~1. Summarizing, X one to one onto Y and \342\200\242 F be given by \302\242(/)= \342\200\242 *)a_1;y_1 e Ker(0). \342\226\240 4>\\G\\
4>-\\y)
= x.
as
second
the
and second
we get function
0.13 Definition
Two sets X X
0.14
Example
and Y have the that is, if there F,
onto
same cardinality if there exists exists a one-to-onecorrespondence
function mapping
a one-to-one
X and Y.
between
\342\226\240
because /(2) = /(-2) = 4 it is not onto E becausethe range is the proper subset of all nonnegative but 2 ^ \342\200\2242. Also, in E. However, g : E \342\200\224>\342\200\242 E defined numbers by g(x) = x3 is both one to one and onto The function
E where / : E ->\342\200\242
= x2 is not
f(x)
to one
one
E.
A
that Z
showed
We by
so that
Ko,
|Z| =
may have the same as a set having this A
we set
downward,
through the
could
have
Z+
Ko.
the same
It is
of elements
number
We denote this cardinal number that a proper subset of an infinite set whole set; an infinite set can be defined
cardinality.
fascinating
as the
property.
We naturally wonder has cardinality Ko if
set
so that for the
of
and
|Z+| =
\"number
and
sets have
all infinite
whether
only
them\"
if all using
of its Z+.
of fractions Q. The square array and contains all members of
Figure extends
Q. We
the
out
and
|Q|
= K0 also.
0.15Figure
to the
infinitely
have
shown
as the
cardinality
be listed
0.15 indicates
this array. Imagine the fractions to be glued to this and pulling to the left in the direction of the all elements of Q appearon it in an infinite row as
string
same
could
elements
that
in
this
row,
is possible
right and
a string
set Z.
an infinite
infinitely
winding its way string. Taking the beginning the string straightens arrow, \342\200\242 \342\200\242 \342\200\242. Thus 0, \\,\342\200\224 ^,1,-1, |,
6
Section
0
Relations
Sets and
set S
the
If
= {x e
E | 0 < x < 1}has
in a column
decimals
as unending
Ko, all
cardinality
its elements couldbelisted perhaps as
downward,
infinitely
extending
\342\200\242 \342\226\240 0.3659663426 \342\200\242
0.7103958453
\342\200\242 \342\200\242 \342\200\242
0.0358493553
\342\200\242 \342\200\242 \342\200\242
0.9968452214---
in S. Surely S contains a such array must omit some number a nth after the decimal number different from 0, from 9, digit point having \342\200\242 from the nth digit of the nth number in this list. For example,r might start .5637and \342\200\242. in 5 rather than 3 after shows r cannot the first S The the decimal be number point r cannot be the The 6 rather than 1 in the second digit shows listed in the array shown. second number listed, and so on. Becausewe could make this argument with any list, 15 indicates we see that S has too many elements to be paired with those in Z+. Exercise the cardinality of E by E has the same number of elementsas S. We just denote that
We now
argue
that
number r
19 indicates
Exercise
|E|.
greater
than
and
disjoint
with
section
partition in
a study
of such breakups, or partitions
of a set
S is a collection of nonempty subsets the subsets. The subsetsare the
exactly
When
of
sets.
one of
discussing
a partition
of a set S, we denote
by x
of S cells
such
of the
the cell
that
every
element \342\226\240
partition.
the
containing
element
S.
x of
0.17 Example
Relations
Equivalence
disjoint
occasion
of S is
even
if no two of them have have any element in common. Later we will to break up a set having an algebraic structure of a notion into addition) (e.g., subsets that become elements in a related algebraic structure. We concludethis
Sets are
A
many different cardinal numbers
are infinitely
there
that
|E|.
Partitions
0.16 Definition
any
as its
subset of even positive integers (thosedivisible by divided integers (those leaving a remainder of 1 when of Z+ into two cells. For example, we can write
Z+
Splitting
into the
of odd positive a partition
14 We
divisible divided
divided
could
Z+
by 3, containing by 3, and the last containing by 3.
into
three
cells,
all positive positive
by
14, 16. 18, \342\200\242\342\200\242\342\200\242}.
= (2,4,6.8,10,12,
also partition another
the subset we obtain 2),
2) and
one consisting
of the
positive
integers
integers leaving a remainder of 1 when integers leaving a remainder of 2 when
Z+ into n cells according Generalizing, for each positive integer n, we can partition \342\200\224 \342\200\242 \342\200\242 the remainder is 0, 1, 2, \342\200\242 n 1 when a , positive integer is divided by n. These cells are the residue classes modulo n in Z+. Exercise 35 asksus to display these and 5. \342\226\262 partitions for the cases n =2,3, to
whether
x,
Each
partition
e S,
let x
y
if and
y
S yields a relation
.M on
S in
only if x
in the
same cell of
a set
of
.M
y are
and
we would write x \342\226\240>\342\202\254 Definition y as (x, y) e ,^\302\276 (see shows that this relation .^\302\276 on S satisfies the three properties of an in the following definition.
Definition
An equivalence
x,
relation .^\302\276 on
(Symmetric)
3.
(Transitive) If x .^\302\276 y and
If x
-jfcy,
x is the
S, the
\342\226\240
need
we
same cell as y
in the reflexive
relation
equality
a partition
only observe (that
that
if y
is, y
of S satisfies the is in the same cell
..M x).
and transitive
properties
= defined
by the subset
We
the
leave
Exercise
to
{(x, x) \\ x
28.
e S]
relation.
of \342\226\262
Let n
The equivalence relation on Z+ corresponding modulo n, discussed in Example 0.17, is denoted by =\342\200\236. Rather than write a=\342\200\236b, we congruence modulo n. It is sometimes write a = b (mod n), read, \"a is congruent to b modulo n.\" For example, we usually have 15 = 27 (mod 4) because both 15 and 27 have remainder 3 when divided by 4. \342\226\262 (Congruence the
partition
Modulo n) of Z+ into
e Z+.
classes
residue
,^/B on the set Z be defined by n .^\302\276 m if ^\302\276 is an equivalence determine whether relation. Reflexive a .M a, because a2 > 0 for all a e Z.
Let a relation
If a
Symmetric
If
we
case
knew b =
not have relation. We
\342\200\2243 ,^2 5.
then
ab
> 0 and b > then ab 0 and be so ba
> 0,
> 0 whence
a
of thought shows that relation c/Z, is not transitive,
moment
Thus the
and only if
>
nm
0, and
let us
.Ma.
Thus ab2c = acb2 > 0. ,M c. We have to examine the \342\200\2243,^0 and 0 .M 5, but we do and hence is not an equivalence > 0.
\342\226\262
observed
that
that
follows
equivalence
b2
0 separately. A
show
(Equivalence
.M b,
\\ia.yib and b .M c, > 0, we coulddeduceac
Transitive
0.22Theorem
three properties for all
x.J?z.
then
,>? correspondingto
condition
set For any nonempty S x S is an equivalence
to
Example
these
satisfies
relation
equivalence
x.
.^
y ,3%z
why the relation in the definition,
illustrate
is, if x .>? y), then (that similar observations to verify
0.21
y
2.
symmetric
Example
then
(Reflexive) x .Mx.
as x
0.20
S is one that
1.
To
0.19 Example
a set
the
S.
z e
y,
way: namely, for In set partition. 0.7). A bit of thought
a natural
notation,
0.18
7
Relations
and
Sets
above that a partition
yields a natural
relation on a set yieldsa states both results for reference.
an equivalence
and
Relations
relation
on
Let S
Partitions)
S. Then ~ yields a
a=
\\x
natural
e
partition
S \\x ~
be a
partition
nonempty
of S,
a}.
relation.
equivalence
where
of the
set
We now
set. Thetheorem
and let
~ be
an
Setsand
Section0
8
Relations
each partition of S gives and b are in the same
Also,
if a
only
~ on S where
to an equivalence relation cell of the partition.
rise
a ~
and
b if
eSdo give a partition cells a = {x eSU~a)fora We must show that the different of S is in some cell and so that if a e b, then a =b. Let of S, so that every element a e S. Then a e a by the reflexive condition (1), soa is in at least one cell. Supposenow that a were in a cell b also. We need to show that a = b as sets; this that is a standard show that a cannot be in more than one cell. There will way to show
Proof
two
sets are
same:
the
Show
that each
set is a subsetof
the
other.
x ~ a. But aefc,soa~i>. We show that a c.b. Let x e a. Then Then,by ~ c a. Let we show that \302\243 Now condition (3), x b. Thus ocj. b, sox \342\202\254 ~ fc. But aefc,sofl~fc and, by symmetry (2), b ~- a. Then by transitivity y = a and our proof is complete. so >' e a. HenceKq also, so \302\243
cell
Each
in the
an equivalence relation is
from
partition
arising
listing
its elements.
an
the
transitive
y
Then e \302\243.
(3),
y
~
a, \342\231\246
equivalence
class.
0
\342\226\240 EXERCISES
In Exercises 1 through
1. {x e
M
3. {m e
Z
5.
{n
e Z+
6.
{n
e Z | n2
7.
{n
e Z
< 0} | 39 < n3
&Q\\x may e Q | x may
Let
elements
the
A = {1,
whether
it
whether the object describedis indeed
= 3}
m2
-
m
<
115}
(is well defined).
a set
Give an
alternate
< 57} an integer}
be written
with denominator greater
be written
with positive
in {a,
2, 3} and
is a
b, c} x
B = {2,
denominator
than
less
100}
than 4}
{1,2, c}. between 4, 6}. For eachrelation A into S. If it is a function,
A and
function mapping
decide
B
given
whether
c. {(1,
b. {(1,4),(2, d. {(2,2), (1,
(2,4), (3, 6)}
a. {(1,4),
6), (1,2), (1,4)}
e. {(1, 6),(2,6),(3, 6)} Illustrate by
|
as a
it is
subset of
one to
one
A x S, decide and whether it is
B.
onto
13.
Z
4. {m e Z | m2
is a large number}
|n
e Q | x is almost
12.
e Z}
set.
8. {x
11. List
for some n 10, decide
9. {x {x
the set by
2. {m e
= 60
| mn
In Exercises 5 through of each description
10.
4, describe
\\x2 = 3}
geometrically
indicating
in Fig.
f.
that two line
0.23
what
point
{(1,2),
6), (3,4)}
6), (3,4)}
(2, 6),
(2,4)}
segments AB and CD of different length have y of CD might be paired with point x of AB.
the
same
number
of points
9
Exercises
p /
A
,B X
c*-
0.23 Figure 14.
that for
Recall Show
the
interval
onto the
first
15. Show
calculusthat the set S.]
[a,
fc]
= {* e R | a < * < \302\243>}. by [a, fc] for a one-to-onefunction / mapping
is denned
in R
a formula
by giving
cardinality
second. 3] and
c. [a, b]
[5, 25]
and
[c, d]
Find an elementary function of < 1} has the same cardinality as R. [Hint: and scale appropriately to make the domain one to one onto R, and then translate
< x \342\202\254'R\\0
{x
an interval
maps
-5^(A).The set
d] e
interval
closed
the same
b. [1,
we denote
set A,
any
{a, b,
b, the
have
[0, 2] S =
that
a <
and
given intervals
a. [0,1]and
For
a, b e R
that the
the collection
SP (A)
by
of
power set of A.
is the
SP(A)
of A.
subsets
all
For example, if A 19 deal with the
16 through
Exercises
= {a, b, c, d}, then notion of the power
set of a set A. 16.
List the
elements of the
a. 0 17. Let
a finite
be
A
For
set A,
any
subset of A
set,
oh
the
make a conjecture
exercise,
preceding
about
{a,
the
b, c}
of
value
let BA be the set of all functions mapping of the set d^(A). [Hint: same as the cardinality
A into Each
B = {0,1}.Show
the set element
that
of BA determines a
natural way.]
set of a set A,
to be able to be put in a one-to-one are an infinite number of infinite cardinal A into .5s(A) to be given. Show that \302\247 numbers. be cannot [Hint: Imagine a one-to-onefunction \302\242) mapping onto d?(A) by considering, for each x & A, whether x e (j>(x) and using this idea to define a subset S of A that is not in the range of \302\242.] Is the set of everything a logically acceptable concept? Why or why not? the power
that
Show
= {1,2}and
20. LetA a.
b.
to decide
i. 3 +
K0,
Illustrate
many
2, 3,
you
would
Continuing
the
three
what
B, you
we consider
with a figure
0, 1,
elements
that there
means
consider
that 2
consider to
be the
we
why
would
+ 3 = 5. Usesimilar
in
many are
consider to be the blanks
text to
value
the
preceding
to give a
list of five
in
of the
there
of
by plotting
decide
0
in the interval
numbers \342\200\242 \342\200\242 9? How \342\200\242,
the idea
3 = 6 2 \342\200\242
that
the
what
you
< 1 can
your
own
+ K0.
in the
expressed
about
and
numbers,
sets of
of A x B in the plane R x R. Use similar consider to be the value of Ko \342\200\242 Ko.
Following
How 10K\302\260.
with
points
would
form .#####?
exercise and using cardinal
be
the
reasoning
of
value
ii. K0 why
How
has too many
intuitively
= {3,4,5}.
A and
using
choice
or infinite,
finite
why this
A. Explain
lets
Illustrate,
reasoning
22.
d.
c. {a,b}
s. Based
=
\\A\\
power set.
your conjecture.
is the
BA
correspondencewith
21.
let
of the
cardinality
or infinite,
finite
in a
and
to prove
try
of
the cardinality
19.
give the
set and
given
b. {a}
\\.3^(A)\\. Then
18.
the
set of
power
12N\"
Exercises
each of
K0.|R|, _,_,_.
this
form
idea,
each # is a digit .##, where and Exercise 15, decidewhat
2N'\302\260?
18 and which
to fill in notation 19, use exponential than the preceding one.
is greater
10 In
27,
23 through
Exercises
and Relations
Sets
0
Section
find
number
the
23. 1 element 26. 4 elements 28. Considera partition
of a
set S.
satisfies
the
In
arising
partition 29. 31.
from each
n M m in Z
=
x.rfymRif\\x\\
33. n J% m
34.
n
35.
Using
,>8
Z+
in
m in Z+
n
a.
relation is
given
an
cell
of
explanations
reflexive
the
why
on the
relation
equivalence
30. x .M y
the
same
number
the
same
final digit in
of the the
x JZ
32.
\\y\\
the
of digits in
the usual
and let ~ be defined for some q e Z.
on
usual
equivalence
on Z.
relation
in M if inMif
x > -
\\x
set. Describethe
if
and
(It is
<
y\\
3
ten notation
base
the residue
if r
only
y
ten notation
classes modulo
= 3
r ~ s
Z by
y
base
\342\200\242 form {#,#. #. \342\200\242 for an infinite set, write \342\200\242} value of n. indicated
- s is divisible
called \"congruence
modulo
c.
n
by
n, that
=
n\" just
n
in Z+
discussed
5
as it
is, was
if
and
only
if
for Z+.
See
modulo
n,
b.)
when restricted of Example 0.20. of this
cells
modulo
37.
same similar
Write
relation.
m have
~ is an
the
in
relation
the
why
equivalence
b. Show that,
c. The
the
whether
b. n
that
Show part
for an equivalence
= 2
36. Let n e Z+ s = nq r \342\200\224
y are
x and
and m have
0.17 for
elements.
relation.
n and
set notation
in Example
a.
if n if
if
0.18 explained
Definition
following
and only
number of
given
25. 3 elements
> 0
nm
if
5 elements
the
also satisifed.
34, determine
29 through
Exercises
27.
.^\302\276 y if
condition
symmetric
and transitive properties are
2 elements
The paragraph
x
of a set having
of different partitions
24.
in
to
partition of Z
Z rather
the
are
subset
residue
than in Z+ using
the
Z+ of Z,
this
~
is vhe
classes modulo notation
n
equivalence
in Z.
\342\200\242 #. #, \342\200\242, {\342\226\240
#.
relation,
congruence
Repeat Exercise 35 for the residue infinite sets.
classes
\342\200\242 \342\226\240 for these \342\200\242}
the concept of a one-to-onefunction I think I know the reason. You (mapping). from A B. It a direction associated with to seems to expect a reasonable it, one-to-one mapping to be a mapping that carries one point of A into one point of B, in the direction simply But of course, every mapping of A into B does this, and Definition indicated by the arrow. 0.12 did not say in mind, make as good a pedagogical that at all. With this unfortunate situation caseas you can for calling the in Definition 0.12 two-to-two it is almost impossible to functions described instead. (Unfortunately, functions get widely used terminology changed.) Students
often
see, a mapping
misunderstand
B has
part
and
Groups
Introduction
1
Section
Section 2
Binary
3
Isomorphic
Section
4
Groups
1
Structures
Cyclic Groups
Section 7
section
Binary
Subgroups
6
Section
and Examples
Operations
Section
Section 5
Subgroups
Generating
Sets
and Cayley
Digraphs
and Examples
Introduction
of abstract algebra. we attempt to give you a little idea of the nature Both addition with addition and multiplication of real numbers. For example, addition combine two numbers to obtain one number. and multiplication to be binary 2 and 3 to obtain addition and multiplication combines 5. We consider sets in which we have one and examine operations. In this text, we abstract this notion, In
this
section,
We are all
familiar
an algebra of a binary operation on a set as giving operations. We think of that algebra. To illustrate are interested in the structural properties with note what we mean by a structural property our familiar set E of real numbers, x = a/2. that the equation x + x = a has a solution x in E for each a e E, namely, a solution x = a does not have However, the corresponding multiplicative equation x \342\226\240 structure than E with in E if a < 0. Thus, E with addition has a different algebraic more
binary
on the
set, and
or
we
multiplication.
Sometimes two different sets with what we naturally regard as very different binary structure. For example, we will see in operations turn out to have the same algebraic structure as the set E+ of Section 3 that the set E with addition has the same algebraic with multiplication! positive real numbers We will This section is designed to get you thinking about such things informally. some in Sections 2 and 3. We now turn to make everything examples. precise numbers of magnitude 1 provides us with several examples that will Multiplicationof complex and with a review of complex numbers be useful and illuminating in our work. We start
their
multiplication.
11
Groups and Subgroups
a +
bi
bi
\342\226\240\342\200\224T I I i
-3
-4
i
-2
\\a
(
-1
0 \342\200\224i
-2i
1.1 Figure
Numbers
Complex A
real
as an
number x-axis.
visualized geometrically as a point on number can be regarded as a point that we label the vertical axis as the Note
can be
1.1. and label the point >'-axis, coordinates (a. Cartesian is defined by in Fig.
shown
one
unit
above the
b) is labeleda C:
+ bi
{a +
origin
I
numbers we write
complexnumber sowe require
i
as i and
developed after the invented to provide a solution were
numbers was
numbers
complex
b e
a.
We consider
ffi to be a subset of the complex the complex number r + Qi.For example, we write 0 + \\i and 0 + 0; as 0. Similarly,
rather than just the with than 1. The point
set C of
1.1. The
in Fig.
bi
yi-axis
we often regard
Euclidean plane, as
in the
i rather
with
with
Complex
a line that
A complex
by identifying a real number it + Qi as Oz' as 3 and \342\200\224
3+
0 +
r \342\200\224 it
si as si.
to the
numbers. The equation x2 = -1,
of real
development
quadratic
that
(1)
Unfortunately,
i
has
generations of students
been
called an imaginary number, the complex numbers with
to view
and
this
more
has led than the real
terminology
skepticism
such as 1,3, it, \342\200\224 a/3, and i are inventions of our minds. is the number 1. If there were, it would There is no physical entity that surely be in a place of honor in some great scientific museum, and past it would file a steady stream of at 1 in wonder and awe. A basic goal of this text is to show how mathematicians, gazing we can invent solutions of polynomial when the coefficients of the polynomial equations may not even be real numbers! numbers.
Actually,
of Complex
Multiplication
The product
all numbers,
{a +
familiar properties
bi){c +
of real
Numbers
di) is defined in arithmetic
the
and require
way
it must
that i2
=
be if we are
\342\200\224 in
1,
to
enjoy
accord with Eq.
the
(1).
Section 1 see that
Namely, we
we
bi){c+ di) = ac+ adi + bci + bdi2 = ac + adi + bci + bd(\342\200\224l) =
Consequently, we define
the form r
is of
which
the
+ bi)(c
+ si
usual
properties z\\Zi all hold for all z\\. Z2, Z3 e C
that
1.2 Example Solution
We
-
(2
Compute
50(8 +
=
establish
absolute
value
Eq. (2), but We have
origin in Fig.
value
1.1. We
+
(ad +
ad +
be. It
to check
is routine
+
andz!(z2
(z\\Z2)z3
(2)
bc)i, Z3) =
+
Z1Z2
the product as
16+ 6;-
40/
15 =
+
31
bi\\ =
-Ja1 +
we did to
Z1Z3
motivate
\342\200\224 34r.
we
first define
\342\226\262
the
b2.
(3)
and is the distance from a + is a nonnegative real number can now describe a complex number z in the polar-coordinate
z = where 6 is the angle z, as shown in Fig.
s =
zi = c + di as
of complexmultiplication,
the geometric meaning \\a + bi\\ of a + bi by
absolute
=
we compute
rather
\\a
This
be and
30-
(2 - 5i)(8+ 30 = To
\342\200\224 a +
\342\200\224 bd and
ac
ZiZ\\. Zifez3)
memorize
don't
that equation.
+ bc)i.
+ (ad
bd)
+ di) = (ac\342\200\224 bd) +
r =
with
\342\200\224
(ac
of z\\
multiplication
= (a
Z\\zi
13
Examples
to have
want
(a +
and
Introduction
measured
1.3. A
famous
|z|(cos# +
i
\342\200\224 cosO
+ i
to the
form (4)
sin#).
from the ;t-axis to the vector counterclockwise formula due to Leonard Euler statesthat e'6
bi
from 0 to
sin#.
Euler's Formula We
ask
e6, cos 6
to derive Euler's and sin 6 in Exercise
you
formula
from the power series expansionsfor this formula, we can expressz in Eq. (4) as
formally
41. Using
y ,,
z = |z|(cos0 + ;'
|c| sin 9
^r^ y^
i Z\\ 1
0
9 V
1.3 Figure
sin 9)
1 1 1 1
1
|z| cos 9
I
1
>
,
14
Part
I
Groups and
z
=
Subgroups set
.Letus
Me
zi = and
their product
compute
hold
with
and
form,
Zl
that
,,1(01+02) = |z, = \\z,\\e^ \\Zi\\e '\\z2\\e'\"' |Zi||Z2|e\"
in the
5 concludes
Eq.
of exponentiation
laws
usual
obtain
= |zil|z2|[cos(0i + 82) Note
\\z2\\e\"
that the
assuming
We
exponents. =
Z\\Zi
this
in
number
complex
i6\\ \\zi\\e'
polar
8\\ polar angle 8 for ziz2 is the sum numbers by multiplying their absolute in Fig. 1.4. Exercise 39 indicateshow to Euler's formula and without recourse
of
form
Eq.
(5)
+ 82)].
i sin(0!
+
4 where
=
\\z\\Zi\\
\\zi\\\\z2\\
the
and
geometrically, we multiply complex values and adding their polar angles, as shown this can be derived via trigonometric identities about complex exponentiation. assumptions
8 =
82. Thus,
+
N7T/2
1.4Figure Note that
i
has
polar
has polar angle 2(n/2) 1.6 Example
Solution
Find all solutions
in
z2 =
the equation
Writing
angle n/2
=u
C of
1.5 Figure
and
|1
the equation z2 i
in polar
Consequently, n yielding solutions
=
values 8 where 0
Eq.
< 8 < 2tt
obtain
(5), we
;sin2<9)=1(0+0-
1, so \\z\\ ~ 1. The angle 8 for 28 = (n/2) + n(2n), so8
=
|z|2
using
1.5. Thus i2
in Fig.
shown
\342\200\224 1.
= i.
form and
|z|2(cos2<9+ Thus
that
i2 =
1, so
1|
1, as
value
absolute
and
\342\200\242 =
z
must
(tt/4)
0 and
are
+
28 =
cos
satisfy
rnt for an 8 =
1, yielding
0
sin2#
and
of integer n. The values tt/4 or 8 = 5n/4. Our
are
Z\\
\342\200\224 1 cos
It
TC . . sin \342\200\224l-1
\342\200\224
4
4
= \302\2432
1
5rt
5tt
. and
cos
1- i
=
-7=(1
V2
+ 0
and
Z2 =
+ \342\200\224p(l
V2
sin
\342\200\224
4
4
or z\\
\342\200\224 1.
0-
1.7
Example
Solution
of z =
all solutions
Find
As in
\342\200\22416.
Example 1.6 we write
the
Consequently, |z|4 =
+ n(2n), so
+ i
2tt
-16 is
2(cos^ In
way, we
a similar
The last by
a +
two
more
in polar
We will
mention
of a
division
+ bi
Let U =
{z
the origin
C \\ \\z\\ \342\202\254
and
c +
=
(a +
A
72(1-0-
c) + (b +
di can be performed
a + bi c \342\200\224 di (ac \342\200\224 c c + di di ac + bd be \342\200\224 ad + -^^TL c2 + d2 c2 + d2
in U
numbers
we can
Thus,
multiplication.
1], so that 1, as shown
U is
=
radius
the product of two
but
we
probably
should
using
+ bd)
(6)
d)i. the device
+ (be \342\200\224 ad)i
c2 +
d2
(7)
view
in
is
the
Fig. again
a number in
multiplication
Euclidean plane with center at relation \\z\\Zi\\ = IZ1IIZ2I shows that in U; we say that U is closed under U as providing algebra on the circle
in the
circle
1.8. The
1.8.
As illustrated number 6 e R
in that
is usually that will be apparent complex
and
on Circles
Algebra
interval
of z4 =
+ -L;-)=V2(l i).
of complex numbers,
division
nonzero
by
a + bi c + di
in Fig.
of 6
values
one solution
that we can find solutions of an equation z\" = a + bi form. There will always be n solutions, provided that 21 ask you to solve equations of this type.
bi) + (c+ di)
(a + and
Thus
0. We find that
is given by
addition
that
sin46\302\273=
and
The different
solutions,
72(-1-0-
/ 0. Exercises16through or not use addition
bi
+
isin^=2(-L
illustrate
examples
the equation
writing
+
+ 0,
72(-1
+ Or).
16(-1
\\z\\
three
find
=
sin46\302\273)
= 2 while cos 46 = -I + n(7t/2) for integers n. (n/4) are tt/4, 3tt/4, 5tt/4, and In/4.
16,so 6 =
6 <
0 <
where
obtained
form, obtaining
in polar
equation
|z|4(cos4<9
40 = n
15
Introduction and Examples
Section 1
Fig. 1.8, lies in the
denoted
numbers is
later. the
by
Recall
sum
6\\
we associate
half-open
with
interval
each where
z =
cos 9 + i sin# 0 < 2n. This
0 <
in U a
real
half-open
but we prefer to denote it by 11\302\276^ for reasons that the angle associated with the product Z1Z2 of two if 6\\ + 62 > 2it + 62 of the associated angles.Of course [0, 2tt),
Part I
16
and
Groups
Subgroups
1.8 Figure
then
the
angle
modulo 2-7T
1.9
Example
on
In R2w, we have
Example
3-f
+2n
was nothing
There
interval
the half-open
1.10
in R2jr associated with z\\Z2 is 8\\ + 62 \342\200\224 2it. This gives us an here by +2n \342\226\240 E2jr. We denote this addition
special
19 =
16 +23
InR23,wehave
lJ\302\243-
about
\342\200\242 We can
R2ir
=
- 2n
=
5-f
any
35 - 23=
A
3f.
number
the
use
addition
2tc that enabled us to define interval
half-open
Rc = {x
8 =
12.InE8.5,wehave6+8.5
on
addition
e ffi|0
14-8.5
< * < c}. 5.5. \342\226\262
multiplication on the circle U where |z| = 1 and addition ffi2.T algebraic properties. We have the natural one-to-one *> in Fig. 1.8. Moreover, we 0 between z e U and 6 e ffi2^ indicated correspondence z Now
modulo
deliberately
number
complex
2tt on
have
defined
if
+2n
z\\
<+ 0\\
the same
so that
and
zi +>
02,
then
z\\-
zi\302\261* (0\\
+2n
(8)
02).
isomorphism
The relation
shows that if we rename each z e U by its corresponding angle 0 then the of two elements in U is renamed by the sum of the 1.8, product for those two elements. Thus U with complex number and ffi2ir angles multiplication with modulo 2it must have the same algebraic properties. They differ addition only in the names of the elements and the names of the operations. Such a one-to-one correspondence the relation of elements and names of Names (8) is called an isomorphism. satisfying (8)
shown
in Fig.
binary
operations are
not
important
in abstract
algebra; we
are interested
in
algebraic
are the
of R2;r
what we
illustrate
We
properties.
mean by
there is exactly one element e The element 0 in R2;r that corresponds e +2n x = x for all x e R23r.
The equation
that
such
e
e U, namely,
z for all z
\342\226\240 z =
to 1 e U is the
only
e =
e in R2;r such
element
1.
that
^
and \342\200\224i. has exactly four solutions, 1, i, \342\200\2241, namely, and 0 e ffi2;r correspond, and the equation x +2jT x +2;r * +2jr x = 0 in R2;r exactly four solutions, namely, 0. tt/2, tt, and 3tt/2, which, of course, correspond
has
lint/ z \342\226\240 z \342\226\240 z \342\226\240 z=
\342\200\224 1, i, 1, and
6 is
angle
interval
equal to
ffi2jT,
put
\342\226\262
\342\200\224i, respectively.
our circle
Because an
algebraic properties of U and
let/
Now
to
that the
saying
same.
1.11Example In U
1.12 Example
17
Introduction and Examples
1
Section
U has
the
the 0 in
of the
length
the
1, it
radius
has circumference
arc down
interval
the
2it
angle
on the 1
on
the
pick
up
our
and wind it
*-axis
of
measure
radian
the
and
If we
subtends.
half-open
around
the
to 1. Moreover, each number way number as the value of the in the interval will fall on the point of the circle that having on in Fig. central angle 6 shown 1.8. This shows that we could also think of addition subtended arcs as of counterclockwise, ffi2;r starting being computed by adding lengths is lit or greater. at z = 1, and 2it if the sum of the lengths subtracting circle
it will reach all
U counterclockwise,
back
the
a starting of addition on a circle in terms of adding lengths of arcs from circle we use circle P on the and can a of radius counterclockwise, point proceeding our has circumference of radius 1. We can take 2, which 4tt, just as well as a circle interval at it will and it around P; counterclockwise, BL^ starting just half-open wrap on cover the whole circle. Addition of arcs lengths us a notion of algebra for points gives this circle of radius is surely isomorphic to ~M^n with addition 2, which However, +4^. if we take as the circle \\z\\ = 2 in Fig. 1.8, multiplication of complex numbers does not circle. The relation \\z\\Z2\\ = \\z\\\\|z21 shows that the product of give us an algebra on this such two numbers absolute value 4 rather than 2. Thus complex number has complex is not this circle. closed on multiplication The preceding paragraphs indicate that a little geometry can sometimes be of help
If we think
to convince ourselves that ffi2;r and BL^ are algebra. We can use geometry stretch out the interval to cover the interval E4^, or, R2;r isomorphic. Simply uniformly if you prefer, use a magnifier of power 2. Thus we set up the one-to-one correspondence a *>2a between a e E2t and 2a e E4JT. The relation (8) for isomorphism becomes
in abstract
a \302\261+ 2a
if
b
and
\302\261* 2b
then
{a
\302\261* (2a
+\302\276b)
(9)
2b).
+4^
isomorphism
This is obvious final
1.13 Example
pairing
x +4^ x which Example1.12.
if a
in the
+ b
< 2tt.
a + b
= 2jt
+ c, then
2a
displayed relation becomes c *> 2c, which
+4^ x +4^x = 0 in
are two
If
times
the
four solutions, exactly found for the analogous
has 1\302\276^
solutions
+ 2b
=
An
2c,
+
and the
is true.
namely, 0, n,
equation
in
2jt, and
ffi2jr
3tt,
in \342\226\262
Part
18
I
Groups and
Subgroups
is nothing
There
Surely, Ec * e Ec with
with
about the numbers to E^ with
special
(d/c)x
the
in
previous
e E+. We
c, d
for all
+j
need
argument. only pair
of Unity
Roots
Theelements
set
of the
1.6
2jt\\ ( m\342\200\224
cos
n J
\\
=
Un
of Examples
technique
and
=
2n\\
(
.
r
n
\\
m =
for
m\342\200\224
\342\200\236 , \342\200\236 \342\200\242 \342\200\242 n \342\200\224 1. \342\200\242,
0. 1, 2,
J
let
If we
C U. 1, so U\342\200\236
=
\302\243 cos
^ +
i
sin
^-,
then these
as
1 = = 1, Because\302\243\"
1} are called the nth roots of unity. Using the we see that the elements of this set are the numbers
1.7,
i sin
+
C | z\"
{z e
They all have absolute value of unity can be written nth roots
n
and An
2n
is isomorphic e Erf.
+c
these n powers
do)
^,^,^,^,---,^-1.
of \302\243 are
For example,with
under multiplication.
closed
\342\200\224
10,
we have
tH = tl0t4 = 1\342\200\242 t4 = t4
t6ts = Thus we see that
we
can compute
i +,,
\302\243' \302\243-7 by computing
j,
-^
viewing
i
j as
elements
addition
modulo
and
ofE\342\200\236.
Let n
1.14
Example
1. 2,
=
Z\342\200\236 {0,
on
is closed
The solution
\342\200\242 n \342\200\224 3. \342\200\242 \342\200\242. 1}. We
see that
and clearly c E\342\200\236 Z\342\200\236
Z\342\200\236.
of the
5 = 3 in
x +
equation
is x
Zg
= 6, because5+86
= 11
\342\200\224 =
8
3. \342\226\262
of
elements
the
each of thenth
rename
If we
all
This Z\342\200\236.
roots
of unity in a one-to-one
gives
(10) by its exponent, we use for names and Z\342\200\236. correspondence between U\342\200\236
Clearly,
if
CJ *+
\302\261\302\261 i and \302\243'
*en
J-
(f'
\342\200\242 ^) \302\261> (i +\342\200\236 j).
(11)
isomorphism
Thus
with [/\342\200\236
complex
number
multiplication
with Z\342\200\236
and
addition
have +\342\200\236
the same
algebraic properties.
1.15Example
It
be shown
can
Under
this
isomorphism,
Exercise
elements
that there
35 asks
of Z8 to
which
is
an
isomorphism
we must
then have
you to
continue
each
of the
of
U%
with
= ei2ir/8 ^> 5, which \302\243 = 5 2. \342\226\262 +8
Z8 in
\342\226\240 = *> 5 \302\243 \302\2432 t;
in Example 1.15, finding six elements of U% correspond. remaining the
computation
the
19
Exercises
1
Section
1
\342\226\240 EXERCISES
In Exercises 1
the given
9 compute
through
arithmetic
the
and give
expression
the form
in
answer
a+
bi
for
a,b\342\202\254R.
1.
/3
2. /4
4.
(-035
5. (4
7.
(2
-
30(4 +
10. Find
/) +
12 through
13.-1 + 16 through
Exercises
21,
20. z6
=
22 through
Exercises
22. 10+17
25. |
31.
*
33. x 35.
29 through
Exercises
29. x
the expression
34, find
the polar
14.12
+ 5/
binomial
theorem.)
form
+ qi)
\\z\\(p
where
\\p
1. + qi\\ \342\200\224
15.-3 + 5/
equation.
given
19. z3
=-8
= -27/
x of
24- 20-5
+25 19-3
no sense.
the
30. x
Z12
addition.
27. 2V2+^32 3V2
^
R6 makes
solutions
modular
indicated
6
+2,
3f
using the
given
t
+2jT
equation.
=
T
32.
x +7 * +7 * =
34.
* +4 x
in
R^ in Z7
5
+4 * +4 * =
0 in
Z4
Zg in which f = e'(n/4)^> 5 and f2 ^> 2. Find six elements f'\" in [/\302\276 for m = 0, 3, 4, 5, 6,
with Example 1.15 asserts that there is an isomorphism of \302\243/g the element of Z$ that corresponds to each of the remaining and
7.
36.
There is
37.
correspond for m = 0,2, 3,4, 5, and 6. Why can there be no isomorphism of U& Derive the formulas
38.
of the
expression
+io
+15 7 = 3 in Zi5 +7 * = 3 in Z7 +12 * = 2 in
z in
18. z3
given
5 +6 8 in all
- /)5 (Use the
(1
-64
26.
why
in C
2/)(3 - /)
4/|.
|6 +
-1
23- 8
+i |
9.
/)3
/
27, compute the
16
28. Explain In
z6 =
21.
1
/23
6. (8 +
3r)
complexnumber
solutions
all
find
17. z4 =
16.z4=l
In
the given
15 write
+
r)(5
11. Find
12.3-4/ In
-
8. (1+
(6- 5/)
|3-4/|.
In Exercises
3.
an
isomorphism
of
C/7
with
Z7 in which
sin(a
=
e'(2-T/7) ^>
Z6 in which f
with
+
t,
b)
=
sin a
4. Find
the
element
= e'(T/3)corresponds to
cosa sin
cos b +
in
Z7
to which
t,m must
4?
b
and
cos(a + by
39.
Euler's
using
formula and
computing
b) =
cosacosb
b.
Derivea
formula
for cos
30
in
terms
of sin
9
and
cos 0 the formula cos 30 = 4 cos3 0-3 in Section 32.) have use for this identity Derive
sin b
e'\"e'b.
+ / sin02). Letzi andz2 = |z2l(cos6>2 = / + + toderiveziZ2 |zi||z2|[cos(0i 92) sin(0i +92)].
= |zi|(cos0i+/sinSi)
40. a.
\342\200\224 sin a
cos from
Use the
9 using part
identities
trigonometric
Euler's
(a) and the
in Exercise
38
formula. identity
sin2
0 +
cos2 0
= 1.(We
will
Part I
20
41. Recall
the
power
and
Groups
series
Subgroups
expansions
e sin,
X2
Derive
calculus.
section
2
x3
=,--
x2
COSX =
from
+ ^ + --<
3!+4!+\"-
---
x1
xA
xb
x0
+
xn
X4
X3
+ .-. +
_
x2\"~] iN\342\200\236_, +
(-1)-^-^
+ -- + r
...,
and
x2n s\342\200\236
{~1)(2ny.+-
l-V.+^.-6i
Euler's formula
Binary
+
=1+*+2!
e'e =
cosS+ ;' sinS
from these
formally
series
three
expansions.
Operations
in a strange world and are observing to a strange civilization a class of fellow creatures in addition world drilling of also that we have not been told that the class is learning to add, but were just placed as observers in the room where this was going on. We are asked to give a The teacher makes noises that sound to us approximately report on exactly what happens. like gloop,poyt. The class responds with bimt. The teacherthen gives ompt, gaft, and the are they doing? We cannot report that are adding with poyt. What class they responds Of course, numbers, for we do not even know that the sounds are representing numbers. with on. All we can is we do realize that there is communication going say any certainty are designated some that these creatures know rule, so that when certain pairs of things like gloop, poyt, they are able to agree on a response, in their language, one after another, in in on drill our bimt. This same procedure addition first classes where a goes grade teacher may say four, seven, and the class responds with eleven. and multiplication of numbers, we are thus led to In our attempt to analyze addition the idea that addition is basically just a rule that people learn, enabling them to associate, with two numbers in a given order, some number as the answer. is also Multiplication that in playing this game with such a rule, but a different rule. Note finally students, careful of what two things teachers have to be a little they give to the class. If a first inserts will be The rule is only teacher the class confused. ten, sky, very grade suddenly from some specified set. defined for pairs of things
Supposethat
we
are visitors
one of the creatures numbers. Suppose
and Examples
Definitions As
of this
mathematicians,
let us
attempt to collect
the
of these
core
and
basic ideas
in
a useful
As we a set. can we to However, attempt attempt be somewhat mathematically precise, and we describe our generalizations as functions than as rules. Recall from Definition 0.10 and Example 0.11)rather 0.4 (see Definition that for any set S, the set S x S consists of all ordered pairs (a, b) for elementsa and b of S. definition,
generalizing
remarked in
2.1 Definition
A
binary
S x S,we
Section
the
0, we
* on operation the will denote
notions
do
not
of addition
of
multiplication
numbers.
to define
a set S is a function element *((a, b)) of
mapping S by
a *
S x
b.
S into
S. For
each (a,
b) e \342\226\240
2
Section
we
Intuitively,
2.2 Example
Our usual
to each regard a binary operation * on S as assigning, an element a * b of S. We proceed with examples.
may
pair (a, b) of elements
ordered
of S,
operation on on E. In this binary operation example, + is a binary
addition
different
21
Operations
Binary
set
the
usual
E. Our
we could
\342\200\242 is a
multiplication
replace E
of the
any
by
sets
C, Z,E+,orZ+. pair (a, b) 2.3 Example
we require a
that
Note
\342\226\262
of elements
Let M(E) be the a
on this
operation
binary
matrices having
different
2.4 Definition
a formal
make
operation on S
* be a binary
Let
under * if for alla,b H given by restricting
2.5 Example
and 0
it means
2.6 Example
you
Let
+
of nonzero
real
E* is
Thus
E*. \302\242.
text, we
a binary
sure
H is the
the set E
+ on
addition
set E*
In our under
a *
have
often
will
and
and
(b)
multiplication.
let
\342\200\242 be the
H =
{n2\\n
numbers
real
because under *.
S,
set
the
H of S
H is closed operation on
subset
H. S is
\342\226\240
closed
under
*, but
does not
2 e E*
and
induce a binary
\342\200\2242 e E*,
but 2
+
operation = (\342\200\2242)0 \342\226\262
a subset H of S is closed to decide whether correct conclusion, we have to know what have trouble here. Be to use this fact. Students
occasion
have
for an element to be in H, and understand the next example.
Z,
* on
of \342\226\262
the binary
of * on
not
pair (A, B)
~
S. The
case,
+ is
addition
shows.
example of
subset of
H. In this
operation
* on S. Toarrive
operation
ordered
every
on a subset
operation
binary
operation
numbers closed
not
b e
induced
definition of a binary not be, as the following may
Our usual on the
* to
H be a
let
and
e //we also
our very
By
a subset
definedfor
S to be
matrices^ with real entries. The usual matrix set since A + B is not defined for an ordered numbers of rows or of columns.
operation on S providesa definition.
a binary
Sometimes
also.We
of all
set
on a set
operation
binary
S.
from
at a
usual binary operations of addition e Z+}. Determine whether H
and
multiplication
is closed under
(a)
on the addition
set and
12 = 1 and 22 = 4 are in H, but that 1 + 4 = 5 addition. For part (b), suppose that r e H and s e H. Using what it means for r and s to be in H, we see that there must be integers n and m in Z+ such that r = n2 and s = m2. rs = n2m2 = (nm)2. of elements in H and the Consequently, By the characterization Forpart
and
5
fact
that
(a), we
H. Thus \302\242.
nm
need
H is
e Z+,
not
only
that
observe
closed
this means that
under
rs
e H,
so H
is closedunder
multiplication.
\342\226\262
T Most students of abstract algebra have studied linear algebra and are familiar with matrices and matrix operations. For the benefit of those students, examples involving matrices are often given. The reader who is with matrices can either skip all references to them or turn to the Appendix at the back of the text not familiar where there is a short summary.
Parti
2.7 Example
and Subgroups
Groups
Let F bethe
real-valued functions as domain the set E of real numbers. f having from calculus with the binary operations +, \342\200\224, and oonF. \342\226\240, Namely, for each orderedpair (/, g) of functions in F, we define for each x e E of all
set
familiar
are
We
f +8 -
/
+ g)(x)
{f
by
\302\243 by
- g)(x)
(/
\342\200\242
/
\342\200\242
(/
8 by
g)(x)
= f{x)
= f(x)
+ g(x)
-
addition,
subtraction,
g(x)
= f(x)g(x)
multiplication,
= f(g(x))
composition.
and
o g by
/
of these
four
All
four
The
In
text,
structural
The most
Example
2.10
Example
F is closedunder
all
\342\226\262
on
multiplication
U and
+c on
on Z\342\200\236, and addition +\342\200\236
addition U\342\200\236,
Ec
1.
Section
On Z+, we define common value ifa
* on a
of describing a particular binary operation a * b assigned to each pair (a, b) by
method
important
set is to characterizethe defined in terms of a and
2.9
E, so
domain
with
above are very familiar to you. operations described in the examples want to abstract basic structural from our familiar concepts algebra. we should illustrate these this concept of abstraction from the familiar, of with unfamiliar operations concepts examples. We presentedthe binary
complex number
2.8 Example
real valued
again
and o. \342\200\224, \342\226\240,
binary we
this
To emphasize
in
are
functions
+,
operations
(/ o g)(x)
element
some
given
property
b.
a binary
= b. Thus
On Z+, we define a binary and 5 *' 5 = 5.
operation * by 2 * 11 = 2; *' by
operation
we define a binary Example 2.8. Thus 4 *\" 7 On Z+,
operation
= 6;25
a * b
*\"
equals
15* 10= a *' b
*\" by a 9 = 11;
*\"
the smaller
10;
3*3
and
a.
\342\226\240 Thus
b =
and 6 *\"
(a * 6 =
2 *'
of a
and
b,
or the
= 3. 3 =
b) + 2, where
\342\226\262
2, 25 *' 10= 25,
* is
defined in
8.
\342\226\262
It may seem that these examples are of no importance, but consider for a moment. Supposewe go into a store to buy a large, delicious chocolatebar. Suppose we see two identical bars sideby side, the wrapper of one stamped $1.67 and the wrapper of the other stamped $1.79.Of course we pick up the one stamped $1.67. Our knowledge of which one we want depends on the fact that at some time we learned the binary operation * of Example2.8.It is a very important operation. Likewise,the binary operation *' of order. Think what a problem we our ability to distinguish Example 2.9 is defined using
not
tried to put about dismissing
if we
have
would
be hasty
course,
our
usual
our
shoes first,
somebinary
of addition
operations
importance well known
on
and
operation
and
multiplication
then
our
as being
socks!
of
little
of numbers
Thus we significance.
should
Of
have a practical
to us.
were chosen to demonstrate that a binary operation may or on the order of the given may not depend pair. Thus in Example 2.8, a * b = b * a for alla,b e Z+ , and in Example 2.9 this is not the case, for 5 *' 7 = 5 but 7 *' 5 = 7.
Examples
2.8
and
2.9
Section2 2.11 Definition
A
set S is commutative
* on a
operation
binary
if
* b = b *a
only if) a
(and
23
Operations
Binary
all
for
a,b&S.
\342\226\240
was
As
and
only
out
pointed
if from
is customary in mathematics to omit the words if Definitions are always understood to be if and only and and no convention such always if only if statements, 0, it
Section
in
a definition.
Theorems are not used for theorems. Now we wish to consider an expression of the form a * b * c. A binary suppose * enables us to and here we have three. The obvious combine two elements, operation only the to combine three elements are to form either (a * b) * c or a * (b * c).With attempts = 2 and then 2*9 = 2. * defined as in Example 2.8, (2 * 5) * 9 is computed by 2 * 5 = = * 2 * is * 5 and 2*5 2. Hence (2 * 5) * 9 = 5 9 then Likewise, (5 9) computed by statements. is ever
2 * (5
and it is
* 9),
to see that
hard
not
is no ambiguity
in writing
(2 *\"
this
b) * c =
(a *
so there
for
*,
a * {b
* c),
a *b
* c. But
5) *\"
9 =
4 *\"
9 =
6,
9) =
2 *\"
7 =
4.
*\" of
for
Example 2.10,
while
2 *\" Thus {a *\" ambiguous.
2.12 Definition
A
b)
*\" c
operation
binary
need
not
on a set
(5 *\"
a *\" {b *\" c),
equal
S is associative
if (a
and
* b)
an
a *\" b
expression
*c =
a * (b
* c)
*\"
for all a,
c may
be
b, c
e S. \342\226\240
such as a * b * if * is associative, then longer expressions in Parentheses be inserted fashion for ambiguous. may any purposes of the final results of two such will be the same. computation; computations of functions mapping E into E was reviewed in Example 2.7. For any Composition set S and any functions define the composition S, we similarly f and g mapping S into o of followed the function S into S such that f g by /as (f o g)(x) = f{g{x)) g mapping for all x & S. Some of the most important we consider are defined operations binary of It is this functions. to know that using composition important composition is always it is defined. associative whenever 2.13
Theorem
Proof
be shown
can
It
c *d
that
not
are
(Associativity
of Composition)
S
/ o (g o h)
into
show
To
S. Then
these
to each x
&
Let S be a set and (/ o
=
let
/, g,
and
h
be functions
mapping
g) o h.
two functions are equal,we must we find that
show
that they
give
the
same
assignment
S. Computing
(/ o {g o h))(x)
= f((g
((/
= {f o g)(h(x)) =
o
h)(x))
= f(g(h(x)))
and
so
the
same
element
O
g)
o h)(x)
f(g(h(x)))
of S is indeedobtained.
f(g(h(x))), \342\231\246
Part I
24
Groupsand
Subgroups
As an example of using Theorem 2.13 to save work, recall that it is a fairly painful exercise in summation of n x n matrices is an notation to show that multiplication that there associative binary show If, in a linear algebra course, we first operation. that is a one-to-one correspondencebetween matrices and linear transformations and of matrices corresponds to the composition of the linear transformations multiplication we obtain this associativity at once from Theorem 2.13. (functions),
Tables For a finite
defined by means of a table in which of columns and at the left side elements set are listedacrossthe top We of the set be listedasheadsacross as headsof rows. always require that the elements the in same the left side. The next illustrates the the order as heads down example top use of a table to define a binary operation. set,
2.14
Example
operation on
a binary
as heads
Table 2.15 definesthe
binary
a
b
c
a
b
c
b
b
a
c
b
c
c
b
a
=
Thus a * b =
2.16 Example
entry
on the left)
c
of the
CompleteTable
d
the top)
on
entry
rule:
the following
table body).
of the
\342\226\262
defined with
by
table and terminates at
lower
the
* is a commutative
is commutative if and
a table
to the
respect
diagonal
on
operation
binary
that
at
starts
corner.
right
the set
S=
c, d).
we
rest of
the
that b *a = d. For * to be commutative, d in the appropriate square defining the diagonal in Table 2.18 from the
Table 2.18 in
a
b
b
d
a
Some
c d
a a
c d b b
Classroomexperienceshows define
square
defining
We
b*a.
to give our solution.
that may result if a on it. Remember that binary operation on a set S we must be sure that the
chaos
some
operation
binary
fashion
a * b,
\342\226\262
of Warning
Words
asked to
this
have a * b = which is located
must
we
see
place
across
symmetrically obtain
c
operation are symmetric
so that
2.17
* (jth
c] by
commutative.
a binary
table
{a,b,
column
jth
a, so * is not
a = that
the
corner
left
d also.Thus b
see in
From Table2.17,we
Solution
Table
a
b *
entries
if the
{a,b,
*
c and
can easily
We only
the upper
2.17
* on
i\\h row and
in the
{entry
S=
operation
(ith
2.15Table *
can be
set
the
of the
the
*
student
in
is given an attempt
a set and to define a
2.18Table \\l
a
exactly one element is assignedto for each ordered pair of elements
1.
a
b
c
d
\\
d
a
a
\\
c
b
Regarding
X
b
of S to \"most\" ordered * is not everywhere
b
\\
attempt
b
d
c d
a
a
c b
2.
could
Condition
assign
1, a
each
of S,
possible
ordered pair
of elementsof
the element assignedto
it is
again
in
S, S.
will an element often make an attempt that assigns but a few element. In this event, for determines no pairs, pairs, defined on S. It may also happen that for some pairs, the In any case any of several elementsof S, that is, there is ambiguity. student
Section 2
of
not well defined. If Condition
* is
ambiguity,
2 is violated,
25
Exercises
S is
then
not closed
under *. of attempts The symbol * is
illustrations
several
are
Following
Some of
are worthless.
them
define
to
used for
the
operations on sets.
binary
in all
operation
attempted
these examples.
