1111569622Abstract.pdf

an

integer. [Part (a) is the case when n

36. Let p, q be primes with p 2::

=

5, q 2:: 5. Prove that 24 I (p2 -

...

........

Prove that

r

q2).

.... .. ...

eap,ngm.20:12�l...umillg.A:l.llitla 11--4.....,-aatn. t:IDJllilrd,. llC...t,, ar�io.wmlliarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ......... q-�� fld.�dlN:t Cl'Na!S---.�c.a.� rigbllD...,,,..��-

...

a.

2.]

Jion1M•Bam:.ndkir�.Bdbmbll_...._ --il......_..:dPLI� ........

2

CHAPTER

Congruence in "lL and Modular Arithmetic

Basic concepts of integer arithmetic are extended here to include the idea of "congruence modulo n." Congruence leads to the construction of the set Z11 of all congruence classes of integers modulo n. This construction will serve as a model for many similar constructions in the rest of this book. It also provides our first example of a system of arithmetic that shares many fundamental properties with ordinary arithmetic and yet differs significantly from it

•

Congruence and Congruence Classes

T he concept of "congruence" may be thought of as a generalization of the equality relation. Two integers a and b

are

equal if their difference is 0 or, equivalently, if their

difference is a multiple of 0. If n is a positive integer, we say that two integers are con­

gruent modulo n if their difference is a multiple of n. To say that a - b

=

nk for some

integer k means that n divides a - b. So we have this formal definition:

Definition

Let a, b, n be integers with n > 0. Then a is congruent to [written "a = b (mod n)"], provided that n divides a - b.

b

modulo n

EXAMPLE 1 17 = 5 (mod 6) because 6 divides 17 - 5 because 7 divides 4 - 25

6-.(-4)

=

Remark

=

-2

12. Similarly, 4 = 25 (mod 7)

10. In the notation

"

a

are really parts of a single symbol; " a

=

1 , and 6 = -4 (mod 5) because 5 divides

= b (mod n)," the symbols " = " and "(mod n)" "a = b" by itself is meaningless. Some texts write

=n b" instead of "a= b (mod n)." Although this single-symbol notation is advanta­

geous,

we

shall stick with the traditional "(mod n)" notation here. 26

CopJrial<2012C...LHng.All ...... _.Mq ... 11o..,,....,..- ....,..._ .. _ ...,...Doo .. -...............____ llo_.._ ....__�·>·--... _ .... ..,_... ,,,__ ... _..,. _ ... _.....,...,_..c.g,..1.Nmlo&---riP<"'____ _,_11..-.-tlajlll-. ....... ll

26

Chapter 2

Congruence in Zand Modular Arithmetic

The symbol used to denote congruence looks very much like an equal sign. This is no accident since the relation of congruence has many of the same properties as the relation of equality. For example,

we know that equality is

reflexive: a =

a for every integer a; symmetric: if a=b, then b =a; transitive: if a= b and b = c, then a= c.

We now

see that congruence modulo n is also reflexive, symmetric, and transitive.

Theorem 2.1 Let n be a positive integer. For all a, b, cEZ, (1) a= a (mod n); (2) if a = b (mod n), then b =a(mod n); (3) if a = b (mod n) and b = c (mod n), then a = c (mod n). Proof ... (1) To prove that a= a(modn), we must show thatn I (a - a). But a - a = 0 andn I 0 (see Example 2 on page 9). Hence, n I (a - a) and a= a(modn).

(2) a= b (mod n) means that a - b = nk for some integer k. Therefore, - a = -(a - b) = -nk = n(-k). The first and last parts of this

b

equation say that n I

(3)

If a= b (modn)

congruence, there

(b

�

a). Hence, b = a(modn).

and b = c (mod n), then by the definition of

are integers k and t such that a - b

b - c =nt. Therefore,

=

nk and

(a - b) + (b - c) = nk + nt

a - c =n(k + t). Thus n I

(a - c) and, hence, a= c (mod n).

•

Several essential arithmetic and algebraic manipulations depend on this key fact: If

a = b and c = d, then a + c = b +

d and

ac

=bd.

We now show that the same thing is true for congruence.

Theorem 2.2

Ifa= b (mod n) and c = d (mod n}, then

(1) a + c = b + d (mod n); (2) ac = bd (mod n). .. ��dml!Olll.....,dllcl....�......��Lamaloa ........riBbtla-....,,.�IDllllll.-..,....jf......._.:lif!,bb�........

�20t2C....-1-mlq.A1�R--4.Mq11Dthlcap.d. IC...:l,,ar�flllt.wtdaarl:aJ*t. 0.10�aeia.-tild_:PMJ'ICOl:llMl:�.,._,......ra:.:..m.111Bom:.ndrot�1).BdbDftlil._...MI

-....ed.--

2.1

Congruence and Congruence Classes

Proof• (1) To prove that a+ c = b + d (mod n), we must show that n (a+ c) -(b+ti). Since a = b (mod n) and c =d (mod n), n J (a-b) and n I (b- ti). Hence, there are i ntegers k and t a - b = nk

and

c

divides

we know that such that

- d = nt.

We use these facts to show that n divides (a+ c)-(h+

ti):

[Aritlunetic]

(a+ c)-(b+ti)= a+ c-b-d = (a-b) + (c-ti) =

27

[Rearrange terms.] [a - b =

nk+ nt

nk a nd c

- d = nt.]

[Factor right side]

(a+ c)-(b+ d) = n(k+t)

The last equation says that n divides (a+ c) -(b+

ti). Hence, a+ c =

b+ d(modn). (2) We must prove that n divides ac-bd *

ac-bd= ac+O-bd = ac-bc+ be- bd

[-be+ be= O.]

= (a - b) c + b(c - d) [Factor first nro terms and last two tenns.] = (nk)c+ b(nt) ac-bd= n(kc

+

[a-b = nk

bt)

and c

-d= nt by(*) above.]

[Factor nfrom each term.]

The last equation says that n

I (ac

- bd). T herefore, ac

With the equality relation, it's easy to

see

= bd (mod 11).

•

what numbers are equal to a given

number a-just a itself. With congr uence, however, the story is different and leads to some interesting consequences.

Definition

Let a and

(denoted

n

be integers with n > 0. The

congruence class of a modulo n

[a]) is the set of all those integers that are congruent to a modulo

n, thatlsi

[a]= {bJbEZ

and

b s a (mod n)}.

To say that b that b

=a (mod n) means that b-a = kn = a+ kn. Thus [a]= {b I b =a

(mod

for some integer k or, equivalently,

n)} = {b I h =a+ kn

with kE.Z}

= {a +kn I kEl}.

*The first two lines of this proof for a suitable expression

are a standard algebraic

technique: Rewrite

0

in the form

-X + X

X.

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28

Chapter 2

Congruence inland Modular Arithmetic

EXAMPLE 2 In congruence modulo 5, we have

[9]

=

=

{9+5k I kEZ}

{9, 9 ± s, 9 ± 10, 9 ± 15,

=

... }

{ .... -11, -6, -1, 4, 9, 14, 19, 24, .. . }.

EXAMPLE 3 The meaning of the symbol

"[ ]" depends on the context.

In congruence

modulo 3, for instance,

[2]

{2 + 3klkEZ}

=

=

{

. . .

, -7,

-

4, -1, 2, 5, 8,

. . . },

but in oongruence modulo 5 the congruence class [2] is the set {2

+

5k I kEZ}

=

{ ... , -13,

-8, -3,

2,

7, 12,

.

. .}.

This ambiguity will not cause any difficulty when only one modulus is under discussion. On the few occasions when several moduli are discussed simultaneously, we avoid confusion by denoting the congruence class of modulo n by

a

[a]n·

EXAMPLE 4 In congruence modulo 3, the congruence class

[2]

=

{.

.

.

, -7, -4, -1, 2, 5, 8,

.

.

}.

•

Notice, however, that [-1] is the same class because

[-1] Furthermore,

=

{-1+3k I kEZ}

2= -1

=

{.

.

. ' -1,

-4, -1, 2, 5, . .

. }.

(mod 3). This is an example of the following theorem.

Theorem 2.3 a= c {mod

n) if and only if [a]

Since Theorem

2.3

=

[c].

is an "if and only if" statement, we must prove two different

things:

1.

If

a= c (mod n),

2.

If

[a]

=

then

[a]

=

[c].

[c], then a= c (mod n).

Neither of these proofs will

use

the definition of congruence. Instead, the proofs will

use only the fact that congruence is reflexive, symmetric, and transitive (Theorem 2.1 ).

�2012.C....,1-mlq.illUPDa--l MaJ"aatbemp.d. KlUOlld,, or�:iawtdlioriaj*t. 0.1o�,....._-1hlm.pmJ-cooim:mayk�ta:.J.._t1&dl::udkx'�l).Bimorilll.......-._ -..d.1lllmy��"'*-ao1.-.d.n,'.n.ctb�._....���---ftgbtn-��-..,..-..��:Dgb&l�...-.it.

2.1

Congruence and Congruence Classes

29

Proof of Theorem 2.3. First, assume that a"" c (modn). To prove that [a]= [c], we first show that [a] !::[c]. To do this, letbE[a]. Then by definitionb= a(modn). Since a= c (modn), wehaveb = c(modn)bytransitivity. Therefore,bE [c] and [a]!:: [c]. Reversing the roles of a andc in this argument and using the fact that c=a by symmetry, show that [c]!:: [a]. Therefore, [a]= [c]. [a] = [c]. Since a=a (mod n) by reflexivity, aE[c]. By the definition of [c] , we see that

Conversely, assume that we have aE[a] and, hence,

a= c (modn).

•

A and Care two sets, there are usually three possibilities: Either A and Care dis­ A = C, or A n C is nonempty but A * C. With congruence classes, however, there are only two possibilities: If

joint, or

Corollary 2.4 Two congruence classes modulo

n are

either disjoint or identical.

Proof• If [a] and [c] are disjoint, there is nothing to prove. Suppose that [a]

b with hE [a] and b E[c ]. b = c (mod n). Therefore, by symmetry and transitivity, a= c (mod n). Hence, [a] = [c] n

[c] is nonempty. Then there is

an integer

B y the definition o f congruence class, b = a (mod n) a n d by Theorem 2.3.

•

Corollary 2.5 Let

n > 1 be an integer and (1)

consider congruence modulo

n.

If a is any integer and r is the remainder when a is divided by n, then

[a] = [r]. (2) There are exactly n distinct congruences classes, namely, [OJ, [1], [ 2], [n - 1], o

o

•

I

Proof•(l) Let aEZ By Thus

the DivisionAlgorithm, a= nq + r, with Os r < n. a - r= qn, so that a = r (mod n). By Theorem 2.3, [a] = [r].

(2) If [a] is any congruence class, then (1) shows 0 s r < n. Hence, [a] must be one of [ O], [l], [2],

that ...,

To complete the proof, we must show that these To do this, we first show that no two of 0, 1, 2, . ..

n

,n

[a] = [r] with [n - 1 ] .

classes are all distinct.

- 1 are congruent

modulo n. Suppose that s and t are distinct integers in the list 0, 1, 2, . .. n

,

- 1. Then one is larger than the other, say t, so that 0s s < t < n.

Consequently, t - s is a positive integer that is less than n. Hence, n does

- s, which means that t ¢ s. Thus, no two of 0, 1, 2, .. . , n - 1 are congruent modulo n. Therefore, by Theorem 2.3, the classes [O], [l], [2], .. . , [n - 1 ] are all distinct. • not divide t

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30

Chapter 2

Definition

Congruence in 1. and Modular Arithmetic

The set of all congruence classes modulo n is denoted Zn (which is read "Z mod n").

There are several points to be careful about here. The elements of

Zh are classes,

not single integers. So the statement [5] EZh is true, but the statement 5 E Zti is not.

F urthermore, every element of Zti can be denoted in many different ways. For example, we know that

2 = 5 (mod 3)

2 =-I

Therefore, by Theorem 2.3, [2] of

=

(5]

==

(mod

2 = 14 (mod 3).

3)

[-I] "" [14] in 71+ Even though each element

Zti (that is, each congruence class) has infinitely many different labels, there are only

finitely many distinct classes by Corollary 2.5, which says in effect that

The set Z� has exactly n elements. For example, the set Z3 consists of the three elements [OJ, [l], [2).

• Exercises A. I. Show that

aP-

I= 1 (modp) for the givenp and a:

(a) a=2,p=5 2.

(a)

If k

=

(c)

(b) a=4,p=7

a

= 3, p=11

1 (mod 4), then what is 6k + 5 congruent to modulo 4?

(b) If r = 3 (mod 10) ands= -7 (mod 10), then what is 2r + 3s congruent to modulo 10? 3. Every published book has a ten-digit ISBN-10 number (on the back cover or the copyright page) that is usually of the form x1-x2x3xrx5xl)X1XsXrX10 (where each xi is a single digit).* The first 9 digits identify the book. The last digit x10 is a

check digit; it is chosen so that

10x1 + 9x2 + 8x3 + 7x4 + 6x5 + 5x6 + 4x1 + 3 x8 + 2x9 + x10 = 0 (mod 11).

If an error is made when scanning or keying an ISBN number into a computer, the left side of the congruence w ill not be congruent to 0 modulo 11, and the number will be rejected as invalid. t Which of the following are apparently valid

ISBN numbers? (a) 3-540-90518--9

(b) 0--031-10559--5

"Sometimes the last digit of an ISBN number is the letter number

10.

X.

(c) 0--385--49596--X

In such cases, treat X as if it were the

trhe procedures in Exercises 3 and 4 will detect every single digit substitution error (for instance,

3 is entered as 8 and no other error is made).They will detect about 90% of transposition errors (for instance, 74 is entered as 47 and no other error is made). However, they may not detect muHiple errors.

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2.1

Congruence and Congruence Classes

31

4. Virtually every item sold in a store has a 12-digit UPC barcode which is scanned at the checkout counter. The first

1 1 digits of a UPC number d1difl3•

• • •

d11d1 2

identify the manufacturer and product. The last digit d12 is a check digit which is chosen so that

If the congruence does not hold, an error has been made and the item must be scanned again, or the UPC code entered by hand. Which of the following UPC numbers were scanned incorrectly?

(a) 037000356691 5.

(a)

(b)

Which of [O], [l], [2], [3] is equal to Theorems 2.2 and 2.3.)

(b) Which of

(c) 040293673034

833732000625

[52000_! in Zt? [Hint: 5 = I (mod 4); use

[O], [1], [2], [3,] [4] is equal to [42001] in�?

6. If

a= b(mod n) and k In, is it true that a = b(mod k)? Justify your answer.

7. If

a E Z. prove that a2 is not congruent to 2modulo 4or to 3 modulo 4.

8. Prove that every odd integer is congruent to

1 modulo 4 or to 3 modulo 4.

9. Prove that

(a) I 0.

(n

- a)2 = a2 (mod n)

(b) (2n

- a)2 = a2 (mod 4n)

If a is a nonnegative integer, prove that a is congruent to its last digit mod 10 [for example, 27= 7 (mod 10)].

B.11. If a, bare integers such that that

a=

a= b(modp) for e very positive prime p, prove

b.

5 and p is prime, prove that [p] [l] or [p] [Hint: Theorem 2.3 and Corollary 2.5.]

12. If p <'!:

=

13. Prove that

=

[5] in �·

a = b (mod n) if and only if a and b leave the same remainder when

divided by n. 14.

(a) Prove or disprove: If ab= 0 (mod n), then a= 0 (mod n) or b= 0 (mod n). (b) Do part (a) when n is prime.

15. If (a, n) 16. If

[a ]

=

=

1, prove that there is an integer bsuch that ab = 1 (mod n).

[ 1 ] in Z,., prove that

(a, n)

=

1.

Show by example that the converse

may be false. 17. Prove that

10" = (-lf (mod 11) for every positive n.

18. Use congruences (not a calculator) to show that

(125698) (23797) ./:- 2891235306. [Hint: See Exercise 21.] 19. Prove or disprove: If [a] 20.

(a)

=

Prove or disprove: If

[b] in Z,., then (a, n) = (b, n).

a2 = b2 (mod n), then a = b(mod n) or

a = -b (mod n). (b) Do part (a) when n is prime.

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32

Chap ter 2

Congruence inland Modular Arithmetic

21. (a) Show that 10"

=

1 (mod 9) for every positive n.

(b) Prove that every positive integer is congruent to the sum of its digits mod 9 [for example, 38 = 11 (mod 9)]. 22.

(a) Give an example to show that the following statement is false: If ah (mod n) and a "1= 0 (mod n), then b = c (mod n). (b) Prove that the statement in part (a) is true whenever (a, n)

=

=

ac

L

EXCURSION: The Chinese Remainder Theorem (Section 14.1) may be desired.

covered at this point if

Ill

Modular Arithmetic

The finite set "11,, is closely related to the infinite set Z. So it is natural to ask if it is possible to define addition and multiplication in "1L,. and do some reasonable kind of arithmetic there. To define addition in Z,,, we must have some way of taking two classes in "1L,. and producing another class-their sum. Because addition of integers is defined, the following tentative definition seems worth investigating: The sum of the classes [a] and [c] is the class containing a+ c or, in symbols,

[a] Ee [c]

=

[a + c],

where addition of classes is denoted by Ef) to distinguish it from ordinary addition of integers. We can try a similar tentative definition for multiplication: The product of [a] and [c] is the class containing ac: [a] 0 [c]

=

[ac],

where 0 denotes multiplication of classes.

EXAMPLE 1 In� we have [3] Ee [4]

=

[3 + 4]

=

[7]

=

[2] and [3] 0 [2]

=

[3 2] ·

=

[6]

=

[l].

Everything seems to work so far, but there is a possible difficulty. Every element of

"1L,. can be written in many different ways. In�. for instance, [3] [13] and [ 4] [9]. In the preceding example, we saw that [3] Ee [4] [2 ] in�· Do we get the same answer if we use [ 13] in place of [3] and [9] in place of [ 4]? In this case the answer is "yes" because =

=

=

[13] ® [9]

=

[13 + 9]

=

[22]

=

[2] .

But how do we know that the answer will be the same no matter which way we write the classes?

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2.2

Modular Arithmetic

33

To get some idea of the kind of thing that might go wrong, consider these five classes of integers: A

=

{.

. .

, -14, -8, -2, 0, 6, 12, 18,

B=

{... ,

c

{. . . ,-9,

D E

=

-5, -1, 3, 7, 11, 15,

. . .

.

•

.

, -16, -10, -4, 2, 8, 14, 20,

=

{. .

=

{... ;

.

-ll, -7, -3, l, 5, 9, 13,

.. .

}

}

}

.

•

-18, -12, -6, 4, 10, 16, 22,

}

.

. ..}.

These classes, like the classes in �. have the following basic properties: Every integer is in one of them, and any two of them are either disjoint or identical. Since 1 is in B and 7 is in C, we could define B + C as the class containing 1 + 7

D.

=

8, that is,B + C =

But Bis also the class containing -3 and C the class containing 15, and so B + C

ought to be theclass containing-3 + 15 = 12. But 12 is in A, so thatB + C = A . Thus you get different answers, depending on which "rep resentatives" you choose from the classes B and

C.

Obviously you can't have any meaningful concept of addition if the

answer is one thing this time and something else another time. In order to remove the word "tentative" from our definition of addition and mul­ tiplication in

Z,.,

we must first prove that these operations do not depend on the

choice of representatives from the various classes. Here is what's needed:

Theorem 2.6 If [a]

=

[b] and [c]

=

[d] in�. then

[a + c]

Proof"' Since [a] [c]

=

=

and

[b + d]

[ac]

=

[bd].

[b], we know that a= b (modn) by Theorem 2.3. Similarly, [d] implies that c = d (mod n). Therefore, by Theorem 2.2, =

a + c = b + d (mod n) Hence,

and

ac= bd(modn).

by Theorem 2.3 again, [a + c]

=

[b + d]

and

[ac]

=

[bd].

