.= MI + M 2 + ... + M n = {Xl + X2 + ... + XnIXkEMd, and

258

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

.JI

e) is a prime ideal if and only if 1 has a single minimal prime. d) If Pis a minimal prinle ideal of 1, then the primary eomponent eorresponding to P is the same for all irredundant primary representations of l. 6. Let 1 be an ideal of a Noetherian ring R in whieh every nontrivial prime ideal is maxima!. Show that 1 is a produet of primary ideals. [Hint: If 1 = ni Qi' then the ideals .jQ¡ are pairwise eomaximal when non trivial ; now use Problem 13, Chapter 1O.J

7. Let R be a Noetherian ring, and 1 and J two ideals of R with 1 f K Prove that l:J = 1 if and only if J is eontained in no assoeiated prime ideal of l. [Hint: Assume that 1 = ni Q¡, Qi prinlary. If J .jQ; for all i, then, by Problem 24(a), Chapter 5, Q¡:J = Q¡. Conversely, let l:J = l. If J s .jQ;., then Jn s (.jQ,Jn S Qk for some n; whenee 1 = 1:J" = ni (Qi :J") = ni'/'k (Qi :J") 2 ni'/'k Q¡ 2 J.J

*

8. Given that R is a Noetherian ring, prove thilt a) An element a belongs to some associated prime ideal of the ideal 1 if and only if there exists some b ~ 1 for whieh ab E l. [Hint: Apply Problem 7 to the ideal J = (a).J b) The set of all zero divisors of R together with zero is the union of the associated prime ideals of {O}. [Hint: Part (a) with 1 = {O}.J e) An ideal 1 of R eonsists entirely of zero divisor s (along with O) if and only if 1 is eontained in some assoeiated prime ideal of {O}. [Hint: Part (b) and Theorem 5-16.J 9. Let 1 be an ideal of the ring KAn element a E R is said to be related to 1 if there exists some r ~ 1 su eh that ar E 1. Prove eaeh of the assertions below: a) An elemerit ti E R is related to 1 if and only if the quotiellt ideal 1 :(a) f l. b) An element a E R is related to 1 if and only if the eoset a + 1 is either zero or a divisor of zero in R/l. e) Every elemen t of the ni! radical .jI is related to l. d) If R is Noetherian and 1.= Q¡ is an irredundant primary representation of 1, then an element a E R is related to 1 if and only if a E Ui .jQ¡. [Hint: Problem 8(a).]

.nr=¡

10. Assume that R is a principal ideal ring with zero prinle radica!. Deduce that the .. zero ideal is the interseetion of a finite number of prinle ideals. 11. Given that 1 is an ideal of the Noetherian ring R, establish the following: a) If 1 S rad R, then 1" = {O}. b) (1 + (rad R)n) = 1. [Hint: Apply part (a) to R/l.J e) If 1 + radR = R, then 1 = K [Hint: R = Rn = (1 + radRf S {l + (rad R)n) = 1.J

n:=l

n:..¡

n:=¡

n:..¡

12. Let R be a Noetherian local ring with maximal ideal M. a) Verify that the interseetion Mn = {O}. b) If 1 is any ideal of R for whieh M = 1 + M 2 , prove that 1 = M. [Hint: From M = 1 + M = 1 + M(I + M 2 ) = 1 + M = "', deduce that M = ¡ (I + Mn) = 1.J

n:..¡

2

3

n:=

259

13. a) Derive the Krull Interseetion Theorem from Theorem 12-10. [Hint: Problem 8(b).J . . b) Show that the set-theoretie eondition .jQ; n (1 - l·) = 0 appearmg In Theorem 12-10 is equivalent to requiring that .jQi + 1 f K 14. Let 1 be a proper ideal of the integral domain K Assume further that, for any ideal J of R, there exists an integer k for whieh lk n J S IJ (when R is a Noetherian domain, every ideal! has this property [33J). Prove that the interseetion 1" = {O}. [Hint: Takea E In and eonsider the prineipalidealJ = (a).J

n:..¡

n:= ¡

15. Suppose that R is a local ring whose maxinlal ideal M is principal, say M = (p). If M is a ni! ideal of R, pro ve that a) M is a nilpotent ideal of K b) For any proper ideal 1 of R, 1 = ann (ann 1). [Hint: By Theorem 12-11, 1 = (p~ for sorne integer k.J 16. Let R be a ring possessing an ideal M whieh is both maximill and ni!. Verify that R is a local ring with unique maximal ideal M. [Hint: Ifthe element a ~ M, show that a is invertible by expanding (ab - l)n.J In Problems 17-26, the term R-module means left R-module.

17. Prove the following statements coneerning submodules of the R-modules M: a) A nonempty subset N !;;; M forms a submodule of M if and only if (i) x, y E N imply x + yE N and (ii) XE N, r E R imply that rx E N .. b) If S is a subring ofthe ring R, then every R-submodule of M can also be regarded as an S-submodule. e) If 1 is an ideal of R and x a fixed element of M, then the set N:x; = {rxlx E l} forms a submodule of M. 18. a) Verify that the submodule [SJ of ~e R-module M genetat~d by a nonempty subset S S;;; M eonsists. of all finite R-linear eombinations''Of elements of S; that is, [SJ = rix;jri E R, Xi E S}.

U::

. .,~. ~

b) Let j; g: M -> N be two R-homomorphisms of the R-IÍl~dule M into the R-module N. Iff(x) = g(x) for every x in a nonempty subs~t $ !;;; M, show that f and 9 agree on the submodule [S]. ., 19. An element x of an R-module Mis said to be a torsion element'if there exists sorne r f O in R for whieh rx = O. Show that the set T of torsion elements of M forms a submoduleof M and that the quotient module M/T is torsion-free (in other words, M/T has no nonzero torsion elements). 20. Let f: M -> N be an R-homomorphism of the sinlple R-module M into the Rmodule N. Establish thatf(M) is a simple submodule of N and thatfis one-to-one whenever f(M) f {O}. 21. Let MI> M 2 , ... , M n be submodules ofthe R-module M. We eall M the (internal) direct sum of MI> M 2 , ... , M n and write M = MI ® M 2 ® ... E9 M n if i) M.= MI + M 2 + ... + M n = {Xl + X2 + ... + XnIXkEMd, and

260

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

ii) M k n (MI + ... + M k- I + Mk+1 + ... + M.) = {O} for all k. 'Prove that M is the direct sum of M ¡, M 2' ••• , M. if and only if each x E M can be expressed uniquely as a finite sum

x=

XI

+

x2

b) TR is an ideal of homRR; e) ,the mapping f(a) =' T;. determines a (ring) homomorphism of R onto Tn ; ,d) if for each O 1= a E R, there exists an element b E R such that ab 1= O, then . R z TR (hence, R can be imbedded as an ideal in a ring with identity); . e) whenever R has a multiplicative identity, then TR = homRR.

+ ... + x.

22 .. Suppose that M is an R-module with submodules M ¡, M 2' ..... M n such that M = M¡ ffi M 2 ffi ... ffi M •. For each k, let N,. be a submodule of M k and set N = NI + N 2 + .. , + N •. Verify that a) N = NI ffi N 2 ffi ... ffi N.; .;; b) M/N ~ M¡jN I ® M 2 /N 2 Et> ..• ® M.IN., as R-modules. " í· ~

,:~.1;. ,,\

'::23.

Let M be an R-module. Pro ve the assertions below: , ",,} i::.\ a) The set A(M) = {r E Rlrx = O for aH X EM} is an ideal of R, known as the ¡'\ ,-;:', . annihi/ator Df M. . ., . ,';\ b) M beeomes an.R/A(M)-module on defining the module product by,' ;;','C: (r + A(M»)x = rX",where rE R, X E M. .h:)1'?; e) Viewed as an R/A(M)-module, M has zero annihilator. ';;S¿y d) A nonempty subset N ~ M is an R-submodule of M if and .orily if it is R/A(M)-submodule of M. e) The length of M as an R-module is the same as its length when eonsidered as an R/A(M)-module.

24. Given an R-module M 1= {O}, establish that a) ,M is a simple module if and only if Rx = M for each O 1=

X E

M; here the set

Rx = {rxlrER}; " b) if N¡, N 2 are submodules of M, with N¡ simple, and if N¡ n N 2 1= {O}, then N¡ ~ N 2 , 25. Derive tbe Second Isomorphism Theorem for Modules: If N ¡ and N 2 are two submodules of the R-module M, then N¡/(N¡ n N;) '" (N¡-j- N 2 )/N 2 . [Hint: Mimic the argument ofTheorem 3-10.] 26. Derive the First lsomorphism Theorem for Modules: lf NI and N 2 are two submodules of an R-module' M with N¡ ~ N 2 , then N 2 /Nl' is a submodule of M/N¡ and (M/N¡)/(Nz/N¡) z M/N 2 •

[Hint: Mimic the argument ofTbeorem 3-9.J 27. An R-module is said to be indecomposable if it is not the direct sum of two nonzero submodules. Let M satisfy both chain conditions on submodules and let f be an R-homoinorpbism of M into itself. Prove that M is indecomposable if and only iffis either nilpotent or an automorphism. [Hint: Use Fitting's Lemma.]. 28. Let R be a eommutative ring (not necessarily with identity) and let homRR be the s~t of aH R-homomorphisms or tbe additive group of R into itself. For each a E R, define T;.: R -> R by setting T;.(x) = ax. lf Tn denotes the set of all such functions, prove !he foHowing: a) homRR forms a ring with identity, where multiplication is taken to be fUllctional composition;

261

.;1.(

',:'

,"

260

FIRST COURSE IN RINGS AND IDEALS

PROBLEMS

ii) M k n (MI + ... + M k- I + Mk+1 + ... + M.) = {O} for all k. 'Prove that M is the direct sum of M ¡, M 2' ••• , M. if and only if each x E M can be expressed uniquely as a finite sum

x=

XI

+

x2

b) TR is an ideal of homRR; e) ,the mapping f(a) =' T;. determines a (ring) homomorphism of R onto Tn ; ,d) if for each O 1= a E R, there exists an element b E R such that ab 1= O, then . R z TR (hence, R can be imbedded as an ideal in a ring with identity); . e) whenever R has a multiplicative identity, then TR = homRR.

+ ... + x.

22 .. Suppose that M is an R-module with submodules M ¡, M 2' ..... M n such that M = M¡ ffi M 2 ffi ... ffi M •. For each k, let N,. be a submodule of M k and set N = NI + N 2 + .. , + N •. Verify that a) N = NI ffi N 2 ffi ... ffi N.; .;; b) M/N ~ M¡jN I ® M 2 /N 2 Et> ..• ® M.IN., as R-modules. " í· ~

,:~.1;. ,,\

'::23.

Let M be an R-module. Pro ve the assertions below: , ",,} i::.\ a) The set A(M) = {r E Rlrx = O for aH X EM} is an ideal of R, known as the ¡'\ ,-;:', . annihi/ator Df M. . ., . ,';\ b) M beeomes an.R/A(M)-module on defining the module product by,' ;;','C: (r + A(M»)x = rX",where rE R, X E M. .h:)1'?; e) Viewed as an R/A(M)-module, M has zero annihilator. ';;S¿y d) A nonempty subset N ~ M is an R-submodule of M if and .orily if it is R/A(M)-submodule of M. e) The length of M as an R-module is the same as its length when eonsidered as an R/A(M)-module.

24. Given an R-module M 1= {O}, establish that a) ,M is a simple module if and only if Rx = M for each O 1=

X E

M; here the set

Rx = {rxlrER}; " b) if N¡, N 2 are submodules of M, with N¡ simple, and if N¡ n N 2 1= {O}, then N¡ ~ N 2 , 25. Derive tbe Second Isomorphism Theorem for Modules: If N ¡ and N 2 are two submodules of the R-module M, then N¡/(N¡ n N;) '" (N¡-j- N 2 )/N 2 . [Hint: Mimic the argument ofTheorem 3-10.] 26. Derive the First lsomorphism Theorem for Modules: lf NI and N 2 are two submodules of an R-module' M with N¡ ~ N 2 , then N 2 /Nl' is a submodule of M/N¡ and (M/N¡)/(Nz/N¡) z M/N 2 •

[Hint: Mimic the argument ofTbeorem 3-9.J 27. An R-module is said to be indecomposable if it is not the direct sum of two nonzero submodules. Let M satisfy both chain conditions on submodules and let f be an R-homoinorpbism of M into itself. Prove that M is indecomposable if and only iffis either nilpotent or an automorphism. [Hint: Use Fitting's Lemma.]. 28. Let R be a eommutative ring (not necessarily with identity) and let homRR be the s~t of aH R-homomorphisms or tbe additive group of R into itself. For each a E R, define T;.: R -> R by setting T;.(x) = ax. lf Tn denotes the set of all such functions, prove !he foHowing: a) homRR forms a ring with identity, where multiplication is taken to be fUllctional composition;

261

.;1.(

',:'

,"

SOME NONCOMMUTATIVE THEOR y

THIRTEEN

SOME NONCOMMUTATIVE THEORY

This, our concluding chapter, is designed primarily for the reader who wishes to know somet,rnng about noncommutative ideal theory. It is not our intention to treat this subject in any exhaustive manner; rather, we have concentrated on those major results which could be fitted into a concise development and which do not require many specialized preliminaries (even with this restraint, some of the theorems are fairly sophisticated). Particular effort is devoted to provingthe far-reaching Wedderburn Structure Theorems for nil-semisimple rings satisfying the descending chain condition on right ideals. These and other related results make intriguing use of the g~neral theory of idempotents, as developed in the present chapter. From this point onward, R will denote a ring with identity element, not necessarily commutative (for most of our work the assumption of an identity is not i:eally essential). In the previous chapters, consider,able progress was made after imposing a chain condition on the ideals oftll.e, ring; this was an entirely natural procedure and it is equally expedie~~:~o do so here. To have a concise statement, we shall can a ring R rig~t\ {1rtinian if it satisfies the descending chain condition on right ideals. This chain condition adniits the usual equivalent formulation: every nonempty set of right ideals of R possesses a minimal member. An important t~~rem of Brauer, which requires only the hypothesis that R be right ArtiniáJ;l. is that each nonnilpotent right ideal of R contains an idempotent element'é We choose to begin our discussion with a proof of this result. Theorem 13-1. (Brauer). In a right Artinian ring R, every nonnilpotent right ideal I contains a nonzero idempotent elemento Proof. The collection of nonnilpotent right ideals of R which are contained in 1 is not empty, for 1 itself is such an ideaL By the minimum condition on right ideals (equivalent to the assumed chain condition), there exists a mínimal member 11 of this collection. In particular, any right ideal of R properly included in 1 1 must be nilpotent. Since li forros a nonnilpotent right ideal contained in 1 1, it follows that = 1 l'

lt

262

263

Now, consider the family of all right ideals J of R with the proper~ies (i) Jl 1 =fo {O} and (H) J 5 1 1 , Such ideals certainly exist, for we have Just seen that 1 1 satisfies the indicated conditions. From among these ideals, a mínimal one can be obtained, call it J l' Using (i), there exists an element u =fo O in J 1 such that ul1 =fo {O}. Since u/ 1 is a right ideal of R contained in 1 1, with (ul 1 )/ 1 = ulf = u/ 1 =fo {O}, the minimality of JI inÍplies that ul 1 = J l' As a result, it is possible to find an element a E 1 1 5 I for which ua = u. Hence, u=ua=ua 2 or u = ua" for all n. The conclusion is that a is not nilpotent and, in consequence, 1 cannot be a nil right ideal. The key to constructing the required idempotent is to consider the right annihilator of u in / 1 , defined by

A(u) = {r E 11 1ur

O}.

Notice that A(u) is a right ideal of R which is properly contained in JI' since ul 1 = JI =fo {O}. By the minimalíty of / 1 , A(u) must be a nilpotent right ideal. Inasmuch as the product u(a 2 - a) = O. a2 - a líes in A(u), and, hence, is a nilpotent element of R. For the final stage of the proof, we propose to show that there exists a polynomialf(x), with integral coefficients, such that e = afea) is a nonzero idempotent in 1. To this purpose, suppose that (al a)" O. Then, upon expanding, one obtains a" = a"+lg(a) for a suitable polynomial g(x) E Z[x]. Itf0Hows that an. = a(a"g(a») = a" + 2g(a)2. Continuing, this process eventuaUy' leads tO,a" = a 2"g(a)". If we now set e = a"g(a)", then surely e belongst6 1, and furthermore satisfies the equation e2 == a 2"g(a)2n = (a 2"g(a)")g(a)" = a"g(a)n e. Were e = 0, this would mean that an == a 2ng(af = a"e = O, a palpable contradiction (a,being non-nilpotent from the first paragraph). Therefore, e serves as the desired nonzero idempotent in 1, which proves the theorem. The attentive reader will ha ve noticed that by proving Brauer's Theorem we have actually obtained a criterion for a right ideal to be nHpotent. Corollary. Let R be a right Artinian ringo Then a right ideal 1 of R is nilpotent if and only if every element of 1 is nilpotent (that is to say, J is a nH ideal).

Proof. Necessity follows from the definition of nilpotent ideal. That the stated condition is also sufficient is a direct consequence of the theorem and the observation that a nonzero idempotent cannot be nilpotent.

SOME NONCOMMUTATIVE THEOR y

THIRTEEN

SOME NONCOMMUTATIVE THEORY

This, our concluding chapter, is designed primarily for the reader who wishes to know somet,rnng about noncommutative ideal theory. It is not our intention to treat this subject in any exhaustive manner; rather, we have concentrated on those major results which could be fitted into a concise development and which do not require many specialized preliminaries (even with this restraint, some of the theorems are fairly sophisticated). Particular effort is devoted to provingthe far-reaching Wedderburn Structure Theorems for nil-semisimple rings satisfying the descending chain condition on right ideals. These and other related results make intriguing use of the g~neral theory of idempotents, as developed in the present chapter. From this point onward, R will denote a ring with identity element, not necessarily commutative (for most of our work the assumption of an identity is not i:eally essential). In the previous chapters, consider,able progress was made after imposing a chain condition on the ideals oftll.e, ring; this was an entirely natural procedure and it is equally expedie~~:~o do so here. To have a concise statement, we shall can a ring R rig~t\ {1rtinian if it satisfies the descending chain condition on right ideals. This chain condition adniits the usual equivalent formulation: every nonempty set of right ideals of R possesses a minimal member. An important t~~rem of Brauer, which requires only the hypothesis that R be right ArtiniáJ;l. is that each nonnilpotent right ideal of R contains an idempotent element'é We choose to begin our discussion with a proof of this result. Theorem 13-1. (Brauer). In a right Artinian ring R, every nonnilpotent right ideal I contains a nonzero idempotent elemento Proof. The collection of nonnilpotent right ideals of R which are contained in 1 is not empty, for 1 itself is such an ideaL By the minimum condition on right ideals (equivalent to the assumed chain condition), there exists a mínimal member 11 of this collection. In particular, any right ideal of R properly included in 1 1 must be nilpotent. Since li forros a nonnilpotent right ideal contained in 1 1, it follows that = 1 l'

lt

262

263

Now, consider the family of all right ideals J of R with the proper~ies (i) Jl 1 =fo {O} and (H) J 5 1 1 , Such ideals certainly exist, for we have Just seen that 1 1 satisfies the indicated conditions. From among these ideals, a mínimal one can be obtained, call it J l' Using (i), there exists an element u =fo O in J 1 such that ul1 =fo {O}. Since u/ 1 is a right ideal of R contained in 1 1, with (ul 1 )/ 1 = ulf = u/ 1 =fo {O}, the minimality of JI inÍplies that ul 1 = J l' As a result, it is possible to find an element a E 1 1 5 I for which ua = u. Hence, u=ua=ua 2 or u = ua" for all n. The conclusion is that a is not nilpotent and, in consequence, 1 cannot be a nil right ideal. The key to constructing the required idempotent is to consider the right annihilator of u in / 1 , defined by

A(u) = {r E 11 1ur

O}.

Notice that A(u) is a right ideal of R which is properly contained in JI' since ul 1 = JI =fo {O}. By the minimalíty of / 1 , A(u) must be a nilpotent right ideal. Inasmuch as the product u(a 2 - a) = O. a2 - a líes in A(u), and, hence, is a nilpotent element of R. For the final stage of the proof, we propose to show that there exists a polynomialf(x), with integral coefficients, such that e = afea) is a nonzero idempotent in 1. To this purpose, suppose that (al a)" O. Then, upon expanding, one obtains a" = a"+lg(a) for a suitable polynomial g(x) E Z[x]. Itf0Hows that an. = a(a"g(a») = a" + 2g(a)2. Continuing, this process eventuaUy' leads tO,a" = a 2"g(a)". If we now set e = a"g(a)", then surely e belongst6 1, and furthermore satisfies the equation e2 == a 2"g(a)2n = (a 2"g(a)")g(a)" = a"g(a)n e. Were e = 0, this would mean that an == a 2ng(af = a"e = O, a palpable contradiction (a,being non-nilpotent from the first paragraph). Therefore, e serves as the desired nonzero idempotent in 1, which proves the theorem. The attentive reader will ha ve noticed that by proving Brauer's Theorem we have actually obtained a criterion for a right ideal to be nHpotent. Corollary. Let R be a right Artinian ringo Then a right ideal 1 of R is nilpotent if and only if every element of 1 is nilpotent (that is to say, J is a nH ideal).

Proof. Necessity follows from the definition of nilpotent ideal. That the stated condition is also sufficient is a direct consequence of the theorem and the observation that a nonzero idempotent cannot be nilpotent.

264

FIRST COURSE IN RINGS AND. IDEALS

SOME NONCOMMUTATIVE THEORY

Before progressing further, we need the following fact about nilpotent ideals, imporlant in itself. Lemma. Let R be a ring which has no nilpotent two-sided ideals, except the zero ideal. TheI1 R possesses no nonzero nilpotent right (left) ideals. Proa! We take l to be any nilpotent right ideal of R, say 1" = {O}. Since lis a right ideal, so also is Rl; at the same time, R being a left ideal implies that RÍ forms one too. More simply put, the,~et Rl comprises a two-sided ' ideal of R. Now, (Rl)" can be written as

Proof By Brauer's Theorem, l contains nonzero idempotent elements. For each such idempotent e, we obtain a right ideal A(e) = {x

+ ei -

e 2 e l = ée1

',"1

=

{O},

so that Rl is a nilpotenÚdeal in its own righ(Íf Rl =1= {O}, a contradiction obviously ensues. Hence, we necessarily hayeRl == {O} ~ l, making l a two-sided ideal of R. Because l is nilpoteiú;'i our hypothesis guarantees that l = {O}, which proves the contention. This lemma prompts us to make a definition. A ring R will be called nil-semisimple when it has no nilpotent ideals different from zero. By what was just proved, every nil-sernisimple ring contains no nilpotent one-sided ideals, other than {O}. A word of caution: Many authors apply the term "semisimph," (standing alone) to any ring R such that (i) R satisfies the descending chain condition on right ideal s and (ii) R has no nonzero nilpotent ideals. The use 'of this nomenclature is justified by the fact that every such ring is the direct sum of finitely many simple rings (Theorem 13-3). Unfortunately, the term would cause difficulty in thepresent text, where semisimple has another meaning. We now restrict the scope of our discussion to nil-semisimple right Artinian rings. Rings satisfying these hypotheses turn out to be qf great importance in the noncommutative theory and the rest ofthe sectior{ centers around their study, Let us also abandon, for the present, the assumption that all rings under consideration must possess a multiplicative identity. (It will be demonstrated shortly that any nil-semisimple right Artinian ring actually does have an identity element.) The coming theorem shows that idempotent element,s occur as an unavoidable part of our theory; in fact, every right ideal is principal, with an idempotent generator. Theorem 13-2. Let R be a nil-semisimple right Artinian ringo Then any nonzero right ideal l of R is generated by an idempotent element, that is, 1 = eR for sorne idempotent e in R.

E

llex = O}

of R. Use the minimum condition to select an idempotent O =1= e El such that A(e)is minimal in this collection. If A(e) =1= {O}, then it has at least one nonzero idempotent, say el' Next, set e~ = e + el - ele. Then e 2 E l and is itselfan idempotent element, since ee l = O (el being a member of A(e)). Furthermore,,:ee2 = e 2 = e, which signifies that A(e 2 ) ~ A(e). The preceding inclusion)s necessarily proper, for ee l = O, while

(Rl)" = R(IR)(IR) ... (IR)]":'.

;,= R(IR)"-l l ~ R1"-U', ~ Rl"

265

el(ee l )

= ei =

el =1= O;

, in particular, we conc1~,cle that e2 =1= O. This leads to a contradjction to the rninimal nature of A(e), thereby forcing A(e) = {O}. 'Now, for any element x E l, the producl e(x - ex) = Oand so x - ex E A(e) '= {O}. It follows that x = ex for~a~l x in l, and consequently, l = el. But then ,"

~

l = el

eR

~

l,

which yields the subsequent equality l = eRo Remark. Notice that the idempotent e acts as a left identity for the right = eRo Indeed, if x E l, then x = ey for sorne y E R; therefore, '

ideal l

ex

=

e2 y

= ey = X.

The foregoing theorem allows us to gather more detailed information concerning the idempotents of R. Corollary 1. Let R be a nil-sernisimple right Artinian ringo If l is any nonzero 'two-sided ideal of R, then l = eR = Re for sorne unique idempotent e =1= O lying in the center of R. Proof By the theorem, we already know that l is idempotent generated as a right ideal; for the sake of argument, suppose that l = eR, e 2 = e =1= O.

Now consider the set J = {x - xelx E I}.

Then J is a leftideal ofR, with J2 ~ JI = J(eR) = fO}. Sin ce R contains no nilpotent left ideal s other than the zero ideal, it follows that J = {O}. As a result, we must have x = xe for all x in l, or, what amounts to the same thing, l = le. Reasoning as in the theorem, this entails that l = Re. . To confirm that e E cent R, simply observe that for each choice of r E R the elements er and re both belong to l; therefore re

=

erre)

=

(er)e

= ero

Finally, if e' is any other idempotent generator of l, then e = ee'

=

e'.

264

FIRST COURSE IN RINGS AND. IDEALS

SOME NONCOMMUTATIVE THEORY

Before progressing further, we need the following fact about nilpotent ideals, imporlant in itself. Lemma. Let R be a ring which has no nilpotent two-sided ideals, except the zero ideal. TheI1 R possesses no nonzero nilpotent right (left) ideals. Proa! We take l to be any nilpotent right ideal of R, say 1" = {O}. Since lis a right ideal, so also is Rl; at the same time, R being a left ideal implies that RÍ forms one too. More simply put, the,~et Rl comprises a two-sided ' ideal of R. Now, (Rl)" can be written as

Proof By Brauer's Theorem, l contains nonzero idempotent elements. For each such idempotent e, we obtain a right ideal A(e) = {x

+ ei -

e 2 e l = ée1

',"1

=

{O},

so that Rl is a nilpotenÚdeal in its own righ(Íf Rl =1= {O}, a contradiction obviously ensues. Hence, we necessarily hayeRl == {O} ~ l, making l a two-sided ideal of R. Because l is nilpoteiú;'i our hypothesis guarantees that l = {O}, which proves the contention. This lemma prompts us to make a definition. A ring R will be called nil-semisimple when it has no nilpotent ideals different from zero. By what was just proved, every nil-sernisimple ring contains no nilpotent one-sided ideals, other than {O}. A word of caution: Many authors apply the term "semisimph," (standing alone) to any ring R such that (i) R satisfies the descending chain condition on right ideal s and (ii) R has no nonzero nilpotent ideals. The use 'of this nomenclature is justified by the fact that every such ring is the direct sum of finitely many simple rings (Theorem 13-3). Unfortunately, the term would cause difficulty in thepresent text, where semisimple has another meaning. We now restrict the scope of our discussion to nil-semisimple right Artinian rings. Rings satisfying these hypotheses turn out to be qf great importance in the noncommutative theory and the rest ofthe sectior{ centers around their study, Let us also abandon, for the present, the assumption that all rings under consideration must possess a multiplicative identity. (It will be demonstrated shortly that any nil-semisimple right Artinian ring actually does have an identity element.) The coming theorem shows that idempotent element,s occur as an unavoidable part of our theory; in fact, every right ideal is principal, with an idempotent generator. Theorem 13-2. Let R be a nil-semisimple right Artinian ringo Then any nonzero right ideal l of R is generated by an idempotent element, that is, 1 = eR for sorne idempotent e in R.

E

llex = O}

of R. Use the minimum condition to select an idempotent O =1= e El such that A(e)is minimal in this collection. If A(e) =1= {O}, then it has at least one nonzero idempotent, say el' Next, set e~ = e + el - ele. Then e 2 E l and is itselfan idempotent element, since ee l = O (el being a member of A(e)). Furthermore,,:ee2 = e 2 = e, which signifies that A(e 2 ) ~ A(e). The preceding inclusion)s necessarily proper, for ee l = O, while

(Rl)" = R(IR)(IR) ... (IR)]":'.