2.19 Example
2.20
* b = a/b. Here* is not everywhere assigned by this rule to the pair (2, 0). On
let a
Q,
a *b =
On Q+, let
Example
a/b. Here
rational
number
2 are satisfied, and
1 and
* is a
binary
operation on Q+. 2.21
On Z+, let
Example
2.22 Example
set
we
* to
\"define\"
f(x)/g(x). Here numbers,
h{x) would
2.23
be as in
Let F
Example
f and g.
both
defined g, 2.24
Let S
Example
* by
in Z+.
Thus *
is not
a
A
*.
soh
and let
not
\342\226\262
f * g = h,
where h is the function greater than worthless. In the first place, we have not to be greater than another. Even if we had, any both being many functions greater than f and
is completely
for one function would result in there
still be
F. \302\243
well
\342\226\262
defined.
be a set consisting
of 20 people, no two of whom are of the same height. Define = c, where c is the tallest person among the 20 in S. This is a perfectly good on the set, although \342\226\262 not a particularly operation interesting one.
a * b
binary
2.25
2.22
Example
This \"definition\"
* would
and
for 1 * 3 is not under
E as in Example with domain real-valued functions 2.7. Suppose = * the of that where usual is, h, h(x) = give f by g, f g quotient in F were to be defined for Condition 2 is violated, for the functions and for some g e F, g(x) will be zero for some values of x in E and be defined at those numbers in E. For example, if f{x) = cos* and
it means
what
2 fails, closed
of all
definition
sensible
Condition
sinceZ+is not
then h(0) is undefined,
= x2,
g(x)
not
a/b. Here
on Z+,
Let F be the all real
\342\226\262
a *b =
operation
binary
is \342\226\262
Conditions
both
on Q, for no
defined
Let S be as in Example c is the shortest person in S who 2.24 and let a * b = c, where is taller than both a and b. This * is not everywhere since if either a or b is the defined, in the a tallest * \342\226\262 b is not determined. set, person
Example
2
\342\226\240 EXERCISES
Computations
Exercises 1 through
1.
Compute
4 concern
the binary operation
c *c,
[(a
b *d,
and
2. Compute
(a * b)
*c
3.
(b * d)
* c and
Compute
and
* c)
a * (b b *
* denned on S =
{a,b,
c, d,
e] by means
of Table2.26.
* e] * a.
* c).
Can you
(d * c). Can you
say
on the
basis of this
computations
say
on the
basis of
computation
this
whether whether
associative? * is associative? * is
Part I
26
and
Groups
Subgroups
*
a
b
C
d
a
a
b
c
b
b
c
c
d
e
4. Is* commutative?
6.
possible and
b
c
b
d
a
a
b
c
a
e
c
b
b
d
c a
b
b
a
c
a
b e d b
b
e d d c
d
c d
a
a commutative
define
completed
8. *
9. * 10.*
11, determine
the
whether
defined on Q by
a
letting
defined
on
Q by letting a * b = ab/2 Z+ by letting a*b-2ah
11. * defined
on
Z+ by letting
the
be a set having if S
question
How
13.
a *b
c d
a
a
b
c d
c
b
b a
c d
b a
c d
c d
c d
binary
of
n
How many
2 elements; exactly
{a,b, c, d). operation * on S = {a,b, c, d).
binary
operation
different
3 elements;
*
defined
Assume this is
* is
and whether
is commutative
be
can be defined
operations
binary
exactly
binary operations can
commutative
on a set
S=
= ab
one element.
exactly
has exactly
different
many
elements?
b
= a \342\200\224 b = *b ab + 1
on
Let S
a
a*b
Z by letting
on
defined
12.
*
d
entries.
associative.
7. * defined
d
* on
operation
binary
an associative
define
to
the missing
compute
7 through
Exercises
In
a
Why?
can be
2.28
Table
e
*
Table 2.27 soas to
5. Complete
2.28Table
2.27 Table
2.26 Table
on
defined
on
SI Answer
elements.
n
of 2 elements? on a set of
a set
3
elements?
Concepts
14 through
In Exercises
is needed, so that
15. A (b
16.
the definition of
the
italicized
term
without reference
to
the
text, if
correction
acceptablefor publication.
if and only if a * b = b * a. operation * is commutative * on a set S is associative and only if, for all a,b,c if binary operation * c) * a = b * (c * a). H of a set S is closedunder subset a binary operation * on S if and only if (a * b) e H for
A
17 through
Exercises
event that are
16, correct
in a form
A binary
14.
In
it is
* is
22,
whether
determine
not a binary operation,
state
the definition
whether
Condition
of * doesgive 1, Condition
a binary
2, or both
e S, all a,b
On Z+,
define
18.
On Z+,
define
19. On
R,
20.
Z+,
On
define define
*b = a \342\200\224 b. * by letting a *b = ab. * by letting a *b = a - b. * by letting a * b = c, where c * by
letting a
is the
smallest
integer
greater
have
e S.
operation on the set. In the conditions on page 24
of these
violated.
17.
we
than both
a
and
b.
21. On
define
Z+,
22. On Z+, define Let H
23.
a 24.
be the
c is at
* by
c is
letting
consisting of all
of M2(M)
subset
of the
each
following
a. If * is
c. If
* is
f.
commutative
product of a and
than the
form
a, b
for
e
Is H
ffi.
b.
closed under
multiplication?
set
a *a
on
set
operation
of
any
importance
any
operation
= a for
if there
defined on a set
all a
S, then a
e S.
* (b
* c) * a
= (b
* c)
all
for
S, thena * (b * c) = (b * c) *aforalla, b, are those defined on sets of numbers.
onanyset
operation
binary
S, then
operation * on a set S is commutative
binary
Every
a + b. less
integer
of the
matrices
any
binary
operations
binary
A binary
e.
on
operation
any associative
d. The only
more than
or false.
true
binary
any
b. If * is any c e S.
least 5
the largest
b matrix
addition?
matrix
Mark
* b = c, where a * b = c, where
* by letting a
27
Exercises
2
Section
exactly
having
exist a,b e S such one element is
a *
that
a, b, c e
S.
b = b *a.
both
commutative
ordered
pair of
and
associative. operation
on a set
h. A binary S.
operation
on a set
binary
operation
on a set
S assigns exactly
binary
operation
on a set
S may
A
of S.
i.
A
of S.
j. 25.
S assigns at
binary
g.
A
elementsof S.
different from
Give
a set
two
different
S assigns at
*
associative
and commutative
your
each
to
ordered
each
text
and
set is well
pair of
ordered
each
element of S to
of the
*' on
and
of S
than one
described in the examples this set. Be sure that
of those
any
of S to
element
one
more
assign
operations
binary
one element
most
of S to
one element
least
some
not a set
elements
elementsof
pair of ordered
elements pair of
of numbers. Define
defined.
Theory
26. Prove
if *
that
is an
(a * b) for
all a, b,
c, d
e S. Assume
the
binary operation on a set
* (c
* d) = [(d * c) * a] * b
law only for
associative
(x * y) for In
all
27 and
27. Every
binary
28.
commutative
Every
Let F be the the
binary
28,
either
on a set
operation
set of
binary
30. Function
= x
* (y
the definition,
that
is,
assume
only
* z)
operations addition
of a
consisting
functions
and o \342\200\224, \342\200\242,
subtraction
+ on
F is
\342\200\224 F
on
or give a counterexample. single element
operation on a set having
all real-valued
+,
the statement
prove
on F.
having In
associative. is commutative
just
as domain
Exercises
counterexample.
29. Function
* z
as in
triples
e S.
x, y,z
Exercises
S, then
in
both
two elements
commutative
and associative.
is associative.
the set ffi of all real numbers. Example 2.7 defined 35, either prove the given statement or give
29 through
a
Part I
28
and
Groups
Subgroups
31. Function
subtraction
32.
multiplication
\342\200\242 on F
multiplication
\342\200\242 on F
33.
Function Function
34. Function 35. If * and
is associative.
\342\200\224 on F
is commutative. is associative.
is commutative.
o on F
composition
any two binary
*' are
a *(b
36. Supposethat Show
that
element 37.
in
* is
H is
an associative
closed
on a set
operations
*' c)
= (a * b) *' (a * c) on a set
operation
binary
*. (We
under
S, then
think
all
for
H = {a e S
S. Let
of H as consisting
of
e S.
a, b, c
of S that
all elements
for
all
x e
with
commute
S}.
every
S.)
Supposethat*isanassociativeandcommutativebinaryoperationonaset5'.ShowthatH under *. (The elements of H are idempotents of the binary operation
is closed
section
= x *a
* x
\\ a
3
the
Table
that
Notice
every c
3.1 for
by
the
*' on
set
the
the set S =
& using
-o-
{a,
b,
c] with
Table 3.2 for
T =
if, in
a we obtain
* on
operation
binary
{#, $, &}. Table 3.1, we replaceall occurrences the one-to-one correspondence
operation
binary
S\\a*a = a)
Binary Structures
Isomorphic Compare
= {a e *.)
#
#, every b by
$, and
c -o- &
-o- $
fc
of a by
differ only in the symbols (or names) 3.2. The two tables * *' and the If we rewrite Table 3.3 and the for denoting symbols operations. with elements in the order y, x, z, we obtain Table 3.4. (Here we did not set up any oneorder outside the heavy one-correpondence; wejust listed the same elements in different in Table 3.1, all occurrencesof a by y, every b by x, and bars of the table.) Replacing, every c by z using the one-to-one correspondence Table
precisely
the elements
a Table 3.4. We
we obtain
These four
tables
differ
-o-
>'
b
of Tables 3.1, in the names only
think
c
x
-o-
-o-
3.2, 3.3, and
(or symbols)
z 3.4 for
as being structurally their elements and
alike. in
the
that those elements arelistedasheadsin the tables. However, Table 3.5 for binary * and Table 3.6 for binary * on the set S = b, c] are structurally operation operation and from each other from Table 3.1. In Table each element three 3.1, different appears times in the body of the table, while the body of Table 3.5 contains the single element b. In Table 3.6, for all 5 eS we get the same value c for 5 * s along the upper-left to lowerin while we three different values Table 3.1. Thus Tables 3.1 right diagonal, get through order
{a,
3.6
give
provided
just three structurally we disregard the
heads in the
of the
binary
elements
operations and
the
on a set of three in which they
order
elements, appear
as
tables.
The situation Germany
different names
learning
to children have just discussed is somewhat akin in France and the operation of addition on the set Z+. The children have different
we
in
3
Section
3.1Table *
a
b
c
*'
#
$
&
*\"
X
y
z
a
c
a
b
#
&
#
$
X
X
y
z
b c
a b c b c a
$
#
$
y
z
X
$
& #
y
&
z
X
y
&
z
3.5Table
3.4 Table
names
3.3 Table
Table
3.2
29
Binary Structures
Isomorphic
Table
3.6
*\"
y
X
z
*
a
c
*
a
b
c
y
z
y
X
a
b b b
a
c a
b
X
y
X
z
b
b
z
X
z
y
c
b b b b b b
b c a a b c
trois,
deux,
(un,
the same
binary
numbers,
so their
\342\200\242 \342\200\242 \342\200\242 versus
b
drei
ein, zwei,
(In this case, they addition tables would appear
structure.
c
but they are learning the same also using symbols for the in the if list the numbers the same they
\342\226\240 \342\200\242 for the \342\200\242)
numbers,
are
same order.) in studying the different We are interested operations types of structures that binary can provide on sets having the same number of elements, as typified 3.4, 3.5, by Tables and 3.6. Let us consider a binary algebraic structure^ {S,*) to be a set S together with a binary structures {S, *) and (S', *') to operation * on S. In order for two such binary be structurally alike in the sense we have described, have to have a one-to-one we would
correspondencebetween x
if one-to-one
A
regard order. structure
Such, an
3.7
Definition
is
function
y ^>
S and
y',
the
mapping
isomorphism.
showing
that
We give a
(f>{x
formal
that
boldface
(1)
* y)
pairing (1),
x ^> x
which
asserts
alike
in
left-to-right
the algebraic
is known
as
definition.
= 4>{y) *'
type indicates that a
for
(f>{y)
homomorphism
Remember
S' such that
*' y'.
y ^> x
4>(x*y) = 0(x)*'0(y). two algebraic systems are structurally
Let (S, *) and {S', *') be binary algebraic structures. one-to-one function 0 mapping S onto S' such that
1
x *
then
the equation 4>{x)= x' as reading the one-to-one In terms of \302\242, the final ^> correspondence in in S' is the same as in S, can be expressed as
a function
x' of
elements
exists if the sets S and S' have the same number of to describe a one-to-one correspondenceby giving a oneS onto S' (see Definition For such a function \302\242, we 0.12).
customary 0
x of
elements
-\302\253\342\226\272 and x
correspondence
elements.It to-one
the
An
all x,
property
term is being
defined.
isomorphism
y
e
S.
of S
with
S'
is a
30
Part
I
Groups and
If
Subgroups
a map S ~
such
denote
S', omitting
You
wonder
may
includes
the
* and
we
labeled
S'
relation
between
S and
is not \302\242)
necessarily
isomorphism. It is apparent
when
which
of isomorphism
notion
definition via
in the
13, we
satisfies the displayed homomorphism then called a homomorphism
showedthat
c e E+. Also,
the
property,
an
than
rather
structures
binary
the
the
discuss
will
is \302\247
1, we
3.7 the
Definition
in
appeared
In Chapter
display.
0:5-^5'
in Section
that
Exercise27 asks
the displayed condition
one to one;
(Ec, +c) are isomorphic for all eachn e Z+.
\342\226\240
isomorphismproperty.The
the
we
which
structures,
notation.
the
correspondence,
before
and onto
but
*' from
than the
rather
of one-to-one
idea
the
one-to-one
words
why
property
homomorphism
S' are isomorphic binary
S and
then
exists, \302\247
by
{U,
and \342\226\240)
for
are isomorphic and \342\226\240) (Z\342\200\236, +\342\200\236} {U\342\200\236,
the for a collection of binary structures, algebraic the Our discussion Definition collection. 3.7 is an equivalence relation on defined by Tables 3.1 definition shows that the binary structures leading to the preceding the w hile Tables 3.5 and 3.6 are 3.4 are in those same equivalenceclass, given by through how to to determine whether in different equivalence classes.We proceed to discuss try
relation ~
are
binary
structures
How
to Show
We
now
isomorphic.
an outline
give
Step
to show that
1
{S,
Define
means
we
that
Step 2 S' and
in
to proceed
how
showing
*} and {S', *') are the
to describe,
in
gives some
the
0 is
that
Show
two
to show that
3.7
Definition
of S
isomorphism
what 0(s) is
fashion,
Show that 0 is a one-to-one function. deduce from this that x = y in S. &
from
isomorphic.
0 that
function
have
Step 3 Show that there does exist s
Step4
Are Isomorphic
That Binary Structures
structures
binary
us
in
That
e S' is given
and
this
Now
for every
be
that
is, suppose
onto S'. That is, suppose that s' that 0(s) = s'. * y) = 4>{x)*' 4>{y)forall x,y
S'.
with
to
s
&
S.
4>{y)
that
show
S such 4>{x
of computation. Compute both
sides
of the
equation
e S.
and
is just
This
see
a question
they are
whether
the
same.
3.8 Example
Let us
that
show
to the isomorphic
the binary
structure
Step 1
We
(R+,
have
multiplication = ex
(j){x)
4>{x)
Step 2
e
where \342\200\242}
to somehow from
Recall
structure (E,
ab+c = of two
for x
+} with
\342\200\242 is the
convert
operation
the usual
addition is
usual multiplication. an
that (ab)(a\342\202\254)
operation
addition
of addition to multiplication. of exponents corresponds to
quantities. Thus we try defining that ex > 0 for all x e E,
e E. Note
E
\342\200\224>\342\200\242 E+ by
so indeed,
E+.
If (j>(x) = 4>{y), then ex = ey. Taking x = y, so 0 is indeedone to one.
the
natural
logarithm,
we see that
3.9
Example
e E+,
Step
3
If r
Step
4
For x, see that
then
e E,
y
e E
ln(r)
31
Isomorphic Binary Structures
3
Section
=
and 0(lnr)
eln
r =
we have 0(x + y) = ex+y = ex an isomorphism.
r. Thus
0 is onto
\342\226\240 ey =
E+.
\342\226\240 Thus
0(x)
0(y).
is indeed \302\247
we \342\226\262
Let 2Z = {2n \\ n e Z}, so that 2Z is the set of all even integers,positive, and negative, zero. We claim that (Z, +} is isomorphic to (2Z, +}, where + is the usual addition. This of a binary structure (Z, +} that is actually isomorphic will give an example to a structure of a proper subset under the induced consisting operation, in contrast to Example 3.8, the operations were totally different. where
Step 1 2
Step
If
Step 3
If
n e
= n so
e Z. The 4>{m +
then shows
=
n. Thus
0 is onto
reverse
to the
turn
How do we
if this
is one \302\247
n/2
e Z.
= 2n
for
n
\342\202\254 Z.
to one.
Hence
equation
n) = 2(m +
=
n)
2m +
2n =
(f>{m)
+
0(n)
A Not
Isomorphic
question, namely: that
demonstrate
isomorphic,
=
2Z.
Show That Binary StructuresAre
How to We now
so m
so n = 2m for m
by
an isomorphism.
0 is
that
2m = 2n
is even
n
2(n/2)
Let m,n
4
then
2Z, then =
(j>{m)
Step
=
(j>{m)
try is given
: Z \342\200\224>\342\200\242 2Z to \302\242)
function
obvious
The
is the
two binary structures
(S, *) and {S',*'}are not
case?
would mean that there is no one-to-one function the property S onto S' with
* >')
3.10 Example
The
binary
while
structures
(Q, +}
|E| ^ K0. (See that Q is a proper the
to say induced A
operation
subset of
following
E. Example
can indeed be isomorphic
Q has cardinality K0 Note that it is not enough Example 0.13.) 3.9 shows that a proper subset with the to the entire binary structure. \342\226\262
not
isomorphic
because
of a binary structure is one that must be shared by any property structure. It is not concerned with names or some other nonstructural of the elements. For example, the binary structures defined by Tables 3.1 and
structural
isomorphic characteristics
and(E, +} are
discussion
are totally different. Also, a structural although the elements with what we considerto be the \"name\" of the binary operation. Example 3.8 showed that a binary structure whose operation is our usual addition can be is our usual multiplication. The number of elements isomorphic to one whose operation in the set S is a structural of (S, *}. property 3.2
are isomorphic, is not concerned
property
Part I
32
and
Groups
Subgroups
In the event that there are one-to-one mappings of S onto (S, *) is not isomorphic to (S', *'} (if this is the case) some structural that the other does not possess. property
3.11
Example
S',
we usually show that one
that
showing
by
has
both have cardinality X0, and there are lots of one-to-onefunctions \342\200\242 and where is the Z+. However, the binary structures (Z, \342\200\242) (Z+, \342\200\242), mapping there are two elements x such that In (Z, \342\200\242) usual multiplication, are not isomorphic. 1. However, in (Z+, \342\200\242), there is only the single element1. \342\226\262 x = x, namely, 0 and x \342\200\242
The sets Z
Z+
and
Z onto
list a few
We
of a binary
structure
examples of possiblestructural (S, *) to get you thinking
PossibleStructural 1. The
set has 4 elements.
2. The
operation
3. x * x
4. The equation x in
solution
element.
operation is called\"addition.' elements of S are matrices. is a subset of C.
c. The d. S
the algebraic notions of commutativity notion that will be of interest
introduced
We
One other
Properties
number 4 is an
b. The
S. a * x = b has a S for all a, b e S. x e
all
properties
line.
the right Nonstructural
a. The
is commutative.
= x for
along
Possible
Properties
and nonstructural
properties
and
in Section
associativity
to us is
2.
Table
3.3, by = = *\" *\" *\" the we on set have x u u x where for the binary u z], {x,y, operation for all choices possiblechoices, y, and z for u. Thus x plays the same role as 0 in = u + 0 = u for all u e E, and the same role as 1 in (E, \342\200\242) where (E, +) where 0 + u 1 \342\200\242 1 = u for all u e E. Because u = u \342\226\240 Tables 3.1 and 3.2 give structures isomorphic to the one in Table 3.3, they must exhibit an element with a similar property. We see that = u*'$ = u in Table 3.1 and that $*'u b * u = u * b = u for all elements u appearing structural
illustrated
x,
for
prove a little 3.12
3.13
Definition
Theorem
a
That
in the
of S
be
s
standard
definition of this
notion
structural
and
unique.
element, have
e of
S is
an
for * if
element
identity
\342\202\254 S.
way
serving as identity
e as an identity element, we must must
all
of Identity Element) is, if there is an identity
(Uniqueness
elements
a formal
give
An element
structure.
binary
e*s = s*e = s for
Proceeding
3.2. We
theorem.
Let {S, *} be
element.
Proof
u in Table
all elements
\342\226\240
A
binary
element,
to show
elements.
structure
it is
{S, *) has
uniqueness, suppose that them compete with
We let
we must have e * e = However, e * e = e. We thus obtain e = e, showing
e.
most
at
one identity
unique.
both each
regarding that
e are
e and other.
Regarding
e as an identity an identity
element
\342\231\246
3
Section
33
Binary Structures
Isomorphic
have a good grasp of the notion of isomorphic it binary structures, * that is an element for indeed a of structural identity having property a structure {S, *}.However, we know from experience that will be unable readers many to see the forest because of all the that have appeared. For them, we now supply a trees If
now
you
be evident
should
careful proof, 3.14Theorem
Suppose {S, *} has an identity {S', *'}, then 0(e) is an
with
Proof
to touch
along
skipping
those trees
elements
that
are
involved.
S' is an isomorphism for*. If 0 : S \342\200\224>\342\200\242 *' on S'. for the binary operation
identity
= s'. Because must show that 0(e) *' s' = s' *'0(e) 0 is an isomorphism, one-to-one map of S onto S'. In particular, there exists s & S such that 0(s) = e is an identity element for * so that we know that e * s = s *e = s. Because Now 0 a function, we then obtain
Lets' e S'.We it is a
(p{e *
Using
3.7 of
Definition
an
0(e)
*' s' = s' *'0(e)
We certain
conclude
*' 4>{s)=
=
s
\342\202\254 S such
0(s * e) = we can
3.14,
that
areindeedstructural.
rewrite
this
as
0(5) *' 0(e)= 0(5). that 0(5)
= s', we obtain
3.17
Example
that
show
The binary
with three more examplesshowing via structural properties that are not isomorphic. In the exercises we ask you to show, as in the structures in these examples the properties we use to distinguish That structure. is, they must be sharedby any isomorphic
structures (Q, +}
the binary
structure
isomorphic
sets have of matrices
relation \342\231\246
Z have
and
(Z,
+} under
Ko, so there are lots
the
(M2(E),
to (E,
are not functions e Q, but this is in Z. We have
addition
usual
of one-to-one
The binary structures (C, \342\200\242} and (E, \342\200\242} under the usual multiplication that C and ffi have the same cardinality.) The equation (It can be shown in ffi. solution x for all c e C, but x \342\200\242 x =-1 has no solution
is not
desired
the
isomorphic. (Both Q cardinality x + x = c has a solution x for all mapping Q onto Z.) The equation = not the case in Z. For example,the x x 3 has no solution + equation exhibited a structural that these two structures. property distinguishes Example
is
s'.
and
3.16
s'.
structures
binary
Theorem
We
we chose
that
Remembering
s) =
isomorphism,
0(e)
3.15 Example
of {S,*}
element
c
\342\226\262
are not
isomorphic. x \342\226\240 x = c has
a \342\226\262
x 2 real matrices with the usual matrix multiplication usual number multiplication. (It can be shown that both of numbers is commutative, but multiplication Multiplication of 2 \342\200\242}
with the \342\200\242}
cardinality |E|.) is not.
\342\226\262
Part
34
Groups and
I
3
\342\226\240 EXERCISES
In
the
all
Subgroups
+ is
exercises,
on the set where
addition
usual
the
it
is specified,
and
\342\200\242 is the
usual multiplication.
Computations
1. What
three
structure
(See Exercise1.)If
the second.
it is
whether the given map 0 is an not an isomorphism, why not?
with
(Z, +) where
Z, +)
with
(Z,
+) where
4. Z, +)
with
(Z,
+) where
0(n) =
2.
Z,
3.
+)
5. Q, +) with
+) where
(Q,
=
\342\200\224n for n
=
2n for n n
=
where \342\200\242) 0(x)
= x2
for
7.
E,
where \342\200\242)
= *3
for* e
8.
M2(E),
9.
Mi(E),
10. E, +) In
0(A)
is the
determinant
of matrix
A
with (E, \342\200\242)
where \342\200\242)
0(A)
is the
determinant
of matrix
A
(E+,
12. (F,
+)
with
13. (F,
+)
with
14. (F,
with
+)
16.
eE
= 0.5'\" for r
11 through 15, let F be the set of all functions the instructions for Exercises 2 through 10.
11. (F, +) with
(F,
(Z,
In
each
17. The
+)
map
In each
18. The
(F,
where \342\200\242)
onto
binary a.
(Q,
\302\242(/)
such
/
by that
of
derivative
give the
0 : Z
-^ Z
: Q-+ \302\242
defined
* on Z
such
case, give
by
that
identity
Q defined
n
+
1 for
n e Z is
the
identity
by
that
of
all
orders.
for
(Z,
*) onto
Z is one to mapping b. (Z, *) onto
n e
isomorphism
and onto
Z. Give the
definition
of a
Z. Give the
definition
of a
Q. Give the
definition
of a
(Z, +).
one
(Z,
and onto
\342\200\242).
on Z.
\342\200\224 1 for x e Q is one to an is 0 isomorphism mapping
b. (Q, *) element for
one
on Z.
= n + 1
for *
one to
mapping
isomorphism
for *
is an \302\242
element
derivatives
/(x)
*), the
have
/(f)A] \302\243[\302\243
is an \302\242
0(n)
that
\342\200\242
0(n) =
element
E
/
b. identity
E into
mapping
/(t)dt
(Z, *),
operation * on Q such +) onto (Q, *),
In each
=
= x
0(/)(x)
defined
* on Z
case, give
map
= ft
+) where 0(/)(*)
onto (Z, \342\200\242)
(Z,
0(/)(*)
+) where
binary operation
a.
/'(0)
(F,
0 : Z -* Z
case,
\302\242(/)
(E,
+) where
binary operation
a.
/', the
=
+} where
with (F, \342\200\242)
The map
=
{F,
with
E
where \342\200\242)
where \342\200\242) 0(r)
structure
e Q
with \342\200\242) (E,
with
binary
binary
* \342\202\254 Q
Exercises
Follow
15.
0(x)
of the first
isomorphism
of a
n e Z
for x
x/2
isomorphism
e Z
1 for
+
S \342\200\224>\342\200\242 5\" is an \302\242:
e Z
6. Q, \342\200\242) with (Q, with \342\200\242) (E,
a function
whether
10, determine
2 through
Exercises
In
things must we check to determine (S, *) with (S\", *')?
* on Q.
onto
one
and onto
(Q, +).
35
Section3 Exercises
19.
The
map
: Q -\302\273 Q defined
* on
binary
operation
a.
onto \342\200\242) (Q,
(Q,
In each
Q
such
by
=
that
\342\200\224
3x
0 is an
1 for
the
identity
for *
element
and onto
to one
Q. Give the
of a
definition
mapping
b. (Q, *)
*),
case, give
Q is one
x e
isomorphism
onto
(Q,
\342\200\242).
on Q.
Concepts
20. The
\"0 must
21 and
In Exercises
needed, so that
it is
21.
A function
22.
Let *
correct
for * for
all
s e
an
of
how
that
sometimes summarized can be viewed
without reference
term
italicized
the
3.7 is condition
in Definition \302\242)
isomorphism
operation(s).\" Explain
binary
the definition
to
the
text, if
this
in
correction is
publication.
if and
isomorphism on a set
operation
binary
the
acceptablefor
5\" is an 0 : S \342\200\224>-
be a
element
22,
commutewith
in a form
for
condition
homomorphism
displayed
saying, manner. by
S.
An
if 0(a
only
e of
element
S
* b)
= 0(a) *' 0(&)-
with
the
5*e
property
= 5=
e*5isan
identity
S.
Proof Synopsis to give a one or two sentence of a proof is your ability synopsis of it, explaining your understanding Note that we said \"sentence\" and not \"equation.\" the proof without all the details and computations. for a synopsis of a proof in the From now on, some of our exercise setsmay contain one or two problems asking Here is our for you what we mean by a synopsis. exceed three sentences. We should illustrate text. It should rarely then the the and our 3.14. statement of theorem of Theorem Read one-sentence now, synopsis. synopsis A
good
test of
idea of
the
an element
Representing
of 0
carry
as 0(5) for
of S'
the computation
of 0(e)
S, use the
s e
some
*' 0(5) back
to a
property
homomorphism S.
in
computation
that one mathematician does the proof go?\" of explanation give another if asked, \"How might the computation or explain of S' as 0(5). To supply why we could represent an element every in our synopsis. written proof. We just gave the guts of the argument detail would result in a completely That
We
did
is the
to
not
kind
make
23. Give a proof synopsis
of Theorem
3.13.
Theory
24.
An
identity
two-sided a. a left
element for a binary operation identity element.\" Using complete
identity
element
Theorem3.13shows
25.
identity
element
the first
place
Continuing element en
impossible.
that
you just where
the ideas
the
ei for *,
sentences,
if a two-sided
identity
by
element
for *
defined? If so,prove it. If not, give proof of Theorem 3.13breaks down. eRl
If so,
Definition
give analogous
b. a right
and
of Exercise24 can
where eL /
* as described
identity
3.12 is
element
exists, it is unique. a counterexample
structure have a left a binary an give example, using an operation
identity
on a
sometimesreferred to
as
\"a
for
definitions
en for
*.
Is the sametrue for a one-sided (S, *) for a finite set S and find element
finite
set
ei and a S. If not,
right
prove
identity it is
that
Part I
36
26.
Recall
=
f(a)
/ :A
if
that
Prove with
appropriate
places
structural
operation
31. For each c
H
' (fc)
(5\", *'),
unique a e A such that is an isomorphism of \302\242^
is the
then
be the
commutative.
* is
associative. x *
equation
b in
S
for a skeptic
proof
3.14, we
x =
c has
H is
closed
a. Show
that
(C,
+) is
b. Show
that
(C,
is isomorphic \342\200\242)
of
to
(H,
to (H,
this
for the
structure property of a binary is an identity element
indicated
the
that
property,
\"There
(S, *) for
*.\
x in S.
~
all matrices matrix
both
under
isomorphic
H is a matrix
*')
in Definition 3.7, is an equivalence relation on any set were asked to prove in the preceding two exercises at
= b.
b*b
that
such
did
a solution
subset of M2(E) consisting that
you
of (S\",
proof.
* is
Section 2 shows
say that
described
quote the results
(In Theorem
an element
: 5\" -\302\273 with (5\", *')andy/5\"' is an isomorphism is an isomorphism of (5, *) with (5\"', *\.
i\\r o 0
isomorphic,
simply
may
32, give a careful
e S,the
32. Thereexists
You
property.
30. The operation
33. Let
with
/
of (S, *)
function
~ of being
in your
29 through
Exercises
of (5, *)
isomorphism
isomorphism
composite
relation
the
structures.
29. The
of
the
addition and
for a,
form matrix
b e
ffi.
Exercise
23 of
multiplication.
+}. \342\200\242).
of the
C.) complex numbers There are 16 possiblebinary structures on the set {a, b} of two elements. How many nonisomorphic (that is, different) structures are there among these 16? Phrased more preciselyin terms of the isomorphism structurally ~ on this set of 16 structures, how many equivalence classes are there? Write down one equivalence relation structure from each equivalenceclass.[Hint: a and b everywhere in a table and then rewriting Interchanging the table with listed in the original elements the one we order does not always yield a table different from (We
34.
the
of binary
is indeed a
S, then
A onto
mapping
(S,*).
*\,") then
(5\"',
function
: S \342\200\224>5\" is an
5\" is an if 0 : S -\302\273
that
28. Prove that
In
Subgroups
one-to-one
\342\200\224>B is a
that if
Prove
fc.
(5\",*') with 27.
Groupsand
started
section
representation
with.]
4
Groups
Letus continue the
the
analysis problems
computational
of our past experience with of addition and multiplication
algebra.
Once
of numbers,
we had
mastered
we were ready
of problems. Often problems lead to operations to the solution is to be determined. The simplest some unknown number x, which and equations are the linear ones of the forms a + x b for the operation of addition, ax =b for multiplication. The additive linear equation always has a numerical solution, and so has the multiplicative one, provideda/0. the need for solutions of Indeed,
to
apply
equations
these
binary
involving
=
such as 5
+ x = 2 is a
for the negative good motivation very is shown by equations such as 2x = 3. the need for rational numbers for us to be able to solve linear equations involving It is desirable our binary is not possible for every binary operations. This however. For example, the equation operation, a * x = a has no solution in S = {a,b, c] for the operation * of Example 2.14. Let us abstract from familiar algebra those properties of addition that enable us to solve the 5 + * = 2 in Z. We must not refer to subtraction, for we are concerned with the equation solution in this case addition. The steps in phrased in terms of a single binary operation, additive
numbers.
linear
equations
Similarly,
Section 4
are as
solution
the
follows:
= 2, =-5+
5+ * (5+ *)
-5+
(-5 +
Strictly speaking, wehave
not
for a solution.
To
given,
2, adding-5, -5 + 2, associative law,
x =
+
5)
0 +
possibility
37
Groups
* = * = x =
\342\200\2245 + 2,
computing
-5 + 2,
property
\342\200\2243,
computing
analysis could be made for the operation of multiplication:
A similar
the
= 3,
2)x
{\\
+ 2.
\342\200\2243 is a
given,
= \\(i), \\{2x) \342\226\240 =
\342\200\224 5
\342\200\2243 is a
that
2x
+ 5,
of 0,
that it is the only solution, but rather solution, one merely computes5 + (\342\200\2243). 2x = 3 in the rational numbers with equation
that
shownhere show
\342\200\224 5
^3,
1 \342\200\242 x = ^3,
43, 2
i,
by
multiplying
^
associative law, computing ^2, of
property
3
1, 13.
computing
have to We can now see what properties a set S and a binary operation * on S would of this procedure for an equation a * x = b for a, b \342\202\254 S. Basic have to permit imitation of an element e in S with the property that e * x = x to the procedure is the existence the role of e, and 1 played the role for S. For our additive example, 0 played for all x \342\202\254 a' in S that has the property that our Then we need an element example. multiplicative the role of a', and \\ played a! * a = For our additive example with a = 5, \342\200\2245 played the role for our multiplicative with a = 2. Finally we need the associative law. example The remainder is just computation. A similar analysis shows that in order to solve the x * a = b (remember that a * x need not equal x * a), we would like to have an equation element e in S such that x * e = x for all x e S and an a' in S such that a * a! = e. With be sure of being able to solve linear equations. all of these properties of * on S, we could Thus we need an associative element e such that an identity binary structure (S, *) with This is precisely the for each a e S, there exists a! S such that a * a' = a' *a
e.
= e.
e
notion of a group,
and Examples
Definition than
Rather
of the who
4.1 Definition
A
describe
a group
preceding paragraph,
picks up
group
axioms Sfx:
this
text
all a,b,c
using terms defined in we
to discover
(G, *) is a set are satisfied: For
define.
we now
which
what a group is
G, closedunder
e G, (a
give
2 and 3 as wedid at definition. This enablesa
Sections
a self-contained
a binary
without
having
operation
to look
*, such
we have
* b) *
c=
a * (b * c).
associativity
of
*
that
the
end
person
up more terms. the
following
Groups and Subgroups
Part I
38
3^:
e
element
is an
There
e * x = x *e = x. a e
to each
Corresponding 3\302\276-.
4.2
see that
We easily
Example
e'e e
U, the
of U has
element
every
that
of
element
every
the
fact that
For z
inverse.
an
of complex numbers
is
for multiplication.
For
identity
e
the computation \302\243/\342\200\236,
= l
\342\226\240=zn
z\"'1
an inverse.
has
Un
(Rc, +c) is isomorphic
Similarly,
\342\226\240
= e2\"1 = 1
\342\226\240 ei{2^e)
z
Because
of a
a'
an
that
G such
in
Multiplication
is
*
computation
that
shows
a'
element inverse
1, which
contain U\342\200\236
and
ew
shows
is an
and are groups. (\302\243/\342\200\236, \342\200\242} \342\200\242}
(U,
and both U
associative
G, there
e for
element
identity
= a *a = e.
a *a
e G,
that for all x
G such
in
to
is isomorphic +\342\200\236} (Z\342\200\236,
Thus (U,
and are groups. (\302\243/\342\200\236, \342\226\240) \342\200\242}
+c) is a group
that (Ec,
to (\302\243/, we see \342\200\242},
for
e M+.
all c
is a (Z\342\200\236, +\342\200\236}
that
shows ([/\342\200\236, \342\200\242}
group
A
forallneZ+.
We point the binary
clarity demands
is of
there
that
we
roots of the theory evident
historical
three
There developmentof
abstract
group
of
mathematical
literature
the theory of and geometry.
algebraicequations,
nineteenth
the
specify
methodswere
considerably
the
theory,
areas used group-
of reasoning,
methods
although more explicit in the
the
One of the
attention
transformations
thought
themes
was
In number century
Leonhard
on division remainders
the
of geometry in the for invariants
search
geometrictransformations.
became which
themselves,
of as
course a binary
refer to
on the
operation
Rather
a group
set G.
* on G, we use the
an operation
focused on the in many cases can be
elements of groups. theory, already in the eighteenth Euler had considered the remainders
of powers a\" have
by a fixed
\"group\"
prime p. These
properties.
Similarly,
his
In
than
use
G, with the the event that \"the group
phrase
G
with
quadratic
in particular of these forms to what amounted
classes
equivalence composition possessed
Arithmetiforms
Disquisitiones
cae (1800), dealt extensively ax2 + 2bxy + cy2, and
showed
that
under group
properties.
the theory
Finally, provided
central
century
under various types of
[
constantly, we often
first area
than in the other two.
Gradually
in notation.
be sloppy
sometimes
Carl F Gauss, in in
century:
number
of these
three
All
theoretic
nineteenth
will
(G, *}
Note
Historical are
notation
that
understanding
that we
now
out structure
concept. initiated
of algebraic equations
explicit prefiguring of the group in fact Lagrange Joseph-Louis (1736-1813) of an the study of permutations of the roots the
most
equation as a tool for solving of course, were ultimately of a group. It was Walter von
Heinrich
Weber
and
give
abstract group.
Dyck
(1842-1913)
able independently roots
it. These considered
to
clear
combine
definitions
permutations, as elements
and (1856-1934) in 1882 were the three historical of the notion of an
who
i
Section 4
*.\" For
under
than write to the group 4.3 Definition
A
without
Zg
if its
abelian
in honor groups are calledabelian of the Norwegian Niels Henrik Commutative mathematician Abel (1802-1829). Abel was interested in the
of solvability in 1828, written
question of
then has
an
functions
of
the
polynomial equations. In a he proved that if all the roots
this
equation
property
permutation groups
led
Camille
algebra
to name
of
of permutations
of commutativity in these associated with solvable equations
Jordan in his 1870 treatise on such groups abelian; the name
4.4 Example
The
4.5
The set
Example
is an
4.7
but
Example
since
Example
addition is not integers
element
0, but no
additive
Q, E,
that
The set
Z,
Z+
under
properties
and C under
Example
he created
papers,
age of 26, two
at the in
of debts.
brilliant
to write
in
Norway
abundance
an
and
to
returned
sets with binary operations
days
a university
finding
that
give
and
groups
a group.
There is no 0) under
(including
for + in
element
identity
is still
addition
not a group.
inverse for 2.
is not a
\342\226\262
There \342\226\262
of integers and of rational, real, are abelian groups. addition
multiplication
Z+.
group. There is
an
and
complex
numbers \342\226\262
1, but
identity
no inverse A
The familiar
multiplicative
the sets Q+ and under
He
Norway.
groups.
of all nonnegative
show
position
Crelle succeeded for him in Berlin.
of some
not give
in
as a teenager
theory
Abel
of tuberculosis
died
position
Z+ under
The familiar
with no
before
do
identity
whose
single-handedly.
of 3. 4.8
refer
groups
teachers in
all his
He neverthelesscontinued that
somethat
set
commutative
to mathematics
attracted
surpassed
of elliptic functions, virtually
1827
Let us give someexamples also of
4.6
free to
to study grant finally received a government travel elsewhere in 1825 and proceeded to Berlin, where he befriended Crelle, the founder of the most August influential German mathematical journal. Abel numerous contributed papers to Crelle's Journal during in the field the next several years, including many
It was
roots.
the
soon
and
f,
group
been applied to
general. Abel was
can be expressed as rational g,... ,h of one of them, say x, and and g{x), the if for any two of these roots, f{x) then the relation f{g{x)) = g{f{x))always holds, Abel showed that equation is solvable by radicals. each of thesefunctions in fact permutes the roots of the these functions are elements hence, equation;
of such
rather
addition
under
ffi
However, we feel
is commutative.
operation
binary
and
i>,
+).
Note
\342\226\240 Historical
paper
Z, (
tedious
more
G is
group
we may refer to the groups (Z, +), (Q, +), and (ffi, specifying the operation.
example,
the
39
Groups
multiplication
E+
properties
of rational,
of positive numbers and are abelian groups.
the
real, sets Q*,
and ffi*,
complex and
numbers
C* of
show that
nonzero numbers \342\226\262
Part
40
I
4.9Example 4.10
4.11
4.12
4.13
Example
Example
Example
Example
Groups and
The
set of all
This
group
Subgroups real-valued
\342\226\26
The set Mmx\342\200\236(E)of all m x matrix with all entries 0 is the The set M\342\200\236(E) of all n x n n x n matrix with all entries
subset S of
Show that the
so that
under
matrices
n
matrix.
identity
be summarized
by asserting
\342\226\26
matrix addition is a group. This group is abelian.
multiplication is
matrix
under
matrices
the axioms
that
note
should
The m
x n
\342\226\26
The
a group.
not
0 has no inverse.
Mn (E)
\342\226\26
of all
consisting
n matrices
n x
invertible
under
matrix
is a group.
S is closed under
by showing that A^1 and B'1
start
We
is a group.
addition
function
(Linear Algebra) Those who have studied vector spaces can for a vector space V pertaining just to vector addition that V under vector addition is an abelian group.
multiplication
Solution
E under
domain
with
functions
is abelian.
exist and
both
(AB)(B'1A~1)=
matrix
=
AA'1
Let A
multiplication.
and
B be
in S,
= In. Then
BB'1
=
A{BB~l)A~l
=.
AlnA'x
I\342\200\236,
invertible and consequently is also in S. acts as the identity element, and matrix multiplication is associativeand /\342\200\236 each element of S has an inverse by definition of S, we seethat S is indeed a group. is not commutative. It is our first example of a nonabelian group. a group
so that
AB is
Since since
This
The group fundamental is usually
a matrix
E\",
defined E\"
into
A
Example
transformations
group
is usually
of
denoted
by GL(W).
be defined on Q+ by
a * b
*c =
ab
abc
2
4
E). Also, matrix all invertible
composition;
GL(W).
,
likewise
Thus * is associative.
a e
Q+, so 2 is an
identity
is an
inverse
abc
2
4
.
= a *2 = a for *.
element
Finally,
=
a a-*a=2, for a. HenceQ+ with the a*-
so a! = A/a
be
= * \342\200\224
that
shows
Computation
2* a all
= a
* c)
operation
\342\200\224
Thus
Then \342\200\224 *c=
T : E\"
transformation
linear
Of course,GL(n,E) ~
= ab/2.
a * (b
for
transformation
in this in GL(n, fashion by some matrix to composition of linear transformations. E\" into itself form a group under function
(a *b)' and
you
linear E) gives rise to an invertible and that conversely, Ax, every invertible
GL(n,
preceding exampleis of
general linear group of degreen, who have studied linear algebra know
is the
algebra. It GL(n, E). Thoseof
corresponds
linear
Let *
in the
described
matrices
n
in linear
by 7\\x) = itself is defined
multiplication
4.14
in
n x
invertible
denoted by
and
that
of
of
importance
* is a group.
this
Section 4
41
Groups
Elementary Properties of Groups As
we proceed
is the
thing
only
to prove our we know
can employ both
Definition
use the
and
definition
first
theorem
about
groups
about groups, we must at the moment. The
and the first theorem; two theorems, and so
4.1
the first
use
Definition
proof of
a second
of a
the proof
4.1, which theorem can
theorem
third
on.
laws. In real arithmetic, we know that establish cancellation 2a = 2b b. We need only divide both sides of the equation inverse of 2. 2, or equivalently, by multiply both sides by |, which is the multiplicative We parrot this proof to establish cancellation laws for any group. Note that we will also use the associative law. Our first
2a =
4.15 Theorem
If
in
a, b, c
Proof
with binary
G, that is,
e G.
a * b
Suppose
a =
that
is a group
G
hold
theorem will
2b implies
= a
operation *,
a * b = a * c implies * c. Then
by
a * (a * b) = By the
definition
the
= c * a implies
b =
laws
c for all
a', and
a' *
(a
* c).
associative law, * a)
(a' By
c, and b * a
exists
there
3\302\276.
right cancellation
and
left
the
then
b =
of a'
a'
in
3\302\276,
* a
*b =
= e,
(a' * a) * c.
so
e * b = e * c. definition
the
By
of e
in
.\302\276
b =
c.
Similarly, from b * a = c *a one can deduce a! and use of the axioms for a group. by Our
next
a group has a solutions
find
4.16 Theorem
If
make use of
unique
of such
4.15.
that we
We show
chose our
right
equation\" in to allow us to
a \"linear
that
properties
group
equations.
group with binary operation *, and if a and b are any a * x = b and y * a = b have unique solutions equations
First we show the existence solution of a * x = b. Note a *
of at
least one
(a' * b)
solution
solution
just
by
of G,
elements
x and
y in
computing
then
= (a * a') * b,
of a
b,
* x = b- In
associative definition property
a similar
that a!
* b is a
law, of a', of e.
fashion,
y
=
b *
the
G.
that
= e* = b, Thus x = a' * b is a of y * a = b.
on the
multiplication
\342\231\246
Theorem
Recall
solution.
= c upon
G is a
linear
Proof
proof can
that b
a' is a solution
42
Part I
and
Groups
Subgroups
To show uniqueness of y, we use the standard method of assuming two solutions, y\\ and yi, so that y\\ * a = b and yi * a = b. Then y\\ The uniqueness of x follows similarly. by Theorem 4.15, y\\ = )\302\276. Of
to prove
in the
the
uniqueness
the definition procedure we used in motivating then x = a! * b. However, we choseto illustrate namely,
same. Note
=
and
* a, )\302\276
\342\231\246
followed the if a * x = b, to prove an object is the standard way two and have such then must be the you objects, prove they suppose x \342\200\224 the solutions a' * b and y = b * a! need not be the same unless * is
course,
unique;
last theorem, we could of a group, showing
we have
that * a
that
have
that
commutative.
Because
that the
identity
a group is a special type e in a group is unique.
of binary
state
We
structure, we know this again as part
from
3.13
Theorem
theorem
the next
of
for easy reference.
4.17Theorem
In
G with binary
a group
operation *, there is only x = x
e *
G. Likewise for
x e
all
for
each aeC,
Proof
element
the identity
summary,
Theorem 3.13 shows that
of the
to
Turning
the
= e and
a * a\" by Theorem
and,
Note
that in a (a
This That
4.18 Corollary
Let
is,
G be a
b)' =
that
for
binary
any
structure
in
a group.
is unique. No
=
use
a *
a'
aeG
has inverses
a'
and
a\"
e. Then \342\200\224 e
{b' * a')
and
a
= a
i
,
group is unique.
group G,
* b) *
equation
(a *
G such
4.15,
of a in a
inverse
in
element are unique
a * a\" =
a
so the
a'
of each
inverse, suppose that a =
a\" *
one element
this.
show
to
of an
uniqueness
= a*a'
a'*a
so that
element
an identity
was required
axioms
group
that
x
is only
inverse
and
e in G such
= a *a = e.
a *a In
*e=
there
element
one
we have
= a * (b 4.17
Theorem
b' * a'. We
group. For all
\342\231\246
state
a,b e
this
G, we
* b')
*a
= (a * e) * a'
that b' * a' is as a corollary.
show
have
(a *
b)' =
V
the
= a *a = e. unique
inverse
of a
* b.
* a'.
we remark that binary algebraic structures For your information, with weaker axioms than those for a group have also been studied quite Of weaker these structures, extensively. the semigroup, a set with an associative binary has perhaps had the most operation, A monoid is a semigroup that has an identity element for the binary attention. operation. is both a semigroup Note that every group and a monoid.
Section4 it is
Finally, be
weaker,
possible to
axioms
give
for a group
(G, *} that
43
Groups
at first
seem
glance
to
namely:
1.
The binary
2.
There
operation
a left
exists
3. For eacha
associative. identity element e in G such G is
* on
e G, there
exists
inverse a! in
a left
that e G such
* x = x for that a'
*a
all
x e
G.
= e.
definition, one can provethat the left identity element is also a right and a left inverse is also a right inverse for the same element.Thus element, identity these axioms should not be calledweaker, since they result in exactly the same structures it called It is conceivable that might be easierin some cases to check these being groups. than to check our axioms. axioms two-sided Of course, by symmetry it is clear that left there are also right axioms for a group. one-sided
this
From
Finite Groups and All
our
after Example 4.2 have been of infinite groups, that number of elements. We turn to finite groups,
examples
the set G has smallest
groups
where with
starting
the
sets.
a group
has to have at
give rise to a
might
is,
infinite
an
finite
Since
Group Tables
on {e} is defined always its own
is a
group
e *
by
e=
one element,
least
one-element
e. The
three
namely, the identity, set {e}.The only possible group axioms hold. The
a minimal binary identity
set that
operation element
* is
in every
group. Let us try to put a group structure on a set of two elements. Since one of the elements must play the role of identity we as well let the set be {e, a}. Let us attempt element, may on to find a table for a binary * that a {e, a} operation gives group structure on {e, a}. we When a table for a shall list the identity first, as in giving group operation, always the
inverse
table.
following
e a
*
e
a
Sincee is
to be
the identity,
so
e *x = x *e = x all
for
x e
{e, a],
*
e
a
e a
e a
a
Also,
a must
we are
to fill in
forced
have an inverse
the table as follows,if
a' such that a *
a =
a *a
= e.
* is
to give a group:
and
Groups
Subgroups
In our case, a! must a! = a, so we have
have
*
e
a
e
e
a
a
a e
be
either
e or
to complete
a. Sincea! = e obviously as follows:
does
not work,
we
must
the table
except possibly the associative property. an operation from a table defining can be a basis Checking associativity = 2 is that under addition modulo tedious we know However, Z2 {0, 1} very process. be the one above with e replaced a group, and by our arguments, its table must by 0 for our and a by 1. Thus the must be satisfied table e associative containing property All
the
group
axioms are
now
satisfied,
on a case-by-case
and a.
With
this
a table
that
example
as background,
giving a binary
a group structure denote by e, that
on the
operation
set. There must
we
should
on a finite be
be able
set
must
to list satisfy
some necessaryconditions for the
of the set, The condition e * x
one element
which
operation to give we may as well
= x means that the row acts as the identity element. the elements appearing e at the extreme left must contain exactly opposite x * e = x across the very top of the table in the same order. Similarly, the condition the e at the very top must contain means that the column of the table under exactly elements appearing at the extreme left in the same order.The fact that every element a has a right and a left inverse means that in the row having a at the extreme left, the element e must appear, and in the column under a at the very top, the e must appear. of
the
table
Thus e must appear in each row and in each column. We can do even better than this, however. By Theorem4.16,not only the equations a * x = e and y * a = e have unique but also the equations a * x = b and y * a = b. By a similar this solutions, argument, means that each element b of the group must appear once and only once in each row and
each
column
of the
table.