•

Because of Theorem 2.6, we know that the following formal definition of addition and multiplication of classes is independent of the choice of representatives from each class:

Definition

Addition and multiplication in Zn are defined by [a] EB [c]

=

[a+ c]

and

[a] 0 [c)

=

[ac].

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34

Chapter 2

Congruence in Zand Modular Arithmetic

EXAMPLE 2 Here are the complete addition and multiplication tables for Zs (verify that these calculations are correct):*

(f)

[OJ

[lJ

[2J

[3J

[4J

8

[OJ

[lJ

[2J

[3J

[4J [OJ

[OJ

[OJ

[I]

[2J

[3}

[4J

[OJ

[OJ

[OJ

[OJ

[OJ

[1]

(lJ

[21

[3J

[4J

[OJ

[1]

[OJ

[l]

[2J

[3J

[4]

[2J

[2]

[3J

[4J

[OJ

[l ]

[2J

[OJ

[2 }

[41

[l]

[3}

[3J

[31

[4]

[OJ

[l]

[2J

[3J

[OJ

[3J

[l J

[4J

[21

[4J

[41

[OJ

[IJ

[21

[3J

[4J

[OJ

[4J

[3J

[2J

[l}

And here are the tables for "14,:

(f)

[OJ

[lJ

[2J

[3J

[4J

[5J

[OJ

[OJ

[lJ

[3J

[4J

[5J

[lJ

[ ll

[2J

[2 1 [3}

[4J

[5J

[OJ

[2]

[2]

[3J

[4]

[5J

[OJ

[l J

[3J

[3]

[4J

[5J

[OJ

[lJ

[2J

[41

[4}

[5]

[OJ

[lJ

[2J

[3J

[SJ

[5J

[OJ

[lJ

[2J

[3J

[4J

8

[OJ

[lJ

[2J

[3J

[4J

[5J

[OJ

[OJ

[OJ

[OJ

[OJ

[OJ

[OJ

[1]

[OJ

[lJ

[2J

[3J

[4J

[5J

[2J

[OJ

[2J

[4J

[OJ

[2J

[4J

[3J

[OJ

[3J

[OJ

[3J

[OJ

[3J

[41

[OJ

[4J

[21

[OJ

[4J

[2J

[5J

[OJ

[5J

[41

[3J

[2]

[IJ

Properties of Modular Arithmetic Now that addition and multiplication are defined in Z,.,we want to compare the properties of these "miniature arithmetics" with the well-known properties of Z The key facts about arithmetic in Z (and the usual titles for these properties) are as follows. For all a, b , cEZ: 1. If

a, bEZ, then a+bEZ

[Closurefor additionJ

(a+b)+ c.

[Associative addition}

2.

a+(b+c)

3.

a+b

4. a+0

""'

=

b +a.

=

a

=

0+a.

[Commutative additionI [Additive ickntity}

*These tables are read like this: If [a] appears in the left-hand vertical column and [c] in the top

[i!] ffi [c] appears at the intersection [a] and the vertical column containing [c].

horizontal row of the addition table, for example, then the sum of the horizontal row containing

eap,rigm.20:12�1..umiq.A:l.lliala 11--4.....,-aathl t:IDJllilrd,. llC...t,, ardufticlMd.io.wmlllarls,_,. 0.1"�dpll.-mkd.��_,,._ ....... 8om.1M11Bam:.ndkir�.Bdbmbll_...._ ...._._q-��._.fld.__...,.a11N:t... �a--.�c...,.� ...... ... rir;bl1a-...,,,..��·..,..._w..._._..:dPLI�...-. ..

2.2

Modular Arithmetic

36

5. For each a E Z, the equation

a+

6. If

x

a,

= 0 has a solution in Z.

7.

a(_bc) =(ab)c.

8.

a(_h+ c) =ab+ ac and

9. 10.

[Closurefor multiplication]

bEZ, then abEZ.

[Associative multiplication]

(a + b)c =ac + be.

[ DistributiVe laws]

ab =ha

[Commutative multiplication] [Multiplicative identity]

a 1 =a =1 a •

11. If

·

ab =0, then a = 0 or b = 0.

By using the tables in the preceding

example,

you can verify that the first ten of

these properties hold in Zs and Z6 and that Property 11 holds in Zs and fails in �·But using tables is not a very efficient method of proof (especially for verify­ ing associativity or distributivity). So the proof that Properties 1-10 hold for any Z,. is based on the definition of the operations in Z,. and on the fact that these properties

are

known to be valid in Z.

Theorem 2.7 For any classes [a], [b], [c] in Z,., 1. If [a]E°Zn and [b]E Z,., then [ a]®[b]E Z,.. 2.

[a]®([ b]®[c]) =([a]®[b]) ®[c].

3.

[a]® [b] =[b] Et3 [a].

4.

[a]®[O] =[a] =[O]®[ a].

5.

For each [a] in Z,., the equation [a] Et3 X =[O] has a solution in Zr,.

6. If [a]E°Zn and [b]E Zn, then [a] 0 [b]E Zn. 7.

[a] 0 {[b] 0 [c]) = ([a] 0 [b] ) 0 [c].

8.

[a] 0 ([b]®[c]) =[a] 0 [b) ®[a) 0 [c] and ([a]®[b]) 0 [c] =[ a] 0 [c] Et3 [b] 0 [c] .

9.

[a] 0 [b]=[b] 0 [ a].

10. [a] 0 [1]=[a]=[1] 0 [a].

Proof" Properties 1

and 6

are

an immediate consequence of the definition of Et3

and0inZ,.. To prove Property 2, note that by the definition of addition,

[a]®([b] EtJ [ c]) = [a]®[b + c] =[a + (h+ c)]. In Z we know that a+

(b+ c) =(a + b)+ c. So the classes of these [a + (h + c)] =[(a+ b) + c). By

integers must be the same in Zn; that is,

the definition of addition in Zr,, we have

[(a+ h) + c ] =[a + b] EtJ [c] =([a]® [bD ® [c ]. �2012.C....,l...Mmiq.AIRqlna-..d.MaJ"mtbll� �-ar....... :towballl«lapd.. 0..W�dalD.-tinl.Jlal;J�a.,.'8....,....m_ta:.:J.beBo'*:.udkx-��----­ dlMm&d.-..:my�-mmllldmmmll___...,.d!Kl. ... �---.�c.g..p�---ftgbtta--.:lditti:rml.ICIDllllnl•_..,.lillll��:Dgbb�...-.:lit.

36

Chapter 2

Congruence inland Modular Arithmetic

2. The proofs of Properties 31 7, 8, and 9 are 10). Properties 4 and 10 are proved by a direct calculation; for instance, [a] 0 [l] [a· l] [a]. For Property 5, it is easy to see that X [-a] is a solution of the [a + (-a)] [O]. • equation since [a] ffi [-a]

This proves Property analogous (Exercise =

=

=

=

=

Exponents and Equations The same exponent notation used in ordinary arithmetic is also used in Zr,. If and k is a positive integer, then

[a]k denotes the product

[a] 0 [a] 0 [a] 0

·

·

·

0 [a]

[a] EZn,

(k factors).

EXAMPLE 3 In Z5,

[3]2

[3] 0 [3]

=

=

[4]

and

[3]4

=

[3] 0 [3] 0 [3] 0 [3]

=

[l].

As noted on page 9, the set 7L11 has exactly n elements. Consequently, any equation in 7L11 can be solved by substituting each of these which ones

are

n

elements in the equation to see

solutions.

EXAMPLE 4 To solve x1 Ee

[5] 0 x

=

[O] in Zt,, substitute each of [O], [1], [2], [3], [4], and [5]

in the equation to see if it is a solution:

x

x2 Ee [5] 0 x

[OJ

[OJ0[0J ffi [5J0[0]

[l]

[1J0[1J ffi [5J0[1]

[2]

[2J0 [2J Ee [5]0[2]

[3]

[3J0[3J ffi [5J0[3]

=

[3J ffi [3J

=

[O]

Yes; solution

[4]

[4J0[4J ffi [5J0[4]

=

[4J ffi [2J

=

[O]

Yes; solution

[5]

[5]0[5] Ee [5]0[5]

=

[lJ Ee [lJ

=

[2]

No

Is =

=

=

=

[OJ ffi [OJ

=

[O]

Yes; solution

[lJ ffi [5J

=

[O]

Yes; solution

[ 4] Ee [4]

So the equation has four solutions: [O], Example

x2 ffi [5] 0 x

=

[2]

[O]?

No

[1], [3], and [4].

4 shows that solving equations in Z,, may be quite different from solving

equations in 7L. A quadratic equation in 7L has at most two solutions, whereas the quadratic equation x1 ffi

[5]0x

=

[OJ has four solutions in Z6•

• Exercises A. I. Write out the addition and multiplication tables for

(a)

(b) �

Z2

(c)

7L7

(d)

Z-12

In Exercises 2--8, solve the equation. 2.

x1 ffi x

=

[O] in �

�2012C...,..1.Nmlmg.Al.1Ua11Da..r..a.V.,.ootbll� �-w�:la11'fdiiwiaJ*I., 0..W�dailD.-1hlinl.:PGQ"�a.,.h�fnml.b•Bo1*:..ab-�1).EdDW.....,._ dlremad.'lmm,-��._ .. .-.m.Dy.n.ctbl�lmmliog�CmgQl-l...eMmog-- .. ftgbtlD--...�OOllll!m·a;J'tlmlo1f..._...._:ligl:U�:MpiNit.

2.3 3. x2 4.

The Structure of ZP (p Prime) and Zn

37

=[lJ in.ls

x4 =[lJ in Zs

5. x2 EB [3J 0 x®[2J = [OJ in Zt,

6. x2 EB [SJ 0 x = [OJin £9 7. x3 EB x2® x®[lJ =[OJ in Zs 8. x3 9.

+

x2 =[2J inZ10

(a) Find an element [aJ in Z7 such that every nonzero element of Z7 is a power of [aJ. (c) Can you do part (a) in�?

(b) Do part (a) in :z'.s.

10. Prove parts 3, 7, 8, and 9 of Theorem 2.7. 11. Solve the following equations.

(a) x®x ®x =[OJinZ

3

(b) x®x ®x ®x =[OJ inZ4 (c) xEBx®x®x®x =[OJ in Zs 12. Prove or disprove: If

[aJ 0 [bJ = [OJin Z,,, then [a] = [OJ or [b] = (O].

13. Prove or disprove: If

[a] 0 [bJ =[a] 0 [cJand [a] :f: [ O ] in Z,,, then [bJ =[c].

B. 14. Solve the following equations.

(a) x1+x=[OJinZs (b) x2 +x =[O] in� (c) If pis prime, prove that the only solutions of x2+ x =[O] in � are [OJand [p - lJ. 15. Compute the following products.

(a) ([aJ ®[b])2 inZ2 (b) ([aJ ®[b])3 inZ3

[Hint: Exercise 1 l(a) may be helpful.]

(c) ([aJ® [b])5 in.ls

[Hint: See Exercise l l(c).J

(d) Based on the results of parts (a)-(c), what do you think ([a]® [b])7 is equal to inZ7?

16.

(a) Find all [a] in Zs for which the equation [a] 0 x =[IJhas a solution. Then do the same thing for

(b) Zi

II

(d) �

The Structure of ZP (p Prime) and Zn

We now present some facts about the structure of Z,, (particularly when n is prime) that will provide a model for our future work. First, however, we make a change of notation. �2012c..pe.i....m.g.u�a_..ilibJ"oi:1thl� me..-t.ar�iowtdlO£�J*I.. 0.10�..-. .... *ild.�cam•OlllJ .. ..,.....S,..._IMllBodt:.ndfl:x'..a.,..(1).:Bdladlll...,...tm -...id.1lm.:Q"��--...-a.o;,-dh:tbt�--.....--..c.g.pu--.--•riPtm-__,_��-..,.--jl���...-. ..

38

Chapter 2

Congruence in Zand Modular Arithmetic

New Notation We have been very careful to distinguish integers in Z and classes in Z,, and have even used different symbols for the operations in the two systems. By now, however, you should be reasonably comfortable with the fundamental ideas and familiar with arithmetic in Z,,. So we shall adopt a new notation that is widely used in mathemat­ ics, even though it has the flaw that the same symbol represents two totally different entities. W henever the context makes clear that we are dealing with ate the class notation

"[a]"

and write simply

"a." In "14,,

Z,,, we shall abbrevi­

for instance, we might say

6 = 0, which is certainly true for classes in � even though it is nonsense if 6 and 0 are ordinary integers. We shall use an ordinary plus sign for addition in Z,, and either a small dot or juxtaposition for multiplication. For example, in

Zs

we may

write things like 3. 4 = 2

or

4+1=0

or

4 + 4= 3.

On those few occasions where this usage might cause confusion, we will return to the brackets notation for classes.

EXAMPLE 1 In this new notation, the addition and multiplication tables for Z3 are +

0

0

0

2

2

CAUTION:

2 1

2

2

0

2

0 0

0

0

0 2

0 2

0

2

0

1

Exponents are ordinary integers-not elements of Z,,. In Z3, for instance, 24 = 2 • 2 · 2 • 2 = 1and21=2, so that24 * 21 even though 4 = 1in Z3•

The Structure of Zp When p Is Prime Some of the

�

do not share all the nice properties of Z. For instance, the product

of nonzero integers in Z is always nonzero, but in � we have2

·

3 = 0 even though

2 * 0 and 3 * 0. On the other hand, the multiplication table on page 34shows that the product of nonzero elements in Z5 is always nonzero. Indeed, Zs has a much stronger property than Z. W hen a

a

* 0, the equation

ax

= 1 has a solution in Z if and only if

= ±1. But the multiplication table for Zs shows that, for any

ax

a

* 0, the equation

= 1 has a solution in Zs; for example, x = 3 is a solution of 2x

x

=

4 is a solution of 4x

=

=

1 I.

eap,ngm.20:12�1..umiq.A:l.lliala 11--4.....,-aatn. cap.d,. � ar�io.wmlliarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ....... ftom.1M•Bam:.ndkir�.Bdbmbll_...._ ...._._q-��._.fld.....,,.dlK:l.._�._.....,.n.:...c.a.� ........ rir;bl1a-...,,,..��·...,. ... w......_..:dPLI�...-. ..

2.3 More generally, whenever

n

The Structure

of

ZP (p Prime) and Zn

39

is prime, Z,. has special properties:

Theorem 2.8 If p > 1 is an integer, then the following conditions are equivalent:*

(1)

p is prime.

(2)

For any a *

O in Zp,

the equation ax= 1 has a solution in

Zp.

(3) Whenever be=O in Zp, then b=O or c=0. The proof of this theorem illustrates the two basic techniques for proving state­ ments that involve Z,,: (i) Translate equations in Z,. into equivalent congruence statements in Z. Then the properties of congruence and arithmetic in Z can be used. The brackets notation fur elements of Z,, may be necessary to avoid confusion. Z,. dim1ly, without involving arithmetic in Z.

(ii) Use the arithmetic properties of

In this case, the b rackets notation in

Proof ofTheorem 2.8 � and [a) #-

use

the first technique. Supposepis prime

[O] in �- Then in Z, a '!/= 0

(modp) by Theorem 2.3. Hence,

a and pis a posi­ 1. Since (a,p) also divides a and p -r a, we must have (a,p)=1. By Theorem 1.2, au + pv=1 for some integers u and v. Hence, au - 1=p(-v), so that au= 1 (modp). Therefore [au]=[l] in� by Theorem 2.3. Thus [a][u]=[au]=[1], so that x = [u] is a solution of [a]x = [l].

p -r

a by

(1 )::::? (2) We

Z,. isn't needed.

the definition of congruence. Now the gcd of

tive divisor of p and thus must be eitherpor

(2) => (3) We use the second technique. Suppose ab = 0 in �· If a=0, there is nothing to prove. If a :/= 0, then by (2) there exists u E 'Ip such that au=1 . Then 0=u

•

0=u(ab)= (ua)b=(au)b=1

In every case, therefore,

we

have

(3) => (1) Back to the first

a=

0 orb

=

•

b=b

0.

technique. Suppose that

b and e are any

integers and that pI be. Then be= 0 (modp). So by Theorem 2.3,

[b][e]=[be]=[O] Hence, by (3),

we

have

in

Zp.

[b] = [O] or [e]=[OJ. Thus, b= 0 (mod p) or e = 0 b orp I c by the definition

(mod p) by Theorem 2.3, which means thatp I

of congruence. Therefore, p is prime by Theorem 1.5.

•

The Structure of� = 1 need not have a solution in Z,.. For instance, = 1 has no solution in �. as you can easily verify. The next result tells when ax=1 does have a solution in Z,,. For clarity, we use brackets notation.

When n is not prime, the equation ax the equation 2x us exactly •see page

508 in Appendix A for the meaning

of "the following conditions are equivalent" and what

must be done to prove such a statement.

.......

�2012.C....,l...Mmiq.AIRqlna-..d.MaJ"mtbll� �-ar :towballl«lapd.. 0..W�dalD.-tinl:pat;JIOOOllm:m.,.'8....,....m_ta:.:J.beBo'*:.udkx-��----­ dlMm&d.-..:my�-mmllldmmmll___...,.d!Kl. �---.�c.g..p�---ftgbttD-__,,,..:ldill.aDlil.�• lillllll��:Dgbb�...-.:lit.

...

..

40

Chapter 2

Congruence inland Modular Arithmetic

Theorem 2.9 Let a and n be integers with The equation

n>

1. Then

[a]x= [1] has a solution in Zn If and only if {a, n)= 1 in Z.

Proof" Since this is an "if and only if" statement, the proof has two parts.

we assume that the equation has a solution and show that (a, n) = 1. [w] is a solution of [a]x= [ l ], then

First If

[ a][w]= [1] [a w]= [1] aw= 1 (mod n) inZ

[Multiplication in z;J [11ieorem 2.3]

aw - 1 = kn for some integer k aw+ n(-k) = 1

[Definition of congruence] [Rearrange terms]

(a, n) by d. Since dis a common divisor of a and n, there are inte­ r ands such that dr= a and ds= n. So we have

Denote gers

aw+ drw +

n(-k) = 1

ds(-k) = 1

d(rw - sk) = 1. So

d 11. Since dis positive by definition, we must have d= 1, that is, (a, n)= 1. Now we assume that (a, n)= 1 and show that [a]x= fl]has a solu­

tion in Z,,. Actually,

we've already

done this. In the proof of (1) � (2)

(a,p)= 1. (a, n)= 1, and shows

of Theorem 2.8, the primeness of pis used only to show that From there on, the proof is valid in any Z,. when that

[a]x= [l] has a solution in Z,,.

•

Units and Zero Divisors Some special terminology is often used when dealing w ith certain equations. An ele­

a in Z,, is called a unit if the equation ax= 1 has a solution. In other words, a is b in Z,. such that ab= 1. In this case, we say that b is the inverse of a. Note that ab= 1 also says that bis a unit (with inverse a).

ment

a unit if there is an element

EXAMPLE 2 Both 2 and 8

are units

8 = 1. 8 is the inverse of 2 and 2 is the L. because 3 3= 1. So 3 is its own inverse.

in Z15 because 2

inverse of 8. Similarly, 3 is a unit in

·

•

EXAMPLE 3 Part (2) of is a unit.

Theorem 2.8 says that whenp is prime, every nonzero element of ZP

Here is a restatement of Theorem 2.9 in the terminology of units.

� 20120.-..i...m.g.A:a� llMlnrld. MaJ"llDtbl-c:iap.d. llCumd,,-ar�:tiawtdilarUtpn.. 0.1��aem.-mim.J1D111t1Dll!Hm.mAJH�fiam:l.m.•Bom:.udkir�).Bdlorilf..._.Mil ....... ..,.��dou.ad........UU,-.dlM:i. �...-..�c.g-.� -rlgbl:ID....,,,.�Oldlllll:-..,.tia:MllE......-i.._.� it.