;,= R(IR)"-l l ~ R1"-U', ~ Rl"

265

el(ee l )

= ei =

el =1= O;

, in particular, we conc1~,cle that e2 =1= O. This leads to a contradjction to the rninimal nature of A(e), thereby forcing A(e) = {O}. 'Now, for any element x E l, the producl e(x - ex) = Oand so x - ex E A(e) '= {O}. It follows that x = ex for~a~l x in l, and consequently, l = el. But then ,"

~

l = el

eR

~

l,

which yields the subsequent equality l = eRo Remark. Notice that the idempotent e acts as a left identity for the right = eRo Indeed, if x E l, then x = ey for sorne y E R; therefore, '

ideal l

ex

=

e2 y

= ey = X.

The foregoing theorem allows us to gather more detailed information concerning the idempotents of R. Corollary 1. Let R be a nil-sernisimple right Artinian ringo If l is any nonzero 'two-sided ideal of R, then l = eR = Re for sorne unique idempotent e =1= O lying in the center of R. Proof By the theorem, we already know that l is idempotent generated as a right ideal; for the sake of argument, suppose that l = eR, e 2 = e =1= O.

Now consider the set J = {x - xelx E I}.

Then J is a leftideal ofR, with J2 ~ JI = J(eR) = fO}. Sin ce R contains no nilpotent left ideal s other than the zero ideal, it follows that J = {O}. As a result, we must have x = xe for all x in l, or, what amounts to the same thing, l = le. Reasoning as in the theorem, this entails that l = Re. . To confirm that e E cent R, simply observe that for each choice of r E R the elements er and re both belong to l; therefore re

=

erre)

=

(er)e

= ero

Finally, if e' is any other idempotent generator of l, then e = ee'

=

e'.

266

SOME NONCOMMUTA TIVE· THEOR y

FIRST COURSE IN RINGS AND IDEALS

Another way of phrasing Corollary 1 is to say that any two-sided ideal of R has a multiplicative identity, namely, the generating idempotent. In the light of the fact that the entire ring R is itself an ideal, we can deduce the fairly remarkable result that R must possess an identityo Corollary 2. A nil-semisimple righ t Artinian ring has an iden tity element. We contipue a little further in this vein by proving LeltIma. LetRbeanil-semisimplerightArtinianringand1 = eR = Re be an ideal of R, e an idempotent. Then any right {left, two-sided) ideal of 1 is also a right (left, two-sided) ideal of R.

Proof. Suppose that J is an arbitrary right ideal of 1, considered as a ringo Since J 5;. Re, each element a E J can be written in the form a = re, with rE R; but then

a= re = (re)e = ae EJe,

leading to the equality J = Jeo Knowing this, one finds that JR

= (Je)R =

J(eR) = JI

5;

J,

which makes J a right ideal of R. This last lemma is considerably deeper than it first appearso For most purposes, its value lies in the corollary belowo Corollary. Let R be a nil-semisimple right Artinian ringo Viewed as rings, 1) each ideal of R is itself a nil-semisimple right ,Artinian ring, and 2) any minimal ideal of R is a simple ringo This preparation brings us to a profound result, the First Wedderburn Structure Theoremo As in the commutative case, the ultimate .aim ls 'to characterize those rings under consideration as a direct sum of certain iings of known typeo ~' Theorem 13-3. (Wedderburn)o Let R be a nil-semisimple right Artinian ringo Then R is the (finite) direct sum of its minimal two-sided ideals, each of which is a simple right Artinian ringo

Proof. Since the minimum condition on right ideal s holds in R, it is possible to find a minimal two-sided ideal 1 1 =1= {O} (simply apply the condition to the collection of aIl nonzero two-sided ideal s of R). With reference to Theorem 13-2, we know that 1 1 = e 1R = Re 1, el being a suitably chosen idempotent in the center of R. Then 1 - el E cent R, from which it follows that JI = (1 - e 1 )R forms an ideal of R. Now, any element x E R can be written as x = e 1 x + (1 - e1 )x, whence the relation R = 1 1 + JI holdso

267

To see that this sum is actually direct, .select an arbitrary x in 1 1 n JI. On the one hand, x = (1 - e 1 )r, so that e 1 x = O, and, on the other hand, x = e l 8., implying that e1x = ets = e1s = x; thus, the element x = O, or equivalently, 1 1 n JI = {O}o In consequence, R = 1 1 EB JI is the direct sum of th~ ideals 1 1 and J lo Furthermore, the ideal 1 1 is simple when regarded as a ring (Corollary 2 aboye). The heart of our argument is the observation that the ideal JI' being an ideal of a nil-semisimple right Artinian ring, inherits these properties (as a ring)o Therefore, if JI =1= {O},the technique of the preceding paragraph may be repeated with JI now replacing R. This yields the decomposition JI = 12 EB J 2' with J 2 an ideal contained in J lo Repeating the process, we arrive at

where each 1i = eiR is a simple, idempotent-generated, minimal ideal of R. Since JI ;2 J 2 ;2 J 3 ;2 o.. , the descending chain condition on right ideals implies that J. = {O} for sorne n. That is to say, at sorne point R is exhibited as the direct sum

R = 1 1 EB 12 EB o.. EB 1.0 To complete the proof, it remains only to show that the 1i include aIl the minimal two-sided ideal s of R. Pursuing this aim, let 1 =1= {O} be any minimal ideal of R. Since R admits the direct sum decomposition R = 1 1 Ei112 EB oo. EB l., we thus have 1 = R1 = 1 11 EB 121 EB ...

~

1.1.

Now, each 1) is an ideal of R which is contained:hdi . By the minimality of 1i , either 1;1 = {O} or else 1i1 = 1. However, if it happened that 1;1 = {O} for every i (i = 1,2, ... ,n), then we would necessarily have 1 = {O}, which is nonsenseo The implication is that 1;1 = 1i for sorne choice of i. But then 1i = 1i1 5; 1 and so the minimal nature of 1 forces 1i = 1, as asserted. This reasoning also allows ustQ concIude that the direct sum decomposition of R is unique, up to the order of occurrence of the summandso Our assertions are now verified. Since the ideals 1i = eiR (i = 1,2, ... ,n) are the only (two-sided) minimal ideals of R, we concIude that any nil-semisimple right Artinian ring R has a finite number of minimal ideals. This observation can be sharpened to a statement regarding the number of ideals of R, minimal or otherwise. Corollary. A nil-semisimple rjght Artinian ring R has 2· ideals for sorne nEZ+o

266

SOME NONCOMMUTA TIVE· THEOR y

FIRST COURSE IN RINGS AND IDEALS

Another way of phrasing Corollary 1 is to say that any two-sided ideal of R has a multiplicative identity, namely, the generating idempotent. In the light of the fact that the entire ring R is itself an ideal, we can deduce the fairly remarkable result that R must possess an identityo Corollary 2. A nil-semisimple righ t Artinian ring has an iden tity element. We contipue a little further in this vein by proving LeltIma. LetRbeanil-semisimplerightArtinianringand1 = eR = Re be an ideal of R, e an idempotent. Then any right {left, two-sided) ideal of 1 is also a right (left, two-sided) ideal of R.

Proof. Suppose that J is an arbitrary right ideal of 1, considered as a ringo Since J 5;. Re, each element a E J can be written in the form a = re, with rE R; but then

a= re = (re)e = ae EJe,

leading to the equality J = Jeo Knowing this, one finds that JR

= (Je)R =

J(eR) = JI

5;

J,

which makes J a right ideal of R. This last lemma is considerably deeper than it first appearso For most purposes, its value lies in the corollary belowo Corollary. Let R be a nil-semisimple right Artinian ringo Viewed as rings, 1) each ideal of R is itself a nil-semisimple right ,Artinian ring, and 2) any minimal ideal of R is a simple ringo This preparation brings us to a profound result, the First Wedderburn Structure Theoremo As in the commutative case, the ultimate .aim ls 'to characterize those rings under consideration as a direct sum of certain iings of known typeo ~' Theorem 13-3. (Wedderburn)o Let R be a nil-semisimple right Artinian ringo Then R is the (finite) direct sum of its minimal two-sided ideals, each of which is a simple right Artinian ringo

Proof. Since the minimum condition on right ideal s holds in R, it is possible to find a minimal two-sided ideal 1 1 =1= {O} (simply apply the condition to the collection of aIl nonzero two-sided ideal s of R). With reference to Theorem 13-2, we know that 1 1 = e 1R = Re 1, el being a suitably chosen idempotent in the center of R. Then 1 - el E cent R, from which it follows that JI = (1 - e 1 )R forms an ideal of R. Now, any element x E R can be written as x = e 1 x + (1 - e1 )x, whence the relation R = 1 1 + JI holdso

267

To see that this sum is actually direct, .select an arbitrary x in 1 1 n JI. On the one hand, x = (1 - e 1 )r, so that e 1 x = O, and, on the other hand, x = e l 8., implying that e1x = ets = e1s = x; thus, the element x = O, or equivalently, 1 1 n JI = {O}o In consequence, R = 1 1 EB JI is the direct sum of th~ ideals 1 1 and J lo Furthermore, the ideal 1 1 is simple when regarded as a ring (Corollary 2 aboye). The heart of our argument is the observation that the ideal JI' being an ideal of a nil-semisimple right Artinian ring, inherits these properties (as a ring)o Therefore, if JI =1= {O},the technique of the preceding paragraph may be repeated with JI now replacing R. This yields the decomposition JI = 12 EB J 2' with J 2 an ideal contained in J lo Repeating the process, we arrive at

where each 1i = eiR is a simple, idempotent-generated, minimal ideal of R. Since JI ;2 J 2 ;2 J 3 ;2 o.. , the descending chain condition on right ideals implies that J. = {O} for sorne n. That is to say, at sorne point R is exhibited as the direct sum

R = 1 1 EB 12 EB o.. EB 1.0 To complete the proof, it remains only to show that the 1i include aIl the minimal two-sided ideal s of R. Pursuing this aim, let 1 =1= {O} be any minimal ideal of R. Since R admits the direct sum decomposition R = 1 1 Ei112 EB oo. EB l., we thus have 1 = R1 = 1 11 EB 121 EB ...

~

1.1.

Now, each 1) is an ideal of R which is contained:hdi . By the minimality of 1i , either 1;1 = {O} or else 1i1 = 1. However, if it happened that 1;1 = {O} for every i (i = 1,2, ... ,n), then we would necessarily have 1 = {O}, which is nonsenseo The implication is that 1;1 = 1i for sorne choice of i. But then 1i = 1i1 5; 1 and so the minimal nature of 1 forces 1i = 1, as asserted. This reasoning also allows ustQ concIude that the direct sum decomposition of R is unique, up to the order of occurrence of the summandso Our assertions are now verified. Since the ideals 1i = eiR (i = 1,2, ... ,n) are the only (two-sided) minimal ideals of R, we concIude that any nil-semisimple right Artinian ring R has a finite number of minimal ideals. This observation can be sharpened to a statement regarding the number of ideals of R, minimal or otherwise. Corollary. A nil-semisimple rjght Artinian ring R has 2· ideals for sorne nEZ+o

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FIRST COURSE IN RINGS AND IDEALS

SOME NONCOMMUTATIVE THEORY

Proo! Accordingtothe.theorem,R = 1 1 @1 2 @ '" @ln ,whereeachl¡ forms a miniinal ideal of R. If 1 is an arbitrary ideal of R, then 1 = IR = l¡I EB 1 2 1 @ ... EB

from which it follows that IR = {O}. Thus, 1 forms a two-sided ideal of R, with 1 2 S IR = {O}. But R is hypothesized to be a nil-semisimple ring, whence 1 ,; {O}. The implication is that e also serves as a right identity for R and so a two-sided identity.

lJ

As before, !he ideal lJ is contained in 1¡, so that either JJ = {O} or lJ = 1¡. In other words, we can express 1 as 1 = 1¡!

+ l¡z + ... +

We pause to summarize what has been proved so far. Theorem 13-4. For any nil-semisimple right Artinian ring' R, there exists a decomposition

l¡k'

where '{i 1,":!2' ... , ik} is a set of distinct integers between 1 and n. It follows that ther~:are exactIy 2n ideal s in R, namely, the ideals 1¡! + 1¡2 + .;. + l¡k'

,"i,

,t.

In th~,'toreg?irig structure theorem, we obtained a decomposition ;.":..

into minimal two-sided ideals. The ideals 1 1,1 2 , by orthogonal idempotents; that is,

'

R

e1R EB e2R EB .. , EB enR, where e~b1i; e· is a nonz~ro idempotent element of R. Let us ne~t show that the e¡ f9Óii' ~n orthogo~al set of idempotents, in the sense that e¡ ej = O whenevér);+ j. This depends on the observation that the intersection of two simple ideals is a two-sided ideal and so must be zero. In the case at hand, we have . (e¡R)(ejR) S e¡R ("'\ ejR = {O}, for i 9= j, =

.

Knowing that any nil-semisimple right Artinian ring can be represented . as a direct sum of simple right Artinian rings, our problem is thus reduced to determining a satisfactory structure theory for simple rings in which the descending chain condition on right ideals holds. It wi11 be found in due course that such rings are isomorphic to the ring of all linear transformations on a suitably defined vector space. For the present, we content ourselves with the observation that any simple ring R for which R 2 9= {O} is automaticalIy nil-semisimple. Indeed, if 1 is any nilpotent two-sided ideal of R, then either 1 = {O} or 1= R. Inasmuch as 1 is nilpotent, the latter possibility implies that Rn = {O} for some n; but this is ruled out by the fact that R 2 = R (since R 2 is an ideal with R 2 9= {O}, necessarily R 2 = R). Hence, the ideal 1 = {O}, as required. . To set the stage for our principal theorem, we next draw attention to certain relations between the structure of eR and that of eRe. One result which we have in mind is the foIlowing.

(r¡.E R). =

e¡r¡; hence, the

+ e2 + ... + en'

Remark. Theorem 13-3 Could be used to establish that any nil-semisimple right Artinian ring R necessarily has an identity element, viz., the idempotent e = el + e2 + ... + en' (In the absence of an identity, the notation (1 - e 1)R occurring in the structure theorem must be interpreted as meaning the set {r - e1rlr E R}.) The reasoning proceeds along the following Hnes. Since R = e1R EB e2R EB ... EB enR, each element l' E R can be represented as l' = e 1r 1 + e2r 2 + ... + enr n for suitable r¡ in R. Thus, the equation = =

Theorem 13-5. Let R be a nil-semisimple ring (no chain conditions) and let e 9= O be an idempotent element of R. Then eR is a minimal right ideal of R if and only if eRe is a division ringo

+ e~r2 + e 1 r 1 + e2 r2 + eir 1

Proo! Before embarking on the proof proper, we note that the set eRe forms a nonzero subring of R with identity element e. Suppose first that eR is a minimal right ideal of R. To show that eRe comprises a division ring, it is enough to find an inverse for each nonzero elemerit. If O 9= ere E eRe,

holds, making e a left identity for R. On the other hand, consider the left ideal 1 = {r - reir E R}. Because R = eR, we have (1' - re)R = (1' - re)eR = r(e - e2)R = {O},

---

-_._------------

----------------

In are.generated ' ,.

and these are such that 1 = el + e2 + ... + en' Furthermore, the direct sum decomposition is unique apart from the order of the summands. '

.

Multiplyirig this equation by e¡, it folIows that e'¡ = e'fr¡ identity element can be expressed more succinctIy as

.... ,

where

which, of course, gives e¡ej = O. Now, let 1 E R be written as

1 = el

269

------

-----

-

---- - - -

268

FIRST COURSE IN RINGS AND IDEALS

SOME NONCOMMUTATIVE THEORY

Proo! Accordingtothe.theorem,R = 1 1 @1 2 @ '" @ln ,whereeachl¡ forms a miniinal ideal of R. If 1 is an arbitrary ideal of R, then 1 = IR = l¡I EB 1 2 1 @ ... EB

from which it follows that IR = {O}. Thus, 1 forms a two-sided ideal of R, with 1 2 S IR = {O}. But R is hypothesized to be a nil-semisimple ring, whence 1 ,; {O}. The implication is that e also serves as a right identity for R and so a two-sided identity.

lJ

As before, !he ideal lJ is contained in 1¡, so that either JJ = {O} or lJ = 1¡. In other words, we can express 1 as 1 = 1¡!

+ l¡z + ... +

We pause to summarize what has been proved so far. Theorem 13-4. For any nil-semisimple right Artinian ring' R, there exists a decomposition

l¡k'

where '{i 1,":!2' ... , ik} is a set of distinct integers between 1 and n. It follows that ther~:are exactIy 2n ideal s in R, namely, the ideals 1¡! + 1¡2 + .;. + l¡k'

,"i,

,t.

In th~,'toreg?irig structure theorem, we obtained a decomposition ;.":..

into minimal two-sided ideals. The ideals 1 1,1 2 , by orthogonal idempotents; that is,

'

R

e1R EB e2R EB .. , EB enR, where e~b1i; e· is a nonz~ro idempotent element of R. Let us ne~t show that the e¡ f9Óii' ~n orthogo~al set of idempotents, in the sense that e¡ ej = O whenevér);+ j. This depends on the observation that the intersection of two simple ideals is a two-sided ideal and so must be zero. In the case at hand, we have . (e¡R)(ejR) S e¡R ("'\ ejR = {O}, for i 9= j, =

.

Knowing that any nil-semisimple right Artinian ring can be represented . as a direct sum of simple right Artinian rings, our problem is thus reduced to determining a satisfactory structure theory for simple rings in which the descending chain condition on right ideals holds. It wi11 be found in due course that such rings are isomorphic to the ring of all linear transformations on a suitably defined vector space. For the present, we content ourselves with the observation that any simple ring R for which R 2 9= {O} is automaticalIy nil-semisimple. Indeed, if 1 is any nilpotent two-sided ideal of R, then either 1 = {O} or 1= R. Inasmuch as 1 is nilpotent, the latter possibility implies that Rn = {O} for some n; but this is ruled out by the fact that R 2 = R (since R 2 is an ideal with R 2 9= {O}, necessarily R 2 = R). Hence, the ideal 1 = {O}, as required. . To set the stage for our principal theorem, we next draw attention to certain relations between the structure of eR and that of eRe. One result which we have in mind is the foIlowing.

(r¡.E R). =

e¡r¡; hence, the

+ e2 + ... + en'

Remark. Theorem 13-3 Could be used to establish that any nil-semisimple right Artinian ring R necessarily has an identity element, viz., the idempotent e = el + e2 + ... + en' (In the absence of an identity, the notation (1 - e 1)R occurring in the structure theorem must be interpreted as meaning the set {r - e1rlr E R}.) The reasoning proceeds along the following Hnes. Since R = e1R EB e2R EB ... EB enR, each element l' E R can be represented as l' = e 1r 1 + e2r 2 + ... + enr n for suitable r¡ in R. Thus, the equation = =

Theorem 13-5. Let R be a nil-semisimple ring (no chain conditions) and let e 9= O be an idempotent element of R. Then eR is a minimal right ideal of R if and only if eRe is a division ringo

+ e~r2 + e 1 r 1 + e2 r2 + eir 1

Proo! Before embarking on the proof proper, we note that the set eRe forms a nonzero subring of R with identity element e. Suppose first that eR is a minimal right ideal of R. To show that eRe comprises a division ring, it is enough to find an inverse for each nonzero elemerit. If O 9= ere E eRe,

holds, making e a left identity for R. On the other hand, consider the left ideal 1 = {r - reir E R}. Because R = eR, we have (1' - re)R = (1' - re)eR = r(e - e2)R = {O},

---

-_._------------

----------------

In are.generated ' ,.

and these are such that 1 = el + e2 + ... + en' Furthermore, the direct sum decomposition is unique apart from the order of the summands. '

.

Multiplyirig this equation by e¡, it folIows that e'¡ = e'fr¡ identity element can be expressed more succinctIy as

.... ,

where

which, of course, gives e¡ej = O. Now, let 1 E R be written as

1 = el

269

------

-----

-

---- - - -

270

then ereR is _a nonzero right ideal of R contained in eRo Since eR is a minima! right ideal, we must have ereR = eRo Therefore,

(ere)(eRe) = (ereR)e = eRe. This relation implies that we can find an element x E eRe for which (ere)x = e. Thus, every nonzero element in eRe has a right inverse with respect t? the identity e. In particular, there is sorne y E eRe satisfying x(ere)y = e; but then e = x(ere)y = xe(ere)y =

=

x[(ere)x](ere)y

At the same time, ere (e - ere) = (e - ere)ere = O. Thus, we have e = ere + (e - ere), where both ere and e - ere are idempotent and orthogonal. From the primitivity of e, it may be conc1uded that one of these is zero; that is, either ere = or ere = e. Conversely, if e is not primitive, then we may write e = u + v, where uand varenonzeroorthogonalidempotents. Hence,u =1= eandeu = ue = u, which implies that the element u = eue is in eRe.

°

We put the finishing touches on our theory of idempotent generated minimal ideals with

x(ere)[x(ere)y] = x(ere)e = x(ere).

This enables us to conc1ude .that each right inverse is also a left inverse, yielding the desired outcome. As regards the con~t'(rse, as sume that eRe constitutes a division ring and let 1 =1= {O} be any right ideal of R contained in eRo This gives el = l. Notice also that le =1= {O}; in the contrary case, 12 ~ leR = {O}, which conflicts with our hypothesis that R has no nonzero nilpotent ideals. Accordingly, there exists an element r E 1 such that re =1= and, :¡;ince 1 = el, we must have ere = re =1= O. Because eRe is taken to be a division ring, ere possesses an inverse s E eRe. But, ere El; hence, e = (ere)s E l. This forces eR ~ 1 and the equality 1 = eR follows.

°

The above theorem is evidentally true with "right" replaced by "left" throughout; this symmetry allows us to add

°

CoroIlary. Let R be a nil-semisimple ring andlet =1= e E R be idempotent. Then.eR is a minimal right ideal if and only if Re is a minimal l.: '~,,' left ideal. It is reasonable to ask ~h~ther the statement of Theorem 1.3-5 could be improved upon by the stipuliltion of a chain condition. In pursuit of an answer, we make the follQwing definition. An idempotent =1= e E R is called primitive if e is not t4~'sum of two orthogonal nonzero idempotents of R; that is, it is not pos§jble to write e = u + v, where u 2 = U =1= 0, 2 v = v =1= and uv = vu ~(O. We can characterize wlÍen an idempotent element of R is primitive in terms of the idempotents of the ring eRe. To be precise:

°

°

°

Lernrna. An idempotent =1= e E R is primitive if and only if e is the only nonzero idempotent in the ring eRe.

Proof. Let e be primitive and assume that ere is idempotent for sorne r E R. Then the element e - ere is also idempotent: (e - ere)2 = e2 - e2re - ere 2 + (ere)2 = -e - ere - ere

+ ere

=

271

SOME NONCOMMUTATIVE THEORY

FIRST COURSE IN RINGS AND IDEALS

e - ere.

Theorern 13-6. Let R be a nil-semisimple right Artinian ringo Then an idempotent O =1= e E R is primitive if and only if eR is a minimal right ideal of R.

Proof. We begin by supposing that eR is not a minimal right ideal. Then eR properly contains a right ideal 1 =1= {O} of R, which is of the form 1 = uR, u a nonzero idempotent. Since u E eR, it follows that u = er for sorne rE R, whence u = e2r = eu. Now, set v = ue and w = e - ue. An easy calculation shows that e = v + w, where v and w are orthogonal idempotents: v2 = (ue)2 = u(eu)e = u2e = ue = v, .

w2

=

vw

= =

wv

e2 - eue - ue 2 + (ue)2 ue(e - ue) = ue 2 - (ue)2 ,;,. ue - ue

(e - ue)2

=

(e - ue)ue

~

eue - (ue)2

=

ue - ue

=

e - ue - ue

+ ue

= 0, =

O.

It is :also important to observe that v and w are both nonzero.

were¡)

=

= w,

Indeed,

0, we would obtain the contradiction u = u2 = u(eu) = (ue)u = vu = O.

On the.other hand, suppose that w = 0, so that e = ue. Then the right ideal 1 = uR will contain the element e; this implies that 1 = eR, which is impossible. Having thus represented e as the sum of two orthogonal nonzero i'dempotents, we infer that e is not primitive. Going in the opposite direction, we as sume that the idempotent e is not primitive. Then e can be expressed as e = u + v, a sum of orthogonal nonzero idempotents. Now, u = eu = ue, whence the right ideal uR = euR ~ eRo This irtc1usion is proper, since the idempotent e E eR, while e fj uRo In fact, if e = ur for a suitable element r of R, then v = ve = vur = 0, an obvious contradiction. Therefore, {O} =1= uR c: eR, so that eR cannot be a minimal right ideal of R. This completes the proof. The two prece<;ling theorems, taken together, yield a single result:

270

then ereR is _a nonzero right ideal of R contained in eRo Since eR is a minima! right ideal, we must have ereR = eRo Therefore,

(ere)(eRe) = (ereR)e = eRe. This relation implies that we can find an element x E eRe for which (ere)x = e. Thus, every nonzero element in eRe has a right inverse with respect t? the identity e. In particular, there is sorne y E eRe satisfying x(ere)y = e; but then e = x(ere)y = xe(ere)y =

=

x[(ere)x](ere)y

At the same time, ere (e - ere) = (e - ere)ere = O. Thus, we have e = ere + (e - ere), where both ere and e - ere are idempotent and orthogonal. From the primitivity of e, it may be conc1uded that one of these is zero; that is, either ere = or ere = e. Conversely, if e is not primitive, then we may write e = u + v, where uand varenonzeroorthogonalidempotents. Hence,u =1= eandeu = ue = u, which implies that the element u = eue is in eRe.

°

We put the finishing touches on our theory of idempotent generated minimal ideals with

x(ere)[x(ere)y] = x(ere)e = x(ere).

This enables us to conc1ude .that each right inverse is also a left inverse, yielding the desired outcome. As regards the con~t'(rse, as sume that eRe constitutes a division ring and let 1 =1= {O} be any right ideal of R contained in eRo This gives el = l. Notice also that le =1= {O}; in the contrary case, 12 ~ leR = {O}, which conflicts with our hypothesis that R has no nonzero nilpotent ideals. Accordingly, there exists an element r E 1 such that re =1= and, :¡;ince 1 = el, we must have ere = re =1= O. Because eRe is taken to be a division ring, ere possesses an inverse s E eRe. But, ere El; hence, e = (ere)s E l. This forces eR ~ 1 and the equality 1 = eR follows.

°

The above theorem is evidentally true with "right" replaced by "left" throughout; this symmetry allows us to add

°

CoroIlary. Let R be a nil-semisimple ring andlet =1= e E R be idempotent. Then.eR is a minimal right ideal if and only if Re is a minimal l.: '~,,' left ideal. It is reasonable to ask ~h~ther the statement of Theorem 1.3-5 could be improved upon by the stipuliltion of a chain condition. In pursuit of an answer, we make the follQwing definition. An idempotent =1= e E R is called primitive if e is not t4~'sum of two orthogonal nonzero idempotents of R; that is, it is not pos§jble to write e = u + v, where u 2 = U =1= 0, 2 v = v =1= and uv = vu ~(O. We can characterize wlÍen an idempotent element of R is primitive in terms of the idempotents of the ring eRe. To be precise:

°

°

°

Lernrna. An idempotent =1= e E R is primitive if and only if e is the only nonzero idempotent in the ring eRe.

Proof. Let e be primitive and assume that ere is idempotent for sorne r E R. Then the element e - ere is also idempotent: (e - ere)2 = e2 - e2re - ere 2 + (ere)2 = -e - ere - ere

+ ere

=

271

SOME NONCOMMUTATIVE THEORY

FIRST COURSE IN RINGS AND IDEALS

e - ere.

Theorern 13-6. Let R be a nil-semisimple right Artinian ringo Then an idempotent O =1= e E R is primitive if and only if eR is a minimal right ideal of R.

Proof. We begin by supposing that eR is not a minimal right ideal. Then eR properly contains a right ideal 1 =1= {O} of R, which is of the form 1 = uR, u a nonzero idempotent. Since u E eR, it follows that u = er for sorne rE R, whence u = e2r = eu. Now, set v = ue and w = e - ue. An easy calculation shows that e = v + w, where v and w are orthogonal idempotents: v2 = (ue)2 = u(eu)e = u2e = ue = v, .

w2

=

vw

= =

wv

e2 - eue - ue 2 + (ue)2 ue(e - ue) = ue 2 - (ue)2 ,;,. ue - ue

(e - ue)2

=

(e - ue)ue

~

eue - (ue)2

=

ue - ue

=

e - ue - ue

+ ue

= 0, =

O.

It is :also important to observe that v and w are both nonzero.

were¡)

=

= w,

Indeed,

0, we would obtain the contradiction u = u2 = u(eu) = (ue)u = vu = O.