Suppose conversely that a table for a binary operation on a finite set is such that there is an element each element acting as identity and that in each row and each column, of the set appears exactly once. Then it can be seen that the structure is a group structure if and only if the associative law holds. If a binary operation * is given by a table, the associative law is usually messy to check. If the operation * is defined by some law is often easy to check.Fortunately, characterizing property of a * b, the associative ' this second case turns out to be the one usually encountered. We saw that there was essentially only one group of two elements in the sense that if the elements are denoted by e and a with the identity e appearing first, the element table must be shown in Table 4.19. Suppose that a set has three elements. As before, we may as well let the set be {e, a, b). For e to be an identity element, a binary operation * on this set has to have a table of the form in Table 4.20. This leaves four shown to be filled in. You can quickly see that Table 4.20 must be completed as shown places in Table 4.21 if each row and each column each are to contain element exactly once. Because there was only one way to complete the table and Z3 = {0, 1, 2}under addition modulo 3 is a group, the associative e, a, property must hold for our table containing andfo.
Exercises
Section 4
that G' is any suppose element identity appearing could be done in only one way,
45
a table for G' imagine = {e,a, b) for the table G filling we see that if we take the for G' and rename the table table for G' e, the next element listed a, and the last element b, the resulting identity must be the same as the one we had for G. As in this Section 3, explained renaming G' with the group G. Definition 3.7 denned the gives an isomorphism of the group notion of isomorphism and of isomorphic binary are just certain structures. Groups definition Thus our work above of so the same to them. structures, types binary pertains with a can be summarized that all element are single by saying groups isomorphic, all two and all with three elements are with elements are groups groups just isomorphic, just Now
with
other
of three
group
elements and
first. Since our
out of
to express this identification use the phrase up to isomorphism using the ~. \"There is of three relation Thus we one elements, only may say, group equivalence up to isomorphism.\" We
isomorphic.
4.21Table
4.20 Table
Table
4.19 *
e
a
e
e
a
a
a e
*
e
a
b
*
e
a
b
e
e
a
b
e
e
a
b
a
a
a
b
b
b
a b
b e
e a
4
\342\226\240 EXERCISES
Computations results,
group
1. Let *
6, determine
1 through
Exercises
In
give the
be defined
on Z
2Z =
2. Let
* be
defined on
3.
* be
defined on R+
Let
4. Let * 5. Let * 6. Let *
be defined
on Q
be defined
on the
be defined
on C
by
by
by
{2\302\253 | n
letting
set R*
e Z} by letting a * b =
of nonzero
a *b
letting
operation
from 3\302\276, 3\302\276
* gives a
Definition
group
4.1
that
given set. If no
not hold.
does
=
a *b =
a + b.
~Jab.
ab. real numbers
by
letting
a * b =
a/b.
\\ab\\.
8.
1000 elements. example of an abelian group G where G has exactly We can also considermultiplication modulo n in Z\342\200\236. For example, 5 -7 6 = 2 \342\200\242\342\200\236 8 is a group. Give the 4(7) + 2. Theset {1,3,5, 7}with multiplication -8 modulo
9.
Show
7.
on the
structure
= ab.
a * b =
letting
binary
order 3^,
a *b
letting
by
the
whether
in the
axiom
first
Give an
10. Let n
that
be a positive
a. Show b. Show
is not isomorphic \342\200\242) (\302\243/,
the group
integer
that
(\302\253Z, +)
that
+) (\302\253Z,
and
let nL
is a group. ~
(Z, +).
=
{nm
to either \\ m
(R,
e Z}.
+) or (R*,
(All three \342\200\242).
in
Z7
because
5
\342\200\242 6 \342\200\224 30 =
for this
group. have groups cardinality table
|R|.)
Part I
46
and
Groups
Subgroups
11 through whether the given set of matrices under the specified matrix 18, determine operation, whose only nonzero entries addition or multiplication, matrix is a square matrix is a group. Recall that a diagonal is a square the upper left to the lower lie on the main diagonal, from right corner. An upper-triangular matrix matrix with only zero entries below the main diagonal. Associated with each n x n matrix A is a number called = det(A) det(S). Also, If A and B are both n x n matrices, then det(AS) the determinant of A, denoted by det(A). = 0. 1 A if and if and invertible is det(7\342\200\236) det(A) ^ only In Exercises
n diagonal
matricesunder
12. All n x
n
diagonal
matrices
under matrix
x
n
diagonal
matrices
with no
11.
13.
n x
All
All n
14.
All
n x
15.
All
n x
16. All n x
x
n n
All n
19. Let S be
\342\200\224 1 under
upper-triangular
under matrix
upper-triangular
matrices
with determinant 1 under
matrices
set of
the
either 1 or
numbers
real
all
addition.
\342\200\224 1 under
\342\200\224 1. Define
except
a.
Show
that
Show
that
c.
Find
the
* gives a
multiplication.
S
by
+ b + ab.
on S.
operation
binary
(S, *) is a group. solution of the equation
matrix
matrix multiplication.
* on
= a
a *b
b.
matrix multiplication.
multiplication.
matrices
with determinant
matrix multiplication.
under
entry
matrices with all diagonal matricesunder matrix
n diagonal
18. All n x
17.
multiplication.
zero diagonal
entries 1 or
n upper-triangular n
addition.
matrix
= 7 in S.
2***3
structures on a set of 4 elements. 20. This exerciseshows that there are two nonisomorphic group Let the set be [e, a, b, c},with e the identity element for the group A group table would then operation. to start in the manner shown in Table 4.22. The square indicated be filled in by the question mark cannot
be
with the
have with
different from both e and a. In this identity latter case, it is no loss of generality to assume that this element is b. If this square is filled in with e, the table can then be completed in two ways to give a group. Find these two tables. (You need not check the associative in with b, then the table can only be completed in one way to give a group. Find this law.) If this square is filled table. (Again, need not the the three check associative Of tables law.) you you now have, two give isomorphic two tables these are, and onto renaming function which is an groups. Determine which give the one-to-one
a.
It must
in either
filled
e or
element
an element
with
isomorphism.
a. b.
table
elements commutative?
of 4
all groups
Are Which
gives
a group
isomorphic
the
to
so \302\243/4,
group
that we
know
the
binary
operation
defined by the
table is associative?
c.
that
Show
the group
one particular
21. According many
given by 12 of
to Exercise
of these
3 elementsgive
give a
so
of n,
value
other tables is structurally defined that the operation
of the we
know
Section 2, there of
structure
a group
one that
are
16 possible of the
a group?
How
for .\302\276
a group.
many
binary
as the
same
the
by that
14 group in Exercise table is associativealso. on a set
operations
19,683 possible
binary
for
of 2 elements.How
operations
on a set
of
structure?
Concepts
22. Considerour orders to
state
axioms the
.^, axioms
and .\302\276
are
We gave them
in the
order
and -^5\302\276\302\276\302\276. .\302\276\302\276\302\276\302\276\302\276. 3\302\276\302\276\302\276^ .\302\276.^.\302\276
Conceivable 5^.\302\276.\302\276\302\276.
Of ^.\302\276\302\276^.
these
other
six possible
Section 4
are acceptable for ask the student to
exactly three instructors
orders,
Most
this.
Which orders are not on at least one test.)
a definition.
define
acceptable,
Exercises and why?
47
(Remember
a group
4.22 Table
23. The
b c
e
a
b c
a
a
?
b
b
c
c
\342\200\242
e
spelling and
of a group are taken verbatim, including too quickly and carelessly.Criticize them.
\"definitions\"
following
students who a.
a
*
a bit
wrote
G is a set of elements
A group
with
together
a binary
operation * such
that
from
punctuation,
the following
papers
of
conditions are
satisfied
associative
* is
e G such
There exists e
that
= x=
e*x=x*e For
G there
a e
every
(inverse) such
an a'
exists
that
a \342\226\240 a = a
b.
A
identity.
\342\226\240 a =
e
G such that on G is associative. identity element (e) in G.
is a set
group
The operation is an
there
a e
for every
c. A
is a
group
the
binary
an
inverse
an
identity
G, there
set
there
each
for
element)
operation such
is defined
operation exists
exists
element
over the binery associative under
a group
* is
operation
Binary
a' (inverse
a binary
with
set G is called
d. A
is an
exist an element
{e}such
operation
a there exists
element
every
an
element
25.
for a
a table
group
but
Mark
each of
not
a.
the
A
true or
may have
b.
Any
c.
In a group,
two
= e*a
=e
= a'*a
set {e,a, b} of
three
Wx.
following
group
on the
operation
binary
axiom
groups
each
false.
more
of three linear
eG
a' such that
a*a'
24. Give
for all a, b
that
a*e Fore
* such that
addition
than
one
identity
element.
elementsare isomorphic.
equation
has a solution.
=e elements
satisfying
axioms
and ,\302\276
,^
for a
Part I
48
Groupsand
d. The as in
Subgroups
proper attitude the text.
is to
a definition
toward
it so
memorize
that
a person gives for a group is correct provided that definition is also a group by the definition in the text.
Any definition
e.
person's
f. Any definition
that satisfies
g.
Every
h.
An
finite
i. The
set
can
group is a binary
he or she can conversely.
group
that
show
word
for
that
by
everything
is abelian.
elements
three
that is a
everything
provided
text and
the
in
form a * x * b = c always be considered a group.
of the
empty
Every
j.
of at most
is correct
the one
satisfies
definition
group
equation
gives for a group
a person
the
it word
can reproduce
you
solution
a unique
has
a group.
in
structure.
algebraic
Proof synopsis
We give an
a
(G, *) is that a
Assuming
that we said
Note
given as
an
notation
as is
synopsis. Here is a one-sentence synopsis
of a proof
example
a group
in
of the
proof
that
the
inverse
of an
element
unique.
* a'
= e and
left
\"the
left cancellation
the
apply
law\" and not a conversation during
cancellation
given
explanation
= e,
a * a\"
4.15.\" We always
\"Theorem at
law to the
lunch,
suppose
no reference
with
a * a'
equation
text
to
our
that
= a
*
a\".
was
synopsis
and as
numbering
little
practical.
26. Givea one-sentence synopsis
of the
27. Give
synopsis of the
at most
a two-sentence
proof of
left
the
proof
cancellation
law in Theorem
in Theorem
4.16
that
an
4.15. ax = b has a
equation
unique
in a group.
solution
Theory
28.
grasp of the is the
intuitive
our
From
isomorphism, then 4>{e) binary structures (S, *) in
also be
should
It
G',
that is, that
and
intuitively
4>{a)'=
G' is a group of isomorphic groups, it should be clear that if cp : G -\302\273 e' of G'. Recall that Theorem 3.14 gave a proof of this for isomorphic the case of groups. (5\", *'). Of course, this covers clear that if a and a' areinverse in G, then 0(a) and
identity
a careful
Give
proof of
a skeptic
for
this
who can't
see the
for all
forest
the
trees.
29. Show
that
if G
such that
a *a
30.
R*
Let
be the
a.
c. Is R* Explain
31. If * has 32.
= e.
set of all
group
real
is a exactly
the
and
except 0. Define binary for
an even
with
* on
operation on
R*
number of elements, then
by letting
a *b =
is a
^ e in
G
\\a\\b.
R*.
inverse for each element
* and a right
there
in R*.
binary
binary operation on a set S, an element one idempotent element. (You may
that every group (a * b) * (a * b).] Show
e
operation a group? of this exercise. significance
this
with
with identity
numbers
Show that * gives an associative Show that there is a left identity
b. d.
is a finite
G
with
identity
e and
S is an
x of use
such
that
for * if
idempotent
any theorems x *
x =
x *x
= x. Prove
proved so far
in the
e
G is abeliaa
for
all
x e
that
a group
text.) [Hint:
Consider
Section 5
33. Let
abelian
be an
G
34. Let
let
and
group
proof that (a
induction
mathematical
c *c
=
c\"
* b)\"
=
n factors
c,
\342\226\240\342\226\240 *\342\226\240 *c for
(a\*")
(b\")
be
all a,
for
Subgroups
49
and n e
Z+. Give a
c & G
where
G.
group with a finite number of elements. Show that for any a e G, there exists an n e Z+ e. See Exercise 33 for the meaning of a\". [Hint: Consider e, a, a2, a3,..., a'\", where m is the of elements in G, and use the cancellation laws.] G be a
a\" =
35. Show
that
if (a *
b)2 = a2 *
a and b
b2 for
in
a group
G, then
(a * b)'
= a'
= b* a.See
a *b
33 for
Exercise
the
such
that
number
meaning
of a2.
G be a group G be a group
36. Let 37.
Let
38. Prove
that
a set
and
G,
G. Show
a,b e
and let
that a
suppose
with
together
that
*b*
a binary
c =
e for
*b'
only if a
and
if
a, b, c e G.Show
operation * on
b * c
that
G satisfying
*b = b*
the left
a.
= e also. axioms 1, 2, and *a
3 given
on
page 43, is a group. 39.
that a nonempty
Prove
set G, together = b
a *x
is a
[Hint:
group.
Use Exercise
be a group. Consider 40. Let (G, \342\200\242)
with
and
y *
an associative
a = b have
for a,b 4>
e G. Show with
41. Let G be a
is an
that
5
for
and let of G
isomorphism
section
a'
in G
* on the
set G defined
a, b
all
G
such
that
e G,
38.] the
(G, *) is a a e G.]
binary
group
g be one fixed with
for
solutions
operation
a * b =
map
* on
operation
binary
and
element
that
b
\342\226\240 a
(G, *) is
of G.
by
Show
isomorphic
actually
that
the
map ig,
to (G,
such that
[Hint: \342\200\242).
ig(x)
Consider
= gxg'
for* e
the
G,
itself.
Subgroups
Notation and
Terminology
notation used in group theory. some conventional and terminology a not use a * to as a rule do denote binary operation different Algebraists special symbol the from with conventional additive or the usual addition and multiplication. stick They or and call notation even the addition multiplication, depending multiplicative operation on the symbol used. The symbol for addition is, of course, +, and usually multiplication of the is denoted by juxtaposition without a dot, if no confusion results. Thus in place notation a * b, we shall be using either a + b to be read \"the sum of a and b,\" or ab to be read \"the product of a and bV There that the is a sort of unwritten agreement to feel should be used commutative + operations. Algebraists symbol only designate when when see a b b a. For this uncomfortable + + reason, they developing our very / not or be in a situation where the we commutative, may theory general operation may shall always use multiplicative notation. element and 0 to denote an additive identity Algebraists frequently use the symbol a the 1 to even denote element, though they may not be multiplicative identity symbol also about if are numbers the 0 and 1. Of course, talking integers actually denoting they at the same time, so that confusion would result, symbols such as e or u are used as It
is time
to explain
Part I
50
5.1 Table
a
b
1
1
a
b
a
a b
b 1
1 a
Subgroups
of three elements might be onelike Table Thus a table for a group 5.1 such a group is commutative, look the table might like Table 5.2. In general we shall continue of a group. to use e to denote the identity element to denote the inverse of an element a in a group by a-1 in customary notation in additive notation. From now on, we shall be using these and by \342\200\224a
elements.
identity
1
b
Groupsand
or, since situations is
It
multiplicative
in place of the symbol a'. Let n be a positive integer.If a is an element
notations
of a group
G,
written
multiplicatively,
we denote the product aaa ... a for n factors a by a\". We let a0 be the identity element .a-1 for n factors a~1a~1a~1.. e, and denote the product by a~n. It is easy to see that = am+n form, n e Z, holds.Form, n e Z+, it is clear. our usual law of exponents, aman another We illustrate type of case by an example:
aaaaa = a (a a)aaaa = a a aaaa = (a a)aaa = eaaa = (ea)aa
= a
-V
5.2Table +
0
a
b
0
0
a
b
a
a b
b 0
0 a
b
In
additive
\342\226\240 \342\226\240 for + (\342\200\224a) + (\342\200\224a) + \342\226\240 + (\342\200\224a) (\342\200\224a)
Be careful:
element.
we prefer to
theory
n summands
appears
in an
Let us explain one more
term
that is
confusion
the n
If G is a
the order any set S, \\S\\
then
group,
Section 0 that,
for
and
Subsets
regarding
\\G\\
(ea)aaa
= a
n summands by let 0a be
and \342\200\224na,
by n
in Z,
is
not
G.
in
denote
na, the
identity
reason
One
notation, even if G is abelian, using multiplicative n as being in G in this notation na. No one ever
caused by when it
is the
= aaa
\342\226\240 \342\226\240 for +a + \342\226\240 +a
na, the number
notation
the
In
present group
misinterprets
5.3 Definition
a+a
we denote
notation,
= a
eaaaa
of
is the
exponent. used so often
it merits
the number of cardinality of S.) G is
a special
elementsin
G.
definition. (Recall
from
\342\226\2
Subgroups
have noticed that we sometimes have had groups contained within larger For example, the group Z under addition is contained within the group Q under in turn is contained in the group ffi under which addition. When we view the addition, as in it is that the contained the (Z, +} (ffi, +}, group group very important to notice operation + on integers n and m as elements of (Z, +} producesthe same element n + m as would result if you were to think of n and m as elementsin (E, +}. Thus we should not regard the group (Q+, \342\200\242} as contained in (ffi, +}, even though Q+ is contained in ffi as a set. In this instance, 2 \342\200\242 3 = 6 in (Q+, \342\200\242}, while 2 + 3 = 5 in (E, +}. We are requiring not only that the set of one be a subset of the set of the other, but also that the group
You may groups.
group operation on to
each
the
subset
pair from
ordered
this
be the subset
induced operation assigned by the
as is
that
assigns
group
the same
operation
on
the
element whole
set.
5.4 Definition
If a subset
H
of
a group
induced operation from H < G or G > H denote H
< G
but
H
/
G.
G is G
closed under
is itself that
the
binary
a group, then H is a subgroup of
operation
of G and if H with the of G. We shall let
is a subgroup G, and H < G or H
G>
H shall
mean
\342\226\2
5
Section
Thus (Z, Q+ C of G.
5.5 Definition
E. Every
If G is a All
group,
other
5.6
Example
turn
Let
E\"
other
of
5.7
Example
Q+ under
5.8
Example
The nth roots
Example
e is the
of G itself
is the
The subgroup
as sets,
though
element
identity
improper subgroup of G. the trivial subgroup
{e} is
\342\226\240
group of all n-componentrow of all of these vectors having
with real number entries. is entry in the first component
vectors 0 as
E\".
\342\226\262
is a
multiplication
under
+}, even
illustrations.
the additive
be
of (E,
and [e], where
a subgroup
G itself
the subgroup consisting are proper subgroups. are nontrivial. subgroups
to some
a subgroup
5.9
is not \342\200\242}
then
The subsetconsisting
numbers
(Q+,
as subgroups
G has
group
subgroups
of G. All We
+} < (E,+} but
51
Subgroups
of unity
in
proper subgroup
C form
a subgroup
of E+ under of
Un
\342\226\262
multiplication.
the group
C* of
nonzerocomplex \342\226\262
multiplication.
There are two different We describe them by
4-group, and the Z4 is isomorphic
notation to
of group structures group tables (Tables V comes from the
types their
the
=
group
\302\243/4
{1, i,
of order
5.10 and German
\342\200\224 \342\200\224 of
1,
i}
4 (see
Exercise 20 of
5.11).
The group
V
word Vier for four. fourth roots of unity
Section
4).
is the
Klein
The
group
under
multiplication.
of Z4 is {0,2}. Note that only nontrivial proper subgroup {0,3 } is not a subgroup For since is not closed under 3 +. + 3=2, and 2 \302\242 Z4, {0, 3}. {0, 3} example, V has three nontrivial proper subgroups, {e,a), and However, the group [e, c). Here {e, a, b] is not a subgroup, since {e, a, b] is not closed under the operation of V because ab = c, and c \302\242 A [e, a, b). The
of
[e,b],
5.10 Table Z4:
+
0
0 1
0 1
2 3
It is often
1
2
3
1
2
3
2
3
0
2
3
0
1
3
0
1
2
of
contains the
e a e a b c
b
c
e a b c a e c b b c e a c b a e of the
of a group. In such subgroups G to a that H is a H means running group G. Thus the larger group is placed nearer the top of the diagram. Figure 5.12 for the groups Z4 and V of Example 5.9. subgroup diagrams useful
a diagram, a line subgroup
5.11Table
to draw a
subgroup diagram
downward
from
a group
Part I
52
and
Groups
Subgroups
Note that if H < have a unique solution,
be viewedas one e of G. A element G, shows
H and
G and
a
the
{0,2}
applied in
H.
G is
to
the
equation
also the
\342\200\224 a must
can also must also be the identity ax = e, viewed in both
But
solution
unique
ax
the equation
4.16,
element of
this
that
then argument inverse a~1 of a
similar
that
Theorem
by
the identity
we see
G, and
in
then
e H,
namely
this
equation
inverse ofa in
the
H.
subgroup
le,a)
{0} (b)
(a) 5.12
5.13
Example
Let F
be the
subset
of F
the sum of
(a) Subgroup
Figure
real-valued
of all
group
consisting of those
x is continuous
and
that
is continuous,
additive
is the
functions
functions
functions
continuous
diagram for
Z4. (b)
E under with domain are continuous is a subgroup
the
element,
identity
Subgroup diagram for
function
f
and if
continuous.
f is
where continuous,
V.
The
addition.
f(x)
of
for
F,
= 0 for all then
\342\200\224 is
f
\342\226\26
to have routine for determining whether a subset of a group G It is convenient steps 5.13 is a subgroup of G. Example indicates such a routine, and in the next theorem, we are available, demonstrate its validity. While more compactcriteria involving carefully this more transparent theorem for a first course. only one condition, we prefer 5.14
Theorem
Proof
A subset H
Example
closed under
1.
H is
2.
the
3.
for all a
identity
the
e H
it
binary
is true
of
subgroup
e of
element
H < G then of definition a subgroup and
G if
operation
G is in
that a^1
and
only
if
of G,
H,
e H
also.
1, 2, and 3 must hold follows at once from the the remarks preceding Example 5.13. G such that Conditions 1,2, and 3 hold. H is a subset of a group suppose Conversely, have 2 we at once that is Also is 3. It to satisfied. remains check By by .^ 5\302\276 satisfied the associative axiom, 8\302\276. for all a, b, c e H it is true that (ab)c = a(bc) in But surely in G, where the associative as an equation law holds. H, for we may actually view this The
that if
fact
Hence H 5.15
G is a
a group
of
Let F the
< G.
be as in
differentiable
constant
differentiable.
Conditions
from
\342\231\24
Example
is a
subgroup function 0 is
of those The subset of F consisting functions that are of F, for the sum of differentiable functions is differentiable, \342\200\224 and if f is differentiable, then differentiable, f is
5.13.
\342\226\262
5
Section
5.16 Example
Recall
from
det(A)
called
linear
that every
algebra
and
its determinant,
square matrix A is
that
invertible
A has
associated
if and
if det(
only
53
Subgroups
with it a number A) / 0. If A and B
\342\200\242 \342\200\224 are square matrices of the same size, thenit can be shown that det(AB) det(A) det(B). x Let G be the multiplicative of n n matrices with entries in all invertible C and group let T be the subset of G consisting of those matrices with determinant 1. The equation = det(A) \342\200\242 shows that T is closed under matrix Recall det(AB) det(B) multiplication. \342\200\242 = matrix From the that the identity has determinant 1. /\342\200\236 det(A) det(A_1) equation = = = = if then 1. Theorem 5.14 we see that det(7\342\200\236) 1, 1, det(AA_1) det(A) det(A_1) \342\226\262 that T is a subgroup of G. then shows
Subgroups
Cyclic
have to be if it contains 3. It would have which is 6. Then it has to contain 6+3, which is 9. 9 and the inverse of 6 is 6.It is easily checked 3. of Z]2, and it is the smallest subgroup containing that H = {0,3, 6, 9} is a subgroup for in a situation. As we remarked Let us imitate this before, reasoning general and let notation. Let G be a group a general use multiplicative argument we always of G containing a must, contain the Theorem result a e G. A subgroup 5.14, an, by of n. These of computing itself for n factors for a and every positive integer products It is possible, of a do give a set closedunder integral powers multiplication. positive Of a a must that the inverse of a is not in this set. course, however, containing subgroup also contain a~m for all m e Z+. It must contain the a-1, and, in general, it must contain the element a must a subgroup identity element e = a0. Summarizing, of G containing contain all elementsan {or nafor additive groups) for all n e Z. That is, a subgroup a must contain {a\" \\n e Z}. Observe that these powers an of a need not be containing in the group distinct. For example, V of Example 5.9,
Let us to
see how
H of Z12 element 0 and 3 Note that the inverse of 3 is large
a subgroup
the identity
contain
a = e, We have 5.17 Theorem
almost provedthe
Let G be a
group
and let
a
a =a, next
a e
subgroup
subgroup
of containing
G and
is
\342\200\224 e,
a~
= a,
and
so on.
theorem.
G. Then ff
is a
would
+ 3,
the
a contains
smallest'
=
K|neZ) subgroup
of G
that
contains
a,
that is,
every
H.
1 We may find occasion to distinguish between to subsets of a set S the terms minimal and smallest as applied A subset H of S is minimal that have some property. with respect to the property if H has the property, and no subset K C H,K ^ H, has the property. If H has the property and H c K for every subset K with the property, then H is the smallest subset with the property. There may be many minimal subsets, but there can subset. To illustrate, {e,a},{e,b], and {e, c} are all minimal nontrivial of the be only one smallest subgroups V contains no smallest nontrivial V. (See Fig. 5.12.)However, group subgroup.
Part
54
I
Proof
5.18Definition
Groups and Subgroups
conditions given in Theorem 5.14 for a subset of a group to give a = ar+s for 5 we that the in of two aras see G elements Since g Z, r, product subgroup. the group of H is again in H. Thus H is closed under operation of G. Also a0 = e, so = e. Hence all the conditions a~rar are satisfied, e e H, and for ar e H, a~r e H and and H < G. that any subgroup of showed Our prior to the statement of the theorem arguments of G containing \342\231\ contain a. G containing a must H, so H is the smallest subgroup G be a
Let in
the three
check
We
Then the
e G.
let a
and
group
5.17, is called
Theorem
the
of G
subgroup
cyclic
Z} of
G, characterized generated by a, and denoted
{a\" \\n e
subgroup
\342\226
by(a>.
5.19
5.20
Definition
Example
G and is a generator for G generates An element a of a group a in G that generates if some element G. G is cyclic there is Let Z4
groups of
be the
7
and
generators, that
Example5.9.Then
5.21
Example
The group
Z
n is
modulo
the
under
cyclic, for trivial
A
group
\342\226
and
both
1 and
only n
subgroups of two
elements.
Example
\342\226\
are generators for this group. Both 1 and \342\200\2241 for n e Z+, the group Z\342\200\236 under addition 1 are generators, then both 1 and n \342\200\224 but there may be
>
Also,
generators.
1,
\342\226\
the
Consider
group
Z under
3. -3
0, In other words,
negative,
5.23
Example
let
addition.
Let us
find
Here
(3).
the notation is additive,
and
contain
(3) must
shall
Of
is a cyclic
addition
cyclic. If
are proper
(c)
element.
of one
others. 5.22
3 are
(3)=24-
{a}, {b},and
subgroup
they are the
and
group,
is not
{e) is
course,
is cyclic
G.
is,
(1) =
However, V
Z4
G if {a}=
and \302\253Z be
the
zero.
3+
3=6, 3 + 3 + -3 = \342\200\2246,-3+-
+ 3
= 9,
and
so on,
= \342\200\2249,and \342\200\2243 H\342\200\2243
so on.
of 3, positive, by 3 consists of all multiples 3Z as well In as a similar by (3). way, we Z. Note that 6Z < 3Z. \342\226\2
generated subgroup We denote this subgroup
cyclic
the cyclic
subgroup
of (\302\253}
For each positive integer n, let U\342\200\236 be the multiplicative group of the nth roots of unity in C. These elements of U\342\200\236 can be represented geometrically by equally spaced points on a circle about the origin, as illustrated in Fig. 5.24. The heavy point the represents number
i;
=
2tt cos
2tc
\\-
n
i sin \342\200\224.
n
Section
The
geometric
Section
1, shows
around
the
landing cyclic group,
group U
of the
as
that
circle,
is a
multiplication
in of multiplication of complexnumbers, explained to powers, it works its way counterclockwise is raised \302\243 on each of the elements of Un in turn. Thus Un under
interpretation
at once
55
Exercises
5
is a \302\243
and
generator.
of all complexnumbers
The
z, where
5.24
is the U\342\200\236
group
=
\\z\\
1, under
cyclic subgroup multiplication.
(\302\243) \342\226\262
Figure
5
EXERCISES
Computations C of
6, determine
1 through
Exercises
In
complex
CR of
set
5. The set7rQ 7.
subset of
given
Which
pure
of the
numbers
imaginary
of rational
sets in
the
is a subgroup
numbers
complex
of the
group
3. 7Z
2. Q+
1. M
4. The
the
whether
addition.
under
numbers
0
including
6. Theset
multiples of n 1 through
Exercises
6 are
of the
subgroups
[it\"
group C*
]\302\253eZ}
of nonzero complexnumbers
under
multiplication?
In Exercises 8 through a subgroup 8.
The n
13, determine
whether the
x
n
with determinant
matrices
9. Thediagonal
n matrices
n x
n x
with
no
on the
with no
n
matrices
with real
number
entries
is
11.
The n x
n
matrices
with determinant
\342\200\224 1
12.
The n x
n
matrices
with determinant
\342\200\224 1 or 1
set of all
the transpose has
n x
of
A,
n
matrices
diagonal
zeros on
The upper-triangular
operation
invertible n x
2 zeros
10.
13. The
set of
given
of GL(n, M).
n matrices A such that (AT)A = is the matrix whose j'th column
the property
(AB)T =
(BT)(AT).]
the
diagonal
7,,. [Thesematrices is the
/th
row
are of A
called
for 1 <
orthogonal.
j < n,
and
Recall that AT, transpose
that the
Part I
56
Let F be the
set of
have a nonzero
that
Subgroups
all real-valued functions value at every point
domain R and let F In Exercises 14 through of the group F under
with in R.
operation is (a) a subgroup
induced
the
with
and
Groups
be the
consisting of those functions
of F
subset
whether the
19, determine
of
(b) a subgroup
addition,
subset
given
the
group
of F
F under
multiplication.
14. The subset
F
The subset
15.
16. The subset
of all
/
of all
/ e F
17. The subset of all
Nine
groups
given
/(1)
= 0
that
/(1)
= 1
that
/(0)
= 1
/(0)
= -1
that
in F.
below. Give a completelist G\\,
groups
subgroup relations,
of all
of the
form
G, <
Gj,
that
exist
by
the
given
\342\200\242 \342\200\242 \342\200\242, Gg.
G?,
addition
12Z under
=
that
functions
constant
are given
between these G\\ = Z under G2
e F such
/
19. The subset of all
20.
such
e F such
/
of all
18. The subset
e F such
addition
G3 = Q+ under multiplication G4 = R under addition
= R+ G6 = {jr\"
G5
under
multiplication
under multiplication
e Z} I \302\253
G7 = 3Z under addition Gg = the set of all integral Gg = {6\" I n e Z} under at least 5
Write
21.
a. 25Z under
elements of each
{tt\"
I
under
Z} under
n e
22 through
In Exercises
addition
of the
cyclic groups.
following
addition
b. {(1)\" I n e Z} c.
of 6 under
multiples multiplication
multiplication
multiplication
25, describeall
in the
elements
the
0
22.
23.
-1
26.
of the
Which
1
1
0
1
Gg
27
3
0
0
2
(Z,
cyclic? For each cyclic +) G2 = (Q, +) G3
{6\"
I
= {a
+
=
G5 =
In Exercises
24.
following groups are G,
35,
through
find the
n e
Z} under
of
GL(2,
R) generated
a, b
fcV21
order of the
The subgroup
of Z4
generated
e Z} under cyclic
by 3
28. The subgroup
of
V generated
29. Thesubgroup
of
U(, generated
30. The
subgroup
of Us
generated
by
subgroup
of
generated
by cos
31. The
U%
by
Table 5.11)
c (see
by cos ^f cos
+
^f- + =y
sin
^f
i sin
^-
i
+ i
sin
0 -2\"
25. group, list =
(Q+,
-2 all
the
G4 \342\200\242)
0 generators = (6Z,
of the
group.
+)
multiplication
element.
27.
cyclic subgroup
matrix.
2x2
=y
addition
subgroup
of the
given
group
generated
by the
indicated
Section 5
32.
The
33. The
subgroup
of
subgroup
of the
by cos ^f
generated
U%
multiplicative
+
sin
i
of the
subgroup
multiplicative
0
1
o\"
0
0
0
1
G of
group
of the
multiplicative
0
\"0
0
0
0
0
10
10
by
0 0
matrices generatedby
invertible 4x4 l\"
0 0 0
1
_0
35. The subgroup
matrices generated
4x4
0
0 10 The
57
^f
group G of invertible
10
34.
Exercises
0^
group G of invertible
matrices generated
4x4
0
1
o
o\"
0
0
0
1
0
0
10
by
10 0 0
36.a.
give the group X(, of 6 elements. (0), (1), (2), (3),(4),and (5) of the group Z6 given in part (a). subgroups c. Which elements are generators for the group X(, of part (a)? of X(,. (We will see later that these are all d. Give the subgroup for the diagram part (b) subgroups Table
Complete
b. Compute
5.25 to
the
the
subgroups
ofZ6.)
5.25 Table
+ 0 1 2
0
1
2
3
4
5
0
1
2
3
4
5
1
2
3
4
5
0
2
3
3
4
4
5
5
Concepts In
37.
it is
that
in a form
of a group inverse of eachof its A subgroup
38. A 39.
38, correct the
37 and
Exercises
needed, so
group
G is
Mark each
G is a
There
subset H
of G that
the identity
contains
to the
element e of
elements. only
following
a. Theassociative b.
reference
without
if
correction
is
also
contains
the
text,
acceptablefor publication.
cyclic if and
of the
italicized term
of the
definition
may be a
if there
true or law group
exists a
e G such
that
G =
false.
holds
in every group. the cancellation
in which
law
fails.
{an
|
n \342\202\254 Z}.
G and
c.
Every
group
is a subgroup
d.
Every
group
has
e.
In every
cyclic
group,
A
group
has a unique
f.
cyclic
Every
h.
A subgroup Z4 is a
of an
means
by
as a
under addition subset of a group.
is a
subgroup under
is also
defined
a group
under
multiplication.
group.
example
group G
in some
solutions
cyclic
generator.
is a group
that
may be
subgroups.
is a generator.
element
every
subset of every
Every
j.
of itself. exactly two improper
set of numbers
g.
i.
40. Show
Groups and Subgroups
I
Part
58
group
with
possible for
it is
that
the
induced
the
operation.
x1 =
equation
quadratic
e
than two
more
have
to
e.
identity
Theory
41 and 42, let 0 : G ^ G' be to convincea proof skeptic of the intuitively In Exercises
of 41. If J? is a subgroup into subgroups. subgroups
G, then
e H}
| h {0(\302\276)
(G', *').
a group
with
out
Write
a
is a
of
subgroup
G'. That
is,
an
carries
isomorphism
If G
of an
flaw in the
the
from 1 and (This
3, for let
46. Prove that 47. Prove that satisfying
48. Repeat
a cyclic
following argument: a e H. Then a~]
group
with
2 of
\"Condition
e H by
3,
and
k
\342\202\254 K]
by
Theorem 5.14 is redundant, 1, aa~] = e is an element
the
n >
47 for the
1 in
49. Show that if 50. Let a nonempty
a e
subgroup of
G.
G be a
an abelian
G,
where
group and let
written
G with
G is a
finite group
fixed
H for \342\202\254
2.\"
all a, b
H. \342\202\254
e, then
element
identity
elements
all
x of G
G.
set H
with
of all
x of the
solutions
element
closed under
of G. Show = {x e
G
there exists
e, then
identity
G be
group
Ha subgroup
proving
2 elements.
most
with
be derived
it can
since
of H,
equation
x\"
=
e for a fixed
identity e.
group
a be one
H of
of the
situation
H of a
at
multiplicatively,
a subgroup
general
subset
finite
can have
one generator
only
is an abelian group, x1 = e form equation
if G
Exercise
integer
is a
h \342\202\254 H and
that a nonempty subset H of a group G is a subgroup of G if and only if ab^1 is one of the more compact criteria referred to prior to Theorem 5.14)
Show
51. Let
I
of G.
is a subgroup
44. Find
G, then
abelian group {hk
45.
(G, *)
statement.
clear
0[ff] =
is cyclic, then G' is cyclic. 43. Show that if H and K are subgroups
42.
of a group
an isomorphism
I
xa
the
binary
n
e Z+
operation
such of G.
that
a\" =
Show
that
e. H
is a
that
= ax}
of G.
Exercise 51, let S be any subset of a group G. of G. H$ = {x e G | xs = sx for all 5 e S} is a subgroup reference to part (a), the subgroup HG is the center of G. Show that
52. Generalizing a. Show that b. In 53.
Let
H be a
equivalence
of
subgroup relation
on
G.
a group
G. For a, b e
G,
let
a ~
b
if
and
only
HG
is an
if ab~]
abelian e H.
group.
Show
that
~ is an
6
Section
54.
H
For sets
and
K,
intersection H n K
we define the
that
if H
< G and
every
cyclic
group
and let
then H
< G,
K
59
by
= {i|xeflandie^).
flni[ Show
Cyclic Groups
(Remember: <
< G.
K
Pi
denotes\"is
a subgroup
of,\"
\"is a
not
subset
of\
55. Prove that 56. LetGbe a
group is abelian. =
Gn
{gn | g
e G}.Under
what
about G can we
hypothesis
that
show
is a G\342\200\236
subgroup
ofG?
57.
Show
that
section
a group
6
no
with
subgroups is cyclic.
nontrivial
proper
Cyclic Groups the
Recall
facts and
following
= {an\\n
H
(Theorem 5.17). This group a given group G and an element
is a subgroup by
a.
G=
group
|
{an
and
a e
G,
then
&Z}
is the cyclic subgroup a in G, if
of G
Also,
5. If G is a
Section
from
notations
n e
(a) of G generated
Z},
the group G = {a)is cyclic.We introduce one new bit of If of a G. the element group cyclic subgroup (a) of G is finite, terminology. of this then we say that a the order of a is the order Otherwise, \\(a) | cyclic subgroup. that if a e G is of finite order m, then m is of infinite order. We will see in this section is the smallest integer such that am = e. positive and all subgroups The first goal of this section is to describe all cyclic groups of an will This is not idle exercise. We see later that serve cyclic groups. cyclic groups as building blocks for all sufficiently small abelian groups, in particular, for all finite are abelian fundamental to the of Cyclic groups understanding groups. groups.
then a is a generator of Let
Properties of CyclicGroups
Elementary
6.1Theorem
with
start
We
Proof Let G be a
a demonstration
cyclic
that cyclic groups
group
and let
a be a G
If gi
and
g2
and g2 =
are any
G is We
(a) =
=
two elements
of G
generator {an
of G, there
|
n e
exist
so
that
Z}. integers
r and 5 such
that
gi
= ar
as. Then glg2
so
are abelian.
abelian.
group is
cyclic
Every
G and
a be an
= aras
= ar+s
= as+r = asar = g2gu
abelian. shall
\342\231\246
continue
groups, even though
they
to use
multiplicative
are abelian.
notation
for
our general work on
cyclic
Part I
60
Groupsand
Subgroups
The division
for
the
algorithm that groups.
is a seemingly
follows
trivial,
but
tool
fundamental
very
of cyclic
study
> 0, q
n
> 0
-\\
\342\200\224|-
n < 0, q
< 0
A\342\200\224
(q + \\)m
n
2m
\342\200\224|-
(q +
qm
2m
\\)m
6.2 Figure
6.3 Division
Algorithm
for
is a positive integer and n
If m
Z
and r such
=
n
Proof
integer, then there
is any
exist unique
q
integers
that
+ r
mq
r <
0 <
and
6.2. On the real x-axis of Fig. diagrammatic explanation, using m the of and the of n. Now n falls either multiples position analytic geometry, on a multiple qm of m and r can be taken as 0, or n falls between two multiples of m. If the latter is the case, let qm be the first multiple of m to the left of n. Then r is as in Fig. 6.2. Note that shown 0 < r < m. Uniquenessof q and r follows since if n is not a multiple of m so that we can take r = 0, then there is a unique multiple qm of m to the in Fig. 6.2. left of n and at distance less than m from n, as illustrated \342\231\ We
an intuitive
give
mark off
6.4 Example
Find
of
the notation
In
the
when n is
algorithm,
we regard q
divided by
m.
remainder r
q and
quotient
division
the
remainder
nonnegative
by 7
is divided
38
when
as the
and r
quotient
according to the
as
the
division
algorithm.
Solution
The positive multiples a nonnegative remainder
of 7
are 7, 14, 21, 28, 35, 42,---. Choosing than 7, we write = 35
38 so the
6.5
Example
quotient
Find the
is q = 5 q and
quotient
the
multiple
to leave
less
r
remainder
7(5) + 3 r = 3.
is
remainder
the
and
+ 3=
is divided \342\200\22438
when
\342\226\
by 7
according to the
division
algorithm.
Solution
The negative multiples of 7 are -7, -14, -21, to leave a nonnegative remainder less than tiple
-38 = -42 + 4 so the We
is also
-6
quotient is q will
cyclic.
use the Think
and
division for
the
remainder
algorithm
a moment
-28,-35,-42, 7, we
=
7(-6)
is r
to show
what we will
. Choosing the
mul-
write + 4
= 4. that a subgroup Hofa to do to prove this.
have
cyclicgroup We
will have
G
to
Section 6
61
Cyclic Groups
of a cyclic group we have proved little since about cyclic groups yet. have to use the fact that G has a generating a. We must then element = in terms of this some c am for H in order to show that exhibit, a, generator generator H is cyclic. There is really only m of a to try. Can you one natural choice for the power guess what it is before you read the proof of the theorem? the
use
definition is, we will
That
6.6 Theorem
Proof
A
Let G be a cyclic H = (e) is cyclic. We claim that
by a
generated
group
in Z+ such
integer
group is cyclic.
of a cyclic
subgroup
If
H ^
that
am
c =
am generates
{e},
then
e H
H;
that
that every b e H is a some n. Find g and r such
show
must
b =
a\" for
n = in
then
= {e}, the
smallest
is,
(am) =
an
=
(c).
of c.
power
Since b
&
H <
and
H
G, we have
that
mq + r
0
for
Then
with the division algorithm.
accord
H be a subgroup of G. If H for some n \342\202\254 Let m be Z+.
e H.
H = We
let
and
a\"
= {am)qar,
amq+r
so ar = Now
since
a\"
H.am
&
& H,
and H is a
(am)\"V e
Since m
was
the
= 0. Thus
have r
fc
noted
As
integer
ft,
the
positive integer such and ^m
The
This corollary two
positive
subgroup which
we
of Z
under
addition
gives
us an
integers of the group may
(am)i
and
ar e
is, am
that
e H
a\"
are
in H.
Thus
ff. and 0
< r < m,
we
must
=cq, \342\231\246
in Examples 5.21 and set nL of all multiples
subgroups
= a\302\273
(am)~q
c.
subgroup generated by ft. Theorem of Z under We addition. subgroups
6.7Corollary
that
smallest n =
power of
is a
both
group,
ff;
b= so
{amyqan.
r and
addition is cyclic and for a positive the cyclic addition, subgroup of Z under 6.6 shows that these cyclic subgroups are the only state this as a corollary.
of ft
is a
the groups
are precisely
elegant
way
to
s. Exercise45 shows
Z under addition. be positive.
choose to
Z under
5.22,
Thus
define that
H must
ftZ
under
the greatest H = {nr +
be cyclicand
addition
for n e
Z.
common divisor of ms | n,m e Z} is a have
a generator
d,
62
Part I
6.8 Definition
and
Groups
Let r
Subgroups
5 be
and
integers. The positive
two positive
H
under d =
is the
addition
+ ms
= {nr
d of
generator
e Z}
n,m
\\
the cyclic group
common divisor (abbreviated gcd) of
greatest
r and
s. We
\342\22
from
Note
and 5
= Or +
the definition that d is a divisor of both in H. Since d e H, we can write
r and
5 since
both r =
\\r
Solution
the must d in
6.8. gcd of
the
Find
Os
nr + ms
n and m. We see that for someintegers every integer dividing both r and 5 divides d and hence must be a divisor of d also.Thus right-hand side of the equation, accounts for the name given to both r and s; this be the largest number dividing
Definition
+
Is are
d =
6.9 Example
write
gcd(r, s).
42
and
72.
The positive divisors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. The positive divisors of 72 is 6. Note are 1, 2, 3, 4, 6, 8, 9, 12, 36, and 72. The greatest common divisor 24, is an algorithm for expressing the greatest that 6 = (3)(72)+ There common divisor d of r and 5 in the form d = nr + ms, but we will not need to make use of it
18, (-5)(42).
here.
\342\22
Two positive integers are relatively prime. Note
of subgroups of cyclic
If r
this.
If r
prime
by
m, we
prime if
need to
andifr
divides sm,
side
Structure
We
can
now
describe
then
then
1. For example,12and 25 In our discussion
common.
following:
r must
divide m.
(1)
we may write a, b
e Z.
obtain
+ bsm.
divides both arm and bsm sincer divides of this equation, so r must divide m.
The
the
some
for
m = arm Nowr
factors in
know
prime,
ar + bs
gcd is
their
no prime
we will
and 5 are relatively 1=
Multiplying
have
they
groups,
ands arerelatively
Let's prove
are relatively that
of Cyclic
sm.
Thus
r is a divisor of the
Groups
all cyclic groups,
up
to an
isomorphism.
right-hand
Section 6
6.10
Theorem
63
CyclicGroups
G be a cyclic group with generator a. If the order of Gis infinite, then G is isomorphic to (Z, +}. If G has finite order ft, then G is isomorphic to (Z\342\200\236, +\342\200\236). Let
Case
Proof
I
For all positiveintegers distinct
exponents
k can
a* and say
aA =
Suppose that
e.
/
m,am and
h
to our
contrary
Case
II
we claim
>
h
that
no
ak of
and
two
G.
k. Then
a h-k =
e,
Case I assumption. Hence every elementof G can be The map \302\242) for a unique meZ. : G \342\200\224> Z given by one to one, and onto Z. Also, well defined,
= 4>{ai+i)= i +
0(a!V) so the
case
this
give equal elementsah
= a\342\200\236h-k a expressed as am (j>(a') = i is thus
In
homomorphism property
j=
is satisfied
+ \302\242(0.1)
0(aJ'),
and 0 is
an
isomorphism.
efor some positive integer m. Let n be the smallest positive integer such that a\" = e.lf s e Z and 5 = nq + r for 0 < r < ft, then
am
n. Thus
the
elements
e, a, a
a0 =
and comprise all elements of G. The \342\200\242 \342\200\242 n - 1 is thus \342\200\242, by i/f(a') = i for *' = 0, 1, 2, = we and Because a\" see that a1 a' onto e, Z\342\200\236. distinct
all
are
given
,a ,
one, k = i +n j-
\\j/
well
: G \342\200\224>\342\200\242 Z\342\200\236
one to
defined,
= ak where
Thus
= i
f(alaj)
so the
map
homomorphism
property
= f(a')
+\342\200\236 j
is satisfied
+\342\200\236 f(aJ),
and
\\j/
is an
isomorphism.
n- 1
6.11Figure 6.13 Example
Motivated by
our
from the start
from
with Un,
work
of a cyclic group element ah is located an~'
of ordern h
of these
bottom where e = ah and
6.12
go k
additional
a0
it
to visualize the elements e= a0, a1, distributed evenly on a circle (seeFig.
is nice
as being
Figure
a2,
\342\226\240 \342\226\240 \342\226\240,
6.11).
The
equal units counterclockwise along the circle, measured is located. To multiply ah and ak diagrammatically, we units around counterclockwise. To see arithmetically
Part
64
I
Groups and
we end
where
The
Subgroups
us all the
takes
nq
q and
find
up,
r such
h +
k =
way
around
that
+
nq
the circle
Figure 6.12 is essentially the same on the generator. The operation exponents
of Finite
Subgroups We
have
completed
6.7gives the
6.14Theorem
us complete
theorem
basic
Let G be a
group
Proof
cyclic
but
these
on
then
we
wind up
is
exponents
A
ar.
at
labeled
the points
with
and
turn to
of infinite
generators of subgroups for the
with n elements and generated H of G containing subgroup
Also, (as) =
divisor ofnands.
common
6.11
as Fig.
our description of cyclicgroups information about subgroups
Then b generatesa greatest
q times, and
< n.
the
with
n.
modulo
addition
Cyclic Groups
regarding
cyclic
0 < r
for
r
their subgroups. Corollary cyclic groups. Let us give
finite
cyclic
by a.
Let b e
n/d
elements,
groups.
G and
where n)
(ar)ifandonlyifgcd(5,
let
b =
d is
as. the
= gcd(t,n).
from Theorem That b generates a cyclicsubgroup H of G is known 5.17. We need show II of that H of Case Theorem we has elements. the 6.10, n/d Following argument only see that H has as many elements as the smallest positive powerm of b that gives the if (as)m = e, or if and if n divides Now b = as, and bm = e if and only only identity. ms. What is the smallest positive integer m such that n divides ms? Let d be the gcd of n and s. Then there exists u and v such that integers d =
Sinced divides
s, we may
n and
both
where
both
find
and
n/d
prime, for smallest
the
s/d any
positive
vs.
write
1 =
relatively
un +
+ v(s/d)
u(n/d)
are integers. integer
dividing
m such
that
This last
equation
shows
both of them must
that n/d
also divide
and
s/d
1. We
are
wish to
.
m(s/d)
is an integer.
(n/d)
From smallest
division
boxed
the
m is
such
n/d. Thus the
concludethat
(1), we
property
n/d
must
divide
m, so
the
is n/d. moment Z\342\200\236 as a model for a cyclic group of H
order
of order n, we see that if d is had n/d elements, and contains all the subgroup (d) of Z\342\200\236 m less than n such that gcd(m, n) = d. Thus there is only one subgroup positive integers of Z\342\200\236 of order the preceding this shows at once that if n/d. Taken with paragraph, a is a generator of the cyclic group G, then (as) = (a'} if and only if gcd(s, n) =
the
for
Taking
a divisor
of n,
then
the cyclic
gcd(f.n). 6.15
Example
\342\231\2
For an example the
greatest
elements,
using
comm common
additive
divisor
consider
notation,
of 3
and
12 is
Z^, with the = 3 3 \342\200\242 1 generates 3,
namely (3)
=
{0,3,6,9}.
a generator a subgroup
= 1. Since of y = 4
Section
Sincethe
12 is 4, 8 generatesa
of 8 and
gcd
= {0,4,
(8}
Since the
of the
whole
The following
6.16Corollary 6.17 Example
If a
elements
Let us
find
cyclic. By
= 3 elements, namely,
of
y
= 12 elements;that
8}.
subgroup
of
all subgroups
5 is
is,
from Theorem 6.14.
immediately
cyclic group G of order n, then form ar, where r is relatively prime
\"L\\%
and
give
their subgroup
1, 5,
elements
7, 11, 13, and
other
the
generators
of G
to n.
diagram.
All
are
subgroups
all generators
17 are
of 7L\\%.