...

.......

......

2.3

The Structure of Z P (p Prime) and Zn

41

Theorem 2.10 Let a and n be integers with n >

1. Then

[a] is a unit in Z n if and only if (a, n)

=

1

in z.

A nonzero element a of Z,. is called a zero divisor if the eq uation

nonzero solution (that is, if there is a

nonzero element

c

ax

in Z,. such that ac

=

=

0 has a

0).

EXAMPLE 4 Both 3 and 5 are zero divisors in Z 15 because 3

in L. because 2 2

divisor

·

=

•

5

=

0. Similarly, 2 is a zero

0.

EXAMPLE 5 Part (3) of Theorem 2.8 says that when pis prime,

there

are

no zero divisors in z . ,

• Exercises A. 1. Find all the units in

(a)

Z7

(b) Zs in

(c) Zg

(d)

Z10•

(b) Zs

{c) Zg

(d)

Z1o·

2. Find all the zero divisors

3. Based on Exercises

1 and 2j

make a conjecture about units and zero divisors

in�. 4. How many solutions does the equation 6x

(b)

Z;,?

=

4 have in

(c)

Zg?

5. If a is a unit andbis a zero divisor in Z,., show that ab is a zero divisor. 6. If

n is composite, prove that

there is at least one zero divisor in�. (See

Exercise 2.) 7. Without using Theorem 2.8, prove that if pis prime and ab a 8.

=

(a)

0 orb =

0. [Hint: Theorem 1.8.]

Give three examples of equations of the form ax

= bin

=

0 in Zp, then

Z12 that have no

nonzero solutions.

(b)

For each of the equations in part (a), does the equation ax

=

0 have a

nonzero solution? B. 9.

(a)

If a is a unit in�. prove that a is not a zero divisor.

( b)

If a is a zero divisor in

Z,., prove that a is not

a unit.

[Hint: Think

contrapositive in part (a).]

�2012.C....,l...Mmiq.AIRqlna-..d.MaJ"mtbll� �m'�:towllalll«la;pld.. 0..W�dalD.-tinl:pat;Joootm:a.,. ... ....,....m_to:.:J.beBo'*:.udkx-��----­ dlMm&d.-..:my�-mmillldmmmll___...,.d!Kl. ... �lmnlliag...,.n-a.C-S.�---ftgbttD__,,,..mddllklDlii.ICDlllllnl•_..,.lillll��:Dgbb�...-.:lit.

42

Chapter 2

Congruence inland Modular Arithmetic

10. Prove that every nonzero element of Z.. is either a unit or a zero divisor, but

not both. [Hint: Exercise 9 provides the proof of "not both".] 11. Without using Exercises 13 and 14, prove: If a,

b E Z.. and a is a unit, then the

equation ax = b has a unique solution in Z.,. [Note: You must find a solution for the equation and show that this solution is the only one.] 12. Let a,

b, n be integers with n > 1 and letd =(a, 11). If the equation [a]x = [b] [r] is a solution, then [ar] [b] so that ar - b kn for some integer k.]

has a solution in Z,,, prove thatd I b. [Hint: If x

=

=

=

13. Let

a, b, n be integers with n > 1. Let d (a, n) and assumed I b. Prove that [b] has a solution in Z.. as follows. =

the equation [a]x

=

(a) Explain why there are integers u,

v, a"

bl> n1 such that au+ nv = d,

a= da1, b = dhto n= dn1•

(b)

Show that each of

[uh1], [uh1+ n1], [uh1+ 2ni] , [uh1o+ 3ni], .. . , [uh1+ (d - l)ni] is a solution of [a]x = [b]. 14. Let a,

b, n be integers with n > 1. Letd= (a, n) and assumed I b. Prove that [b] has d distinct solutions in Z,, as follows.

the equation [a]x

=

(a) Show that the solutions listed in Exercise 13 (b) are all distinct.

[Hint: [r]

(b)

=

[s] if and only if n I (r - s).]

If x = [r] is any solution of [a]x= [b] , show that [r]

==

[uh1+ kn1] for some

integer k with 0 s ks d - 1. [Hint: [ar] - [auh i]= [O] (Why?), so that

n I (a(r - uh 1)). Show that n1 I (a1(r - uh1)) and use Theorem 1.4 to show that n1 I (r - uh1).] 15. Use Exercise 13 to solve the following equations.s

(a) 1 5x = 9 in .l18

(b) 25x=

10 in "145•

a ""- 0 and bare elements of Z,, and ax== b has no solutions in Z,,, prove that a is a zero divisor.

16. If

17. Prove that the product of two units in Z,, is also a unit. 18. The usual ordering of Z by< is transitive and behaves nicely with respect to

addition. Show that there is (i) if a < (ii) if

a

ordering of Z,, such that

no

b and b< .c, then a <

< b, then a+

c


c

c; for every

c

in Z,,.

[Hint: If there is such an ordering with 0 < 1, then adding 1 repeatedly to both < n - 1 by (ii). Thus 0 < n - 1 by (i). Add 1 to each side and get a contradiction. Make a similar argument when 1 < O.]

sides shows that 0 < 1 < 2 < ·

·

·

APPLICATION: Public Key Cryptography (Chapter 13) may be covered at this point if desired.

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CHAPTER

3

Rings

ALTERNATE ROUTE: H you want to cover groups before studying rings, you should read Chapters 7 and 8 now.

We have seen that many rules of ordinary arithmetic hold not only in Z but also in the miniature arithmetics Zn. You know other mathematical systems, such as the real numbers, in which many of these same rules hold. Your high-school algebra courses dealt with the arithmetic of polynomials. The fact that similar rules of arithmetic hold in different systems suggests that it might be worthwhile to consider the common features of such systems. In the long run, this might save a lot of work: If we can prove a theorem about one system using only the properties that it has in common with a second system, then the theorem is also valid in the second system. By " abstracting" the com­ mon core of essential features, we can develop a general theory that includes as special cases Z, Zno and the other familiar systems. Results proved for this general theory will apply simultaneously to all the systems covered by the theory. This process of abstraction will allow us to discover the real reasons a particular statement is true (or false, for that matter) without getting bogged down in non­ essential details. In this way a deeper understanding of all the systems involved should result. So we now begin the development of abstract algebra This chapter is just the first step and consists primarily of definitions, examples, and terminology. Systems that share a minimal number of fundamental properties with Z and Zn are called rings. Other names are applied to rings that may have additional prop­ erties, as you will see in Section 3.1. The elementary facts about arithmetic and algebra in arbitrary rings are developed in Section 3.2. In Section 3.3 we consider rings that appear to be different from one another but actually are "essentially the same" except for the labels on their elements.

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44

Chapter 3

Ill

Rings

Definition and Examples of Rings

We begin the process of abstracting the common features of familiar systems with this definition:

Definition

A ring is a nonempty setRequippedwith two operations* (usually written as addition and multiplication) that satisfy the fo llowi ng axioms. For all a, b,cER: 1. If a ERandbER, then a+ bER.

[Closure for addition]

2. a+ (b+ c) = (a+ b) + c.

[Associative addition]

3. 8+ b = b+ 8.

[Commutative addition]

4. There is an element On in R such

[Additive identity

that a+ OR = a = On + a for every a ER.

or zero element]

5. For each a ER, the equation

a+ x =On has a solution in R.t 6. lfaERandbER,thenabER.

[Closure for multiplication]

7. a(bc)

[Associative multiplication]

=

(ab)c .

8. a(b+ c)

(a +

=

b)c =

ab + ac and

[Distributive laws]

ac + be.

These axioms are the bare minimum needed for a system to resemble Z and Zn. But Z and Zn have several additional properties that are worth special mention:

Definition

A commutative ring is a rin g Rthat satisfies this axiom: 9. ab = bafor all a, b ER.

Definition

A ring with identity is a ring axiom:

(Commutative multiplication]

R that contains an element 1n satisf ying this [Multiplicative identity]

*"Operation" and "closure" are defined in Appendix trhose who have already read Chapter

B.

7 should note that Axioms 1-5 simply say that a ring is an

abelian group under addition.

.......

....

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3.1

Definition and Examples of Rings

45

In the following examples, the verification of most of the axioms is left to the reader.

EXAMPLE 1 With the usual addition and multiplication, Z (the integers)

R (the real numbers)

and

are commutative rings with identity.

EXAMPLE 2 The set Z11, with the usual addition and multiplication of classes, is a commuta­ tive ring with identity by Theorem

2.7.

EXAMPLE 3 Let Ebe the set of even integers with the usual addition and multiplication. Since the sum or product of two even integers is also even, the closure axioms (1 and

6) hold. Since 0 is an even integer, Ehas an additive identity

element (Axiom 4). If a is even, then the solution of

a+

x =

0 (namely-

also even, and so Axiom 5 holds. The remaining axioms (2, 3, hold for

all integers and, therefore, are true whenever a, b, care even.

Consequently, Eis a commutative ring. Edoes not have because no integer

a) is

7, 8, and 9)

even integer e has the property

that

ae = a

an

=

identity, however,

ea for every even

a.

EXAMPLE 4 The set of odd integers with the usual addition and multiplication is not a ring. Among other things, Axiom 1 fails: The sum of two odd integers is not odd.

Although the definition of ring was constructed with 7L and 7L11 as models, there are many rings that aren't at all like these models. In these rings, the elements may not be numbers or classes of numbers, and their operations may have nothing to do with "ordinary" addition and multiplication.

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46

Chapter 3

Rings

EXAMPLE S The set T= {r, s, t, z} equipped with the addition and multiplication defined by the following tables is a ring: +

z

z

z

r

s

r

r

z

t

s

s

s

t

z

t

s

r

r

s

z

r

s

z

z

z

z

z

r

z

z

r

r

r

z

z

s

s

z

z

z

You may take our word for it that associativity and distributivity hold (Axioms 2, 7, and 8). The remaining axioms can be easily verified from the operation tables above. In particular, they show that Tis closed under both addition and multiplication (Axioms 1 and

6) and that addition is commuta­

tive (Axiom 3). The element z is the additive identity-the element denoted OR in Axiom 4. It be­ haves in the same way the number 0 does in Z (that's why the notation 0R is used in the axiom), but z is not the integer 0-in fact, it's not any kind of number. Nevertheless, we shall

call z the "z.ero element" of the ring T.

In order to verify Axiom 5, you must show that each of the equations r+x=z

s+x=i

has a solution in T. This is easily seen

z+x=z

t +x =z

to be the case from the addition table; for

example, x= r is the solution of r + x = z because r + r = z. Finally, note that Tis not a commutative ring; for instance, rs= rand sr = z, so that rs -:¢: sr.

EXAMPLE 6 Let M(IR) be the set of all 2 X 2 matrices over the real numbers, that is, M(IR) consists of all arrays where a, b, c, dare real numbers. Two matrices are equal provided that the entries in corresponding positions are equal; that is,

(; �) (; �) =

R>r example, 0

1

) ( =

if and only if

2 + 2 1

-

4

�)

b ut

a = r, b = s, c = t, d =

G

3 2

) ( *

3

5

1

2

)

u.

.

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....... my�CDlllllll.dmmoot.......,.�... �--....,m-..c.pp� ...... -rigbt10__,,.. ...... QXllslll:lll..,. .... il�:ds:f:lb� ........

3.1

Definition and Examples of Rings

47

Addition of matrices is defined by

( )

)

(a' + d c'

a

b'

b

c

=

(a+a'

d'

c+c'

)

b + b'

d+d'.

R>r example,

) (

-2

1

+

4

7)

6

0

=

(

-2+ 7) = ( 7

3+4

l+O

5+6

5)

11

1 ·

Multiplication of matrices is defined by (a c

)

b)(w

d

x

y

R>r example,

=

-5) 7

(2.

=

0

(aw +by cw+dy

z

•

)

ax+ bz .

ex+dz

1 +3. 6

2(-5)+3·7 )

1+(-4)6

0(-5)+( -4)7

(-� -� }

Reversing the order of the factors in matrix multiplication may produce a different answer, as is the case here:

)(

-5

7

2

()

3) ( =

-4

1. . 2+( -5)0

1. 3+(-5)(-4))

6. 2 + 7. 0

6. 3 + 7(-4)

)

23 -10 . So this multiplication is not commutative. With a bit of work, you can verify that M(�) is a ring with identity. T he zero element is the 1..ero matrix

which is denotedOandX=

(

-a

-c

-b'\. - d} is a so1ut10n of .

Weclaim thatthemultiplicative identity e lement(Axiom l O)isthematrix/ = To prove this claim, we first multiply a typical matrix in

(

a

c

b)( l

d

0

) (

0 = 1

a·

1+b·0

c·l+d·O

G ) 0 I

.

M(R) on the right by I:

a·0+b·1) c·O+d·l

=

( J a

c

b ·

d

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48

Chapter 3

Rings

Since multiplication is not commutative here, we also need to check left multiplication by las well;

b

) ( •a =

d

1 + 0 ·c O•a+I·c

l·b+O· tf\ O·b + l ·d)

-(

a c

This proves that I satisfies Axiom 10. * Consequently, I is called the identity matrix.

Note that the product of nonzero elements of

M(R) may be the :zero element; for

example,

6·2 -9)6 (4(-3) 2(-3) + 3·2 +

=

4(-9) + 6•6 2(-9) + 3·6

0

) ( =

0

EXAMPLE 7 If R is a commutative ring with identity, then

M(R) denotes the. set of

all

2 X 2 matrices with entries in R. With addition and multiplication defined as in Example 6,

M(R) is a noncommutative ring with identity, as you can read­

ily verify. For instance, M(Z) is the ring of 2 X 2 matrices with integer entries, M(O) the ring of 2 X 2 matrices with rational number entries, and M(lL,J the ring of

2 X 2 matrices with entries from.l,,.

EXAMPLE 8 Let The the set of all functions from IR to R, where R is the set of real numbers. As in calculus,/+

(f + g)(x)

=

g and fg are the functions defined by

f(x)

+

g(x)

and

(fg)(x)

=

f(x)g(x).

You can readily verify that Tis a commutative ring with identity. The zero ele­ ment is the function h given by h(x) 0 for all x E !R. The identity element is the =

function

e

given by

e(x)

=

1 for all

x E IR. Once again the product of nonzero

elements of Tmay tum out to be the zero element; see Exercise

36.

We have seen that some rings do not have the property that the product of two nonzero elements is always nonzero. But some of the rings that do have this property, such as .l, occur frequently enough to merit a title.

Definition

An integral domain is a commutative ring R with identity 1R * OR that satisfies this axiom: 11. Whenever a, bER and ab= OR, then a= OR orb= OR.

•checking a possible identity element under both right and left multiplication is essential. There are rings in which an element acts like an identity when you multiply on the right, but not when you multiply on the left. See Exercise

11.

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3.1

Definition and Examples of Rings

49

The condition lR if:. OR is needed to exclude the zero ring (that is, the single-element ring {OR}) from the class of integral domains. Note that Axiom 11 is logically equiva­ lent to its contrapositive.* Whenever a '# OR and b

*

OR, then ab * OR.

EXAMPLE 9 The ring z of integers is an integral domain. If pis prime, then zp is an integral domain by Theorem 2.8. On the other hand, Z6 is not an integral domain because 4

·

3 = 0, even though 4 *

0 and 3 * 0.

You should be familiar with the set fractions

a/b with a, b EZ

Q

of rational numbers, which consists of all

and b '¢: 0. Equality of fractions, addition, and mul tiplica­

tion are given by the usual rules: a

r

b

s

if and only if

as= hr a

-

b

·

c

-

d

�

ac

-

bd

Q is an integ ral domain. But Q has an additional property that does not hold in Z: Every equation of the form ax = 1 (with a * 0) has a solution in

It is easy to verify that

Q. Therefore, Q is an example

Definition

A field

of the next definition.

is a commutative ring

R with

identity 1,q :f. OR that satisfies this

axiom: 12. For each a* 08 in

R, the equation ax= 1R has a solution in R.

Once ag ain the condition IR * OR is needed to exclude the zero ring. Note that Axiom 11 is not mentioned explicitly in the definition of a field. However, Axiom 11 does hold in fields, as we shall see.in Theorem 3.8 below.

EXAMPLE 10 The set� of real numbers, with the usual addition and multiplication, is a field. If p is a prime, then zp is a field by Theorem 2.8.

EXAMPLE 11 The set IC of complex numbers consists of all numbers of the form a + bi, where a, b E� and i2 = -1. Equality in IC is defined by a+ bi= r + si

if and only if

a= rand b

=

s.

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50

Chapter 3

Rings

The set C is a field with addition and multiplication given by

(a+ bt)+ (c +di) =(a+ c) +(b + d)i (a+ b1)(c+ di) =(ac - bd) + (ad+ bc)i. The field R of real numbers is contained in C because � consists of all complex numbers of the form a+ Oi. If a+ bi * (a + bl)x = 1 is x =c+ di, where

c =a/(a2+ /l-) ER

0 in C, then the solution of the equation

d = -b/(a2+ b'-)E IR (verify!).

and

EXAMPLE 12 Let K be the set of all 2 X 2 matrices of the form

)

b a ' where a and b are real numbers. We claim thatK is a field. For any two matrices in K,

(

a -b

(

a -b

) (

( a+ -b -d c} d\ ( ac b) ( =

b

d'\

c

+

a

c

=

-

- bd

c

a

-d

·

d

b +·d'\ a+ c } ad

-ad - be

c}

+be)

ac - bd

·

In each case the matrix on the right is in K because the entries along the main diagonal (upper left to lower right) are the same and the entries on the opposite diagonal (upper right to lower left) are negatives of each other. Therefore, K is closed under addition and mu ltiplication. K is commutative because

) (

b a

-

ac-bd d - be

-a

) (

ad+bc ac - bd

-

a -b

Clearly, the zero matrix and the identity matrix I are in K. If

A=

(

a -b

)

b a

is not the zero matrix, then verify that the solution of

X=

(aid bid

-bid aid

)

EK,

AX =I is

where

d = a2 +b1•

W henever the rings in the preceding examples are mentioned, you may assume that addition and multi plication are the operations defined above, unless there is some specific statement to the contrary. You should be aware, however, that a given set (such as Z) may be made into a ring in many different ways by defining different addition and multiplication operations on it. See Exercises 17 and 22-26 for examples.

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3.1

Definition and Examples of Rings

61

Now that we know a variety of different kinds of rings, we can use them to produce new rings in the following way.

EXAMPLE 13 Let Tbe the Cartesian product Zr, X Z, as defined in Appendix B. Define addition in Tbythe rule

(a, z) +(a', z') =(a+ a', z + z'). The plus sign is being used in time ways here: Jn the first coordinate on the right-band side of the equal sign,+ denotes addition in Zr,; in the second coordinate+ , denotes adlition in Z; the+ on the left of the equal sign is the addition in Tthat is being defined.

Si.tx:e Zr, is a ring and a, a' E Zr,, the first coordinate on the right, a+

a', is in Zr,. Similarly z+ z' E Z. Therefore, addition in Tis closed. Multipli:ation is defined similarly: (a, z)(a', z') = (aa', zz'). (3, 5) + (4, 9) (3-+ 4, 5 + 9) (1, 14) and (3, 5)(4, 9) (3 4, 5 9) = (0, 45). You can readily verify that Tis a commutative ring with identity. The zero element is (0, 0), and the multiplicative identity is (1, 1). What For example, •

=

=

=

•

was done here can be done for any two rings.

Theorem 3.1 Let R and S be rings. Define addition and multiplication on the Cartesian product R x S by

(r, s) + (r', s') = (r + r', s + s')

and

(r,

s)(r', s')

=

(rr', ss').

Then Rx Sis a ring. If Rand Sare both commutative, then so is Rx S. If both Rand S have an identity, then so does R x S.