On the.other hand, suppose that w = 0, so that e = ue. Then the right ideal 1 = uR will contain the element e; this implies that 1 = eR, which is impossible. Having thus represented e as the sum of two orthogonal nonzero i'dempotents, we infer that e is not primitive. Going in the opposite direction, we as sume that the idempotent e is not primitive. Then e can be expressed as e = u + v, a sum of orthogonal nonzero idempotents. Now, u = eu = ue, whence the right ideal uR = euR ~ eRo This irtc1usion is proper, since the idempotent e E eR, while e fj uRo In fact, if e = ur for a suitable element r of R, then v = ve = vur = 0, an obvious contradiction. Therefore, {O} =1= uR c: eR, so that eR cannot be a minimal right ideal of R. This completes the proof. The two prece<;ling theorems, taken together, yield a single result:

272

FIRST COURSE IN RINGS AND ,IDEALS

SOME NONco;MMUTA TIVE THEOR y

Tbeorem 13-7. In a nil-semisimple rigbt Artinian ring R, an idempotent e' =1= O is primitive if and only if eRe forms a division ringo We continue our development after a few preparatory remarks about modules. Let M, N be nonzero rigbt R-modules, wbere R is any ringo (The reader is reniinded that, wbenever tbere is reference to a' ring R, it is tacitly assumed tbat R possesses an identity elem~t.) We shall bereafter use the notation bomR(M, N) to denote the set of all R-bomomorpbisms from M into N; in other words, tbe set of all mappin~s, f: M -+ N sucb tbat ~

f{x

+

y) = f(x)

+ f(y),

f(xr)

..

~.

= f(x)r'::,;;

(x, y E M ; r' E R).

Under tbe usual pointwise addition of functio'rls, bomR(M, N) forms, a subgroup of tbe commutative group bom (M;,!,,). EqualIy important .is tbe observation that bomR(M, M) turns out td"h~ a subring (witb identity) oftbe ring hom (M, M):ring multiplication being::pomposition of mappings. . In fact, for arbitrary 1, 9 E bomR(M, M), x E M,'~d r E R, (f - g)(xr) = f(xr) - g(xr)'" :' = f(x)r

= (j{x) wbence f

-

g(x)r - g(x»)r .;" (f

g)(x)r,

9 E bomR(M, M). It ls just as routine that

(f o g)(xr)

= f(g(xr») = f(g(x)r)

(f o g)(x)r,

and so f o 9 aIso lies in bomR (M, M). One ca/Is bomR(M, M) tbe ring of R-endomorphisms of M, or sometimes tbe centralizer of tbe R-module M. The reason for tbis latter cboice of terminology is that tbe elements of bomR(M, M) are precisely tbose additive endomorpbisms f of M wbicb cornmute witb tbe rigbt multiplications determined by elements of R. Indeed, with every elemimt r ER there is assoCÍated a mapping T,. of M into itself given by T,.x = xr (as an irnmediate consequence of its definition, T,. E homR(M, M)). Tben the condition f(xr) = f(x)r can be reformulated as (f o T,.)(x)

=

(T,. o f)(x)

(x E M),

or, equivalently, as f o T,. T,. o f It is natural to ask whetber homR(M, M) can be turned into an Rmodule. In answering this question, let us suppose for tbe moment that R is a commutative ring and that f belongs to homR (M, M). Given a fixed r in R, define fr by means of (fr)(x)

Witb this definition, one finds that bomR(M, M) (regarded merely as a commutative group) becomes a rigbt R-module. Clearly, fr isan additive bomomorpbism from M into M. Moreover, for any S-E R, (fr)(xs)

f(x)r

(x EM).

= f(xs)r

f(x)sr

= f(x)rs

=

(fr)(x)s.

Here, rs = sr, because R is taken to be cornmutative. Tbus, fr actually does form ap R-endomorpbism. To recapitulate; in tbe cornmutative case, bomR(M, M),is botb ~:ring and an R-module, with ring multiplication and module mu1tiplication)~nked by tbe relation (f o g)r = f o (gr) ;:(fr) o 9

(1, 9 E bomR(M, M); rE R).

In p~ticular, when M =tR,: a clear distinction must be made between tbe R-homomorpbisms of R(cbhsidered as module over itself) and its bomomorpbisms as a ringo >,t~ . Before turning to the¡-,~~xt tbeorem, it is necessary to recalI that any right ideal 1 of tbe ring :RJs:a r1gbt R-module; the module product being given by the ,ordinary ringo multiplication xr, where x E 1 and rE R. Tbeorem 13-8. Let R be a simple ring witb minimal rigbt ideal 1 e =1= O an idempotent. Then bomR(I, 1) ::::: eRe, viewed as, rings. Proof As remarked aboye, we consider 1 as a right R-module. fE bomR(I, 1) implies tbat fea

= f(g(x»)r

273

+

b) = fía) +.f(b), f(ar) = f(a)r

eR, Thus,

(a, bE 1; rE R).

The first task is to characteríze tbesé R-endomorphisms in sorne convenient way. We handle tbis problem as folIows. LetfE bomR(I, 1) and tbe element a E 1 = eR, saya = ero Then fea) = f(er) = f(e)r, so tbat tbe function f is completely determined once tbe value fíe) is known. Since e is an idempotent of R, we bave fíe) f(e Z):= f(e)e. But fíe) E eR, whence fíe) = es for sorne choice of s in R, and so fíe) = f(e)e = ese. It readily folIows that fea) = f(e)r = (ese)r (ese 2 )r = (ese)er ~ (ese)a._ In brief, f acts on the elements of 1 as left multiplication by ese, wbere f(e) = ese. We shall utilize this information presently. Now, defin~ a mapping cp; homR(I, 1) -+ eRe by means of the rule cp(f¡' = fíe). It is iminediately apparent that cp(f + g)

=

(f + g)(e)

fíe)

+ g(e) =

cp(f)

+

CP(g).

If f is caused by a left multíplication by tbe eIement ese and g is a left muHiplícation by ete, then cp enjoys the further property

cp(f o g) = (f o g)(e) =

=

j(g(e») = f(ete)

esete = (ese)(ete) = f(e)g(e)

= CP(f)CP(g).

272

FIRST COURSE IN RINGS AND ,IDEALS

SOME NONco;MMUTA TIVE THEOR y

Tbeorem 13-7. In a nil-semisimple rigbt Artinian ring R, an idempotent e' =1= O is primitive if and only if eRe forms a division ringo We continue our development after a few preparatory remarks about modules. Let M, N be nonzero rigbt R-modules, wbere R is any ringo (The reader is reniinded that, wbenever tbere is reference to a' ring R, it is tacitly assumed tbat R possesses an identity elem~t.) We shall bereafter use the notation bomR(M, N) to denote the set of all R-bomomorpbisms from M into N; in other words, tbe set of all mappin~s, f: M -+ N sucb tbat ~

f{x

+

y) = f(x)

+ f(y),

f(xr)

..

~.

= f(x)r'::,;;

(x, y E M ; r' E R).

Under tbe usual pointwise addition of functio'rls, bomR(M, N) forms, a subgroup of tbe commutative group bom (M;,!,,). EqualIy important .is tbe observation that bomR(M, M) turns out td"h~ a subring (witb identity) oftbe ring hom (M, M):ring multiplication being::pomposition of mappings. . In fact, for arbitrary 1, 9 E bomR(M, M), x E M,'~d r E R, (f - g)(xr) = f(xr) - g(xr)'" :' = f(x)r

= (j{x) wbence f

-

g(x)r - g(x»)r .;" (f

g)(x)r,

9 E bomR(M, M). It ls just as routine that

(f o g)(xr)

= f(g(xr») = f(g(x)r)

(f o g)(x)r,

and so f o 9 aIso lies in bomR (M, M). One ca/Is bomR(M, M) tbe ring of R-endomorphisms of M, or sometimes tbe centralizer of tbe R-module M. The reason for tbis latter cboice of terminology is that tbe elements of bomR(M, M) are precisely tbose additive endomorpbisms f of M wbicb cornmute witb tbe rigbt multiplications determined by elements of R. Indeed, with every elemimt r ER there is assoCÍated a mapping T,. of M into itself given by T,.x = xr (as an irnmediate consequence of its definition, T,. E homR(M, M)). Tben the condition f(xr) = f(x)r can be reformulated as (f o T,.)(x)

=

(T,. o f)(x)

(x E M),

or, equivalently, as f o T,. T,. o f It is natural to ask whetber homR(M, M) can be turned into an Rmodule. In answering this question, let us suppose for tbe moment that R is a commutative ring and that f belongs to homR (M, M). Given a fixed r in R, define fr by means of (fr)(x)

Witb this definition, one finds that bomR(M, M) (regarded merely as a commutative group) becomes a rigbt R-module. Clearly, fr isan additive bomomorpbism from M into M. Moreover, for any S-E R, (fr)(xs)

f(x)r

(x EM).

= f(xs)r

f(x)sr

= f(x)rs

=

(fr)(x)s.

Here, rs = sr, because R is taken to be cornmutative. Tbus, fr actually does form ap R-endomorpbism. To recapitulate; in tbe cornmutative case, bomR(M, M),is botb ~:ring and an R-module, with ring multiplication and module mu1tiplication)~nked by tbe relation (f o g)r = f o (gr) ;:(fr) o 9

(1, 9 E bomR(M, M); rE R).

In p~ticular, when M =tR,: a clear distinction must be made between tbe R-homomorpbisms of R(cbhsidered as module over itself) and its bomomorpbisms as a ringo >,t~ . Before turning to the¡-,~~xt tbeorem, it is necessary to recalI that any right ideal 1 of tbe ring :RJs:a r1gbt R-module; the module product being given by the ,ordinary ringo multiplication xr, where x E 1 and rE R. Tbeorem 13-8. Let R be a simple ring witb minimal rigbt ideal 1 e =1= O an idempotent. Then bomR(I, 1) ::::: eRe, viewed as, rings. Proof As remarked aboye, we consider 1 as a right R-module. fE bomR(I, 1) implies tbat fea

= f(g(x»)r

273

+

b) = fía) +.f(b), f(ar) = f(a)r

eR, Thus,

(a, bE 1; rE R).

The first task is to characteríze tbesé R-endomorphisms in sorne convenient way. We handle tbis problem as folIows. LetfE bomR(I, 1) and tbe element a E 1 = eR, saya = ero Then fea) = f(er) = f(e)r, so tbat tbe function f is completely determined once tbe value fíe) is known. Since e is an idempotent of R, we bave fíe) f(e Z):= f(e)e. But fíe) E eR, whence fíe) = es for sorne choice of s in R, and so fíe) = f(e)e = ese. It readily folIows that fea) = f(e)r = (ese)r (ese 2 )r = (ese)er ~ (ese)a._ In brief, f acts on the elements of 1 as left multiplication by ese, wbere f(e) = ese. We shall utilize this information presently. Now, defin~ a mapping cp; homR(I, 1) -+ eRe by means of the rule cp(f¡' = fíe). It is iminediately apparent that cp(f + g)

=

(f + g)(e)

fíe)

+ g(e) =

cp(f)

+

CP(g).

If f is caused by a left multíplication by tbe eIement ese and g is a left muHiplícation by ete, then cp enjoys the further property

cp(f o g) = (f o g)(e) =

=

j(g(e») = f(ete)

esete = (ese)(ete) = f(e)g(e)

= CP(f)CP(g).

274

SOME NONCOMMUTATIVE THEOR y

FIRST COURSE IN RINGS AND IDEALS

This perinits us to conclude that cp, at the very least, is a homomorphism from the ring homR(l, 1) into the ring eRe. We next proceed to show that ker cp {O}. If cpU) = O, then f(e) = O, which can be applied to yield

fea) = f(er)

f(e)r

=O

for any a E 1 = eRo This of course mean s tbat f O, making cp a one-toone fundion. Finally, we select an arbitrary x in eRe (for instance, x = ere) and define the endomorphism hE homR(I, 1) by setting h(a) xa = (ere)a, where a E 1. Then we will ba ve

cp(h) = h(e) = (ere)e

ere

x,

so that cp maps onto eRe. The proof that cp serves as a ring isomorphlsm between homR(l, 1) and eRe is now complete. In combination with'Theorem 13-5, one obtains as a corollary CoroJlary. If R is a simple ring with minimal right ideal 1, then homR(I,1) forms a division ringo The foregoing corollary is a special case of a much wider theorem due to Schur. In accordance with the terminology of Section 12, we shaJl call an R-module simple provided it is nonzero and has no nontrivial submodules. Tbeorem 13-9. (Schur's Lemma). If M is a simple R-module, then homR(M, M) forms a division ringo

Proo! What we must pro ve is that any nonzerO element homR(M, M) has an inverse in homR(M, M). Since the image f(M) is a nQnzero submodule of M, it folJows that f(M) = M (M being a simple module by hypothesis) and therefore f maps onto M. In addition, f is.~e.cessarily a

one-to-one function. For ker f is certainly a submodule of M'·a:nd cannot equal M, becausef =1= O; the implication i8 tbat kerf = {O}.''Áll told, the inverse f-1: M --+ M exists. The reader is left the task of éerifying that f -1 is actuaHy an R-homomorphism. ' ... This might be the proper place to characterize aH simpie. R-modules of a given ring R. To this end, observe that if 1 is a right ideal of R, then we can certainly form the quotient group R/l; that is, R/l is the additive group whose elements are the cosets x + 1, x E R, with addition defined by (x

+ 1) +

(y

+ 1)

= x

+ y + l.

We can, however, do more than this. For, setting (x

+ l)r = xr +

1

(r E R),

R/l carries a well-defined structure as a right R-module. Indeed, the verification tbat the module moms are satisfied presents no problem and will

275

therefore be omitted. The reader may also conftrm that the submodules of R/l are of the form J/l, where J is a right ideal of R and J ;;;2 l. These remarks lead us to a precise identification of aH simple R-modules. Theorem 13-10. A right R-module M is simple if and only if it is Risomorphic to a quotient module R/l for some maximal right ideal 1 of R.

Proo! Assume that M is a simple R-module. For a fixed nonzero element x E M, define the right ideal 1 by 1 = {a E Rlxa = O}. To see that 1 is maximal, take J to be any right ideal of R satisfying 1 e J S;;; R. Now, the set xJ forros a nonzero submodule of M. From the supposition that M is simple, it follows that xJ = M = xR. (Actually, any element O =1= y E M is cyclic in the sense tbat yR = M.) Thus, for arbitrary rE R, there exists some bE J such that xb xr. But then, x(r - b) = O, giving r - b E 1 S;;; J; this says that rE J, whence the equality J = R, and 1 is a maximal right ideal of R. Next, consider the module homomorphism f: R --+ M in whichf(a) xa. The elements of the kernel off are precisely the elements of l. Furthermore, the condition xR = M assures us that f maps onto M and so, invoking Theorem 12-13, there is induced an Risomorphism R/l ;::: M. The other direction of the theorem relies on the fact that R/l is a simple right R-module if and only if 1 is a maximal right ideal of R; the reader may supply the necessary details. If M is a right R-module, then the set

A(M) = {r E RIMr

{O}}

is called the annihilator of the module M.As is easily verified, A(M) forms a two-sided ideal of R. We shall refer to Mas afaithful R-module or speak of Ras acting faithful1y on M whenever A(M) {O}; tbat is to say, whenever Mr = {O} implies that r = O. Since.the ring R is assumed to pqssess an identity element, R is evidently Caithfulas a module over itself. To round out our studies, let us introdüce a notíon of radical which is meaningful for general rings and which agrees with the previously defined . Jacobson radical in the event that the ringJs conuilUtative. The definition we give is phrased in terms of annihilators oC modules. Definition 13-1. The radical of the. ring R is the set J(R) = fl A(M), where the intersection varies over all simple right R-modules M; if there are no simple right R-modules, put J(R) = R. In short, the radical of a ring R may be described as the annihilator of all simple right R-modules. Notice, too, tbat J(R), being the intersection of two-sided ideals oC R, itself is an ideal of R. A final point to which attention should be drawn is that if R happens to have a faithful simple right R-

274

SOME NONCOMMUTATIVE THEOR y

FIRST COURSE IN RINGS AND IDEALS

This perinits us to conclude that cp, at the very least, is a homomorphism from the ring homR(l, 1) into the ring eRe. We next proceed to show that ker cp {O}. If cpU) = O, then f(e) = O, which can be applied to yield

fea) = f(er)

f(e)r

=O

for any a E 1 = eRo This of course mean s tbat f O, making cp a one-toone fundion. Finally, we select an arbitrary x in eRe (for instance, x = ere) and define the endomorphism hE homR(I, 1) by setting h(a) xa = (ere)a, where a E 1. Then we will ba ve

cp(h) = h(e) = (ere)e

ere

x,

so that cp maps onto eRe. The proof that cp serves as a ring isomorphlsm between homR(l, 1) and eRe is now complete. In combination with'Theorem 13-5, one obtains as a corollary CoroJlary. If R is a simple ring with minimal right ideal 1, then homR(I,1) forms a division ringo The foregoing corollary is a special case of a much wider theorem due to Schur. In accordance with the terminology of Section 12, we shaJl call an R-module simple provided it is nonzero and has no nontrivial submodules. Tbeorem 13-9. (Schur's Lemma). If M is a simple R-module, then homR(M, M) forms a division ringo

Proo! What we must pro ve is that any nonzerO element homR(M, M) has an inverse in homR(M, M). Since the image f(M) is a nQnzero submodule of M, it folJows that f(M) = M (M being a simple module by hypothesis) and therefore f maps onto M. In addition, f is.~e.cessarily a

one-to-one function. For ker f is certainly a submodule of M'·a:nd cannot equal M, becausef =1= O; the implication i8 tbat kerf = {O}.''Áll told, the inverse f-1: M --+ M exists. The reader is left the task of éerifying that f -1 is actuaHy an R-homomorphism. ' ... This might be the proper place to characterize aH simpie. R-modules of a given ring R. To this end, observe that if 1 is a right ideal of R, then we can certainly form the quotient group R/l; that is, R/l is the additive group whose elements are the cosets x + 1, x E R, with addition defined by (x

+ 1) +

(y

+ 1)

= x

+ y + l.

We can, however, do more than this. For, setting (x

+ l)r = xr +

1

(r E R),

R/l carries a well-defined structure as a right R-module. Indeed, the verification tbat the module moms are satisfied presents no problem and will

275

therefore be omitted. The reader may also conftrm that the submodules of R/l are of the form J/l, where J is a right ideal of R and J ;;;2 l. These remarks lead us to a precise identification of aH simple R-modules. Theorem 13-10. A right R-module M is simple if and only if it is Risomorphic to a quotient module R/l for some maximal right ideal 1 of R.

Proo! Assume that M is a simple R-module. For a fixed nonzero element x E M, define the right ideal 1 by 1 = {a E Rlxa = O}. To see that 1 is maximal, take J to be any right ideal of R satisfying 1 e J S;;; R. Now, the set xJ forros a nonzero submodule of M. From the supposition that M is simple, it follows that xJ = M = xR. (Actually, any element O =1= y E M is cyclic in the sense tbat yR = M.) Thus, for arbitrary rE R, there exists some bE J such that xb xr. But then, x(r - b) = O, giving r - b E 1 S;;; J; this says that rE J, whence the equality J = R, and 1 is a maximal right ideal of R. Next, consider the module homomorphism f: R --+ M in whichf(a) xa. The elements of the kernel off are precisely the elements of l. Furthermore, the condition xR = M assures us that f maps onto M and so, invoking Theorem 12-13, there is induced an Risomorphism R/l ;::: M. The other direction of the theorem relies on the fact that R/l is a simple right R-module if and only if 1 is a maximal right ideal of R; the reader may supply the necessary details. If M is a right R-module, then the set

A(M) = {r E RIMr

{O}}

is called the annihilator of the module M.As is easily verified, A(M) forms a two-sided ideal of R. We shall refer to Mas afaithful R-module or speak of Ras acting faithful1y on M whenever A(M) {O}; tbat is to say, whenever Mr = {O} implies that r = O. Since.the ring R is assumed to pqssess an identity element, R is evidently Caithfulas a module over itself. To round out our studies, let us introdüce a notíon of radical which is meaningful for general rings and which agrees with the previously defined . Jacobson radical in the event that the ringJs conuilUtative. The definition we give is phrased in terms of annihilators oC modules. Definition 13-1. The radical of the. ring R is the set J(R) = fl A(M), where the intersection varies over all simple right R-modules M; if there are no simple right R-modules, put J(R) = R. In short, the radical of a ring R may be described as the annihilator of all simple right R-modules. Notice, too, tbat J(R), being the intersection of two-sided ideals oC R, itself is an ideal of R. A final point to which attention should be drawn is that if R happens to have a faithful simple right R-

276

FIRST COURSE IN RINGS 'AND IDEALS

SOME NONCOMMUTATlVE THEORY

module, then J(R) = {O}. ,(This observation is made in preparation for Theorem 13':'13.) The relevance of J(R) to sorne of our éarlier investigations is clarified by the next theorem. What one can prove is Tbeorem 13-11.. In a ring R with identity,

J(R) = (I{IIJ is a maximal right ideal of R}. Proo!, If the elernent rE /(R), then r annihilates every simple right. Rmodule. In particular, fo(any maxirnal right ideal J of R, the quohent module RJJ forms a simple'R-module (we remark that maximal right ideals exist by Zorn's Lernma).J;Ience, (RJJ)r = {O} or, in other terms, Rr c:;::; J. Since R possesses an ideritity element, it may be inferred that.r = Ir E J. Thus r Hes in every maxiInal right ideal of R and (with self-explanatory nota~ion) J(R) c:;::; Í I J. \;'", For proof of the oppósite inclusion, let r E K = Í I J, where J runs through aH maximal right:ideals of R. We contend that the right ideal (1 - r)R = R. In the contrary case, (1 - r)R would be contained in sorne maximal right ideal I' of R; frorn r E K c:;::; I' foHows 1 '7 (1 - r) + rE I', which is irnpossible. The next step is to consider any simple right R-module M. If Mr 1 {O}, then xr 1 for a suitable element x of M. Since xK is a submodule of M, this yields xK. = M. Accordingly, there exists an element s E K for which xs = x: As was pointed out a moment ago, s E K secures that (1 -. s)R = R; hence, (1 - s)t = - s for sorne t in R. This relation gives

°

.°

:00 x(s

+

t - st) = xs

+

xt - xst = x

+

xt -:- xt = x,

leaving us with a contradi<;:tion, The implicatioll is that Mr = {O} for every simple R-module M. Thus, the element r E J(R), which settles the proof...

Remark, For the reader's guidance, it needs to be' statedthat, when R lacks an identity element, the analog of Theorem 13..,.11 asserts that J(R) is the intersection of the modular maximal right ideals of R. (Of course, in a ring with identity, a1l ideals aretriviaHy modular.) Furthermore, the intersection of the modular maximal right ideals of R always coincides with the intersection of aH modular maxirnalleft ideals [10]. In order to extend Schur's Lemma, we introduce a further concept, which, to begin with, may appear unrelated. Let (G, +) be a commutative group having more than one element and let E(G) denote the collection of endomorphisms of G. (Remember that by an endomorphism of G is meant a homomorphisrn of G into itself.) Dnder the operations of pointwise addition and functional composition, E(G) forms a ring with identity. Any

277

subring S of the ring E(G) is called a 'primitive ring of endomorphisms of.G if for aH x, y E G, with x 1 0, there exists sorne fES such that f(x) = y. As regards notation, we shaH hereafter write C(S) for the centralizer of S in E(G); that is, C(S) consists of those endomorphisms of G which cornmute, under composition, with every member of S. It is readily verified that C(S) is a subring of E(G) containing the identity endomórphism. We;are now ready to prove the following theorem. Theorem 13-12. Let G be a commutative group with more than one element.· Ir S is' a prirnitive ring of endomorphisms of G, then the centralizer C(S) forms a division ringo

°

'.' ¡

.

,

.

Proop'.Given that 1 fE C(S), let the element x E G be such thatf(x) 1 o. For e#h y E G, the prirnitivity' of S assures us that there exists an endomorphi~m g E S satisfying g(J(x») = y; hence, f(g(x») = g(J(x») = y.

,-,o'

As this'holds ·for arbitrary y in G, we conclude thatfmaps G onto itself. Now, assume thatf(x 1 ) = f(x 2 ), where Xl' x 2 E G, so thatf(x 1 - x 2 ) = O. If the element Xl - X2 1 0, then for any y E. G we can choose sorne h in S with h(x 1 - x 2 ) = y. (Once again, this is possible by the primitivity of S.) But then, .

f(y)

= f(h(x 1

-

x 2 »)

=

h(J(x 1

-

h») = h(O)

=

O,.

which means' thatf = 0, an óbvious contradiction. In other words, we must have Xl - 'x 2 = 0, or, rather, Xl = Xl' ensuring that the mapping f is a one-to-one function. As a result, there exists a well-defined multiplicative inversef-l: G ....;. G, which is also an endomorphism of G. For any g E S,

f- 1 0(g°f)of- 1 =f- 10 (f0g)of-1, leaving us with f-l ° g = g ° f- 1 ; therefore, f- 1 E C(S) and aH is proved. Note that this theorem actually does generalize the one due to Schur. For, let M be any simple right R-module. In the ring E(M) of endomorphisms of the additive group of M, consider the subring S of rightmultiplication functions induced by elements of R : S ={T,:lr E R; T,:(x) = xrfor x E M}. As was seen earlier, the centralizer of S in E(M)is exactIy the ring homR (M, M) of R-endomorphisms of M. We contend further that S is a primitive ring of endomorphisms of the module M. Indeed, if x, y E M, where X 1 0, ·then the set {T,:xlr E.R} forms a nonzero submodule of M and so .must be M itself; this guarantees that T,x = y for sorne choice of l' E R. As the conditions ofTheorem 13-12 hold, we find that homR(M, M) = C(S) is a division ringo

o,,:}:

276

FIRST COURSE IN RINGS 'AND IDEALS

SOME NONCOMMUTATlVE THEORY

module, then J(R) = {O}. ,(This observation is made in preparation for Theorem 13':'13.) The relevance of J(R) to sorne of our éarlier investigations is clarified by the next theorem. What one can prove is Tbeorem 13-11.. In a ring R with identity,

J(R) = (I{IIJ is a maximal right ideal of R}. Proo!, If the elernent rE /(R), then r annihilates every simple right. Rmodule. In particular, fo(any maxirnal right ideal J of R, the quohent module RJJ forms a simple'R-module (we remark that maximal right ideals exist by Zorn's Lernma).J;Ience, (RJJ)r = {O} or, in other terms, Rr c:;::; J. Since R possesses an ideritity element, it may be inferred that.r = Ir E J. Thus r Hes in every maxiInal right ideal of R and (with self-explanatory nota~ion) J(R) c:;::; Í I J. \;'", For proof of the oppósite inclusion, let r E K = Í I J, where J runs through aH maximal right:ideals of R. We contend that the right ideal (1 - r)R = R. In the contrary case, (1 - r)R would be contained in sorne maximal right ideal I' of R; frorn r E K c:;::; I' foHows 1 '7 (1 - r) + rE I', which is irnpossible. The next step is to consider any simple right R-module M. If Mr 1 {O}, then xr 1 for a suitable element x of M. Since xK is a submodule of M, this yields xK. = M. Accordingly, there exists an element s E K for which xs = x: As was pointed out a moment ago, s E K secures that (1 -. s)R = R; hence, (1 - s)t = - s for sorne t in R. This relation gives

°

.°

:00 x(s

+

t - st) = xs

+

xt - xst = x

+

xt -:- xt = x,

leaving us with a contradi<;:tion, The implicatioll is that Mr = {O} for every simple R-module M. Thus, the element r E J(R), which settles the proof...

Remark, For the reader's guidance, it needs to be' statedthat, when R lacks an identity element, the analog of Theorem 13..,.11 asserts that J(R) is the intersection of the modular maximal right ideals of R. (Of course, in a ring with identity, a1l ideals aretriviaHy modular.) Furthermore, the intersection of the modular maximal right ideals of R always coincides with the intersection of aH modular maxirnalleft ideals [10]. In order to extend Schur's Lemma, we introduce a further concept, which, to begin with, may appear unrelated. Let (G, +) be a commutative group having more than one element and let E(G) denote the collection of endomorphisms of G. (Remember that by an endomorphism of G is meant a homomorphisrn of G into itself.) Dnder the operations of pointwise addition and functional composition, E(G) forms a ring with identity. Any

277

subring S of the ring E(G) is called a 'primitive ring of endomorphisms of.G if for aH x, y E G, with x 1 0, there exists sorne fES such that f(x) = y. As regards notation, we shaH hereafter write C(S) for the centralizer of S in E(G); that is, C(S) consists of those endomorphisms of G which cornmute, under composition, with every member of S. It is readily verified that C(S) is a subring of E(G) containing the identity endomórphism. We;are now ready to prove the following theorem. Theorem 13-12. Let G be a commutative group with more than one element.· Ir S is' a prirnitive ring of endomorphisms of G, then the centralizer C(S) forms a division ringo

°

'.' ¡

.

,

.

Proop'.Given that 1 fE C(S), let the element x E G be such thatf(x) 1 o. For e#h y E G, the prirnitivity' of S assures us that there exists an endomorphi~m g E S satisfying g(J(x») = y; hence, f(g(x») = g(J(x») = y.