2,
(2) =
is
-j
A
follows
Corollary 6.16, the
with
Starting
of
of a finite
of the
65
Cyclic Groups
Z^.
group
corollary
is a generator
are the
and 5 is 1, 5 generatesa
of 12
gcd
a generator
subgroup
6
{0,2,4,6,8, 10,12,14,16}.
of the form h2, where h is relatively has as generatorselements prime = and h2 so to 9, namely, 1, 2, 4, 5, 7, 8, 2,4, 8, 10,14,and 16. The element 6 of of this subgroup. (2) generates {0, 6, 12}, and 12 alsois a generator We have thus far found all subgroups generated by 0, 1, 2, 4, 5, 6, 7, 8, 10,11,12, 15 to consider. and 17. This leaves just 3, 9, and of
order
9 and
h =
13,14,16,
12,15},
(3) ={0,3,6,9, and
15 also
generates
this
group
of order
6, since
15 =
5 \342\200\242 3, and
the gcd
of 5 and
6 is
1.
Finally,
(9}
The subgroup
diagram
for these
= {0,
9}.
subgroups of Zig (1)
6.18 Figure
=
is given
in Fig.
6.18.
zlg
Subgroup
diagram
for Z18.
we are afraid we wrote it out in such detail This example is straightforward; some The exercises look may give practice along these lines. complicated.
that
it \342\226\262
Groups and
I
Part
66
Subgroups
6
\342\226\240 EXERCISES
Computations 1 through
Exercises
In
4,
the
find
and remainder, according to
quotient
division
the
when n is
algorithm,
divided
by m.
1. n =
42, m
3. n =
-50,
m
In Exercises 5
5. 32 and
=
9 =
8
6. 48
24
In Exercises 8 through
the number of
11, find
13.
In
is an
17. The cyclic
14.
Z6
21,
17 through
Exercises
generated
by
25
of Z42
generated
by
30
the group
subgroup
19.
The cyclic
subgroup
(;') of
subgroup
of the
subgroup
of the
20. The cyclic 21. Thecyclic
22 through
In Exercises
all subgroups
indicated
of the given
25 through 26.
Z6
of the
28.
27. Z12
Z8
the
an automorphism?]
under multiplication
by
(1 +
by
1 + ;'
and draw 24.
orders of subgroups
find all
29,
16, find
cyclic group.
numbers
group,
23. Z36
In Exercises
under
12 through
Exercises
16. Z12
C* of nonzero complex C* ofExercise 19 group generated C* of Exercise 19 group generated
24, find
22. Z,2 25.
of the group. In
the
in
order.
given
11. 60
image of a generator
of elements
420
and
360
the
having
15. Z
of Z30
The cyclic
two integers.
cyclic group
automorphism
subgroup
18.
8
7. of a
generators
Z8
number
the
find
=
9
and 88
group. be the must
use ofExercise44.What
12. Z2
50, m
=
m
10. 12
of a group with itself An isomorphism number of automorphisms of the given
[Hint: Make
4. n = of the
9. 8
8. 5
-42,
commondivisor
the greatest
7, find
through
2. n =
i)/V2
the subgroup
diagram
for the
subgroups.
Z8
given group. 29.
Z20
Z,7
Concepts
30 and
In Exercises
needed, so that 30.
An element
31. Thegreatest 32.
Mark
31,
a of a of the
of two positive
divisor
following
cyclic group abelian
d.
Every
addition
f. Every
group
without reference
least
and
integers
only
if a\" =
to
the
text, if
correction is
is the
e.
largest positive
is abelian. is cyclic.
group
is a cyclic
group. cyclic group generates abelian group of every <4 is cyclic.
of every
element
e. Thereis at
term
italicized
or false.
true
a. Every under
order n e Z+ if
G has
b. Every
c. Q
the
acceptablefor publication.
group
common
each
the definition of
correct
in a form
it is
one
of order
the group. finite
order
>0.
integer
that
divides
both of them.
Exercises
Section6
g.
All
H and
i. If
33 through
Exercises
are
^f
prime
numbers.
G n G' is a
then
37,
either
of a
D K
H
isa
group.
two distinct generators.
at least
an example
give
group.
G, then
of a group
subgroups
group of order >2 has
cyclic
Every
j. In
and G' are groups,
If G
h.
of Z20 are
generators
67
described, or
the property
with
group
explain
why
no
roots
of
example exists.
33.
A
34.
An
finite
35. A
that is not
group
cyclic
not cyclic group group having only one generator that is
infinite
cyclic
36.
An infinite
37.
A
finite
cyclic
four generators
having
group
group having four
cyclic
generators
of all nth roots of unity of the cyclic multiplicative The generators U\342\200\236 group of unity for the 41, find the primitive nth roots unity. In Exercises 38 through
38.
n
are the primitive value of n.
in C given
nth
= 4
39. n =
6
40.
8
n =
41. n =
12
ProofSynopsis
42. Give
synopsis of
a one-sentence
43. Give at
proof
6.1.
of Theorem
synopsis of the
a three-sentence
most
the
of Theorem
proof
6.6.
Theory
44. Let
group
show
\342\200\224> G' and
G
to G. If
be a cyclic
G
isomorphism,
xff
that,
Let r
and
5 be
positive integers.
a
and
b be
elements of a
47. Let r
and
s be
positive integers.
45.
46. Let
a. Define the c.
part
Generalizing
and 5 is
of r
48. Show
that
49.
by
Show such
50. Let
that
51. Let
p
has
a counterexample
G be a
Show
(b), show
group
that
ax =
and
q be
5 as a
and
the product
that
\\
e Z} is a
n, m
that if ab
common
least
ms
finite
has
order
generator of a
of Z.
subgroup
n, then
certain
ba alsohas
cyclic
order
n.
group.
multiple of r and 5 their product, rsi of the greatest commondivisor and of the least commonmultiple
rs.
a group that
every
of r
multiple
condition is the
{nr +
G. Show
group
common
least
b. Under what
that
Show
only that
a finite the
proper subgroup is and suppose for all x e
xa
number of
\"converse\" of
following
cyclic, then
G is
a e G generates G. [Hint: Consider
distinct prime numbers.
Find
must
subgroups
the
be a finite
group.
Theorem 6.6is not
a theorem:
\"If
a group
G is
cyclic.\"
a cyclic
subgroup of order
2 and
is the
unique such
(xax'])2.] number
of generators
of the
cyclic
group
7Lpq.
element.
52. Let 53.
p be a
solutions
prime number. in a finite
that
Show
x in
54.
With
55.
Show that 1p
56.
Let G be an
b.
G
cyclic each
if r
that
7
and
n
m does
divide
=
integer
e has
> 1.
exactly
m
n?
number. with
\\H\\
= r
and
of order
cyclic subgroup
subgroup of order the
a cyclic
not
xm
an
least
common
\\K\\
= s.
rs. multiple of r
and
s.
And Cayley Digraphs
Sets
Generating
the equation
subgroups
G contains a
then
G contains
that
finite cyclic
K be
and
prime,
relatively
(a), show
part
H
<
subgroups if p is a prime
nontrivial and let
situation if 1 < m
is the
Zpr, where r is
group
cyclic
order n, written multiplicatively, integer m that divides n.
53, what
and 5 are
the
G of
positive
group
of generators of
number
group
no proper
has
abelian
Generalizing
section
for
Subgroups
the
Find
to Exercise
reference
a. Show
Groups and
I
Part
68
G be a group, and let a e G. We have described the cyclic subgroup (a) of G, which the element a. Suppose we want to find as of G that contains is the smallest subgroup that contains both a and b for another element b in G. By as possible small a subgroup Theorem 5.17, we see that any subgroup containing a and b must contain a\" and bm for all finite products of such powers ofa and b. all m, n e Z, and consequently must contain For example, such an expression might be a2b4a~3b2a5. Note that we cannot \"simplify\" this expression by writing first all powers of a followed by the powers of b, since G may of the same not be abelian. However, products of such expressionsare again expressions e = a0 and the of such an expression is again of the same inverse type. Furthermore, of a2fo4a~3fo2a5 is a~5b~2a3b~4a~2. By Theorem5.14, type. For example,the inverse of a and b form a subgroup of G, this shows that all such products of integral powers which surely must be the smallest containing both a and b. We call a and b subgroup If this subgroup should be all of G, then we say that {a, b) generators of this subgroup. generates G.Of course, there is nothing sacred about taking just two elements a, b e G. We could of elements of G, have made similar arguments for three, four, or any number
Let
as long 7.1 Example
Klein
The It
as we
is also
take only
The group
so that also
{e, a, b, c] of {a, c], {b,c], and
by
then every subsetof 7.2Example
generated
since (2) =
by {3,
{0,2, 4}
{1}
containing
subgroup
Example
generated by {a, b] since ab = c. group G is generatedby a subset S,
5.9 is
G.
\342\226
and {5}. It is also generatedby {2,3}since2 and 3 must contain 5 and must therefore
2
4}, {2, 3, 4}, {1,3},and
contains
integral powers.
{a, b, c}. If a
S generates
G containing
Z6 is generated by
any
of their
products
V =
4-group
generated
finite
2 and
4.
{3,
5}, but it is
not
generated
+ 3 = be Z6- It
5,
by {2,
4}
is
\342\226
of a group G generated by given an intuitive explanation of the subgroup follows in of G. What is a detailed exposition of the same idea approached of subgroups. After we get an intuitive grasp of via intersections another way, namely as possible. We give a set-theoretic a concept,it is nice to try to write it up as neatly a theorem that was in Exercise definition and generalize 54 of Section 5. We
a subset
have
Generating Setsand
Section7
7.3 Definition
I
Let {5,-| i
e
DieISiof
the
1} be a collection of sets.Here may be any set sets St is the set of all elements that are in all n S{ iel
If
= {1,2, ...,
finite, I
I is
= {x\\x
we
n],
Theorem
Proof
Let us fo
Let a
closure.
show
\342\226\240
of a group
//,
G for
2
e
/ is
again a subgroup
of
G.
so that a e //, for all 2 e I and and fo e n,-e///,-, 2 since for all e //,/, //, is a group. Thus afo e n,-e///,-. all 2 e /, and hence all i e /, we have e e for for
e n,-e///,
//, for all; e /. Thenafo Since //, is a subgroup
e
e /}.
e Si for all;
ns2n---0\302\276.
of somesubgroups
The intersection
of indices. Theintersection the sets St; that is,
denote n,-e/S,- by
may Si
7.4
69
Digraphs
Cayley
e
//,
e e n,-e///,-.
Let containing
if we
a a,-
7.5 Definition
take the //
subgroup
for
2'
group
let
and
e //,
for all
2
e /,
so a-1
e
for all //,\342\226\240
2
e
intersection of all This
/, \342\231\246
a,- e
at for i
the elements of G.
a
have
e n,e///,-.
a-1
that
G be a all
n,-e///,-, we
a e
for
Finally,
which implies
subgroups H
subgroup
There is at least one subgroup of G itself. Theorem 7.4 assuresus that namely of G containing all a,- for z' e /, we will obtain is the smallest subgroup of G containing all the
G for
i
e /.
e /,
G is
e /.
of G containing be a group and let a,- e G for 2 e /. The smallest subgroup If this is all of G, then is the e e {a,- 12 /} subgroup by {a,- 12 /}. subgroup generated of G. If is a finite set {a,- 12 e /} G and the are there e a,/} generates {a,- 12 generators \342\226\240 that generates G, then G is finitely generated.
Let G
definition is consistentwith our previous definition of a generator for also that the statement a is a generator of G may mean either that G = {a} or that a is a member of a subset of G that generates G. The context in which the statement is made should indicate which is intended. Our next theorem gives the into the subgroup of G generatedby {a,- 12 e /} that we discussed for structural insight two before Example 7.1. generators Note
that this
a cyclicgroup.
7.6 Theorem
Proof
Note
If G is a group and a,- e G for 2 e /, then the subgroup // of G generated by {a,- 12 e /} has as elementspreciselythose elements of integral powers of G that are finite products of the a,-, where powers of a fixed a,- may occur several times in the product.
Let K denote the We
need
only
set
observe
of all
containing at for 2 e again in K. Since (a,-)0= the
product
finite
of integral
products
that K is a subgroup /, we will be done.
giving k a
new
e, we have product
and
powers of since H is
Observethat
e & K.
with
then,
For every
the order
a,-.
smallest
a,-
K c
H.
subgroup
of elements in K is if we form from and the opposite reversed
a product
element fc
of the
Then
the the
in ^T,
Part I
and
Groups
sign on all
Subgroups
exponents, we have
is
thus
in K.
For example,
in K.
is again
which
1, which
k
\342\2
Digraphs
Cayley
For each generating group in terms of
S of a
set the
digraph. These visual referred to as Cayley
finite
the G, there is a directed graph representing The term directedgraph is usually abbreviated as and are also of groups were devised by Cayley,
group
in S.
generators representations
diagrams in the literature. consists of a finite number of points, called vertices of the a digraph Intuitively, a direction denoted by an arrowhead) joining and some arcs (each with vertices. digraph, a generating G using set S we have one vertex, represented by In a digraph for a group a dot, for each elementof G. Each generator in S is denotedby one type of arc. We and paperwork. colors for different arc types in pencil Since different coulduse different are not available in our text, we use different colors arcs, like solid, dashed, and style dotted,
to denote
a this
With xa
= y.
by
different generators. Thus if
\342\226\240
and
bby---*
might denote
c] we
cby
\342\200\242
an occurrence of x\302\273*y in a Cayley digraph means that an arc in the direction of the arrow indicates that multiplication at the start of the arc on the right by the generator corresponding element at the end of the arc. Of course, since element of arc yields the group = x. Thus traveling we know immediately that ya~l an arc in the group,
notation,
That is, traveling
of the group to that type we are in a direction oppositeto
of the
corresponding
denote
this by
For example,if 7.7Example
S = {a,b,
the
corresponds to multiplication If a generator in S is its own arrowhead from the arc, rather
arrow
generator.
omitting b2 =
the
e, we
might
denote
on
the right by
the inverse
it is customary to using a double arrow.
inverse, than
b by
= {1} BothofthedigraphsshowninFig.7.8representthegroupZ6withgeneratingset5 and shape of an arc nor the angle between arcs has any significance the length
Neither
(b)
7.8 Figure
Two digraphs for
Ze with
S =
(1}using
\342\226\240
7
Section
Sets and
Generating
(b)
(a)
Two digraphs
7.9 Figure
7.10
Example
Both of
the
for Z6
{2,3}.
3 is
how
different
group.
The difference is due
its
to
satisfy
these
1. The digraph we can get
that is,
is connected,
g to by traveling along any vertex
from
h
vertex
consecutive arcs, starting 2.
3. Each vertex
of each
g has
type
of each type 4. If two different
starting
from the can
digraph
for
reasons
the
\342\226\262
indicated.
from
vertex
same
vertex
any
equation
in a
group.
The
solution
gx
= h
has a
=
is unique.
solution
e G and
of gx
arc
For g
one
can compute gb, and
each
h
generator
b we
{gb~l)b
= g.
vertex
of arc g lead
h, then
those
arc
types
u will
lead
If gq ug~lh
=
h
and
gr
= h,
then
uq
=
= ur.
starting to
vertex v.
same
that, conversely,every digraph group. Due to the symmetry
some
like a, b, c for
Every
and
a vertex
exactly one at g, and
sequences
starting
be shown for
four properties
of generators.
at g.
ending
samesequences of
It
at g
ending at h. At most one arc goesfrom g to a vertex h.
types to the
set
the
Reason
Property
any
\342\226\240 *\342\200\224 and-
choice for
different
the
for a group must
digraph
= (2, 3}using
7.9 representthe group Z& with generating set S = own inverse, there is no arrowhead on the dashed arcs representing 3. in Fig. 7.8 for the same those these Cayley diagrams look from
Notice
Every
with S
in Fig.
shown
digraphs
Since
71
Cayley Digraphs
the
various
name each other vertex by to attain that vertex starting
constructed (found)
using
from
these
of such a
digraph,
four properties is a Cayley we can choose labels
name any vertex e to represent the identity, and of arc labels and their inverses that we can travel the one that we named e. Somefinite were first groups
arc types, a product
digraphs.
satisfying
Part I
72
and
Groups
Subgroups
\342\226\240
,, //
/
;
\\ \\
\\
(a)
7.11 Figure
7.12 Example
A
digraph
7.11
Fig.
the four
satisfying
(b), we selected
the
properties on page 71 is
in Fig.
shown
7.11 (a). To obtain
labels
and
>
a
,
b
a vertex e, and then the other vertices as shown. We have a group named we named of eight elements. Note that the vertex that {e,a,a2,a3,b,ab,a2b,a3b} well be named ba~l, the vertex that we named a3 couldbenamed ab could equally a~l, etc. It is not hard to compute products of elementsin this group. To compute (a3b)(a2b), we just start at the vertex labeled a3b and then travel in succession two solid arcs and one dashed arc, arriving at the vertex a, so (a3b)(a2b)= a. In this fashion, we could write out the table for this \342\226 eight-element group. named
7
EXERCISES
Computations In
Exercises
1 through
6, list
1. The subset {2, 3}of
the
elements
of the
Z12
subgroup generated by the given subset. 2. The subset {4, 6} of Z]2
subset {8, 10}of Zi8 The subset {12, 42} of Z
subset {12, 30}of Z36 The subset {18, 24, 39} of
3. The
4. The
5.
6.
7. For
the
group
described
in Example
7.12 compute these
b.
a. (a2b)a3
-Ta
products,
using
Fig. 7.11(b). c.
(ab)(a3b)
'
Z
b(a2b)
*
bid
(a)
f (b)
7.13 Figure
(c)
Exercises
Section 7
In
8 through 10, give List the identity
Exercises
your answers
8. Thedigraph 9. The digraph 10. The digraph
easy to
be
will
for the
table
the
e first
element.
identity
group
the
having
in your table,
list
and
indicated
digraph. In each
elements
the remaining
73 e as
take
digraph,
so
alphabetically,
that
check.
in Fig.
7.13(a)
in Fig.
7.13(b)
in Fig.
7.13(c)
Concepts
11.
How
12.
Referring
from a Cayley digraph
we tell
can
whether
11, determine
to Exercise
the
whether
or not
the
group
corresponding
group
corresponding
to the
is commutative?
in Fig. 7.11(b) is
digraph
Cayley
commutative.
13. Is it
14. The large triangle
outside
triangle
in Fig.
similarly
exhibit
a cyclic
S =
{1, 2} for
15. The generating group.
digraph of a
a Cayley
from
obvious
set
we can
Nevertheless,
or not
whether
group
the
group
is cyclic?
the cyclic subgroup {0, 2, 4} of 7.9(b) exhibits subgroup of Zg? Why or why not? Zg
Z6.
at
Fig.
7.9(b).]
the smaller inside
Does
since 1 is a generator for the more generators than necessary, such a Cayley digraph for Z6 with this generating set S. Draw
contains
a Cayley
draw
[Hint: Look
digraph.
16.
Draw
17.
A
digraph for
a Cayley
on a set
relation
their inverses
taking
S of generators
the
to
Z8
identity
e of
one relationisaba~lb~l= e.If,
as generating of a
G. For
set S =
group G is
its
b is
own
{2,5}.
equation
if S
example,
moreover,
an
that equates some product G is commutative so
= {a, b} and inverse,
then another
a. Explain how we can find some relations on S from a Cayley digraph = b. Find three relations on the set S for the group {a,b}of generators 18. Draw
digraphs
possible
in
two possible case. You need not
of the each
different
structurally
label
groups
of order
relation is
and
of generators that
=
b2
ab
= ba,
then
e.
of G.
describedby 4, taking
Fig.
as small
7.11 (b). a generating
set as
vertices.
Theory
19.
Show
that
order 2.
for n
> 3, there
exists
a nonabelian
group
with
2n
elements
that is generated
by
two
elements
of
part
8
Section
Cosets and
Section 11 12
Section
Alternating
Theorem
the
Generated
and Finitely
Direct
Products
fPlane
Isometries
Groups
of Lagrange Abelian
Groups
of Permutations
Groups We
and the
Cycles,
Orbits,
10
Section
of Permutations
Groups
Section 9
8
Products
Direct
and
IJ
section
Cosets,
Permutations,
seen
have
examples
of groups of
We have also introduced
addition.
like
numbers,
groups
groups Z, Q, and E under like the group GL(2, E). Each plane E2 into itself; namely, if we the
of matrices,
of the element A of GL(2, E) yields a transformation vector. x then Ax is also a 2-componentcolumn vector, regard as a 2-componentcolumn in that its elements is of the most useful The group GL(2, E) of many groups typical act on things to transform them. Often, an action produced by a group element can be can be regarded of the group ^function binary operation whose we finite construct some elements, called section, groups composition. us with examples of finite permutations, act on finite sets. These groups will provide the same as some is nonabelian shall show that finite We structurally any groups. group as a function,
regarded
In
group
turn
and
the
this
of permutations. out to be particularly
this
Unfortunately, useful
You may be familiar with of the set. Thus for
elements
be given
schematically as in
the
column
the notion the
result,
which
sounds
very powerful,
does not
to us.
set
of a permutation
{1,2,3,4,5},
of a
set as a rearrangement of the
of the
a rearrangement
elements
could
8.1, resulting in the new arrangement Fig. {4, 2, 5, 3, 1}. Letus think of this schematic diagram in Fig. 8.1 as a function mapping of eachelement element from the same in the left column into a single listed (not necessarily different) set listedat the right. Thus 1 is carried into 4,2 is mapped into 2, and so on. Furthermore, each element of the set, this mapping must be such that to be a permutation appears in right
once
and only
once. For example,the
1 does a permutation, for 3 appears twice while to be such a mapping. now define a permutation '
Section
12 is not used
in
the remainder
of
diagram
not appear
at
in Fig. all in
8.2 does not
the right
give
column. We
the text.
75
Part II
76
Products
Direct
and
Cosets,
Permutations,
l->4
l->3
2^2
2^2
3^5 4^3
3^4
4^5
5->l
5^3
8.1Figure 8.3
Definition
of a set A
A permutation
Permutation now
We
is both
\342\200\224>\342\200\242 A that
: A \302\247
that function composition o is of a set A. We call this operation
show
and let a
mapping
function
one to
one
and onto.
Groups
permutations
set,
is a
8.2 Figure
and
x be
of
permutations
A
so
A onto A. The compositefunction
a
binary
operation
on the
collection of all
multiplication. that a and x are both one-to-one a ox defined schematically by permutation
Let
A
a
be
functions
A\\A^A, o for permutation gives a mapping of A into A. Rather than keep the symbol we will denote a o x by the juxtaposition multiplication, ax, as we have done for general groups. ax will be a permutation if it is one to one and Now onto A. Remember that the action of ax on A must be read ax is one to one. If
in
first apply x and
order:
right-to-left
(ax)(ai)
= (ax)(a2),
a(x(ai))
= a(x(a2)),
then
a.
Let us
show
that
then
since a is given to be one to one, we know and that x{a\\) = x{a2). But then, since one to one, this gives a\\ \342\200\224 ax is one to one. To show that ax is onto A, a2. Hence a e A. Since a is onto exists a' & A such that a(a') = a. Sincex is onto A, there
exists
there
a\"
e
x(a\") =
A such that a =
so
8.4 Example
ax
is onto
a{a')
a'. Thus
= a{x{a\
= (ax)(a\,
A.
Supposethat A = and
that
changing
a is
the
permutation
the columns
given
to rows
in
{1,2, 3,4,
by Fig. parentheses
5}
8.1. We write and omitting
a
in a more standard the arrows, as
notation,
x is let A,
Section8
of the permutations
of Creation, sometime
in the
written
by
in
the letters
of
the
counting the
Hebrew
Sefer
Jewish
century.
various
ways can be
alphabet
author
was
author
The
ibn al-Banna a mathematician (1256-1321), from Marrakech in what is now Morocco, and Levi ben Gerson, a French rabbi, philosopher, and Abbas
Book
or
Yetsirah,
an unknown
the eighth
of
studies
recorded
earliest
occurs
before
interested
was
number
n!,
arranged.
in some
so that
a (I)
= 4,
were able to give rigorous proofs that the of permutations of any set of n elements is as well as prove various results about counting
mathematician,
which
in
sense a mystical one. It was that the letters had magical believed could suitable powers; therefore, arrangements actual The text of the the forces of nature. subjugate \"Two letters build Yetsirah is Sefer very sparse: four build 24 six words, two words, three build words, five build 120, six build 720, seven build 5040.\" enough, the idea of counting Interestingly also the arrangements of the letters of the alphabet occurred in Islamic mathematics in the eighth and in both ninth centuries. By the thirteenth century, idea the Islamic and Hebrew cultures, the abstract root so that both Abu-1-' of a permutation had taken
The question
77
Permutations
Note
\342\226\240 Historical
One
Groupsof
combinations.
Levi and his with
concerned
predecessors, however, were as simply arrangements of solutions of and others
permutations
a given finite set. It was the search for polynomial equations that led Lagrange late
the
in
eighteenth century from a finite
to
think
of
the set of a being given equation. And it was Augustin-Louis Cauchy (1789-1857) who in theorems of permutation detail the basic developed and the standard who introduced notation theory used in this text. functions
permutations as
of
that
cr(2) = 2, and
to itself,
the roots
Let
so on.
3 4
'12
T=
set
5
2
1
5
4
3
4
5
4
2 1
,3
Then
1 2
ax For
2
4
3
4
5
3
3
5
order,
in right-to-left
multiplying
example,
12
5 1
(crr)(l) = cr(T(l))= cr(3) We
8.5 Theorem
Proof
that
show
now
group under
this
a nonempty
We have
shown
under
that
we showed that
function
set,
let
and
of
a nonempty
set
A
forms
a
SA be
the collection of all
of A.
permutations
Then Sa
multiplication.
of
composition multiplication
composition that
two
permutations
multiplication. is defined as
permutation
The permutation i such satisfied.
A
multiplication.
permutation
so Sa is closed under Now permutation
5.
of all permutations
the collection
permutation
Let A be is a group
=
i(a)
of A
yields a
permutation
function composition, and
of
in Section
is associative. Hence S^J is satisfied. \342\200\224 a, for all a e A acts as identity. Therefore
A,
2,
^ is
78
Part
II
Cosets, and DirectProducts
Permutations,
a permutation
For
direction of the
existenceof a
one
one to one and
is both
inverse
the
a, that
mapping
exactly
that reverses the a~l, is the permutation function, -1 a' of A such that a = a (a'). The is, a (a)is the element such element a' is a consequenceof the fact that, as a function, onto. For each a e A we have
a,
cr(a')= cr(a~l(a))
a=
i(a) =
=
(crcr~l)(a)
and also
= a~\\a) = a~\\a{a'))
L(a') =a' so that
Thus
= /x(cr(a)).
{o\\L){a)
refer to
the
(CT_1o-)(a'), is (\302\276
a \\\302\261 of permutations get for
they
permutation
on this
text
another
the
i. Thus
permutation
us to check in two material, be sure to
51 asks
Exercise \\\302\261o.
by computing
If you
are both
crcr-1
texts compute a product
Some
Warning: that
and
<7_1<7
=
a\\i
ways check
satisfied.
\342\231\24
in left-to-right order, so is the one we would get that we still get a group. its order for permutation
multiplication.
was nothing in our definition of a permutation to require that the set A be will be concerned with most of our examples of permutation groups of finite sets. Note that the structure of the group Sa is concerned only permutations of elements in the set A, and not what the elements in A are. If setsA and with the number There
finite.
However,
B have the same cardinality, then Sb- To define an isomorphism \302\242: Sa \342\200\224>\342\200\242 Sb , we Sa \342\200\224 B be a one-to-one let f : A \342\200\224> function mapping A onto B, which establishes that A and B have the same cardinality. For a e Sa, we let 4>{a) be the permutation a e Sbsuch that = f(a(a)) for all a e A. To illustrate this for A = {1, 2, 3} and B = {#, $, %} a(f(a)) function
the
and
/ :A
as
\342\200\224>\342\200\242 B defined
/(2) = $,
/(1) =#,
/(3) =
%,
\302\242) maps
We
8.6Deflnition
simply renaming
{1,
2, 3,
\342\200\242 \342\200\242to \342\200\242, n}
that
(,3
2
3\\
8.7Example
An
letters,
Sn has
interesting
We list
%\\
$ #J-
ij11110^
in B using in our two-row notation by elements thus of to those elements be of We can take Sa Sbf, renaming be a prototype for a finite set A of n elements.
finite set
Important the
$
of A
n\\
{1,2,
\342\200\242 \342\200\242 \342\200\242 The
,n}.
and is
denoted
elements,
where
by
nl=n(n-
Two
/#
.
function
Let A be the group on n Note
2
the elements
rename
the
/1
group of all permutations
of A
is the
symmetric
\342\226\24
Sn.
-2)---
\\){n
(3)(2)(1).
Examples
example for us is the of A and permutations
group
63 of
assign to
3!
each
\342\200\224 6 elements.
a subscripted
Let the
set A
be
{1,2,3}.
Greek letter for a name.
Section8 The reasons for
will be
of names
choice
the
Mo
Mi
92 =
1 1
2 2
1
2 3
2
3
3
3
3
clear later. Let 2 Ml 1 3 M2
1
2
3
1
2
Groupsof
M3
79
Permutations
3
2
3
2 2
2
2 3 1 3
3 1
8.8 Table Po
Pi
P2
Mi
M2
M3
Pa
Po
Pi
P2
Mi
M2
M3
Pi
Pi
Pi
Po
M3
Mi
M2
Pi
Pi
Po
Pi
M2
M3
Mi
Mi
Mi
M2
M3
Po
Pi
P2
M2
M2
/U-3
Mi
P2
Po
Pi
M3
M3
Mi
M2
Pi
P2
Po
The multiplication table for 53 is shown in Table 8.8. Note that have seen that any group of at most 4 elementsis abelian. a group of 5 elements is also abelian. Thus S3 has minimum
We
this
is not
group
Later
we will
order for
any
abelian! see
that
nonabelian \342\226\262
group.
of S3 in Example 8.7 and the correspondence between the elements of an equilateral triangle with vertices 1,2, and 3 (see Fig. 8.9 can be placed, one covering the other with vertices on top of vertices. For this reason, we used p, 53 is also the group D3 of symmetries of an equilateral triangle. Naively, for rotations and jxt for mirror images in bisectors of angles. The notation for D3 stands the third dihedral group. The nth dihedral group Dn is the group of symmetries of the See Exercise 44.' regular n-gon. Note that we can consider the elements of S3 to act on the triangle in Fig. 8.9. See the discussion at the start of this section. There
ways
8.9Figure
8.10
Example
is a natural
in which
Let us form the copies
vertices of
the
Many
two
dihedral
of a square
on square.
top
copies
of
with
vertices
It is
to the ways that two D4 of permutations corresponding 4 and can be one 1, 2, 3, placed, covering the other with will then be the (see Fig. 8.11). D4 group of symmetries
group vertices
also called the
people denote the nth dihedral
group
octic
group.
Again,
we choose seemingly arbitrary
since the order of the group is 2n. by D2,, rather than by D\342\200\236
Part II
80
Permutations, Cosets,and
that we shall explain later.
notation
images
in
involved
are using pt
we and
<5,
for
for
diagonal
/x,- for
rotations,
flips.
12
3
4
12
3
4
2
3
3
4
12
Pi
4
pi 2
4
12
1
4
3 4
12
P2 =
2 14
3 4
12 Pi
Si =
12 3
12
3
12 Mi
3
12
3
4
2
1
3
2
3
4
3
4
14
4
12
3
4
3
14
3
2
8.12'Table Po
P\\
Pi
Pi
Mi
M2
\302\253i
Si
Po
Po
Pi
Pi
Pi
Mi
M2
\302\253i
Si
Pi
Pi
Pi
Pi
Po
\302\253i
Si
M2
Mi
Pi
Pi
Pi
Po
P\\
M2
Mi
Si
s,
Pi
Pi
Po
Pi
Pi
Si
s,
Mi
M2
Mi
Mi
Si
M2
\302\253i
Po
Pi
Pi
Pi
M2
M2
Si
Mi
Si
Pl
Po
P\\
Pi
\302\253i \302\253i Mi
Si
M2
Pi
Pi
Po
Pi
Si
\302\253i
Mi
Pi
Pi
Pi
Po
Si
M2
{PO'^ll
{Po-
8.13 Figure
Subgroup
diagram
for D4.
mirror
There are
here. Let
Po
Figure
Naively,
of sides,
bisectors
perpendicular
permutations
8.11
Products
Direct
h
eight
Section 8
Groups of
81
Permutations
in Table 8.12. Note that D4 is again nonabelian. This group The table for D4 is given is simply beautiful. It will provide us with nice for many concepts we will examples in that table! Finally, introduce in group theory. Look at the lovely symmetries we give in Fig. 8.13 the subgroup for the subgroups of D4. Look at the lovely symmetries diagram in that
\342\226\262
diagram! Theorem
Cayley's
Look at any group table in the text. Note how each row of the table gives a permutation as listed at the top of the table. Similarly, each column of the set of elements of the group, of the group set, as listed at the left of the table. In view of the table gives a permutation that at least every finite G is isomorphic of these observations, it is not surprising group for infinite of G. The same is true to a subgroup of the group So of all permutations that is to some states theorem every group isomorphic group consisting Cayley's groups; This is a nice and intriguing of permutations under permutation result, multiplication. At first glance, the theorem and is a classic of group theory. might seem to be a tool to of groups of shows is the generality answer all questions about groups. What it really of all permutation groups Sa for sets A of all sizes subgroups permutations. Examining exists theorem does show that if a counterexample would be a tremendous task. Cayley's will then some group of permutations to some we have made about groups, conjecture provide
the counterexample. to the proof of Cayley's theorem, proceed in its own right. lemma that is important
We now then
* Historical
different
set and
gave an of a group in a
(1821-1895) definition
sounding
1854:\"A
starting
with
a definition
and
Note
Cayley
Arthur
a
abstractpaper of
\342\226\240 of symbols, all of them 1, a, f), \342\226\240 \342\226\240, of such that the product of any two
them (no matter in what order) or the product of one of them into itself, belongs to the set, is any said to be a group.\" He then proceeded to define a
unable to find a suitable teaching post. In 1863, he In 1878, finally became a professor at Cambridge. he returned to the theory of groups by publishing four papers, in one of which he stated Theorem 8.16 of this text; his \"proof was simply to notice from
table that multiplication by any group the group elements. However, he of wrote, \"this does not in any wise show that the best group table and note that every line and column \" \342\226\240 \342\226\240.or the the table \"will contain all the symbols 1, a, /3, \342\226\240 easiest mode of treating the general problem is thus to however, always represented [of finding all groups of a given order] Cayley's symbols, operations on sets; it does not seem that he was aware regard it as a problem of [permutations]. It seems of any other kind of group. He noted, for instance, clear that the better course is to consider the general the
that
= /S the
four
matrix
transposition, non-cyclic
his definition
group went
operations 1, a = inversion, and y = a/S, form, abstractly, of four elements. In any case, unnoticed for a quarter of a
century.
paper of 1854 was one of about 14 yearsCayley the was practicing during This
300
written
law, being
the
group
element
permuted
problem
in
itself.\"
the earlier one, papers of 1878, unlike found a receptive audience;in fact, they were an on Walter Van Dyck's 1882 important influence axiomatic definition of an abstract group, the definitionthat led to the development of abstract group
The
theory.
Part II
82
8.14 Definition
8.15
Lemma
Permutations, Cosets,and Let f : A
->
{f(h) | h
e //}
Let G
function and and is denoted by
B be a
0(x)0(y)forallx, of G
Proof
and let
be groups
G'
and
Products
Direct
The
A.
under f is
of H
image
\342\2
/[#].
G
e G. Then0[G]
y
subset of
H be a
let
-> G' be a one-to-one function such is asubgroup of G'and^providesan
=
that 0(xy) isomorphism
with0[G]. for a
the conditions
We
show
Let
x', y' e
exist
8.14 are satisfied by 0[G]. = x and 4>{y) = y'. By shown that e 0[G]. We have
in Theorem
given
subgroup
0[G]. Then there
such
ijeG
hypothesis,
that
that
\302\242^)
x'y'
0[G] is
e'$(e) =
0(e) = 0(ee)= 4>{e)$(e). e 0[G].
shows that e' = 0(e) so e' 0[G] where x' = 0(x), we have in G'
Cancellation
e
x'
For
which shows is a subgroup
x'_1
that
4>{xx~l)=
4>{e)=
e' =
= 4>{x~l)
e 0[G].
)
= x'$(x~l),
This completes
that 0[G]
demonstration
the
G'.
of
That 0 provides an isomorphism a one-to-one map of G onto
provides
of G with 0[G]
0[G]
because
at once
follows
now
that 4>{xy) =
such
for
4>{x)4>{y)
all x,
y
e G.
\342\23
8.16
Theorem
Proof
Theorem)
(Cayley's Let G
be a group. We show
needonly to define G.Forx e ije of Xx
all c
Every
eG
of SG. By Lemma 8.15, we isomorphic to a subgroup = function 0 : G \342\200\224>\342\200\242 So such that
G is
that
a one-to-one :
GJetA^
as performing
left
of permutations.
to a group
is isomorphic
group
G
defined by kx(g) by x.) The equation
->\342\200\242 G be
multiplication
= xgforallg
=
Xx(x~1c)
then Ax maps G onto G. If kx(a) kx(b), cancellation. Thus AA- is also one to one, and is a permutation Sg by defining 0(x) = Ax for all x e G. 4> : G \342\200\224>\342\200\242
To show one. It g e
is one \302\247
that
G
mapping
that
shows
into
G.
remains
only
to show
G,wehaveAX},(g):
(XxXy)(g) = For the
Xx(Xy(g))
proof of the
px of G defined
to one,
In particular that
(xy)g.
Permutation
for g e these
G. (We
multiplication
x(yg).Thus
we could
by
= xb
xa
Then
of G.
We
Xx =
Xy
x Xxy
have considered equally
so a
= c for
=b
- XxXy.
by
define
now
as functions
y. Thus 0 is
is function
associativity,
G.(Wethink
Now
one to
for any
composition, so
A\342\226\240xy XyXV well
the permutations
by
P.Ag)
that
4>{y)-
~ ye and Ax(e) Xy(e), \342\226\240 that is, that (j>(xy) 4>{x)4>{y), so xe
: ^x(yg) = theorem,
suppose that
e
= x(x~lc)
can
permutations
think
= gx
of px as meaning right multiplication by x.) Exercise 52 shows form a subgroup of SG,again to G, but provided by isomorphic
a map /x
:
So defined
->
G
by
=
M(*)
8.17
Definition
The map the
8.18
Example
Let us
compute
the
left
the
Here the
table.
group
a
Xe=(e
a
\\e
The table for this representation in Table 8.20. For example,
8.19
is just
like
given
group
left
\342\226\240
table, by the group representation
regular
the
\\b
=
b)
a a
(e \\e
a)
with x
table
original
e
Xb =
and
a
Ib b)(l e)
a (l \\b
b\\.
e
a)
renamed Xx,
as seen
?W.
A
b)
8.20 Table
Table
e
a
b
e
e
a
b
a
a
b
e
b
b
e
a
For a finite group corresponding to their notations
pa
K
K
K
K
K
K
K
K
K
K
K
h
h
K
K
by a group table, pa is
given
in the
order
to the
permutation corresponding
left. The
the
Ib b), e)
\\a
Kh = (e \\a
of
and
of G.
elements are
K = (e
f), b)
of G,
representation
regular
regular representation
elements for the
mean
we
left
right
representation give the
regular
\"compute\"
is the
\342\226\240
Px->
8.16is the
proof of Theorem the preceding comment
in the
8.19. By
Table
and
/x in
map
83
Exercises
8
Section
column
under
the
of the
permutation
the very top, and in the row opposite
a at
order of the elements chosen to suggest right
and Xa were
and
a
elements
is the
ka at
the
left multiplication
extreme
by a,
respectively.
8
\342\226\240 EXERCISES
Computation In
1 through
Exercises
/12345 CT~V3
1456
1.
Exercises
6. |(cr)|
6 through
the
indicated
product
involving the
/12345
6\\
r~l2
2j' 2.
xa
In
5, compute
3.
x2a
9, computethe
7.
|(r2)i
expressions
/12345
6\\
4136
5J'
4.
for the permutations 8.
ct100
in S(,: 6\\
2 4 3 1 6J'
^(5
/x<72
shown
permutations
following
5. a'ha
a~2x a,
x and
/x
defined
9. /x100
prior to
Exercise 1.
Part II
84
10.
the
Partition
collection of groups
following
elements of the
all nonzero
addition
Z under
17Z under
addition
O under
addition
3Z under
addition
R under
addition
Let
A
be
let a
a set and
multiplication
Q* under
multiplication
C* under
multiplication
The subgroup
{it)
The subgroup
G of S$ generated a e
a fixed
SA. For
under
of R*
a.
a under
{a\"(a)\\n&Z} orbit
the
of 1 under
13.
With
16.
Find
the
number
of elements
in the set
17.
Find
the
number
of elements
in the
Find
the
cyclic
b.
Find
all
subgroups,
proper and
then
{a e
S4
set {a e
will
be
a. Verify that
Let
a{2)
= 5}.
in Table
8.12.
the
cyclic
them
of S3.
(/xi)
and give
the
of S5 generated
subgroup
12
3
4 5
2 4
5
13
be p, p2,
p3, p4,
p3
and
=
p\302\260
for them.
diagram
subgroup
in Fig. 8.13 is correctby finding by two elements, etc.
generated
subgroups
six elements.
= 3}.
of S3
improper,
P =
all (cyclic)
subgroups generated
by
p6. Is
this
group
isomorphic
to S3?
the six matrices
1 0
0\"
0
1
0
0
0
1
a group
think
how the
column
\"0
1
0\"
0
f
\"1
0
0\"
0
0
1
1
0
0
0
0
1
1
0
0
0
1 0
0
1 0
,
under matrix
form
b. What
/x
for the 8 elements of D4
labeling
cr(3)
|
S51
diagram for D4 shown
all
table for
multiplication
alternative
(pi), (p2), and
subgroups
the subgroup
element,
the
There
prior to
8.7
S3 of Example
group
a.
that
14, give a similar
to Exercise
18. Consider the
one
defined
permutation
of S3. Some authors use the notations Mi, M2, M3 as the names of the 6 elements these elements, where their e is our identity po, their P is our pi, and their 0 is our/^. that their six expressionsdo give all of S3.
geometrically
Verify
the
p4>, p24> for
reference
21.
4
5
8.8, we usedpo:Pi. P2,
15.
20. Give
3
r
e, p, p2, 4>,
by
3
12 by
12.
14. In Table
19.
multiplication
13, find
11 through
Exercises
In
11. a
Verify
means
superscript
set
the
A,
Oa.a =
is the orbit of Exercise 1.
Herea *
multiplication
R+ under
&
of isomorphic groups.
subcoUections
into
set.
S2 R* under
Z6 Z2 S6
Products
Direct
and
Cosets,
Permutations,
group discussedin
[Hint:
multiplication.
vector
this
\"0
is transformed section
Don't
is isomorphic to
this
group
1 0
, 1 0 0 1 0 0 0 0 1
0
try to computeall
by multiplying
0
f
0
\"0
it on of six
1
0
of these
products
the
left
by each
matrices?
matrices. Instead,
of the
matrices.\"
Section 8
(Consider this
part to
85
(c)
(b)
(a)
Exercises
continue infinitely
to the left and right.)
(d)
8.21
22. After In this
21, write
Exercise
working
isomorphicto
down
section we discussed the
matrix
under
multiplication
that is
of symmetries of an equilateral triangle and of a square. In Exercises discussed in the text that is isomorphic to the group of symmetries of on the figure, write some permutations label some special points corresponding
group
want
to
and compute some products
23. The
in Fig.
8.21 (a)
in Fig.
8.21 (c)
the
group
we have
that
to symmetries,
25. Thefigure 27. Compute
that form a
matrices
eight
D4.
through 26, give a group indicated figure. You may
figure
Figure
of permutations.
24. Thefigure 26. Thefigure
representation of Z4. Compute
left regular
23 the
the
in Fig.
8.21 (b)
in Fig.
8.21 (d)
right regular
representation of S3 using
the notation
8.7.
of Example
Concepts In
29, correct the
28 and
Exercises
of the
definition
acceptablefor publication. of a set S is a one-to-onemap from 28. A permutation 29. The left regular representation of a group G is the of G that carries each ieG into gx.
needed, so
-> ->
34,
R defined R defined
32. /3 : R
-> R
33. /4 : R
->
defined
-> R defined 35. Mark each of the a. Every b. Every
c. Every d. Every
by by
determine = x
f2(x)
= x2
without
reference
to the
text,
if
correction
= ex
by
f5(x)
= x3 true or
-
x2
into So whose value is a
function
permutation
- 2x
false. function.
is a one-to-one
permutation function
is a permutation
function
from G is
of G
map
= -x3
fA{x)
group
S.
+ l
by
following
S to
whether the given
fl(x)
by f2(x)
R defined
34. /5 : R
italicized term
in a form
30 through
In Exercises
30. /1 : R 31. /2 : R
it is
that
if and only
a finite set isomorphic to a
onto
itself
subgroup
if
it is
must
one to
one.
be one to one.
of Sq \342\226\240
at g
of R.
e G is
the
permutation
is
II
Part
86
e.
of an abelian group element of a group generates
g. The
symmetric
h. The symmetric j.
36.
by an
Show
group
S10 has
group
S3 is
a cyclic
subgroup of the
cyclic.
is not
example
that
proper
every
some
group
of permutations.
subgroup of a nonabelian
38. Indicate schematically a Cayley radians and a reflection (mirror
digraph
for Dn
using a
may be
group
37. Let A be a nonempty set. What type of algebraic structure mentioned of all functions A into itself under function mapping composition?
Proof
group.
10 elements.
cyclic for any n. Every group is isomorphicto Sn
is abelian.
subgroup
Every
f. Every
i.
Cosets, and Direct Products
Permutations,
abelian. in the
previously
set
generating
image). SeeExercise44.
text is given
of a rotation
consisting
set
by the
through
2jt/n
Synopsis
Give a two-sentence
39.
of the
synopsis
proof of
Cayley's theorem.
Theory
40
In Exercises
e
40. {a
SA
42. {a e SA 44. In analogy
| |
a[B]
let b be
and
the induced
one
of B. Determine = cr[B] {a{x) \\x e B}. element
particular
Here
operation.
41. {a
e
SA
|
a{b)
e B}
43. {a
e
SA
|
a[B]
= B}
a regular plane n-gon for n > 3. Each way that two copies of of the vertices. the other, correspondsto a certain permutation in a group, the nth dihedral Find group Dn, under permutation multiplication. permutations elements Dn. Argue geometrically that this group has a subgroup having group just half as many
with Examples
whole
of SA under
c B}
can be
The set of these the order of this as the
be a set, B a subset of A,
be a subgroup
= b}
a(b)
an n-gon
such
let A
43,
through
given set is sure to
the
whether
8.7 and
placed, with
8.10,
one
consider
covering
has.
group
cubical box. As correspond to a certain group
8.10, the ways in which the of the vertices of the cube. This is the group of rigid motions (or rotations) of the cube. (It should not be confused with the group of group of the which will be discussed in the exercises Section 12.) How many elements does this symmetries of figure, this group has at least three of order 4 and at least have? that different group Argue geometrically subgroups four different subgroups of order 3. 46. Show that Sn is a nonabelian group for n > 3.
45.
Consider
a cube
that
cube can be placed
47.
e 5,, is cr
= 1, the
48. Orbits were defined in common, then
49. If
fills a certain
the box
46, show
Exercise
Strengthening
y
exactly
into
identity
> 3,
then
of permutations
element of
a of S\342\200\236 satisfying
the
only
and
a e SA \342\226\240 Show that ifOaM
11. Let
a, b e
A
H of SA is transitive on A if for each a, b set, then a subgroup = b. Show that if A is a nonempty finite set, then there exists a finite IHI = \\A\\ that is transitive on A. A
is a
e
cr(a)
50.
if and 51.
to the
Referring only
(See the define
a.
for
all
definition before Exercise 11and = A for some a \342\202\254 A.
to Exercise
49, show
that
A
there
cyclic
for
a e
and Ob.a have exists subgroup
an element
a e H such H of SA
Sa, (o) is transitive
that with
on A
if Oa,a
on page 78). Let G be a group with binary operation *. Let G' be the same set as G, and operation *' on G' by x *' y = y * x for all x, y e G'. that G' under *' is a group.) the front wall of your room were made class argument Suppose and that all possible a * b = c and all possible instances a * (b * c) = glass, products
warning
a binary
(Intuitive
of
cry = ycr
permutation.
Exercise = Ob.a-
before OaM
if n
that
8.7 and
in Examples
transparent
Section 9
(a * b) * c of b.
52. Let a
the
from
Show
53. A
matrix permutation n x n permutation matrix
a.
at the
group. Prove that group isomorphic to G.
same
Show
that every matrix
that
finite
can
from
be obtained
n matrix
n x
any
of the
rows of A
of order n
group
elements
four
and
as the
matrix
an identity C = PA,
What
G, do form
its rows. If P is an from A by making produced P from /\342\200\236.
by reordering
then C can
of the
reordering
and x e
= iaforaeG
pa{x)
is isomorphicto a group
and c in
e,a,b,
to it under
corresponds
9
A is
a magic marker. front of yours?
be obtained
which
rows
matrices
consisting
ofnxn
permutation
the
V, give a
specific 4x4
multiplication.
each of the
section
that
and
-> G, where
pa:G
permutations
is one
reordering
under
b. For
the
87
Groups
Alternating
property
see when
G be a
precisely the
the
* were written on the wall with for G under of wall from the next room in other the side looking *' that G' is a group under *'. mathematical definition of
associative
the
a person
would
Orbits, Cycles,and
group
matrix
Groups
and the Alternating
Cycles,
Orbits,
5.11 for
Table
the
an isomorphism.
such
Orbits Each
a of a
permutation
property
that
a, b e
establish
this
partition
For
a,
check
now
We
b e
A are in
A, let a
that ~ defined
n,
be
cr
9.2 Example
9.3 Example
Find
relation
e Z.
so a
~ c.
one-element
the orbits
indeed
some
for
cr\"(a)
an
n e
Z.
(1)
relation.
equivalence
an{a)
for some
e Z.
n
But then
a =
so b ~ a. ~
Substituting,
of a set A. (1) are the orbits
permutation i of of A.
the identity
Since the
b =
then
e Z,
m
=
~ a sincea = i(a)= cr\302\260(a).
a permutation
equivalence
(1) is
Condition
by
if b
only
Supposes~ foandfo
Transitive
Let
and
\342\200\224 n
and
partition of A into cells with the if b = an{a) for some n e Z. We
a natural
determines
~ b if
Clearly a If a ~ b,
Symmetric
Definition
A
the same cell if and only an appropriate equivalence relation:
using
Reflexive
9.1
set
c, thenfo
= an(a)
we
that
find
The
equivalence of a.
A
each
leaves
c =
and
crm(b) for
am(a\"(a)) =
in A
classes
c =
the
element
\\h
of
A
fixed,
the orbits
867415
8\\
2)
inS8. Solution
To
find
the orbit
the
of i are \342\226\262
234567
_/l
by
\342\226\240
permutation
~ \302\260
some
a\"+m(a),
determined
subsets
of
a~n(b)
containing 1, we apply
a
repeatedly,
obtaining
symbolically
\342\226\240 \342\226\240 \342\226\240 L \342\200\224> L \342\200\224> 1\342\200\224> i \342\200\224> o \342\200\224> i \342\200\224> o \342\200\224> i \342\200\224> .
Part
88
II
Cosets, and Direct Products
Permutations,
Sincea'1 the
orbit
say 2, orbit
would
1 is
and
we see that
4 is
the
{1, 3, 6}.We that the
find
similarly
containing
{4, 7,
choose
now
orbit
containing
5}. Sincethese
list of
complete
in this chain, we see that arrows an integer from 1 to 8 not in {1, 3, 6}, 2 is {2, 8}. Finally, we find that the orbits include all integers from 1 to 8,
of the
the directions
reverse
simply
containing
three
is
of a
orbits
{4,5,7}.