Proof• Exercise 33.

•

Subrings If R is a ring and S is a subset of R, then Smay or may not itself be a ring under the operations in R. In the ring Z of integers, for example, the subset E of even integers is a ring, but the subset 0 of odd integers is not,

as

we saw in Examples 3 and 4. When

a subset S of a ring R is i tself a ring under the addition and multiplication in R, then we say that Sis a subring of R.

EXAMPLE 14 Z is a subring of the ring Q of rational numbers and Q is a subring of the field R of all real numbers. Since Q is itself a field, we say that Q is a subfield of Ill.

Similarly, ll is a subfield of the field C of complex numbers.

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62

Chapter 3

Rings

EXAMPLE 15 The matrix rings M(Z) and M(O!) in Example 7 are subrings of

M(lll).

EXAMPLE 16 The ring Kin Example 12 is a subring of

M(lll).

EXAMPLE 17 Let Tbe the ring of all functions from Ill to R in Example 8. Then the subset S consisting of all continuous functions from R to R is a subring of T. To prove this, you need one fact proved in calculus: T he sum and product of continuous functions are also continuous. So Sis closed under addition and multiplication (Axioms l and 6). You can readily verify the other axioms.

Proving that a subset S of a ring Ris actually a subring is easier than proving directly that Sis a ring. For instance, since a + b

=

b + a for all elements of

R, this fact is also true

when a, b happen to be in the subset S. Thus Axiom 3 (commutative addition) automati­ cally holds in any subset S of a ring. In fact, to prove that a subset of a ring is actually a subring, you need only verify a few of the axioms for a ring, as the next theorem shows.

Theorem 3.2 Suppose that Ris a ring and that S is a subset of R such that (i) Sis closed under addition (if a,

b ES, then a+ bES);

(ii) Sis closed under multiplication (if a, bES, then ab ES); (iii} OR ES; (iv) If a ES, then the solution of the equation

a+ x

= OR is in

S.

Then S is a subring of R. Note condition (iv) carefully. To verify it, you need not show that the equation a+ x = OR has a solution-we already know that it does because Ris a ring. You need only show that this solution is an element of S (which implies that Axiom 5 holds for S).

Proof of Theorem 3.2 ... As noted before the theorem, Axioms 2, 3, 7, and 8 hold for all elements of R, and so they necessarily hold for the elements of the subset S. Axioms 1, 6, 4, and 5 hold by (i}-(iv).

•

EXAMPLE 18 The subset S = {O, 3} of� is closed under addition and multiplication

(0 + 0

= O; 0 + 3 = 3; 3 + 3 = O; similarly,

0 0 •

=0 =0

·

3; 3

•

3 =

3). By

the

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3.1

Definition and Examples of Rings

53

definition of S we have 0 ES. Finally, the equation 0 + x = 0 has solution x = 0 ES, and the equation 3 + x = 0 has solution x = 3 ES. Therefore, Sis a subring of z6 by Theorem 3.2.

EXAMPLE 19 Let S be the subset of M(lll) consisting of all matrices of the form

Then Sis closed under addition and multiplication because

r (ab Oc) + (s' 0t) - (a+ b+s

0+0 - a+r O c+t b+s c+t

)es

) (

(ab O )(' 0) (hr+cs ct) ar

c

s

The identity matrix is in S (let

(: )

t

-

0

0

.

c

and

ES.

a = 1, b = 0, c = 1) and the solution of 0

-c

)

ES.

Hence S is a subring by Theorem 3.2.

EXAMPLE 20 The set that

Z[Vl]

=

{a + bVl I a, b EZ} is a subring of

(a + bVl}(c + dv'2)

= ac

=

So

R You can easily verify

+adVl + bc'\/2+bdVl v'2 ·

(ac + 2hd) + (ad+hc)\12) e Z[V2].

Z [\12] is closed under multiplication. See Exercise 13 for the rest of the proof.

• Exercises A. 1. Th e following subsets of Z (with ordinary addition and multiplication) satisfy

all but one of the axioms for a ring. In each case, which axiom f ails? (a) The set S of all odd integers and

0.

(b) The set of nonnegative integers. 2. Let R

=

{O, e,

,b c} with addition and multiplication defined by the tables on

page 54. Assume associativity and distributivity and show that R is a ring with

identity. Is R commutative? Is R a field?

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54

Chapter 3

Rings

3.

+

0

e

b

c

0

0

e

b

c

0

e

b

c

0

0

0

0

0

e

e

0

c

b

e

0

e

b

c

b

b

c

0

e

b

0

b

b

0

c

c

b

e

0

c

0

c

0

c

Let F= {0, e, a, b} with operations given by the following tables. Assume associativity and distributivity and show that Fis a field. +

0

0

0

e

a

b

e

e

0

b

a

a

a

b

0

b

b

a

e

e

0

e

0

0

0

0

0

e

0

e

a

b

e

a

0

a

b

e

0

b

0

b

e

a

b

a

a

b

4.

Find matrices A and C in M(ll) such that AC= 0, but CA =#- 0, where 0 is the zero matrix. [Hint: Example 6.]

5.

Which of the following six sets are subrings of M(R)? Which ones have an identity? (a) All matrices of the form (b) All matrices of the form (c) All matrices of the form (d) All matrices of the form (e) All matrices of the form (f) All matrices of the form

6.

(� (� (: (: (� (�

�) with rEQ.

!) �) �) �) �)

with a, b, c EZ. with a, b, c ER with a E II!. with a ER with a E ll.

(a) Show that the set R of all multiples of 3 is a subring of Z. (b) Let k be a fixed integer. Show that the set of all multiples of k is a subring of Z.

7.

Let K be the set of all integer multiples of v2, that is, all real numbers of the form nVl with n EZ. Show that K satisfies Axioms 1-5, but is not a ring.

8.

Is the subset {l,-1, i, -i} a subring of Cl

9.

Let R be a ring and consider the subset R* of RX R defined by R* = {(r, r) Ir ER}. (a) If R

=

Z6, list the elements of R*.

(b) For any ring R, show that R* is a subring of R

x

R.

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3.1 10. Is S= {(a, b)

I a+ b =O}

11. Let Sbe the subset of

Prove that Sis a ring.

(b)

Show thatJ = every A in S).

(c)

1

( ) 0

O

Show thatJ is not a

JB 1" B.

55

a subringof Z X Z? Justify your answer.

M(R)

(a)

Definition and Examples of Rings

1

consistingof all matrices of the form

is a

(: :)

.

right identity in S(meaningthat AJ =A for

left identity in Sby

findinga matrix B in S such that

For more information about S , see Exercise 4 . 1 {a + bi la, bEZ}. Show thatZ[i] is a subringof C.

12. Let Z[1] denote the set

13. Let Z Vl denote the set of

[Ii.

(

)

[See Example 20.]

{a+ bVl I a, bEZ}.

14. Let Tbe the ringin Example 8. Let S = subringof T.

Show that Z \12 is a subring

(

{/E Tlf(2)

=

)

O}. Prove that Sis a

15. Write out the addition and multiplication tables for

(a) Z2 16. Let A

X Z3

(b) Z2

C !)

=

and 0

x Z2

(� �)

=

such that AB = 0.

(a) (b)

(c) Z3

X Z3

in M(R). Let S be the set of all matrices B

List three matrices in S. [Many correct answers are possible.] Prove that Sis a subring of

M(R).

[Hint: If B and Care in S, show that

B + C and BC are in Sby computingA(B + C) and A(BC).]

17. Define a new multiplication inZ by the rule: ab = 0for all

a, b,EZ

Show that

with ordinary addition and this new multiplication Z , is a commutative ring.

18. Define a new multiplication in Z by the rule: ab

=

I for all

a, b,EZ. With

ordinary addition and this new multiplication , isZ is a ring?

19. Let S:

{a, b, c} and let P(,S) be the set of P(,S) as follows:

elements of

S={a, b,

A=

{a};

D = { a, b};

c};

B = {b};

C=

M

+

N= (M - N)

U

E={a, c};

{c};

Define addition and multiplication in

all subsets of S; denote the

0=

P(S) by

(N - M)

F= {b, c};

0.

these rules:

and

MN=MnN.

Write out the addition and multiplication tables for P(S). Also, B. 20. Show that the subset R an identity?

:

{O, 3, 6, 9, 12,

21. Show that the subset S= {O, 2, identity?

see

Exercise 44.

15} of Zl8 is a subring. Does R have

4, 6, 8}of Zio is a subring.

Does Shave an

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56

Chapter 3

Rings

22. Define a new addition ® and multiplication 0 on Z by a® b

:=:

a+ b

and

1

-

a0 b

=

a + b - ab,

where the operations on the right-hand side of the equal signs are ordinary addition, subtraction, and multiplication. Prove that, with the new operations ® and 0, Z is an integral domain.

23. Let Ebe the set of even integers with ordinary addition. Define a new multiplication *on Eby the rule "a*b

=

ab/2" (where the product on the

right is ordinary multiplication). Prove that with these operations Eis a commutative ring with identity.

24. Define a new addition and multiplication on Z by a® b

=

a+b

-

and

1

a0b

ab

=

-

(a + b) + 2

Prove that with these new operations Z is an integral domain.

25. Define a new addition and multiplication on Q by r® s = r + s + 1

and

Prove that with these new operations

r 0 s = rs + r + s.

Q is a commutative ring with identity. Is

it an integral domain?

26. Let L be the set of positive real numbers. Define a new addition and multiplication on L by a® b = ab

and

a® b

=

J<>rP.

(a) Is La ring under these operations? (b) Is La commutative ring? (c) Is La field?

27. Let S be the set of rational numbers that can be written with an odd denominator. Prove that Sis a subring of Q but is not a field. 28. Let p be a positive prime and let R be the set of all rational numbers that can be written in the form r/p' with r, iEZ, and i � 0. Note that Z � R because each n EZ can be written as n/JJ°. Show that Ris a subring of Q. 29. The addition ta ble and part of the multiplication table for a three-element ring are given below. Use the distributive laws to complete the multiplication table. +

r

s

r

r

s

s

s r

r

s

r

r

r

r

s

r

s

t

r

r

30. Do Exercise 29 for this four-element ring: +

w

x

y

z

w

x

y

z

w

w

x

y

z

w

w

w

w

w

x

x

y

z

w

x

w

y

y

y

z

w

x

y

w

w

z

z

w

x

y

z

w

w

y

�20-l2C.....1-:*g.Al.1UB11Da--d..lb;J"DDtt.� �ar�ia.1'tdlleckaJllfl. 0..'ID�dBID.-aiird.:Pmt;J�a.J'ile......._thim.1bll•Bodl:��).:lidlmW...W-t..

-...d.'lm:mJ"��._alll.....UO,.dlk1.'1119�..,..�Cmg.Qei...mos--a..:rigM1D__,_mdllllli:lml.romim•..-tilll9V....:DafUllWlrictims ... -..n:11t.

3.1

Definition and Examples of Rings

31. A scalar matrix in M(IR) is a matrix of the form

numberk.

(� �)

57

for some real

(a) Prove that the set of scalar matrices is a subring of M(IR). (b) If Kis a scalar matrix, show that KA= AKfor every A

in M(�).

(c) If Kis a matrix in M(R) such that KA = AK for every A in M(IR), show that Kis a scalar matrix. fact that KA

[Hint: If K

= AKto show that b

argument with A

=

(� �) )

(: !).

0 and

t o show that

c=

a=

let A

=

G �).

Use. the

0. Then make a similar

d.]

{a E R I ar = ra for every r ER}. In other R that commute with every other element of R. Prove that Z(R) is a subring of R. Z(R) is called the center of the ring R. [Exercise 31 shows that the center of M(IR ) is the subring of scalar

32. Let

R be a ring and let Z(R

=

=

=

words, Z(R) consists of all elements of

matrices.]

33. Prove Theorem

3.1.

34. Show that M(Z2)(all

2 2

X matr ices with entries in Z2) is a 16-element noncommutative ring with identity.

35. Prove or disprove:

(a) If Rand Sare integral domains, then R (b) If R and Sare fields, then R

X Sis an integral domain.

X S is a field.

36. Let T be the ring in Example 8 and let f, g be given b y

ifx :<;;2 ifx >

2

g(x)

{2 =

x

ifx:s;2 ifx > 2.

0

Show that/, gE Tand that/g =Or. Therefore T is not an integral domain. 37.

(a) If Ris a ring, show that the ring M(R ) of Ris a ring.

all 2 X 2

matrices with entries in

(b) If R has an identity, show that M(R) also has an identity. 38. If R is a ring and a ER, let AR=

of

{r ER I ar= OR}. Prove that AR is a subring R. AR is called the right annihilator of a. {For an example, see Exercise 16 in

which the ring Sis the right annihilator of the matrix A.]

Q(V2) = (r + sv'2 I r, s E Q}. Show that O(v'2) is a subfield of R. [Hint: To show that the solution of (r + M lx= 1 is actually in O(V2), multiply 1/ ( r + sv'2) by (r M)/(r

39. Let

-

- s\12).]

40. Let dbe an integer that is not a perfect square. Show that

{a +

bW I a, b E Q} is a subfield of

C.

O(\ld)

[Hint: See Exercise

..

39.]

=

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......

tm

58

Chapter 3

Rings

11.

41. Let S be the ring in Exercise

(a)

Verify that each of these matrices is a right identity in S:

(� k) (b)

1

I

2

2

x+y=l. x

+y

= 1, show that

.7 .3 'and

)

{x \y

Prove that the matrix

(c) If

'

)

(·7.3 x y

-�}

(-12

is a right identity in S if and only if

G ;)

is not a left identity in S.

42. A division ring is a (not necessarily commutative) ring R with identity I R¢ OR that satisfies Axioms

11 and 12 (pages 48 and 49). Thus a field is a

commutative division ring. See Exercise 43 for a noncommutative example.

b are nonzero elements of R. (a) If bb = b, prove that b = IR. [Hn i t: Let be the solution of bx = lR and Suppose R is a division ring and

a,

u

no te that bu

(b)

= b2u.]

If u is the solution of the equation ax= IR, prove that u is also a solution

of the equation xa

=

IR. (Remember that R may not be commutative.)

[Hn i t: Use part (a) with

b=

ua.

]

43. In the ring M(C), let

1=

G �)

i

= (i0

�)

J •

-1

0 = (-1

�)

k= e

�)

The product of a real number and a matrix is the matrix given by this rule:

r(�

;) (: :) =

The set Hof real quaternions consists of all matrices of the form al

+ bj + cj +dk

=

( ) +b(;

a

1

0

o O\ 1) +d(o.1• 0;) ( + 0 -i} 1 0 0) + (-c0 0 +(di0 di0 -bi)

o 1

c

_

)

c

-

where

(a)

(b)

a,

b,

c,

bi ( a+ + di -c

c a

)

+di) - bi '

and dare real numbers.

Prove that

jl = j2 = kl = -1

ij =-ji = k

jk =-kj =

ki =-ik = j.

i

Show that His a noncommutative ring with identity.

eap,ngm.20:12�1..Mmiq.A:l.llialaa--&.....,-aatn.t:IDJllilrd,. llC...t,,ar�io.wtdliarl:a,_,. 0..1"�dpll.-mkd.�1r1C11Hm.�M .ftom.1M•Bam:.ndkir�.Bdbmbll_...._ ........ q-�� fld.�dlN:t Cl'Na!S---.�C...� rigbtlD...,,,..��-...,.--il......_,.:dPLI� ........

...

...

.......

......

3.2

Basic Properties of Rings

59

[Hint: If M al + bi+cj + dk, then verify that the solution of the equation Mx = 1 is the matrix tal - tbi - tcj - tdk, where t 1f(a2 + I} + c2 + d2).]

(c) Show that His a division ring (defined in Exercise 42).

=

=

(d) Show that the equation

x2 -1 has infinitely many solutions in H. [Hint: Consider quaternions of the form 01 +bi+cj - dk, where b2 + c2 + d2 = l.] =

44. Let S be a set and let P<..S) be the set of all subsets of S. Define addition and multiplication in P(S) by the rules

M + N = (M - N) U (N - M) (a)

and

MN= Mn N.

Prove that P(S) is a commutative ring with identity. [The verification of additive associativity and distributivity is a bit messy, but an informal discussion using Venn diagrams is adequate for appreciating this example. See Exercise

19 for a special case.]

(b) Show that every element of P(S) satisfies the equations x

+x = Or
x2 = x and

C. 45. Let C be the set Iii X R with the usual coordinatewise addition (as in Theorem

3.1) and a

new multiplication given by

(a, b)(c, d) = (ac - bd, ad+be) Show that with these operations C is a field. 46. Let

I

r

ks + I for some k with , (s - l)r} of .Z,. is a ring with

and s be positive integers such that r divides

s k 181

r.

Prove that the subset

{O, r, 2r, 3r,

. •

.

identity ks+ l under the usual addition and multiplication in Z,4• Exercise 21 is a special case of this result.

APPLICATION: Applications of the Chinese Remainder Theorem (Section 14.2) may be covered at this point if desired.

Ill

Basic Properties of Rings

When you do arithmetic in Z, you often use far more than the axioms for an integral domain. For instance, subtraction appears regularly,

as

do cancelation and the various

rules for multiplying negative numbers. We begin by showing that many of these same properties hold in every ring.

Arithmetic in Rings Subtraction is not mentioned in the axioms for a ring, and we cannot just assume that such an operation exists in

an

arbitrary ring. If we want to define a subtraction

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60

Chapter 3

Rings

operation in a ring, we must do so in terms of addition, multiplication, and the ring axioms. The first step is

Theorem 3.3 For any element a in a ring R, the equation a + x =OR has a unique solution.

Proof ... We know that a+ x = 0 R has at least one solution u by Axiom 5. If

vis

also a solution, then a+u=OR and a+v=OR, so that

v=�+v=�+aj+v=�+aj+v=u+�+�=u+�=L Therefore, u is the only solution.

•

We can now define negatives and subtraction in any ring by copying what happens in familiar rings such as Z. Let R be a ring and aE R. By Theorem 3.3 the equa­ tion a+ x =OR has a unique solution. Using notation adapted from Z, we denote this

unique solution by the symbol "-a." Since addition is commutative,

-a is the unique element of R such that a +(-a) = OR = (-a) + a. In familiar rings, this definition coincides with the known concept of the negative of an element. More importantly, it provides a meaning for "negative" in any ring.

EXAMPLE 1 In the ring z6, the solution of the equation 2 + x = 0 is 4, and so in this ring -2=4. Similarly, -9 = 5 in Z14 because 5 is the solution of 9 + x = 0. Subtraction in a r ing is now defined by the rule b

- a means b +(-a).

In Z and other familiar rings, this is just ordinary subtraction. In other rings we have a new operation.

EXAMPLE 2

In�wehavel- 2= 1+(-2)= 1+4=5. In junior high school you learned many computational and algebraic rules for deal­ ing with negatives and subtraction. The next two theorems show that these rules are valid in any ring. Although these facts are not particularly interesting in themselves, it is essential to establish their validity so that we may do arithmetic in arbitrary rings.

Theorem 3.4 If a + b = a

+c

in a ring R, then

b = c.

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B as ic Properties of Ri ngs

3.2

61

Proof"' Adding - a to both sides of a + b=a + c and then using associativity and negatives show that

-a + (a + b) = -a + (a + c) (-a

+

a) + b= (-a + a) + OR + b =OR+ b = c.

c

c

•

Theorem 3.5 For any elements a and b of a ring R, (1) a ·OR= OR=OR· a. In particular, OR· OR= OR. (2) a(-b) = -ab

and

(-a)b = -ab.