,-,o'

As this'holds ·for arbitrary y in G, we conclude thatfmaps G onto itself. Now, assume thatf(x 1 ) = f(x 2 ), where Xl' x 2 E G, so thatf(x 1 - x 2 ) = O. If the element Xl - X2 1 0, then for any y E. G we can choose sorne h in S with h(x 1 - x 2 ) = y. (Once again, this is possible by the primitivity of S.) But then, .

f(y)

= f(h(x 1

-

x 2 »)

=

h(J(x 1

-

h») = h(O)

=

O,.

which means' thatf = 0, an óbvious contradiction. In other words, we must have Xl - 'x 2 = 0, or, rather, Xl = Xl' ensuring that the mapping f is a one-to-one function. As a result, there exists a well-defined multiplicative inversef-l: G ....;. G, which is also an endomorphism of G. For any g E S,

f- 1 0(g°f)of- 1 =f- 10 (f0g)of-1, leaving us with f-l ° g = g ° f- 1 ; therefore, f- 1 E C(S) and aH is proved. Note that this theorem actually does generalize the one due to Schur. For, let M be any simple right R-module. In the ring E(M) of endomorphisms of the additive group of M, consider the subring S of rightmultiplication functions induced by elements of R : S ={T,:lr E R; T,:(x) = xrfor x E M}. As was seen earlier, the centralizer of S in E(M)is exactIy the ring homR (M, M) of R-endomorphisms of M. We contend further that S is a primitive ring of endomorphisms of the module M. Indeed, if x, y E M, where X 1 0, ·then the set {T,:xlr E.R} forms a nonzero submodule of M and so .must be M itself; this guarantees that T,x = y for sorne choice of l' E R. As the conditions ofTheorem 13-12 hold, we find that homR(M, M) = C(S) is a division ringo

o,,:}:

r 278

I

FIRST COURSE IN RINGS AND IDEALS

(a

1 '1

I

of nght mUltiplicatioris T,., rE R, forms a primitive ring of (group) endomorphisms of M. Furthermore, the supposition that M is faithful implies that T,. = O if and only if r = O. The mapP!ng p: r ~ thus determines a ring-isomorphism of R onto a primitive ring of endomorphisms of M, givin~ rise to the conc1usion that R is a primitive ringo On the other hand, suppose that R is isomorphic by means of the mapping r ~ r' to a primitive ring R' of endomorphisms of a commutative group M. Define a module structure on M by setting xr = r'(x), where x E M, rE R. It follows at once that Mr = {O} if and only if r'(x) = O for all x E M; that is to say, if and only if the endomorphism r' = O, or, equivalently, r = O. Thus, R acts faithfully on M as a right R-module. To see that the module in question is simple, notice that the primitivity of R' implies that M;= {r'(x)lr' E R'} for each nonzero x E M; this in its turn tells us that M '= ')¡:R for' any O =1= x E M, which makes M a simple Rmodule.

r..

= xr.

But xR = M, whence Mab = M; in other terms, 'M(ab - 1) = {O}. Since M is a faithful R-module, this gives ab = 1, so that a is an invertible element of R.

+ 1 = 1,

~

A(Rjl)

=

A(M)

= {O}.

Since {O} is a maximal ideal of the ring R, it follows that R must be a field. In this connection, we should observe that every simple ring R (hence, any field) is primitive. For, if 1 is a maximal right ideal of R, then Rjl is a simple right R-module. Its annihilator A(RjI) forms a two-sided ideal of R, which is necessarily proper since 1 rt A(Rjl). By hypothesis, we thus have A(Rjl) = {O}, so that Rjl is a faithful simple R-module; this assures that R is a primitive ringo The theory of primitive rings is extensive but somewhat specialized from our present point of view. We shall therefore break offthe discussion in order to turn to more important matters.

Lemma. For any finite set Xl' x 2 , ... , XII in M, linearly independent with respect to D, there exists an element a E R, such that

Corollary. Fqr any primitive ring R, the radical J(R) = {O}.

= x(ab)r

ar

Let M =1= {O} be a simple right R-module, where R is a given ring with identity, and consider the cJivision ring D = homR(M, M) of all R-endomorphisms of M. For any fE D, we can define a (left) module product fx by fx = f(x), where x E M. With this definition M beco mes in a natural way a left D-module (that is, a vector space over the division ring D); the verification is routine. Bearing this in mind, we next present a technical lemma, the value oí . ,~which will become c1ear as we proceed.

Note that as ap immediate consequence of the theorem we get

xr(ab)

+ l)r =

which says that the element r E A(Rj1). As a result, we deduce that

Proo! Let M be any faithful simple R-module. Since Mis simp'le, we have M = . xR for each O =1=., x E M. The force of this o bservation is that the ring

Now that Theorem 13-13 is available, let us make immediate use of it to prove that a cpmmutative primitive ring R is a field. (Theorem 13-12 could just as well be employed.) To set this fact in evidence, suppos'e that O =1= a E R and that M is any faithful simple right R-module. Then there exists sorne x E M such that xa =1= Oand, as a result, xaR = M. It is therefore possible to select an element b in R satisfying xab = X. The commutativity of R now implies that, for every rE R,

279

It might be of interest to present a somewhat differ~nt proof of the fact that a commutative primitive ring R forms a field. As befo re, let M be any faithful simple right R-module. Theorem 13-10 asserts that M ~ Rjl, viewed as R-modules, for sorne maximal (two-sided) ideal 1 of R. Now, if r E 1 and a E R, then

It is appropriate to call attention here to a very special c1ass of rings which are caJled primitive rings. A primitive ring is, by definition, isomorphic to a primitive ring of endomorphisms of sorne commutative group. That any division ring D is primitive should be c1ear. (The required primitive ring of endomorphisms is just the ring of tight multiplication functions on the additive group of D.) One frequently c1assifies rings according to the module~ which they admit; in this scheme, the primitive rings are perhaps the simplest type, for these admit a faithful simple module. Theorem 13-13. A ring R is primiti've if and only if it has a faithful simple R-module.

SOME NONCOMMUTATIVE THEORY

xla

= ... = xII-la = O,

xlla

=1=

O.

Proo! The proof argues by induction of n. To get the induction started, take n = 1; in this case, we have simply to establish that xlR =1= {Ol: Suppose that it happened that xlR = {O}. Since Xl =1= O, the se~ ,

¡'" I

,.'

N={XEMlxR= {Ol}

would form a nonzero R-submodule of M. Taking stock ofthe hypothesis that M is simple, we infer that N = M. It then follows that MR = {O}, which is definitely impossible when R has an identity. Now, as sume inductively that n > 1 is arbitrary and that the assertion of the lemma is already proved for n - 1. Put 1

=

{aERlxla = ...

=

x ll _ 2 a = O}.

r 278

I

FIRST COURSE IN RINGS AND IDEALS

(a

1 '1

I

of nght mUltiplicatioris T,., rE R, forms a primitive ring of (group) endomorphisms of M. Furthermore, the supposition that M is faithful implies that T,. = O if and only if r = O. The mapP!ng p: r ~ thus determines a ring-isomorphism of R onto a primitive ring of endomorphisms of M, givin~ rise to the conc1usion that R is a primitive ringo On the other hand, suppose that R is isomorphic by means of the mapping r ~ r' to a primitive ring R' of endomorphisms of a commutative group M. Define a module structure on M by setting xr = r'(x), where x E M, rE R. It follows at once that Mr = {O} if and only if r'(x) = O for all x E M; that is to say, if and only if the endomorphism r' = O, or, equivalently, r = O. Thus, R acts faithfully on M as a right R-module. To see that the module in question is simple, notice that the primitivity of R' implies that M;= {r'(x)lr' E R'} for each nonzero x E M; this in its turn tells us that M '= ')¡:R for' any O =1= x E M, which makes M a simple Rmodule.

r..

= xr.

But xR = M, whence Mab = M; in other terms, 'M(ab - 1) = {O}. Since M is a faithful R-module, this gives ab = 1, so that a is an invertible element of R.

+ 1 = 1,

~

A(Rjl)

=

A(M)

= {O}.

Since {O} is a maximal ideal of the ring R, it follows that R must be a field. In this connection, we should observe that every simple ring R (hence, any field) is primitive. For, if 1 is a maximal right ideal of R, then Rjl is a simple right R-module. Its annihilator A(RjI) forms a two-sided ideal of R, which is necessarily proper since 1 rt A(Rjl). By hypothesis, we thus have A(Rjl) = {O}, so that Rjl is a faithful simple R-module; this assures that R is a primitive ringo The theory of primitive rings is extensive but somewhat specialized from our present point of view. We shall therefore break offthe discussion in order to turn to more important matters.

Lemma. For any finite set Xl' x 2 , ... , XII in M, linearly independent with respect to D, there exists an element a E R, such that

Corollary. Fqr any primitive ring R, the radical J(R) = {O}.

= x(ab)r

ar

Let M =1= {O} be a simple right R-module, where R is a given ring with identity, and consider the cJivision ring D = homR(M, M) of all R-endomorphisms of M. For any fE D, we can define a (left) module product fx by fx = f(x), where x E M. With this definition M beco mes in a natural way a left D-module (that is, a vector space over the division ring D); the verification is routine. Bearing this in mind, we next present a technical lemma, the value oí . ,~which will become c1ear as we proceed.

Note that as ap immediate consequence of the theorem we get

xr(ab)

+ l)r =

which says that the element r E A(Rj1). As a result, we deduce that

Proo! Let M be any faithful simple R-module. Since Mis simp'le, we have M = . xR for each O =1=., x E M. The force of this o bservation is that the ring

Now that Theorem 13-13 is available, let us make immediate use of it to prove that a cpmmutative primitive ring R is a field. (Theorem 13-12 could just as well be employed.) To set this fact in evidence, suppos'e that O =1= a E R and that M is any faithful simple right R-module. Then there exists sorne x E M such that xa =1= Oand, as a result, xaR = M. It is therefore possible to select an element b in R satisfying xab = X. The commutativity of R now implies that, for every rE R,

279

It might be of interest to present a somewhat differ~nt proof of the fact that a commutative primitive ring R forms a field. As befo re, let M be any faithful simple right R-module. Theorem 13-10 asserts that M ~ Rjl, viewed as R-modules, for sorne maximal (two-sided) ideal 1 of R. Now, if r E 1 and a E R, then

It is appropriate to call attention here to a very special c1ass of rings which are caJled primitive rings. A primitive ring is, by definition, isomorphic to a primitive ring of endomorphisms of sorne commutative group. That any division ring D is primitive should be c1ear. (The required primitive ring of endomorphisms is just the ring of tight multiplication functions on the additive group of D.) One frequently c1assifies rings according to the module~ which they admit; in this scheme, the primitive rings are perhaps the simplest type, for these admit a faithful simple module. Theorem 13-13. A ring R is primiti've if and only if it has a faithful simple R-module.

SOME NONCOMMUTATIVE THEORY

xla

= ... = xII-la = O,

xlla

=1=

O.

Proo! The proof argues by induction of n. To get the induction started, take n = 1; in this case, we have simply to establish that xlR =1= {Ol: Suppose that it happened that xlR = {O}. Since Xl =1= O, the se~ ,

¡'" I

,.'

N={XEMlxR= {Ol}

would form a nonzero R-submodule of M. Taking stock ofthe hypothesis that M is simple, we infer that N = M. It then follows that MR = {O}, which is definitely impossible when R has an identity. Now, as sume inductively that n > 1 is arbitrary and that the assertion of the lemma is already proved for n - 1. Put 1

=

{aERlxla = ...

=

x ll _ 2 a = O}.

280

FIRST COURSE IN RINGS AND IPEALS

SOME NONCOMMUTATIVE THEORY

Then 1 is a right ideal of R and so 11 constitutes an R-submodule of M. By the induction hypothesis, X"_11 {O}, which implies that x n - l l ='M (M, of course, being simple). We wish to establish the existence of sorne element a E 1 with X,,_la = O, while x"a O. Let us assumethat this 1S not the case; in other words, for all a E 1, whenever x,,_la 0, then necessarily X"a =:' O. To derive a contradiction from ihis last sentence, We tentatively define a mapping f: M M by setting

+

--)o

(a

f(X"_la) = x"a

E

1)

',.

and proceed to show thatfis an R-endomorphism of M. First, one should check to see that f is well-defined. If X"..., X"_lb two elements a, b E 1, then x.-la - b) = O. But the aS$umption made'above'assures us that, in this event, xn(a - b) = O, whence x"a x"b, ptoducing á welldefined fun~tioÍl,; The following simple calculation confirms that f is an endomorphlsm of M: ......

la

ror

281

Theorem 13-14. (Wedderburn-Artin). Let R be a simple right Artinian ring with identity. Then R ,::: homD(I, 1) (thought of as rings) for sorne finite dimensional vector space 1 over a dívision ring D.

Proof. By virtue of the minimum cóndition, as applied to the colléction of all nonzero right ideals of R, there exists a minimal right ideal 1 of R; furthermore, 1 is oí the form 1 eR, e a nonzero idempotent. The next step is to observe that, since (eRe)eR = eR, the ideal 1 = eR can be viewed as a left vector space over the division ring D eRe. (Theorem 13-5 affirms that the set eRe comprises a divisiorúing.) 1 Now, let Rrl denote, the set of alI r[~ht multiplications T,,: 1 determined by elements of R; that is, , --)o

R rt = {T.;I a E R;

Ya x = (:x:'a: for x E 1}.

There is no special difficulty in verifying thát::R. rt is a subring of the endomorphism ring homD(I, 1). At the same tiÍÍle~ the correspondence a Ya sets up in a .natural way a nonzero homomoi'phism of R onto R rl ,. whose kernel is a two-sided ideal of R. Inasmuch as R is assumed to be simple, this kernel is {O}, whence the isomorphism R ,::: Rrl follows. The main contention of our 'proof is that R rt homD(I, To settle this point, note first that the set --)o

f(x,.-la

+

x"_lb)

=

f(x,,-l(a

= x,,(a =

xna

+

+ b) + xnb

b»)

=

f(x n- 1a)

+ f(X"-lb).

Using the fact that 1 is a right ideal of R, we note further that if a E Iand rE R, then

f(xn-1ar)

=

xn(ar)

=

(x"a)r = f(X"_la)r.

By these considerations, f becomes an R-endomorphism of M; that is, a member of the division ring D. This means that the module productfx"_la = xna for al! a E 1. Thus, ' for each a in 1, we have

xla

=

...

= x,,_za

=

(fxn':' l

~

x.)a

=

O.

-+ fE D)

must be linearly dependent over D.As aresult, the elements Xl' are also dependent and we arrive at the desired coñtradiction,

. ReR =

X2,'" , XII

Armed with this rather intricate machinery, we can now derive the fundamental structure theorem for simple right Artinian rings (the so-called Second Wedderburn Structure Theorem).. From the many ways of proving this result, we select a module-theoretic approach essentially due to Henderson [40].

fl:: a¡eb¡la¡, b¡ E R}

(here I represents an arbitrary finite sum) constitutes a two-sided ideal of R different from zero, since O e = lel E ReR. Hence, we must have ReR == R. In particular, the identity 1 E ReR, so that it is possible to select elements al' b¡ E R satisfying 1 = I a¡eb¡. Now, choose any D-endomorphism fE homD(I, 1) and any element x = el' E 1. A straightforward computation gives

+

f(x)

Again, from our induction supposition that the ¡emma holds for n - 1 elements, it follows that

(O

n

=

f(erl)

=

f(er

I

a¡eb¡) =

f(I era¡eb¡)

= If(era¡eb,)

= I era¡ef(eb;) = er I a¡ef(eb¡)

(since eraje

E

D)

xI a¡ef(eb¡'). Fromthis formula, it appears thatf(x) = T.x, where the element s (which does not depend on x) is given by s = I a¡ef(eb¡}. Therefore, fE R rt , confirming that Rrl = hom D(I, 1). Puttirig our remarks together, we obtain the isomorphism

280

FIRST COURSE IN RINGS AND IPEALS

SOME NONCOMMUTATIVE THEORY

Then 1 is a right ideal of R and so 11 constitutes an R-submodule of M. By the induction hypothesis, X"_11 {O}, which implies that x n - l l ='M (M, of course, being simple). We wish to establish the existence of sorne element a E 1 with X,,_la = O, while x"a O. Let us assumethat this 1S not the case; in other words, for all a E 1, whenever x,,_la 0, then necessarily X"a =:' O. To derive a contradiction from ihis last sentence, We tentatively define a mapping f: M M by setting

+

--)o

(a

f(X"_la) = x"a

E

1)

',.

and proceed to show thatfis an R-endomorphism of M. First, one should check to see that f is well-defined. If X"..., X"_lb two elements a, b E 1, then x.-la - b) = O. But the aS$umption made'above'assures us that, in this event, xn(a - b) = O, whence x"a x"b, ptoducing á welldefined fun~tioÍl,; The following simple calculation confirms that f is an endomorphlsm of M: ......

la

ror

281

Theorem 13-14. (Wedderburn-Artin). Let R be a simple right Artinian ring with identity. Then R ,::: homD(I, 1) (thought of as rings) for sorne finite dimensional vector space 1 over a dívision ring D.

Proof. By virtue of the minimum cóndition, as applied to the colléction of all nonzero right ideals of R, there exists a minimal right ideal 1 of R; furthermore, 1 is oí the form 1 eR, e a nonzero idempotent. The next step is to observe that, since (eRe)eR = eR, the ideal 1 = eR can be viewed as a left vector space over the division ring D eRe. (Theorem 13-5 affirms that the set eRe comprises a divisiorúing.) 1 Now, let Rrl denote, the set of alI r[~ht multiplications T,,: 1 determined by elements of R; that is, , --)o

R rt = {T.;I a E R;

Ya x = (:x:'a: for x E 1}.

There is no special difficulty in verifying thát::R. rt is a subring of the endomorphism ring homD(I, 1). At the same tiÍÍle~ the correspondence a Ya sets up in a .natural way a nonzero homomoi'phism of R onto R rl ,. whose kernel is a two-sided ideal of R. Inasmuch as R is assumed to be simple, this kernel is {O}, whence the isomorphism R ,::: Rrl follows. The main contention of our 'proof is that R rt homD(I, To settle this point, note first that the set --)o

f(x,.-la

+

x"_lb)

=

f(x,,-l(a

= x,,(a =

xna

+

+ b) + xnb

b»)

=

f(x n- 1a)

+ f(X"-lb).

Using the fact that 1 is a right ideal of R, we note further that if a E Iand rE R, then

f(xn-1ar)

=

xn(ar)

=

(x"a)r = f(X"_la)r.

By these considerations, f becomes an R-endomorphism of M; that is, a member of the division ring D. This means that the module productfx"_la = xna for al! a E 1. Thus, ' for each a in 1, we have

xla

=

...

= x,,_za

=

(fxn':' l

~

x.)a

=

O.

-+ fE D)

must be linearly dependent over D.As aresult, the elements Xl' are also dependent and we arrive at the desired coñtradiction,

. ReR =

X2,'" , XII

Armed with this rather intricate machinery, we can now derive the fundamental structure theorem for simple right Artinian rings (the so-called Second Wedderburn Structure Theorem).. From the many ways of proving this result, we select a module-theoretic approach essentially due to Henderson [40].

fl:: a¡eb¡la¡, b¡ E R}

(here I represents an arbitrary finite sum) constitutes a two-sided ideal of R different from zero, since O e = lel E ReR. Hence, we must have ReR == R. In particular, the identity 1 E ReR, so that it is possible to select elements al' b¡ E R satisfying 1 = I a¡eb¡. Now, choose any D-endomorphism fE homD(I, 1) and any element x = el' E 1. A straightforward computation gives

+

f(x)

Again, from our induction supposition that the ¡emma holds for n - 1 elements, it follows that

(O

n

=

f(erl)

=

f(er

I

a¡eb¡) =

f(I era¡eb¡)

= If(era¡eb,)

= I era¡ef(eb;) = er I a¡ef(eb¡)

(since eraje

E

D)

xI a¡ef(eb¡'). Fromthis formula, it appears thatf(x) = T.x, where the element s (which does not depend on x) is given by s = I a¡ef(eb¡}. Therefore, fE R rt , confirming that Rrl = hom D(I, 1). Puttirig our remarks together, we obtain the isomorphism

282

FlRST COURSE IN RINGS AND IDEALS PROBLEMS

283

To c1inch the argument, let us show that 1 is a finite dimensional (left) vector space over D. We suppose this to be falseo Then 1 possesses an infinite basis, from which can be extracted a sequence XI' X , ... of linear1y 2 independent elements. That is, for each integer n, the set {XI' X , ... , XII} 2 is independent with respect to D. Given an integer n, define

1rr = {a E R/xla

T~en 11

= ... =

= O}

X' I a J.

These relations make clear that

12 ;2 ••. ;2 1" ;2 ' .. forms a descending chain of right ideals of .R. Sin ce the ideal 1 = eR can be regarded as a simple (right) R-module, a direct appeal to the last lemma is permissible. Thus, there exists sorne element a E R for which ;2

j

jo g

If Vis an n-dimensional vector space over the division ring D, then the ring homD(J~ V) is well-known to be isomorphic to the ring M,,(D) of n x n matrices over D. To spell out somedetails, let Vhave the basis {XI' X , '" , x,,}. 2 If j E homD(v, V), each of thése basis elements will be mapped by j into a uniquely determined linear combination of XI' x 2 , ... , x". In other words, there exist elements a¡i E D, uniquely defined by j, such that j(x ) can be expressed in the form . i ~-".

aijx¡

¡= I

(j = 1,.2, ... , n). \ J1'.

~.

(Observe that the summation is over the first i:ndex.) Thus, to each endomorphism jE.homD(v, V) there corresponds #.·i;unique n x n matrix (aij) with entries from D. ¡: , There is no problem in showing that the ~flPj -+ (aij) yields a one-toone function from homD(v, V) onto the matrixring M,,(D). Indeed, starting withanarbitrary(a¡) E M,,(D),onedefinesaD-endomorphismj E homD(V, V) by first setting j(x) = ¿í= I aUx¡ and then extending linearly to all of V; it is evident that (a¡) is precisely the matrix identified with the resulting endomorphism. If (a u ) and (bu) are the matrix representations of two elements of homD(v, V), say j and g, then (f + g)(x) = j(xi ) and

+ g(x)

=

¿ n

¡=I

aijx¡

(aij)

+ (bu),

-+( ¿ 11

aikbkj) = (a¡j)(b¡J

k=1

leading to the conc1usion that 1" is proper1y contained in In-l' The point which we wish tp make is tha t 1 I ::J 12 ::J ••• ::J 1" ::J ... is a strictly descending chain, in violation of the assumption that R is a right Artinian. Accordingly, the dimension dim D l < co and this finally ends the proof of Theorem 13-14.

= ¿"

-+ (a¡j + bu) =

and, by definition of the product of two matrices, that

xla = ... = xII-la = O,

j(xi )

+g

+ ¿" i=1

b¡jx¡ =

¿ "

i=1

(aij

+

b¡)x¡

. . . h' ch associates with eachj E hom D (V, V) The conc1usIOn IS that the mappmg w .1 th fixed basis) induces a ring its matrix representation {a u) (relatlve to e isomorphism homD(V, V) ~ M,,(D) ..

. .. Its may be collected ?n thedstre~gtt~ to glve a escnp IOnof ~7.e:.I_:::;~~feurri~~7°~:tf~::n rings in terms of matrix rings. .. . Th Theorem 13-15. Let R be a nil-sernisimple right Artmta.n nng. e~ there exist division rings Di and suitable integers ni (1 = 1, 2, ... , r such that R ~ M",(D I ) Etl M",(D 2 ) E!7 ... E!7 M,,/D r ).

Although we must now c~ose this ~~aP::~iS:d a~~h;~Oi~~n~~!: ~~: presentation of the theory.of nngs,. w~ a p ~ore thorough treatment could start delving d~eply mto the ,sub~ect. (;~:~ account by Herstein [15J of the noncommutatlve aspects, ~ee t e exce we have merely scratched and the reference~ cited. the.re·)bNeed~essf t~;:~;a' nonetheless, the reader the surface of tbis fascmatmg . ranc o . '. d he difficulties. should now be in a better POSiti~l'l to apprectate the detatls an t

PROBLEMS Unless spedlied otherwise, R always (jenotes an arbitrary ring with identity.

1. Let e be an idempotentelement of the ring R. For any two-sided ideal l of R, show that the subring ele = l n (eRe). 2. If R is a right Artinian ring, prove that ,,:,henever therede~ists ~:~ ~~n~n!~;,;E.~ with ah = 1, then ba = 1. [Hint: Conslder the deseen mg e _ ;2 b"R ;2 ... of right ideals of R.]

282

FlRST COURSE IN RINGS AND IDEALS PROBLEMS

283

To c1inch the argument, let us show that 1 is a finite dimensional (left) vector space over D. We suppose this to be falseo Then 1 possesses an infinite basis, from which can be extracted a sequence XI' X , ... of linear1y 2 independent elements. That is, for each integer n, the set {XI' X , ... , XII} 2 is independent with respect to D. Given an integer n, define

1rr = {a E R/xla

T~en 11

= ... =

= O}

X' I a J.

These relations make clear that

12 ;2 ••. ;2 1" ;2 ' .. forms a descending chain of right ideals of .R. Sin ce the ideal 1 = eR can be regarded as a simple (right) R-module, a direct appeal to the last lemma is permissible. Thus, there exists sorne element a E R for which ;2

j

jo g

If Vis an n-dimensional vector space over the division ring D, then the ring homD(J~ V) is well-known to be isomorphic to the ring M,,(D) of n x n matrices over D. To spell out somedetails, let Vhave the basis {XI' X , '" , x,,}. 2 If j E homD(v, V), each of thése basis elements will be mapped by j into a uniquely determined linear combination of XI' x 2 , ... , x". In other words, there exist elements a¡i E D, uniquely defined by j, such that j(x ) can be expressed in the form . i ~-".

aijx¡

¡= I

(j = 1,.2, ... , n). \ J1'.

~.

(Observe that the summation is over the first i:ndex.) Thus, to each endomorphism jE.homD(v, V) there corresponds #.·i;unique n x n matrix (aij) with entries from D. ¡: , There is no problem in showing that the ~flPj -+ (aij) yields a one-toone function from homD(v, V) onto the matrixring M,,(D). Indeed, starting withanarbitrary(a¡) E M,,(D),onedefinesaD-endomorphismj E homD(V, V) by first setting j(x) = ¿í= I aUx¡ and then extending linearly to all of V; it is evident that (a¡) is precisely the matrix identified with the resulting endomorphism. If (a u ) and (bu) are the matrix representations of two elements of homD(v, V), say j and g, then (f + g)(x) = j(xi ) and

+ g(x)

=

¿ n

¡=I

aijx¡

(aij)

+ (bu),

-+( ¿ 11

aikbkj) = (a¡j)(b¡J

k=1

leading to the conc1usion that 1" is proper1y contained in In-l' The point which we wish tp make is tha t 1 I ::J 12 ::J ••• ::J 1" ::J ... is a strictly descending chain, in violation of the assumption that R is a right Artinian. Accordingly, the dimension dim D l < co and this finally ends the proof of Theorem 13-14.

= ¿"

-+ (a¡j + bu) =

and, by definition of the product of two matrices, that

xla = ... = xII-la = O,

j(xi )

+g

+ ¿" i=1

b¡jx¡ =

¿ "

i=1

(aij

+

b¡)x¡

. . . h' ch associates with eachj E hom D (V, V) The conc1usIOn IS that the mappmg w .1 th fixed basis) induces a ring its matrix representation {a u) (relatlve to e isomorphism homD(V, V) ~ M,,(D) ..

. .. Its may be collected ?n thedstre~gtt~ to glve a escnp IOnof ~7.e:.I_:::;~~feurri~~7°~:tf~::n rings in terms of matrix rings. .. . Th Theorem 13-15. Let R be a nil-sernisimple right Artmta.n nng. e~ there exist division rings Di and suitable integers ni (1 = 1, 2, ... , r such that R ~ M",(D I ) Etl M",(D 2 ) E!7 ... E!7 M,,/D r ).

Although we must now c~ose this ~~aP::~iS:d a~~h;~Oi~~n~~!: ~~: presentation of the theory.of nngs,. w~ a p ~ore thorough treatment could start delving d~eply mto the ,sub~ect. (;~:~ account by Herstein [15J of the noncommutatlve aspects, ~ee t e exce we have merely scratched and the reference~ cited. the.re·)bNeed~essf t~;:~;a' nonetheless, the reader the surface of tbis fascmatmg . ranc o . '. d he difficulties. should now be in a better POSiti~l'l to apprectate the detatls an t

PROBLEMS Unless spedlied otherwise, R always (jenotes an arbitrary ring with identity.

1. Let e be an idempotentelement of the ring R. For any two-sided ideal l of R, show that the subring ele = l n (eRe). 2. If R is a right Artinian ring, prove that ,,:,henever therede~ists ~:~ ~~n~n!~;,;E.~ with ah = 1, then ba = 1. [Hint: Conslder the deseen mg e _ ;2 b"R ;2 ... of right ideals of R.]