{2,8},
{1,3,6},
Cycles elements.
of this section, we considerjust permutations of a finite set A of \342\200\242and that we are We may as well supposethat A = {1, 2, 3, \342\226\240 \342\200\242, n) dealing with
elements
of the
For
n
remainder
the
symmetric
back to
Refer
Sn.
group
Example 9.3. The 12 3
of
orbits
8
3
4
6
7
5 6 7 4 15
8
(2)
2
1 to 8 on in Fig. 9.4. That is, a acts on each integer from on it the circle traveled into the next integer one of the circles by carrying For example, the leftmost circleindicates that in the direction of the arrows. counterclockwise, = = = a to the 1. Figure 9.4 is nice way visualize structure of 3, (7(3) 6, and cr(6) cr(l) are
indicated
the
permutation
graphically
a.
9.4 Figure Each example,
circle in Fig.
individual the leftmost
9.4 also defines,
circle corresponds to 12 M
that acts on 1, 3, and fixed. In summary,
9.5 Figure
6 just /x has
3
as a
2
by
itself,
the
permutation
3
4
5 6 7
6
4
5
does, but
one three-element
leaves
17
in
S%
For
8
(3)
8
the remaining
orbit {1,
a permutation
3, 6}
and
integers 2, 4, 5, 7, and five
one-element
8
orbits
described graphically by a single circle, {2}, {4}, {5}, {7},and{8}.Such a permutation, is called a cycle (for circle). We consider the identity to be a cycle since it permutation can be represented a circle the as shown in Fig. 9.5. We now by having only integer 1, define
the term
cycle
in
a mathematically
precise
way.
Orbits, Cycles,and the
Section 9
9.6 Definition
A
a e The length
is a S\342\200\236
permutation
element.
cycle if it has
of a cycleis the
the cumbersome notation, notation. In cyclic notation, cyclic
To
avoid
Eq. (3), for a cycle, we introduce cycle in Eq. (3) becomes
as in the
=
number
into
the
first number
be left our
9.7 Example
3 into
second
next
the
a single-row
(1,3,6). the first
carries
/x
\342\226\240
number
6, etc.,
number 1 into
until
finally
the
the last
second number 3, number 6 is carried /x is understood to
1. An integer for not appearing in this notation Of course, the set on which /x acts, which is {1, 2, 3, 4, 5, 6, 7, 8}in must be made clearby the context.
fixed
by /x.
example,
see
S5, we
within
Working
that
notation
this
by
the
one
than
of elements
number
/x
We understand
orbit containing more in its largest orbit.
one
most
at
89
Groups
Alternating
that
12 (1,3,5,4)
2
3
4
14
5
3
5
that
Observe
5, 4)
(1, 3,
= (3, 5, 4, 1)= (5,
4,
1, 3)
= (4,
1,3, 5).
since cycles are special types of permutations, course, they can be multiplied just two permutations. The product of two cycles need not again be a cycle,however. a in Eq. (2) can be written as a Using cyclic notation, we see that the permutation Of
as any
of cycles:
product
12 3
no
appears its orbits, and
8
15 that
one number
7
6
5
are disjoint, meaning
These cycles
cycles;thus
4
3
6 7 4
8
the
in
(4)
(1,3,6)(2,8)(4,7,5).
2
any integer is moved notations of two different
by
at most
one of
these
cycles.
Equation
(4)
one-line description of Fig. 9.4. Every permutation in Sn can be expressed in a similar fashion as a product of the disjoint to its orbits. We state this as a theorem and write out the proof. corresponding
exhibits a
9.8 Theorem
Proof
Every
terms
in
a
permutation
Let B\\,
of
\342\226\240 \342\226\240be Br \342\226\240,
B?,
of a finite
the orbits
set is a product of disjoint
of a,
/x,(x)
Clearly distinct
a = equivalence
mm
is a
and
let
/x,- be
cycles.
the cycle defined
a{x)
for* e
x
otherwise.
cycles
by
S,
\342\200\242 \342\200\242 \342\200\242 Since
Mr-
classes,
\342\226\240 \342\226\240 the equivalence-class orbits B, being B\\, B2, \342\226\240, \342\200\242 \342\226\240are are disjoint, the cycles \342\200\242, /xj, /x2, /xr disjoint also.
\342\231\246
in general is not commutative, it is readily seen commutative. Since the orbits of a permutation is multiplication cycles the representation none of are unique, of a permutation as a product of disjoint cycles, which is the identity is unique up to the order of the factors. permutation, While
that
permutation
multiplication
of disjoint
Part
90
9.9
II
Example
Permutations, Cosets,and DirectProducts
the
Consider
permutation
12
6 Let us
it as a
write
(1, 6).
the cycle
(2, 5, 3).
12 of
may
7 through 9.10
Example
4
3
2 4
5
6
3
1
5
6
3
1
1, giving
to 2, or
(1,6)(2,5,3).
so
is commutative,
cycles
disjoint
of the
order
the
factors (1,6)
and
\342\226\2
permutations
multiplying
practice not be disjoint.
or may
5
4
important.
should
You
4
First, 1 is moved to 6 and then 6 to product of disjoint cycles. to 3, which 2 is moved to 5, which is moved is moved care of all elements but 4, which is left fixed. Thus
6 Multiplication (2, 5, 3) is not
3 2
Then
takes
This
5
give an example
We
in
and
where the cycles practice in Exercises
notation
cyclic
further
provide
9.
Consider
the
(1,4,5,6)
cycles
and (2,1,5)
in
1,5) =
(1,4,5,6)(2,
we find that
S6. Multiplying,
3 4
12
5
3
4
6
5
6
2
1
and
Neither of thesepermutations
It
is a
12
reordering of the of repeated interchange positionsof pairs of
by
2 6
5
cycle.
that every
reasonable
seems
6
5
4
3
13
4
Permutations
Odd
and
Even
=
1,5)(1,4,5,6)
(2,
2,..., n can We discuss this
1,
sequence
numbers.
be achieved a bit
more
formally.
9.11 Definition
A
cycle
of length
Thus the
other.
2 is a transposition.
elements but
a transposition leaves all A computation shows that
cycle
any
two
fixed,
and maps
\342\200\242 \342\226\240 = \342\200\242\342\200\242-,\302\253\342\200\236) (ai,a\342\200\236)(ai,a\342\200\236_i)- (a\\, a3)(ai,
{a\\,a2,
Therefore
\342\226\2
is a product
of transpositions. We
then
have
each of these onto
a2). the
following
as a
corollary to Theorem9.8.
9.12
Corollary
Any
permutation
Naively, by
successively
this
of a finite set
corollary interchanging
of at
least
two
elements
just states that any rearrangement pairs of them.
is a product of transpositions. of n objects
canbe achieved
Section9 9.13Example
(1, 6) 9.14
Example
the
Following
In Sn for
>
n
prior to
remarks
(2, 3) (2,5) of
Orbits,Cycles, the
the
and
we see that
corollary,
91
Groups
Alternating
(1, 6)
(2, 5, 3) is the
product \342\226\262
transpositions.
2, the identity
permutation is
the
product
(1, 2)
(1, 2)
of transpositions. \342\226\262
of a finite set with at least two elements is a We have seen that every permutation not be disjoint, and a representation product of transpositions. The transpositions may of the permutation in this way is not For example, we can always insert at the unique. (1,2) twice, because (1, 2) (1,2) is the identity beginning the transposition permutation. to represent a given permutation used What is true is that the number of transpositions be odd. This is an important fact. We will give must either always be even or always of determinants from linear algebra. The second two The first uses a property proofs. M. Bloom. orbits and was suggested by David involves counting
9.15 Theorem
Proof 1 (From linear algebra)
Proof 2 (Counting orbits)
No permutation
in
transpositions and
as a product
We work
can be S\342\200\236
of
an
expressed
in Section 8 that of the permutations
remarked with
~
SA n
both as a product of an of transpositions.
even
number
of
number
odd
SB if A and of the n
rows
x
n
B have
the
matrix
identity
same
We cardinality. than of the
rather /\342\200\236,
matrix 1. Interchanging has determinant numbers 1, 2, ..., n. Theidentity any two rows of the of a square matrix the sign determinant. Let C be a matrix obtained by a changes could be obtained from /\342\200\236 by both an even number permutation a of the rows of I\342\200\236.lfC to be both 1 and of rows, its determinant would have an odd number of transpositions \342\200\224 a Thus of an even and is impossible. cannot be expressed both as a product 1, which of number and an odd number transpositions.
Let a e Sn of a and of
and
xa
Case I
= (z, j) differ by 1.
let r
Supposei and disjoint
contains
product
cycles,
be a j
are in the first
i, symbolized
of these two
the symbols
We claim that
the
number
of orbits
different orbits of a. Write a as a product of of which contains j and the second of which in Fig. 9.16. We may write by the two circles
cycles
(b, j, where
in Sn.
transposition
symbolically
as
x, x, x)(a,i, x, x)
x denote possibleother
9.16 Figure
elements
in these
orbits.
the
Part
92
II
Cosets, and Direct Products
Permutations,
the
Computing
of the
product
three
first
=
in xa
cycles
(2,
we
j)a,
obtain
j, x, x, x)(a,i, x, x)
(2, j)(b,
=
x, x,
(a, j,
x,b,i, x, x).
have been joined to form just one in xa as the computation Exercise 28 asks us to repeat symbolized Fig. to show that the same thing happens if either one or both of i and j of their orbit in a. should be only element The
2 orbits
original
9.16.
in
Case
Suppose 2 and j are in the same orbit of a. We can with the first cycle of the product of disjoint cycles
II
x, x,
(a, i,
x, b, j, x, x) the product of
by the circle in Fig. 9.17. Computing we obtain two cycles in xa = (2, j)a,
shown
symbolically first
the
(2, j)(a,
The original
x, b,
x,
x,
2,
single
j, x,
x) = (a,j, x, x)(b,
has been
orbit
a as a
write
then
form
split
two
into
x, x).
i, x,
as symbolized
in
9.17.
Fig.
of of xa differs from the number n is 1has each element because orbits, identity permutation the number of orbits of a member of its orbit. Now the given only permutation but not both. Thus it is a e Sn differs from n by either an even or an odd number, to write impossible
We have orbits of a by
the
orbits
1. The
a = where
of
the number
that
shown
are 7\302\276
\342\226\240 \342\226\240 \342\226\240
xmL
X\\X2xi
in two ways,
transpositions
once with
and once
m even
m
with
odd. 9.18
Definition
\342\231
of a
A permutation
as a
of an
product
set
finite
or odd
is even
even number of
accordingto or
transpositions
it can
whether
the product
of
an
be expressed
odd
number
transpositions, respectively. 9.19
Example
\342\22
The identity permutation 1 in Sn is an even If n = 1 so that we cannot form this product,
the
(1,
permutation
4, 5,
6) (2,
1,5) in
(1, 4, 5, 6)(2,1,5) has
which
The
what
S6 can =
we define 1 to
be written
(1, 6)(1,
is an
since
we have 1 =
be even.On
the
(1,2)(1, 2). other
hand,
as
5)(1, 4)(2, 5)(2, 1)
odd permutation.
\342\226
n >
2,
the
number
of even
permutations in Sn equally and both numbers
is the
same as the
number
is, permutation; !)/2. To show this, be the set of even permutations in S\342\200\236 and let B\342\200\236 be the set of odd permutations A\342\200\236 n > 2. We proceedto define function from A\342\200\236 onto a one-to-one Bn. This is exactly is needed to show that A\342\200\236 and Bn have the same number of elements.
of odd
for
for
this
permutation
Groups
Alternating
We claim that let
so
five transpositions,
of
that
is split S\342\200\236
are (n
Let x be any
fixed
in
transposition
(1,2). We define a
x =
that
Orbits, Cycles,and the
9
Section
since n >
it exists S\342\200\236;
93
Groups
Alternating
2. We
may
as well
suppose
function
->\342\200\242 Bn
XT : An
by
XT{a) =
(1,
e
a
thatis,
into (1, 2)<7by
is mapped A\342\200\236
can
2)cT
be
transpositions,
that since
XT. Observe
as a product of a (1 + indeed in Bn. If for a and
expressed is
so (1, 2)<7
xa,
even
a is even,the
permutation
number, of
or odd
number),
it is true /x in A\342\200\236
that
X
T{a)
= Ar(/x),
then
(1,2)ct = and
group, we have
Sn is a
since
a
= /x. Thus
T =
so if
p
e
(1.2)//,
(1,2)
XT
is a
one-to-one
function.
Finally,
= T_1,
then B\342\200\236,
and
= x{x~lp)
A.r(T\"V)
= p.
is the same as the number the number of elements in A\342\200\236 in onto B\342\200\236. Hence between the elements of the sets. there is a one-to-onecorrespondence is again even. Also sincen > 2, S\342\200\236 Note that the product of two even permutations = is an and i even has the transposition 2) (1, 2)(1, permutation. (1,2) Finally, note that the product of the same transpositions if a is expressed as a product of transpositions, if a is an even permutation, a^1 taken in just the opposite orderis a _1.Thus must also be even. Referring statement. to Theorem 5.14, we see that we have proved the following Thus
Ar is
Bn since
9.20
Theorem
If n >
2, then
of order nl/2 9.21
Definition
The subgroup group
on n A\342\200\236
of all even
collection
the of
the
of
Sn
permutations
symmetric
group
S\342\200\236.
consisting
of the
even
letters.
permutations
of {1,2,3,\342\200\242\342\226\240-,\302\253} forms a
of
n letters
is
the
subgroup
alternating \342\226\240
groups. Cayley's theorem shows that every finite for n = \\G\\. It can be shown subgroup of S\342\200\236 structurally for solution of polynomial that there are no formulas involving radicals just equations of degree n for n > 5. This fact is actually due to the structure of A\342\200\236, as that surprising seem! may Both
and are very S\342\200\236 A\342\200\236
group G is
important
identical
to some
Part II
94
Direct
and
Cosets,
Permutations,
Products
9
\342\226\240 EXERCISES
Computations Exercises
In
1 through
13
5 3-
:Z
9.
of the given
permutation.
6.
the indicated
9, compute
of cycles
product
: Z
-^ Z
where
+
= n
a(n)
that are permutations
l
\342\200\224 3
3,4, 5,
of {1,2,
6, 7, 8}.
8. (1, 3, 2, 7)(4,8, 6)
8)(2, 5,7)
5)(7,
17
3
8
4.CT:Z-.ZwhereCT(n)= n
+ 2
n
4
2
6
\\5
A)
a(n) =
where
8
/1234567
6 8 7)
7 through
In Exercises
all orbits
find 6\\
4
1
5
-\302\273 Z
7. (1,4,
5
6 2
3
(2
5. a
4
3
12
6,
(1,2)(4,7,8)(2,1)(7,2,8,1,5)
10 through
In Exercises
1ft 10-
3
2
/1
(,8 2 6 ./12345678 * 14
^3
13. Recall power
12, express
the permutation
of {1, 2, 3, 4, 5, 6, 7, 8} as
of disjoint
a product
cycles,
and
of transpositions.
as a product
then
4 5 6 7 4
8\\
5
3
7
7
2 5 8 6
that
element
of a
is
the
a of a
l)
identity element the group S&.
G with
group
Consider
identity.
What is the order of the cycle (1,4, 5, 7)? b. State a theorem by part (a). suggested c. What is the order of a = (4,5)(2,3,7)? of
4
/12
3
V
6 4 1
5
6 7 8
8
2
e has order r > 0 if
= e
ar
7
5
and no smaller positive
a.
d.
the
Find into
e.
a theorem
common
In Exercises
14. n
=
looking at its
12 by
words
decomposition
are looking
you
for are least
multiple.]
14 through
the maximum
18, find
15. n =
5
19. Figure
(1,4)(3, 5,7, 8)? of the permutations given in Exercises 10 through of disjoint cycles. suggested by parts (c) and (d). [Hint: The important of each
order
a product
State
x =
9.22 shows
16. n
6
=
labeling
group with
of Sn for the
given value of n.
10
18. n
element
17. n =
7
digraph for the alternating the other nine vertices
a Cayley
(1,2)(3,4)}.Continue
possible order for an
the
A4 using
the
elements
of A4,
generating
=
15
set S
= {(1,2, 3),
expressed as a
product
of
disjoint cycles.
Concepts In
Exercises
20 through 22, correct the definition of the it is in a form acceptablefor publication.
italicized term
without
reference
to the
text, if correction
is needed, so that
a of a
20.
For a permutation
21.
A cycle is a permutation
22. Thealternating
group
set A, having
is the
an orbit
of a is a
only one
group of all
nonempty
orbit. even
permutations.
minimal
subset of
A
that
is mapped
onto itself by
a.
95
Exercises
9
Section
(1,2,3)
(1,2)(3,
4) \342\226\240
9.22
each of the
23. Mark
a.
is a cycle.
permutation
Every
b. Every cycle c. The definition
false.
true or
following
Figure
is a permutation. of
and odd
even
have
could
permutations
been given
well
equally
before
Theorem9.15.
subgroup H
d. Every nontrivial
e. A 5 has
f. g. h.
SV
is
cyclic for a commutative
of the permutations
Which
n >
any
a transposition.
contains
to the
subgroup
of all
those elements
subgroup
of all
those elements
in S$
in S3
1.
group.
to the
is isomorphic
i. SV is isomorphic j. Theoddpermutations 24.
odd permutation
some
containing
120 elements.
is not S\342\200\236 A3
of Sg
8.7 areeven
that
leave
the number 8 fixed.
that
leave
the number 5 fixed.
of S%.
form a subgroup
of Example
of S% of S%
permutations?
Give the
table
for
the
alternating
group
A3.
Proof Synopsis
25. Give 26.
a one-sentence
Give a two-sentence
synopsis of synopsis
of
9.15. Proof 1 of Theorem Proof 2 of Theorem 9.15.
Theory
27.
Prove
a.
the following
about
> 3.
if n S\342\200\236
permutation
in
can be S\342\200\236
b. Every
permutation
in
Sn that
c. Every
odd permutation
Every
as a product
of 2n
written as a product
is not a
in S\342\200\236 can be
written as a product
+ 8 transpositions.
of at
cycle can be written
most as a
of 2n
n\342\200\224\\ transpositions.
product of
+ 3 transpositions,
at
most
n and
\342\200\224 2
transpositions.
every even
permutation
Part II
28. a.
a figure
Draw
like Fig.
9.16 to
part (a) if o(j) = Show that for every subgroup them are even.
b. Repeat 29.
30. Let
32.
be
Let A
an
Consider a product
> 2, either
34.
Show
35.
Following
of a
some
2 and
G be a g e G, is a
group
and let
to Exercise
38.
to Exercise
Referring
immediate
in
that
Sn
if a
A\"
a. If A
(a) ^
is a
half
exactly
finite set,
of
how
many
a e
SA
that the
such
number of elements moved
by
a (see
a e
SA
(see Exercise
move
that
30) at
most
50 elements
of
A.
fixed odd
cycle
that every
Show
permutation.
odd
in Sn
permutation
is
A\342\200\236.
is a cycle.
the following
34, complete
n,
of length
a be a fixed element set G.
36, show 49 of of one
corollary
a e
or
are even
H
in
also a cycle if
cr'' is
then
Show
of G.
the
that
map
and
only
: G
Xa
involving n and
a condition
with
if...
-+ G,
given
Afl(g) =
by
ag for
of the
permutation
37. Referring
(1, 2,
of all
let cr be a
permutation
If a is a
39. Show
set
cycle ofoddlength, then a2 the line of thought opened by Exercise the resulting statement is a theorem:
36. Let
the
of S\342\200\236.
that if a is a
r so that
then
Why?
SA1
fixed n >
and
= i,
a(i)
n?
of all
set
the
the permutations
all
moves
\"a
is a subgroup
K be the
set. Let
infinite
for a S\342\200\236
H be that H
Show
subgroup of
Is K a
33.
of Sn for n
H
set. Let
and
also.
permutation of a set A. We shall say are moved by a cycle cr e 5/1 of length
Let A be an infinite Exercise 30) is finite.
orbits of a
j are in different of a.
and
i
number of orbits
than the
less
a be a
elements 31.
j
that if
illustrate
is one
of (i, j)o
of orbits
number
Products
Direct
and
Cosets,
Permutations,
that
H
=
{Afl
|
a e
G} is a
the group
of So,
subgroup
Section 8, show that H of Exercise 37 is transitive theorems in Section 4.]
on
of all
the set
of G.
permutations
G.
[Hint:
This
is an
of the
\342\200\242 \342\200\242 \342\200\242 [Hint: Show that as r varies, (1,2, 3, \342\200\242 \342\200\242, n)}. 2) nj{\\, by {(1,2), (1,2, 3, \342\200\242, - 1, \342\200\242 \342\200\242 Then show that all the \342\200\242, (n, 1). (1,2), (2,3),(3,4), (n n), gives any transpositions is a product of someof these transpositions and use Corollary 9.12]
is generated
\342\200\242 \342\200\242, 3, \342\200\242 n)\"~r
transposition
section
10
Cosets and the You
may
noticed
have
to be a divisor of the
Theoremof Lagrange that the
a partition of will have
exhibiting
order of a
of G.
order
G into
This is
cells,
H of a
subgroup the
theorem
all having
finite
the same
G seems always We shall prove it by H. Thus if there are r
group
of Lagrange.
size as
such cells, we
r(order from which
the
cosetsof
and they
H,
of H)
=
(order
of
G)
The cells in the partition will be called immediately. in their own right. In Section 14, we will see that if then each coset can be regardedas an element of a group in this section to give some indication of such coset groups
follows
theorem
are
important
H satisfies a certain property, in a very natural way. We a feel for help you develop
the
topic.
Cosets
Let H two
be a subgroup
partitions
of G
of by
a group
defining
G,
which
may
two equivalence
be of
finite
or infinite
relations, ~L
and
order. We
~8onG.
exhibil
10.1 Theorem
Let H be a
be defined
Let ~jj
Then ~l Proof
show
We
Let
of G.
subgroup
Cosets and the Theorem
10
Section
~L be
relation
the
97
of Lagrange
defined on G by
a ~L
b
if and
only
if
a~lb
a ~#
b
if
only
if
ab~x e
e H.
by
and
~j?
are both
that
~l
is
When reading the a subgroup of G.
an
and leave relation, how we must constantly
notice
Leta e G.Then a ~L a.
Reflexive
Symmetric
Let a
and
= e
a~la
~-Lb
and
is a subgroup,
fe
c. Then
~l
H is a
Since His a
so fc_1a is in
H and
a~lb e is
in
H,
H
that
and
fe
is
Thus
subgroup.
(a~1b)~l
subgroup,
H
26.
Exercise
to ~\302\253
of the fact
use
e H since
= a~lc
{a~xb){b^c)
for
proof make
e H.
= b~la,
{a~xb)~l
the
and e
fe.Thena-1^
~L
Supposes
is in H Transitive
H.
equivalence relations on G.
equivalence
proof,
and
\302\253\342\200\242
~l
fe_1c
e H.
so a
~l c.
Since H \342\231\246
a partition of G, as described like. Suppose a e G. The partition a consists of all x e G such that a ~~l x, which cell containing means all x e G such that if a~xx e H. Now a~lx e H if and only if a-1* = h for some h & H, or equivalently, = the cell which a/jforsome/j a is e e H], andonlyif;t containing {ah \\h relation we denote by aH. If we go through the same reasoning for the equivalence ~j? defined by H, we find the cell in this partition containing a e G is Ha \\h e H}. {ha Since G need not be abelian, we have no reason to expectaH and Ha to be the same subsetof G. We give a formal definition. relation ~l in Theorem equivalence 0.22. Let's see what the cells in
The
in
Theorem
10.1 defines
this
look
//.Therefore
=
10.2
Definition
be a
Let H coset
of H
Example
Solution
the
Exhibit
containing
G. The
the subset
subset aH = {ah Ha =
{ha
|
e H]
h
H] is
| /j e
the
of G is right
coset
the
left
of H \342\226\240
cosets
left
Our notation m
a group
a, while
a.
containing
10.3
of
subgroup
= 0,
here
and the
right cosets of the so
is additive,
the
left
of 3Z
coset
3Z of
subgroup
containing
Z. m
is m
+ 3Z.
Taking
we see that 3Z=
-3,0,3,6,9,---}
{---,-9,-6,
is itself one of its left cosets, the coset containing 0. To find another left we select coset, an element of Z not in 3Z, say 1, and find the left coset containing it. We have 1+ These
two left
of them.
The
cosets, 3Z
left
coset
3Z = and
1 +
containing
2+
\342\200\242 \342\200\242, -8, {\342\200\242
-5,
3Z, do 2 is
not
\342\200\242 -2, 1, 4, 7,10, \342\200\242 \342\200\242}.
yet
exhaust
Z. For
example, 2 is in
\342\200\242 \342\200\242 3Z = {\342\200\242 \342\200\242, -7, -4, -1, 2, 5, 8, 11, \342\200\242 \342\200\242}.
neither
98
Part
II
Cosets, and Direct Products
Permutations,
It
that these three left
is clear
of Z
partition
Since Z is so We
the
partition
two things
observe
cosets we have
For
a subgroup
and
the
the
the
constitute
coset 3Z
right
+
m
the same,
are
\342\22
from Example10.3.
H of an abelian group G, the into right cosets are the same.
partition
Z, so they
do exhaust
found
of 3Z. the left coset m + 3Z and abelian, of Z into right is the same. cosets cosets
left
into
of
partition
G into
left cosets of H
relation ~j? 0.17 and 0.20, we see that the equivalence Also, looking back at Examples n. Recall for the subgroup that nZ of Z is the same as the relation of congruencemodulo is h = k (mod n) in Z if h \342\200\224 as saying that h + (\342\200\224k) k is divisible by n. This is the same the partition of in nZ, which is relation ~# of Theorem Thus 10.1 in additive notation. of Z into residue classesmodulo n. For that reason, Z into cosets of nZ is the partition as cosets modulo nL. Note that we do not have we often refer to the cells of this partition the this to specify or since are same for abelian cosets they group Z. right left
10.4 Example
Solution
The group
is {0,
coset
One
abelian. Find
3} itself.
2 + {0, 3} = the cosets. 2 is
containing are
Z6 is
all
the
The coset containing
{2,5}.
Since
point out a fascinating to Example 10.4, Table
order they
listed in the
according to their
shownin
we
obtain
3}, {1,
4},
+ {0, 3} = {1, {2, 5} exhaust
4}.
The
we will
that
thing
10.5gives in the
appear
the
develop
binary
in
detail
the
operation
cosets {0, 3}, {1,4},(2,
5}.
with
We shaded
elements
the table
we denote thesecosetsby LT(light), and DK(dark) MD(medium), Table 10.5 then defines a binary operation on these shadings, as 10.6. Note that if we replace LTby 0,MD by 1, and DK by 2 in Table 10.6, of shadings forms a group! We will see in table for Z3. Thus the table
10.6Table
10.5 Table +6
0
3
0
0
3
3
3
0
1
1
4
4
4
1
2
2
4
\"1
1
4
2
5
4
1
5
2
^
0
1
1
5
0
3
4
1
'
4
LT
MD
DK
LT
MD
DK
MD
MD DK
LT
DK
DK
MD
LT
\342\200\224
5
Z6, these
14. Referring
in Section
for 2,(, but
shading.
Table
coset
all of
and
{0, 3}.
cosets.
these
Suppose to according
{0,
1 is 1
H =
subgroup
\342\226
We back
cosets of the
of X(, into
partition
.1
\342\226\240
5
2~
0
0
1
LT
Cosetsand
Section 10
the
for a
14 that
Section
group
table
abelian
of an
partition
to
according
group into cosets
99
of Lagrange
Theorem
in the
elements
the
the
cosets
of a
always
gives
subgroup, reordering rise to such a coset
group.
10.7Example
Solution
Table
shows Table 8.8 for
10.8 again
(mi) = into right
the
subgroup
the
partition
For
the partition
into
left
=
P\\H
p2H = into
H =
partition into left example, the left coset
the
partitions
{po,Ml), {P1P0, M1M1} = {p2po,
{pi, /x3},
= {P2,
M2M1}
M2l-
cosets is
right
The
group S3 on three letters. Let H be of S3 into left cosets of H, and
symmetric
we have
cosets,
H =
The partition
the
{p0, /xj} of S3.Find cosets of H.
= {Pi, M2},
Hpi
=
{poPu MiMi)
Hpl
=
{A)P2, M1M2} =
cosets
of H
containing
{pi, Mil- This doesnot
{Mo,Mil,
surprise
is
us since
{p\\, /13},
the
group
M3l-
the partition
from
different
is
p\\
{/02,
while S3 is
the
into coset
right
cosets.
right
containing
not abelian.
For
px is \342\226\262
in S3. The 10.7, Table 10.9gives permutation multiplication order they appear in the left cosets {po,Mil, {p\\, /13}, {/\302\276. M2} and dark according found in that example. Again, we have shaded the table light, medium, between this table and to the coset to which the element belongs.Note the difference Table 10.5. This time, the body of the table does not split up into 2x2 blocks opposite under the shaded cosets at the left and the top, as in Table and 10.5 and we don't get a cosetgroup. The product of a light element and a dark one may be either dark or medium. = Table 10.8is shaded according to the two left cosets of the subgroup {p\\) \302\260f the two is These are also even not abelian. S3 right cosets, though {po, p\\, /\302\276} ^3Referring
elementsare
to Example in the
listed
10.9Table
10.8 Table PO
Pi
Pi
Mi
M2
M3
Po
Po
Pi
Pi
Mi
M2
M3
Pi
Pi
Pi
Po
M3
Mi
Pl
Pi
Po
Pi
M2
>1
Ml
M2
M.i
M2
M2
M3
Ms
Ms
Mr
Po
Mi
Mi
M3
Po
Po
Mi
Mi
Ms
Mz
Mi
Mi
Po
^3
P\\
MJ
Mi
Pi
Pi
m
A\"
11 '
PO
Pi
P2
M3
M3
Mi
Pi
Po
Pi
M2
P\\
P2
Po
mi
*1
jjjl
iigsij
Mi
Po
Ml
fli
Po
Mi
Ml
M3
M3
Pi
Mi
Po
Part
II
Permutations, Cosets,and Direct Products
see in
will
We
is clear that Section 14 that
10.8 it
Table
From
we the
do have left cosets
a coset group of a subgroup
this case.
to Z2 in
isomorphic
group G give of H is the same
of a
H
to
rise
the partition of G into left cosets as the cosets of H. In such a case, we may simply speak of the cosets of H, partition into right left or right. We discusscosetgroups in detail in Section 14, but omitting the adjective them if think it will easier for to understand then we be you you experiment a bit with a coset
when
precisely
group
them
now. Some
The
Theorem
of the
have
onto a
will show
of such (See
that
coset
left
gH
Definition
same number of elementsas H. If for equality of the size definition
has the
the existence
is infinite,
H
as the
is taken
a map
experimentation.
coset group G. We claim that every left coset and every right of elements as H. We show this by exhibiting a one-to-one this order, gH of H for a fixed elementg of G. If H is of finite
of a
number
same
the
of H
map
such
of Lagrange
Let H bea subgroup of H
section are designedfor
in this
exercises
of H
and
the
of gH.
size
0.13.)
one. Let
choice
Our
coset
Every
summarize as follows:
27.) We
Exercise
(See
(left or
of a
right)
subgroup
H of a group
G has
of
the same number
elementsas H.
We
10.10
Theorem
Proof
now
can
prove the
theorem of Lagrange.
of Lagrange) (Theorem H is a divisor of the order
of
be
that this elegant and important and the number of elementsin
We continue somethingl regarded as a counting Corollary
Proof
finite group
G. Then the
Every
Let G be of the cyclic
prime
subgroup
to
derive
order {a)
theorem each
coset.
consequences
comes from the simple Never underestimate results
of Theorem
shows partition
\342\231
of
counting
10.10,which
count
that should
be
theorem.
order is
of prime
group
of
order
the order
Note
cosets
10.11
of a
of G, and let H have order m. The preceding boxed statement H also has m elements. Let r be the number of cells in the coset of every G into left cosets of H. Then n = rm, so m is indeed a divisor of n.
Let n that
Let H bea subgroup of G.
cyclic.
p, and let of G
a be an
generated
by
element a has
of G different from at least two elements,
the
Then
identity.
a and
e.
But
by
Theorem 10.10,the m =
p
and
{a)
every
group
a theorem
underestimate
examination
favorite
10.12 Theorem
The order of
Proof
Remembering
from
follow easily
result
elegant
counts
that
have
must
\342\231\246
is
to Zp, we see that is only there order Now doesn't this p. prime
isomorphic
of a
given
a counting theorem? Never the preceding corollaryis
of Lagrange,
theorem
Proving
something.
a
question.
of a
element
an
the
p. Thus we
the prime
divide
must
cyclic group of order p structure, up to isomorphism,
Since
one
m > 2 of (a) so G is cyclic.
order
= G,
101
Exercises
10
Section
finite
of the
the order
divides
group
group.
is the same as the order of the cyclic subgroup that the order of an element that theorem the we see this follows directly from Theorem 10.10. element, by
generated
\342\231\246
10.13
Let H
Definition
be a subgroup
(G :7/)
a group
of
G. The number
of
(G : H)
The
index
(G : H)
is finite
defined
just
(G : H)
and
=
may be
every
77 and
Suppose
G
(77: K) and
(G
K are subgroups of a are both finite. Then
: 77)
G
group
(G : K)
such
is finite,
finite,
is the
then
H contains
coset of well
is
If G
or infinite.
finite
since
\\G\\/\\H\\,
35 shows the index (G : 77) could be equally cosets of 77 in G. We state a basictheorem concerning to the exercises (see Exercise38). the proof Theorem
in
index \342\226\240
Exercise
10.14
of H
left cosets
HinG.
of
obviously
\\H\\
elements.
as the number
defined
of right leave
indices
of subgroups, and
that K
<
and
77
(G : K)
G, and suppose = (G : 77)(77:K).
<
Theorem shows that if there is a subgroup 77 of a finite 10.10 G, then the group order of 77 divides the order of G. Is the converse of order true? That is, if G is a group a subgroup of order ml We will see in the next section n, andm divides n, is there always that this is true for abelian of groups. However, A4 can be shown to have no subgroup order 6, which gives a counterexample for nonabelian groups.
10
\342\226\240 EXERCISES
Computations
1.
Find
all cosets
of the
subgroup
4Z
of Z.
2.
Find
all cosets
of the
subgroup
4Z
of 2Z.
3.
Find
all cosets
of the
subgroup
(2) of
4.
Find
all cosets
of the
subgroup
(4) of Z^.
5.
Find
all cosets
of the
subgroup
(18)
6.
Find
all left
7. Repeat
the
cosets of preceding
the
subgroup
exercise,
but
Z12.
of Z36. {po, find
M2}
the
of the
right
group D4
cosets this
given
by Table
time. Are
they
8.12. the same
as
the
left
cosets?
Part II
102
Permutations, Cosets,and
Products
Direct
8. Rewrite Table 8.12in the order exhibited by the left cosets order 4? If so, is it isomorphic to Z4 or to the Klein 4-group
9. Repeat Exercise 6 for
the
10. Repeat the preceding 11. Rewrite Table 8.12in 4? If so, is
order
12.
Find
13. 14.
of (3) in
Find the
index
of
Find the
index
of (/x2)
the
a = (1,
Let (1 =
order
the
the
the
find
to the
group
S3, using
in the
group
D4
2, 5, 4)(2,3)
in S5.
(1,2,4, 5)(3,
6) in S6.
seem to get a cosetgroup
of
in Exercise
cosets
left
Klein
as the
the same
they
9. Do you
to get
seem
coset?
left
a coset group
of
VI
4-group
Z24-
group
in the
(/^)
time. Are
cosets this
right
exhibited by the to Z4 or
isomorphic
index
15. Let 16.
it
but
exercise,
6. Do you
V?
of -D4.
{po, Pi\\
subgroup
Exercise
in
of Example
the notation in Table
given
8.12
Find the index
of (a) in
Find the
of
index
10.7
S5.
(/x) in S6.
Concepts In
Exercises
needed, so
that
of the italicized 17 and 18, correct the definition for publication. it is in a form acceptable
17. Let G be a
G be a group Mark each of the
18. Let
and
19.
following
a.
H c
and let
group
let H
G. The left coset < G. The index of
true or
of H H in
term
a is
containing
G is the
aH = {ah of right
number
to the
reference
without
\\h
text,
if
e H}.
cosets of H in
G.
false.
group has left cosets. of a subgroup of a finite divides the order of the group. b. The number of left cosets group c. Every group of prime order is abelian. . d. One cannot have left cosets of a finite subgroup of an infinite group. e. A subgroup of a group is a left coset of itself. f. Only subgroups of finite groups can have left cosets. for n > 1. g. An is of index 2 in S\342\200\236 of Lagrange is a nice result. . h. The theorem the order of the group. i. Every finite group contains an element of every order that divides order that divides the order of the j. Every finite cyclic group contains an element of every In Exercises
20
it is
impossible.
20.
A
of every
subgroup
Every
give an
24,
through
abelian
example of the
desired
left cosets
and group
subgroup
if
of an
21. A
subgroup
of a
one cell
22.
A
subgroup
of
group into
23. A
subgroup
of
subgroup
of a
A
group
G whose
and
right
give different partitions
cosets
group G whose left cosets give a partition of G into just a group of order 6 whose of the left cosets give a partition a group of order 6 whoseleft cosets give a partition of the group of order 6 whoseleft
cosets
give a partition
If impossible,
possible.
subgroup
24.
is
correction
of the
group.
say
why
of G
6 cells
group
into 12
group
into 4
cells
cells
Proof Synopsis
25. Give
synopsis of
a one-sentence
the
proof
of Theorem
10.10.
Theory
26. Prove
that
the relation
27. Let H bea subgroup is one to one and
~/? of Theorem 10.1is an of a
is onto
group G and Hg.
let g
equivalence
relation.
e G. Define a one-to-one
map
of H
onto Hg. Prove that
your
map
28.
a subgroup of a group G such that g same as the right coset Hg.
H be
Let
gH is the
29. Let
partition
a group
subgroup of
be a
H
into
Exercise28.) be a
H
Let
subgroup of
G.
of H,
cosets
right
a group
then
G
that if the e H for all
Prove g~lhg
G. In
a, b e
and let
e H for
lhg
e G and
all g
G and
g e
30 through
Exercises
cosets of
H. (Note
/j e
all
Show that every left coset
all /j e ff.
G into left
of
partition
103
Exercises
Section 10
H is the
33 prove the
as the
same
is the
this
that
converse of or give
statement
a
counterexample.
30. If
aH
= bH,
Ha
= Hb,
= Hb. then b e Ha.
=
then
31.
If
32.
If aH
then Ha
fcff,
33. If aH = bH, then 34. Let G be a group
= Hb'1.
Ha~l
= b2H.
a2ff
pq, where
of order
p
prime numbers.
q are
and
that
Show
proper subgroup of
every
G is
cyclic.
35. Show
same number
that there are the a one-to-one map of the
obvious 36.
for finite
counting
by
groups.
Exercise 29 of Section4 showed of Lagrange, show the theorem
cosets
Your
proof
H of a group G; that is, exhibit of right cosets. (Note that this result is
subgroup
collection
onto
the
must
hold for
group.)
any
group of evenorder 2n contains an abelian group of order 2n
every finite if n is odd, then
that that
cosets of a
left as right
of
of left
collection
of order 2. Using one element
an element contains
precisely
of order 2.
37. Show
a group
that
at least
with
two elements
with
but
no proper nontrivial
subgroups
be
must
and of
finite
prime order. \342\200\242 38. Prove Theorem10.14[Hint: Let {atH | i \342\200\224 1, \342\200\242 \342\200\242, r} be the collection of distinct = \342\200\242 \342\200\242 the of be collection distinct left cosets of K in H. Show that 1, \342\200\242, s} {bjK | j
{(aibj)K\\i = is
the
39.
Show
that
40.
Show
that
Show
that
41.
x such
42. Show
H
of index
of K in
2
in
in
G and
coset
of H.
G.] group G, then
a finite
group G with identity e has finite order n, then a'1 = Z of the additive group of every left coset of the subgroup 0 < x < 1.
of H is alsoa right
left coset
every
e for
all a
e G.
real numbers
contains
exactly
one
element
sine assigns the same value to each elementof any fixed left coset of the subgroup (2jt) real numbers. (Thus sine induces a well-defined function on the set of cosets; the the function on a cosetis obtained we choose an element x of the coset and compute sin x.) when the function
and K k e
some
is a subgroup
cosets
left
of H
cosets
= l,---,s}
1,---,rj
if a
additive
of
value
43. Let
if H
that
that
of the
of distinct
collection
left
M of
group
be
of a
subgroups
group G. Define
~ on G by
a ~ b if
and only if a
=
hbk
for
some
h e
H
and
K.
a. Prove that b. Describe
~ is an on G. equivalence relation the elements in the equivalence class containing a
e G. (These
equivalence
classes
are called
double cosets.)
44. Let
SA be
the group
of all
a. Show that {a e SA d ^ c be another
b. Let
\\ a{c)
permutations
= c}
particular
of the
set
is a subgroup element
of A.
A,
let c
and
be one particular
element
Sc.c ofSA.
Is
Sc_d
=
{a e SA I a (c) =
d} a subgroup
not?
c. Characterize
the set Sc,d
of part
(b) in
of A.
terms of the
subgroup
Scx of
part (a).
of SA ?
Why or why
Part II
104
45.
Show
are all
that
a finite
the
subgroups
cyclic group it has.
46. TheEuler phi-function less
than or
is defined
equal to
exactly one subgroup
of each
order d
integers n by ip(n) = s, where s prime to n. Use Exercise 45 to show
for positive
are relatively
that
n
n has
of order
Products
Direct
and
Cosets,
Permutations,
is the
that these
n, and
dividing
number of positive
integers
that
d\\n
Note that the number of generators sum being taken over all positive integers d dividing n. [Hint:
of Zj
the
section
11
is
equation G must
Groups Direct Products and Finitely GeneratedAbelian
Products
Direct
Let us take a
moment to review
we have
our
stockpile
present
of groups.
Starting
with
finite
and the S\342\200\236,
alternating group group the dihedral groups Dn of Section 8, and the that subgroups of these groups exist. Turning to Klein sets of numbers under the usual addition or infinite groups, we have groups consisting of elements as, for example, Z, ffi, and C under addition, and their nonzero multiplication, We the U of numbers of 1 under under multiplication. have group complex magnitude groups,
the
the symmetric Z\342\200\236,
group
cyclic
positive integer n. We also V. Of course we know 4-group
for each A\342\200\236
have
Ec under addition modulo c, groups have the group set A, as Sa of all permutations of an infinite formed from matrices. well as various groups as building blocks One purposeof this section is to show a way to use known groups to form more groups.The Klein 4-group will be recovered in this way from the cyclic this procedure with the cyclic groups gives us a largeclassof abelian groups.Employing to all possible structure for a finite abelian group. include groups that can be shown types Definition We start by generalizing 0.4.
multiplication,
c e ffi+.
where
11.1 Definition
We
by
an),
of the
also
The Cartesian product (\302\253i, 02,---,
to each
is isomorphic
which
where
of
a,- e
sets S,-
for
is the Si,S2,---.Sn \342\226\240 i = 1, 2, \342\226\240 n. The \342\226\240,
set
of
Cartesian
all ordered
product is
n-tuples
denoted
either
Si x
S2 x
\342\200\242 \342\200\242 \342\200\242 x S\342\200\236
or by
n
s,
\342\2
1=1 the Cartesian of an infinite number of sets, but the product more sophisticated and we shall not need it. \342\226\240 be let G\\, G%, \342\226\240 notation for all \342\226\240, Gn groups, and let us use multiplicative the G,- as sets, we can form Y[\"=i ^,- Let us show that Regarding operations. of multiplication make J~[\"=1 Gt into a group by means of a binary operation by
We could definition Now
the
group
we
can
also define
is considerably
components. Note again that group as for 11.2
Theorem
Proof
the
of
the
Abelian
Generated
Finitely
are being
we
of elements
set
and
Products
Direct
11
Section
sloppy
we
when
use the
same
a
for
notation
group.
\342\226\240 \342\226\240be \342\226\240 and Let Gl5 G2, \342\226\240 For (^, a2, \342\226\240 ,G\342\200\236 groups. \342\226\240, a\342\200\236) (b\\, \342\226\240 \342\226\240 \342\226\240 to be the element define {a\\, a2, \342\226\240 \342\226\240, \342\226\240, a\342\200\236)(bi, b2, b\342\200\236) (a\\b\\, the direct product of the groups G;, under Yi\"=i Gi is a group,
Note that
105
Groups
\342\226\240 \342\226\240 in \342\200\242, b\342\200\236) YYi=i
b2,
^/,
\342\226\240 \342\226\240 Then \342\226\240, a\342\200\236b\342\200\236).
a2b2, this
binary
operation.
a,- e G,, bt e G,, and G,- is a group, we have a,-fc,- e G,. Thus the makes binary operation on ]~~[;=i Gt given in the statement of the theorem sense; that is, Yl'-=i Gi is closed under the binary operation. in each The associative law in \\Yl=i Gt is thrown back onto the associative law
definition
since
of the
as follows:
component
{ax,a2,\342\226\240\342\226\240\342\226\240 ,b\342\200\236)(cuc2, \342\200\242\342\200\242\342\200\242,(:\342\200\236)] ,a\342\200\236)[(bi,b2,--\342\226\240\342\226\240 \342\226\240, an)(bicu
\342\200\224
(a\\,a2,
e2,
{e\\,
the
=
(aiihci),
a2(b2c2),
\342\226\240\342\226\240\342\226\240, a\342\200\236(b\342\200\236c\342\200\236))
{{a\\bi)ci,
(a2b2)c2,
\342\226\240\342\226\240\342\226\240, (anb\342\200\236)cn)
=
(aibua2b2, [(\302\2531,02,
\342\226\240 \342\226\240 \342\200\242\342\200\242\342\200\242, \342\226\240, c2, anbn)(cu c\342\200\236) \" - \"
, a\342\200\236)(i>l, fc-2,
\" \342\200\242 \"
, fc\"n)](Cl, C2,
---,0,,).
with multiplication in G,-, then by components, clearly, \342\226\240 is an inverse of {a\\, a2, \342\226\240 in Gi\342\226\240, an) Finally, identity YYI=i Hence is a group. the compute product by components. n?=i Gj
element
identity
\342\226\240 \342\226\240 is \342\226\240, e\342\200\236)an
(aj-1, a^1,
\342\200\242 \342\200\242 \342\200\242, b\342\200\236c\342\200\236)
=
=
If a is
b2c2,
\342\226\240 \342\226\240 \342\226\240, a~l);
\342\231\246
we sometimes use additive G, is commutative, refer to ]~[\"=1 G, as the direct sum of the groups G,. The notation @\"=1 G,- is sometimes used in this case in place of YYi=i Gi, especially with \342\226\240 \342\226\240 \342\226\240 abelian groups with operation +. The direct sum of abeliangroups ,G\342\200\236 G2, Gj, may \342\226\240 \342\200\242 We leave be written G\\ @ G2 @ \342\226\240 to Exercise 46 the proof that a direct product @ G\342\200\236. is again abelian. of abeliangroups \342\200\242 seen It is quickly that if the 5; has r,- elements for i = 1, \342\200\242 \342\200\242, n, then F]\"=1 5,- has \342\200\242 \342\200\242 \342\200\242 for in an n-tuple, there are r\302\261 choices for the first component from rn elements, r\\ri are r2 choices for the next from S2, and so on. Si, and for each of thesethere component In the
11.3 Example
event that the
in
notation
J-[\"=1
Consider the
(1,
0),
(1,
group
1), and
G,
operation of each
and
Z2 x
Z3,
(1, 2). We
which
has
claim
that
2
\342\200\242 3 =
Z2 x
Z3
6 elements, namely is cyclic. It is only
generator. Let us try (1, 1). Here the operations in we do the same in the direct product Z2 x Z3.
Z2
and
Z3 are
(0, 0), (0, 1),(0,2), necessary
written
(1,1) = (1,1)
2(1,1) =
(1,1)
+ (1,1)
= (0,2)
3(1,1) = (1,1)+ (1,1)+ (1,1)= (1,0) 4(1, 1) = 3(1, 1) + (1.1) = (1, 0) + (1, 1) = (0, 1) 5(1,1)= 4(1, 1) + (1, 1) = (0, 1)+ (1,1) = (1, 2) 6(1, 1) = 5(1,1)+ (1,1)= (1, 2) + (1, 1) = (0,0)
to find a
additively,
so
Part II
106
Thus (1, 1) generates all structure of a group given 11.4
Example
x
The
Z2
order,
Consider
the
As our
5.17.
one
cyclic
\342\226
notation
(1,
1) that involve
the
first
and the
gives
has the
adding 1 e Zm
component
so on. For
them
and n.
Zm
x
Z\342\200\236 generated
shown,
the order
Here
(0, 0).
identity
(1, 1) to
are
relatively
cyclic
of (1,
a power
taking
m summands,
after
n
1) as describedby Theorem is the smallest subgroup
by (1,
of this
itself repeatedly.Under
yields 0 only
second component 1 e
of
1.
\342\226
4-group.
theorem:
following
is
n
of
subgroup
cyclic
previous work
will
powerof
and
of m
gcd
Klein
the
if and only if m and and is isomorphic to Zm\342\200\236
is cyclic Z\342\200\236
the
the
to
be isomorphic
must
Z2
illustrate
examples
preceding
ThegroupZm x
x
cyclic. Thus Z2
is not
prime, that is, Proof
x Z3. Sincethere is, up to isomorphism, only we see that Z2 x Z3 is isomorphic to Zg.
of Z2
Consider Z3 x Z3. This is a group ofnine elements. We claim that Z3 x Z3 is not cyclic. Since the addition is by components, and since in Z3 every element addedto itself three the times gives the identity, the same is true in Z3 x Z3. Thus no element can generate could only give the identity after nine successively group, for a generator added to itself shows summands. Wehave found another argument group structure of order 9. A similar that Z2
11.5 Theorem
Products
Direct
and
Cosets,
Permutations,
1) in
addition
our
additive
by components,
2m summands, and
so on,
2n summands, and 0 only after n summands, the number of summands must be a multiple
Z\342\200\236 yields
0 simultaneously, The smallest number that is a multiple of both m and n will be mn if if the gcd of m and n is 1; in this case, (1, 1) generates a cyclic subgroup of and only is cyclic of order mn, which is the order of the whole group. This shows that Zm x Z\342\200\236 to Zm\342\200\236 if m and n are relatively order mn, and hence isomorphic prime. For the converse, suppose that the gcd of m and n is d > 1. The mn/d is divisible we have by both m and n. Consequently, for any (r, s) in Zm x Z\342\200\236, both
m
to yield
(r, s)
*
+ (r, s) +
Hence no element(r, s) in Zm and therefore not isomorphic
x
not cyclic
can be
theorem
arguments. We state this 11.6
Corollary
The for
11.7Example
The
group
i =
Y[l=i
1,
preceding
numbers,
^m, is
\342\200\242 \342\200\242 n are \342\200\242,
the entire
can generate Z\342\200\236
to
cyclic
corollary and
such that the
so
group,
Zm x
is Z\342\200\236
\342\231
Zm\342\200\236.
without
going
to
isomorphic
gcd of any
two
shows that if n is written
corollary
= (0, 0).
extended to a product of more
as a
through
Zmim2...m;i
of them as a
two
than
factors
the details
if and
of
by similar the
proof.
only if the numbers
mt
is 1.
product
of powers of distinct
prime
as in n =
then
(r, s) '
+
summands
mn/d
This
\342\226\240 \342\226\240 \342\226\240
,
is isomorphic Z\342\200\236
^(f>l)\"'
In particular,
(Pir(P2)n2---(Pr)nr,
to
Z72 is
^(Pl)\"2 X
X
isomorphic
to
7L%
x
\342\200\242 \342\200\242 \342\200\242 X
Z9.