(3) -(-a)=a. (4) -(a + b} =(-a) + {-b). (5) - {a - b} = -a + b. (6) (-a)(-b) =ab. If R has an identity, then

Proof"' (1) Since OR + OR= OR, the distributive law shows that a· OR + a ·OR = a(OR + ORl

=

a OR =a· OR + OR. ·

Applying Theorem 3.4 to the first and last parts of this equation shows that a·OR= OR. The proof that

OR· a= OR is similar.

(2) By definition, -ab is the unique solution of the equation ab +

=OR,

x

and so any other solution of this equation must be equal

to -ab. But x =a(-b) is a solution because, by the distribution law and (1),

ab + a(_-b) = a[b + (-b)] = a[OR ] =OR. Therefore,

a(_-b) = -ab. The other part is proved similarly.

(3) By definition,

a is a

-(-a) is the unique solution of (-a) + x =0R· But

solution of this equation since

(-a) + a =OR. Hence, -(-a)=a

by uniqueness. (4) By definition, -(a+ b) is the unique solution of (a + b) + x = OR, but (-a) + (-b) is also a solution, because addition is commutative, so that

(a+ b) + [(-a) + (-b)] =a + (-a} + b + (-b) =

OR

+

OR

=OR.

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62

Chapter 3

Rings

Therefore, -(a+ b) =(-a)+ (-b) by uniqueness. (5) By the definition of subtraction and (4) and (3),

-(a - b) =-(a+(-b)) =(-a)+(-(-b)) =-a+ b. (6) ( -a)(-b) = -(a (-b)) [By th£ second equation in (2), with -h in place of b] -(-ab) =ab

[By th£first equation in (2)]

=

[By (3), with ab in place of a]

(7) By (2), (-1.R)a

=

-(lRa)= -(a) = -a.

•

When doing ordinary arithmetic, exponent notation is a definite convenience, as is its additive analogue (for instance, a+ a +a =3a). We now carry these concepts over to arbitrary rings. If R is a ring, a ER, and n is a positive integer, then we define

a•= aaa ••·a

(n factors).

It is easy to verify that for any a E R and positive integers

m

and n,

and If R has an identity and a * OR, then we define r./l to be the element lR. In this case, the exponent rules are valid for all m, n 2!: 0. If R is a ring, a ER, and n is a positive integer, then we define

= a + a + a + • ·+a. (n summands) =(-a)+(-a)+(-a)+···+(-a). (nsummands) na

-na

•

Finally, we define Oa = 0R· In familiar rings this is nothing new, but in other rings it gives a meaning to the "product" of an integer n and a ring element a.

EXAMPLE 3 Let R be a ring and a, b ER. Then

(a + b)2 =(a+b)(a +b) =a(a+ b)+b(a+b) =aa+ab+ha+bh =ti- +ab+ ha+b1. Be careful here. If ab * ha, then you cant combine the middle terms. If R is a com­ mutative ring, howevei; then ab ha and we have the familiar pattern =

(a+b)2=rl-+ab+ba+�=ri-+ab+ ab+�=ri-+�+� For a calculation of (a +b)" in a commutative ring, with n > 2, see the Binomial Theorem in Appendix E.

It's worth noting that subtraction provides a faster method than Theorem 3.2 for showing that a subset of a ring is actually a subring.

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..

..

3.2

Basic Properties of Rings

63

Theorem 3.6 Let S be a nonempty subset of a ring R such that (1)Sis closed under subtraction (if a, bES, then a - bES}; (2)S is closed under multiplication (ifa, bES, then abES). Then Sis a subring of R.

Proof"' We show that S satisfies conditions (i)-(iv) of Theorem 3.2 and hence is a subring. The conditions will be proved in this order: (ii), (iii), (iv), and (i). (ii) Hypothesis (2) here is identical with condition (ii) of Theorem 3.2 . Hence, S satisfies condition (ii). (iii) Since Sis nonempty, there is some element c with (with a= c and b = c), we

see that c

cES.

- c =OR is in S.

Applying (1)

Therefore, S

satisfies condition (iii) of Theorem 3.2. (iv) If a is any element of S, then by (1), OR -a is

the solution of a +

x = OR,

- a=

-ais also in S. Since

condition (iv) of Theorem 3.2 is

satisfied. (i) If a, bE S, then -b is in S by the proof of (iv). By (1), a a + b is in S. So S satisfies condition Therefore, Sis a subring of R by Theorem 3.2.

(i) of Theorem

-

( -b) =

3.2.

•

Units and Zero Divisors Units and

zero

divisors in Z,. were introduced in Section 2.3. We now carry these con­

cepts over to arbitrary rings.

Definition

An element a in a ring R with identity is called a unit if there exists u ER such that au= 1n= ua. l n thiscase theelemerit u iscalled the(multiplica­ tive) inverse of a and is denoted

.,-1•

EXAMPLE 4 The only units in Z

are

1 and -1.

EXAMPLE 5 By Theorem 2.10, the units in Z15 are 1, 2, 4, 7, 8, 11, 13, and 14. For instance, 8 = 1, so 2-1 = 8 and s-1 = 2.

2

·

..........

..

..flBcd:udhr�l).Bdlaftlll........ ....:Dgbl.!lllWtrktkJas ... .......it.

CopJftglli.20t2�l...umlill.g.Al.1li9iib�Mqoatbe� ICUDild.ar�iawfdil«blJll"I. 0.10� tinl.p:dJCCIGl mAJM__....fmn. ....... my���aol�dGl.b�a.mliag-.m---�l...Amiof;--•rilht1u_,,,.��-..,.1imllljf

64

Chapter 3

Rings

EXAMPLE 6 Every nonzero element of the field �is a unit: If

a

-;; = 1. The same 1

:f. 0, then a

•

thing is true for every field F. By definition, F satisfies Axiom 12: If a # Op, then the equation ax = 1, has a solution in F. Hence,

Every nonzero element of a field is a unit.

EXAMPLE 7

. (a �)

Amatr1x

c

in M(�) such that ad - be -:/= 0 is a unit b ecause, as you can

(

easily verify,

0

1

)

and

d ad-be -c

ad-be

In particulat; each of these matrices is a unit: A

=

G �).

B

=

-b ad-be a ad-be

c=

C-� �).

)(

a

=

!) G �).

c

(1;3 �).

Units in a matrix ring are called inYertible matrices.

EXAMPLE 8 Let Fbe a field andM(F) the ring of 2 X 2 inatrm> with entries in F. If

A

=

(: !)

EM(F) and ad-be

'¢ OF> then ad-be is a unit in Fby Example 6.

Thecom]:Utations inF.xample7,with d 1 a

aninvertiblematrix[unit inM{F)]with inverse

Definition

rqiacedb y(ad-bet1,showthatAis

(-cd(ad - bc)-1 (ad bc -1

- bc

_

-

)

-b(ad bc)-1 a(ad- bc)-1 ·

)

An element a in a ring Risa zero divisor provided that

(1)

a ;e- OR.

{2} There exists a nonzero element c in R such that ac = OR or ca = OR.

Note that in requirement (2), the element c is not unique: Many elements in the ring

may satisfy the equation

ax=

OR or the equation

xa

=OR (Exei:cise

.......8om.1M•Bam:.ndkir�.Bdbmbll_...._ ...--if�:dpa.R!Jlltril:tlima......iL

eap,ngm.20:12�1..umiq.A:l.llialall--4.....,-aatn.t:IDJllilrd,.llC...t,,ar�io.wtdaarls,_,.o.1"�dpll.-mkd.�lrlDlllllmmll)'M ....... q-�� fld.__...,.dlN:t �a--.�c.a.� rigbtlD---��-

...

...

........

6). Furthermore,

3.2

Basic Properties of Rings

ac

in a noncommutative ring, it is possible to have Section 3.1).

66

= OR and ca :!- OR (Exercise 4 in

EXAMPLE9 Both 2 and 3 are zero divisors in Z6 because 2 • 3 zero divisors in Z12 because 4

9

•

=

=

0. Similarly, 4 and 9 are

O.

For a zero divisor A in a matrix ring, it is possible to find a matrixC such that

AC=OandCA=O.

EXAMPLE 10 Let F be a field. A nonzero matrix zero divisor because,

as

(: �)

you can easily

inM(.F) such that ad

verify,

-

be=OF is a

In particular, each of the.ie matrices is a zero divisor in the given ring:

A= G �)

inM(R),

B=

4/3

(

-2

��) inM(Q), and C =

(� !)

inM(Z6).

EXAMPLE 11 1 1: If

Every integral R domain satisfies Axiom

ab=OR, then a=OR orb =OR.

In other words, the product of two nonzero elements cannot be 0. Therefore, An integraJ domain contains no zero divisors.

F inally, we present some useful facts about integra l domains and fields.

Theorem 3.7 Cancelation is valid in any integral domain R:

If a

#OR and ab= ac in R, then

b =c. Cancelation may fail in rings that are not integral domains. In Z12, for instance, 2

•

4= 2

·

10, but 4 :!-

Proof ofTheorem 3.7

10.

...

If ah = be, then ah

a :!- OR, we must haveb

dicting Axiom

... .......

-

c =OR

-

be = OR, so that a(b

(if not,

1 1). Therefore,b =c.

-

c)=OR.

Since

then a is a zero divisor, contra­

•

...

�2012.C....,l....e-*g.AIRqlna-..d.MaJ"mtbll� �-ar :towballl«lapd.. 0..W�dalD.-tinl:pat;Joootm:a.,. ....,....m_to:.:J.beBo'*:.udkx-��----­ dililmad.*'-l:my�-mmillldmmmll.....mllJ'd!Kl. �---.�c.g..p�---ftgbttD__,,,..mddllklDlii.ICDllllnl•_..,.limlo��:Dgbb�...-.:lit.

66

Chapter 3

Rings

Theorem 3.8 Every field

F is

an integral domain.

Proof • Since a field is a commutative ring with identity by definition , we need

only show that Fsatisfies Axiom 11: If ah = 0p, then a = 0For h = Op. So suppose that ah = Op If h = Op, there is nothing to prove. If b #: Op, then b is a unit (Example 6). Consequently, by the definition of unit and part (1) of Theorem 3.5, a= alp= abh-1 = Opb-1 = Op

So in every case, a =Op orb= Op. Hence, Axiom 11 holds and Fis an integral domain . • The converse of Theorem 3.8 is false in general (Z is an integral domain that is not a field), but true in the finite case.

Theorem 3.9 Every finite integral domain Risa field.

Proof• Since Ris a commutative ring with identity, we need only show that for

each a#: OR, the equation ax= lR has a solution. Let ah a,;, , a,, be the distinct elements of Rand suppose a,#: OR. To show that a,x= 1R has a solution, consider the products a1ai. a,a,;, a,a3, , a,a,.. If a1#: a1, then we must have a,a1 #: a,a1 (because ata1= l4a1 would imply that 0i = a1 by cancelation). Therefore, a1a1, a1a'b . . . , a,a,. are n distinct elements of R. However, Rhas exactly n elements all together, and so these must be all the elements of Rin some order. In particular, for some}, C4a1= 1R. Therefore, the equation arJC = 1R has a solution and Ris a field. • • • •

• • •

• Exercises A. 1. Let R be a

(a)

ring and a,b ER.

(a+ h)(a-b) =?

(b) (a+ b)3 =?

(c) What are the answers in parts (a) and (b) if R is commutative? 2.

Find the inverse of matrices A, B, and C in Example 7.

3.

An element e of a ring R is said to be idempotent if e1= e.

(a) Find four idempotent elements in the ring M(R). (b) Find all idempotents in Z12• �2012C...,.1.Nmlmg.Al.1Ua11Da..r..a.V.,.ootbll� �-w....... :lawtdiiwia:r-t. O..to�dpm.-1blinl.:PGQ"�a.,.h�fnml.b•Bo1*:..ab-�1).EdDW.....,._ a...ad.'lmm,-��._ .. .-.m.Dy.n.ctbl�...-...,.._...�IA.mog-- .. :dgbtm-__,_�OOllll!m·a;J'timlo1f..._...._:ligl:U� ...... it.

3.2

Basic Properties of Rings

4. For each matrix A find a matrix Csuch that AC = 0 or CA A= 5.

(

= 0:

-10)·

5 -2

67

1/4

4 '

)

3/2 .

(a) Show that a ring has only one zero element. [Hint: If there were more than one, how many solutions would the equation OR+ x =OR have?] (b) Show that a ring R with identity has only one identity element. (c) Can a unit in a ring R with identity have more than one inverse? Why?

6. (a) Suppose A and Care nonzero matrices in M(IR) such that AC= 0. If k is any real number, show that A(kC) = 0, where kC is the matrix C with every entry multiplied by k. Hence the equation AX= 0 has infinitely

many solutions.

(b) If A = 7. Let

G �)

R be a ring

.find four solutions of the equation AX= 0.

with identity and let S

= {nlR I nE Z}. Prove that Sis a a ER is on page 62. Also see

subring of R. [The definition of na with n E Z, Exercise

27.]

8. Let R be a ring and

b a fixed element of R. Let

is a subring of R. 9. Show that the set S of matrices of the form numbers is a subring of I 0. Let

M(IR).

T = {rb

(: �)

I rER}.

Prove that T

.with a and b real

R and S be rings and consider these subsets of RX S: R=

{(r,Os) I rER}

(a) If R = Z3 and S

""

and

S ={(OR,

s) J SES}.

Z5• What are the sets Rand S?

(b) For any rings R and S, show that Ris a subring of RX S. (c) For any rings Rand S, show that Sis a subring of RX S.

11.

Let

R be a ring and ma fixed integer. Let S = {r ER J mr =OR}· Prove that S

is a subring of R.

12.

Let

a

and b be elements of a ring

R.

(a) Prove that the equation a +

x

= b has a

unique solution in

R.

(You

must prove that there is a solution and that this solution is the only one.)

(b) If Ris a ring with identity and a is a unit, prove that the equation ax = b has a unique solution in R.

13.

Let Sand The subrings of a ring

R. In (a) and (b), if the answer is "yes,"

prove it. If the answer is "no," give a counterexample.

(a) Is Sn Ta subring of R? (b) Is SU Ta subring of R? ..

�2012c..pe.�A.t�a_..ilibJ"oi:1thl� me..-t.ar�iowtdlOl!�J*I.. 0..10�..-.--*ild.�caal-OlllJ ..,.....tfam.M•Boi:*ndi!IX'..a.,..(1).lldladlll. .-..id.--my��--ad�dh:t--�--.....--..c.g.pu--.--•Dgbtm-__,_��-..,.--il......_.:Datu�...-. ..

...... tm

68

Chapter 3

Rings

14. Prove that the only idempotents in an integral domain Rare OR and Exercise 15.

lR. (See

3.)

(a) If a and b are units in a ring Rwith identity, prove that ab is a unit whose inverse is (ab)-1 b -1a- 1• (b) Give an example to show that if a andb are units , then a-1b-1 need not be =

the multiplicative inverse of ab. 16. Prove or disprove: The set of units in a ringRwith identity is a subring ofR. 17. If u is a unit in a ring Rwith identity, p rove that u is not a zero divisor. 18. Let

a be a nonzero element of a ring Rwith identity. If the equation ax=lR u and the equ ation ya=lR has a solution u, prove that u=u.

has a solution

19. LetR and Sbe rings with identity. What are the units in the ring R X S? 20. LetR and Sbe nonzero rings (meaning that each of them contains at least one nonzero element). Show thatR X S contains zero divisors. 21. Let Rbe a ring and let

a be a nonzero element ofRthat is not a zero divisor.

Prove that cancelation holds for a; that is, prove that

(a) If ab= acinR, thenb=c. (b) If ha = ca inR, thenb = 22.

c.

(a) If ab is a zero divisor in a ring R, prove that a orb is a zero divisor. (b) If a orb is a zero divisor in a commutative ring Rand ab# OR, prove that ab is a zero divisor.

23. (a) LetR be a ring and

a, b ER. Let m and n be nonnegative integers and

prove that (i)

(m+ n)a=ma+ na.

(ii)

m(a+ b)=ma+ mb.

(iii)

m(ab) = (ma)b = a(mb).

(iv)

(ma)(nb) == mn(ab).

(b) Do p art (a)when m and n are any integers. m and n be p ositive integers. "<+" " (a) Show that a"'a =a and (aj" =a-.

24. LetR be a ring and a, b ER. Let

(b) Under what conditions is it true that (abf=a"b"? 25. Let Sbe a su bring of a ringRwith identity.

(a) If Shas an identity, show by example that ls may not be the same as lR. (b) If both Rand Sare integral domains , prove that ls= lR. B. 26. Let She a subring of a ring R. Prove that the equation a +

Os=OR. [Hint: For a E S, consider

x=a.]

27. LetRbe a ring with identity and h a fixed element ofRand let S= Is Snecessarily a subring of Kl [Exercise 7 is the case when b

eap,ngm.20:12�1...umiq.A:l.lliala 11--4.....,-aatn. t:IDJllilrd,. � ar�:illwmlliarm,_,. 0.1"�dpll .-mllnl.�1r1C11111m�M ....._._q-��._.fld.__...,.dkt._owadl._.....--..c...� ........ rir;bl1a-...,,,..��·...,.

=

{nb [ nEZ}. lR.]

.......8om.1M•Bam:.ndkir�.Bdbmbll_...._ ... w......_..:dPLI�...-...

3.2 28. Assume that R=

Basic Properties of Rings

69

{OR, lR, a, b} is a ring and that a and bare units. Write out

the multiplication table of R. 29. Let Rbe a commutative ring with identity. Prove that Ris an integral domain if and only if cancelation h olds in R (that is ,

b=

a if: OR and ab = ac in Rimply

c).

30. Let Rbe a commutative ring with identity and b ER. Let Tbe the subringof all multiples of b (as in Exercise 8).

If u is a unit in Rand u ET, prove that T

31. A Boolean ring is a ring Rwith identity in which x2 = x for every examples, (a)

see

x ER.

=

R.

For

Exercises 19 and 44 in Section 3.1. If Ris a Boolean ring, prove that

a+ a= OR for every a ER, which means that a= (a + a)2.]

(b) Ris commu tative.

-a.

[Hint: Expand

[Hint: Expand (a+ b)2.]

32. Let Rbe a ring without identity. Let Tbe the set R X Z. Define addition and multiplication in Tby these rules:

(r, m) + (s, n) = (r + s, m + n). (r, m)(s, n) = (rs

+

ms

+ nr, mn).

(a) Prove that Tis a ring with identity. (b) Let R consist of all elements of the form (r, 0) in T. Prove that Ris a subring of T. 33. Let R be a ring with identity. 34. Let Fbe a field and A

=

If ab and a are units in R, prove that b is a unit.

(: !)

a matrix in M(F).

(a) Prove that A is invertible if and only if

8, and 10 and Exercise I7].

(b) Prove that A is a zero divisor 35. Let A (a) If

=

( ) a

b

e

d

ad

-

be if;

O_,... [Hint: Examples 7,

if and only if ad - be= o_,...

. . . . b e a matrix with mteger entries.

ad - be= ±I, show that A is invertible in M(T). [Hint: Example 7.]

(b) If ad - be * 0, M(Z).

1, or

-1, show that A is neither a unit nor a zero divisor in

[Hint: Show that A has an inverse in M(lli) that is not in M(Z ); see

Exercise S(c ). For zero divisors , see Exercise 34(b ) and Example 10.] 36. Let R be a commutative ring with identity. Then the set M(R) of 2 X 2

(: �)

matrices with entries in R) is a ring with identity by Exercise 37 of Section If A

=

M(R).

�

3.1.

E M(R) a d ad - be is a unit in R, show that A is invertible in

[Hint: Replace ad_ be by (ad - be)-1 in Example 7.]

37. Let Rbe a ring with identity and a, bER. Assume that

a is not a zero divisor. ab = IR, if and only if ha= IR. [Hint: Note that both ab = IR and ha= IR imply aha= a (why?); use Exercise 21.]