284

FIRST COURSE IN RINGS AND IDEALS

3. Given a right Artinian ring R, establish that a) The sum N of all nilpotent right ideals of R is again a nilpotent right ideal. [H int: If N is not nilpotent, it contains a nonzero idempotent by Theorem 13-L] b) N fOnTIsa two-sidedideal ofR. [Hint: The ideal RN is sueh that (RN)" whence RN ~ N.] e) The ideal N contains any nilpotent left ideal 1 of R. [H int: 1 ~ RN ~ N.]

m",

= {O} or 12 = 1; in thé second case, deduce' that J = eR = el for some idempotent e =1= O in l. [Hint: See the proofs ofTheorems 13-1 and 13-2; first, establish the existence of an element a e J ror whie~.ál = J.]

4. Let 1 be a minimal right ideal of the ring R. Show that either J2

t

1·,

5. Assume !hat 1. and J are/t~o minimal right ideals of the ring R which are iso1, prove that morphie as right R-modul~8: Ir J2 a) Any R-isomorphismj)~ J is orthe fornlf(x) = ax for some a ej. b) The produet JJ J.. [Hint: By part (a),. J al, where a E J.] 6. For nonzero idempotents'é;'u of the ring R, prove that a) eR = uR ir and onlylf é4 = u and ue = e. b) eR and uR are isomorphié as R-modules if and only ifthere.exist x, y E R sueh tbat xy = u and yx = e. [Hint: Sinee xe ux and yu "" ey, the funetion f(er) = x{er) = u(xr) defines an R-isomorphism f: eR ..... uR, with f-I(US) = y(us) = e(ys).] e) eR ~ uR, as R-modules, if and only if Re ~ Ru.

7. Let eR and uR be two minimal right ideals of the ring R, where e and u are nonzero idempotents of R. Show tbat eR and uR are R-isomorphíe ifand only iftheir produet uReR =1= {O}. [Hint: Ifure =1= Ofor some r E R, define the R-isomorphisnl f: eR ..... uR by f(es) = (ure)s.] .

8. Let R bea nil-semisimple ring without identity. Ifthe element a E Rand aR {O}, establish that a O. [Hint: The ideal A(R) = {r E RlrR = {O}} satisfies A(R)2 ~ A(R)R = {O}.] 9. Establish the statements below: a) A right ideal J of the ring R is a direct summand of R if and oruy· ir 1 = eR for some idempotent e e R ; b) a mininlal right ideal J of the ring R is a dÍreet summand of R if and only ir 12 =1= {O}. 10. Prove that a right Artinian fing R is nil-semisinlple ir and only ir 12 each mínimal rigbt ideal 1 of R.

285

PROBLEMS

=1=

{O} ror

11. Assuming tbat R is a nil-semisimple right Artinian ring, veriry the following assertions: a) R is rigbt Noetherian; that is, R satisfies the ascending ehain condition on right ideals. [Hint: By Theorem 13-2, the right ideals of R are finiteIy generated.] b) The mapping e ..... eR defines a one-to-one eorrespondence between the set of all idempotent elements e e cent R and the set of two-sided ideaIs of R. c) For any two-sided ideal 1 of R, ann,1 = ann¡1 and so R = 1 EB annl. d) Every right ideal J of R is a direct sumÍnand of any right ideal containing it.

12. a) Suppose that the ring R is a finite direct sum of right ideals J¡ =1= {O}, say R = 1¡ fE) J 2 EB ... EB Jn • Ir 1 e¡ + e2 + ... + e", where e¡el" prove that the elements e¡ fornl a set of orthogonaI idempotents and that 11 = eóR (i = 1, 2, .. , , n). b) If the ideals 1¡ of part (a) are aH two-sided, show that e¡ E cent R an4 so serves as an identity element for 1¡.

13. a)' Prove that an idempotent e =1= O of the ring R is prinlitive if and only if R contains no idenlpotent u =1= e sueh that eu = ue = u. bl,'Establish that any idempotent elenlent e =1= O of a nil-semisimple right Artinian nlng R is the sum of a finite number of orthogonal primitive idempotents. ;.CHint: There exists a minimal rigbt ideal 1 S;; eRo Write eR = 1 fE) J, where :~ •.the right ideal J 1. Now, either J = {O}, or else e ;,. el + ez, with e¡ e J .:/aprimitive idempotent. If ez J is not primitive, repeat this proeess as applied J e2R.]

e

/:ro

14. ~)áf M

=1= {O} is a right R-modulé, verify that M becomes a left homR(M, M)"'!:.,rllodule on defining the module productfx by

fx

='

f(x)

(fE homR(M, M); x e M).

b) Let M and N be right R·modules which are R-isomorphic under the mapping a: M -> N. Show that homR(M, M) ~ homR(N, N), as rings, by means of the funetion that éames fe homR(M, M) toa. fo a-l. 15. Let F be a field and M 2(F) denote the ring of 2 x 2 matrices over F. Prove thát a) The matrices el =

(~ ~) and e2 = [~ ~). are orthogonal idempotents;

b) 1¡ = e¡Mz(F)(i = 1,2)isaminimalrightidealofM2 (F),withM2(F) e) e¡Mz(F)ei ~ F for i = 1,2.

1 1 EB lz;

16; Use Theorem 13-14 to deduce that any commutative semisinlple (in the usual sense) Artinian ring is a finite direct sum of fields. 17. Prove tbat a right Artinian ring R is a regular ring if and only if R is nil-semisinlple. [Hint: Problems 19 and 20, Chapter 9.J

18. Let M

=1= {O} be a simple right R-module and so, by Theorem 13-9, a vector space over the division ring D homR(M, M). Prove ihe following version of the Jacobson Density Theorem. Given any x¡, X2' ... , x. e M which are lineady independent' with respect to D aÍld arbitrary y¡, Yz, ... , Y. e M, there exists some element a E R (equivalently, some D-endomorphism Ya e Rr¡) such that Xka = Yk for k = 1,2, ... , n. [Hint: From the lenlllla preeedíngTheorem 13-14, it is possíble to choose elements aj e R for which

= Ofor j =1= i xja¡ {=1= Ofor j = i . Let r¡ E R be an elenlent such that x¡a¡r¡ = Yi' Now, consider a

= 2::i=¡ a¡r¡.]

284

FIRST COURSE IN RINGS AND IDEALS

3. Given a right Artinian ring R, establish that a) The sum N of all nilpotent right ideals of R is again a nilpotent right ideal. [H int: If N is not nilpotent, it contains a nonzero idempotent by Theorem 13-L] b) N fOnTIsa two-sidedideal ofR. [Hint: The ideal RN is sueh that (RN)" whence RN ~ N.] e) The ideal N contains any nilpotent left ideal 1 of R. [H int: 1 ~ RN ~ N.]

m",

= {O} or 12 = 1; in thé second case, deduce' that J = eR = el for some idempotent e =1= O in l. [Hint: See the proofs ofTheorems 13-1 and 13-2; first, establish the existence of an element a e J ror whie~.ál = J.]

4. Let 1 be a minimal right ideal of the ring R. Show that either J2

t

1·,

5. Assume !hat 1. and J are/t~o minimal right ideals of the ring R which are iso1, prove that morphie as right R-modul~8: Ir J2 a) Any R-isomorphismj)~ J is orthe fornlf(x) = ax for some a ej. b) The produet JJ J.. [Hint: By part (a),. J al, where a E J.] 6. For nonzero idempotents'é;'u of the ring R, prove that a) eR = uR ir and onlylf é4 = u and ue = e. b) eR and uR are isomorphié as R-modules if and only ifthere.exist x, y E R sueh tbat xy = u and yx = e. [Hint: Sinee xe ux and yu "" ey, the funetion f(er) = x{er) = u(xr) defines an R-isomorphism f: eR ..... uR, with f-I(US) = y(us) = e(ys).] e) eR ~ uR, as R-modules, if and only if Re ~ Ru.

7. Let eR and uR be two minimal right ideals of the ring R, where e and u are nonzero idempotents of R. Show tbat eR and uR are R-isomorphíe ifand only iftheir produet uReR =1= {O}. [Hint: Ifure =1= Ofor some r E R, define the R-isomorphisnl f: eR ..... uR by f(es) = (ure)s.] .

8. Let R bea nil-semisimple ring without identity. Ifthe element a E Rand aR {O}, establish that a O. [Hint: The ideal A(R) = {r E RlrR = {O}} satisfies A(R)2 ~ A(R)R = {O}.] 9. Establish the statements below: a) A right ideal J of the ring R is a direct summand of R if and oruy· ir 1 = eR for some idempotent e e R ; b) a mininlal right ideal J of the ring R is a dÍreet summand of R if and only ir 12 =1= {O}. 10. Prove that a right Artinian fing R is nil-semisinlple ir and only ir 12 each mínimal rigbt ideal 1 of R.

285

PROBLEMS

=1=

{O} ror

11. Assuming tbat R is a nil-semisimple right Artinian ring, veriry the following assertions: a) R is rigbt Noetherian; that is, R satisfies the ascending ehain condition on right ideals. [Hint: By Theorem 13-2, the right ideals of R are finiteIy generated.] b) The mapping e ..... eR defines a one-to-one eorrespondence between the set of all idempotent elements e e cent R and the set of two-sided ideaIs of R. c) For any two-sided ideal 1 of R, ann,1 = ann¡1 and so R = 1 EB annl. d) Every right ideal J of R is a direct sumÍnand of any right ideal containing it.

12. a) Suppose that the ring R is a finite direct sum of right ideals J¡ =1= {O}, say R = 1¡ fE) J 2 EB ... EB Jn • Ir 1 e¡ + e2 + ... + e", where e¡el" prove that the elements e¡ fornl a set of orthogonaI idempotents and that 11 = eóR (i = 1, 2, .. , , n). b) If the ideals 1¡ of part (a) are aH two-sided, show that e¡ E cent R an4 so serves as an identity element for 1¡.

13. a)' Prove that an idempotent e =1= O of the ring R is prinlitive if and only if R contains no idenlpotent u =1= e sueh that eu = ue = u. bl,'Establish that any idempotent elenlent e =1= O of a nil-semisimple right Artinian nlng R is the sum of a finite number of orthogonal primitive idempotents. ;.CHint: There exists a minimal rigbt ideal 1 S;; eRo Write eR = 1 fE) J, where :~ •.the right ideal J 1. Now, either J = {O}, or else e ;,. el + ez, with e¡ e J .:/aprimitive idempotent. If ez J is not primitive, repeat this proeess as applied J e2R.]

e

/:ro

14. ~)áf M

=1= {O} is a right R-modulé, verify that M becomes a left homR(M, M)"'!:.,rllodule on defining the module productfx by

fx

='

f(x)

(fE homR(M, M); x e M).

b) Let M and N be right R·modules which are R-isomorphic under the mapping a: M -> N. Show that homR(M, M) ~ homR(N, N), as rings, by means of the funetion that éames fe homR(M, M) toa. fo a-l. 15. Let F be a field and M 2(F) denote the ring of 2 x 2 matrices over F. Prove thát a) The matrices el =

(~ ~) and e2 = [~ ~). are orthogonal idempotents;

b) 1¡ = e¡Mz(F)(i = 1,2)isaminimalrightidealofM2 (F),withM2(F) e) e¡Mz(F)ei ~ F for i = 1,2.

1 1 EB lz;

16; Use Theorem 13-14 to deduce that any commutative semisinlple (in the usual sense) Artinian ring is a finite direct sum of fields. 17. Prove tbat a right Artinian ring R is a regular ring if and only if R is nil-semisinlple. [Hint: Problems 19 and 20, Chapter 9.J

18. Let M

=1= {O} be a simple right R-module and so, by Theorem 13-9, a vector space over the division ring D homR(M, M). Prove ihe following version of the Jacobson Density Theorem. Given any x¡, X2' ... , x. e M which are lineady independent' with respect to D aÍld arbitrary y¡, Yz, ... , Y. e M, there exists some element a E R (equivalently, some D-endomorphism Ya e Rr¡) such that Xka = Yk for k = 1,2, ... , n. [Hint: From the lenlllla preeedíngTheorem 13-14, it is possíble to choose elements aj e R for which

= Ofor j =1= i xja¡ {=1= Ofor j = i . Let r¡ E R be an elenlent such that x¡a¡r¡ = Yi' Now, consider a

= 2::i=¡ a¡r¡.]

286

FIRST COURSE IN RINGS AND IDEALS

19. Given a right R-module M, set M* = homR(M, R). Prove the statements below: a) M* caD. be made into a left R-module (known as the dual module of M) by defining the module product rf as

(rf)(x)

= rf(x)

(rER,x

E

APPENDIX A

M).

b) When R is a division ring, so that M* forms a vector space over R, then [M*:R] = [M:R]. [Hint: Ifx¡, x z, ... , x. is a basis for M, then the n functions j¡t' E M* (i = 1, 2, ... , n) prescribed by

J1 for for i = j -lO i 0/= j

j¡i *(Xj ) -_ bij _

RELATIONS

serv.e as a basis for M*.] c) M and M** are naturally isomorphic as right R-modules, where M** = (M*)* = homR(M~, R). [Hint: For each x E M, defind E M** by x(g) = g(x), g E M*; then M ~!\ll** under the R-isomorphism that sends x to x.] 20. Let M be a right R-module and 1 a right ideal of R. Prove that a) The set MI = {¿ Xir;jXi E M, ri El}, where ¿ is an arbitrary finite sum, constitutes a submodule of M. b) If M and 1 are both simple (as R-modules) with MI 0/= {O}, then M ~ i. [Hint: Since xl = M for some x E M, an R-isomorphism f: 1 -> M can be given by f(a) = xa, where a E l.] 21. Let R be a nil-semisimple right Artinian ringo Verify that, up to R-isomorphism, there exist only a finite number of sinlple right R-modules. [Hint: R is a direct sum R = 1¡.EB lz EB .. · EB l. offinitely many two-sided ideals. If M 0/= {O} is any simple right R-module, use Problem 20(b) to conc1ude that M ~ li for some i.] 22. Prove that the ring R is simple if and only if every simple right R-module is faithful. [Hint: Ir R has non trivial two-sided ideals, it possesses a maxin:ial one 1 by Zorn's Lemma, Let J be any maximal right ideal of R with J ;2 1 and obtain a contradictionby considering the simple right R-module A(RfJ) ;2 l.] 23. Prove'each of the following assertions: a) 1'he radical J(R) = (1 :,R), where the intersection runs over all maximal right ideals 1 of R. b) ~py nonzero ideal of a prinlitive ring of endomorphisms of a commutative group G;j!i also a primitive ring of endomorphisms of G. c) A ring R is primitive if and only if it contains a maximal right ideal 1 such that thequotient ideal (1:,R) = {O}.

n

24. An ideal 1 of the ring R is said to be a primitive ideal if R/l is a primitive ringo Establish that the radical J(R) can be represented as J(R) = P, where the intersection is taken over a1l primitive ideals P of R. [Hint: If P = A(M), where M is a simple R-module, then M is a faithful simple module over the ring R/P.]

n

We herein append a few definitions and general results concerning certain types of relations that can be imposed on a set. For the most part, our attention is confined to two relations of particular utility, namely, equivalence relations ancl order relations. Intuitively, a (binary) relation on a set S pro vides a criterio n such that for each ordered pair (a, b) of elements of S, one can determine whether the statement "a is related to b" is meaningful (in the sen se of being true or false according to the choice of elements a and b). The relation is completely characterized once we know the set of all those pairs for which the first component stands in that relation to the second. This idea can best be formulated in set-theoretic language as Definition A-l. A (binary) relation R in a nonempty set Sjs any subset of the Cartesian product S x S. (' :. If R is a relation, we express the fact that the pair {a, b) e,i{'by saying that. a is related to b with respect to the relation R, and we wrfte'aRb. F or instance, the relation < in R can be represented by aH points iri' the plane lying aboye the diagonalline y = x; it is customary to write 3: < 4, rathet 'l ! than the awkward (3, 4) E <. Our immediate concern is with equivalence relations. In pi~ctice, these arise whenever it is desirable to identify, as a single entity, aH eIements of a set that have sorne preassigned characteristic.

Definition A-2. A relation R in a set S is an equivalenee relation in S provided that it satisfies the three propertit';s, . 1) aRa for aH a E S (reflexive property), 2) if aRb, then bRa (symmetric property), 3) if aRb and bRe, then aRe (transitive property). Equivalence relations are usually denoted by the symbol ~ (pronounced "tilda") rather than by R as heretofore. With this change in notation, the conditions of the aboye definition may be recast in a more familiar form: ,)~7

286

FIRST COURSE IN RINGS AND IDEALS

19. Given a right R-module M, set M* = homR(M, R). Prove the statements below: a) M* caD. be made into a left R-module (known as the dual module of M) by defining the module product rf as

(rf)(x)

= rf(x)

(rER,x

E

APPENDIX A

M).

b) When R is a division ring, so that M* forms a vector space over R, then [M*:R] = [M:R]. [Hint: Ifx¡, x z, ... , x. is a basis for M, then the n functions j¡t' E M* (i = 1, 2, ... , n) prescribed by

J1 for for i = j -lO i 0/= j

j¡i *(Xj ) -_ bij _

RELATIONS

serv.e as a basis for M*.] c) M and M** are naturally isomorphic as right R-modules, where M** = (M*)* = homR(M~, R). [Hint: For each x E M, defind E M** by x(g) = g(x), g E M*; then M ~!\ll** under the R-isomorphism that sends x to x.] 20. Let M be a right R-module and 1 a right ideal of R. Prove that a) The set MI = {¿ Xir;jXi E M, ri El}, where ¿ is an arbitrary finite sum, constitutes a submodule of M. b) If M and 1 are both simple (as R-modules) with MI 0/= {O}, then M ~ i. [Hint: Since xl = M for some x E M, an R-isomorphism f: 1 -> M can be given by f(a) = xa, where a E l.] 21. Let R be a nil-semisimple right Artinian ringo Verify that, up to R-isomorphism, there exist only a finite number of sinlple right R-modules. [Hint: R is a direct sum R = 1¡.EB lz EB .. · EB l. offinitely many two-sided ideals. If M 0/= {O} is any simple right R-module, use Problem 20(b) to conc1ude that M ~ li for some i.] 22. Prove that the ring R is simple if and only if every simple right R-module is faithful. [Hint: Ir R has non trivial two-sided ideals, it possesses a maxin:ial one 1 by Zorn's Lemma, Let J be any maximal right ideal of R with J ;2 1 and obtain a contradictionby considering the simple right R-module A(RfJ) ;2 l.] 23. Prove'each of the following assertions: a) 1'he radical J(R) = (1 :,R), where the intersection runs over all maximal right ideals 1 of R. b) ~py nonzero ideal of a prinlitive ring of endomorphisms of a commutative group G;j!i also a primitive ring of endomorphisms of G. c) A ring R is primitive if and only if it contains a maximal right ideal 1 such that thequotient ideal (1:,R) = {O}.

n

24. An ideal 1 of the ring R is said to be a primitive ideal if R/l is a primitive ringo Establish that the radical J(R) can be represented as J(R) = P, where the intersection is taken over a1l primitive ideals P of R. [Hint: If P = A(M), where M is a simple R-module, then M is a faithful simple module over the ring R/P.]

n

We herein append a few definitions and general results concerning certain types of relations that can be imposed on a set. For the most part, our attention is confined to two relations of particular utility, namely, equivalence relations ancl order relations. Intuitively, a (binary) relation on a set S pro vides a criterio n such that for each ordered pair (a, b) of elements of S, one can determine whether the statement "a is related to b" is meaningful (in the sen se of being true or false according to the choice of elements a and b). The relation is completely characterized once we know the set of all those pairs for which the first component stands in that relation to the second. This idea can best be formulated in set-theoretic language as Definition A-l. A (binary) relation R in a nonempty set Sjs any subset of the Cartesian product S x S. (' :. If R is a relation, we express the fact that the pair {a, b) e,i{'by saying that. a is related to b with respect to the relation R, and we wrfte'aRb. F or instance, the relation < in R can be represented by aH points iri' the plane lying aboye the diagonalline y = x; it is customary to write 3: < 4, rathet 'l ! than the awkward (3, 4) E <. Our immediate concern is with equivalence relations. In pi~ctice, these arise whenever it is desirable to identify, as a single entity, aH eIements of a set that have sorne preassigned characteristic.

Definition A-2. A relation R in a set S is an equivalenee relation in S provided that it satisfies the three propertit';s, . 1) aRa for aH a E S (reflexive property), 2) if aRb, then bRa (symmetric property), 3) if aRb and bRe, then aRe (transitive property). Equivalence relations are usually denoted by the symbol ~ (pronounced "tilda") rather than by R as heretofore. With this change in notation, the conditions of the aboye definition may be recast in a more familiar form: ,)~7

288

FIRST COqRSE IN RINGS AND IDEALS

for every a, b, C E S, (1) a ~ a, (2) a ~ b implies b ~ a, (3) a ~ b and b ~ C together imply a ~ c. We say that a is equivalent to b if and only if a ~ b holds. In the following set of miscellaneous examples, we leave to the reader the task of verifying that each relation described actually is an equivalence relation. .. '

~xample/A-1.

.

Let S be a nonempty set and define, for a, b E S; a ~. b if and only if a = b; that is, a and b are the same element. (TechnicalIy 'speaking, ~ is the subset {(a, a)la E S} of S x S.) Then ~ satisfies the :requirements of Definition A-2, showing that theequivalence concept is a ;g~neralization of equality. .

::,~xample A-2. In the Cartesian product R # X R #, let (a, b) ~ (c, d) signÚy. ,t:liat a - c and b - d are both integers. A simple calculation reveals that . ",".-;,!, so definee!, is an equivalence relation in R # X R #. g'E'Xample A-3. As another illustration, consider the set L of all lines in a pnme. Then the phrase "is parallel to" is meaningful when applied to the" elements of L and may be used to define a relation in L. If we a'gree that . any line is paralIel to itself, this yields an eq,tlÍvalence relation. Examp]e A-4. Let f: X ..... Y be a given mapping. Take for ~ the relation a ~ b if and only if f(a) = f(b); then ~ is an equivalen ce relation in X, cal1ed the equivalence relation associated with the mapping f. More general1y, if fF is an arbitrary family offunctions from X into y, an equivalence relation . can be introduced in X by interpreting a ~ b to meanf(a) = f(b) for every fE fF. (The underlying feature in the latter case is that any intersection of equivalen ce relations in X is again an equivalence relation, for ~ = nJEff {(a, b)lf(a) = f(b)}.) One is often led to conc1ude, incorrectly, that the reflexive property is redundant in Definition A-2. The argument proceedslike this: if a ~ b, then the symmetric property implies thát b ~ a; from a ~ b and b ~ a, . together with the transitivity of ~,it follows that.a ~ a. Thus, there appears to be no necessity for the reflexive condition at all. The flaw in this reasoning lies, of course, in the fact that, for sorne element a E S, there may not ex.ist any b in S such that a ~ b. As a result, we would not have a ~ a for every member of $, as the reflexive property requires. Any equivalence relation ~ determines a separation of the set S into a collection of subsets of a kind which we now describe. For each a E S, let [a] denote the subset of S consisting of all elements which are equivalent to a: [a] = {bESlb ~ a}. This set [a] is referred to as the equivalence class determined by a. (The reader should realize that, in general, the eq uivalence c1ass [a] is the same

RELATIONS

289

as the c1ass [al] for many elements al E S.) As a notational device, let us henceforth use the symbol SI ~ to represent the set of al1 equivalence c1asses of the relation ~; that is,

S/,..; = {[a]la E S}. Sorne of the basic properties of equivalence dasses are listed in the theorem below. Theorem A-lo Let ~ be ah equivalence relation iti the set S. Then,

1) for each a E S, the c1ass [a]

=1=

0 ;i?, . ;

2) if bE [a], then [a] = [b]; in other words,',ány element of aÍl . ":\;' equivalen ce c1ass determines that c1ass;

3) for all a,

bE S, [a]

= [bJ.if and only if

a~

br"} ,,'~.

4) foralla,bES,either[a] n [b] = 00r[a].;=;,:lb];

5) U [a] = s. :~"~'~'.'. aeS Proof. Clearly, the element a E [a], for a ~ a. Tci:'prove the' second assertion, let b E [a], so that b ~ a. Now, suppóse that x E [a], which mean s x ~ a. Using the symmetric and transitive properties of -, we thus obtain x ~ b, whence x E [b]. Since x is an arbitrary member of [a], this establishes the inc1usion [a] S;; I b]. A similar argument yields the reverse inc1usion and equality folIows. As regards (3), first assume that [a] = [b]; then .aE [a] = [b] and so a ~ b. Conversley, if we let a ~ b, then the element a E [b]; hence, [a] = [b] from (2). To derive (3), suppose that [a] and [b] have an element in common, say, e E [a] n [b]. Statement (2) then informs us that [a] = [e] = [b]. In brief, if [a] n [b] =1= 0, then we must have [a] = [b]. Fin ally, since each c1ass [a] S;; S, the inc1usion u {[a]la E S} S;; Scertainly holds. For the cípposite inc1usion, it is enough to show that each element a E S belongs to sorne equivalence c1ass; but this ís no problem, for a E [a]. As evidenced by Example A-4, any mapping determines an eq uivalence re1ation in its domain. The foIlowing corolIary indica tes that every equivalence relation arises in this manner; that is to say, each eq uivalence relation is the associated equivalen ce relation of sorne function. Corol1ary. Let~· be an equivalence relation in the set S. Then there exists a set T and a mapping g: S ..... T such that a ~ b if and only if

q(a)

=

g(b).

Prooj: Simply take T = SI ~ and g: S ..... T to be the mappíng defined by g(a) = [a]; in other words, send each element of S onto the (necessarily unique) equivalen ce c1ass to which it belongs. By the foregoing theorem, a ~ b if and only if [a] = [b], or equivalently, g(a) = g(b).

288

FIRST COqRSE IN RINGS AND IDEALS

for every a, b, C E S, (1) a ~ a, (2) a ~ b implies b ~ a, (3) a ~ b and b ~ C together imply a ~ c. We say that a is equivalent to b if and only if a ~ b holds. In the following set of miscellaneous examples, we leave to the reader the task of verifying that each relation described actually is an equivalence relation. .. '

~xample/A-1.

.

Let S be a nonempty set and define, for a, b E S; a ~. b if and only if a = b; that is, a and b are the same element. (TechnicalIy 'speaking, ~ is the subset {(a, a)la E S} of S x S.) Then ~ satisfies the :requirements of Definition A-2, showing that theequivalence concept is a ;g~neralization of equality. .

::,~xample A-2. In the Cartesian product R # X R #, let (a, b) ~ (c, d) signÚy. ,t:liat a - c and b - d are both integers. A simple calculation reveals that . ",".-;,!, so definee!, is an equivalence relation in R # X R #. g'E'Xample A-3. As another illustration, consider the set L of all lines in a pnme. Then the phrase "is parallel to" is meaningful when applied to the" elements of L and may be used to define a relation in L. If we a'gree that . any line is paralIel to itself, this yields an eq,tlÍvalence relation. Examp]e A-4. Let f: X ..... Y be a given mapping. Take for ~ the relation a ~ b if and only if f(a) = f(b); then ~ is an equivalen ce relation in X, cal1ed the equivalence relation associated with the mapping f. More general1y, if fF is an arbitrary family offunctions from X into y, an equivalence relation . can be introduced in X by interpreting a ~ b to meanf(a) = f(b) for every fE fF. (The underlying feature in the latter case is that any intersection of equivalen ce relations in X is again an equivalence relation, for ~ = nJEff {(a, b)lf(a) = f(b)}.) One is often led to conc1ude, incorrectly, that the reflexive property is redundant in Definition A-2. The argument proceedslike this: if a ~ b, then the symmetric property implies thát b ~ a; from a ~ b and b ~ a, . together with the transitivity of ~,it follows that.a ~ a. Thus, there appears to be no necessity for the reflexive condition at all. The flaw in this reasoning lies, of course, in the fact that, for sorne element a E S, there may not ex.ist any b in S such that a ~ b. As a result, we would not have a ~ a for every member of $, as the reflexive property requires. Any equivalence relation ~ determines a separation of the set S into a collection of subsets of a kind which we now describe. For each a E S, let [a] denote the subset of S consisting of all elements which are equivalent to a: [a] = {bESlb ~ a}. This set [a] is referred to as the equivalence class determined by a. (The reader should realize that, in general, the eq uivalence c1ass [a] is the same

RELATIONS

289

as the c1ass [al] for many elements al E S.) As a notational device, let us henceforth use the symbol SI ~ to represent the set of al1 equivalence c1asses of the relation ~; that is,

S/,..; = {[a]la E S}. Sorne of the basic properties of equivalence dasses are listed in the theorem below. Theorem A-lo Let ~ be ah equivalence relation iti the set S. Then,

1) for each a E S, the c1ass [a]

=1=

0 ;i?, . ;

2) if bE [a], then [a] = [b]; in other words,',ány element of aÍl . ":\;' equivalen ce c1ass determines that c1ass;

3) for all a,

bE S, [a]

= [bJ.if and only if

a~

br"} ,,'~.