^{prf
\342\226\240
\342\226
We remark
and
Products
Direct
11
Section
Abelian
Generated
Finitely
107
Groups
the order of the factors in a direct product yields a group one. The names of elements have simply been changedvia a isomorphic original the in the of permutation components n-tuples. Exercise 47 of Section6 askedyou to define the least common multiple of two r and 5 as a of a certain It is to positive integers generator cyclic group. straightforward that of Z r the subset of all that are of both and 5 is prove consisting integers multiples a subgroup of Z, and hence is a cyclic group. Likewise, the set of all common multiples \342\226\240 of n positive integers r1; r2, \342\226\240 of Z, and hence is cyclic. rn is a subgroup \342\226\240, that
changing
to the
11.8 Definition
Let rl,
\342\226\240 \342\226\240 \342\226\240, rn be
r2,
is the positive the cyclic group
smallest
11.9
Theorem
Proof
11.10
Example
Solution
divisible
multiples of
the
r;,
that
\342\200\242 = 1, 2, \342\200\242 n. \342\200\242,
rt for i
each
by
1cm) is,
(abbreviated
multiple
all common
of
\342\226\240
\342\226\240 on cyclic groups,we seethat the 1cm ofr1; r2, \342\226\240 \342\226\240, \342\200\242 \342\200\242 = of for each i hence r,n, 1, 2, \342\200\242, multiple
is a
that
multiple.
\342\226\240 e Let (ax, a2, \342\226\240 \342\226\240, a\342\200\236) YYi=i Gi- ^ \342\226\240 \342\226\240 in \342\226\240, a2, (a\\, an) n?=i ^/ is e(3ual
a<
order
fimte \302\260f
*s
in
r,-
G,,
the order
then
multiple of all the
common
least
t\302\260 the
of
rt.
in the proof of Theorem 11.5.For a This follows by a repetition of the argument used \342\226\240 \342\226\240 to \342\226\240 \342\226\240 the of be a \342\226\240, (flj, a2, \342\226\240, an) en), power give {e\\, e2, power must simultaneously of so that this of the will first component r\\ a\\ multiple power yield e\\, a multiple of r2, so that this power of the second componenta2 will yield e2, and so on. \342\231\246
Find the
Since
order of (8,4, 10)in
the
gcd we
Similarly,
of 3, Zm.
11.11 Example
cyclic group
positive integer
least common
the name
integers. Their least common
11.8andourwork
FromDefinition is the r\342\200\236
positive
of the generator of all integers
The
15, and
group
the
Z12 x
group
x
Z24.
we see that 8 is of order ^ = 3 in Zj2. (See Theorem 6.14.) is of order 15in Z60 and 10 is of order 12 in Z24. The 1cm 5 \342\200\242 3 \342\200\242 4 = 60, so (8, 4, 10)is of order 60 in the group Zj2 x Z60 x A
of 8andl2is4, find that 4 12 is
Z2 is generated
Z x
by
the
elements
of n cyclic groups, each integer m, is generated by the n n-tuples direct
Z60
product
(0,1,0,---,0),
(1,0,0,---,0),
(1, 0)
of which
is
and
(0,0,1,---,0),
Such a direct product might also be generated by fewer Z4 x Z35 is generatedby the single element (1, 1, 1). that if
Note
Yil-i Gi is the
direct
Gi = {(
the
set
of all
subgroup of n^i just rename
n-tuples
Gt.
It
is also
of groups
product
\342\226\240 \342\226\240 \342\226\240,
with
the
(0,
at, ei+i,
identity
clear that this
(eue2, --/,-1,^,^1,
subgroup
Zm
for
generally, some
the
positive
\342\200\242\342\200\242-, (0,0,0,---,1).
elements. For example,Z3
x \342\226\262
G,,
then
\342\226\240 \342\226\240 \342\226\240, en) \\ at
elements
1). More
Z or
either
the
subset
e G,},
in all places
G,- is naturally
but
the
/th,
is a
isomorphic to G,-;
\342\200\242 \342\226\240-,\302\243\342\200\236) by a,-.
Part II
108
Permutations, Cosets,and
The group G,- is mirrored the other components just
Products
Direct
in
the
/th
of
component
ride along.
the
of G,, and the the internal
elements
consider
e,- in
direct is called the as applied to
Yl1=i Gi The direct product given by Theorem 11.2 and external direct product of the groups The terms internal G,-. external, we are regarding the a direct product of groups, whether or not (respectively) just reflect of the We shall the words as omit subgroups product group. usually component groups and just say direct product.Which term we mean will be clear from external and internal
product of these
We
to
be
G,.
subgroups
the context.
= Historical his
Note Carl Gauss Arithmeticae, in what is today the results
Disquisitiones
Indemonstrated
various
of abelian groups in Not only did he classes of quadratic equivalence
theory
number theory.
considered residueclassesmodulo Although he noted that results
were similar,
of
context
the
deal extensively
with
also
but he
forms,
a given integer. in these two areas
attempt to develop an abstract of abelian groups. theory In the 1840s, Ernst Kummer in dealing with ideal numbers noted that his results were in complex many respects
analogous to thoseof
Historical Note student in
Section
did not
he
in
Leopold
Section
Kronecker
29) who
finally
The
Gauss.
(See
the
Kummer's the Historical Note
26.) But it was
(see realized
an abstract
that
Some
theorems
be quite
abstract realm of ideas.It is therefore appropriate their development from all unimportant so that one can spare oneself from restrictions,
to free
necessity different cases. development simplicity,
if it
is given
manner, since the with
clarity.\"
the basic
groups
of abstract
technical
and
in
most
most
the
important
principles of the
and
was
algebra are easy to
theory
to state
able
general admissible features stand out
then proceeded to
Kronecker
Theorem 11.12restricted
time-consuming
the
in of repeating the same argument This advantage already appears in the and the presentation gains in itself,
of Finitely Generated
Structure
may
be developed out of the analogies. As could theory he wrote in 1870, \"these principles [from the work of Gauss and Kummer] belong to a more general,
and
to finite
develop
of finite prove
abelian of
a version
groups.
Abelian Groups
understand
to present.
and use, although their This is one sectionin
proofs the
text
and significance we explain the meaning of a theorem but omit its proof. The whose our understanding, and meaning of any theorem proof we omit is well within with it. It would we feel we should be acquainted be impossible for us to meet someof these to insist on wading through facts in a one-semester course if we were fascinating all theorems. The theorem that we now state of complete proofs gives us complete about all sufficiently small abelian groups, in particular, structural information about all
where
finite
11.12 Theorem
abelian
groups.
Abelian Groups) Generated (Fundamental Theorem of Finitely abelian of cyclic group G is isomorphic to a directproduct
generated
Z(piyi
X
Z(p,)-2
X
\342\200\242 \342\200\242 \342\200\242 X X Z Z(P(|)r\342\200\236
X Z
X
Every
finitely
groups in the
\342\200\242 \342\200\242 \342\200\242 X
Z,
form
where
Direct Products and Finitely
11
Section
the
number
not necessarily
are primes,
pt
direct product is
of G)
the
and
distinct,
r,-
of factorsZ is unique
and
the
109
Groups
are
integers.
positive
rearrangement of
for possible
except
unique
number
(Betti
Abelian
Generated
the
factors;
The is, the
that
{pif are
prime powers
unique.
Proof
11.13
Example
The
Find all abelian groups,
signifies to one of the that
Solution
here.
proof is omitted
of Theorem 11.12. Since our appear in the direct product
use
make
We
factors Z
of
groups
will
we express 360 as a as possibilities
First
we get
1.
Z2
x
Z2 x
2.
Z2
x
Z4
Z2 x x Z3 x
3.
Z2 x
Z2
x
4.
Z2
x
Z4 x Z9
Z8
x
Z3
Z8
x
Z9 x
5. 6.
Z3
x
Z3 x
Z3
x
Z5
be
groups
are to
to isomorphism
up
identical
structurally
in the
shown
of prime
product
360. The phrase
of order
to isomorphism,
group of order 360should order 360 exhibited.
abelian
any
up
\342\231\246
be of
the
statement
powers 23325.Thenusing
360, no
order
finite
of
(isomorphic)
theorem.
the
Theorem
11.12,
Z5
Z2 x Z9 x Z5
x Z5 x Z3 x Z5 Z5
Thus there are six different
abelian
(up to
groups
isomorphism) of order
360.
\342\226\262
Applications
this section with
We conclude
abelian
regarding
11.14 Definition
A
group
Theorem
The
finite
power of a Proof
Proof
we could
theorems
now prove
groups.
to a direct product of two decomposable if it is isomorphic Otherwise G is indecomposable. abelian
indecomposable
groups
are exactly
the
proper
groups
cyclic
nontrivial \342\226\240
with order a
prime.
Let G be a finite indecomposable abelian group. Thenby Theorem 11.12,Gis isomorphic to a direct product of cyclicgroups of prime power order. Since G is indecomposable, this direct product must consist of just one cyclic group whose order is a power of prime number. Conversely, let p be a prime. Then Zpr is indecomposable, for if Zp>- were = then every element would have an order at most isomorphicto Zpi x ZpJ, where i + j \342\200\236max(rj) <
11.16 Theorem
many
G is
subgroups. 11.15
of the
a sampling
r,
r
If m divides the
a
^ order
By Theorem11.12,we
of a finite can
think Z0>,)\">
abelian of G
group
G,
then
G has a
subgroup
as being
xZ(kP
*\342\226\240\342\226\240\342\200\242 xZ0>\342\200\236)''\">
of order
m.
II
Part
110
Cosets, and Direct Products
Permutations,
pt need be distinct.
all primes m must be of the not
where then
form
a cyclic
cyclic
(p, )r> ~s'.
the cyclic
denotes
{a)
(pnY\" < r,-.
st
to the
~s- is
is the
By Theorem
(pt f- ~s>.
Thus
(pt f
~s>
= (piyi-
[(pi)r'i/[(piY'-Sii
Recallingthat
<
of Z^.y,- of order equal subgroup But the gcd of (pt f' and (pt fsubgroup Z(Piy, of order
{Pi)r'~s>generatesa gcd of (pi f- and
the
generates
{piY^iPiY1 where 0 (PnY\",
(piY1 (P2)Sl
order of G, 6.14, quotient of (Pi)ri by
\342\226\240 \342\226\240 \342\226\240
Since
\342\226\240 \342\226\240 \342\226\240
we see that
by a,
generated
subgroup
tors,)x((ftr-)x-xten
is 11.17
the
of order
subgroup
required
m.
\342\231\
that is, m is is a square free integer, of m is cyclic. order abelian every group
If m
Theorem
Let G be an
Proof
abelian
of square
group
by the
divisible
not
free orderm. Then
square of
Theorem
by
then
prime,
any
11.12, G is
to isomorphic
X
Z(piyi
m =
where
all pi
\342\226\240 \342\200\242 \342\200\242 \342\226\240 Since
(PnY\"
(P2Y2
{p\\)ri
distinct
, Z(pn)r\342\200\236
m is
we must
square free,
11.6 then shows
Corollary
primes.
\342\200\242 \342\200\242 X \342\200\242 X
Z(p2yi
G is
that
have
all
r,-
=
1 and
to ZpiP2...Pa,so G
isomorphic
is cyclic.
\342\231\
11
EXERCISES
Computations
1. List the
2.
the group
7, find
through
x Z12
Z4
x
the
Z3
nontrivial
10. Find
all
proper
nontrivial
11. Find
all
subgroups
12.
all subgroups
Find
Find
the
Disregarding
resulting
14. Fill
in the
product
order
of each
of the
the order of
the
element
given
4-
(2*
7.
(3, 6,
among the orders of
subgroups of Z2 subgroups of Z2 x
of the
direct
product.
5. (8, 10)in
12, 16) in
all
the
Z4
cyclic
x Z,2
x
Z20
of
subgroups
Z12
x Z18
x Z24 Z&
x 7L%>. of Zj2
x Z15?
x Z2.
of Z2
x
Z4
of order
of Z2
x
Z2
x Z4
Z2
x Z2.
4.
that are
isomorphicto
direct of the factors, write products as to in Z60 many ways as isomorphic
blanks.
a. The cyclicsubgroup b. Z3 x Z4 is of order
group cyclic?
Z6 x Z15
3)in
order is
Is this
elements.
x Z4.
Zi5
order
largest
all proper
13.
x Z4. Find
(2, 6) in Z4 x Zi2
6. (3, 10,9) in 8. What is the
9.
1 for
Exercise
Repeat
In Exercises 3 3.
of Z2
elements
of Z24
generated
by
18 has
order
the
Klein
4-group.
of two or
possible.
more groups
of the
form
so that Z\342\200\236
the
Section 11
c.
element
The
d. The e.
15.
Klein
Z4
possible order for
maximum
the
16.
Are
17.
Find the maximum
the
Z2 x
groups
Zl2
and
x
to Z
x
Z4
Z
order.
of finite
elements
has
order
has
Z8
is isomorphic
4-group
Z2 x Z x
Find
of Z12 x
(4, 2)
some element
%&
x Z6.
of Z4
or
Why
isomorphic?
possible order for someelement
why
not?
x Z10 x
of Z8
Z24.
x
Z40 isomorphic? groups Z8 Why or 19. Find the maximum possible orderfor someelement of Z4 x Z18 x Z15. 20. Are the groups Z4 x Z18 x Z]5 and Z3 x 1^ x Z10 isomorphic?Why or
Are the
18.
x Z10
21 through
In Exercises
25,
x Z24 and
x Zi2
Z4
11.13to find
as in Example
proceed
111
Exercises
not?
why
why not?
groups, up to isomorphism,of the
all abelian
given
order.
22- 0rder 16
21. Order 8 24. Order
25. Order
720
0rder
23-
32
1089
isomorphism) are there of order 24? of order 25? of order (24)(25)? the idea suggested in Exercise 26, let m and n be relatively 27. Following prime integers. Show that if positive r abelian groups of order m and 5 of order n, then there are (up to isomorphism) there are (up to isomorphism) rs abelian groups of order mn. 26.
How
abelian
many
28. UseExercise27
29. a. Let up
p be a
groups
(up to
the number
to determine
prime number.
in the
Fill
of abelian groups
second row
of the
(up
table
of order (10)5.
to isomorphism)
to give
the number
of abelian
groups of order
p\",
to isomorphism.
n
2
3
4
6
5
7
8
find
the
number of groups b. Let up
i. 30.
p, q,
and
of the
Consider
the Figure
a Cayley
of two
vertices 7.11(b)
a.
with
Under
n-gons,
regular
of the outer shows one
counterclockwise)direction group
such a Cayley
for
digraph
with two arc
digraphs
Cayley
and consisting
Use the
table
you created
to
iii.
n-gon with
as those
Zm x
types,
for the Z\342\200\236
a solid
one
generating with
an
will
of abelian
groups,
set S =
arrow
q5r4q3
{(1,0),(0, 1)}.
and a dashed
one with
no
arrow,
> 3, with solid arc sides, one inside the other, with dashed arcs to the inner one. Figure shows such a Cayley digraph with n = 3, 7.9(b) n = 4. The arrows on the outer n-gon may the same (clockwise or have on the inner n-gon, or they may have the opposite direction. Let G be a for n
digraph.
what circumstances
number
order.
given
ii. (qr)1
schematically
31.
and
distinct prime numbers.
p3q4r7
Indicate
joining
r be
to isomorphism,
G be
abelian?
b. If G is abelian, to what familiar group is it isomorphic? under c. If G is abelian, what circumstances is it cyclic? d. If G is not abelian, to what group we have discussedis it
isomorphic?
Permutations, Cosets, and DirectProducts
II
Part
112
Concepts
32.
Mark each of the
a. If Gi and
b.
true or
following
are any
Gj
groups,
then
is always
x G2
G\\
external direct
in an
Computation
false.
product
to G2
isomorphic
is easy
of groups
if
you
know
x G]. how to compute in
group.
component
c. Groups of finite order must be used to form an external direct product. d. A group of prime order could not be the internal direct product of two proper e. Z2 x Z4 is isomorphicto Zg. f. Z2 x Z4 is isomorphicto S%. to S4. g. Z3 x Zg is isomorphic h. Every element in Z4 x Zg has order 8. i. The order of Z12 x Zj5 is 60. has mn elements whether m and n are relatively prime or not. j. Zm x Z\342\200\236 33.
Give an
nontrivial
34. a.
How
many
subgroups
of Z5 x
b. How
many
subgroups
of Z
Mark
Zg
are
isomorphic
x Z are isomorphicto
of a nontrivial
example
nontrivial
36.
not every nontrivial
that
example illustrating subgroups.
35. Give an
group
to Z5 x Z x Z?
is the
group
internal direct
of the
following
a. Every
d.
Zg
e. All
and
is not
the internal direct
f.
Any
group
of prime
abelian
group
of prime power order is
6}. 5, 6}. groups are classifiedup generated abelian groups
generated
by {4,
is generated
by {4,
finite
abelian
two
finitely
g.
h.
Every
abelian
i.
Every
abelian
j.
Every
finite
an abelian
G be
39. LetGbe an
say
the
of two
group
by
group
of order
divisible
by 4
of order
divisible
group has a Betti
by Theorem
the same
with
divisible
group
cyclic.
to isomorphism
of order
abelian
group
how many
say how many abelian
order
of order
5 contains
a cyclic a cyclic
6 contains
by
number
Betti
contains a cyclic
number
11.12. are
isomorphic.
subgroup of order 5. subgroup
of order
Show of G.
43
of the
deal with
torsion
that
subgroup of order 6.
groups
of order
pr
72.
the
elements
the concept of the
subgroup
4.
of 0.
subgroups of order 8 G has?Why, subgroups of order 4 G has? Why,
group.
subgroup
Exercises40 through Find
product
order is cyclic.
abelian
Every abelian
a. Can you b. Can you
40.
proper
or false.
true
p and q be distinct prime numbers. How does the number (up to isomorphism)of abelian comparewith the number (up to isomorphism)of abelian groups of order qr ?
torsion
of two
product
37. Let
the
subgroups.
Z6?
of prime order
is not
that
abelian
nontrivial
subgroups.
each
b. Every c. Z8 is
38. Let
each
of Z4
of
finite
torsion
or why
not?
or why
not?
in G
order
subgroup
x Z x Z3; of Zn
form a subgroup.
just defined.
x Z x Z12.
This
subgroup
is called
11
Section
41. 42.
torsion
the
Find
subgroup of
the
E* of nonzero real numbers.
group
multiplicative
113
Exercises
subgroup T of the multiplicative group C* of nonzero complexnumbers. 43. An abelian group is torsion free if e is the only element of finite order. Use Theorem 11.12 to show that abelian group is the internal and of a torsion-free direct product of its torsion every finitely generated subgroup that {e} may be the torsion subgroup, and is also torsion free.) (Note subgroup. 44. Thepart of the decomposition of G in Theorem to the subgroups of prime-power order 11.12 corresponding \342\200\242 \342\200\242 \342\200\242 r \342\200\224 in the form Zm, x Zm, x \342\200\242 can also be written x Zmr, where m, divides mi+\\ f or / = 1, 2, \342\200\242 1. The \342\200\242, to be unique, and are the torsion of G. numbers m; can be shown coefficients torsion
the
Find
a.
the
Find
b. Find
torsion
the torsion
c. Describean
coefficients
of
coefficients
of Z6
to find
algorithm
x Z9.
Z4
x Z12 x Z20
the
torsion
coefficients
the
proof
of Theorem
of a direct
product of cyclicgroups.
Proof Synopsis
45. Give
synopsis of
a two-sentence
11.5.
Theory
46. Prove
is abelian. product of abelian groups the all elements subset of G consisting of the identity e together with of 47. Let G be an abelian group. Let H be G of order 2. Show that H is a subgroup of G. for every abelian group 48. Following whether H will always be a subgroup up the idea of Exercise47 determine G if H consists of the identity e together with all elements of G of order 3; of order 4. For what positive for every abelian group G, if H consists n will H always be a subgroup of the identity e together with integers all elements of G of order n? Compare with Exercise 48 of Section 5.
49. Find
a direct
that
47
of Exercise
a counterexample
with
the
hypothesis
and K be subgroups of a group G. Exercises 50 and for G to appear as the internal direct product of H and K.
Let H
50. Let H and way.
a.
51.
52.
is of the
form
e K.
and K be subgroups
of a
kh for
all
H x
H (actually
H and k
=
that
{e})and
some
hk for
51 ask
h
e H
both
omitted.
abelian
you to establish
necessary
and sufficient
criteria
H and K appear as subgroups of G in a natural {e} x K) have the following properties.
K (actually
e K.
and k
c. H n K
=
{
exercise. Show the three properties listed in the preceding group G satisfying that for each g \342\202\254 Then let each g be renamed G, the expression g = hk for h e H and k e K is unique. (h, k). G becomes structurally identical (isomorphic)to H x K. Show that, under this renaming,
Let H
Show
that
some prime 53.
H x K.Recall
G=
h e
of G
element
Every
b. hk
groups and let that these subgroups
K be
Show
that G is
a finite
abelian
is not
group
if
and
only if it
contains a
subgroup
isomorphic
to Zp x Zp
for
p.
Prove that if a finite abelian group is a power of p. Can the hypothesis
54. Let
cyclic
G, H, and
K
be finitely
has
order
a power
of commutativity
generated
abelian
of a prime p, then be dropped?Why,
groups.
Show
that
if
order
the
or why G x
of every
element
in
the
group
not?
K is isomorphic
to
H
x K,
then
114
Part II
SECTION 12
and
Cosets,
Permutations,
Products
Direct
^PlANE IsOMETRIES of E2 is a permutation : E2 \342\200\224> E2 plane E2. An isometry \302\242) distance, so that the distance between points P and Q is the same as the points (f>(P)and 4>(Q) for all points P and Q in E2. If ij/ is the distance between also an isometry of E2, then the distance between i/f(0(P)) and VWK2)) must be the in turn is the distance P same as the distance between and \302\242(0), which between (f>(P) that the composition of two isometries is again an isometry. Since the and Q, showing of an isometry is an isometry, we see that the identity map is an isometry and the inverse isometries of E2 form a subgroup of the group of all permutations of E2. of E2 that carry S onto itself form a Given any subset S of E2, the isometries of isometries. This subgroup is the group of the group of symmetries of S in subgroup of symmetries of an equilateral the group E2. In Section 8 we gave tables for triangle and for the group of symmetries of a square in E2. we have defined in the two preceding paragraphs could equally well Everything have been done for n-dimensional EuclideanspaceE\", but we will concern ourselves with chiefly plane isometries here. that every isometry of the plane is one of just four types (see Artin It can be proved that can be carried [5]). We will list the types and show, for each type, a labeled figure of In of a nd that each into itself Figs. 12.1, 12.3, 12.4, consider the type. by an isometry be to extended infinitely the left and to the right. We also give line with spikes shown to Consider
that
an
the Euclidean
preserves
example
of each
type
of coordinates.
terms
in
r: Slide every 12.1. (Example:r(x, y) Fig. translation rotation
(Example:p(x, y) origin (0, 0).) /x:
reflection
L, each
the plane
Rotate
p:
point
point = (x,
=
about
(\342\200\224y, x) is
the
same
a point
a rotation
Map each point into of which is left fixed
its
distance
(2, -3)
y) +
P through
through
mirror
the
in
an
direction.
same
= (x + 2,y-
3).)
6. See
angle
image
(/x for mirror)
12.3.The by = (y,x) is a reflectionacrossthe
line
Fig. 12.2. about
90\302\260 counterclockwise
/x. See Fig.
See
the
across a line the axis of
L is
y =
x.) a aline and reflection across glide reflection y: product mapped into itself is a by the translation. See Fig. 12.4. (Example: y(x, y) = (x + 4, \342\200\224y) the reflection a:-axis.) glide along (Example:
reflection.
n(x,
y)
line
of a translation
The
that is carried into another arrow in each of curved translation and rotation, the counterclockwise directions of the curved arrows remain the same, but for the reflection and glide reflection,the a clockwise counterclockwise arrow is mapped into arrow. We say that translations and while the reflection and glide reflectionreverse rotations preserve orientation, as any definite orientation. We do not classify the identity one of the four isometry types well be considered to be a translation it could equally listed; by the zero vector or a of 0\302\260. We always rotation about consider a glide reflection to any point through an angle be the product of a reflection and a translation that is different from the identity isometry.
Notice
the little curved
Figs. 12.1through
12.4.
arrow
For the
This section is not used in the remainder
of the text.
115
Plane Isometries
12
Section
T(P)
r(R)
r(R)
Q
12.1Figure
Translation
P
lAQ) i
>
P-KQ)
r-\\Q)
x.
12.2 Figure
Rotation
p.
R \342\226\240
Y(P)
y-HP) s
i
Q
IMP)
12.3 Figure
The theorem that full
12.5 Theorem
group
Proof Outline
G.
G =
12.4
/x.
describes
/ \\
\\KG)
P
y2(P)
Glide reflection y.
Figure
the possible
group G of isometries of the for some positive integer n. D\342\200\236
plane
of finite
structures
subgroups
of
the
is isomorphic
to either
or to Z\342\200\236
a dihedral
P in the plane that is left fixed by every isometry that there is a point can be done in the following way, using coordinates in the plane. Suppose \342\200\242 \342\200\242. 0m} and let {0i. 02- \342\200\242
show
This
(*;.)';) = Then
\\ y-2(P)
n(K>
finite
we
First in
follows
o.
group.
isometry
Every
Reflection
\342\226\240
g
l
y
the
0/(0.0).
point Cc, J)
centroid of
^
'xi +x2-\\
h x\342\200\236, Ji +
)'2
H
H )'\342\200\236
m
S =
{(Xj. }'i)\\i =
\342\200\242 1. 2. \342\200\242 \342\200\242. m}. The
isometries in G permute then points 0,0j fa(Xj, >'j) = 0,-(0,-(0,0)] = = shown that a set of points can be the centroid of It is uniquely 0,(.(0.0) (\302\276.yk). and since each isometry the points, in G just permutes determined by its distances from the set S, it must leave the centroid rotations (x. y) fixed. Thus G consistsof the identity, across a line through P. about P, and reflections in G form a subgroup The orientation-preserving isometries H of G which is either in the same way that we showed that all of G or of order m /2. This can be shown the of S\342\200\236 the are a subgroup half elements of even permutations S\342\200\236. (See containing just in G. If we choose a Exercise 22.) Of course H consistsof the identity and the rotations as small an angle 0 > 0 as possible,it can be rotation in G that rotates the plane through shown to generate the subgroup H. (See Exercise 23.) This shows that if H = G, then =\302\243 to Z\342\200\236,. G is cyclic of order m and thus isomorphic H G so that G contains Suppose is the the
in S
the
set
among themselves, sinceif
= fa
116
Part II
and
Cosets,
Permutations,
Products
Direct
\342\226\240 some reflections.Let H = {i,p\\, \342\226\240 with n = m /2. If /x is a reflection in G, then \342\226\240, p\342\200\236_i} coset Hfi consists of all n of the reflections in G. Consider now a regular n-gon in the plane having P as its center and a vertex with on the line through P left fixed by /x. The elements of H rotate this n-gon through lying
the
all positions,
turning this
the
of H/x first reflect and then rotate through
all
of Dn, so G is isomorphic
theorem preceding to some infinite
The
in an
elements
over,
is the action
n-gon
We turn
the
and
n-gon
gives
the
axis
a vertex, effectively Thus the action of G on
through
positions.
to D\342\200\236.
about
story
complete
finite
\342\231
plane
isometry
groups.
isometries that arise naturally in decorating and art. Among these are the discrete frieze groups. A discrete friezeconsistsof a pattern of finite width and height that is repeated endlessly in both directions along its baseline a strip of infinite a to form but of it as think border strip finite decorative length height; to the ceiling on wallpaper. We consider that goes around a room next those isometries that carry each basic pattern onto instance itself or onto another of the pattern in the frieze. The set of all such isometriesis called the \"frieze All discrete frieze group.\" and have a subgroup groups are infinite isomorphic to Z generated by the translation that slides the frieze lengthwise until the basic pattern is superimposed on the position in that direction. of its next neighbor pattern As a simple exampleof a discrete frieze, consider integral distances and continuing signs spaced apart equal infinitely to the left and right, indicated schematically like this. now
of plane
groups
\342\226\240\342\226\240///////////////////////////////////-
consider the
The symmetry group of this frieze integral signs to be one unit apart. the plane one unit to the right, and by a rotation p by a translation x sliding of 180\302\260about a point in the center of some integral There are no horizontal or sign. vertical is nonabelian; we can frieze reflections, and no glide reflections.This group
Let us
is generated
check that do
= px~l. The a rotation commute,
not
xp
n-th p\\
dihedral through
group 360/n\302\260
Dn is of
2 satisfying /?i/x = /xp^1. Thus it is natural to use frieze group generated by x of infinite order and p As another consider the frieze given example,
generated by two elements that n and a reflection /x of order the notation D^ for this nonabelian of order 2. by an infinite string of D's. order
DDDDDDDDDDD Its
by a translation x one stepto the right and by a vertical reflection /x line cutting the middle of all the D's. We can check that these through generators commute this time, that is, t/x = /xt, so this frieze group is abelian x to Z Z2. isomorphic can be shown that if we classify such discretefriezes only by whether or not their
group
group is
and
is generated
a horizontal
across
It
groups
contain
a
rotation axis reflection
vertical then there group in
that
are a total
horizontal
axis
nontrivial
glide
reflection reflection
of seven possibilities. A nontrivial in a symmetry glide reflection that is not equal to a product of a translation in that group and a reflection The group for the string of D's above contains glide reflectionsacross group.
is one
Section12 the horizontal line
in the
is also
that group.
The frieze group
of each
reflections
in
D
D
D
D
D
D
D
is not an element of component glide reflection whose translation exercises exhibit the seven possible cases, and ask you to tell, for each of the four types of isometriesdisplayed above in the symmetry which appear We do not obtain seven obtained different group structures. Each of the groups a nontrivial
contains
The
group.
case, group.
glide
for
D
D
component
so they
group
D
the
of the D's, but the translation are all considered trivial
centers
the
through
glidereflection
117
Isometries
Plane
can be
to be
shown
to one of
isomorphic
Z,
Z
\302\243>oo,
x Z2,
or
Doo x Z2.
a pattern in the shape of a square, when study of symmetries or hexagon is repeated by translations along two nonparallel that on wallpaper. These to fill the entire plane, like patterns vector directions appear a or the plane crystallographicgroups.While groups groups are called the wallpaper itself frieze could not be carriedinto by a rotation through a positive angle less than of 60\302\260, for some of these to have rotations 180\302\260 90\302\260, 120\302\260,and 180\302\260,it is possible is the
interesting
Equally
rhombus,
parallelogram,
where the pattern consists of 12.6 provides an illustration of plane isometries that carry this square onto the group itself or onto another are given by two translations Generators for this group square. a square to the next neighbor to the right and one to the next (one sliding above), by a rotation through 90\302\260 about the center of a square, and by a reflection in a vertical (or line along the edges of the square. The one reflection is all that is needed to horizontal) the plane \"turn After over\"; a diagonal reflection can also be used. being turned over, a
the
square.
Figure
patterns.
plane-filling
We are
translations
in the pattern unit translations
interested
rotations
and
plane to
in
surely
can be
used again.
a subgroup
contains
The
isometry
isomorphic
to
group for this periodic Z generated by
Z x
the
and upward, and a subgroup to \302\243)4 isomorphic generated by into itself. one square (it can be any square) carry to be filled with as in Fig. 12.7, we do not the plane parallelograms
the
right
those isometries that If we consider all the types of isometriesthat get
we
did for
12.6Figure
Fig. 12.6. The symmetry
group
this time
is
Part
118
II
Permutations,
Cosets, and DirectProducts
12.7
Figure
about translations indicated through 180\302\260 by the arrows and a rotation of a parallelogram. of wallpaper patterns when It can be shown that there are 17different they are types classified according to the types of rotations, reflections, and nontrivial glide reflections and that they admit. We refer you to Gallian [8] for pictures of these 17 possibilities a few of them. The situation a chart to help you them. The exercises illustrate identify is more complicated; it can be shown that there are 230 three-dimensional in space in space. The final exercise we give involves rotations crystallographic groups. M. C. Escher (1898-1973)was an artist whose work included plane-filling patterns. of his works of this The exercises include reproductions of four type.
by the
generated
any vertex
12
\342\226\240 EXERCISES
1. This exercise shows that of the space in dimension a. Describe
all
the group of symmetries of a certain which we consider the figure to lie. of a point in
symmetries
the real line R; that
of geometric
type
is, describe
figure may depend on
all isometries
of R that
leave
the
one point
fixed.
b. Describeall c. Describeall
symmetries
(translations,
symmetries
of a line
reflections,
segment
in
d. Describe
all symmetries of a line segment
in
e. Describe
some symmetries of a line
2. Let P with
stand
P or R
for
an orientation
to denote
the
R2.
etc.) of a point
in the
plane R2.
in R3.
segment
preserving plane
orientation
R.
isometry
preserving
P p R
and
R for an
or reversing
R
property
orientation of a
reversing
product.
one.
Fill in
the table
3.
119
Exercises
12
Section
of plane isometries given by a product table to give all possible types of two types. For example, a a rotations be or it may be another Fill in the box corresponding to pp with rotation, product may type. both letters. Use your answer to Exercise 2 to eliminate from consideration. some types. Eliminate the identity Fill
in the
of two
X
p
y
M
T P M
y
4. Draw a plane
5. 6. 7.
group as its
of symmetries
in R2.
of symmetries
in R2.
figure
that
has a one-element
Draw
a plane
figure
that
has
a two-element
Draw
a plane
figure
that
has
a three-element
group as its group group as its group
Draw
a plane
figure
that
has
a four-element
group isomorphicto
8. Draw a
plane
has a four-element
that
figure
group
in R2.
of symmetries
group isomorphicto
its group of symmetries
Z4 as
Klein
the
4-group
group of symmetries
as its
V
R2.
in
inR2.
9.
isometries (other than the identity), give the possibilities for the order group of plane isometries. 10. A plane isometry cp has a fixed point if there exists a point P in the plane such that 4>{P) = P. Which four types of plane isometries (other than the identity) can have a fixed point? of the
four
isometry of that
type
For
each
in the
11.
Referring to Exercise 10,which
12.
Referring
13.
types
of plane
isometries, if
any,
have
10, which
types
of plane
isometries,
if
any,
have exactly
Referring to Exercise 10,which
types
of plane
isometries,
if
any,
have an
to Exercise
14.
Argue
15.
Using Exercise 14, show
algebraically
that is, if
for noncolinear
that a plane
geometrically
16. Do the
cj>(Pj)
=
rotations,
of an
of plane
types
if (Pi)
isometry
that if points
with the identity
together
two plane isometriescp Pi, P2, and P3, thcncp a subgroup
form
map,
one fixed point?
exactly
three noncolinear
leaves
that
two
of the
^ are
of fixed be the
on three
points? map.
identity
noncolinear
points,
the same map.
of plane
group
points?
must
fixed
t/t agree
and
fixed
number
infinite
points
and
of the
isometries?
or why
Why
not?
17. Do the
translations,
together
with the
map,
identity
form
a subgroup
of
the
of plane
group
isometries?
Why
or
group
of
why not?
18. Do the
about
rotations
plane isometries?
19. Does the of plane
20. Do the or why
21.
Which
22.
Completing
Why
reflection
isometries?
glide
one particular or why not?
across Why
reflections,
P,
point
one particular not?
line
together
L,
together
with the
identity
with the
identity
map,
map,
form
a subgroup
form
of the
a subgroup
of
the
group
or why
together
with
the identity
map, form a
subgroup
of the
group of plane
isometries?Why
not? of the
the
rotations
IGI
=2\\H\\.
of plane isometriescan be elements
subgroup of the group of plane isometries? of the proof of Theorem 12.5,let G be a finite group of plane isometries. Show that a detail in G, together H of G, and that H = G or with the identity isometry, form a subgroup either that we used to show that \\S\342\200\236\\ [Hint: Use the same method =2\\A\342\200\236\\.] four types
of a finite
Part
120
23.
II
Permutations,
Cosets, and Direct Products
in the proof of Theorem let G be a finite of the identity 12.5, group consisting isometry about one point P in the plane. Show that G is cyclic, generated by the rotation in G that turns the plane counterclockwise about P through the smallest angle 9 > 0. [Hint: Follow the idea of the proof that a subgroup of a cyclic group is cyclic] a detail
Completing
and rotations
Exercises
24
symmetries.
frieze always of the
through
Imagine contains
30 illustrate the seven different the figure shown to be continued For each of these translations.
when
exercises answer
these
are
they
and
to the right
infinitely
left.
questions
classified The
according to their group of a
symmetry
about the
symmetry
group
frieze.
a. Doesthe group b. Does the group c. Does
the
d. Does the e. To which frieze
24.
of friezes
types
contain
a rotation?
contain
a reflection
across a horizontal
group
contain
a reflection
across a vertical line?
group
contain
a nontrivial
glide reflection?
of the
possible groups Z,
D^, Z x Z2,
line?
or D^,
x Z2
do you
think
the symmetry
group of the
is isomorphic?
FFFFFFFFFFFFFFF
25. TTTTTTTTTT
26.EEEEEEEEEEEE
12.8
Figure
of Regular Division Escher Foundation-Baarn-Holland.
The Study
of the Plane
with
All rights
Horsemen
reserved.)
1946 (\302\251
M. C.
121
Exercises
Section 12
27zzzzzzzzzzzz
28HHHHHHHHHH J J J 29.
30.
J
J
11111
n
u
n
u
n
n
u
u
to be used to fill 37 describe a pattern 31 through in each case. these the specifiedvectors.Answer questions Exercises
a. Doesthe
symmetry
group
contain
any
rotations?
the
plane
If so,
by translation
through
what
in the
two
angles
possible
directions
where 6\302\273
given 0 <
by
6 <
180\302\260?
r-.^su
/i- -ft
'# \342\226\240 \342\226\240\342\226\240A
i~m. \342\226\240-\"i
12.9 Figure
The 1936
i.',
.''
'-\342\200\224 \"V-'\"~-
Study of Regular Division of M. C. Escher Foundation-Baarn-Holland.
the
;*tT,--
Plane
. ji-
with Imaginary Human All rights
reserved.)
Figures (\302\251
Part II
122
12.10Figure
The
Products
Direct
and
Cosets,
Permutations,
Study of Regular
Division of the
b. Does the c. Does
31. A
square
32. A
the
symmetry
group
contain
any
symmetry
group
contain
any nontrivial
and
with horizontal
vertical
using
as in
Exercise 31
using
square
as in
Exercise 31
with
the
letter L at
its center
(0,
1).
34. A
square
as in
Exercise 31
with
the
letter E at
(0,
1).
35. A
square
as in
Exercise 31
with
the
letter H at
(0,
1). with a vertex at
the
37.
hexagon
with a vertex
the
regular
of the
at the center Exercises and
assume
38
through
41
the markings
invisible due to shading. 36, and also answer this
at
using
hexagon,
top
each
Answer part
given by vectors
(1, 0)
(0,
Escher
by vectors
given
(1, 1/2) and
(0,
and
1).
1).
using
translation
directions
given by vectors
(1,0)
and
its center
using
translation
directions
given by vectors
(1,0)
and
its center
using
translation
directions
given by vectors
(1,0) and
translation
using
directions
given by
top containing an equilateral triangle translation directions given by vectors
with
(1, 0)
vectors
vertex and
(1,
(1,0)
and (1,
at the
top
and
V3). centroid
V3).
C. Escher. Neglectthe shading in the figures or are the even horseman same, figure, reptile, though they may be the same questions were asked for Exercises 31 through (a), (b), and (c) that
are concerned in
directions
translation
directions
translation
square
hexagon
M. C.
glide reflections?
33. A
A
1939 (\302\251
reserved.)
reflections?
edges
36. A regular
with Reptiles
Plane
All rights
Foundation-Baarn-Holland.
with
art
works
of M.
human
(d).
coordinate axes with d. Assuming horizontal and vertical allel directions of vectors that generate the translation of these vectors.
equal
scales
as usual,
subgroup. Do not
give vectors in
concern
yourself
two nonparwith the length
the
12.11
Figure
The
Study of Regular Division of the Escher Foundation-Baarn-Holland.
Plane All
Study
of Regular
Plane
with Horsemen
in Fig.
39. The
Study
of
Division of the Regular Division of the
Plane
with Imaginary
Human Figures
Division of the
Plane
with Reptiles in
Fig. 12.10.
Regular Division of the
Plane
with Human
40. The Study
41. The
Study
of Regular
of
42. Show that the rotations the diagonals through
of a cube the
center
in
space
of the
form
cube.]
a group
Figures
in
isomorphic
with Human
1936 Figures (\302\251
M. C.
rights reserved.)
12.8.
The
38.
123
Exercises
Section 12
Fig. to
in
Fig.
12.9.
12.11.
S4. [Hint:
A rotation of the
cube
permutes
PART
Factor
III
section
and
Homomorphisms
13
Groups
Section
13
Homomorphisms
Section
14
Factor
Section
15
Factor-Group Computations
Section
16
-Group
Section 17
^Applications
Groups Simple
Groups
a Set
on
Action
and
to Counting
of G-Sets
Homomorphisms
Structure-Relating Maps one
G -*'\342\226\240 \302\247
groups. We
to
of G
structure about
G' be
and
G
Let
the
the group G and group structure of
general structure-relating by no longer requiring the
purely
13.1 Definition
do with
the
which
A
map
G', for it
G
to
often
G'
relate the group gives us information that
is the
group G
the
a group
into
G'. The
in this binary
of G. We now from thoseof
know all consider
all
about
more
an isomorphism
those conditions are to isomorphism, and have nothing are what us the binary operations give onto.
You see,
of an
definition
and of
study
related to
just a copy the conditions
be one to one and
our
of G
focus of our
an isomorphism 0 of a
of
we immediately
isomorphism,
is structurally
the maps
portion
operations
binary
an
weakening
maps, that
0 is
that
know
set-theoretic
algebra property of
in maps from of G'. Such a map
interested
of the groups from known structural of the other. An isomorphism properties of such a structure-relating map. Ifwe know G', if one exists, is an example
about the
are
structure
group
text. We keepjust the homomorphism for the definition we operations
G' is a
homomorphism if
the
now
make.
homomorphism
property 4>{ab) =
holds for all a, b 1
Section
4>{a)(j>{b)
e G.
16 is a prerequisite
Section 17 is not required
(1)
B Sections 17 and 36. remainder of the text.
only for for the
125
Part III
Factor
and
Homomorphisms
Let us
idea behind the ab on the product
the
examine
now
Groups
: G \342\200\224>\342\226\240 G'. In Eq. (1), the \302\242)
(1) for a homomorphism takes place in G, while the
requirement side
left-hand
side takes place in G'. Thus Eq. (1) gives a relation between the two group structures. operations, at least one homomorphism
the right-hand
on
4>{a)4>{b)
betweenthesebinary
and hence
homomorphism. We give onto G' may give structural 13.2 Example
Let \302\247 : G
how a illustrating about G'.
example information
group homomorphism be abelian. Let a!,b'
e G'.
G', there exist a,b e G such that we have ab = ba. Using property (1), = b'a', so G' is indeedabelian? 4>{b)4>{a) onto
will give
13.16
Example
13.3 Example
of G onto
\342\200\224>\342\226\240 G' be a
must
G'
then
an
give information
about
homomorphisms
for
be the Let S\342\200\236
G via
showing : G \302\242
Solution
how information about G' may now give examples of
\342\200\224>\342\226\240 G''. We
group on
n
and let
letters,
even
: Sn \302\242
denned
\342\200\224>\342\226\240 Z2 be
show operation on the in the group Z2.
that
\302\242(0(/)
a odd
(i even,
and
a even
and
(i odd,
a even
and
(i even.
the
first then
Thus \302\242(0(1) =
case,
and
if a
cr/x can
0 and
for \302\242((/,)
\302\242(0)+
this
permutation,
odd
permutation.
can
both
as the
\302\242(0)+
/x
=
\302\242((/)
is written
1 +
be written
0
Note S\342\200\236.
since it takes just four cases:
in
Z2.
\342\226\
(Evaluation ffi
into
0C : Recall value
the
place
as a product of an odd number of even number of transpositions. The other cases can be checked
similarly.
Example
that
of an
product
1 =
e
additively
to checking
amounts
be written
of a, (i
all choices
equation
this equation
Verifying (i odd,
=
side of
right-hand
a odd and
transpositions,
13.4
by
is a homomorphism.
We must
Checking
is abelian,
4>{ab) = 4>{ba) =
=
is Since\302\242
groups.
_ JO if a is an 11 if a is an Show that 0
G is abelian,
that a'b' = b'a!. = b'. Since G
=
G
mapping
\342\226\
a homomorphism
specific
symmetric
We claim that if
G'.
4>{a) = a' and we have a'b'
illustration
an
show
must
We
homomorphism
ffi,
F ^that, at x
Let F be the additive group of all functions mapping and let c be any real number. Let group of real numbers, = ffi be the evaluation for e defined F. f(c) f homomorphism by 0C(/) the sum of two functions and is the function whose g f + g by definition, f we have is f(x) + g(x). Thus Homomorphism)
let
ffi
be
the additive
4>c(f + and Eq.
8) =
(1) is satisfied,so we
(/ + 8)(c)= f(c)+ g(c) have
a homomorphism.
=
\302\242,(/)
+
$c{g),
\342\226\
13
Section
13.5 Example
Let E\" n
with n real-number components. (This vectors group of column course isomorphic to the direct product of E under with itself for addition Let A be an m x n matrix of real numbers. : E\" \342\200\224>\342\226\240 Em be denned by Let \302\242) for each column vector v e E\". Then 0 is a homomorphism, since for v, w e
the additive
be is of
group
factors.)
0(v) = Av E\", matrix algebra shows that 0(v + w) = A(v + linear such a map computed by multiplying algebra, matrix A is known as a linear transformation. 13.6
Example
Let GL(n, E) be the matrix A is invertible for
if and
only if its
determinant,
Z and
+
wehave0r(m Note that 0O
13.8
Example
= Gi
Example
Let F
G
in
+
r(m
denned
n) =
rm
+
homomorphism,
the map
Z,
G2 x
x
additive
group
a
the
group
multiplicative
useful for studying into of a group homomorphism
is not \302\242,.
by
=
\302\242r(n)
rn for all n
rn = 4>T(m)+ is the
map,
identity
structure
e Z. For all m,
n
e Z,
is a homomorphism.
so0,
4>r{n)
the
itself.
and
Z onto
maps
0_i
onto Z.
\342\226\262
\342\200\242 \342\200\242 \342\226\240 \342\200\242 \342\200\242 x G,\342\226\240 x \342\200\242 x G\342\200\236 be a
\342\226\240 \342\226\240 \342\226\240 \342\226\240 \342\226\240 = \342\226\240 \342\226\240, gn) gi-
of continuous functions with domain [0, 1] and numbers. The map a : F \342\200\224>\342\226\240 E defined by a{f) = fQ
group
of real
\302\260U+
8)=
8){x)dx = I
f
(f +
/
f(x)dx
Jo
Jo
for all f, g
Modulo
where r is the is a
We need to
Jo
n)
remainder
Let given
y
be
s, t
[f{x)+
map of Z
the natural
for
g(x)]dx
by the division
algorithm
into
when
Z\342\200\236 given
m is
by y(m)
divided
by
= r,
n. Show
homomorphism. show
that
= y(s)
y(s+t) for
f(x)dx
the
e F.
(Reduction
y
E be
g(x)dx= o(f) + o(g)
+ /
Jo
that
let
homomorphism, for
f efisa
Solution
E) into
are often
itself
into
\342\200\224>\342\226\240 G, where
G
be the
additive
13.10Example
Recall that Recall also that
matrices.
is nonzero.
direct product of groups.The projection map Ttj 7tj(gi. gi. gi is a homomorphism for each \342\200\242 \342\226\240 i = n. from the fact that the binary operation of G 2. \342\200\242. This follows 1, immediately coincidesin the i\\h component with the binary operation in G,\342\226\240. \342\226\262 Let G
:
13.9
=
is the trivial
Z. For all other r
x n
n
det(fi).
mapping GL(n,
a nontrivial
->\342\226\240 Z be
: Z \302\247r n)
by a
\342\226\262
of a group
let
In
left
\342\226\262
det(A),
= det(A)
is a homomorphism
Our next examplegives
Let r e
0(w).
numbers.
Homomorphisms
13.7 Example
Aw = 0(v) + vector on the
Av +
a column
have
det(AB)
Thismeans that det E* of nonzeroreal
=
w)
of all invertible
group
multiplicative
e GL(n. E) we
A, B
matrices
of G.
127
Homomorphisms
e Z. Using
the
division
algorithm,
5=
+ y(t)
we let qxn
+
ri
(2)
Part III
128
Homomorphismsand Factor Groups
and
t =
0 < r,
where
<
i =
for
n
n, then
<
the sum r\\
=
of the
Each
Consider, for
q2 +
is equal Z\342\200\236
y(s) =
to r3
also.
y(s)
+
+ qi)\302\253
the
illustration,
and
r\\
r3,
= r2.
y(t)
do indeedhave
so we
y(t),
Equation
shows
(4)
a homomorphism.
that
\342\226\2
three examples is a many-to-one map. preceding of the map may be carriedinto the same point.
homomorphisms in the points of the domain
different
is,
we see that
(3)
(qi +
we see that
(3)
Consequentlyy(s + t) =
That
=
(4)
+ r3
qin
r3.
(2) and
r2 in
+
t
=
r2
(2) and
adding Eqs. 5+
so that y(s + t) From Eqs.
(3)
r2
1, 2. If
n + for 0 < r3
q2n +
it\\ :
homomorphism
x
\"L2
Z4
->\342\226\240 Z2 in
Example
13.8 We
have
0) =
7^(0,
so
Composition
y :
\342\200\224>\342\226\240 G' and
: G \302\242)
(y 0
Properties turn
We
to
G'
0 in
=
13.11
Definition
we
Let 0
be a
of
A
Tt\\.
y{4>{g))
49.)
of Homomorphisms some structural features of G and
G'
axe preserved by a homomorphism Note the use of square brackets
that
review set-theoretic definitions. function to a subset of its domain.
: G \342\200\224>\342\200\242 G''. First we \302\242) when
= 0,
a group
again
both group
(y 0 0)(^)
X2 by
3)
That is, if homomorphism. homomorphisms then their composition for g e G, is also a homomorphism.
is
\342\200\224>\342\226\240 G\" are
\342\200\224>\342\226\240 G\", where
: G \302\242)
(See Exercise
7^(0,2) = 7Ti(0,
in Z2 x Z4 are mappedinto of group homomorphisms
elements
four
7Tl(0, 1) =
apply
a
of a set X
mapping
in Y under
0 is {0(a)| a
0_1[B]ofBinX is
{.x e
into a set Y,
e A}.
X | 0(*)
The
and
let A
c
is
the
set 0[X]
X
and
range
BcF. of 0.
The image 0[A] The inverse image
e B}.
\342\226\2
first three properties of a homomorphism in the theorem that follows stated have been of an encountered for the case in Theorem isomorphism; namely, already special 28 of Section4, and Exercise 41 of Section 5. There they were 3.14, Exercise really obvious because the structures We will now see that of G and G' were identical. they if the maps are not one to one and onto. hold for structure-relating maps of groups, even We do not consider them obvious in this new context. The
13.12 Theorem
Let 0 be a 1. 2.
homomorphism
If e is the If a
identity
G
into
in G,
then
of a group element
e G,then0(a_1) = 0(a)\"1.
a group 0(e)
G'.
is the identity
element e'
in
G''.
Section13 If H
4.
If K' is a
Let 0 be a
G, then
of
subgroup
preservesthe of G
homomorphism
0[H] is a
of
subgroup
G'.
0_1 [K'] is a subgroup
G', then
of
subgroup
speaking, 0
Loosely
Proof
is a
3.
of
inverses,
element,
identity
129
Homomorphisms
G.
and subgroups.
Then
into G'.
0(a) = 0(ae)= 0(a)0(e). on
Multiplying
element e'
G'.
in
e' = that
shows
the
Then
4>[H].