Prove that

CllpJliglll2012.C.....,LAmag.AIRqlaa-wd.lbJ"mtbll��Ol'�:iDwldm«ia:PKL0.10�dala,.-tinl��_,-119____..fa:m:J.1ll9t1Book.nilloc�:Mlmilil......- ...

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70

Chapter 3

Rings

38. Let R be a ring with identity and a, b ER. Assume that neither a nor b is a zero divisor. If ab is a unit, prove that a and bare units. [Hint: Exercise 21.] 39. (a) If Ris a finite commutative ring with identity and a ER, prove that a is either a zero divisor or a unit. [Hint: If a is not a

zero

divisor, adapt the

proof of Theorem 3.8, using Exercise 21.]

(b) Is part (a) true if

Ris infinite? Justify your answer.

40. An element a of a ring is nilpotent if a" = OR for some positive integer n. Prove that R has no nonzero nilpotent elements if and only if OR is the unique solution of the equation

x'- =OR.

TIU! following ckfinition is needed for Exercises 41-43. Let R be a ring with ickntity. OR, then R is said to have R is said to have characteristic zero.

If there is a smallest positive integer n such that nl R characteristic n. If no such n exists,

7L,, has characteristic

n.

Prove that a finite ring with identity has characteristic n for some

n

41. (a) Show that Z has characteristic zero and

(b) 42.

=

What is the characteristic of

Zi X 7L6? > 0.

43. Let R be a ring with identity of characteristic n > 0. (a) Prove that

(b)

na

=OR for every a ER.

If R is an integral domain, prove that n is prime.

C. 44. (a) Let a and b be nilpotent elements in a commutative ring R (see Exercise 40). Prove that a + band ab are also nilpotent. [You will need the Binomial Theorem from Appendix E.]

(b)

Let N be the set of all nilpotent elements of R. Show that N is a subring of R.

3 45. Let R be a ring such that x = x for every x ER. Prove that

R is commutative.

46. Let R be a nonzero finite commutative ring with no zero divisors. Prove that Ris a field.

II

Isomorphisms and Homomorphisms

If you were unfamiliar with roman numerals and came

across a discussion of

integer

arithmetic written solely with roman numerals, it might take you some time to realize that this arithmetic was essentially the same as the familiar arithmetic in Z except for the labels on the elements. Here is a less trivial example.

EXAMPLE 1 Consider the subset S = {O, 2, 4, 6, 8} of Z10• With the addition and multiplica­ tion of Z10, Sis actually a commutative ring, as can be seen from these tables:*

•The reason the elements of Sare listed in this order will become clear in a moment

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3.3

Isomorphisms and Homomorphisms

71

+

0

6

2

8

4

0

6

2

8

0

0

6

2

8

4

0

0

0

0

0

0

6

6

2

8

4

0

6

0

6

2

8

4

2

2

8

4

0

6

2

0

2

4

6

8

8

8

4

0

6

2

8

0

8

6

4

2

4

4

0

6

2

8

4

0

4

8

2

6

4

A careful examination of the tables shows that Sis a field with five elements and that the multiplicative identity of this field is the element 6. We claim that Sis "essentially the same" as the field Zs except for the labels on the elements. You can

see

this as follows. Write out addition and multiplication tables

for Zs.* To avoid any possible confusion with elements of S, denote the elements of

Zs by 0, T,

2, 3, 4. Then relabel the entries in the Z5 tables according to this scheme: Relabel 0

as

relabel

0,

relabel 3 as 8,

T as 6,

relabel

2 as 2,

relabel 4 as 4.

Look what happens to the addition and multiplication tables for Zs: 0

+

j 1 -,. -

0 6 2 8

J

�

4

6

1

jJ 0

j

2 2

z

8

J �

l j 8

:z

4

Ji 4

JJ

4

0

j

0

1 6

r 6

2

j

g 0

j 6

4

4

A

0

j

.

�

j

Ji

4

� 8

j 8

2

6

1

j 2

6

j

8

2

l

2

z 2

1

-,. -

0 6 2 8

J 8

�

4

jJ j g ;a j j

0 0 0 0 0 0

j j 1 "J. 1 �

6 0 6 2 8 4

l j l Ji j 1

2 0 2 4 6 8

:z j :z r Ji j

8 0 8 6 4

ff j A j -

4 0 4 8 2

z 2

j

6

By relabeling the elements of Zs, you obtain the addition and multiplication tables

for S. Thus the operations in Zs and S work in exactly the same way-the

only difference is the way the elements are labeled. As far as ring structure goes, Sis just the ring Z5 with new labels on the elements. In more technical terms, Zs and S are said to be isomorphic. In general, isomorphic rings are rings that have the same structure, in the sense that the addition and multiplication tables of one are the tables of the other with the ele­ ments suitably relabeled, as in Example 1. Although this intuitive idea is adequate for small finite systems, we need a rigorous mathematical definition of isomorphism that agrees with this intuitive idea and is readily applicable to large rings as well. There are two aspects to the intuitive idea that rings R and S are isomorphic: relabeling the elements of R and comparing the resulting tables with those of S to verify that they are the same. Relabeling means that every element of R is paired with a unique element of S (its new label). In other words, there is a function f.R-+ S that *The "11..5 tables (in congruence class notation) are shown in Example 2 of Section 2.2.

eupJIWil2012.C.....,LAmag.AIRqliba-wd.lbJ"mtbll� ---a,.Ol'�:iawldmOl'ia:PKI. 0.ID�dala.-tinli-ll;Jruu.ma,-119�fil:m:J.1118eBOOll:.nilloc�:Blb:nlll......- ... �--my�awmrdl-.alll.....mllydlN:l.._O'llmd._.....,.n.c..c.g.,..i...iag--•ftgM11.1--.noa�CD1111B1S:•_,...._��:ligl!U�....-.it.

72

Chapter 3

Rings

assigns to each

r ER

its new label/(r) ES. In the preceding example, we used the rela­

beling function f: Z5 -+ S, given by /(0) = 0

/(l)= 6

/(2)= 2

/(3)= 8

/(4)= 4.

Such a function must have these additional properties: (i) Distinct elements of R must get distinct new labels: If r '# r' in R, then/(r) '# /(r') in S. (ii) Every element of S must be the label of some element in R:* For each Statements (i) and

s ES,

there is an rE R such that/(r) =

(ii) simply say that the functionf must

tive, that is,/must be a

bijection. t

In order for a bijection (relabeling scheme)/to be

an

s.

be both injective and surjec­ isomorphism, applying/to

the addition and multiplication tables of R must produce the addition and multiplica­ tion tables of S. So if

a+ b =

c

in the R-table, we must have/(a)+ f(b) = f(c) in the

S-table, as indicated in the diagram:

�ft)) R� I � S +

/(a)

a,

�

However, since

f{c)

a+ b = c, we must also have/(a + b) = f(c). Combining this

fact that/(a) + f(b) = f(c),

we

with the

see that

f(a + b)

=

f(a) + f(b).

This is the condition that f must satisfy in order for f to change the addition tables

of R into those of S. The analogous condition on f for the multiplication tables is

/(ab)= f(a)f(b). We now can state a formal definition of isomorphism:

Definition

A

ring R is isomorphic to a ring S (in symbols, R = S) if there is a function f:R-+ S such that (I) f is injective; (ii) f is surjective; {iii) f{a + b) = f(a}+ f(b)

and

In this case the function f is called an

f{ab)= f(a) f(b) for all a, b ER. isomorphism.

*otherwise, we couldn't possibly get the complete tables of S from those of R.

f1njective, surjective, and

bijective functions are discussed in Appendix

B.

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�'lm:mJ"nw--l..,.._.'*-alll.....UO,.dllK.1."lle�---.�CmgQ&i...mog--a..:rigbt1D__,_mdllllklml.�•..-ttm.V........_:Dgl:U�----:it.

3.3

CAUTION:

Isomorphisms and Homomorphisms

73

In order to be an isomorphism, a function must satisfy all three of the conditions in the definition. It is quite possible for a function to satisfy any two of these conditions but not the third; see Exercises 4, 25, and 32.

EXAMPLE 2 In Example 12 on page 50, we considered the field K of all 2 the form

x

2 matrices of

b

where a and are real numbers. We claim that K is isomorphic to the field C of complex numbers. To prove this, define a functionfK-+ C by the rule

I(-� !) a+ bi. =

To show that/is injective, suppose

( b) = i( s). -b a -s b =as. bi = si f

Then by the definition off, we must have rand

a=

r

a

r

+

r

in C. By the rules of equality in C,

+

Hence, in K

so that/ is injective. T he function/is surjective because any complex number f + is the image under of the matrix

a

bi

in K. Finally, for any matrices A and Bin K, we must show that/(A + B) f(A) + f(B) andf(AB) f(A)f(B). We have

=

=

I

b +cf\ [( a b) + ( cf\ ] ( -b a -d c} -b - d a+ c} = (a+ c) + (b + d)i (a + bi) + ( + di) ( b) +i( d) -b a -d c c

_

i

a+

c

=

=

c

i

a

c

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74

Chapter 3

Rings

and

i[(

a -b

b a

)(

c -d

d\] i ( c}

_

= =

=

)

ac - bd -ad - be

ad+be ac - bd

(ac - bd) + (ad+bc)i (a+bi)(c + di)

1( : !)1 ( ; J· _

_

Therefore,/is an isomorphism.

It is quite possible to relabel the elements of a single ring in such a way that the ring is isomorphic to itself.

EXAMPLE 3 Letf"C-+ C be the complex conjugation map given byf(a + function fsatisfies

l[(a t bi)+ (c +di)]

=

=

=

bi)

= a-bi.* The

f[(a+c) + (b+d)i] (a+ c) - (b+d)i (a - bi)+ (c - di) f(a+ bi)+f (e+di) =

and

/[(a+ bi)(c +di)]= f[(ac - bd) + (ad+ bc)i] (ac - bd) - (ad+bc)i (a - bi)(c - di) f(a + bi)f(c+di). =

=

=

You can readily verify that/is both injective and surjective (Exercise 17). Therefore/ is an isomorphism.

EXAMPLE 4 If R is any ring and t.R:R-+ R is the identity map given by t.R(r)

=

r,

then for

any a, bER

"R (a+ b) Since

=a+b

=

t.R(a) + 1.Jb)

and

t.R(ab)

=

ab

= 1.Ja)1.Jb).

'R is obviously bijective, it is an isomorphism.

Our intuitive notion of isomorphism is symmetric: "R is isomorphic to S" means the same thing as "Sis isomorphic to R". The formal definition of isomorphism is not

"The function fhas a geometric interpretation in the complex plane, where a+ bi is identified with the point (a, b): It reflects the plane in the x-aKis.

eap,ngm.20:12�1..umiq.A:l.lliala 11--4.....,-aatn. t:IDJllilrd,. llC...t,, ar�io.wtdaarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ....... 8om.1MelkK*..tkir..ai.p.r(•).Bdbmbll_...._ ...._._q-��._.fld.__...,.a11N:t... �a--.�c.a.� ........ rir;bl1a-...,,,..��·...,. ... w....._.,....� ........

3.3

Isomorphisms and Homomorphisms

75

symmetric, however; since it requires a function from R onto S but no function from S onto R. This apparent asymmetry is easily remedied. If f.R -+S is an isomorphism, then/is a bijective function of sets. Therefore,fhas an inverse function g:S-+ R such that go f = t.R (the identity function on R) and/• g = Ls-* It is not hard to verify that the function g is actually an isomorphism (Exercise 29). Thus R = S implies that S = R, and symmetry is restored. Homomorphisms Many functions that are not injective or surjective satisfy condition (iii) of the definition of isomorphism. Such functions are given a special name.

Definition

LetR and S be rings. A function f:R-+S is said to be a homomorphism

f(a + b)

=

f(a) + f{b) and

f(ab)

=

if

f(a)f(b) for all a, b ER.

Thus every isomorphism is a homomorphism, but as the following examples show, a homomorphism need not be an isomorphism because a homomorphism may fail to be injective or surjective. EXAMPLE 5 For any rings R and S the zero map z:R-+ S given by z(r) = Os for every r ER is a homomorphism b ecause for any a, b ER z(a + b) = Os = Os+ Os = z(a) + z(b)

and z(ab) = Os= Os· Os= z(a)z(b�

When both R and S contain nonzero elements, then the zero map is neither injective nor surjective. EXAMPLE 6 The function/:Z-+ "Z
::=

f(a) + f(b)

and f(ab) = [ab]

=

[a][b] ""'f(alf(b ).

The homomorphism/is surjective, but not injective (Why?).

*See Appendix B for details.

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76

Chapter 3

Rings

EXAMPLE 7 The map g:R--+. M(R) given by

g(r) ( 0 0\r} r, s g(r) g(s) (-� �) (-� �) ( s r � s) ) g( ) ( - -(r s) r =

-r

is a homomorphism because for any +

+

=

_7o_

=

0

-

and

ER

0

+

-

+s

r+s

g(r)g(s) (-� �)(-� �) (-�s �) g(rs). =

=

=

The homomorphism g is injective but not surjective (Exercise 26).

CAUTION:

Not all functions are homomorphisms. Th e properties

f(a + b)

=

f(a) + f(b)

f(ab)

and

=

fail for many functions. For example, if fR --+. f(x) x + 2, then

f(a)f(b)

� given by

=

/(3 + 4)

=

/(7)

=

9

but

/(3)

+ f(4)

so that/(3 + 4) * /(3) + /(4). Simil arly,/(3 because /(3

•

4)

=

/(12)

=

14,

but

/(3)/(4)

=

·

5 +6

=

11

4) o:fo /(3) /(4)

=

5

•

6

=

30.

Theorem 3.10 Let f:R--+.S be a homomorphism of rings. Then

(1) {{OR)

=

(2) f(-a)

Os.

=

(3) f(a - b}

-f(a) for every a ER. =

f{a} - f(b) for all a, b ER.

If R is a ring with identity and f is surjective, then

(4) S is a ring with

identity f(1R)·

{5) Whenever u is a unit In R, then f(u) is a unit in Sand f(ur1

..

..

=

f(u-1).

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3.3

Isomorphisms and Homomorphisms

Proof .. (t) f(oiJ+ f(oiJ= f(oR + oR)

77

[/is a homomorphism.]

/O ( R) +/(OR) = f(OR)

O ( R+OR= ORinR]

f(OiJ + f(OiJ= f(OiJ +Os

( /O ( R) +Os= f(OiJ in SJ

[Suhtract f(OiJfrom both sides.].

/O ( R)= Os

(2) First, note that f(a) +/(-a) = f(a

+

(-a))

[/is a homomorphism.] [a+ (-a)= OR)

= f(OR)

= Os [Part (I)]. Therefore,/(-a) is a solution of the equation/(a) + x =Os. But the unique solution of this equation is-/a ( ) by Theorem 3.3. Hence f(-a) = -f(a) by uniqueness.. ( ) f(a - b) = f(a + (-b)) 3 = f(a) + f(-b) )

[Definition of subtraction] (f iSahomomorphism.]

= f(a) + (-f(b))

[Part (2)]

= f(a) - f(b)

[Definition of subtraction].

(4) We shall show that/l ( iJE S is the identity element of S. Lets be any element of S. Then since/is surjective, s = f(r) for some rER. Hence, s

·/(IR) = f(r')f(IR) = f(r lR) = f(r) = s •

and, similarly,fl ( _,J • s = s. Therefore, Shas/1 ( _,J as its identity element. ( 5) Sinceu is a unit in R, there is an element v in R such that uv = 1R =vu. Hence, by (4)

f(u)f(v)

=

f(uv)= f(liJ=

ls-

Similarly ,vu= IR implies thatf(v)f(u) 15• Therefore,/(u) is a unit in S, with inverse/(v). In otherwords,f( u)-1 = f(v). Sincev = u-1, we see that/(u)-1 = f(v) = /(u-1). • ==

Iff.R 4 S is a function, then the image of /is this subset of S: Im/= {SES Is= f(r) for some rER} = if(r) I rER}. If f is surjective, then Im f = S by the definition of surjective. In any case we have:

Corollary 3.11 Iff:R 4 S is a homomorphism of rings, then the image off is a subring of S.

Proof .. Denote Im/byL Iis nonemptybecause05 =f(OiJEibyl ( )ofTheorem31 . 0. The definition of homomorphism shows that I is closed under multiplica­ tion: Iff(a),f(h)El, tltenf(a)f(b) = f(ab)El. Similarly, [�closed under subtraction because/a ( ) - f(b)= f(a - b)E/ by Theorem 3. 10. Therefore, I is a subring of S by Theorem 3.6. • ...._._my�mmal-*-oot...uu:rlflKl.b�a.mliag-.m---�l...Amiot;--•sight:D_,,,.mddlitkxllll�•_,.tiullljf........_:Dgbl.!lllWtrktioas ...... it.

CopJftglll.20t2C,....l...umlill.g.Al.1li9iiba_...a.Uqoatbe� ICUDlld.ar�ia.wtdil«blJll"I. 0..10�....._...,.-.._p:llJ'Ced...__,.,.__....tmn.-.aBcd:udhr�1).Bdlaftlll.

.... ._

78

Chapter 3

Rings

Existence of Isomorphisms If you suspect that two rings are isomorphic, there are no hard and fast rules for finding a function that is an isomorphism between them. However the properties of homomorphisms in T heorem 3.10 can sometimes be helpful.

EXAMPLE 8 If there is an isomorphism/from Z12 to the ring Z3 X Z4, thenf(l) = (1, 1) by part (4) of Theorem 3.10. Since/is a homomorphism, it has to satisfy /(2) =/(1+ 1)

=

/(1)+ /(1)

=

(1, 1) + (1, 1)

=

(2, 2)

/(3) =/(2+ 1) =/(2) + /(1) = (2, 2) + (1, 1) = (0, 3) /(4)

==

/(3+ 1) = /(3)+ /(1) = (0, 3) + (1, 1) = (1, 0).

Continuing in this fashion shows that if/is an isomorphism, then it must be this bijective function: /(1) = (1, 1)

/(4) = (1, 0)

/(7) = (1, 3)

/(10) = (1, 2)

/(2) = (2, 2)

/(5) = (2, 1)

/(8) = (2, 0)

/(11) = (2, 3)

/(3)

/(6) ""'(0, 2)

/(9)

=

(0, 3)

=

(0, 1)

f(O)

=

(0, 0).

All we have shown up to here is that this bijective function/is the only possible isomorphism. To show that this/actually is an isomorphism, we must verify that it is a homomorphism. This can be done either by writing out the tables (tedious) or by observing that the rule off can be described this way: f([a]1 2) = ([a]3, [a]4) , where [afo denotes the congruence class of the integer a in Z12, [a]3 denotes the class of a in Z3, and [a]4 the class of a in z_.. (Verify that this last statement is correct.) Then f([a]i2+ [b]i:z) =/([a+ b]ii)

[Definition of addition in Zn]

= ([a+ bh, [ a+ b14)

[Definition offl

= ([a]3 + [b]3, [a]4 + [b]4)

[Definition of addition in Z1 and Z4]

= ([a]3, [a]4)+ ([bh, [b]4)

[Definition of addition in Z3 X Z4]

= f([a]ii) + f([b]1i)

[Definition offl.

identical argument using multiplication in place of addition shows that f([a]t2[bfo) = /([a]tz)f( [b]ll). T herefore,fis an isomorphism and Z12 = Z3 X Z,..

An

Up to now we have concentrated on showing that various rings are isomorphic, but sometimes it is equally important to demonstrate that two rings are not isomorphic. To do this, you must show that there is no possible function from one to the other satisfying the three conditions of the definition. �20:12�J.....i.g.A:l.1Uala 11--4.....,-aatn. t:IDJllilrd,. llC...t,, ar�io.wmlliarls,_,. 0.1"�dpll.-mkd.�lrlDlllllm�M ....... ftom.1M•Bam:.ndkir�.Bdbmbll_...._ ...__.___,.��._.fld.__...,.a11N:t... Cl'Na!Sa--.�c.a.� ...... .. rir;bl1a-...,,,..��·...,. ... w......_..:dPLI�...-. ..