4) foralla,bES,either[a] n [b] = 00r[a].;=;,:lb];

5) U [a] = s. :~"~'~'.'. aeS Proof. Clearly, the element a E [a], for a ~ a. Tci:'prove the' second assertion, let b E [a], so that b ~ a. Now, suppóse that x E [a], which mean s x ~ a. Using the symmetric and transitive properties of -, we thus obtain x ~ b, whence x E [b]. Since x is an arbitrary member of [a], this establishes the inc1usion [a] S;; I b]. A similar argument yields the reverse inc1usion and equality folIows. As regards (3), first assume that [a] = [b]; then .aE [a] = [b] and so a ~ b. Conversley, if we let a ~ b, then the element a E [b]; hence, [a] = [b] from (2). To derive (3), suppose that [a] and [b] have an element in common, say, e E [a] n [b]. Statement (2) then informs us that [a] = [e] = [b]. In brief, if [a] n [b] =1= 0, then we must have [a] = [b]. Fin ally, since each c1ass [a] S;; S, the inc1usion u {[a]la E S} S;; Scertainly holds. For the cípposite inc1usion, it is enough to show that each element a E S belongs to sorne equivalence c1ass; but this ís no problem, for a E [a]. As evidenced by Example A-4, any mapping determines an eq uivalence re1ation in its domain. The foIlowing corolIary indica tes that every equivalence relation arises in this manner; that is to say, each eq uivalence relation is the associated equivalen ce relation of sorne function. Corol1ary. Let~· be an equivalence relation in the set S. Then there exists a set T and a mapping g: S ..... T such that a ~ b if and only if

q(a)

=

g(b).

Prooj: Simply take T = SI ~ and g: S ..... T to be the mappíng defined by g(a) = [a]; in other words, send each element of S onto the (necessarily unique) equivalen ce c1ass to which it belongs. By the foregoing theorem, a ~ b if and only if [a] = [b], or equivalently, g(a) = g(b).

290

FIRST COURSE IN RINGS AND IDEALS RELATIONS

Ex~mple Á-5. This example is given to illustrate that any mapping can be wntten as the composition of a one-to-one function and an onto function. Letf; X -+ Ybe an arbitrary mapping and consider the equivalence relation ~ associated with! If the element a E X, then we have

[a]

=

{bEXlf(b) =f(a)} =f-l(J(a)).

!neffect" the equivalence c1asses for the relation ~ are just the inverse lmages.off-1(y), where y Ef(X) 5; y. Now, ~efine the f~nctionl: X/~_ -+ Yby the ruleJ([a]) = f(a). Since [a] = [b]lfand only lfj(a) = f(b),fis well-defined. Observe that whereas the original ~unction L may no.! have been one-to-one, 1 happens to be one-to-one; mdeed, f([a]) = f([b]) implies that f(a) = f(b), wherice [a].= [b]. At this point, we intro~uce the ~nto function g;)( -+ X/~ by sett!ng g(a) = [a]. Tht(nf(a) =J([a]) = f(g(a)) = (Jo g)(a) for all a in X, m consequence of which f = f o g. This achieves our stated aim. We next connect the notion of an equivalence relation in S with that . of a partition of S. Definition Á-3. By a partition ofthe set S is meant a family of S with the properties 1) 0 ~ [l},

2) for any A, B 3) u [l} = S.

E [l},

either A = B or A

11

B

=0

{!J

of subsets

(pairwise disjoint),

Expressed otherwise, a partition of S)s a collection {!J of nonempty subsets of S such th~t every element of S be~0rt&s to one and ónly one member of {!J. The set Z of mtegers, for instance, c::t.n/be partitioned into the subsets of od~ ~nd. even integers; another partiti~~ ;~f Z might consist of the sets of poslÍlve mtegers, negative integers and {Q}~ Theorem A-1 may be viewed as assertirig that each equivalence relation ~ ~n a set S yields a partition of S, na.D:l:ely, the partition SI ~ into the eqUlvalen~e c1asses fo!~. (In this connection, notice that, for the equivalence relatlOn ofequahty, the corresponding.~lasses contain only one element each ~ hen.ce, the resulting partition is the finest possible.) We now reverse the S.ltu~tlOn and show that a given partition of S induces an equivalence relatlOn m S. But first a preliminary lemma is required. Lemma. Two equivalence relations ~ and ~ in the set S are the same if and only if SI ~ and SI ~, are the same.

291

but not under the other; say a ~ b, but not a ~' b. By Theorem A-1, there is an equivalence c1ass in SI ~ containing both a and b, while no such c1ass appeats in SI ~'. Accordingly, SI ~ and SI ~., differ. Theorem Á-2. If [l} is a paitition of the set ·S, then there is a unique equivalence relation in S whose equivalel}ce c1asses are precisely the members of [l}. Proo! Given a, b E S, we write a ~ b if and only if a and b both belong to the same subset in [l}, (The fact that [l} partitions S guarantees that each element of S lies in exactly one member of [l}.) The reader may easily check that the relation ~, defined in this way, is indeed an equivalence relation in S. Let us prove that the partition [l} has the form SI ~. If the subset P E [l}, then a E P for .some a inS. Now, the element bE P if and only if b ~ a, or, what amounts to the same thing, if and on!:y if b E [a]. This demonstrates the equality P = [a] E SI ~. Since this holds for each P in [l}, it follows that [l} 5; Sl~. On the other hand, let fa] be an arbitrary equivalence class and P be the partition set in !?J> to \vhich the element a belongs. By similar reasoning, we conc1ude that [a] = P; hence, S/~ 5; [l}. Thus, the set of equivalence c1asses for ~ coincides with the partition {!J. The uniqueness assertion is an immediate consequence of the lernma. To surnmarize, there is a natural one-to-one correspondence between the equivalence relations in a set and the partitions of that set; every equivalence relation gives rise to a partition and vice versa. We have a single idea, which h~~ been considered from two different points of view. Another type of'relation which occurs in various branches of mathematics is the so-called partial order relation. Just as equivalence generalizes equality, this relation'(as we define it below) generalizes the idea of "Iess than or equal to" on the realline. Definition Á-4. 'A. relation R in a nonempty set S is called a partial order in S if the (ollowing three conditions are satisfied; 1) aRa (reflexiveproperty), 2) if aRb and bRa, then a = b (antisymmetric property), 3) if aRb and bRe, then aRe (transitive property), where a, b, e denote arbitrary elements of S.

I

Proo! If ~ and ~' are the same, then surely SI ~ = SI ~'. So, suppose that ~ and ~' are distinct equivalence relations in S. Then there exists a pair of elements a, b E S which are equivalent under .qne of the relations,

From now on, we shall follow custom and adopt the symbol ::;; to represent a partial order relation, writing a ::;; b in place of aRb; the foregoing axioms then read; (1) a ::;; a, (2) if a ::;; b and b ::;; a, then a = b, and {3) if a ::;; b and b ::;; e, then a ::;; e. As a linguistic convention, let us also agree to say (depending on the circumstance) that "a precedes b" or

290

FIRST COURSE IN RINGS AND IDEALS RELATIONS

Ex~mple Á-5. This example is given to illustrate that any mapping can be wntten as the composition of a one-to-one function and an onto function. Letf; X -+ Ybe an arbitrary mapping and consider the equivalence relation ~ associated with! If the element a E X, then we have

[a]

=

{bEXlf(b) =f(a)} =f-l(J(a)).

!neffect" the equivalence c1asses for the relation ~ are just the inverse lmages.off-1(y), where y Ef(X) 5; y. Now, ~efine the f~nctionl: X/~_ -+ Yby the ruleJ([a]) = f(a). Since [a] = [b]lfand only lfj(a) = f(b),fis well-defined. Observe that whereas the original ~unction L may no.! have been one-to-one, 1 happens to be one-to-one; mdeed, f([a]) = f([b]) implies that f(a) = f(b), wherice [a].= [b]. At this point, we intro~uce the ~nto function g;)( -+ X/~ by sett!ng g(a) = [a]. Tht(nf(a) =J([a]) = f(g(a)) = (Jo g)(a) for all a in X, m consequence of which f = f o g. This achieves our stated aim. We next connect the notion of an equivalence relation in S with that . of a partition of S. Definition Á-3. By a partition ofthe set S is meant a family of S with the properties 1) 0 ~ [l},

2) for any A, B 3) u [l} = S.

E [l},

either A = B or A

11

B

=0

{!J

of subsets

(pairwise disjoint),

Expressed otherwise, a partition of S)s a collection {!J of nonempty subsets of S such th~t every element of S be~0rt&s to one and ónly one member of {!J. The set Z of mtegers, for instance, c::t.n/be partitioned into the subsets of od~ ~nd. even integers; another partiti~~ ;~f Z might consist of the sets of poslÍlve mtegers, negative integers and {Q}~ Theorem A-1 may be viewed as assertirig that each equivalence relation ~ ~n a set S yields a partition of S, na.D:l:ely, the partition SI ~ into the eqUlvalen~e c1asses fo!~. (In this connection, notice that, for the equivalence relatlOn ofequahty, the corresponding.~lasses contain only one element each ~ hen.ce, the resulting partition is the finest possible.) We now reverse the S.ltu~tlOn and show that a given partition of S induces an equivalence relatlOn m S. But first a preliminary lemma is required. Lemma. Two equivalence relations ~ and ~ in the set S are the same if and only if SI ~ and SI ~, are the same.

291

but not under the other; say a ~ b, but not a ~' b. By Theorem A-1, there is an equivalence c1ass in SI ~ containing both a and b, while no such c1ass appeats in SI ~'. Accordingly, SI ~ and SI ~., differ. Theorem Á-2. If [l} is a paitition of the set ·S, then there is a unique equivalence relation in S whose equivalel}ce c1asses are precisely the members of [l}. Proo! Given a, b E S, we write a ~ b if and only if a and b both belong to the same subset in [l}, (The fact that [l} partitions S guarantees that each element of S lies in exactly one member of [l}.) The reader may easily check that the relation ~, defined in this way, is indeed an equivalence relation in S. Let us prove that the partition [l} has the form SI ~. If the subset P E [l}, then a E P for .some a inS. Now, the element bE P if and only if b ~ a, or, what amounts to the same thing, if and on!:y if b E [a]. This demonstrates the equality P = [a] E SI ~. Since this holds for each P in [l}, it follows that [l} 5; Sl~. On the other hand, let fa] be an arbitrary equivalence class and P be the partition set in !?J> to \vhich the element a belongs. By similar reasoning, we conc1ude that [a] = P; hence, S/~ 5; [l}. Thus, the set of equivalence c1asses for ~ coincides with the partition {!J. The uniqueness assertion is an immediate consequence of the lernma. To surnmarize, there is a natural one-to-one correspondence between the equivalence relations in a set and the partitions of that set; every equivalence relation gives rise to a partition and vice versa. We have a single idea, which h~~ been considered from two different points of view. Another type of'relation which occurs in various branches of mathematics is the so-called partial order relation. Just as equivalence generalizes equality, this relation'(as we define it below) generalizes the idea of "Iess than or equal to" on the realline. Definition Á-4. 'A. relation R in a nonempty set S is called a partial order in S if the (ollowing three conditions are satisfied; 1) aRa (reflexiveproperty), 2) if aRb and bRa, then a = b (antisymmetric property), 3) if aRb and bRe, then aRe (transitive property), where a, b, e denote arbitrary elements of S.

I

Proo! If ~ and ~' are the same, then surely SI ~ = SI ~'. So, suppose that ~ and ~' are distinct equivalence relations in S. Then there exists a pair of elements a, b E S which are equivalent under .qne of the relations,

From now on, we shall follow custom and adopt the symbol ::;; to represent a partial order relation, writing a ::;; b in place of aRb; the foregoing axioms then read; (1) a ::;; a, (2) if a ::;; b and b ::;; a, then a = b, and {3) if a ::;; b and b ::;; e, then a ::;; e. As a linguistic convention, let us also agree to say (depending on the circumstance) that "a precedes b" or

292

FIRST COURSE IN RINGS AND IDEALS

"a is a predecessor of b", or "b succeeds a", or "b is a successor of a" if a ::5;; b and a' =F b. By a partially ordered set is meant a pair (S, ::5;;) consisting of a set S and a partial order relation ::5;; in S. In practice, one tends to ignore the second component and simply speak of the partially ordered set S, or, when moreprecision is required, say that S is partially ordered by ::5;;. If A is a subset of a partially ordered set S, then the ordering of S restricted to A is a partial ordering of A, called the induced partial order; in this se/nse, any subset ofa partially ordered set beco mes a partially ordered set in lts own right. When consiqering subsets of a partially ordered set as partially ordered sets, it is always th~ induced order that we have in mind. Let S be a set partially ordered by the relation ::5;;. Two elements a and b of S such that either a ::5;; for b ::5;; a are said to be comparable . . In view of the refiexivity of a partiai.·order, each element of S is comparable to itself. There is nothing, howev'er, in Definition A-3 that .en sures the comparability of every'two elemen.ts of S. Indeed, the qualifying adverb "partially" in the phrase "partiallY'ordered set" is intended to emphasize that there may exist pairs of eleniénts in the set which are not comparable. Definition A-S. A partial order ::5;; in a set S is termed total (sometimes simple, or linear) if any two elements of S are comparable; that is, a ::5;; b or b ::5;; a for any two elements a and b of S. A partially ordered set (S,::5;; ) whose reiation ::::; constitutes a total order in S is called a totally ordered set or, for short, a chain. Let us pause to illustrate sorne of the preceding remarks. Example A-6. In the set R# of real numbers, the relation ::5;; (taken with the usual meaning) is the most natural example of a total brdering. Example A-7. Given the set Z+ of positive integers, define a ::5;; b if and only if a divides b. This affords a partial ordering of Z +, which is not total; for instance, the integers 4 and 6 are not comparable, since neither divides the other. Example A-S. Let S be the collection of all real-valued functions defined on a nonempty set X. If f::5;; 9 is interpreted to mean f(x) ::5;; g(x) for all x E X, then ::5;; partially, but not totally, orders S. Example A-9. For a final illustration, consider the set P(X) of all subsets of a set X. The relation A ::5;; B if and only if A ~ B is a partial ordering of P(X), but not a total ordering provided that X has at least two elements. For example, if X = {1,2, 3}, and A = {l,2}, B = {2,3}, then neither A ::5;; B nor B :$; A holds. As regards terminology, any family of sets ordered in this manner will be said to be ordered by inc/usion. Let (A, :$; ) and (B, ::5;;) be two partially ordered sets (when there is no danger of confusion, we write :$; for the partial orders in both A and B).

RELATIONS

293

A mappingf: A - Bis said to be order-preservingor an or~er-homomorphism if for all a, b E A, a ::5;; b implies f(a) ::5;; f(b) in B. A one-to-one orderhomomorphism f of the set A onto B whose inverse is als.o an order-homomorphism (from B onto A) is ·an ordá-isomorphism. If such 'a function exists, we say that the two partially ordered sets (A,::5;; ) and (B,::5;;) are order-isomorphic. When th~ partial order is the primary object of interest and the naturé of the elements plays no essential role, order-isomorphic sets can be regarded as identical. The corrting theorem emphasizes the fundamental role of our last example on ¡{~rtially ordered sets (Example A-9), for it allows us to represent any partially'prdered set by a family of sets. ";/::.:'

Theoreg):A-3. Let A be a,' set partially ordered by the relation ::5;; • Then A,is'prder-isomorphic to a family ofsubsets of A, partially ordered . by inclusibn.

Proof. Fo?¡~~~h a E A, let la = {x EAlx ;; a}. It is not hard to verify that the mapping f: A - P(A) defined by f(a) = la is an order-homomorphism of A into P(A). Indeed, if a ::5;; b, then the condition x ::5;; a implies x ::5;; b and therefore la ~ lb' or, equivalently,f(a) ~ f(b). To see thatfis one-to-one, suppose a, b E A are such thatf(a) = f(b). Then the element a E la = lb' and, hence, a::5;; b by definition ofIb ; likewise, b ::5;; a, from which it follows that a = b. Finally, the inverse f-1 is also orderpre~rving. For, if the inc1usion la ~ lb holds, then a E lb and so a ::5;; b. These calculations make it clear that A is order-isomorphic to a certain set of subsets of P(A). Corollary. For no set A is A order-isomorphic to P(A).

Proof. We argue that ifj: A - P(A) is any order-homomorphism from A -into P(A); then f cannot map onto P(A). For purposes of contradiction; assume that f does carry A onto P(A). Define B = {a E Ala fjf(a)} and B* = {cEAlc ::5;;aforsomeaEB} . . By supposition, the set B* = f(b) for someelement bE A. If b fj B*, then, according to the definition of B, b E B ~ B*, a contradiction. Hence, b E B* and so b ::5;; a for sorne a in B. From the order-preserving character of f, B* = f(b) ~ f(a) . . But then, a E B' ~ B* ~ f(a). The implication is that a fj B, which is again a contradiction. In an ordered set, there are sometimes elements with special properties that are worth mentioning. Definition A-6. Let S. be a set partially ordered by the relation ::5;;. An element x E S is said to be a minimal (maximal) element of S if a E S and a ::5;; x (x ::5;; a) imply a = x.

292

FIRST COURSE IN RINGS AND IDEALS

"a is a predecessor of b", or "b succeeds a", or "b is a successor of a" if a ::5;; b and a' =F b. By a partially ordered set is meant a pair (S, ::5;;) consisting of a set S and a partial order relation ::5;; in S. In practice, one tends to ignore the second component and simply speak of the partially ordered set S, or, when moreprecision is required, say that S is partially ordered by ::5;;. If A is a subset of a partially ordered set S, then the ordering of S restricted to A is a partial ordering of A, called the induced partial order; in this se/nse, any subset ofa partially ordered set beco mes a partially ordered set in lts own right. When consiqering subsets of a partially ordered set as partially ordered sets, it is always th~ induced order that we have in mind. Let S be a set partially ordered by the relation ::5;;. Two elements a and b of S such that either a ::5;; for b ::5;; a are said to be comparable . . In view of the refiexivity of a partiai.·order, each element of S is comparable to itself. There is nothing, howev'er, in Definition A-3 that .en sures the comparability of every'two elemen.ts of S. Indeed, the qualifying adverb "partially" in the phrase "partiallY'ordered set" is intended to emphasize that there may exist pairs of eleniénts in the set which are not comparable. Definition A-S. A partial order ::5;; in a set S is termed total (sometimes simple, or linear) if any two elements of S are comparable; that is, a ::5;; b or b ::5;; a for any two elements a and b of S. A partially ordered set (S,::5;; ) whose reiation ::::; constitutes a total order in S is called a totally ordered set or, for short, a chain. Let us pause to illustrate sorne of the preceding remarks. Example A-6. In the set R# of real numbers, the relation ::5;; (taken with the usual meaning) is the most natural example of a total brdering. Example A-7. Given the set Z+ of positive integers, define a ::5;; b if and only if a divides b. This affords a partial ordering of Z +, which is not total; for instance, the integers 4 and 6 are not comparable, since neither divides the other. Example A-S. Let S be the collection of all real-valued functions defined on a nonempty set X. If f::5;; 9 is interpreted to mean f(x) ::5;; g(x) for all x E X, then ::5;; partially, but not totally, orders S. Example A-9. For a final illustration, consider the set P(X) of all subsets of a set X. The relation A ::5;; B if and only if A ~ B is a partial ordering of P(X), but not a total ordering provided that X has at least two elements. For example, if X = {1,2, 3}, and A = {l,2}, B = {2,3}, then neither A ::5;; B nor B :$; A holds. As regards terminology, any family of sets ordered in this manner will be said to be ordered by inc/usion. Let (A, :$; ) and (B, ::5;;) be two partially ordered sets (when there is no danger of confusion, we write :$; for the partial orders in both A and B).

RELATIONS

293

A mappingf: A - Bis said to be order-preservingor an or~er-homomorphism if for all a, b E A, a ::5;; b implies f(a) ::5;; f(b) in B. A one-to-one orderhomomorphism f of the set A onto B whose inverse is als.o an order-homomorphism (from B onto A) is ·an ordá-isomorphism. If such 'a function exists, we say that the two partially ordered sets (A,::5;; ) and (B,::5;;) are order-isomorphic. When th~ partial order is the primary object of interest and the naturé of the elements plays no essential role, order-isomorphic sets can be regarded as identical. The corrting theorem emphasizes the fundamental role of our last example on ¡{~rtially ordered sets (Example A-9), for it allows us to represent any partially'prdered set by a family of sets. ";/::.:'

Theoreg):A-3. Let A be a,' set partially ordered by the relation ::5;; • Then A,is'prder-isomorphic to a family ofsubsets of A, partially ordered . by inclusibn.

Proof. Fo?¡~~~h a E A, let la = {x EAlx ;; a}. It is not hard to verify that the mapping f: A - P(A) defined by f(a) = la is an order-homomorphism of A into P(A). Indeed, if a ::5;; b, then the condition x ::5;; a implies x ::5;; b and therefore la ~ lb' or, equivalently,f(a) ~ f(b). To see thatfis one-to-one, suppose a, b E A are such thatf(a) = f(b). Then the element a E la = lb' and, hence, a::5;; b by definition ofIb ; likewise, b ::5;; a, from which it follows that a = b. Finally, the inverse f-1 is also orderpre~rving. For, if the inc1usion la ~ lb holds, then a E lb and so a ::5;; b. These calculations make it clear that A is order-isomorphic to a certain set of subsets of P(A). Corollary. For no set A is A order-isomorphic to P(A).

Proof. We argue that ifj: A - P(A) is any order-homomorphism from A -into P(A); then f cannot map onto P(A). For purposes of contradiction; assume that f does carry A onto P(A). Define B = {a E Ala fjf(a)} and B* = {cEAlc ::5;;aforsomeaEB} . . By supposition, the set B* = f(b) for someelement bE A. If b fj B*, then, according to the definition of B, b E B ~ B*, a contradiction. Hence, b E B* and so b ::5;; a for sorne a in B. From the order-preserving character of f, B* = f(b) ~ f(a) . . But then, a E B' ~ B* ~ f(a). The implication is that a fj B, which is again a contradiction. In an ordered set, there are sometimes elements with special properties that are worth mentioning. Definition A-6. Let S. be a set partially ordered by the relation ::5;;. An element x E S is said to be a minimal (maximal) element of S if a E S and a ::5;; x (x ::5;; a) imply a = x.

294

FIRST COURSE IN RINGS AND IDEALS

In other words, x is a mínimal (maximal) element of S if no element of S precedes (exceeds) x. It is not always the case that a partially ordered set possesses a mínimal (maximal) element and, when such an element exists, there is no guarantee that it will be unique. Example A-lO. The simplest illustration of a partially ordered set without minimal or maximal elements is fumished by the set R# with the ordering ::;; in thé usual sen se. Example A-U. In the collection P(X) -' {0} of all nonempty subsets of a nonempty set S {partially ordered by set-theoretic inclusion), the minimal elements are those subsets consisting of a single element.

RELATlONS

We emphasize that a lower (upper) bound for a subset A of a partially ordered set is not required to belong to A itself. If A happens to have a first (last) element, then the same element is a lower (upper) bound for ~; conversely, if a lower (upper) bound for A is contained in the set A, then 1t serves as the first (last) element for A. Notice, too, that a lower (upper) bound for A is a lower (upper) bound for any subset of A. A subset of a partially ordered set need not have upper or lower bounds (just consider Z S; R # with respect to ::;; ) or it may have many. For an example of this latter situation, one may turn to the family P(X) of all subsets of a set X, with the order being given by the inclusion relation; an upper bound for a subfamíly d ~ P(X) is any set containing u d, while a lower bound is any set contained in n d.

Example A-12. Consi.der the set S of all integers greater than 1 and the partial order ::;; defined, by a ::;; b if and only if a divides b. In this setting, the prime numbers serve as minimal elements. It is technically convenient to distinguish between the notion of a mínimal (maximal) element and that of a first (last) element.

Definition A-7. Let S be a set partially ordered by the relation ::;;. An element x E S is called the first (last) element of S if x ::;; a (a ::;; x) for all a ES. Let us point out immediately the important distinction between first (last) elements and mínimal (maximal) elements. Definition A-7 asserts that the first (last) element of a partially ordered set S must be comparable to every element of S. On the other hand, as Definition A-6 implies, it is not ' required that a mínimal (maximal) element be comparable to every element . of'S, only that there be no element in S which precedes (exceeds) it. A, lninimal (maximal) element has no predecessors (successors), whereas a first ' , (iá:st) element precedes (succeeds) every element. Clearly, any first (last) " element is a mínimal (maximal) element, but not conversely. , First (last) elements of partially ordered sets are unique, if they exist at all. Indeed, suppose that the partially ordered set (S, ::;; ) has two first " elements, say x and y; then, x ::;; y and y ::;; x, so that x = y by the antisymmetric property. Thus, x is unique and we are justified in using the definite article when referring to the first element of S. A similar argument holds for last elements. ' Let us introduce some additional terminology pertaining to partially ordered sets. Definition A-8. Let S be a set partially ordered by the relation ::;; and let A be a subset of S. An element x E S is said to be a lower (upper) bound for A if x ::;; a (a ::;; x) for all a E A.

295

, ~" :

294

FIRST COURSE IN RINGS AND IDEALS

In other words, x is a mínimal (maximal) element of S if no element of S precedes (exceeds) x. It is not always the case that a partially ordered set possesses a mínimal (maximal) element and, when such an element exists, there is no guarantee that it will be unique. Example A-lO. The simplest illustration of a partially ordered set without minimal or maximal elements is fumished by the set R# with the ordering ::;; in thé usual sen se. Example A-U. In the collection P(X) -' {0} of all nonempty subsets of a nonempty set S {partially ordered by set-theoretic inclusion), the minimal elements are those subsets consisting of a single element.

RELATlONS

We emphasize that a lower (upper) bound for a subset A of a partially ordered set is not required to belong to A itself. If A happens to have a first (last) element, then the same element is a lower (upper) bound for ~; conversely, if a lower (upper) bound for A is contained in the set A, then 1t serves as the first (last) element for A. Notice, too, that a lower (upper) bound for A is a lower (upper) bound for any subset of A. A subset of a partially ordered set need not have upper or lower bounds (just consider Z S; R # with respect to ::;; ) or it may have many. For an example of this latter situation, one may turn to the family P(X) of all subsets of a set X, with the order being given by the inclusion relation; an upper bound for a subfamíly d ~ P(X) is any set containing u d, while a lower bound is any set contained in n d.

Example A-12. Consi.der the set S of all integers greater than 1 and the partial order ::;; defined, by a ::;; b if and only if a divides b. In this setting, the prime numbers serve as minimal elements. It is technically convenient to distinguish between the notion of a mínimal (maximal) element and that of a first (last) element.

Definition A-7. Let S be a set partially ordered by the relation ::;;. An element x E S is called the first (last) element of S if x ::;; a (a ::;; x) for all a ES. Let us point out immediately the important distinction between first (last) elements and mínimal (maximal) elements. Definition A-7 asserts that the first (last) element of a partially ordered set S must be comparable to every element of S. On the other hand, as Definition A-6 implies, it is not ' required that a mínimal (maximal) element be comparable to every element . of'S, only that there be no element in S which precedes (exceeds) it. A, lninimal (maximal) element has no predecessors (successors), whereas a first ' , (iá:st) element precedes (succeeds) every element. Clearly, any first (last) " element is a mínimal (maximal) element, but not conversely. , First (last) elements of partially ordered sets are unique, if they exist at all. Indeed, suppose that the partially ordered set (S, ::;; ) has two first " elements, say x and y; then, x ::;; y and y ::;; x, so that x = y by the antisymmetric property. Thus, x is unique and we are justified in using the definite article when referring to the first element of S. A similar argument holds for last elements. ' Let us introduce some additional terminology pertaining to partially ordered sets. Definition A-8. Let S be a set partially ordered by the relation ::;; and let A be a subset of S. An element x E S is said to be a lower (upper) bound for A if x ::;; a (a ::;; x) for all a E A.