0[H] is closedunder completes
0(e). Thus
0(e)
be
must
the identity
0(e) = 0(aa_1)= 0(a)0(a_1)
(3), let
to Statement
Turning
e' =
= 4>{a)~x.
0(a_1)
two elements in
we see that
the left by 0(a)-1, The equation
proof
the
H be a
operation
that 0[H] is a
of G, and let 0(a) and 0(fo) be any e 0[H]; thus, we see that 0(a)0(fo) of G'. The fact that e' = 0(e) and 0(a_1) = 0(a)-1 =
subgroup
subgroup
G'.
K' be a
of G'. Suppose a and b are subgroup = The equation
Going the other
K' must 0(a) e
so
Hence0_1[K1] 0 : G
Let {e'}
0(a)-1 is a
(f> :
G
{x e
G
(j){x)
|
^
and let e'
homomorphism
of G',
so 0_1[{e'}] is a
is critical
be a
G' = e'}
is
the
to the
homomorphism
kernel
of 0,
then e 0_1[^'].
e 0_1[.Sr'],
a-1
\342\231\246
\342\200\224>\342\226\240 G' be a
is a subgroup
Let
= 0(e), so e e 0_1[^'].If a = 0(a_1),so we must have
element e' e Z'. But 0(a)-1 subgroup of G.
13.12. This subgroup 13.13 Definition
(4), let
the identity
contain 7T,
for Statement
way
study
be the
H of
subgroup
identity
G by
element
Statement
of G'.
(4) in
Now
Theorem
of homomorphisms.
of
denoted
groups. by
Ker(0).
The
subgroup
0_1[{e'}] = \342\226\240
the homomorphism 0 : E\" \342\200\224>\342\226\240 Em given by 0(v) = A\\ In this context, Ker(0) is called the null space of A. It consists of all v e ffi\" such that Av = 0, the zero vector. Let H = Ker(0) for a homomorphism G''. We think of 0 as \"collapsing\" 0 : G ->\342\200\242 onto e'. Theorem 13.15 that follows H down shows that for g e G, the cosets gH and Hg are the same, and are collapsed onto the single element is 0(g) by 0. That = ,?# = Hg. (Besure that you understand the reason for the uses of (), [], 4>~l [{0(g)}] and to symbolize We have attempted this collapsing in Fig. 13.14, {} in 0_1[{0fe)}]-) where the shaded rectanglerepresents G, the solid vertical line segments representthe of H = Ker(0), and the horizontal cosets line at the bottom represents G'. We view of G, which are in the shaded 0 as projecting the elements rectangle, straight down of G', which onto elements are on the horizontal line segment at the bottom. Notice the downward arrow labeled at G and ending at G'. Elements of 0 at the left, starting H = Ker(0) thus lie on the solid vertical line segment in the shaded box lying over e', as labeled at the top of the figure. Example
where
discussed
13.5
A is an m
x
n
matrix.
Part
130
III
and Factor Groups
Homomorphisms
l,H
-'[(\"'ll
H
Hx
<*>-%'}]
ftx)
y'
G'
a'
\302\242(6)
13.14 Figure
13.15
Theorem
Let
0
:
G
be a group
G'
e'
of H
Cosets
homomorphism,
collapsed by
Let a
\342\200\224
H
let
and
\302\242.
Ker(0).
e G.
Then the
set
is the
left
Proof
coset of
partitions
We want to
aH
G into
of H, and is left cosets and
the right coset Ha of H. right cosets of H are the
also into
There is a standard
way
G |
to show that
=
0(a)}
= aH.
two sets are equal;
show
that each
is a subset
other.
Suppose
that
\342\200\224
Then
4>{a).
0(a)~V(*)
where e' is so we
the two
Consequently, same.
that
show
{x e
of the
= 0(fl)}
= {*eG|0(x)
0-1[{0(a)}]
the
identity
By Theorem
of G'.
= e',
13.12, we
know
that
4>{a)~x
\342\200\224
4>{a~x),
have
0(0-1)^)
Since0 is a
homomorphism,
we have
4>{a~l)(p{x) = But this shows that a~lx ah e aH. This shows that
= e.
is in
{x
4>{a~lx), H =
&G\\
so
Ker(0), so a~lx = h 00)
= 0(a)}
=
4>{a~lx)
c aH.
for
e'.
some
h e
H,
and
x \342\200
13
Section
To h
other direction, let
in the
containment
show
131
Homomorphisms
y
so that
e aH,
=
y
ah
for some
Then
& H.
y e {x We leave
so that
= 4>{a)e'=
= 0(a)0(/j)
\342\200\224
(p(y)
(p(ah)
e G | \302\242^)=
0(a)}. demonstration
similar
the
4>{a),
4>{x)=
that {ieG
=
4>{a)}
Ha
to Exercise
52. 13.16
Example
\342\231\246
5 of Section 1 shows This means that the absolute numbers of nonzerocomplex
that
Equation
under
numbers
again
under
numbers
complex
IZ1HZ2I
{1} is a
of magnitude
numbers
f\302\260r complex
| is a homomorphism onto
multiplication
Since
multiplication.
the
that
=
\\z\\Z2\\
value function |
subgroup
1 form
the
of
of
M+, Theorem
of intersection
point
We
13.17
Example
additive
additive
: D map \302\242) for \302\242(/+
of Theorem13.15from
an illustration
give
be the
Let D be the
of
by origin. Each circleis collapsed with the positive real axis.
group of all
group
functions mapping \342\200\224 \302\242(/) /'for / e F.
\342\200\224*F, where
+ g)'
\342\200\224
g)
(/
of all differentiable
+ g'
\342\200\224
/'
=
that
that
the
the magnitude
the cosets
homomorphism
of U are onto its \342\226\262
calculus.
M into
R, and let F differentiation gives us a see that 0 is a homomorphism, We easily the derivative of a sum is the sum 0(g); functions
M
\302\242(/)+
this
and
z2. C*
positive real 13.12 shows
U of C*.Recall
a subgroup
at the
center
M+ of
group
the coordinate plane, as filling complex numbers can be viewed from the origin. Consequently, of a complex number is its distance
circleswith
z\\ and the group
mapping
M. Then
into
derivatives.
the
/' = 0, the zero constant function. Thus Ker(0) consistsof all constant functions, which form a subgroup C of F. Let us find in G mapped into x2 by \302\242, all functions that is is, all functions whose derivative x2. Now we know that x3/3 is one such function. 13.15, all such functions By Theorem Ker(0) consists
Now
the coset
form
will
We
13.18Corollary Proof
A
x^/3 +
often use
= {e},then elements of the left
into
the
following
0 :
G ^-
familiar?
look
corollary
G' is
\342\226\262
of Theorem
13.15.
map if and
a one-to-one
if Ker(0)
only
\342\200\224
{e}.
a e G, the elements into \302\242(0)are precisely the mapped shows that \302\242 is one to one. a{e] = {a}, which = e', is one to one. Now that 0(e) 13.12, we know Conversely, suppose\302\242 by Theorem element is one to one, we see that of G'. Since \302\242 e is the only element mapped identity = {e}. so Ker(0) e' by \302\242, \342\231\246
If Ker(0)
the
/ such that
C. Doesn'tthis
homomorphism
group
of all functions
In showing
view that
groups G
and
for
of Corollary a map 0 is G'.
every
coset
13.18, we modify an
isomorphism
the
outline
of binary
given prior to
structures
when
Example 3.8 for
the
structures
are
III
Part
132
Homomorphisms and Factor Groups
To Show
Stepl Step
2
Step
3
G' Is an Isomorphism
homomorphism. Show Ker(0) = {
G onto
\302\242) maps
G whose
G'.
that the
shows
13.15
H of
kernel of a group homomorphism coincide, so that gH = right cosets coincide,we
and right cosets left and when
left
Section 14 that
see in
will
We
G ->
Show 0 is a
Theorem
subgroup
: G \302\247
discussedintuitively in Section 10. Furthermore, we will of G onto this coset group appears as the kernel of a homomorphism
can
form
see
that
group, as Such
way.
a group,
subgroups
Normal Galois in radicals.
1831
polynomial Galois
are
and
given
H whose left and right cosets coincide are very a special name. We will work with them a lot
noted
by Evariste for deciding whether equation was solvable by that a subgroup H of a group introduced
were
as a tool
of
proper decomposition,
equation if one to the
corresponding
Galois'sworkin ideas considerably.
called the
subgroups,
normal
a coset H then
in a very natural in studying
useful in
14.
Section
Thus a subgroup
proper.
decomposition
a proper
is
decomposition
subgroup. Galois stated
13.19 Definition
A
H
that
Proof
If This
: G \302\242)
follows
Definition
call
the
group
subgroup
as on
his commentaries on elaborated on these He also defined normal without using the term, essentially and likewise gave the first definition
G is normal if
in
group
its
and 1869,
1865
page
of a simple
and
cosets.
Jordan,
although this
the
a
the given
an equation then an equation
solve
H
has
equation
left
149).
(page
and right
cosets coincide,that
if
is,
eG.
\342\226
all subgroups
->\342\200\242 G' is
a
we
of a group
tfgforallg
Note
13.20 Corollary
if
that
subgroup
gH =
what
Camille
an
can solve
one
then
can first
corresponding to
of
the roots
of
permutations
of permutations induced two decompositions G into what we call left cosets and cosets. right If the two decompositions coincide,that is, if the left cosets are the same as the right cosets, Galois G of
giving
a
e G.
Note
\342\226\240 Historical
a given
subgroups
\342\200\224*G' is
Hg for all g
a group
immediately 13.19.
of abeliangroups
homomorphism, from the last
are
then
normal.
Ker(0)
sentence in
the
is a normal
statement
For any group homomorphism
subgroup of
of Theorem
G.
13.15 and
\342\231\
are of primary indicated the
importance: importance
of
13
Section
Ker(0).Section that
show
to
indicate the importance then |0[G]| is finite
14 will
if
is finite,
\\G\\
Exercises
133
Exercise
44 asks us
of the
0[G]. image is a divisor of
and
\\G\\.
13
EXERCISES
Computations The straightforward 1 through the given map 0 is a homomorphism. [Hint: 15, determine whether = a the if we should whether and b in domain to is to check for all of However,
In
1. Let0
Z
2. Let0 R
addition be
\342\200\224>\342\200\242 Z under
3. Let0 R*
4. 5.
Let0
\342\200\224y 7Li be
Zg
6. Let0
R
7. Let0; G; ^* Gt and ej is the identity 8. Let G be any group given by 10. Let
R is
= \302\242(/)
F be the
let
and
R* is
G be : G ->\342\200\242
of functions f\", the second derivative
cp
group
of
< x.
divided
by 2,
divided
by
R into
mapping
=
by
Compare
g~'
for g
as in
2, as in
= (ex.
0;(g,)
by
map.
injection
given by
the
division
algorithm.
the
division
algorithm.
gi,..., er), where
ei....,
with
gt
e G,
13.8.
Example
&G,
R having derivatives
of all
Let
orders.
F be : F \342\200\224>\342\200\242
cp
of /.
mapping R
functions
continuous
all
R be : F \342\200\224y
integer
multiplicative, be given
G, be given
of Gj. This is an
when
of x
\342\200\242 \342\200\242 \342\200\242 x
group
additive
and
numbers,
and let
remainder
\342\200\242 \342\200\242 \342\200\242 x
element
additive
the
additive
remainder of x when
the
G, x
x
x Gi
= the
given by
\342\200\224y R*, where
9. LetF be
given
cp{x)
by
n.
= the greatest by
multiplication
given
cp(ri) =
by
given
\342\200\224>\342\200\242 R* under
\"Lb \342\200\224y 1j be
Let0
addition be
\342\200\224y R under
R.
into
be
Let R
additive
the
group
of real
given by =
\302\242(/)
f/o/
f(x)dx.
Jo
11. LetF be the 12. Let
additive
be the M\342\200\236
Let
of
all
=
the
det(A),
be as in Exercise on the main diagonal
and R M\342\200\236
elements
functions
of all
additive group
numbers. Let 0(A) 13.
group
R into R,
mapping
n matrices determinant of A, n x
real
with for
and
let
entries,
and let R
be the
additive
group
3/.
of real
A e M\342\200\236.
12. Let
tr(A)
is the
sum of
the
14. Let GL(n, R) be the multiplicative group of invertible n x n matrices, and let R be the additive group R be given by cp{A) = tr(A), where tr(A) numbers. Let cp : GL(n, R) \342\200\224y is defined in Exercise 13.
of real
15. Let
x &R.
F be
the multiplicative
Let R* be the In Exercises
16. Ker(0) for0
17. Ker(0) 18. Ker(0)
: S3
24, -+
Z2 in
and
0(25)
for
and
0(18)
for 0
the indicated
compute
: Z
functions
continuous
group of nonzero
multiplicative
16 through
group of all
real numbers. quantities
Example 13.3 -+
: Z -+
Z7 Zi0
such such
that 0(1) that 0(1)
= 4 = 6
mapping R Let
for
into
R that
R* be cp : F \342\200\224y
the given
are nonzero
at
every =
given
by
\302\242(/)
homomorphism
cp.
(See
f0 f(x)dx.
Exercise
46.)
Part III
134
19. Ker(0) and
20. Ker(0)
and
21. Ker(0)
and
: Z -+
0(20)
for 0
0(3)
for 0 :
Zl0
Factor
and
Homomorphisms
= (1, 4,2,
that 0(1)
such
S&
Groups
-\342\226\272 Z20 such
6)(2,5, 7)
= 8
that 0(1)
Z24 -\342\226\272 Sg where 0(1) = (2, 5)(1,4, 6, 7) 22. Ker(0) and 0(-3, 2) for 0 : Z x Z -* Z where 0(1, 0) = 3 and 0(0,1) 23. Ker(0) and 0(4, 6) for 0 : Z x Z -* Z x Z where 0(1, 0) = (2, -3)
24.
0(14)
and 0(3,
Ker(0)
for 0 :
10) for 0 : Z x Z ^
0(1, 0) = (3,
where
Sw
25. How
many
homomorphisms
are there
of Z
onto
26. How
many
homomorphisms
are there
of Z
into
Z?
homomorphisms
are there
of Z
into
Z2 ?
How
27.
many
= -5 and
5)(2,4)
1) =
0(0,
and
(\342\200\2241,5)
= (1,
0(0,1)
7)(6, 10, 8,9)
Z?
G. Let 0g : G \342\200\224>\342\200\242 G be defined G be a group, and let g \342\202\254 a homomorphism? 0g G be defined 29. Let G be a group, and let g & G. Let 0g : G ->\342\200\242 is 0g a homomorphism?
28. Let
by
by
= gx
for x e
G. For
which
= gxg~l
for x e
G. For
reference
to the
text,
g e
which
G is
geG
Concepts In
30.
31, correct the
30 and
Exercises
needed, so
it is
that
in a form
is a map such
A homomorphism
:G
31. Let0
\342\200\224y G' be
italicized term
of the
definition
without
if
correction
is
acceptablefor publication. that
a homomorphism
of
0(*)0(;y). The
groups.
o/ 0 is
kernel
{* e
G\\
= e'}
where e' is the
identity
inG'.
32. Mark
each of the a.
following
true or
false.
subgroup of S\342\200\236. G and G', there exists a homomorphism of G into b. For any two groups G'. c. Every homomorphism is a one-to-onemap. is one to one if and only if the kernel consists of the identity element alone. d. A homomorphism e. The image of a group of 6 elements under some homomorphism have 4 elements. (See Exercise may is a normal A\342\200\236
44.) f. The image
of a group
h.
In Exercises
exists. If
no
There
i.
A
j.
It
33
33.
Z12
35.
Z2 x Z4
37.
Z3^S3
39.
Z x
under
a homomorphism
may
12 elements.
have
43,
->\342\200\242 Z5 ->\342\200\242 Z2 xZ,
Z^
2
elements
possible to have a nontrivial give an
homomorphism
41.
of some
is a
is not
through
such
of 6 elements
into some group of 12 elements. group of 6 of 6 elements into some group of 10 elements. homomorphism of some groups homomorphism may have an empty kernel.
g. There is a homomorphism
homomorphism
example of a nontrivial
exists, explain
homomorphism
that
why
of some
is so. You
34.
Z12
-+ z4
36.
z3-
*
38.
0
z^
z
S3
40.
may
use
finite
group
0 for the Exercises
into some
given groups, 44 and 45.
infinite
if an
group.
example
135
Factor Groups
Section 14
Theory
44.
Let
: G
be a group
homomorphism.
Show
that if
G' be a group
homomorphism.
Show
that
-+ G'
\\G\\
is finite,
then |0[G]| is
is finite,
then,
is a divisor
and
finite
of|G|.
45. Let
G -+
if
|G'|
is finite
\\cp[G]\\
and is a
divisor
of|G'|.
46.
47.
Let a group G be generated set and a,- e Gforalh' e /.Let : G \342\200\224y G' some indexing by {a,- \\i e /}, where/is = G' be two homomorphisms from G into and /x : G \342\200\224>\342\200\242 a group G', such that (\302\253,) I. Prove /x(a,) for every i \342\202\254 that cp = /x. [Thus, for example, a homomorphism of a cyclic group is completely determined by its value on a of the group.] [Hint: Use Theorem7.6and, of course, Definition 13.1.] generator Show that any group homomorphism or a one-to-one map.
48. The sign of
an
even
G' where | G | is a prime
G -+
is +1
permutation
\342\200\224>\342\200\242 defined : 5\342\200\236 {1, \342\200\2241} sgn\342\200\236
cp :
the
and
is
permutation
trivial
homomorphism
that the
\342\200\224 1. Observe
map
by
sgnn(ff) = is a homomorphism
odd
of an
sign
be the
either
must
the of S\342\200\236 onto
of a
sign
kernel? Compare with
is the
What group {1, \342\200\2241}.
multiplicative
Example
13.3.
49. Show
that
if G,
G', and
are
G\"
groups
and if
: G
compositemap y
51. Let
that
-+ G'
is abelian
0[G]
-+
y : G'
and
if
are
G\"
and
only if
of G. Let
52.
Let
G be
G -+ = Ha.
(j) :
G' be a homomorphism
kernel
with
H and
let a
e G. Prove
the
then the
homomorphisms,
for
all
x, y
= a\".
0(\302\253)
set equality
e G, we have
Show
{x e
that
0 is a
G | (j>{x)
=
e G and let
53. Let
G be a
group,
Let h,k
sufficient
54.
55. Let
G be a
0
group,
< n. Give
h an
element of
= h' for -+ G be defined by 0(i) : Z\342\200\236 G, and n a positive integer. Let \302\247 and sufficient condition (in terms of h and n) for 0 to be a homomorphism. Prove
a necessary
assertion.
section
14
Groups
Factor
Let H be a of G,
H.
subgroup
listing
of a finite heads
element
group G. Supposewe
at the top
in Section
and
at the
10. The body
write
left as
a table they
for
occur
the
group
in the left
operation
cosets of
table may break up into blocks to the cosets a on the cosets, or they corresponding (Table 10.5), giving group operation that not break start this section We up (Table may way 10.9). by showing that if H is the kernel of a group homomorphism that left G', then the cosets of H (remember 0 : G ->\342\200\242 and right cosets then coincide) are indeed elements of a group whose binary operation We
this
illustrated
is derived from
the
group
operation
of G.
of
the
136
Part III
Factor
and
Homomorphisms
Groups
from Homomorphisms Let G be a group and let S be a set having the same cardinality to-onecorrespondence*>between S and G. We can use *> to Factor
Groups
on S, making
if x
The direction
the
y *>
z *>
and
g2
S as
x,y e
of xy for
computation
and
\302\261* gi
operation
g\\g2,
then
follows: xy
= z.
(1)
s & S and g e G correspondence s -^ g between ^- of *> S onto G. (Of course,the direction /x_1). Expressed in terms of /x, the computation (1) of xy
one-to-one
\342\200\224\302\273\342\200\242 of the
function
a one-to-one
gives us the for
a binary
a group isomorphic to G. Naively, use the correspondence we simply of G by the name of its corresponding element in S. element (under \302\253\302\273)
to renameeach
us
Then there is a one-
define
S into
We can describeexplicitly
gives
as G.
function
inverse
/x
mapping
becomes
ijei
if/i(x) = gi
and
/x(;y)
(i(z) =
and
= g2
gig2,
then
*;y
= z.
(2)
an isomorphism mapping the group S The map /x : S \342\200\224*G now becomes = /x(z) = g\\g2 = /x(*)/x(;y), the (2), we obtain ^{xy) groupG. Noticethatfrom
onto the required
property. G' be a homomorphism, and let H = Let G and G' be groups, let
Ker(0). have
a
of the
x = ah for some h & H, then that if x e aH, so that subgroup 0[G] of G'. Remember of 0[G] 0(;t) = 0(a/j) = 0(a)0(/j)= 0(a)e'= 0(a), so the computation of the element corresponding to the coset aH = xH is the same whether we compute it as 0(a) or as (We read G/H as \"G over H\" or (j>{x).Let us denotethe set of all cosets of H by G/H. as \"G modulo H\" or as \"G mod H\" but never as \"G divided by H\ In the preceding a homomorphism G' having with : G \342\200\224*\302\242) paragraph, we started kernel H, and we finished with the set G/H of cosets in one-to-one correspondence with the elements of the group In our work above that, we had a set S with elements 0[G]. in one-to-one correspondence with a those of a group G, and we made S into a group
to G with
isomorphic
0[G] that
in
that
an isomorphism /x. we can consider /x. In terms of G/H and
construction,
isomorphism
yH e
for xH,
(xH)(yH)
then
4>(x)4>(y)\\
and
0 is
a
that
the product xy on our depend
the
we
the
and
(j>(y)
G,
fi(zH)
= (j>(x)(j>(y),
can easily
here
of xH
and
z e
find
G such
that
ii{zH)
=
and find that =
ftxy)
y in
and
again
in
n(xyH) =
choices x from xH
We demonstrate it xh\\ is an element
=
H)
of two (xH)(yH) G. While this
product
of x
G by G/H and replacing isomorphic to 0[G] with (2) of the product computation
a group
(3)
z = xy
fiizH) = This shows
ix(y
homomorphism,
we take
namely,
0[G],
be
= zH.
(xH)(yH)
But because
to
G/H becomes
= (j>(x)
if fi(xH)
S by
Replacing
G/H
cosets computation
0W0(y). is the
coset (xy)H
of (xH)(yH)
that
contains
may seem
to
our work above shows it does not. because it is such an important e H so that point. lfhi,h2 there exists hj, e H such yh2 is an element of yH, then and
y from
yH,
Section 14
that
= yh3
h\\y
because Hy =
(xhl)(yh2) = obtain
yH by =
x(Iny)h2
Thus we have
13.15.
Theorem
= (xy)(h3h2)
x(yh3)h2
of
137
Factor Groups
\342\202\254 (xy)H,
cosets is accomplished of the cosets, the coset in G of the choices. that contains the product Any time we define something (like a to show in terms of it is that it is well defined, which means choices, important product) that it is independent of the choices made. This is precisely what we have just done. We this work in a theorem. summarize so we
choosing
by
14.1 Theorem
Let
: G \302\242)
a factor
same
the
coset.
\342\200\224*G' be
a group
homomorphism where (aH)(bH) =
G/H.
group,
Computation
the
from each coset and
an element
of two
product
as product
taking,
H. Then
kernel
with
{ab)H.Also,
the
the cosets of H form
map
/x
= 4>{a) is an isomorphism. Both cosetmultiplication defined by /i(aH) a and b from the cosets. defined, independent of the choices
14.2Example
13.10 considered the Example m is divided by n in accordance
= nZ. By Theorem 14.1, homomorphism. Of course, Ker(y) of nL are the residue The cosets group Z/nZ is isomorphicto Z\342\200\236. example,taking n = 5, we see the cosets of 5Z are
-10,-5,0,5, {\342\200\242\342\200\242\342\200\242
5Z=
5Z
smallest
its
It is
and
them
Example
(4
1 + in 4
For
10
-8,-3,2,7, 12,
:
-6,-1,4,9,14,
{\342\200\242\342\200\242\342\200\242 \342\200\224*Zg of
Z/5Z
That is,
Theorem
/x(5Z)
=
0,
14.1 assigns to each /x(l + 5Z) = 1, etc.
coset
of \342\226\262
very important that we learn how to compute in a factor group. We can multiply cosets by choosing any two representative elements, multiplying (adding) the resulting product (sum) lies. finding the coset in which
Considerthe +
factor
{\342\200\242\342\200\242\342\200\242
element.
nonnegative
the
modulo n.
two
(add)
14.3
/x
classes
{\342\200\242\342\200\242\342\200\242
5Z=
the isomorphism
that
Note
we see that
-7,-2,3,8, 13,
3+5Z= 4 +
well
/x are
and
1,6, 11,
-9,-4,
1+5Z={---
2 +
-*- 0[G]
G/H
where Z\342\200\236, y : Z \342\200\224>\342\226\240 y{m) is the remainder when with the division algorithm. We know that y is a
map
5Z=
:
5Z)
by
factor
group
choosing 2
Z/5Z and
4,
5Z. We couldequally well + 5Z; the sum 27 + (-16)
with the finding add these
cosets shown = 6, and two cosets by
2+4
= 11is alsoin
the
coset
We can
above. noticing choosing
1 +
5Z.
add (2 +
that
6 is
in
27 in
2+
5Z and
the
5Z) + coset \342\200\22416 \342\226\262
factor groups L/nL in the preceding example are classics. Recallthat we refer n. Two integers cosets of nL as residue classesmodulo in the same coset are is carried over to other factor groups. A factor congruent modulo n. This terminology called the factor group of G modulo in the same H. Elements group G/H is often H. By abuse of notation, coset of H are often said to be congruent modulo we may as the additive group of residueclasses and think of Z\342\200\236 of sometimes write Z/nZ = Z\342\200\236 The
to the
Z modulo
(n), or
abusing
notation
further,
modulo
n.
138
Part III
Factor
and
Homomorphisms
Factor
Groups
from Normal Subgroups
Groups
So far, we have obtained factor groups from homomorphisms. Let G be a group and only let H be a subgroup of G. Now H has both left cosets and right cosets, and in general, a left coset aH need not be the same set as the right coset Ha. Suppose we try to define
operation on
a binary
cosets
left
by defining
= (ab)H
(aH)(bH)
as in by it
a
b from
and
choosing representatives a well-defined gives operation,
chosen from
cosets.
the
if and
operation
binary
14A Theorem
14.1Equation
of Theorem
statement
the
Let H be a
subgroup
4 attempts
to define left
the cosets.
4 is
independent of
The theorem
only if H is a normal G. Then
of a group
Equation
the
that Eq.
coset
unless
meaningless
a and b
elements
4 gives a well-defined
of G.
subgroup
left
coset multiplication
representative
shows
follows
that
(4)
is well defined
multiplication
by
the
equation
= (ab)H
(aH)(bH)
if and Proof
is a normal
if H
only
Suppose first cosets. Let
that
(aH)(bH)
a e G. We
want
technique of showing
standard
of
subgroup
G.
= (ab)H does give a well-definedbinary to show that aH and Ha are the same that each is a subset of the other.
on left
operation
set.
use
We
the
we have Let x&aH. a^&a^H, Choosing representatives x e aH and (xH)(a~lH) = (xa~1)H.On the other hand, choosing representatives a e aH and = eH = H. Using our assumption a'1 e a~lH, we see that (aHXa^H) that left coset = well we h e H. Then is must have xa~l defined, by representatives multiplication x = ha, sox & Ha and aH c Ha. We leave the symmetric proof that Ha c aH to Exercise 25. turn now to the converse:If H is a normal We then left coset multiplication subgroup, is well-defined. Due to our hypothesis, we can simply say cosets, by representatives to compute (aH)(bH). omitting left and right. Suppose we wish Choosing a e aH and b e
bH, we
bh2
\342\202\254 we bH,
Now
h\\b
obtain
the
coset
obtain the
= \342\202\254 Hb
bH, so
h\\b
(ahi)(bh2) and
(ab)(h3h2)
14.4
Theorem
well-defined with
14.5 Corollary
Let
such
H
binary
\342\202\254 (ab)H.
binary
{ab)H.
Choosing
coset ahibh2H. bhj, for
=
=
a(h\\V)hi
Therefore,
must
some
hj,
=
ah\\bh2
e
that these H. Thus
in
are the
e aH same
and
coset.
= {ab){h^hi)
a(bh3)h2
is
ah\\
representatives
show
\342\231\24
(ab)H.
shows that if left and right cosets on cosets. We wonder operation
coset multiplication.
of H whether
coincide, the
cosets
Eq. 4 gives a do form a group
then
This is indeed true.
G. Then = (ab)H. operation (aH)(bH)
be a normal
different
We
subgroup of
the
cosets
of H form a
group
G/H
under
the
\342\226\26
Proof
= [a(bc)]H,
= (aH)[(bc)H]
(aH)[(bH)(cH)]
Computing,
= [(ab)c]H, so associativity = (ae)H = aH = (aH)(eH)
[(aH)(bH)](cH) G.
Because
the
identity
Definition
The group
G by 14.7
Example
14.8
Example
G/H
in
the
in =
(ea)H
139
Groups
and
have
we
similarly,
from associativity in we see that eH = H is (eH)(aH), = (a~la)H = eH = (aa^)H = follows
G/H
(a~lH)(aH)
Finally,
= {aHTx.
thata^H
shows
(aH)(a~lH)
14.6
in G/H.
element
Factor
14
Section
\342\231\246
is
corollary
preceding
factor
the
group) of
(or quotient
group
H.
\342\226\240
nZ is a normal subgroup. Corollary 14.5allows us to Since Z is an abelian group, construct the factor group with no reference to a homomorphism. As we observed Z/nZ \342\226\262 in Example 14.2, Z/nZ is isomorphic to Z\342\200\236.
Consider the abelian group of E contains as elements
under
\342\200\242 \342\200\242 \342\200\224 \342\200\242 0, c, 3c, \342\200\2242c, \342\200\224c,
one elementof
of (c) contains just
coset
Every
elements as representativesof
cosets
the
discussedfor
sum
modulo
c as
5.37,then
the
is
contains 8.07 + (5.37), which coset elements x where 0 <
8.07
coset
the
with
Example
4.2 is isomorphicto
all x e
Ec. Of
seen that the
have
We =
E/ (c)
1, 3,
group \342\200\242 \342\200\242 n \342\200\224 \342\200\242, 1}, the
4,
under
an
Z/(\302\253)
set of
is
that 0
the
in
c. If
in Ec
computation
these
choose
we
we find that
E/(c),
in
we are 1.
Section
cosets 4.65 + (5.37)and 3.42 + (5.37) - 5.37 = 2.7,which is 4.65 +5.37 3.42.
x < c, we
isomorphism
is then also isomorphicto E/(c) 1 under multiplication.
course,
of magnitude
Z\342\200\236 {0,
of the
these
Working
numbers
sum
(c)
subgroup
cyclic
\342\200\242 \342\200\242 \342\200\242.
when computing
c=
their
2c, 3c,
x such
For example, if
computing
E+. The
c e
and let
addition,
-f where -fy (x) the circle group
Ec
group
of
= x
+ (c) for
U of
complex \342\226\262
to the
isomorphic
integers
nonnegative
the
see that
thus
as a set,
and Z\342\200\236,
group
less
than n.
Example 14.8
E/(c) is isomorphic to the group Ec. In Section 1, we choosethe than the conventional [0, c) for the half-open interval of nonnegative less than c. We did that to bring out now the comparison real numbers of these factor groups of Z with these factor groups of E. that
shows
notation
the group
Ec rather
Homomorphism Theorem
The Fundamental We
have
G' gives rise to a natural : G \342\200\224*homomorphism \302\242) We now that each factor group show 14.1), namely, G/Ker(0). as H kernel. homomorphism having
seen
(Theorem
to a
natural
14.9Theorem Let H be a homomorphism
Proof
Let x, y
e
that every
normal
of G. Then
subgroup
with kernel
y
: G
G/H
given
H.
G. Then
y(xy) =
(xy)H = (xH)(yH)
=
y(x)y(y),
by
factor G/H
y(x)
group
gives
rise
= xH is a
Part
III
Homomorphisms and Factor Groups
G/H
14.10 Figure
so
is a
y
= H if
Since xH
homomorphism. H.
an
if x
only
see that the
\342\202\254 H, we
kernel of
y is indeed
We
have seen
in
\342\231\2
/j,:
where y is a homomorphism as a theorem.
14.11
Theorem
(The Fundamental homomorphism
and
is
jx
/i(gH) = 0(g) y(g) = gH, then
is an 0(g)
Let
Theorem)
Homomorphism
and 0[G] is a group, If y : G \342\200\224*is G/H for each g e G.
H. Then
kernel
with
with kernel H, G' is a homomorphism 0 : G \342\200\224>\342\226\240 = 4>{g)is an isomorphism. 14.9 shows Theorem = gH is a homomorphism. Figure 14.10 shows the homomorphism \302\247 can be factored, 0 = fiy, an isomorphism of G/H with 0[G]. We state this
14.1 that if
Theorem
\342\200\224>\342\226\240 where n(gH) G/H 0[G] that defined by y{g) y : G -*- G/H these groups and maps. We see that then
isomorphism.
= /xy(g)
G' be : G \342\200\224*-
/x : G/H
a group
\342\200\224>\342\226\240 0[G] given
the homomorphism
given
by by
to as a natural 14.11 is referred or canonical to the are used describe homomorphism y. There isomorphism, but be other and for these the same may isomorphisms homomorphisms groups, maps with are uniquely determined by Theorem 14.11. /x and
and
Homomorphisms
14.12Example
the
Classify
generated
Solution
The
groups
map
onto Z4 with
Normal We
factor
some an
easier
decompositions.
related. We
Z2)/({0} x Z2)according (Theorem 11.12).
it\\ : Z4 kernel
Subgroups
derive with
are closely
groups
an example
give
alternative way to
indicating
can be.
x
\342\200\224>\342\200\242 Z4 given
by
Z2
{0} x
to the
fundamental
it\\{x,
y)
= x
Z2. By Theorem14.11,we
theorem of finitely
is a homomorphism of that the given factor
know
to Z4.
is isomorphic
group
us
and
(Z4 x
group abelian
projection x Z2
Z4
/x in Theorem same adjectives
this relationship
useful
how
the
isomorphism
\342\226\2
and
Inner
Automorphisms
characterizations of normal
check normality
than
finding
subgroups,
which
both the left
and
often provide right coset
the
Section 14
Then gHg~l
gHg~l
= H. We
Consequently, Suppose and
all
ghig~l = Hg gH
that
Hg.
But
c gHg~l
e H, we
=
/i
obtain
e gHg~l,
all g
G. Theng/j
=
then
G,
g~'/jg
e G. We
claim
e G. Let h = g~lhg
/ii g,
this means
paragraph,
= H for all g & = H giving g~lHg
also,
for
e G
e H for all g
ghg~l
g-'/iC?-1)-1 we are done.
and
for all g e
preceding
that
such
c H for all g
e H}
that H
show
must
//. By the if gHg~1
A e
Conversely,
gH c
= {ghg~l\\ h
relation ghg~l
in the
g-1
of G
is a subgroup
H
that
Suppose
heH.
141
Factor Groups
& H.
=
hi
= H
gHg^1
ghg~1 = hi = h2, so
actually Replacing g by where hi e H.
e H for
so ghg^1
that
g/j =
so that
e
h\\g
e G
all g
for all g
= gh2
/jg
all
and
that
G.
&
and
Hg,
c
and Hg
gH.
14.13Theorem
The
our work
summarize
We
are three
following
normal
H for 2. gHg'1 = H for 1.
ghg~l e
3.
gH = Hg for all geG.
Condition
(2)
H of a group
14.14Example
for
all h
e H
all all
G
g e g
and
G
group
to be a
H.
A e
G.
&
14.13 is
of Theorem
taken
often
as the
definition of a
normal
subgroup
G.
H of an
subgroup
Every
H of a
for a subgroup
conditions
equivalent
G.
of
subgroup
as a theorem.
abelian
all g & G,
and
group
so, of
G
We need
is normal.
course, ghg^1 =
h
e H
note
only
for all g
that
e G and
gh =
hg
h
H.
all
&
\342\226\262
Exercise
29 of
Section 13 shows that
the
map
G ^- G denned gag~l = gbg^1 if ^ :
is a homomorphism of G into itself. We see that is one to one. Since g(g~lyg)g~1 = y, we see that ig of G with itself.
14.15
Definition
An isomorphism 0 automorphismof
:G
ig : G
automorphism
G by
\342\200\224*G of
a group
\342\200\224*G, where
^(x)
ig
is onto
by and
G, so it
^(x)
only ifa = b, so is an isomorphism
G with itself is an automorphism = gxg-1 for all x e G, is the inner
g. Performing ^ on x is calledconjugation
of x
= gxg-1
by g.
of G. The \342\226\240
(1) and (2) in Theorem 14.13 shows that gH = Hg = for all g e G if andonly if ig[H] Hforallg e G, that is, if and only if His invariant = H is an all inner automorphisms of G. It is important to realize that ig[H] under = in sets; we need not have h for all h e H. That is ^ may perform a equation ig{h) G are nontrivial permutation of the set H. We see that the normal subgroups of a group K of G is under all inner automorphisms. A subgroup precisely those that are invariant = a conjugate subgroup of H if K ig[H] for someg & G. The
equivalence
of conditions
Homomorphisms and Factor Groups
III
Part
142
14
\342\226\240 EXERCISES
Computations 1. 3.
8, find
1 through
Exercises
In
group. (Z4 x
2.
(Z4 x Z2)/<(2,
4.
1))
(Z2
x Z4)/((l,
1))
7.
(Z2
x 53)/<(l,
A))
15, give
9. 5 + (4) in (2, 1) +
element
of the
order
the
factor
the
in
<2\302\273
x Z5)
x Z5)/({0}
(Z3
x
Z12)/\302\2532)
6. (Z12x 8- (Zn x
9 through
Exercises
11.
factor
given
Zfi/<3)
5.
In
the order of the
Z18)/((4, 3)) Z15)/<(1, 1))
group.
26 + (12) in Z60/(12) 12. (3,1) + ((1,1)) in (Z4 14. (3, 3) + ((1, 2)) in (Z4 10.
Z12/(4)
((1,1)) in
x Z6)/((l,
(Z3
1))
13. (3, 1)+ ((0,2))in (Z4 x Z8)/((0, 2)) 15. (2, 0) + ((4,4)) in (Z6 x Z8)/((4,4)) = 16. Compute iP] [H] for the subgroup H
/xi}
{po,
group S3 of Example
of the
x Z4)/((l,
1))
x Z8)/((l,
2))
8.7.
Concepts
In Exercises
17 through
is needed, so that 17.
A normal
18.
A
normal
19.
An
20.
What
subgroup
H
subgroup
H
21. A
a. Why b. c.
22.
to
is asked
starts
for
= Gh
hG
e H
g~lhg
for
all
H.
h e
G into
H and
all
g e
G.
G.
group G?
of a
subgroup
h e
all
theoremsabout
factor
groups.
of an
abelian
The next
two
exercises
are designed
that
show
normal
is abelian. Let a
the instructor the student
should
is a
if H
subgroup
group
G/H
is abelian. The
G, then
as follows:
that G/H
show does
What
reading have written?
this
and
proof
b be
two elements
expect
to
find
of G/H.
nonsense
from
here on
in
the
torsion
23. Mark
each of the a.
It
following
sense
makes
true or
false.
to speak
of the
factor
abelian
group
group
G/N if and only
if N
is a normal
G.
b.
student's
paper?
the proof.
Complete
A group is torsion order. group is a group all of whose elements have finite the only element of finite order. A student is asked to prove that if G is a torsion group, H of G. The student writes every normal subgroup We must show that each element of G/H is of finite Let x e G/H. order. Answer the same questions as in Exercise 21. A
text, if correction
publication.
nonsense when first proving basic type of error.
proof
We must
the
to
to one
student
student's
reference
term without
italicized
the
G is a homomorphismmapping
of a normal
importance
write
often
Students
to call attention
acceptable of G is one satisfying of G is one satisfying
for
of a group
automorphism is the
the definition of
19, correct
in a form
it is
Every
c. An
subgroup inner
of an
automorphism
G is a
of an abelian
group
normal subgroup must
be just
of G.
the identity
map.
free if then
subgroup
the
so is
identity
is
G/H for
of the
group
d. Every
factor
e. Every
f.
Every
g. Every
factor
h. Every
factor
i.
j.
group
of a
finite
factor
group
of a
torsion group is a
factor
group
of a torsion-free
group
of an abelian
group
of a nonabelian
Z/nZ
is cyclic
of ordern.
R/nR
is cyclic
of order
order.
finite
torsion
(See Exercise
group.
free. (See
is torsion
group
22.)
Exercise 22.)
is abelian.
group
group is
nonabelian.
{nr | r e
nR =
where
n,
of
is again
group
143
Exercises
14
Section
R] and
R is
under addition.
Theory
24. Show
normal subgroup of
is a A\342\200\236
that
find a known
and compute S\342\200\236
that is, S\342\200\236/A\342\200\236;
to which
group
is S\342\200\236/A\342\200\236
isomorphic.
25. Complete
the
14.4
of Theorem
proof
is a
if if
that
showing
by
= (ab)His well defined, then Ha c aH. subgroup T of an abelian group G is a normal
G
a group
of
subgroup
coset
if left
and
multiplication (aH)(bH)
26. Prove
the torsion
that
and
of G,
subgroup
that
T is
G/
torsion
free. (SeeExercise22.)
27.
A
H is conjugate to a subgroup = K. Show that conjugacy is an
subgroup
that
ig[H]
28. Characterize
normal
the
29.
to Exercise
Referring
30.
Let H
31.
Show
be a normal
32. Given
any
cells where
(Example 8.7) that m = (G
are
: H). Show
G is
of a group
again
H
am e
that
in the
ig of G G.
partition
such
given
to {p0, /x2}-
conjugate
a normal
of
subgroups
appear
they
sense to speak of the
that it makes
automorphism
of
for
every
a e
G.
subgroup of G. normal
smallest
subgroup
that
31.]
of G that can be expressed in the form aba~lb~l for somea, b e G is a exercise shows that there is a smallest normal C of a group G subgroup the C is of the commutator G. Show that G/C is an subgroup subgroup
G.
group.
Show
that
if
35.
Show
that
if H
G be a
x~lHx
and
of G
G,
normal
N is
and
order,
given
normal
is a normal
then H
G, then
in
H n
N
of G.
subgroup
is normal
in H.
Show
in G.
normal
one subgroup
of a
of G.
subgroup
fixed
order
finite
[Hint: Use the
fact
s. Show that if
that
the
of
intersection
H has order s,
then
all
so does
G.]
the inner function
under
be
H of a
subgroup
of a group
at least
all automorphisms
b. Show that
one
exactly
subgroups need not
HON
of order 5 is a
x e
all
are
containing
group
that
Show
N
that
of G
for
group G has
a finite
an example
subgroups
37. a.
let
and
subgroups
show
G,
group
G,
of the
terms
inner
an
collection
on the
relation
The preceding in G; all commutators
34.
36. Let
of a group
of normal
Exercise
of 53
subgroups
An element
group. in
containing
by
Use
[Hint:
G be a
commutator
abelian
S of a
all
find
subgroup
subset
contains S. 33. Let
27,
an intersection
that
of a group G in exercise. preceding
the
in
there exists
G if
group
equivalence
subgroups
relation
by the conjugacy
K of a
of a group
G form
automorphisms
composition.
under
a group
G form a [Warning: Be sure
of a group
normal to
show
function
composition.
subgroup of the group of all automorphisms that the inner automorphisms do form a
subgroup.]
38. Show that
the
of a group
G.
39. Let G and
G'
homomorphism
H'. (This fact
set of all
g e
G such
that
ig : G
\342\200\224*G is
the
be groups, and let H and H' be normal of G into G'. Show that 0 induces a natural is used constantly in algebraic topology.)
identity
inner
automorphism
subgroups of G homomorphism
and
G',
: (G/H) 0\342\200\236
ie is
a normal
respectively.
subgroup
Let 0
be a
\342\200\224\302\273\342\200\242 {G1 /H')\\i(j)[H]
c
III
Part
Use
a.
the
properties
The n
b. Then
x
n
x n
a. b.
c.
det(AB)
= det(A)
\342\200\242
matrices
with determinant
matrices
with
G be a group, subset AB = \\ab\\a
Let
Homomorphisms and Factor Groups
and
let
det(S)
determinant
=
det(7\342\200\236) 1
1 form a
normal
\302\2611 form
a normal
be the set e B}.
dP{G)
e A,b
and
of all
subsets
x
for n
n
to show
matrices
the
following:
of GL(n,
W). subgroup of GL(n, W).
subgroup
of G.
For
any
e 3?{G), let us
A, B
define
the
product
this multiplication of subsets is associativeand has an identity element, but that i/\\ G) is not a under this group operation. of G, then the set of cosetsof N is closed under the above Show that if N is a normal subgroup operation on 2P(G), and that this operation agrees with the multiplication given by the formula in Corollary 14.5. Is Show (without using Corollary 14.5) that the cosets of N in G form a group under the above operation. its identity element the same as the identity element of 3\\ G)? Show
section
that
15
Factor-Group Factor
groups
Computations and SimpleGroups
can be a
tough topic for students
to strengthen
to grasp.
There is
nothing
We start by
like
a bit
of
understanding improve attempting concerning factor groups. Sincewe will be dealing with normal subgroups a subgroup of a group this section, we often denote G by N rather than throughout by H. In the N Let N be a normal of G. the acts as factor subgroup subgroup group G/N, element. as to a either to 0 in We N identity single element, may regard being collapsed additive notation or to e in multiplicative notation. This collapsing of N together with the algebraic structure of G require that other subsets of G, namely, the cosets of N, also collapse into a single element in the factor group. A visualization of this collapsing is providedby Fig. 15.1. Recall from Theorem 14.9 that y : G \342\200\224>\342\226\240 defined by G/N = aN a for e G is a G onto is of 15.1 similar to y(a) homomorphism G/N. Figure very but in the 15.1 the under is formed Fig. 13.14, Fig. image group homomorphism actually at the bottom of the figure as obtained by collapsing from G. We can view the \"line\" G/N each coset of N in another to a point of G/N thus corresponds copy of G. Each point to a whole vertical line segment in the shaded portion, representing a coset of N in G. It is crucial to remember that multiplication of cosets in G/N can be computed by in of the cosets as shown elements in the figure. G, multiplying using any representative computation
mathematics.
in
our intuition
i'
I I
I I
N
bN
I I
/~
aN
15.1
(cN)(bN) = (cb)N Figure
(ab)N
=(aN)(bN)
cN
c
to
Factor-Group Computations and SimpleGroups
15
Section
145
two elements of G will collapse into the same element of G/N if they an element of N. a and b collapse together if ab~v is in N. by Multiplicatively, The degree of collapsing can vary from nonexistent to catastrophic. We illustrate the two extreme cases by examples. Additively,
differ
15.2
Example
The
trivial
Since N =
Solution
the
Z/{0} 15.3
Example
Solution
{0}has
are of
cosets
{0} of
N =
subgroup
one
only
the form {m}
~ Z. Each m
e
Z
element, every coset of N has only for m e Z. There is no collapsingat
simply renamed
Z is
a normal subgroup.
of course,
is,
positive integer. The set nE = {nr normal since E is abelian. Compute
in
{m}
Let
n be a
\\ r
and
it is
E/nl
A bit
of
shows
thought
that actually
and x/n e E. Thus E/nE a trivial group consisting
has only only
of the
nE =
e
Compute Z/{0}.
one all,
That is,
element.
and consequently, \342\226\262
Z/{0}. .
E, becauseeach
one element,the
of E under
is a subgroup
x
e E nE.
subgroup
addition,
is of the form The factor
n{x/n) is
group
identity element.
\342\226\262
~
G As illustrated in Examples 15.2 and 15.3 for any group G, we have G/{e] of the e. where is the element and G/ G ~ {e}, trivial identity {e} group consisting only We would like These two extremes of factor groups are of little importance. knowledge about the structure of G. If N = {e}, of a factor group G/N to give someinformation G the factor as G and we might as well have tried to study group has the same structure = no structure to If N the factor has information G, supply directly. significant group about of G, then is a G. If G is a finite and N / {e}is a normal group subgroup G/N have a structure than smaller group than G, and consequently more G. The may simple multiplication of cosetsin G/N reflects the multiplication in G, since products of cosets can be computed in G representative elementsof the cosets. by multiplying We give two examples that even when has order 2, we may be able to G/N showing If G is a finite deduce some useful results. and then group G/N has just two elements, H containing we must have \\G\\ =2\\N\\. Note that every subgroup half the elements just of a finite group G must be a normal subgroup, since for each element a in G but not in both aH and Ha must the left coset the coset consist of all elements in G that H, right are not in H. Thus the left and right of H coincide and H is a normal subgroup cosets
of G.
15.4Example
Because
\\Sn\\
Let a be
an
An
\"even\"
=
2
\\A\342\200\236 |, we
see that An is a normal subgroup of Sn, and Sn/An = in S\342\200\236, so that S\342\200\236/A\342\200\236 {An, a A\342\200\236}. Renaming
odd permutation and the element
a
A\342\200\236 \"odd,\"
the multiplication
has order
2.
the element
in
shown in S\342\200\236/A\342\200\236
=
odd
Table 15.5
becomes
15.5Table (even)(even)
An
aAn
An
aAn
An
oAn
oAn
An
= even
(even)(odd) = odd
Thus the Sn.
factor
group
reflects these
multiplicative
(odd)(even)
(odd)(odd) properties
= even. for all
the
permutations
in
A
Part III
146
and Homomorphisms
Factor
Groups
that while knowing the product of two cosets in G/N does Example 15.4 illustrates us what the product of two elements of G is, it may tell us that the product in G of two types of elementsis itself of a certain type. not
15.6 Example
tell
states if 77 is a subgroup We show that it is false
H of G having
of Lagrange
theorem
The
group G, then the order of 77 divides the order of G. there must exist a subgroup that if d divides the order of G, then no that d. Namely, we show A4, which has order 12, contains a finite
of
order
of order
subgroup
of Lagrange)
of the Theorem
Converse
the
of
(Falsity
6.
77 were a subgroup of A4 having order 6. As observed before in of A4. Then A4/H it would follow that H would be a normal subgroup only two elements, H and crH for some a e A4 not in H. Since in a group we would have 7777 = 77 and the of each element is the identity, square = H. Now in a factor group can be achieved by computing computation
Supposethat 15.4,
Example have
would
of order
2,
(crH)(crH)
with representatives
a e H we
the
in
H
every elementin
square of
A4
so (1, 2, 3) and
(1,
8 elements
least
Example
Let
x
Z4
are
in
4),
and
in H,
(2,
several examplesthat and abelian,
H = {(0.0), Since
factor
its
(0,
has 24 elements and and (Z4 x Z(,)/H must
(Z4 x Zg)/7/ (remember, from the original group). // =
we
compute
In additive
(0,0) + 7/,
1)). Here
((0, 1))is
the
6 elements, all cosets 4. Since Z4 x Zg
order
in a factor notation,
(1,0)+
start
we
a
such
subgroup
cyclic
2), (0, 3), (0, 4),(0,5)}.
has
H have
group
fundamental theorem (Theorem11.12).
to the
1), (0.
groups. If the group will be also. Computing
factor
compute
then
according
x Zg
Z4
computation shows that (1, 2, 4), (1, 4, 2), are all in H. This shows that there must be at 6. \342\226 order fact that H was supposed to have
the factor group (Z4 x Zg)/((0, generated by (0, 1). Thus
Z(,
= (1,2,3)2
(1,3,2)
4, 3) the
means classifying it
6 elements,
and
A similar
H.
contradicting
generated
us compute
H of
3, 2) (2, 3,
turn to
now
We
with is finitely factor group
15.7
3),
for
and
= (1,3.2)2
(1,2,3)
(1,3,4), (1,4,
group. Thus, computing in A4, we find that for each each P e oH we must have is, the P2 e H. That must be in H. But in A4, we have
original
a2 e
have
must
7/,
the
group
cosets (2,0)
by
means
of H is
must
abelian,
have
so is
of representatives
are
(3,0) +
+ 7/,
7/.