3.3

Isomorphisms and Homomorphisms

79

EXAMPLE 9 � is not isomorphic to Z12 or to Z because it is not possible to have a surjective func­ tion from a six-element set to a larger set (or an injective one from a larger set to "4,). To show that two infinite rings or two finite rings w ith the same number of elements are

not isomorphic, it is usually best to proceed indirectly.

EXAMPLE 10 The rings Z4 and Z2 X Z2 are not isomorphic. To show this, suppose on the contrary that/:,l4�Z2 X Z2is an isomorphism. ThenJlO) ( 0, 0) and =

/(1) = (1, 1) by T heorem 3.10 . Consequently, /(2)

=/(1

+

1) = f(l)

Since f is injective and /(0)

�

+ f( l)

= (1, l)

+

(1, l) = (0, O}

/(2), we have a contradiction. Therefore, no

isomorphism is possible. Suppose that/:R�Sis an isomorphism and the elements a,

b, c, . .. of Rhave a par­

ticular property. If the elements/(a),f(b),f(c), ... of Shave the same property, then say that the property is preserved by isomorphism. According to parts (1),

we

(4), and (5) of

Theorem 3.10, for example, the property of being the zero element or the identity element or a unit is preserved by isomorphism. A property that is preserved by isomorphism can sometimes be used to prove that two rings are IWt isomorphic, as in the following examples.

EXAMPLE 11 In the ring Zs the elements l, 3, 5, and 7 are units by Theorem 2.10. Since being a unit is preserved by isomorphism, any isomorphism from Z8 to another ring with identity will map these four units to four units in the other ring. Consequently, Z8 is not isomorphic to any ring with less than four units. In particular, ls is not isomorphic to l4 X Z2 because there are only two units in this latter ring, namely

(1, 1) and (3, l) as you can readily verify.

EXAMPLE 12 None of fields

0, R, or C is isomorphic to Z because every nonzero element in the 0, R, and C is a unit, whereas Z has only two units (1 and -1).

EXAMPLE 13 Suppose Ris a commutative ring andfR�Sis an isomorphism. Then for any

a, b ER, we have ab = ba in R. Therefore, in S f(a)f(b) =/(ab) = f(ba) =f(b)f(a). CnpJIWll2012.C.....,LAmag.AIRqliba-wd.lbJ"mtbll� �Ol'�:iawidmOl'kaJIKL O.W�dalD.-tinl��_,-119�fa:m:J.1119eBOOll:.nilloc�:Blb:nlll......- ... �--mJ'�-l:llWmldl-.alll.....mllydlN:l.._O'llmd._.....,.n-c..c.q..p�---ftgbtn__,,,.�CDllllll:•_..,...._��:ligl!U�....-.it.

80

Chapter 3

Rings

Hence, Sis also commutative because any t wo elements of Sare of the for

f(b ) (since/is surjective). In other

mf a

( ),

words, the property of being a commutative ring is preserved by isomorphism. Therefore, no commutative ring can be iso­ morphic to a noncommutative ring.

• Exercises A. 1. Let/:Z�-+Z2 X Z3 be the bijection given by

0-+(0,0),

2-+(0,2),

1-+(l,1), 5-+ (1, 2).

4-+ (0, 1),

3-+(1,0),

Use the addition and multiplication tables of "14 and Z2

X Z3 to show that/is

an isomorphism. 2. Use tables to show that ll.2 X Z2 is isomorphic to the ring Rof Exercise 2 in

Section 3.1. 3. Let Rbe a ring and let R* be the subring of R X Rconsisting of all elements

of the form (a, isomorphism.

a). Show that the functionf:R-+ R* given byf(a)

4. Let Sbe the subring {O,

.

2,4, 6, &} of ll.10 and let ll.5

=

=

a,

( a) is an

{O, T, 2, 3, 4,} (notation

as in Example 1) Show that the following bijection from 71..5 to Sis not an isomorphism:

o

____,,.

____,,.

T

o

2

2

____,,.

4

3

_____,,.

6

4

____,,.

s.

5. Prove that the field R of real numbers is isomorphic to the ring of all

matrices of the form by f(a)

=

G �)

.with a E Iii.

2X2

[Hint: Consider the function/given

(� �).]

6. Let Rand S be rings and let

R be the subring of R X Sconsisting of all

elements of the for m (a, Os)· Show that the functionf:R-+ R given by f(a) (a, Os) is an isomorphism. =

7. Prove that !fl is isomorphic to the ring S of all 2

(� �)

X 2 matrices of the form

.whereaER.

O(v2) be as in Exercise 39 of Section 3.1. Prove that the function /:Q(Vl)-+ O(Vl) given by /(a + bVl) = a bVl is an isomorphism.

8. Let

-

9. If /:Z -+ 7l. is an isomorphism, prove that/is the identity map.

are/(1 ) ,/( 1 + 1),

..

[Hint: What

. ?]

10. If Ris a ring with identity and/:R-+ Sis a homomorphism from Rto a ring S, prove thatf(IR) is an idempotent in S. [Idempotents were defined in Exercise 3 of Section

......

..

....

3.2.]

......

..

.......

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3.3 11.

Isomorphisms and Homomorphisms

State at least one reason why the given function is not a homomorphism.

(a) /:Ill-+ R and/(x)

==

Vx.

(b) g:E-+ E, where Eis the ring of even integers and/(x) (c) h:R-+ R and/(x)

=

2'".

(d) k:Q 4 0, where k(O ) 12.

81

=

0

(�) �

and k

=

=

3x.

if a'/: 0.

Which of the following functions are homomorphisms?

(a) f Z -+ .l, defined byf(x)

= -x.

(b) fZ2-+Z2, defined byf(x) = -x. (c) g:CI! -+ 0, defined by g(x)

=

(d) h:R-+ M(R), defined by h(a)

x2 =

I +

1

°

( : �} -

(e) /:Z12 -+.l.i, defined by/([x]12) = [x1, where [u],. denotes the class of the integer u in Z,.. 13.

Let R and S be rings.

(a) Prove that/:R X S-+Rgiven byf((r , s)) = r is a surjective homomorphism. (b) Prove thatg:R X

S-+.S given by g((r, s))

=sis a surjective homomorphism.

(c) If both Rand Sare nonzero rings, prove that the homomorphisms/ and g are not injective. 14.

Letf:Z -+.Z6 be the homomorphism in Example 6. Let K = {aEZ lf(a) = [O]}. Prove that K is a subring of Z.

15.

Let/:R-+ S be a homomorphism of rings. If r is a zero divisor in R , isf(r) a zero divisor in S?

B. 16. Let T, R, and Fbe the four-element rings whose tables are given in Example 5 of Section 3.1 and in Exercises 2 and 3 of Section 3.1. Show that no two of these rings are isomorphic. 17.

Show that the complex conjugation function/:C -+ C (whose rule is

f(a +bi)= a- bi) is a bijection. 18.

Show that the isomorphism of Zs and Sin Example 1 is given by the function whose rule is/([x]s) [6x]10 (notation as in Exercise 12(e)). Give a direct proof (without using tables) that this map is a homomorphism. =

19.

Show that S {O, 4, 8, 12, 16, 20, 24} is a subring of Z28• Then prove that the map/:Z7-+ S given by/([x ],) = [8xh8 is an isomorphism.

20.

Let E be the ring of even integers with the • multiplication defined in Exercise 23 of Section 3.1. Show that the map f:E-+ Z given byf(x) = x/2 is an isomorphism.

21.

Let Z* denote the ring of integers with the EB and 0 operations defined in Exercise 22 of Section 3.1. Prove that L is isomorphic to L*.

=

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82

Chapter 3

Rings

22.

Let 71_ denote the ring of integers with the ffi and 0 operations defined in Exercise 24 of Section 3.1. Prove that 71_ is isomorphic to Z.

23.

Let C be the field of Exercise 45 of Section 3.1. Show that C is isomorphic to the field C of complex numbers.

24.

(a) Let R be the set� X �with the usual coordinatewise addition, as in Theorem 3.1. Define a new multiplication by the rule (a, b)(c, d) (ac, be). Show that R is a ring. =

(b) Show that the ring of part (a) is isomorphic to the ring of all matrices in M(R) of the form 25.

(: �}

Let L be the ring of all matrices in M(Z) of the form

A: 0)

functio�/L : -+ � given by 1\ not an 1somorph1sm.

c

= a

(: �)

.Show that the

is a surjective homomorphism but

26.

Show that the homomorphism gin Example 7 is injective but not surjective.

27.

(a) If gR : -+ SandfS-+ Tare homomorphisms, show that/0g:R-+ Tis a homomorphism.

(b) If fandgare isomorphisms, show that fogis also an isomorphism. 28.

(a) Give an example of a homomorphismfR -+ S such that R has an identity but S does not. Does this contradict part (4) of Theorem 3.10? (b) Give an example of a homomorphismf:R-+ S such that Shas an identity but Rdoes not.

29.

Let/:R-+ S be an isomorphism of rings and let gS : -+ R be the inverse function of / (as defined in Appendix B). Show that gis also an isomorphism. [Hint: To show g(a + b) g(a) + g(b), consider the images of the left- and right-hand side under/and use the facts that/is a homomorphism and/ogis the identity map.] =

30.

Let/:R-+ S be a homomorphism of rings and let K = {rER lf(r) =Os }· Prove that K is a subring of R.

31.

Let/:R-+ S be a homomorphism of rings and Ta subring of S. Let P = {rER lf(r)ET}. Prove that Pis a subring of R.

32.

Assume n == l (mod m). Show that the function f: Z,,, -+ lm11 given by f([x],,,) = [nx]""' is an injective homomorphism but not an isomorphism when n <2: 2 (notation as in Exercise 12(e)).

33.

(a) Let The the ring of functions from IR to� as in Example 8 of Section 3.l. Let (}T : -+ R be the function defined by fJ(/) =/(5). Prove that(}is a surjective homomorphism . Is(}an isomorphism?

(b) Is part (a) true if 5 is replaced by any constant c ER? 34.

If f:R-+ Sis an isomomorphism of rings, which of the following properties are preserved by this isomorphism? Justify your answers. (a) aERis azero divisor.

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-...d.'lm:m,-��*-alll.....mllJ"dllK.1.b�._,..�c.g..gei...mos--a.:rigM1D__,_�romim•..-tilll9V.._...:DafUllWlrictims-.n-:11t.

3.3

Isomorphisms and Homomorphisms

83

(b) a ERis idempotent.* (c) Ris an integral domain . 35. Show that the first ring is not isomorphic to the second.

(a) EandZ

(b) � X �

(c) �

(d) 0 and�

(e) Z 36.

X Z 14 and li6

X R X �andM(R)

(t) Z.. X Z4 and Z16

X Z2andZ

(a) Iff:R--+ Sis a homomorphism of rings, show that for any n EZ,f(nr) = nf(r).

r

ER and

(b) Prove that isomorphic rings with identity have the same characteristic. [See Exercises 41-43 of Section 3.2.]

(c) If/:R--+Sis a homomorphism of rings with identity, is it true that Rand S have the same characteristic? 37.

(a) Assume that e is a nonzero idempotent in a ring Rand that e is not a zero divisor.* Prove that e is the identity element of R. [Hint: e2 = e (Why?). If a ER, multiply both sides of e2 = e by a.] (b) Let

S be a ring with identity and Ta ring with no zero divisors. Assume

that/:S...+Tis a nonzero homomorphism of rings (meaning that at least one element of Sis not mapped to OT)· Prove that/(18) is the identity element of T. [Hint: Show that/(18) satisfies the hypotheses of part (a).] 38. Let Fbe a field andf:F--+ R a homomorphism of rings.

(a) If there is a nonzero element c of Fsuch that/(c) =OR, prove that/is the zero homorphism (that is,/(x) =OR for every xEF). [Hint: c-1 exists (Why?). If xEF, consider/(xcc-1).] (b) Prove that/is either injective or the zero homomorphism. (Hint: If/is not the zero homomorphism and/(a) = f(b), then/(a - b) OR.] =

39. Let Rbe a ring without identity. Let Tbe the ring with identity of Exercise 32

in Section 3.2. Show that Ris isomorphic to the subring R of T. Thus, if Ris identified with R, then R is a subring of a ring with identity. C. 40. For each positive integer k, let kZ denote the ring of all integer multiples of k (see

Exercise 6 of Section 3.1). Prove that if m # n, then mZ is not isomorphic to nZ.

41. Let m, n EZ with (m, n) = 1 and let/:Z_--+ Zm X Zn be the function given by /([aJ-) = ([a]m, [a]n). (Notation as in Exercise 12 (e). Example 8 is the case m 3,n 4.) =

=

(a) Show that the map/is well defined, that is, show that if [a],,,,,= [bJmn in Z...,., then [a]m [bJm in Zm and [a]n= [b]n inZ,,. =

(b) Prove that/is an isomorphism. [Hint: Adapt the proof in Example 8: the difference is that proving/is a bijection takes more work here.]

42. If (m, n) # 1, prove thatZmn is not isomorphic to Zm

x Z,,.

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..

...... tm

CHAPTER

4

Arithmetic in f[x] In Chapter 1 we examined grade-school arithmetic from an advanced standpoint and developed some important properties of the ring Z of integers. In this chapter we follow a parallel path, but the starting point here is high-school algebra-­ specifically, polynomials with coefficients in the field IR of real numbers, such as x2 - 3x - 5,

6x3

-

3x2 + 7x + 4,

x12 -1.

Dealing with polynomials means dealing with the mysterious symbol "x", which is used in three different ways in high-school algebra First, xoften "stands for'' a number, as in the equation 12x - 8

O, where xis the number

�

· Second, xsome­ times doesn't seem to stand for any particular number but is treated as if it were a =

number in simplification exercises such as this one:

x3+ x

x(x2 + 1) x - x2+1 - · x2+1-

--

Third, xis also used as the variable in the rules of functions such as f(x) 3x + 5. Now that you know what rings and fields are, we shall consider polynomials =

with coefficients in any ring and attempt to clear up some of the mystery about the nature of x. In Sections 4.1-4.3, we shall see that when xis given a meaning similar to the second way it is used in high school, then the polynomials with coef­ ficients in a field Fform a ring (denoted F[x]) whose structure is remarkably similar to that of the ring Z of integers. In many cases the proofs for Z given in Chapter 1 carry over almost verbatim to F[x]. In Sections4.4-4.6 we consider tests to determine whether a polynomial is irre­ ducible (the analogue of testing an integer for primality). Here the development is not an exact copy of what was done in the integers. The reason is that the polyno­ mial ring Ff x] has features that have no analogues in the ring of integers, namely, the concepts of the root of a polynomial and of a polynomial function (which cor­ respond to the first and third uses of xin high school). 86 °'l'Jrill":!Ol20...Loomlog.Allllla"'..__Mor,..llooopiod._or..,..._ill_«ia,.i.DmlD-dPD,..., _____ llo_.._.,.•Bo<*-�1il!dlmlll..-i. _ .... ..,_.... __ ... _..,. ................. ..,.....Co... og l.olmlo&---� ..---·..,-11..-.-�-·...-.11.

86

Chapter 4

•

Arithmetic in

F[x]

Polynomial Arithmetic and the Division Algorithm

The underlying idea here is to define "polynomial" in a way that is the obvious exten­ sion of polynomials with real-number coefficients. Let R be any ring. A polynomial with coefficients in R is an expression of the form ao +

a1x

+ a,.i'- +

·

·

·

+ a,,x",

where n is a nonnegati ve integer and ai ER. This informal definition raises several questions: What is XI Is it an element of R1 If not, what does it mean to multiply

x

by a ring element? In order to answer these

questions, note that an expression of the form ao

+

a1x

+ ¥1 +

·

·

•

+ a,.X' makes

sense, provided that the a1 and x are all elements of some larger ring. An analogy might be helpful here. The number 3

-

41T +

l27T2 +

1T� and 8

1T is not in the ring Z of integen;, but expressions such as 1T2. + fru5 make sense in the real numbers. Furthermon;

-

it is not difficult to verify that the set of all numbers of the form ao

+ a11T + a2TT2 +

·

·

·

+ a,.11',

is a subring of � that contains both Z and

with n 2!: 0 and a1EZ

1T (Exercise 2).

For the present we shall think of polynomials with coefficients in a ring R in much the same way, as elements of a larger ring that contains both R and a special element x

that is not in R. This is analogous to the situation in the preceding paragraph with

R in place of Z and the element answer,

x

x

in place of 1T, except that here we don't know anything about

or even if such a larger ring exists. The following theorem provides the

as well

as

a definition of "polynomial".

Theorem 4.1 If R is a ring, then there exists

a

ring T containing an element

x that is not in

R and has these properties: (i) R Is a subring (ii) xa = ax for

of T .

every a ER.

(iii} The set R[x] of all elements of T of the form (where n 2!: O and is a subring

a1ER)

of T that contains R.

(iv) The representation of elements of R[x] is unique: If n ::!'> m and l1o

then {v} aa

+ a1x + a.,x2 +

·

•

·

+ a/' =b0 + b1x + b.,x2 +

·

·

·

+ b,.,x"',

a1=b1fori=1, 2, .. . , n and b1=OR for each i> n.

+ a1x + a�2 +

·

·

·

+ anJ(I =OR if and only If a1 =OR for every I.

Proof" See Appendix G. We shall assume Theorem 4.1 here. The elements of the ring

R[x] in Theorem 4.1 (iii)

are

•

called polynomials with

coefficients in R and the elements a1 are called coefficients. The special element

x

is

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4.1

Polynomial Arithmetic

and

the Division Algorithm

sometimes called an indeterminate.* To avoid any misunderstandings in Theorem

87

4.1,

please note the following facts. I. Property (ii) of T heorem 4.1 does not imply that the ring Tis commutative, but only that the special element

x commutes with each element of the subring R

(whose elements may not necessarily commute with each other). 2, Property (v) is the special case of property (iv) when each b,

=

OR.

3. The first expression in property (v) is not an equation to be solved for x. In this context, asking what value of x makes ao + a1x + a-iX2 + + a,;<' OR is as meaningless as asking what value of 1T makes 3 + 51T - 77T2 0 because x (like 1T) is a specific element of a ring, not a variable that can be assigned values.t ·

·

·

=

=

EXAMPLE 1 The rings Z[x],

Cl![x], and n[x] are the rings you are familiar with from high 3 + 5x - 7x'- is in all three of these rings, but 3 + 7 .5x'- is only in Q[x] and �x] because the coefficient 7.5 is not an integer. Similarly, 4.2 + 3x + V5x4 is in R[x] but not in the other two rings since V5 is not a school . For instance,

rational number. Terms with zero coefficents are usually omitted , as they were in the preceding sentence.

EXAMPLE 2 4 - 6x + 4x3 E E[x]. However, the Eix], because it cannot be written with even coefficients.

Let Ebe the ring of even integers. Then polynomial xis not in

Polynomial Arithmetic The rules for adding and multiplying polynomials follow directly from the fact that

R[x] is a ring.

EXAMPLE 3 If f(x)

=

1

+

5x - x1 + 4x3 + 2x4 and g(x)

=

4

+

2x + 3r + x3 in Z7[x], then

the commutative, associative, and distributive laws show that

f(x) + g(x)

=

=

=

(I

+

5x - x'- + 4x3 + 2x4) + (4 + 2x + 3x2 + x3 + Ox4)

(1 + 4) + (5 + 2)x + ( 1 + 3)x2 + (4 + l )x3 + (2 + O)x4 5 + Ox + 2x2 + 5x3 + 2x4 5 + 2x2 + 5x3 + 2x4• -

=

G, there T and is

*AHhough in common use, the term "indeterminate" is misleading. As shown in Appendix is nothing undetermined or ambiguous about x. It is a specific element of the larger ring

not an element of R.

tvariables and equations will

be dealt with

in Section

4.4.