295

, ~" :

ZORN'S LEMMA

APPENDIX B

297

A fundamental axiom of set theory, which has a surprising variety of 10gicalIy equivalent formulations, is the so-caIled Well-Ordering Theorem of Zermelo (1904). The designation "theorem" notwithstanding, we take this to be an axiom (assumed and un pro ven) of our system. We ,state: Zermelo's Theorem. Any set S can be well-ordered; that ¡s, there is a partial ordering ~ for S such that (S, ~) is a well-~rdered set. Accepting th~ existence of such orderings, we do not pretend at all to be able to specify them. Indeed, nobody has ever "cons,tructed" an explicit function that well-orders an uncountable set. More'over, the promised well-ordering may bear no relation to any other order1:dg that the given set may alrcady possess; the well-ordering of R#, for instahce, cannot coincide with its customary ordering. . ..; , ( Zermelo based the "proof" of his classical Well-0idering Theorem on a seemingly innocent property whose validity had iÍ,f?ver been questioned and which has since become known as the axiom oFéMice. To state this axiom, we first need the delliliÍ:ionof a choice function".\¡(·

ZORN'S LEMMA:t:: '/

~'

"

;'::\1n this Appendix, we give a brief áccount of sorne of the axioms of set theory¡!l"i, ,}with the primary purpose of introducing Zom's Lemma. Our presentatioQ::í l descriptive and most of the facts are merely stated. The reader who is not content with tbis bird's-eye view should consult [12J for the details. As we know, a: given partially ordered set need not have a first element and, ifit does, sorne subset could very well faíl to possess one. This prompts the foIlowing definition: ti partially ordered set (S, ~) is said to be wellordered if every nonempty subset A f; S has afirst element ("with respect to ~"being understood). The set Z + is well-ordered by the usual ~ ; each nonempty subset has a first element, iJamely, the integer of smallest magnitude in the set. Notice that any well-ordered set (S, ~ ) is in fact totally ordered. For, each subset {a, b} f; S must have a first element. According as the first element is a or b, we see that a ~ b or b ~ a, whence the two elements a and b are comparable. Going in the other direction, any total ordering of a finite set is a weIl-ordering of that set. Let it also be remarked that a subset of a well-ordered set is again well-ordered (by the restriction of the ordering).

ts

We parExample B-l. Consider the Cartesian product S = Z+ x tialIy order S as follows: if (a, b) and (a', b1 are ordered pairs of po~itive integers, (a, b) ~ (a', b' ) means that (1) a < a' (in the usual sense) or (2) a a' and b ~ b' , (This is called the lexicographíc order of Z + x Z +, because of its resemblance to the way words are arranged in a dictionary.) For instance, (4, 8) ~ (5,2), while (3, 5) ~ (3,9). To confirm that ~ is a well-ordering of S, let 0 =1= A f; S and define B = {aEZ+ICa,b)EA}. Since A is a nonempty subset of Z +, it has a first element, call it ao. Now, let e == {b E Z + I(a o, b) e A}. Again, the well-ordering of Z + under ~ guarantees that e has a first eIement, say bo' We Ieave it to the reader to convince himselfthat the pair (a o, boj serves as the first element of A, thereby making S a well-ordered set relative to ~. 296

Definition B-l. Let ce be a (nonempty) colIection of nonempty sets. A function f: ce -+ u ce is called a choice function for ce if feA) E A for every set A in ~. Informally, a choice function f can be thought of as "selecting" from each set A E ce a certain representative elementf(A) of that set. As a simple illustration, there are two distinct choice functions f1 and f2 for the family of nonempty supsets of {1, 2} :

=

fl({1,2})

1,

fl({l})

f2({1,2})

2,

f2({1}) = 1,

1,

fl({2})

=

2,

f2({2}) = 2.

The question arises whether tms selecHon"process can actually be carried out when ce has infinítely many members. The possibUity of making such choices is handled by the axiom mentioned above: Axiom of Choice. Every collection ce ofnonempty sets has at least one choice function,

l'

~~;

Since this general principie of choice has a way of slipping into proofs unnoticed, the reader should become familiar with its disguised forms. For instance, one often encounters the following wording: if {X¡} is a family of nonempty sets indexed by the nonempty set .1, then the Cartesian product X i•.1 Xi is nonempty (it should be clcar that the e1ements of X XI are precisely the choice functions for {Xi})' For another common phrasing, which again expresses the idea of se1ection, let ce be a collectionof disjoint, nonempty sets. The axiom of choice, as we have stated it, is equivalent to

ZORN'S LEMMA

APPENDIX B

297

A fundamental axiom of set theory, which has a surprising variety of 10gicalIy equivalent formulations, is the so-caIled Well-Ordering Theorem of Zermelo (1904). The designation "theorem" notwithstanding, we take this to be an axiom (assumed and un pro ven) of our system. We ,state: Zermelo's Theorem. Any set S can be well-ordered; that ¡s, there is a partial ordering ~ for S such that (S, ~) is a well-~rdered set. Accepting th~ existence of such orderings, we do not pretend at all to be able to specify them. Indeed, nobody has ever "cons,tructed" an explicit function that well-orders an uncountable set. More'over, the promised well-ordering may bear no relation to any other order1:dg that the given set may alrcady possess; the well-ordering of R#, for instahce, cannot coincide with its customary ordering. . ..; , ( Zermelo based the "proof" of his classical Well-0idering Theorem on a seemingly innocent property whose validity had iÍ,f?ver been questioned and which has since become known as the axiom oFéMice. To state this axiom, we first need the delliliÍ:ionof a choice function".\¡(·

ZORN'S LEMMA:t:: '/

~'

"

;'::\1n this Appendix, we give a brief áccount of sorne of the axioms of set theory¡!l"i, ,}with the primary purpose of introducing Zom's Lemma. Our presentatioQ::í l descriptive and most of the facts are merely stated. The reader who is not content with tbis bird's-eye view should consult [12J for the details. As we know, a: given partially ordered set need not have a first element and, ifit does, sorne subset could very well faíl to possess one. This prompts the foIlowing definition: ti partially ordered set (S, ~) is said to be wellordered if every nonempty subset A f; S has afirst element ("with respect to ~"being understood). The set Z + is well-ordered by the usual ~ ; each nonempty subset has a first element, iJamely, the integer of smallest magnitude in the set. Notice that any well-ordered set (S, ~ ) is in fact totally ordered. For, each subset {a, b} f; S must have a first element. According as the first element is a or b, we see that a ~ b or b ~ a, whence the two elements a and b are comparable. Going in the other direction, any total ordering of a finite set is a weIl-ordering of that set. Let it also be remarked that a subset of a well-ordered set is again well-ordered (by the restriction of the ordering).

ts

We parExample B-l. Consider the Cartesian product S = Z+ x tialIy order S as follows: if (a, b) and (a', b1 are ordered pairs of po~itive integers, (a, b) ~ (a', b' ) means that (1) a < a' (in the usual sense) or (2) a a' and b ~ b' , (This is called the lexicographíc order of Z + x Z +, because of its resemblance to the way words are arranged in a dictionary.) For instance, (4, 8) ~ (5,2), while (3, 5) ~ (3,9). To confirm that ~ is a well-ordering of S, let 0 =1= A f; S and define B = {aEZ+ICa,b)EA}. Since A is a nonempty subset of Z +, it has a first element, call it ao. Now, let e == {b E Z + I(a o, b) e A}. Again, the well-ordering of Z + under ~ guarantees that e has a first eIement, say bo' We Ieave it to the reader to convince himselfthat the pair (a o, boj serves as the first element of A, thereby making S a well-ordered set relative to ~. 296

Definition B-l. Let ce be a (nonempty) colIection of nonempty sets. A function f: ce -+ u ce is called a choice function for ce if feA) E A for every set A in ~. Informally, a choice function f can be thought of as "selecting" from each set A E ce a certain representative elementf(A) of that set. As a simple illustration, there are two distinct choice functions f1 and f2 for the family of nonempty supsets of {1, 2} :

=

fl({1,2})

1,

fl({l})

f2({1,2})

2,

f2({1}) = 1,

1,

fl({2})

=

2,

f2({2}) = 2.

The question arises whether tms selecHon"process can actually be carried out when ce has infinítely many members. The possibUity of making such choices is handled by the axiom mentioned above: Axiom of Choice. Every collection ce ofnonempty sets has at least one choice function,

l'

~~;

Since this general principie of choice has a way of slipping into proofs unnoticed, the reader should become familiar with its disguised forms. For instance, one often encounters the following wording: if {X¡} is a family of nonempty sets indexed by the nonempty set .1, then the Cartesian product X i•.1 Xi is nonempty (it should be clcar that the e1ements of X XI are precisely the choice functions for {Xi})' For another common phrasing, which again expresses the idea of se1ection, let ce be a collectionof disjoint, nonempty sets. The axiom of choice, as we have stated it, is equivalent to

298

ZORN'S LEMMA

FIRST COVRSE IN RINGS AND IDEALS

Theorem B-l. Every partially ordered set contains a maximal chain; that is, a chain which is not a proper subset of any other chain.

asserting the existence of a set S with the property that A 11 S contains exactIy one element, for each A in ~. Granting Zermelo's Theorem, it is c1ear that a choice function f can be defined for al1y collection ~ of'nonempty sets: having well-ordered u ~, simply take f to be the function which assigns to each set A 'in ~ its first element. As indicated earlier, Zermelo's Theorem was originally derived from theaxiom of choice, so that these are in reality two equivalent principies (a1though seemingly quite different). A1though the axiom of choice may strike the reader as being intuitively obvious, the soundness of this principIe has aroused more philosophical discussion than any other single question in the foundations ofmathematics. At the heart of the controversy is the ancient problern of existence. Soine mathematicians believ€; that a set exists only if each of its elements can be de~ignated specifically, qr at the very least if there is a rule by which each Of1tS members can be constructed. A more liberal school ofthought is that an axiom about existence of sets may be used if it does not lead to a contradiction. In 1938 Gode1 demonstrated that the axiom of choice is not in contradiction with the other genera:lly accepted axioms of set theory (assuming that the latter are consistent with one another). It was subsequentIy established by Cohen (1963) that the denial of this axiom is also consistent with the rest of set theory. Thus, the axiom of choice is in fact an independent axiom, whose use or rejection is a matter of personal inc1ination. The feeling among most mathematicians today is that the axiom of choice is harmless in principIe and indispensable in practice (provided that one calls attention to the occ;asions of its use). It is also valuable as an heuristic tool, since every proof,by means of this assumption represents a result for which we can then seek ,p'roofs along other lines. A non-constructive criterion f6'V'the existence of maximal e1ements is given by the so-called "maximality principIe", which general1y is cited in the literature under the name Zorn'sLemma. (From the point of view of priority, this principIe goes back t9' Hausdorff and Kuratowski, but Zorn gave a formulation of it which is patticularly suitable to algebra; he was also the first to state, without proof, ili.at a maximality principIe implies the axiom of choice.)

Proa! Consider the collection ~ of all chains of a partially ordered set (S, ~ ); ~ is nonempty, since it contains the chains consisting of single elements of S. Partially order ~ by inc1usion and let d be any chain of ~ (for the ordering s). We maintain that the union u d belongs to C(f. Given elements a, b E U d, we have a E A E d and bE BE d, for sorne A, B. As d is a chain, either A S B or B S A; suppose, for convenience, that A S B. Then a, b both lie in B and, since B is itself a chain (in S), i t follows that a ~ b or b :5: a. In con sequen ce, any two elements of u d are comparable, making u d a chain in S. Since u d is c1early an upper bound for d in ~, Zorn's Lemma implies that (~, s) has a maximal mernber. As another brief application of Zorn's Lemma, consider the following assertion: if (S, :5:) is a partially ordered set every chain of which has an upper bound, then for each a E S there exists a maximal element x E S with the property that a ~ x; in other words, there exists a maximal element !arger than the given element For a proof of this, nrst observe that the set J a = {y E Sla :5: y} satisfies the hypotheses of Zorn's Lemma (under the restriction of :5:); hence, has a maximal e1ement x. But x is maxirnal in S, not merely in J a • For, suppose that s E S with x :5: s. Then a ~ s (since both a ~ x and x :5: s) and so s E J a • From the maximality of x in J a , it then follows that s = x, completing the argument. Needless to say, we could just as well have phrased Zorn's Lemma in terms of lower bounds and minimal elements. The assertion in this case is that there exists at l{¡ast one minimal element in S. Before cÓtlcluding, let us sta te Theorem' B,:;;2. Zermelo's Well-Ordering Theorem, the Axiom of Choice and Zor~~~l,Lemma are aH equivalent. The deduction of these equivalences is somewhat involved and the argument is 'not presented here; the interested reader can find the proofs in any numb:ef:of texts on set theory. 'F'"

Zorn's Lemma. Let S be a nonempty set partially ordered by ~ . Suppose that every subset A S S which is total1y ordered by ~ has an upper bound (in S). Then S possesses at least one maximal element. Zorn's Lemma is a particularly handy tool when the underlying set is partially ordered and the required object of interest is characterized by maximality. To demonstrate how it is used in practice, let us prove what is sometimes known as Hausdorff's Theorem (recall that by a chain is meant a totally ordered set):

299

I

~

298

ZORN'S LEMMA

FIRST COVRSE IN RINGS AND IDEALS

Theorem B-l. Every partially ordered set contains a maximal chain; that is, a chain which is not a proper subset of any other chain.

asserting the existence of a set S with the property that A 11 S contains exactIy one element, for each A in ~. Granting Zermelo's Theorem, it is c1ear that a choice function f can be defined for al1y collection ~ of'nonempty sets: having well-ordered u ~, simply take f to be the function which assigns to each set A 'in ~ its first element. As indicated earlier, Zermelo's Theorem was originally derived from theaxiom of choice, so that these are in reality two equivalent principies (a1though seemingly quite different). A1though the axiom of choice may strike the reader as being intuitively obvious, the soundness of this principIe has aroused more philosophical discussion than any other single question in the foundations ofmathematics. At the heart of the controversy is the ancient problern of existence. Soine mathematicians believ€; that a set exists only if each of its elements can be de~ignated specifically, qr at the very least if there is a rule by which each Of1tS members can be constructed. A more liberal school ofthought is that an axiom about existence of sets may be used if it does not lead to a contradiction. In 1938 Gode1 demonstrated that the axiom of choice is not in contradiction with the other genera:lly accepted axioms of set theory (assuming that the latter are consistent with one another). It was subsequentIy established by Cohen (1963) that the denial of this axiom is also consistent with the rest of set theory. Thus, the axiom of choice is in fact an independent axiom, whose use or rejection is a matter of personal inc1ination. The feeling among most mathematicians today is that the axiom of choice is harmless in principIe and indispensable in practice (provided that one calls attention to the occ;asions of its use). It is also valuable as an heuristic tool, since every proof,by means of this assumption represents a result for which we can then seek ,p'roofs along other lines. A non-constructive criterion f6'V'the existence of maximal e1ements is given by the so-called "maximality principIe", which general1y is cited in the literature under the name Zorn'sLemma. (From the point of view of priority, this principIe goes back t9' Hausdorff and Kuratowski, but Zorn gave a formulation of it which is patticularly suitable to algebra; he was also the first to state, without proof, ili.at a maximality principIe implies the axiom of choice.)

Proa! Consider the collection ~ of all chains of a partially ordered set (S, ~ ); ~ is nonempty, since it contains the chains consisting of single elements of S. Partially order ~ by inc1usion and let d be any chain of ~ (for the ordering s). We maintain that the union u d belongs to C(f. Given elements a, b E U d, we have a E A E d and bE BE d, for sorne A, B. As d is a chain, either A S B or B S A; suppose, for convenience, that A S B. Then a, b both lie in B and, since B is itself a chain (in S), i t follows that a ~ b or b :5: a. In con sequen ce, any two elements of u d are comparable, making u d a chain in S. Since u d is c1early an upper bound for d in ~, Zorn's Lemma implies that (~, s) has a maximal mernber. As another brief application of Zorn's Lemma, consider the following assertion: if (S, :5:) is a partially ordered set every chain of which has an upper bound, then for each a E S there exists a maximal element x E S with the property that a ~ x; in other words, there exists a maximal element !arger than the given element For a proof of this, nrst observe that the set J a = {y E Sla :5: y} satisfies the hypotheses of Zorn's Lemma (under the restriction of :5:); hence, has a maximal e1ement x. But x is maxirnal in S, not merely in J a • For, suppose that s E S with x :5: s. Then a ~ s (since both a ~ x and x :5: s) and so s E J a • From the maximality of x in J a , it then follows that s = x, completing the argument. Needless to say, we could just as well have phrased Zorn's Lemma in terms of lower bounds and minimal elements. The assertion in this case is that there exists at l{¡ast one minimal element in S. Before cÓtlcluding, let us sta te Theorem' B,:;;2. Zermelo's Well-Ordering Theorem, the Axiom of Choice and Zor~~~l,Lemma are aH equivalent. The deduction of these equivalences is somewhat involved and the argument is 'not presented here; the interested reader can find the proofs in any numb:ef:of texts on set theory. 'F'"

Zorn's Lemma. Let S be a nonempty set partially ordered by ~ . Suppose that every subset A S S which is total1y ordered by ~ has an upper bound (in S). Then S possesses at least one maximal element. Zorn's Lemma is a particularly handy tool when the underlying set is partially ordered and the required object of interest is characterized by maximality. To demonstrate how it is used in practice, let us prove what is sometimes known as Hausdorff's Theorem (recall that by a chain is meant a totally ordered set):

299

I

~

BIBLIOGRAPHY

18. lACOBSON, N., SlrUClure of Rings, Rev. Ed. Providence: American Mathematica1 Society, 1964. 19. JANS, J., Rings and Homology. New York: Hol!, 1964. 20. KUROSH, A., General Algebra. New York: Chelsea, 1963. 21. LANG, S., Algebra. Reading, Mass.: Addison-Wesley, 1965. 22. LAMBEK, J., Lectures on Rings and Modules. Waltham, Mass.: B1aisdell, 1966. 23. MCCARTHY, P., Algebraic Extensions of Fields. Waltham; Mass.: BlaisdeIl, 1966. 24. McCoy, N., Rings andldeals. (Carus Monographs). Menascha, Wis.: Mathematical Association of America, 1948.' 25. McCoy, N., TheoryofRings. NewYork: Macmillan, 1964. 26. NAGATA, M., Local Riñgs. NewYork: lnterscience, 1962 .. 27. NORTHCOTT, D. G., Ideal Theory. Cambridge, Eng1and: Cambridge UniversityPress, 1953. . . .

BIBLIOGRAPHY

,.--!'

GENERAL REFERENCES .

28. NORTHCOTT, D. G., Lessonson Rings, Modulesand Multiplicities. Cambridge, England: Cambridge Uriiversity Press, 1968. 29. REDEl, L., Algebra, VoLI. 'oxford, Eng1and: Pergamon, 1967. 30. SAH, C.-H., Abstract')i.igebra. New York: Academic Press, 1967. 31. W ARNER, ~., Modern1!Algebrti, 2 Vols. Englewood Cliffs, New Jersey: Prentice-Hall, 1965. 32. WEISS, E., Algebraic'Nu~ber Theory. New York: McGraw-Hill, 1963. 33. ZARISKI, O. and P. SAMUEL, Commutative Algebra, Vol. 1. Princeton: Van Nostrand, 1958.

.

Our purpose bere is to present a list Of sugg~sÜons for collateral reading and further study. Tbe specialized sources wiUtarry tbe reader considerably beyond tbe point attained in tbe final pages'óf tbis work. 1. ADAMSON, T., Introduction 10 Field Theory. New York: lnterscience, 1964. 2. ARTIN, E., Galois Theory, 2nd Ed. NotreDame, lnd.: University ofNotre Dame Press, 1955. 3. AATIN, E., C. NESBITT, and R. THRALL~ Rings wilh Mínimum Condítion. Ann Aibor, Mich,: University of Michigan Press, 1944. 4. AUSLANDER, M., Rings, Modules and Homology, Chapters l and n. Waltham, Mass.: Department of Mathematics, Brandeis University (lecture notes), 1960. 5. BARNES, W., Introduction 10 Abstract Algebra. Boston: Heath, 1963. 6. BOURllAKI, N., Algebra, Chapter 8. Paris: Hermann, 1958. 7. BOURBAKI, N., Algebra Commutative, Chapters 2, 4 and 5. Paris: Hermann, 1961. 8. BURTON, D. ,'Introductíon to M odern Abstract Algebra. Reading, Mass : Addison-Wes1ey, 1967. . 9. CURTIS, C. and 1. REINER, RepresentalÍon Theory of Finíte Groups and Associatíve Algebras. NewYork: lnterscience, 1962. lO. DIVINSKY, N., Rings and Radicals. Toranto: University ofToronto Press, 1965. 11. GOLDIE, A., Rings with Maximum.Condition. New Haven: Department of Mathematics, Yale University.(lecture notes), 1961. 12 .. HALMOS, P., Naive Set Theory. Princeton: Van Nostrand, 1960. 13. HERSTEIN, 1. N., Topies in Algebra.. New York: Blaisdell, 1964. 14. HERSTEIN, l. N., Theory of Rings. Chicago: Department of Mathematics, University of Chicago (lecture notes), 1961. ,15. HERSTEIN, 1. N., Noncommutative Rings. (Carus Monographs). Menascha, Wis.; Mathematical Association of America, 1968. 16. HEWITT, E. and K. STROMBERG, Real and Abstract Analysis. New York: SpringerVerlag, 1965. . 17. JACOBSON, N., Lectures in AbstractAlgebra, Vol. l, Basic Concepts. Princeton: Van Nostrand, 1951.

300

301

JOURNAL ARTICLES

1;

',1

'f'

1:

34. BROWN, B. and N. McCoy, "Radicals and Subdirect Sums," Am. J. Math. 69, 46-58 (1947). . 35. BUCK, R. c., "Extensions ofHomorphisms and Regular ldeals," J. Indian Math. Soco 14, 156-158 (1950). 36. COHEN, 1. S:, "Commutative Rings with Restricted Minimum Condition," DukeMath. J. 17, 27-42 (1950). 37. DIYINSKY, N., "Cornmutative Subdirectly Irreducible Rings," Proc. Am. Math Soco 8, 642~648 (1957). . 38. FELLER, E., "A Type of Quasi-FrobeniusRings," Canad. Math. Bull. lO, 19-27 (1967). 39. GIFFEN, c., "Unique Factorization of Polynomials," Proc. Am. Math. Soco 14, 366 (1963). 40. HENDERSON, D., HA Short Praof ofWedderburn's Theorem," Am. Math. Monthly 72, 385-386 (1965). 41. HERSTEIN, 1. ·N., "A Generalízation of a Theorem of Jacobson, l," Am. J. Math. 73, 756-762 (1951). 42. HERSTEIN, 1. N., "An Elementary Proof of a Theorem of Jacobson," Duke Math. J. 21,45-48 (1954). 43. HERSTEIN, 1. N., ~'Wedderburn's Theorem and a Theorem cif Jacobson," Am. Math. Monthly 68,249-251 (1961). 44. JACOBSON, N" "The Radical and Semi-Simplicity for Arbitrary Rings," Am. J. Math. 67,300-320 (1945). 45. KOHLS, c., "The Space ofPrime ldeals of a Ring," Fund. Math. 45, 17-27 (1957).

1: l'

·u -

_

..... ....

__..

_.

--~-

-------~

- - - - - - - - - - - - _.

. ~~-.-~----------.-._._-----

BIBLIOGRAPHY

18. lACOBSON, N., SlrUClure of Rings, Rev. Ed. Providence: American Mathematica1 Society, 1964. 19. JANS, J., Rings and Homology. New York: Hol!, 1964. 20. KUROSH, A., General Algebra. New York: Chelsea, 1963. 21. LANG, S., Algebra. Reading, Mass.: Addison-Wesley, 1965. 22. LAMBEK, J., Lectures on Rings and Modules. Waltham, Mass.: B1aisdell, 1966. 23. MCCARTHY, P., Algebraic Extensions of Fields. Waltham; Mass.: BlaisdeIl, 1966. 24. McCoy, N., Rings andldeals. (Carus Monographs). Menascha, Wis.: Mathematical Association of America, 1948.' 25. McCoy, N., TheoryofRings. NewYork: Macmillan, 1964. 26. NAGATA, M., Local Riñgs. NewYork: lnterscience, 1962 .. 27. NORTHCOTT, D. G., Ideal Theory. Cambridge, Eng1and: Cambridge UniversityPress, 1953. . . .

BIBLIOGRAPHY

,.--!'

GENERAL REFERENCES .

28. NORTHCOTT, D. G., Lessonson Rings, Modulesand Multiplicities. Cambridge, England: Cambridge Uriiversity Press, 1968. 29. REDEl, L., Algebra, VoLI. 'oxford, Eng1and: Pergamon, 1967. 30. SAH, C.-H., Abstract')i.igebra. New York: Academic Press, 1967. 31. W ARNER, ~., Modern1!Algebrti, 2 Vols. Englewood Cliffs, New Jersey: Prentice-Hall, 1965. 32. WEISS, E., Algebraic'Nu~ber Theory. New York: McGraw-Hill, 1963. 33. ZARISKI, O. and P. SAMUEL, Commutative Algebra, Vol. 1. Princeton: Van Nostrand, 1958.

.

Our purpose bere is to present a list Of sugg~sÜons for collateral reading and further study. Tbe specialized sources wiUtarry tbe reader considerably beyond tbe point attained in tbe final pages'óf tbis work. 1. ADAMSON, T., Introduction 10 Field Theory. New York: lnterscience, 1964. 2. ARTIN, E., Galois Theory, 2nd Ed. NotreDame, lnd.: University ofNotre Dame Press, 1955. 3. AATIN, E., C. NESBITT, and R. THRALL~ Rings wilh Mínimum Condítion. Ann Aibor, Mich,: University of Michigan Press, 1944. 4. AUSLANDER, M., Rings, Modules and Homology, Chapters l and n. Waltham, Mass.: Department of Mathematics, Brandeis University (lecture notes), 1960. 5. BARNES, W., Introduction 10 Abstract Algebra. Boston: Heath, 1963. 6. BOURllAKI, N., Algebra, Chapter 8. Paris: Hermann, 1958. 7. BOURBAKI, N., Algebra Commutative, Chapters 2, 4 and 5. Paris: Hermann, 1961. 8. BURTON, D. ,'Introductíon to M odern Abstract Algebra. Reading, Mass : Addison-Wes1ey, 1967. . 9. CURTIS, C. and 1. REINER, RepresentalÍon Theory of Finíte Groups and Associatíve Algebras. NewYork: lnterscience, 1962. lO. DIVINSKY, N., Rings and Radicals. Toranto: University ofToronto Press, 1965. 11. GOLDIE, A., Rings with Maximum.Condition. New Haven: Department of Mathematics, Yale University.(lecture notes), 1961. 12 .. HALMOS, P., Naive Set Theory. Princeton: Van Nostrand, 1960. 13. HERSTEIN, 1. N., Topies in Algebra.. New York: Blaisdell, 1964. 14. HERSTEIN, l. N., Theory of Rings. Chicago: Department of Mathematics, University of Chicago (lecture notes), 1961. ,15. HERSTEIN, 1. N., Noncommutative Rings. (Carus Monographs). Menascha, Wis.; Mathematical Association of America, 1968. 16. HEWITT, E. and K. STROMBERG, Real and Abstract Analysis. New York: SpringerVerlag, 1965. . 17. JACOBSON, N., Lectures in AbstractAlgebra, Vol. l, Basic Concepts. Princeton: Van Nostrand, 1951.

300

301

JOURNAL ARTICLES

1;

',1

'f'

1:

34. BROWN, B. and N. McCoy, "Radicals and Subdirect Sums," Am. J. Math. 69, 46-58 (1947). . 35. BUCK, R. c., "Extensions ofHomorphisms and Regular ldeals," J. Indian Math. Soco 14, 156-158 (1950). 36. COHEN, 1. S:, "Commutative Rings with Restricted Minimum Condition," DukeMath. J. 17, 27-42 (1950). 37. DIYINSKY, N., "Cornmutative Subdirectly Irreducible Rings," Proc. Am. Math Soco 8, 642~648 (1957). . 38. FELLER, E., "A Type of Quasi-FrobeniusRings," Canad. Math. Bull. lO, 19-27 (1967). 39. GIFFEN, c., "Unique Factorization of Polynomials," Proc. Am. Math. Soco 14, 366 (1963). 40. HENDERSON, D., HA Short Praof ofWedderburn's Theorem," Am. Math. Monthly 72, 385-386 (1965). 41. HERSTEIN, 1. ·N., "A Generalízation of a Theorem of Jacobson, l," Am. J. Math. 73, 756-762 (1951). 42. HERSTEIN, 1. N., "An Elementary Proof of a Theorem of Jacobson," Duke Math. J. 21,45-48 (1954). 43. HERSTEIN, 1. N., ~'Wedderburn's Theorem and a Theorem cif Jacobson," Am. Math. Monthly 68,249-251 (1961). 44. JACOBSON, N" "The Radical and Semi-Simplicity for Arbitrary Rings," Am. J. Math. 67,300-320 (1945). 45. KOHLS, c., "The Space ofPrime ldeals of a Ring," Fund. Math. 45, 17-27 (1957).

1: l'

·u -

_

..... ....

__..

_.

--~-

-------~

- - - - - - - - - - - - _.

. ~~-.-~----------.-._._-----

1;

302

BIBLIOGRAPHY

46. KOVACS, L., ;'A Note on Regular Rings," Publ. Math. Debrecen 4,465-468 (1956). 47. LUH, J., "On the Cornrnutativity of J-Rings," Canad. J. Math. 19, 1289-1292 (1967). 48. McCov, N., "Subdirectly Irreducible Cornrnutative Rings," Duke Math. J. 12, 3lH-387 (1945). 49. McCov, N.; "Subdirect Sums ofRings," Bull. Am. Math. Soco 53, 856-877 (1947). 50. McCoy, N., HA Note on Finite Unions of Ideals and Subgroups," Proc. Am. Math. Soco 8, 633-637 (1957). 51. NAGATA, M., "On the Theory of Radicals in a Ring," J. Math. Soco Japan 3, 330-344 (1951). 52. VON, NEUMANN, J., "On Regular Rings," Proc. Nat!. Acad. Sci. U.S. 22, 707-713 (1936). 53. NORTHCOIT, D., HA Note on the Intersection Theorern for Ideals," Proc. Cambridge Phil. Soco 48, 366-367 (1952). 54. PERL1S, S. HA Characterization of the Radical of an Algebra," Bul!. Am. Math. Soco 48, 128-132.(1942). . 55. SAMUEL, P., "On Unique F~ctorization Dornains," Illinois J. Math, S, 1-17 (1961). 56. SATYANARAVANA, M., "Rings with Prirnary Ideals as Maximal Ideals," Math. Scand: 20; 52-54 (1967). 51. SATYANARAYANA, M.,. "Characterization ofLocal Rings," TohokuMath.J.19,411-416 (1967). 58. SNAPPER, E., "Cornpletely Prirnary Rings, 1," Ann. Math. 52, 666-693 (1950). 59. STONE, M. H., "The Theory of Representations of Boolean AIgebras," Trans. Am. Math. Soco 40, 37-Ill (1936).

'.

INDEX OF SPECIAL SYMBOLS

The following is by no means a complete list of all the symbols used in the text, but is rather a listing of certain symbols which occur frequently. Numbers refer to the page where the symbol in question is first found.

{a} [a] a+l

(a) aR annS alb, atb a == b (mod n) alb aob AAB AxB A(M)

e

C(a)

centR char R contf(x) degf(x)

f(A)

1

1

(A) F(a) F[aJ

[F':F] gcd (a,b) GF(p") hom(R,R')

set consisting of the element a, 8 congruence class determined by the element a, 4 coset of the ideal 1, 39 smallest (two-sided) ideal containing the element a, 19 smallest right ideal of R containing the element a, 19 annihilator of the set S, 36 a divides (does not divide) the element b, 90 integer a is a congruent to integer b modulo n, 4 formal fraction of elements a and b, 61 circ1e-product of elements a and b, 171 symmetric difference of sets A atifl.,#, 3 Cartesian product of sets A an(!t;, ·9 annihilator of the module M, iJ5, field of complex numbers, 53 ; . , centralizer of the element a, 14;Ló' center of the ring R, 9 characteristic of the ring R, 11 d content of the polynomialj{x), c:i~9 degree ofthe polynomialj{x), 119 direct image of the set A under J,. 27 inverse imageofthe set A underf, 27 field generated by the element a over F, 137 set of polynomials in the element a, 120 degree ofthe field over the subfield F, 140 greatest common divisor ofthe elements a and b, 92 Galois field with pD elements, 191 set of ring homomorphisms from R into R', 26

r

303

1;

302

BIBLIOGRAPHY

46. KOVACS, L., ;'A Note on Regular Rings," Publ. Math. Debrecen 4,465-468 (1956). 47. LUH, J., "On the Cornrnutativity of J-Rings," Canad. J. Math. 19, 1289-1292 (1967). 48. McCov, N., "Subdirectly Irreducible Cornrnutative Rings," Duke Math. J. 12, 3lH-387 (1945). 49. McCov, N.; "Subdirect Sums ofRings," Bull. Am. Math. Soco 53, 856-877 (1947). 50. McCoy, N., HA Note on Finite Unions of Ideals and Subgroups," Proc. Am. Math. Soco 8, 633-637 (1957). 51. NAGATA, M., "On the Theory of Radicals in a Ring," J. Math. Soco Japan 3, 330-344 (1951). 52. VON, NEUMANN, J., "On Regular Rings," Proc. Nat!. Acad. Sci. U.S. 22, 707-713 (1936). 53. NORTHCOIT, D., HA Note on the Intersection Theorern for Ideals," Proc. Cambridge Phil. Soco 48, 366-367 (1952). 54. PERL1S, S. HA Characterization of the Radical of an Algebra," Bul!. Am. Math. Soco 48, 128-132.(1942). . 55. SAMUEL, P., "On Unique F~ctorization Dornains," Illinois J. Math, S, 1-17 (1961). 56. SATYANARAVANA, M., "Rings with Prirnary Ideals as Maximal Ideals," Math. Scand: 20; 52-54 (1967). 51. SATYANARAYANA, M.,. "Characterization ofLocal Rings," TohokuMath.J.19,411-416 (1967). 58. SNAPPER, E., "Cornpletely Prirnary Rings, 1," Ann. Math. 52, 666-693 (1950). 59. STONE, M. H., "The Theory of Representations of Boolean AIgebras," Trans. Am. Math. Soco 40, 37-Ill (1936).

'.

INDEX OF SPECIAL SYMBOLS

The following is by no means a complete list of all the symbols used in the text, but is rather a listing of certain symbols which occur frequently. Numbers refer to the page where the symbol in question is first found.

{a} [a] a+l

(a) aR annS alb, atb a == b (mod n) alb aob AAB AxB A(M)

e

C(a)

centR char R contf(x) degf(x)

f(A)

1

1

(A) F(a) F[aJ

[F':F] gcd (a,b) GF(p") hom(R,R')

set consisting of the element a, 8 congruence class determined by the element a, 4 coset of the ideal 1, 39 smallest (two-sided) ideal containing the element a, 19 smallest right ideal of R containing the element a, 19 annihilator of the set S, 36 a divides (does not divide) the element b, 90 integer a is a congruent to integer b modulo n, 4 formal fraction of elements a and b, 61 circ1e-product of elements a and b, 171 symmetric difference of sets A atifl.,#, 3 Cartesian product of sets A an(!t;, ·9 annihilator of the module M, iJ5, field of complex numbers, 53 ; . , centralizer of the element a, 14;Ló' center of the ring R, 9 characteristic of the ring R, 11 d content of the polynomialj{x), c:i~9 degree ofthe polynomialj{x), 119 direct image of the set A under J,. 27 inverse imageofthe set A underf, 27 field generated by the element a over F, 137 set of polynomials in the element a, 120 degree ofthe field over the subfield F, 140 greatest common divisor ofthe elements a and b, 92 Galois field with pD elements, 191 set of ring homomorphisms from R into R', 26

r

303

304

INDEX OF SPECIAL SYMBOLS

hornR(M,M') IJ J+J JrJ;JJ (J ;J)

Ll i

.Jl

J(R) kerf I(M)

lcrn(a,b) ; Mn(R) rnap(X,R) , nat¡ ordf(x)! :'<

o

,",'f:.

4J(n)

4Jr P(X)

Q Q(.jñ) Qc¡(R) R#

R*

R

V

R[x] R[x,y] R[[x]] R/J rad R Rad R Etl R¡ IrJ;JR i Z'Ze Z+

Is

Zl Z(i)

Zn ' Z(.jñ)

+",

*n

set of R-rnodule hornornorphisrns frorn M into M', 272 product of the ideals J and J, 22 surn of the ideals J and J, 21 internal direct sum of the ideals J and J, 21 quotient of the ideal J by the ideal J, 23 surn of a set ofideals Ji' 21 nil radical ofthe ideal J, 79 J-radical of the ring R, 172 kernel ofthe homomorphisrnj, 28 length ofthe module M, 252 least common multiple ofthe elements a and b, 94 ring of n x n matrices oyer R, 3 ring of mappings from X into R, 4 natuflil mapping determined by the ideal J, 40 orderbfthe power seriesf(x), 115 the empty set; 3 Euler phi-function, 57 substitution homomorphisrn induced by the element r, 120 power set ofthe set X, 3 field of rational numbers, 2 ,quadratic number field, 105 cIassical ring of quotients of R, 60 field of real numbers, 2 ' set of in vertible elernents of R, 2 heart ofthe,ring R, 212 polynomial ring in one indeterminant ,oyer R, 118 polynomial ring in two indeterminants over R, 134 power series ring in one indeterminant oyer R, 114 quotient ring of R by the ideal J, 40 ' Jacobson radical of R, 157 prime radical of R, 163 su bdirect sum of a set of rings R¡, 206 complete direct sum of a set of rings R¡, 204 ring of integers (eyen integers), 2,9 set of positiye integers, 12 ring of multiples ofthe identity element, 12 domain of Gaussian integers, 91 ring of integers modulo n, 4 a domain of quadratic integers, 106 addition (multiplication) modulo n, 5 is isomorphic to, 29

INDEX o',:.,

'. ",'

additive group of a ring, 1 comparable elements, 292 , ", adeal, ' 38 component rings (in a direct sum), 204 adjunction (of an element to a field), 137 component projection, 206 ,,' algebraic element, 138 'coIi1mutati~~ diagram, 43 extension fiele, 140 commutative ring, 2 number field, 155 complete dir~t sum, 204 algebraically c10sed field, ' 156 composition series, 251 annihilator of a su bset, 36 , congruence moduló n, 4 Artinian ring, 223 congruen'ce cIass, 4 ascending chain condition, 217 representation of, 4 associated elements, 91 conjugate of an element, "105 prime ideal of a primary ideal, '81 content of a polynomia1, 129 priÍne ideal in a N oetherian ring, 236 Correspondence Theorerti: 30 atom in a Boolean ring, 200 coset of an ideal, 39 au tomo rphism, 25 degree, of an extension field, 140 axiom of choice, 297 of a polynomial, 119 derivative functiQn, 153 Bezout identity, 93 descending chain condition, 223 binomial equation, 13 direct sum, complete, 204 Boolean ring, 14 discrete, 205 external, 33 cancellation law, 7 internal, 21 center of a ring, 9 ofmodules, 259 centralizer, of an element, 194 direct summand, 34 of a set of endomorphisms, 277 divides (divisor), 90 chaiIl (in a partially ordered set), 292 clivision ring, 52 chain conditions, 217,223, finite, 194 characteristic of a ring, 11 divisor of zero, 7 choice function, 297 domain, Euelidean, 102 classical ring of quotients, 60 integral, 7 coefficien ts of a power series, 114 principal ideal, 20 comaximal ideals, 211 unique factorization, 100

305

304

INDEX OF SPECIAL SYMBOLS

hornR(M,M') IJ J+J JrJ;JJ (J ;J)

Ll i

.Jl

J(R) kerf I(M)

lcrn(a,b) ; Mn(R) rnap(X,R) , nat¡ ordf(x)! :'<

o

,",'f:.

4J(n)

4Jr P(X)

Q Q(.jñ) Qc¡(R) R#

R*

R

V

R[x] R[x,y] R[[x]] R/J rad R Rad R Etl R¡ IrJ;JR i Z'Ze Z+

Is

Zl Z(i)

Zn ' Z(.jñ)

+",

*n

set of R-rnodule hornornorphisrns frorn M into M', 272 product of the ideals J and J, 22 surn of the ideals J and J, 21 internal direct sum of the ideals J and J, 21 quotient of the ideal J by the ideal J, 23 surn of a set ofideals Ji' 21 nil radical ofthe ideal J, 79 J-radical of the ring R, 172 kernel ofthe homomorphisrnj, 28 length ofthe module M, 252 least common multiple ofthe elements a and b, 94 ring of n x n matrices oyer R, 3 ring of mappings from X into R, 4 natuflil mapping determined by the ideal J, 40 orderbfthe power seriesf(x), 115 the empty set; 3 Euler phi-function, 57 substitution homomorphisrn induced by the element r, 120 power set ofthe set X, 3 field of rational numbers, 2 ,quadratic number field, 105 cIassical ring of quotients of R, 60 field of real numbers, 2 ' set of in vertible elernents of R, 2 heart ofthe,ring R, 212 polynomial ring in one indeterminant ,oyer R, 118 polynomial ring in two indeterminants over R, 134 power series ring in one indeterminant oyer R, 114 quotient ring of R by the ideal J, 40 ' Jacobson radical of R, 157 prime radical of R, 163 su bdirect sum of a set of rings R¡, 206 complete direct sum of a set of rings R¡, 204 ring of integers (eyen integers), 2,9 set of positiye integers, 12 ring of multiples ofthe identity element, 12 domain of Gaussian integers, 91 ring of integers modulo n, 4 a domain of quadratic integers, 106 addition (multiplication) modulo n, 5 is isomorphic to, 29

INDEX o',:.,

'. ",'

additive group of a ring, 1 comparable elements, 292 , ", adeal, ' 38 component rings (in a direct sum), 204 adjunction (of an element to a field), 137 component projection, 206 ,,' algebraic element, 138 'coIi1mutati~~ diagram, 43 extension fiele, 140 commutative ring, 2 number field, 155 complete dir~t sum, 204 algebraically c10sed field, ' 156 composition series, 251 annihilator of a su bset, 36 , congruence moduló n, 4 Artinian ring, 223 congruen'ce cIass, 4 ascending chain condition, 217 representation of, 4 associated elements, 91 conjugate of an element, "105 prime ideal of a primary ideal, '81 content of a polynomia1, 129 priÍne ideal in a N oetherian ring, 236 Correspondence Theorerti: 30 atom in a Boolean ring, 200 coset of an ideal, 39 au tomo rphism, 25 degree, of an extension field, 140 axiom of choice, 297 of a polynomial, 119 derivative functiQn, 153 Bezout identity, 93 descending chain condition, 223 binomial equation, 13 direct sum, complete, 204 Boolean ring, 14 discrete, 205 external, 33 cancellation law, 7 internal, 21 center of a ring, 9 ofmodules, 259 centralizer, of an element, 194 direct summand, 34 of a set of endomorphisms, 277 divides (divisor), 90 chaiIl (in a partially ordered set), 292 clivision ring, 52 chain conditions, 217,223, finite, 194 characteristic of a ring, 11 divisor of zero, 7 choice function, 297 domain, Euelidean, 102 classical ring of quotients, 60 integral, 7 coefficien ts of a power series, 114 principal ideal, 20 comaximal ideals, 211 unique factorization, 100

305

lNDEX

element(s), algebraic, 138 associa te, 91 conjugate, 105 idempotent, 14 identity, 2 ; invertible, 2 irreducible, 97 nilpotent, 14 prime, 97 quasi-regular, 170 related to an ideal, 258 relatively prime, 93 transcendental:, 138 torsion, 259 zero, 1 Eisenstein irreducibility criterion, 133 endomorpbism of a module, 272 ofa ring, 25 evaluation homorphism, 26 equivalence class, 288 relation, 287 Euler pbi-function, 57 Euclidean domain, 102 valuation, 102 extension, algebraic, 140 simple, 137 extension ring, 31 ; faithful module, 275,,;,,'; field, 52 algebraically c10sedj ;, 156 extension, 136 " Galois, 191 •, obtained by adjoining an element, of algebraic numbers; 155 of complex numbers, 53 of quadratic numbers, 105 of ratioÍlal functions, 138 skew, 52 splitting, 148 finite division ring, 194 integral domain, 56 field, 187 ring, 2 finitely generated, 19 first element, 294

137

307

306

semiprime, 80

fixed field, 69 formal fraction, 61 formal po.wer series, 112 Frobenius automorphism, 202 Fundamental Homorphism Theorem, 44 Fundamental Theorem of Algebra, 128

idempotent Hoolean ring, 200 element, 14 orthogonal, 268 primitive, 270 imbedding, 31 induced partíal order, 292 irreducible element, 97 ideal, 235 polynomial, 126 irredundant primary representaion, subdirect sum, 214 isomorphism ofmodules, 250 of partiaUy ordered sets, 293 ofrings, 25

Galois field, 191 Gaussian integers, 91 gcd-property, 95 generators (of an ideal), 19 greatest COmInon divisor, 92 group ofinvertible elements (of a ring) , 2 group ring, 15 H-ring, 203 heart of a ring, 212 Hilbert ring, 178 homomorphism, 25 evaluation, 26 kernel of, 28 ofmodules, 250 of partially ordered sets, 293 of rings, 25 reduction, 130 substitution, 120 trivial, 26 homomorphic image, ' 25 associated prime, 236 comaximai, 211 commutator, 50 finitely generated, 19 ideal, 16 irreducible, 235 left (right), 16 maxímal, 71 mínimal, 86 minimal prime, 84 modular, 173 modular maximal, 174 nil, 47 nilpotent, 47 primary, 81 príme, 76 product of, 22 quotient, 23 regular, 179 sum of, 20

INDEX

J-radical, 172 J-ring, , 196 Jacobson radical,

236

157

kernel of a homomorphism, 28

i

last element, 294 Icm-property, 95 leasi common multiple, 94 left annihilator, 36 ideal, 16, leroma, FittirÍ$'~, 256 Gauss" ,,130 Nakayama's, 243 Schur's, 274 length of an,element, 109 of a normal series, 252 of a module, 252 lexicographic order, 296 lifting idempotents, 167 local ring, 88 localization, 88 lower bound (for a partially ordered set), 294 maximal element, 293 ideal, 71 maximum condition, 218 , minimal element, 293 ideal, 86 prime ideal of an ideal, 84

prime ideal of a ring, 84 minimum condition, 223 polynomial, 139 modular ideal, 173 module, 247 annihilator of, 275 centralizer or, 272 direct sum, 259 dual, 286 endomorphism or, 272 faithful, 275 homomorphism of, 250 indecomposable, 260 isomorpbism, 250 quotient, 249 simple, 249 submodule, 249 torsíon-free, 259 monic polynomial, 119 multiplicatively closed set, 70 multiplicative sernigtoup of a ring, natural mapping, 41 nil ideal, 47 nilpotent element, 14 ideal, 47 nil radical of an ideal, 79 ofa ring, 79 nil-sernisimple ring, 264 Noetherian ring, 219 nontrlviai subring, 8 subdir~t sum, 206 non-zero-divisor, 60 norm, 105 normal series, 251 order homomorphism, 293 order isomorphic, 293 order of a power series, 115 Ore condition, 69 orthogonal idempotents, 268 partial order, 291 partition, 290 poiynomial, 118 content, 129 cyclotomic, 133

lNDEX

element(s), algebraic, 138 associa te, 91 conjugate, 105 idempotent, 14 identity, 2 ; invertible, 2 irreducible, 97 nilpotent, 14 prime, 97 quasi-regular, 170 related to an ideal, 258 relatively prime, 93 transcendental:, 138 torsion, 259 zero, 1 Eisenstein irreducibility criterion, 133 endomorpbism of a module, 272 ofa ring, 25 evaluation homorphism, 26 equivalence class, 288 relation, 287 Euler pbi-function, 57 Euclidean domain, 102 valuation, 102 extension, algebraic, 140 simple, 137 extension ring, 31 ; faithful module, 275,,;,,'; field, 52 algebraically c10sedj ;, 156 extension, 136 " Galois, 191 •, obtained by adjoining an element, of algebraic numbers; 155 of complex numbers, 53 of quadratic numbers, 105 of ratioÍlal functions, 138 skew, 52 splitting, 148 finite division ring, 194 integral domain, 56 field, 187 ring, 2 finitely generated, 19 first element, 294

137

307

306

semiprime, 80

fixed field, 69 formal fraction, 61 formal po.wer series, 112 Frobenius automorphism, 202 Fundamental Homorphism Theorem, 44 Fundamental Theorem of Algebra, 128

idempotent Hoolean ring, 200 element, 14 orthogonal, 268 primitive, 270 imbedding, 31 induced partíal order, 292 irreducible element, 97 ideal, 235 polynomial, 126 irredundant primary representaion, subdirect sum, 214 isomorphism ofmodules, 250 of partiaUy ordered sets, 293 ofrings, 25

Galois field, 191 Gaussian integers, 91 gcd-property, 95 generators (of an ideal), 19 greatest COmInon divisor, 92 group ofinvertible elements (of a ring) , 2 group ring, 15 H-ring, 203 heart of a ring, 212 Hilbert ring, 178 homomorphism, 25 evaluation, 26 kernel of, 28 ofmodules, 250 of partially ordered sets, 293 of rings, 25 reduction, 130 substitution, 120 trivial, 26 homomorphic image, ' 25 associated prime, 236 comaximai, 211 commutator, 50 finitely generated, 19 ideal, 16 irreducible, 235 left (right), 16 maxímal, 71 mínimal, 86 minimal prime, 84 modular, 173 modular maximal, 174 nil, 47 nilpotent, 47 primary, 81 príme, 76 product of, 22 quotient, 23 regular, 179 sum of, 20

INDEX

J-radical, 172 J-ring, , 196 Jacobson radical,

236

157

kernel of a homomorphism, 28

i

last element, 294 Icm-property, 95 leasi common multiple, 94 left annihilator, 36 ideal, 16, leroma, FittirÍ$'~, 256 Gauss" ,,130 Nakayama's, 243 Schur's, 274 length of an,element, 109 of a normal series, 252 of a module, 252 lexicographic order, 296 lifting idempotents, 167 local ring, 88 localization, 88 lower bound (for a partially ordered set), 294 maximal element, 293 ideal, 71 maximum condition, 218 , minimal element, 293 ideal, 86 prime ideal of an ideal, 84

prime ideal of a ring, 84 minimum condition, 223 polynomial, 139 modular ideal, 173 module, 247 annihilator of, 275 centralizer or, 272 direct sum, 259 dual, 286 endomorphism or, 272 faithful, 275 homomorphism of, 250 indecomposable, 260 isomorpbism, 250 quotient, 249 simple, 249 submodule, 249 torsíon-free, 259 monic polynomial, 119 multiplicatively closed set, 70 multiplicative sernigtoup of a ring, natural mapping, 41 nil ideal, 47 nilpotent element, 14 ideal, 47 nil radical of an ideal, 79 ofa ring, 79 nil-sernisimple ring, 264 Noetherian ring, 219 nontrlviai subring, 8 subdir~t sum, 206 non-zero-divisor, 60 norm, 105 normal series, 251 order homomorphism, 293 order isomorphic, 293 order of a power series, 115 Ore condition, 69 orthogonal idempotents, 268 partial order, 291 partition, 290 poiynomial, 118 content, 129 cyclotomic, 133

~ I~

,I INDEX

q.egree of, 119 function, 153 in two indetenninants, 134 irreducible, 126 leading eoefficient, 119 minimum, 139 monie, 119 primitive, 129 root of, 121 primary component, 236 ideal, 81 representation, 236 ring, 169 prime element, 97 field, 65 . ideal, 76 radical, 163 primitive idempotent, 270 ideal, 286 polynomial, 129 ring, 278 principal ideal, 19 ideal ring, 20 proper su bring, 8 pseudo-inverse, 25 quadratie num ber field, quasi-inverse, 170 quasi-regular, 170 quatemions, 54 quotient ideal, 23 fie1d of, 60 module, 249 ring, 40

105

radical, J-, 172 J aeo bson, 157 nil, 79 prime, 163 . rational function, 138 reduction homomorphism, 130 regular ring, 24 relation (binary), 287 antisymmetric, 291 associated with a function, 288 compatible equivalence, 49 congruence modulo n, 4

308

defined by a partition, 291 equivalence, 287 partialorder, 291 refiexive, 287 symmetric, . 287 transitive, 287 relatively prime elements, 93 . Remainder Theorem, 123 refinemen t of a normal series, 251 'ring, I 'Artinian, 223 Boolean, 14 . commutative, 2 . 'divisible, 233 division, 52 \:ti.nite, 2 H-, 203 Hil bert, 178 J-, 196 local, 88 nil-semisimple, 264 Noetherian, 219 of endomorphisms ofá module, 272 of extended power series over R, 152 of formal power series over R, 113 of functions between a set and ring, 4 of integers modulo n, 5 ofpolynomials over R, 118 of matriées over R, 3 primary, 169 quotient, 40 regular, 24 right Artinian, 262 semisimple, 157 simple, 18 subdirectIy irreducible, 211 with identity, 2 without radical, 157, 163 zero, 13 ring of quotients, classical, 60 generalized, 70 relative to a set, 70 root of a polynomial, 121 multiple, 153 saturated, 178 semisimple ring,

157

309

INDEX

series, composition, 251 _~quiva1ent, 252 normal, 251 length of, 252 refinement of, 251 simple extension field, 137 simple module, 249 .. 1 ring, 18 skew field, 52 splítting field, 148 square-free integer, 105 subdirectly irreducible, 211 subdirect sum, 206 subfield, 59 submodule, 249 subring, 8 ... ti generated by a set, 14 .~ proper, 8 trivial, 8 substitution homomorphism, 120 in a polynomial, 120 ., symmetric difference, 3 Theorem, Akizuld-Hopldns, 255 Birkhoff's, 212 Brauer's, 262 Chinese Remainder, 211 Cohen's, 241 Dorroh Extension, 31 Euler-FeIlflat, 58 Euclid's, 101 Fermat's little, 68 Hausdorff's, 299 Herstein's, 199

Hilbert Basis, 220 Jordan-HOlder, 252 Jacobson's. 198 Jacobson Density, 285 Kroneeker's, 144 Krull Intersection, 244 Krull-Zorn, 74 Levitski's, 222 McCoy's, '212 Noether's, 236 Stone Representatíon, 183 Wilson's, 188 Wedderbum's, 194, 266, 230 yYedderbum-Artin, 281 Zermelo's, 297 total ordering, 292 torsion element, 259 torsión-free module, 259 transcendental element (over a field), trivial subring, 8 homomorphism, 26

138

unique factorization domain, 100 upper bound (for a partially ordered set), 294' valuation ring,

88

we1l-ordered set, 296 without radical, 157, 163 zero divisor, 7 zero e1ement of a ring, zero ring, 13 Zorn's lemma, 298

ABCDE79876543210

~ I~

,I INDEX

q.egree of, 119 function, 153 in two indetenninants, 134 irreducible, 126 leading eoefficient, 119 minimum, 139 monie, 119 primitive, 129 root of, 121 primary component, 236 ideal, 81 representation, 236 ring, 169 prime element, 97 field, 65 . ideal, 76 radical, 163 primitive idempotent, 270 ideal, 286 polynomial, 129 ring, 278 principal ideal, 19 ideal ring, 20 proper su bring, 8 pseudo-inverse, 25 quadratie num ber field, quasi-inverse, 170 quasi-regular, 170 quatemions, 54 quotient ideal, 23 fie1d of, 60 module, 249 ring, 40

105

radical, J-, 172 J aeo bson, 157 nil, 79 prime, 163 . rational function, 138 reduction homomorphism, 130 regular ring, 24 relation (binary), 287 antisymmetric, 291 associated with a function, 288 compatible equivalence, 49 congruence modulo n, 4

308

defined by a partition, 291 equivalence, 287 partialorder, 291 refiexive, 287 symmetric, . 287 transitive, 287 relatively prime elements, 93 . Remainder Theorem, 123 refinemen t of a normal series, 251 'ring, I 'Artinian, 223 Boolean, 14 . commutative, 2 . 'divisible, 233 division, 52 \:ti.nite, 2 H-, 203 Hil bert, 178 J-, 196 local, 88 nil-semisimple, 264 Noetherian, 219 of endomorphisms ofá module, 272 of extended power series over R, 152 of formal power series over R, 113 of functions between a set and ring, 4 of integers modulo n, 5 ofpolynomials over R, 118 of matriées over R, 3 primary, 169 quotient, 40 regular, 24 right Artinian, 262 semisimple, 157 simple, 18 subdirectIy irreducible, 211 with identity, 2 without radical, 157, 163 zero, 13 ring of quotients, classical, 60 generalized, 70 relative to a set, 70 root of a polynomial, 121 multiple, 153 saturated, 178 semisimple ring,

157

309

INDEX

series, composition, 251 _~quiva1ent, 252 normal, 251 length of, 252 refinement of, 251 simple extension field, 137 simple module, 249 .. 1 ring, 18 skew field, 52 splítting field, 148 square-free integer, 105 subdirectly irreducible, 211 subdirect sum, 206 subfield, 59 submodule, 249 subring, 8 ... ti generated by a set, 14 .~ proper, 8 trivial, 8 substitution homomorphism, 120 in a polynomial, 120 ., symmetric difference, 3 Theorem, Akizuld-Hopldns, 255 Birkhoff's, 212 Brauer's, 262 Chinese Remainder, 211 Cohen's, 241 Dorroh Extension, 31 Euler-FeIlflat, 58 Euclid's, 101 Fermat's little, 68 Hausdorff's, 299 Herstein's, 199

Hilbert Basis, 220 Jordan-HOlder, 252 Jacobson's. 198 Jacobson Density, 285 Kroneeker's, 144 Krull Intersection, 244 Krull-Zorn, 74 Levitski's, 222 McCoy's, '212 Noether's, 236 Stone Representatíon, 183 Wilson's, 188 Wedderbum's, 194, 266, 230 yYedderbum-Artin, 281 Zermelo's, 297 total ordering, 292 torsion element, 259 torsión-free module, 259 transcendental element (over a field), trivial subring, 8 homomorphism, 26

138

unique factorization domain, 100 upper bound (for a partially ordered set), 294' valuation ring,

88

we1l-ordered set, 296 without radical, 157, 163 zero divisor, 7 zero e1ement of a ring, zero ring, 13 Zorn's lemma, 298

ABCDE79876543210