�20l2C...S.J...umlil.g.At��....,.oathl� me..-t.ar�iowtdlOl!�J*I.. 0.10�..-.--mkd.�caal-OlllJ .-,.....tfam.M1118oi:*ndfix'�1).Bdladlll. �--mDJ'�..-a.o;,-dh:tbt�'-uiag..,.n-._c.g.pu--.--•Dgbtm-__,,,.��-..,.--il......_.:ligtu�...-. ..

.....

...

....... tm

88

Chapter 4

Arithmetic in

F[x]

EXAMPLE 4 The product of

1 - 1x + x2 and 2 + 3x in Q[x] is found

tive law repeatedly:

(1 - 1x + x2)(2

+

3x)

by using the distribu­

=

1(2 + 3x) - 1x(2 + 3x) + x2(2 + 3x) 1(2) + 1(3x) - 7x(2) - 7x(3x) + x2(i) + x2(3x) 2 + 3x - 14x - 2lx2 + 2x2 + 3x3

=

2 - llx - 19x2 + 3xl,

=

=

The preceding examples are typical of the general case. You add polynomials by adding the corresponding coefficients, and you multiply polynomials by using the

distributivelaws and collecting like powers of x. Thus polynomial addition is g iven by the rule:*

(ao + a1x + a-i.:C + + a;,X") + (ho + h1x + b2x2 + +b;,X") + (a,, + b1.)::t!' (Oo + ho) + (a1 + bi)x + (a + bi)xi- + 2 ·

·

·

·

=

·

·

·

·

·

and polynomial multiplication is given by the rule:

+ bmx!") (ao + a1x + a,;x2 + + a,,x")(b0 + h1 x + h x2 + 2 2 + + + (aJJ + 1 a1bo)x (aob aobo a1h1 + ¥o)X + + a,,bmxi+"'. 2 ·

·

·

•

·

=

·

·

·

·

For each k � 0, the coefficient of ;/
aohk + a1bk-1 + ¥k-2. +

·

·

·

+ ak-2.b,, + ak-1h1 + a,)Jo

k =

�aA-e. t=O

whereai =OR if i> n andb1

=

ORifj>

m.

R[x] that if R is com­ R has a multiplicative identity also the multiplicative identit y of R[x] (Exercise 8).

It follows readily from this descrip tion of multiplication in

mutative, then so is

lR, then lR is

Definition

R[x] (Exercise 7).

F urthermore, if

Let f(x) a0 + a1x + l!i!X2 + + ilnx" be a polynomial in R[x] With an oft OR. Then a11 is called the leading coefficient of f(x). The degree of f(x) is the integer n; it is denoted "deg f(x)". In other words, deg f(x} is the largest exponent of x that appears with a nonzero coefficient, and this coefficient is the leading coefficient. =

·

·

·

EXAMPLE 5 3 - x + 4x2 - 7x3 ER[x] is 3, and its leading coefficient is -7. 1 (3 + 5x) 1 and deg (x 2) 12. The degree of 2 + x + 4x2 ox3 + Ox5 is 2 (the largest exponent of x with a nonzero coefficient); its leading coefficient is 4.

The degree of Similarly, deg

=

=

*We may assume that the same powers of x appear by inserting zero coefficients where necessary. eap,ngm.20:12�1..umiq.A:l.lliala 11--4.....,-aatn. t:IDJllilrd,. llC...t,, ar�io.wtdaarls,_,. 0.1"�,.._.-mkd.JIBQ'ICClllUmt�M ....... ftonb•Bam:.ndkir�.Bdbmbll_...._ ...._._q-��._.fld.__...,.a11N:t... �a--.�c.a.� ...... .. ftebtb....,.. ....... � • ...,. ... w......_..:dPLI�...-. ..

4.1

Polynomial Arithmetic and the Division Algorithm

89

The ring R that we start with is a subring of the polynomial ring R(x]. The elements of R, considered as polynomials in R[x], are called constant polynomials. The polyno­ mials of degree 0 in R[x] are precisely the nonzero constant polynomials. Note that the constant polynomial OR does not have a degree

(because no power of x appears with nonzero coefficient).

Theorem 4.2 If R is an Integral domain and f(x), g(x) are nonzero polynomials in R[x], then deg[f(x)g(x) ]

=

de g f (x)

+

deg g(x).

Proof,. Supposef(x) = ao + aix + a2X'- + + a,.x' and g(x) = b0 + b1x + �+ + bmx!" with a,,* OR and bm :FOR, so that degf(x) = n and · · ·

· · ·

deg g(x) = m. Then

f(x)g(x)

=

aJJo + (aoh1

+

a1bn)x + (a.jl0 + a1b1 + � +

·

· ·

+

aHb,,.X'+"'.

The largest exponent of x that can possibly have a nonzero coefficient is But a,.b,,. :F OR because R is an integral domain and aH :F OR and b,,. :F OR. Therefore,f(x)g(x) is nonzero and deg[/(x)g(x)] = n + m = deg/(x) + deg g(x). •

n + m.

Corollary 4.3 If R is an integral domain, then so is R[x].

Proof"' Since R is a commutative ring with identity, so is R[x] (Exercises 7 and 8).

The proof of Theorem 4.2 shows that the product of nonzero polynomials is nonzero. Therefore, R[x] is an integral domain. •

in R[x]

The first five lines of the proof of Theorem 4.2 this conclusion.

are

valid in any ring and lead to

Corollary 4.4 Let R be a ring. If f(x), g(x), and f(x)g(x) are nonzero in R[x], then deg [f(x)g(x)] � de g f(x)

+

deg g(x).

EXAMPLE 6 In �[x], let/(x) = 2x4 and g(x) = 5 x. Thenf(x)g(x) = (2x4)(5x) = 4x5, so deg [f(x)g(x)] deg/(x) + deg g(x). However, if g(x) = 1 + Ji'-, then =

f(x)g(x) = 2x4(1 + 3x3) = 2x4 + 2 3X' = 2x4 + OX'= 2x4, •

which has degree 4. But degf(x) + deg g(x) = 6. So deg [f(x)g(x)] < deg/(x) + degg(x). CopJDaM2012C..-.l...Mmimg.A1Ripb:a-..d.-...,.autbll� KlUD91d.«�:Mt.1"ldliw:-lapld.. O.'lo�dalD.-lbinlpat;Je�a.J'h�fmm._1111kd:.udll;x'�).�..w...rm.

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90

Chapter 4

Arithmetic in F[x]

For information on the degree of the sum of polynomials,

see

Exercises 4 and 12.

Corollary 4.5 LetR be an integral domain and f(x)ER[x]. Then f(x) is a unit inR[x] if and only if f(x) is a constant polynomial that is a unit inR. In particular, if Fis a field, the units in F[x] are the nonzero constants inf. Remember that the proof of an "if and only if" statement requires two separate proofs.

Proof of Corollary 4.5 ... First, assume thatf(x) is a unit in R[x]. Thenf(x)g(x) =lR for some g(x) in R[x]. By Theorem 4.2, deg/(x) + degg(x) =deg [f(x)g(x)] =deg lR =0. Since the degrees of polynomials are nonnegative, we must have deg/(x) =0 and degg(x) =0. Therefore,/(x) and g(x) are constant poly­ nomials, that is, constants in R. Sincef(x)g(x) =lR,f(x) is a unit in R. Conversely, assume that.f(x) is a constant polynomial that is a unit in R, say f(x) =b, with b a unit in R. Let h(x) =b-1• Thenf(x)h(x) =bb-1 =lR. Therefore,f(x) is a unit in R[x]. The last statement of the corollary follows immediately since every nonzero element of a field is a unit in the field (see Example 6 in Section 3.2). •

EXAMPLE 7 The only units in Z[x] are 1 and -1, since these are the only units in Z. The units in R[x] (or in Q[x] or in C[xD are all nonzero constants, since

R, 0, and C are fields.

Corollary 4.5 may be false if R is not an integral domain (Exercise 11).

EXAMPLE 8 5x + 1 is a unit in Z25[x] that is not a constant because (as you should verify)

(Sx + l)(20x + l) =l.

The Division Algorithm in F[x] Our principal interest in the rest of this chapter will be polynomials with coefficients in

a field F (such as Q or IR or Z5). As noted in the chapter introduction , the domain F[x] has many of the same properties as the domain Z of integers, including the Division Algorithm (Theorem 1.1), which states that for any integers a and b with b positive, there exist unique integers q and r such that a=bq+r

and

0�

r

< b.

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Polynomial Arithmetic and the Division Algorithm

4.1

For polynomials, the only changes

are

to

require the divisor to

be

91

nonzero and

to

replace the statement "O $ r < b'' by a statement involving degrees. Here is the formal state­

ment (with/(x) in place of a,g(x) in plaoe of b, and q(x), r(x) in plaoe of

q,

r

respectively).

Theorem 4.6 The Division Algorithm in f[x] Let F be a field and f(x}, g(x) EF{x] with g(x) # Op Then there exist unique polynomials q(x) and r(x) such that

f(x)

=

g(x)q(x)

+

r(x)

and either r(x)

=

or

OF

deg r(x) < deg g(x).

Example 9 shows how polynomial division works and why the Division Algorithm is valid in one particular case. EXAMPLE 9

dividef(x)

We shall 3x5 + 2x4 + 2x3 + 4x2+ x - 2 by g(x) 2x3 italic column on the right keeps track of what happens at each step.* =

=

+

1.

The

dhlisorg(_x)

! 113�ir+ +

+-quotient q(x) 2i' 2x1 4x2 2 +- diVidendf(x) 3� + �x2 +-(ir)g(x) 2 2x"'+2x'!+ �+x-2 +-f(x)-(�x2)g(x) 2 +x 2x4 +-xg(x) 2x3+�x2 +-f(x) - (�x2)g(x) xg(x) 2x3 + 1 +-lg(x) 5 - 3 +-f(x)-(iryx)-xg(x) - lg(x) der r(x) 2x2 f(x) -g(x) (ir x 1) x

2J!-+

+

1

+

+

+x

2

remain

-

-2

-

-----+

+

+

=

=

f(x) -g(x)q(x) The last line on the left side and the last three lines on the right side show that

f(x)

g(x)q(x) + r(x). f(x) - g(x)q(x) r(x) or equivalently, So the Division Algorithm holds for the polynomialsf(x) andg(x). =

•Division Re(esher:Thefir&tterm of the quotient dividend divisor

j-r

is obtained by dividing the leading term ofthe

(at") by the leading term of the divisor (2x"):at"/2x3

( (irpCx))

=

=

ixt.

The product of this term and the

is then subtracted from the dividend resuHing in 2x' +

2x"

+

j-r x +

-

2, as

shown. The process is repeated, using this last expression as the dividend and the same divisor, and continues until you reach a polynomial with degree smaller than the degree of the divisor. �2012.C....,l...Mmiq.AIRqlna-..d.MaJ"mtbll� �-ar....... :towballl«laJ*I.. 0..tD�fiBID.-tinl:pat;Joootm:a.,.'8....,....m_ta:.:J.beBo'*:.udkx-��----­ dlMm&d.-..:my�-mmllldmmmll...-...,..act.b�---.�c.a.i...mog--miftgbttD__,,,..mdICDl dllklDlil. llllnl•_..,.lillll��:Dgbb�...-.:lit.

92

Chapter 4

Arithmetic in

F[x]

Of course, an example is not a proof, even though you can readily convince your­ 5).

self that the same procedure works with other divisors and dividends (Exercise

Consequently, skipping the proof until you are familiar with mathematical induc­ tion, would be quite reasonable. Th at's why the proof of Theorem

4.6 is marked

optional.

Proof of Theorem 4.6 The Division Algorithm

(Optional)"

q(_x) and r(x). degf(x)
We first prove the existence of the polynomials Case 1: If f(x) =Op or if

=

Assume inductively that the theorem is true whenever the dividend has degree less than n. This part of the proof is presented in two columns. The left-hand column is the formal proof, while the right-hand column refers to Example 9. The example will help you understand what's being done in the proof.

PROOF

EXAMPLE9

We must show that the theorem is true whenever the dividendf(x) has degree n, say

a x + ao 1 Op. The divisor g(x) must have the

f(x) with a,, * form

=

a,,:x!' +

· •

g(x) =b;,,X"' + with h,.. * Op and m $

· ·

n.

+

·

·

+

hx 1

+

h0

n=5 f(x) 3x5 =

,.......,-.,

t

2x3 +

4x2 + x

-

2

a,.x"

m=3 g(x) =2x3

+ l

,.......,-.,

b,.x1"

We begin as we would

in the long division of g(x) intof(x). Since Fis a field and b,,. * Op, bm is a unit. Multiply the divi­ sor g(x) by

+ 2x4

a,,b,,.-1:x!'-m to obtain

a,,bm-l:x!'-"'g(x) = a,,b,,.-1x"-"'(h;,,X"' +

· ·

·

+

h1x

+

h0)

3

2x2g(x)

3

=

=

2x2{2x1

first term of the quotient

+ l)

3 3x5 +-x2 2

*We use the Principle of Complete Induction; see Appendix C.

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4.1

Polynomial Arithmetic and the Division Algorithm

93

Since a,)J,,.-1X'-"'g(_x) and f(x) have the same degree and the same leading coefficient, the difference f(x) - a,.bm -tx"�,.g(x) is a polynomial of degree less than n (or possibly the zero polynomial). Now apply the induction hypothesis with g(x) as divisor and the poly­ 1 nomial .f(x) - a,;,,,.- x!'-"'g(x) as dividend (or use Case 1 if this dividend is zero). By induction there exist polynomials q1(x) and r(x) such that

fourth line of long dMsion

f(x) -a,.b,,.-Jxr-"'g(x) = g(x)q1(x)+r(x) and r(x) = Op or deg r(x) < deg g(_x).

q1(x)

=

x

+ 1

r(x)

last part of

5

=

-r - 3 2

remainder

the quotient Therefore,

f(x)

=

r(x)

=Op

g(x)[a,.b,,. -tr--m+ q1(x)] or

deg r(x)

<

+

r(x)

and

deg g(x).

Thus the theoremistruewith q(x) = a,.bm-1x"-"'+q1(x)whendegf(x) = n. This completes the induction and shows that q(x) and r(x) always exist for any divisor and dividend. To prove that q(x) and r(x) are unique, suppose that q2(x) and r.J.x) are polynomials such that f(x)

= g(x)q1(x)+ r2(x)

and

Then

g(x)q(x)+r(x)

=

r2(x)

= Op

or deg r2(x) < deg g(x).

f(x) = g(x)q2(x) + r1(x),

so that g(x)[q(x) - q2(x)]

= r2(x)

-r(x).

q(x) - q2(x) is nom.ero, then by Theorem 4.2 the degree of the left side is deg g(x)+ deg[q(x) - qz(x)], a number greater than or equal to deg g(x). However, bothrz(x) and r(x) have degree strictly less than deg g(x), and so the right-hand side of the equation must also have degree strictly less than deg g(x) (Exercise 12). This is a contradiction. Therefore q(x) - q2(x) = Op, 01; equivalently, q(,x) = q.J..x). Since the left side is zero, we must have r1(x) - r(x) = O.F> so that r1(x) = r(x). Thus the polynomials q(,x) and r(x) are unique. • If

• Exercises NOTE: R denotes a ring and F afield A. 1.

Perform the indicated operation and simplify your answer: 2 (a) (3x4+ 2x3 -4r+ x+ 4)+ (4x3+ x + 4x+ 3) inZ5[x] (b) (x + 1)3 inZ3[x] (c) (x - 1)5 inZ5[x] 2 (d) (x - 3x +2)(2x3 - 4x+ 1) inZ7[x]

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94

Chapter 4

Arithmetic in F[x] 2.

Show that the set of all real numbers of the form a0

2

+ aJ'lT + a27r +

·

·

·

+ a,,'TT",

with n � 0 and a1 E Z

is a subring of R that contains both Z and 7r. 3.

(a) List all polynomials of degree 3 in Z2[x]. (b) List all polynomials of degree less than 3 in Z3[x].

4.

In each part, give an example of polynomials/(x), g(x) E Q[x] that satisfy the given condition:

(a) The deg of f(x) + g(x) is less than the maximum of deg/(x) and deg g(x). (b) Deg [f(x) + g(x)] =max {deg/(x), degg(x)}. polynomials q(x) and r(x) such that/(x) =g(x)q(x) + r(x), and r (x) =0 or deg r(x) < deg g(x):

5. Find

(a) f(x) = 3x4 - 2x3 + 6x2 - x + 2 and g(x) (b) f(x) = x4 - 7x + l and g(x) =2x2 + l

in

==

x2 + x +

1

in Q[x].

Q[x].

(c) f(x) =2x4 + x?- - x + l andg(x) =2x - 1 in Zs[x]. (d) f(x) =4x4 + 2x3 + 6x2 + 4x + 5 andg(x) =3x?- + 2 in .l7[x]. 6.

Which of the following subsets of R(x] are subrings of R[x]? Justify your answer:

(a) All polynomials with constant term OR. (b) All polynomials of degree 2. (c) All polynomials of degrees

k,

where k is a fixed positive integer.

(d) All polynomials in which the odd powers of x have zero coefficients. (e) All polynomials in which the even powers of x have zero coefficients. 7.

If R is commutative, show that R[x] is also commutative.

8. If R has multiplicative identity lR, show that lR is also the multiplicative identity of R[x]. 9.

If c E R is a zero divisor in a commutative ring R, then is c also a zero divisor in R[x]?

I 0. If F is a field, show that F[x] is not a field. [Hint: Is x a unit in F[ x]?] B. 11. 12.

Show that 1 + 3x is a unit in .l.i[x]. Hence, Corollary 4.5 may be false if R is not an integral domain. If f(x),g(x) E R[x] and/(x) + g(x) *- OR, show that deg[f(x) + g(x)] s max {deg/(x), degg(x)}.

13.

Let R be a commutative ring. If a,, *- OR and f(x) 0() + a1x + a2x?- + + a,,x!' (with a"*- O� is a zero divisor in R[x], prove that a,, is a zero divisor in R.

14.

(a) Let R be an integral domain and/(x),g(x)

=

· ·

E R[x]. Assume that the leading coefficient of g(x) is a unit in R. Verify that the Division Algorithm holds forf(x) as dividend andg(x) as di'\isor. [Hint: Adapt the proof of Theorem 4.6. Where is the hypothesis that F is a field used there?]

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...

·

...

........

4.2

(b) 15. Let

Divisibility In

F[x]

95

Give an example in Z[x] to show that part (a) may be false if the leading coefficient of

R

g(x) is not a unit. [Hint: Exercise 5(b)

be a commutative ring with identity and

with Zin place of

Q.]

a E R.

(a) If d OR, show that lR+ax is a unit in R[x]. [Hint: Consider a2x2.] (b) If a4 OR, show that IR+ax is a unit in R[x]. =

l

-

ax+

=

16. Let

R be a commutative ring with identity and a E R . If 1R+ ax is a unit in R[ x], show that d' OR for some integer n > 0. [Hint: Suppose that the inverse of lR+ax is b0+b1x+b2i1'+ +bd'. Since their product is lR, b0 lR (Why?) and the other coefficients are all OR.] =

· · ·

17. Let

R

==

be an integral domain. Assume that the Division Algorithm always

holds in R[x]. Prove that

R

is a field.

18. Let 1p:R[x]-+ R be the function that maps each polynomial in constant term (an element of

R[x] onto its R). Show that 'Pis a surjective homomorphism

of rings.

19. Let 1p:Z[x]-+ Zn[x] be the function that maps the polynomial ao+a1x+ ·

+ ad' in Z[x] onto the polynomial [ao]+[a1]x+ +[a.k]x", where [a] denotes the class of the integer a in Z,.. Show that