Sincewe can compute by choosing the representatives (0, 0), (1, 0), (2,0), and (3, 0), it is clear that (Z4 x Zg)/7/ is isomorphic to Z4. Note that this is what we would expect, since in a factor group modulo 7/, everything in 7/ becomes the identity that is, element; in we are essentiallysetting 7/ to zero. Thus the whole second factor everything equal \342\226 the first factor Z4. Zg of Z4 x Zg is collapsed, leaving just 15.7 is a
Example
We
should
factors
acquire
to the
identity
special case of a general feeling for this
an intuitive element.
theorem theorem
that we in terms
now
state
and prove.
of collapsing one of
the
15
Section
15.8
Theorem
normal
Consider
15.9 Theorem Proof
it
prove the
following
A
factor
be cyclic aN
coset
computing, in
15.10
Example
in
Let us
= k. (See Example subgroup of H x K. Because
K)/H ~
K.
\342\231\246
compute
To illustrate
groups.
group, we
whole
the
in
theorem.
with
G,
H = {{h, e)\\h e 7/} natural way. Similarly,
Jt2(h, k)
is a normal
H
(H x
that
us
->\342\200\242 where K,
is cyclic.
group
a, and let N
generator
be a
G/N. We must compute powers of the representative
all
the powers
compute the
of aN
factor
(Z4 x
group
of G. We claim aN. But this amounts to powers give all elements
subgroup
of N
all cosets
give
certainly
normal
all powers of a and all these
generates
Hence
G.
we see that
14.11tells
of a cyclic
group
Let G the
is to
K in a
to
with additional computations of abelianfactor if we can compute in a factor group
continue
We
how easy
x K
7t2 : H
Ker(7T2) = H,
K, Theorem
is onto
Then
K.
and
147
way.
homomorphism
Because
13.8). n2
of G.
in
the
product of groups H Also G/H is isomorphic
the direct
be
subgroup a natural
~ H
G/K
Proof
= HxK
Let G
is a
Computations and SimpleGroups
Factor-Group
Z6)/((0, 2)). Now
and
is cyclic.
G/N
the
2) generates
(0,
\342\231\246
subgroup
= ((0,0),(0,2),(0,4)}
7/
of Z4 x Z(, of order 3. Here the first factor Z4 of Z4 x Z(, is left alone. The Z(, factor, on the other hand, is essentially collapsed by a subgroup of order 3, giving a factor group in the second factor of order 2 that must be isomorphicto Z2. Thus (Z4 x Zg)/((0, 2)) to Z4 x Z2. \342\226\262 is isomorphic 15.11
Example
Let us
compute the to
temptation
say that
factor
(Z4 x Z6)/((2. 2 of Z4 and
group
we are setting
the
Z4 is collapsed to a factorgroup isomorphic a total factor group isomorphic to Z2 x
is of
order 2, so (Z4
doesnot
make
x
(2, 0)
{(0,
order 12, not
to zero
to one
to
zero,
so that
to Z3, giving
isomorphic that
Note
3)}
(2,
0),
is a great
There
both equal
3 of Zg
and Zg This is wrong]
Z3.
3)) has
Z6)/((2.
and (0, 3) equal
the
to Z2
((2, 3)) =
7/ =
3)). Becarefull
6.
(2, 3) equal to zero do not collapse
Setting
so
the
factors
Z3 and
Z2 x
individually,
separately.
The possibleabelian easily
distinguished
not. We To find must, the
to which
decide
must
the by
that
claim
smallest
choosing
subgroup
of order
groups
one
in that Z4 x Z3 the coset (1. 0) +
power of a
7/ is
coset giving find the
representatives,
12 are
Z4
x
group is isomorphic. These has an element of order 4, and
factor
our
of
two
Z2
Z2
4 in the factor group in a factor group identity
order
the
smallest powerof
x
Z3,
x Z2
x Z3 does
(Z4 modulo
a representative
and we are most
groups
x Ze)/H. 7/, we that is in
7/. Now,
4(1. 0) =
(1,
0)
+
(1, 0)
+ (1, 0) +
(1,
0)
=
(0, 0)
is the first time that (1, 0) addedto itself gives an element of 7/. Thus of order 4 and is isomorphic has an element to Z4 x Z3 or Zj2.
(Z4
x Z6)/((2,
3)) \342\226\262
Part
148
15.12
Example
III
and Factor Groups
Homomorphisms
11.12 the group (Z x Z)/((l, We (that is, classify as in Theorem may Z x Z as the points in the plane with both coordinates as indicated integers, of those points that lie on the the dots in Fig. 15.13. The subgroup ((1, 1)) consists by 45\302\260 line through the origin, indicated in the figure. The coset (1, 0) + of 1)) consists the point (1, 0), also shown those dots on the 45\302\260 line through in the figure. Continuing, lines we see that each coset consists of those dots lying on one of the 45\302\260 in the figure. Letus
1)).
compute
visualize
((1,
We may
the representatives
choose
\342\200\242 \342\200\242 \342\200\242, (-3,
of these cosetsto to the
precisely is
0), (-2,
in the
compute of Z
points
0), (-1,0), (0, 0), (1,0),
on the
(2,
0), (3,
0),---
factor group. Since these representatives
*-axis, we seethat
the
factor
correspond
group (Z x
Z)/((l, 1)) A
to Z.
isomorphic
15.13Figure
Groups
Simple
mentioned in the about crude information
As we
may
be
a group
no nontrivial
of prime
order
section, preceding the structure of the
one feature of a factor group is that it gives whole there Of sometimes course, group.
proper normal subgroups. For example,Theorem can have no nontrivial of proper subgroups
10.10 any
sort.
shows
that
15
Section
15,14
15.15
Definition Theorem
Proof
A
if it is
is simple
group
The alternating
An
group
Computations and SimpleGroups
Factor-Group
nontrivial
is simple
and
has no
for
n
>
proper
normal
nontrivial
149
\342\226\240
subgroups.
5.
See Exercise 39. There
\342\231\246
are many simple A& is of order
order 60 and
168, betweenthese
groups other than those given above. For example,Ag is of of nonprime 360, and there is a simple group order, namely
orders.
determination and classification of all finite were simple groups Hundreds of mathematicians worked on this task from 1950 to 1980. completed. recently It can be shown that a finite has a sort of factorization into group simple groups, where are unique up to order. the factors The situation is similar to the factorization of positive The new knowledge of all finite integers into primes. simple groups can now be used to some problems of finite solve group theory. We have seen in this text that a finite abelian simple group is isomorphic to %p for some prime p. In 1963, Thompson and Feit [21] published their proof of a longstanding that is of even every finite nonabelian simple group conjecture of Burnside, showing strides toward the complete classification were made order. Further great by Aschbacher in the 1970s. Early in 1980, Griess announced that he had constructed a predicted \"monster\" simple group of order The
complete
808, 017,424, 794, 512,875,886,459,
904,
710, 757,
961,
005, 754,368,
000,000, 000. of the classification in August 1980. The research entire fill classification 5000 journal pages. paperscontributing roughly We turn to the characterization of those normal Nofa subgroups group G for which a is First we state an addendum to on properties of Theorem 13.12 simple group. G/N added
Aschbacher
the
final details
to the
a group 15.16 Theorem
The
homomorphism.
Let \302\247 : G
->\342\200\242 G' be
is a normal
subgroup
is a
subgroup
normal
proof is left
to
35 and
Exercises
a group homomorphism. If N is of 0[G]. Also, if N' is a normal
a normal subgroup
36.
subgroup
of
of 0[G],
G, then 0LV] then 0_1 [N']
of G.
Theorem 15.16 should
be viewed as saying that a homomorphism G' : G \342\200\224*\302\247 subgroups between G and 0[G]. It is important to note that 0[iV] may in G', even though N is normal in G. For example, not be normal : Z2 \342\200\224*S3, where \302\242) 0(0) = po and \302\242(1)= /xi is a homomorphism, and Z2 is a normal subgroup of itself, but of 53. {p0, /xi} is not a normal subgroup We can now characterize when G/N is a simple group.
preserves
15.17Definition
normal subgroup that there is no proper normal
A maximal
such
normal
of a group G is a subgroup
N of
normal
subgroup
G properly
containing
M not M.
equal
to
G \342\226\240
Part
150
III
15.18
Theorem
Proof
and Factor Groups
Homomorphisms
normal
is a maximal
M
Let M be a y :
G
of
subgroup
given
G/M
so
maximal,
this
Conversely,
containing M,
G if
and
normal subgroup of by Theorem 14.9. Now
maximal
\342\200\224*G/M
of
subgroup
is simple.
if G/M
only
G. Considerthe y~l
canonical homomorphism of any nontrivial proper normal G properly containing M. But M
is a proper normal subgroup of can not happen. Thus G/M is simple. Theorem 15.16 shows that if N is a normal subgroup If also N / G, then then y [N] is normal in G/M. y[N]
/
Thus, if G/M is simple so that
G/M
and
such
no
/
y|W]
y[N]
of
G properly
{M}.
can exist, no such N
can exist, and
M
maximal.
G
Commutator
and
nonabelian
Every
G has
group
the commutator
and
two
C
subgroup
zentrum, meaning center.) The center Z(G) =
Exercise 52 of
Gandz e Z(G) of
subgroup
15.19
Example
we
G. If
the
group
case of
Exercise
Subgroups normal
important
subgroups,
(The letter Z comes from
of G.
is denned
Z(G)
{z&G\\zg = gz for
to
Z(G) of
German
word
by all
g e
G}.
gzg-1
= {e}, It may be that Z(G) examination of Table 8.8 example, the of is a so center is trivial. ,\302\276 (This {/?oK special of every nonabelian group of order pq the center
G always contains the identity the center of G is trivial.
that
us that Z(&) = which shows that 38, is trivial.) Consequently, q
shows ,\302\276
the
elements.
For
center
of S3
x
Zg
must
be
x {/\302\276}
^5,
\342\22
to the commutator subgroup, recall that in forming a factor group a normal subgroup element in G that N, we are essentially putting every in the factor group. This indicatesanother e, for N forms our new identity
Turning
equal
the
5 shows that
have
for primes p and which is isomorphic to Z5. modulo
the center
Z(G) is an abelian subgroup of G. Since for eachg e = zgg-1 = ze = z, we see at once that Z(G) is a normal = G; in this case, the center is not useful. G is abelian, then Z(G)
Section
The center of a group in which case we say for
is
\342\23
Center
The
is
Suppose, for example,that
of is in use
G
N for
the structure of a nonabelian information about the structure of all 11.12 group G. Since Theorem gives complete sufficiently small abelian groups, it might be of interest to try to form an abelian group as much like G as possible,an abelianized of G, by starting with G and then version that ab = ba for all a and b in our new group To require that ab = ba structure. requiring is to say that aba~lb~l = e in our new group. An element aba~lb~l in a group is a commutator of the group.Thus we wish to attempt to form an abelianized version of G of G by e. By the first observation of this paragraph, we by replacing every commutator to form the factor group of G modulo the smallest normal should then attempt subgroup we can find that contains all commutators of G. factor groups.
15.20
Theorem
we
commutator subgroup) of if N is a normal subgroup Furthermore, (the
G generates a subgroup normal of subgroup subgroup then G/N is abelianif and only if C < fora,
LetGbeagroup.Thesetofallcomrnutatorsafca-1^1 C
are studying
G. This of
G,
b e
C is a
G.
N.
Section15 Exercises
for all g
it is normal in C; we must show that certainly generate a subgroup of a commutator is a (aba~lb~l)~l again commutator, namely, Also e = eee~1e~lis a commutator. Theorem 7.6 then shows that C consists of all finite products For x e C, we must show that g~lxg e C of commutators. so is g~lxg e G, or that if x is a product of commutators, for all g e G. By
inserting
e =
commutators
The
Proof
151
G. Note
the
that
bab~la~l.
precisely
is sufficient
inverse
between each product of commutators for each commutator cdc~ld~l
gg~l
to
show
occurring that
in x,
g~1(cdc~1d~l)g
we see that it is in C. But
= (g-lcdc-l)(e)(d-lg)
g-\\cdc-ld-l)g
= (g-lcdc-l)(gd-ldg-l)(d-lg) =
is in C. Thus
which
The rest
of the
C is normal
[(g-1c)d(g-lc)-ld~l][dg-1d-1g],
G.
in
if we have way, but writing
is obvious
theorem
groups. Onedoesn'tvisualize
in this
that G/
out
= (abb~lcTl)baC= baC Furthermore, (b-lN)(a-lN); C < N, then
if N is a normal subgroup that is, aba~lb~lN
of G
and G/N
= N, so
=
is abelian,then e N,
aba~lb~l
= (abb~la~l)baN= baN
(a~1N)(b~lN)
and C < N. Finally,
=
Pi^iPi^i
8.8, we find
commutator S3/A3
subgroup is abelian, Theorem
15.20 shows that
= A3.
C
\342\226\262
15
\342\226\240 EXERCISES
Computations 1 through
Exercises
In
abelian
12, classify
x (\302\276
Z4)/((0,1))
3.
x (\302\276
Z4)/((l,
9.
2))
(Z4 x
(Z
given
according
group
to
the
fundamental
x Z
x Z4)/((3,
11. (Z x Z)/((2,2))
4.
x 24)/((0, 2)) (Z4 x Z8)/((l, 2))
6.
(Z
8.
(Z x Z x
2. (Z2
Z4 x Z8)/((l, 2,4)) x 7. (Z Z)/((l, 2)) 5.
the
groups.
1.
0, 0))
if
(bN)(aN).
S3 in Table group = M3M2 /?2-We similarly
=
=
that one commutator is /?i/xi/?j~'/x^' = = M2M3 = Pi- Thus the find that paMiP^Vi\"1 = P2M1P1M1 C of S3 contains A3. Since A3 is a normal subgroup of S3 and
For the
15.21 Example
(bC){aC).
= ab{b~laTlba)N
= abN
(aN)(bN)
for factor feeling proper C is abelian follows from
= ab{b~la~lba)C
= abC
(aC){bC)
acquired the
10. (Z 12.
x Z)/((0,
x Z
(Z x Z
1))
Z)/((l, 1, 1)) 0))
x Z8)/((0,4,
x Z)/((3,
3, 3))
theorem
of
finitely
generated
Part III
152 13.
the center
both
Find
Factor
and
Homomorphisms
Z{D^)
commutator
the
and
Groups
C of the
subgroup
symmetries of the
D4 of
group
square
in
Table 8.12.
14.
Find
both
the center
and
the
commutator
15. Find
both
the center
and
the
commutator
16. Describe the
subgroup subgroup
of Z3 x
53.
subgroup
of S3 x
D4.
of order < 4 of Z4 x Z4, and in each case classify the factor group of Z4 x Z4 modulo subgroups of Z4 x Z4 modulo by Theorem 11.12. That is, describe the subgroup and say that the factor group is isomorphic to Z2 x Z4, or whatever the case may be. [Hint: Z4 x Z4 has six different cyclic
all
the
subgroup
of subgroups of order 4 that
order 4. Describethem to the is isomorphic
a generator, such as the subgroup There are three subgroups 4-group.
((1, 0)).
by giving Klein
of order
There is one subgroup
2.]
Concepts In
17.
The center
18. The
18, correct the
17 and
Exercises
needed, so that
it is
in a form
of a
group
commutator
19. Mark
a.
factor
of a cyclic
group
Exercises
20. Let
K
of
group
with every
commute
element of G.
e G}.
\\a,b
is cyclic.
group
the commutator
then
be the of F
elements
subgroup of F
be the
is
group is again noncyclic. of order 2. has elements of order n for all n e Z+. has an infinite number of elements of order 4. G is {e}, then G is abelian. subgroup C of a group
23, let F all
correction
of a simple
nontrivial
20 through
multiplicative
if
subgroup of C of G contains H. subgroup group G must be G itself. of a nonabelian subgroup simple group G must be G itself. finite simple groups have prime order.
h. The commutator i. The commutator
In
text,
no element
has
d. R/Z under addition e. M/Z under addition f. If the commutator g. If G/H is abelian,
All
to the
reference
without
a noncyclic
c. R/Z
j.
of G that
G is {a~lb~lab or false.
true
factor group of under addition
b. A
italicized term
of a group
following
Every
all elements
G contains
subgroup
each of the
of the
definition
acceptablefor publication.
additive that
the constant
of
consisting
of all functions mapping assume the value 0 at any point
group
do not
functions.
into
R
a subgroup
Find
R, and let
F* be the
of R. of F
to
which
F/K
is
isomorphic.
21. Let
K*
F*/K*
22. Let K or why 23.
be the
be the
having
order
24. Let zo 25. To what 26. Let
subgroup of
F* consisting
of the
in F.
functions
continuous
nonzero constant
Can
you
Find
functions.
an element
find
a subgroup
of F/K
having
of F* order
to
which
2? Why
not?
Let K*
In Exercises
27.
the subgroup of is isomorphic.
be
24 through U. Show \342\202\254 group
= f\342\200\236
or why
To what group
26,
that
we have
cos(2jr/n)
consisting of the
of F*
subgroup
2? Why
let U
zqU =
be
the
{zqz
mentioned
+ i sin(2jr/n)
mentioned
continuous
in F*.
functions
Can you
find
an element
of F*/K*
not?
in the
group
multiplicative \\ z \342\202\254 U) is in
the
n e
where
text is
the
a subgroup
text is
C \342\202\254 | \\z\\
of U,
and
=
1}. U/zoU.
compute
U/{\342\200\224 1) isomorphic?
Z+. To what
additive
{z
group
group
we have
mentioned is
R/Z isomorphic?
isomorphic? U/{!;\342\200\236}
Section 15
28.
an
Give
29.
Let H
30.
Describe the
G/H
31.
are of
G having
no
of finite
elements
order > 1 but
a factor
having
group G/H, all of
order.
finite
and K be normal subgroups of a is not isomorphic to G/K.
a. abelian
b.
of a group
example elements
whose
153
Exercises
center of every
G.
group
Give an
example
that
showing
we may
~
have H
K
while
simple
group
nonabelian
group.
Describe the
commutator
of every
subgroup
simple
a. abelian group b.
nonabelian
group.
Proof Synopsis
32. Give
synopsis of
a one-sentence
33. Give at
a two-sentence
most
the
15.9.
of Theorem
proof
synopsis of the
proof
of Theorem
15.18.
subgroup of
index
Theory
34. Show 35.
Let
if a finite
that cj> :
group G contains
G' be a
G ->
a nontrivial
and let N
homomorphism,
group
be a normal
2 in
G, then G is not of G.
subgroup
simple.
Show
that
cj>[N]
is normal
subgroup of 0[G].
36. Let
37. Show
39.
Prove
a.
b.
37, show
Exercise
Using
that
is simple A\342\200\236
that
a
of G'.
Show
that
is a
[N1]
group G/Z{G) is not cyclic. [Hint: Show the equivalent G is abelian (and hence Z(G) = G).] cyclic nonabelian p and q are primes has a trivial group G of order pq where
contra-
then
for n > 5, following
the
every 3-cycle if n > 3. Show is generated A\342\200\236 by the 3-cycles for {a, c){a, b) = {a, b, c).] 5 be
r and
3-cycles by
subgroup
steps
center.
and hints given.
contains A\342\200\236
Show
c. Let
N' be a normal
the factor
then
namely, that if G/Z{G) is
positive,
38.
is nonabelian,
if G
that
and let
homomorphism,
of the
n >
3. [Hint:
that
Note
{a, b)(c,
d) =
\342\200\242for n > 3. Show that is generated fixed elements of {1,2, \342\200\242 \342\200\242, A\342\200\236 n} form (r, s, i) for 1 < i < n [Hint: Show every 3-cycle is the product
(a, c, b){a,c, d) by the
n
of \"special\"
and
\"special\"
3-cycles
computing
(r. s, jf(r,
(r,s, j)(r,s,i)2,
(r, s,if,
s,i),
and (r,
5. i)2{r,
s, k)(r,
s, j)2(r,s, i).
Observe that these products give all possible types of 3-cycles.] for n > 3. Show d. Let N be a normal subgroup of A\342\200\236 that if N contains a 3-cycle,then \342\200\242 n by computing that (r, s, i) e N implies that (r, s, j) e N for j = 1, 2, \342\200\242 \342\200\242,
N =
A\342\200\236. [Hint:
Show
((r,s)(i,mr,s,i)2((r,s)(ij))-1.]
e. Let N
be a
nontrivial normal in each case that
and conclude
subgroup
N
=
of A\342\200\236 for n A\342\200\236.
> 5.
Show
that
one of the
following cases must
hold,
III
Part
154
N contains
I
Case
Casell
Af
a 3-cycle.
and N
the
N
product of the
a disjoint
Case
subgroup
that
e K.
a2, a3) where
product of disjoint
(i is a
2-cycles.
a disjoint
the
also be a
let M
exercise,
preceding
normal
\\h e
Show
of G.
subgroup
H,n
&
Show
N}.
is again a
NM
that
that
of G.
subgroups of a
and ^f are normal [Hint: Consider the
if H
have
of a
seen
triangle
acting on ffi\", set. The next
S. The
into
square, group this section, we give will give an application to
of a defines
function
to yield an element generally, for defining a \"multiplication,\" some
element
section, we will be a map *:GxX->I Let
X be a
set
and
e ff
of a cube, the
linear
general
notion of
general
of symmetries
the group
like
things,
the
group
action
group
on a
counting.
a binary operation * on a * gives us a rule for \"multiplying\"
More
value
/z
Group Action
si *
in S
52
may act on of rotations
groups
the
section
2.1
Definition
how
so on. In
and
Notion
The
of
examples
all
a Set
on
or of a
that H n K = {e},then M = kh for = (hkh~l)k~l = h{kh'lk~1).]
G such
group
hkh~lk~x
commutator
Action
tGROup
We
Definition
= (i(a\\,
of G. Let HN = {hn subgroup of G and let H be any subgroup of G, and is the smallest subgroup containing both N and H.
a normal subgroup
section 16
16.1
02, aA)
o~l(a\\,
of disjoint
41. With reference to 42. Show and fc
Show
product a of the former = (i(a3, 04)(01,02), where /x is a product of an even 2-cycles. [Hint: Show that o~Y(ai, C12,a3)a(a\\, ai, a3)~lis in N, and compute it to deduce that a = (02, 04)(01, a3) is in N. Using n > 5 for the first time, find i ^ a\\, a.2, a3, a4 \342\200\242Let = (ai,a3, 0- Show that j}~xafia e A7, and compute it.] in {1, 2, \342\200\242 \342\200\242, \302\253}. /J
number
normal
0^)(01,02, a3). [Hint:
it.]
a disjoint product of the former Show a2 e N and compute it.]
N contains
V
40. Let N be HN is a
= (1(04,0$,
former
and compute
N,
contains
[Hint:
03)-1 isin Af,
02, a3)o{a\\,a.2,
ShowCT-1(ai,
/x(\302\253i, 02,---,0.,.).
it.]
compute contains
a(a\\, a.2,a4)_1 is in
CaselV
a =
product
disjoint
Suppose
[Hint:
containsaproductofdisjointcycles,atleastoneofwhichhaslengthgreaterthan3.
N contains
Caselll
and Factor Groups
Homomorphisms
any
52 in
sets
be
a function
an element
and C,
any element
we can a of
A
view times
a map
C. Of course, we write a * b = c, or simply with the case where X is a set, G is a group, We shall write *(g, x) as g * x or gx.
c of
group. An
xS
an element
and
* :A
any element
concerned
G a
mapping S
si in S
S.
A, B,
where
set S to
action of G on X is a
map
*:GxI^I
x
B
\342\200\224>\342\20 C as
b of B has as ab = c. In this and
we have
such
that
\342\226
1.
ex =
x for all x e X,
2. (gig2)(x)= gi(g2x) Under these
t
This section is a
for
conditions, X
prerequisite only
all x
is a
X and \342\202\254
G-set.
for Sections
17 and 36.
all gu
g2
e
G.
Section 16
16.2 Example
Let X be any
set,
Then X is an
//-set, where
be a
let H
and
the group
of
subgroup
e H on
of a
action
the
Group Action
X
Sx of all is
155
a Set
on
as
X.
of
permutations
its action
of
element
an
for all x e X. Condition 2 is a consequence of the of definition as function and Condition 1 is immediate from permutation multiplication composition, the definition of the identity as the identity function. Note that, in particular, permutation \342\200\242 \342\200\242is an \342\226\262 set. Sn {1, 2, 3, \342\200\242, n} Sx, so
= o{x)
ax
that
that for every G-set X and each g e G, the map = of X, and that there is a homomorgx is a permutation by ag{x) -+ G the G on X is the Example 16.2action such that action : of Sx phism essentially \302\242) of Sx on X of the image subgroup H = 0[G] of Sx on X. So actions of subgroups all possible actions on When the set actions describe X. X, studying subgroups group using G via a group action of G on X. of Sx suffice. However, sometimesa set X is used to study 16.1. Thus we need the more general concept given by Definition
16.3 Theorem
Proof
theorem
next
Our
ag : X
-+
X
show
will
defined
For each g e G, the function ag : X -+ X defined for x \342\202\254 X is a permutation of X. Also, the map 0 : G -+ Sx defined by with the property that
Let
a G-set.
be
X
by
og{x)
0(g)
=
= gx is a crg
show that it is a one-to-one map ag is a permutation of X, we must X. Then itself. Suppose that ag{x\\) = ag(xi) for x\\, X2 \342\202\254 gx\\ = gX2= 2 Condition in Definition 16.1, we see that g~l(gx\\) g~1(gX2)- Using Consequently, = = so 1 of definition Condition the then ex2xi, (g~1g)x2, ex\\ (g_1g)xi yields x\\ \342\200\224 so ag is one to one. The two conditions of the definition show that for x e X, we have = g(g~l)x = (gg~l)x = ex = x, soag maps X onto X. Thus ag is indeed a ag(g~Xx)
To show of
that
X onto
permutation.
To
permutations Using
the
: G -+ Sx defined
that $ 4>{g\\g2) =
show
that
show
two
in
Sx
4>{g\\)4>{gi)
by showing
conditions
by
g2
they both
carry
in Definition
ag is a homomorphism, we must the equality of these two into the same element. anieX
4>{g) =
all gi,
for
16.1 and
We show
G.
&
rule
the
for
composition, we
function
obtain
=
0(gig2)(*)
=
agig2(x)
= (gig2)x
(\302\2760\302\276)^)
= gi(g2x)= gi(Tg2(x)
=
agt(ag2(x))
=
=
(\302\276\302\276)\302\276 (0(\302\243l)0(,g2))(*).
Thus 0 is a homomorphism. The stated we have 4>{g){x)= og{x) definitions,
property
of
= gx.
follows \302\242)
at once
since
by
our \342\231\246
from the preceding theorem and Theorem13.15that if X is G-set, then of G leaving every element of X fixed is a normal subgroup NofG, we and of a coset gN on X is given by (gN)x = gx can regard X as a G/N-set where the action for eachx e X. If N = {e}, then the identity element of G is the only element that leaves every x & X fixed; we then say that G acts faithfully on X. A group G is transitive on It follows
the
subset
a G-setX if transitive
Exercise We
on
49
for
each
x\\,X2
and only of Section 8. X if
continue
with
\342\202\254 X, there
if the
exists g & G
subgroup
more examples
such that gx\\ of is transitive Sx 0[G]
of G-sets.
=
X2-
on
Note
X, as
that
G is
defined
in
Part III 16.4 Example
16.5
Example
an
where
//-set,
a subgroup
Let//be
g e G
for
*(gi, g2) = *(/i, g) = hg.
That is,
multiplication.
as
/j e
and
If H
gi
before the
=
H on G by cause
would
definition
is given
e G
G, we can
also
left
by
G
regard
A
(hih2)g(hih2r1
action of
this
write
always
described
gig2-
on g2 e G by is a subgroup of
action
the
of G. Then G is an //-set under conjugation //. Condition 1 is obvious, and for Condition
g) =
*(hih2,
We
Groups
itself a G-set, where
G is
group
Every
Factor
and
Homomorphisms
=
hx{h2gh~^)h~x
confusion
The the
hgh~x
that
g)).
*(hu*(h2,
with
g) =
*(h,
2 note
as hgh\"x.
conjugation
terrible
where
abbreviation
hg
operation
group
ofG. 16.6
Example
16.7
Example
16.8
Example
\342\2
For students who have studied vector spaces with real (or complex) scalars, we mention that the axioms (rs)\\ = r(s\\) and lv = v for scalars r and 5 and a vector v show that for the multiplicative group of nonzero the set of vectors is an R*-set (or a C*-set) \342\2 scalars. Let H
be a
of
subgroup
a G-set,
where
Observe
that
the
this
action
action
and g{yH)
= (gy)H
shows
that
every
G-sets
as building
Let G
be
described
in Example 8.11. We also
in
Fig.
horizontal that
pi
the
G-set
let L# be the set of all left cosets of H. Then LH is = (gx)H. e G on the left coset xH is given by g(xH) = xh for some h e //, is well defined: if yH = xH, then y G, and of g
= (gxh)H= (gx)(hH)
=
is isomorphic to
one that
may
blocks. (SeeExercises14through Z)4 =
group
{p0, p\\,
pi, /xi, [i2, 16.9 we show
p2,
= g(xH).
(gx)H
be formed
A
series
of exercises
using these
left
coset
\342\2
17.)
h\\,
of of symmetries with vertices square
S2]
the
square,
1, 2, 3, 4 as sides the and vertical and si, 52, s3, s4, d\\ d2, diagonals axes m\\ and sides center the Recall of 5,-. point C, midpoints P,to rotating the square counterclockwise corresponds through ni/2 radians, /x,-
8.10. In label the and m2, the
Fig.
16.9
Figure
the
16
Section
Group
157
a Set
on
Action
16.10 Table 2
3
Po
1
Pi
2
2 3
4
4 1
Pi
3
1
Mi M2 Si S2
3
4 1 1 3 2
4 2 4 3 1
P-i
4
ThenX can
X
G
with
c
P3 P4
Pi
Pi
P4
Pi Pi P3
di
dx
P3
Pi
Pi
di
di
Pi
Pi
di
di
P4
p3
S3
S4
S4
Si
3
Si
S4
Si \342\200\242S3
2 1 3
1
s3
Si
m\\ \342\200\242S4 \342\200\24251
4
Si
Si
S4
ffjT \342\200\242S3
2
S4
S3
Si
mi \342\200\24251
nil
on
S2 \342\200\242Si
as a
mi
m,, and 5,- to
axis
the
3, 4, si,
{1, 2,
53, 54,
52,
in \302\243>4-set
di di
on
flipping
mi, mi,
d\\,
di.
way. Table
a natural
c
c
c
c
c
Pi P3 P4
P4
p3
P4
P4
P4
Pi Pi P3 Pi P4 P3
Pi
Pi
di. We let
the diagonal
P2, P3, Pn).
C, P\\,
16.10 describescompletely
the
is given to provide geometricillustrations of ideas to be introduced. this table is formed before continuing. \342\226\262 that we understand how
and
X
Let x
&
{x e
X
Let X
be
Let x
&
I
X
courseex and
gx
G. It
will
be important
and
= x]
GA
=
to
{g e
when
know
G
I
gx
gx =
x.We
= x}.
Example 16.8, we have XIM={sl,S3,mum2,C,PuP3}
GJ3={po,Mi}-
lp0,S2}, of the
computation
other Xa
the subsets Gx given
is
true
in
and
the
Gdl Gx
= {p0,p2,Si,S2}.
to Exercises
preceding
example
G for
each ieX,
1 and
were,
2. in
\342\226\262
each
case,
in general.
Then Gx is a
a G-set.
g\\{g2x) =
g e
X and
XPI={C],
G. This
of
subgroups
Definition
Pi
P3
= D4,
that
Note
16.13
Pi
c
2
X in
Z)4-set
We leave the
g~lx,
C
di
di di d2 di di
3
Gi =
Proof
di d2
S4
\342\226\24054 S\\
Xpo=X,
16.12 Theorem
tn2
m2
S3
S3
2
a G-set.
be
For the
Also,
mi
S2
Si
Pi
Subgroups
Isotropy
Let let
Pi
4
sure
be
C
S\\
be regarded
We should
di
m\\ \342\200\24254
mi mi \342\200\24252 \342\200\242S3
of D4 on
action
di
S3
mi m2 mi mi m2 mi
=
X
m2
Si \342\200\24251
to flipping
corresponds
16.11 Example
4
1
subgroup
of
= Then g\\x = x and g2x = x. Consequently, (gig2)* and is e closed under the of G. Of induced Gx, Gx x,sogig2 g\\x operation = = = If g e Gx,then gx = x, sox Gx. (g~lg)x g~l(gx) Thus is a e of \342\231\246 G. Gx Gx. g~l consequently subgroup
Let X be a
and
let
g2
g\\,
e Gx.
=
= ex =
x, soe e
G-set
and
let
x
&
X.
The
subgroup
Gx is
the
isotropy
subgroup
of x.
\342\226\240
Part
158
III
and Factor Groups
Homomorphisms
Orbits the
For
X of
ZVset
Furthermore,
each of
the subset by
the
Theorem
Let
X
=
gX\\
Proof
Then
X2-
Suppose
(g~lg)xl = Then
16.15
Definition
~
x\\ ex\\
if*
Finally,
of D4. We
=
^2 ~
=
~
x\\
\302\2432*2
applications
relationship. Recall (G : H) is the index
Let X be a divisor of
*3, so *i
x\\
left
coset
e
Then
leX.
of Gx. We e G such that
g\\ Gx
\\Gx\\
g\\x
from
Gx e
can
X
such
that
of the
If x e
g~lX2
~ xi for some g{, g2 transitive.
equivalence
X, the
cell
=
= g~~l(gxi)
\342\202\254 G.
\342\231\2
in
described
relation
x is
containing
orbit
the
\342\226\2
the group structure 17. The following
of G
X and
=
(G
the
:
Gx).
number
If
\\G\\
theorem
of elements
is finite,
lies
at
the
gives
this
in
and
then
X,
is a
\\Gx\\
=
which
onto
the collection
G such that
gi* =
of left cosetsof
*i. We
define
we
show
that each
left
is a divisor
is finite,
of IGI.
then the
equation
the
\\G\\
^(\302\276).
=
,\302\276
Then
^1 g
for
G is of the form ^f(^i) for some = x\\, we have g\\Gx i/r(xi). Thus ^ = so \\Gx\\ cosets (G : \\GX\\(G : Gx) shows that \\Gx\\ = (G : Gx)
of Gx in
coset
Gx. Let giG* be a left coset. Then if g\\x maps Ga:one to one onto the collection of right
a'i e
\\G\\
to be
that this
and giGA = g\\'Gx. = To show the map -f is one to one, suppose & Gx, and \\jr(x\\) x\\,X2 = = there exist g\\, g2 e G such that x 1 e gi Gx. Then gi x. *2 ft^, and ,\302\276 some g e Gx, so ^2 = ^2-^ = gi(gx) = g\\x = x\\. Thus \342\226\240f is one to one. Finally,
in G,
Gx
i/{x\\)
map ty is well defined, independent of the = g\\x, so xi. Suppose also that g\\x = xi.Then, g\\x = we deduce x g^x g{ e Gx, (gilgi')x. Therefore Thus the map ^ is well defined.
show
must
e giG.r,
If
G-set
exists geG
Then
X3
X, we use |X| for subgroup H in a group G,
a one-to-one map -f from Then there exists g\\
= gil(gi'x),
8i\\gix)
so g\\
of
elements
is reflexive.
= X2 and g2X2 ~ and ~ is
in Section
appear
Gx.
of g\\
choice
Z)4.
by
\\G\\.
We define
Let
of a
in
orbits
if there
only
some g&G. symmetric.
a set
for
that
and let
G-set
the
that
~
and
thengi*i
X3,
=
between
relationship
of the
action
carried into all the other to show that every proceed
xi if and equivalence relation on X.
let
X2 & X,
X2 and
in the
elements
under
type.
ex = x, so x ~ x so X2, gx\\=X2 for ~ ~ is so X2 x\\, x\\, and
~
1
16.10, the
subset
same
this
4 is
and
Let X be a G-set. Each cell in the partition in X under G. Theorem 16.14 is an orbit of x. We let this cell be Gx.
The
Proof
elements
= g2(gixi)
(g2gi)xi
heart
16.16 Theorem
x\\,
3,
1, 2,
X, we have
x e
each
For
For
~ is an
of
elements
various
a G-set.
be
the
in Table
table
action
elements
into
carried
into subsets of this
be partitioned
16.14
Example 16.8with
2, 3, 4} are
subset {1,
=
=
Gx).
\342\231\2
Let X be the
16.17 Example
(G : GO
=
We
16.8,
Gx.
given by Table 16.10. With = 8, we have \\G\\ |G1| = A
table
action
with
2, 3, 4}andG[= {p0,
= {1,
S2}. Since
4. not only
remember
should
x of G carrying if g e G^then^g)* = x so (gil g2)x =
the elements
that g\\
in Example
Z)4-set
= Z)4,wehaveGl
G
Namely,
then gi\\g2x)
the into
precisely
= gi(gx) = g\\x.
x. Thus
16.16
in Theorem
equation
cardinality
g\\X are
159
Exercises
16
Section
On
the other
g~lg2 e Gx so g2
but
of the left
elements
the
hand, if g2x
=
also
coset g\\x,
e g^Gx.
16
\342\226\240 EXERCISES
Computations In
1 through
Exercises
3, let X
be the
of Example
ZVset
1. The
sets Xa
fixed
2. Theisotropy 3. The
{1, 2,
3, 4,
16.8 with
action
=
in X
that
is, X^
&X,
P2, P3, P4)
C, Pu
dud2,
16.10.Find
in Table
table
Gv for each x
subgroups
orbits
e D4,
for each a
S].s2,s3,s4,m\\,m2,
the
where G
following,
= D4.
,XPi,---,X^
that
Gj,
is,
G2,
\342\200\242 \342\200\242 \342\200\242, GP,, Gpt
under D4
Concepts In
4 and 5, correct the definition of the italicized term in a form acceptablefor publication.
Exercises
so that 4. A
it is
group G actsfaithfully G is transitive group
5. A 6. LetX be a G-set
X
7.
Characterize
8.
Mark
on X if
X
a G-set
on
and
if
and let S c X. If Gs c of orbits in X and G.
G-set
in terms
G-set
a transitive
of the
each
and only if gx
following
in
c. If every element of a d. Let X be a G-set with X
be a
f. Each orbit g. Let X
G-set of
X
that is if
such
and
one
Y be
with
to
one,
onto
an isomorphism
sub-G-set
of a
orbits.
fixed
is left
G-set
by the
fixed by
x\\,x2
& X
and g
jel
and
g\\,
a G-set X is a
be a G-set and
G-sets
e.
some g e G, gx can be every other x. a eS, then S is a sub-G-set.Characterize
all s
let
transitive
H <
same
e G. If
g2 e
of G.
identity
the
G. If
gxi g\\x
element g of = gx2, =
G, then =
thenx\\
g2x, then gi
g must
be the
identity
e.
x2.
= g2.
sub-G-set.
G. Then
X
can
be regarded
h. With reference to (g), the orbits in X under H are the same i. If X is a G-set, then each element of G acts as a permutation X Let x X. be a G-set and let If G is finite, then \\G\\ = e j. 9. Let
is needed,
correction
or false.
true
a. Every G-set is alsoa group. b. Each element of a G-set is left
e. Let
g =
that
text, if
only if, for
S for
of its
terms
= x implies
to the
reference
without
in a
natural
as the
way
orbits
in
as an
ff-set.
X under
G.
of X.
\\Gx\\
\342\226\240
|G.V |.
between 7 G-sets X and 7 is a map cj> : X \342\200\224>\342\200\ group G. An isomorphism Two G-sets are isomorphic for all * e X and geG. 7, and satisfies g0(.x) = cpigx) between them exists. Let X be the ZVset of Example 16.8.
with
the
same
Part III
160
a.
Find
b.
Show
of G that
10.
the Z)4-set in
a. Does
b.
orbits
the
X be
Let
acts
act
Z)4
sub-Z)4-sets. are
,\302\25321*31 *4}
n\302\260t isomorphic
on the
fashion
to part
(a) the
[Hint: Find an element
sub-Z)4-sets.
two orbits.] isomorphic sub-Z)4-setsof XI
two different
only
16.8.
Example
on XI
faithfully
in X on
all orbits
Find
Groups
are isomorphic
that
X
{1, 2, 3, 4} and {s\\ in an essentially different you gave for your answer
the orbits
that
c. Are
orbits of
distinct
two
Factor
and
Homomorphisms
faithfully as a
Z)4 acts
which
sub-Z)4-set.
Theory
11. Let
G-set. Show
X be a
each element
action on 12. Let
and let
a G-set
be
X
the
=
GY
if and
only
{g e
G | gy
=
Let the
of real numbers. about the origin
additive group counterclockwise
plane
X. Let
c
Y
G be the
13. Let
X
elements of G have
two distinct
no
if
the
same
y for
y e
all
Y}. Show
is a
GY
subgroup
of G,
16.12.
Theorem
generalizing
faithfully on
G acts
that
of X.
action
e G on the real plane M2 be given by rotating Let P be a point other than the origin in the
of 9
9 radians.
through
plane.
a. Show R2 is a G-set. b. Describe geometrically
c.
the
Find
group
Exercises 14 through G. the group 14.
Let
\\i
{Xt
a.
Show
that
b.
Show
that
15. LetX be a
16. Let
=
X'i
all
P. up to
G-sets,
possible
collectionof sets,so Xj
can
GXo,
gGXo. Be
G-set, described
sure
to
and let
an obvious
way.
Xj for;
j.)
fl Xj
isomorphism(seeExercise9), can = 0 for
i
^ j.
Let each Xt
be
a G-set
be
formed
for
the
from
same
elements
(The
X. Show \342\202\254
xo
of
1J.\302\243/Z-
Xj
have
is the
and
the sets X\\
the sets
suppose are
been tagged by union of the G-sets
to a disjoint union of left coset G-sets, every G-set is isomorphic 17. The preceding exercisesshow that every G-set X is isomorphic question then arises whether left coset G-sets ofdistinct subgroups Note that the map defined in the hint of Exercise 15 depends on the by goXQ and if form distinct
GXo
^
G-sets
a. Let X be a
G-sets Xt.
of the
X-,
still
all by
disjoint. Let
not necessarily
are
and each can
disjoint,
simply
disjoint
union
that X is isomorphic (see Exercise9) to the G-set L of For igl, suppose x = gxo, and define
group G,
same
7. Then
the
as a G-set,
way
orbits.
in Example 16.7. [Hint: show
be G-sets for the e Xj] for each ;' e
The G-set
natural
of its
union
i e 7
{{x, i)\\x 7^
be viewed in a
every G-set X is the
transitive
for
Xt
=
be a disjoint
\\Ji\342\202\254lXj
of
cosets
how
17 show
e 7}
orbit containing
G.
group
left
the
Gp.
as a G-set in
be regarded
to distinguish of them from the elements Xt. Using Exercises 14 and 15, show that as described in Example 16.7. i
a disjoint union of left H and KoiG can themselves choice
of xq as
\342\200\224and
G-sets.
coset
to
The
be isomorphic.
\"basepoint.\"
L% of left
cosets
K =
GgQXo
If xq
is replaced
of K
GXo Gffii0,thenthecollectionsL^ofleftcosetsofi? must be isomorphic, since both LH and Lk are isomorphic to X.
\342\200\224
GgoXo
that
transitive
G-set
and let
xq
& X
and go
e G. If
H
=
GXo
describe
in terms
of H
and go-
b. Basedon and
K are
c. Prove your
(a), conjecture isomorphic.
part
conjecture
in part
conditions (b).
on
subgroups
H and
K of G such that
the
left coset
G-sets of H
Section 17
18. Upto
19. Repeat Exercise18for 20. Repeat
listing the set
the
Exercise 18for
X. 1,(,.
group
the
161
to Counting
transitive Z4 sets X are there? (Use the preceding exercises.)Give an example an action table of eachas in Table 16.10. Take lowercase names a, b, c, and
how many
isomorphism,
of each isomorphism type, so on for the elements in
of G-Sets
Applications
&. List
group
the
of & in
elements
the order (, (1,2, 3), (1,3,2),(2,3),(1,3),
(1,2).
section
17
of G-Sets
Applications
to Counting
an application of our work with G-sets to counting. Suppose, for how many distinguishable ways the six faces of a cube can be marked with from one to six dots to form a die. The standard die is marked so that the 6 is on when placed on a table with the 1 on the bottom and the 2 toward the front, top, the 3 on the left, the 4 on the right, and the 5 on the back. Of course,other ways of section
This
presents
example,we wish
to count
a distinguishably different die are possible. between the faces of the cube for the moment and call them the can have any one of six marks bottom, top, left, right, front, and back. Then the bottom from one dot to six dots, the top any one of the five remaining marks, and so on. There the cube faces can be marked in all. Some markings yield the same are 6! = 720 ways can be carried into another die as others, in the sense that one marking by a rotation die described of the marked cube. For example, if the standard above is rotated 90\302\260 as we look down counterclockwise on it, then 3 will be on the front face rather than 2, but it is the same die. There are 24 possiblepositions of a cube on a table,for any one of six faces can be \342\200\242 and then any one of four to the front, down, placed giving 6 4 = 24 possible positions.
the cube to
marking
us
Let
Any
give
distinguish
position
form a group
can G,
be achieved from is isomorphic
which
any to
by a
other
rotation of of
a subgroup
S%
(see
the
die.
Exercise
rotations
These
45 of
Section 8).
the 720 possibleways of marking the cube and let G act on X by rotation of two markings to give the same die if one canbe carried into the other under action by an element of G, that is, by rotating the cube. In other words, we consider each orbit in X under G to correspond to a single die, and different to orbits of the number of distinguishable dice thus leads give different dice. The determination to the question of determining the number of orbits under G in a G-set X. The following theorem of orbits in a Gthe number gives a tool for determining set X under G. Recall that for each jeGwe let Xg be the set of elements of X left = {x e X | = x}. Recall also that fixed for each x e X, we let by g, so that gx Xg = = G | gx Gx x], and Gx is the orbit of x under G. {g \342\202\254 X be
We
let
the
cube.
We consider
17.1Theorem (Burnside'sFormula) of orbits in X
Let
under G,
G be
a
finite
group
and X
r-\\G\\
=
J2\\Xg\\g\342\202\254G
i
This section is
not used in
a
finite
G-set.
If r
is the
number
then
the remainder
of
the text.
(1)
Part III
162
Proof
Homomorphismsand Factor Groups
We
all pairs (g, x) where gx = x, e there G are \\Xg\\ pairs having g as g
consider
each
N =
the number of such pairs. For
N be
let
and
member.
first
Thus, (2)
J2\\Xg\\-
?eG
On the other
x e
each
for
hand,
are | Gx |
there
X
pairs
x as
having
second member. Thus
we also have =
n
Y,\\\302\260*
x&X
By Theorem
16.16 we have
so we
\\GX\\
obtain
=
=
\\Gx\\
:
(G
k\\G*\\ Now
l/\\Gx\\
has the
know
But we
Gx).
that
(G
:
Gx)
=
\\G\\/\\GX\\,
Then
\\G\\/\\Gx\\.
\\k\\Gx\\)
same value for all x
in
the
orbit, and if
same
we let O be any
orbit,
then
yz_^ xeO (4)
Substituting
in
\342\200\224 =
17.2
Corollary
of Eq.
is a finite
If G
2 and
Proof
The proof of this
orbits
follows
corollary
r. G) = |G \\ \342\226\240
(5)
\342\231\2
then
G-set,
in X
under G)
immediately
our computation of
us continue
Let
(4) w
Eq. 1.
is a finite of
(number
in X under
of orbits
5 gives
Eq.
and X
group
i-
\\n\\ x<=0 ^1
we obtain
(3),
N = | G | (number Comparison
\342\200\224 =
y\302\243-j
X] IIrj-v-1
\342\200\224\342\200\224 \342\200\242 T^ IX,,I.
from
number
the
=
the
theorem.
preceding
of distinguishable
\342\231\2
dice as our first
example.
17.3
Example
We let
X
be
the set
dots. Let G be the number
of 720
group
markings
e G where
element
changes
any one
identity
element
of the
720
number =
0,
markings
of faces
cube
leaves all 720 markings (number
so there are 30 distinguishable
of a cube using from one to six above. We saw that the = 24. of orbits in X under G. Now \\G\\ because any rotation other than the identity into a different one. However, | Xe | = 720
rotations of the
of distinguishable dice is the g ^= e, we have \\Xg\\
For g
sincethe
different
of 24
dice.
of orbits) =
as discussed
fixed. Thenby
\342\200\224 \342\200\242 = 720
Corollary
17.2,
30,
\342\226\26
17
Section
the number
course
Of
of
machinery of the preceding in a freshman finite taught
assume
rotation
if necessary,
the top
(opposite) face. By rotating
the
remaining
choices
are3
dice
distinguishable
be counted
could
The
in
the
using
the
face
all.
next two means.
elementary
examples appearin some finite We use Corollary 17.2so that
math texts and are easy to solve we have more practice thinking
by
in
of orbits.
terms
seven people be seated at a round table, where table? Of course there are 7! ways to assign peopleto the different chairs. We take X to be the 7! possible assignments. A rotation of each person to move one placeto the right results in the same people achieved by asking Such a rotation generates a cyclicgroup G of order 7, which we consider arrangement. How
ways can
distinguishable
many
there is no
to the
\"head\"
distinguishable
to act on X in the obvious way. Again, only the identity e leaves and it leaves all 7! arrangements fixed. 17.2 By Corollary
(number of orbits) Example
without
using elementary combinatoricsas often marking a cube to make a die, we can, marked 1 is down. There are five choices the die as we look down on it, any one of
four faces could be brought to the front position, so there are no different involved for the front face. But with respect to the number on the front face, there \342\200\242 2 \342\226\240 1 possibilities forthe remaining three side faces.Thus there are 5 \342\200\242 3 \342\200\242 2 \342\200\242 1 = 30
possibilities
17.5
163
of G-Sets to Counting
but by corollary, In math course.
for
by
17.4 Example
Applications
=
-
6! =
\342\200\242 7! =
720.
any
arrangement
fixed,
\342\226\262
no clasp) How many distinguishable necklaces (with can be made using seven differentcolored beads of the same size? Unlike the table in Example the necklace can be 17.4, turned over as well as rotated. Thus we considerthe full dihedral of order group Dq 2 \342\200\242 7 = 14 as acting on the set X of 7! possibilities.Then the number of distinguishable
necklaces is
(number
of orbits)
=
\342\200\224 \342\200\242 7! =
\\Xg | is
not as trivial
knowledge of very 17.6 Example
Let us find
\342\226\262
\\Xg\\ for each g e G. In the Let us give an example where problem. the preceding examples.We will continue to assume
17.2, we have to compute using Corollary and the exercises, | G | will pose no real examples In
360,
to compute
as in
elementary
combinatorics.
\\G\\
and
of distinguishable ways the edges of an equilateral triangle can colors of paint are available, assuming one color is used only on each edge,and the same color may be used on different edges. Of course there are 43 = 64 ways of painting the edges in all, since each of the three be of four colors. We one considerX to be the set of these 64 edges may any The G is on X the of of the acting possiblepainted triangles. group symmetries group which is we to and which consider use to be We the for notation .\302\276 .\302\276. triangle, isomorphic
be painted
the
number
if four
different
Part III
164
and
Homomorphisms
elements
in
53 given
Factor
Groups
in Section
8. We
to compute
need
\\Xg
|
for
each
of the
six elements(
inS3. -^ Pol
= 64
xPl\\
=4
Xpi\\
=
triangle is left fixed by p0. To be invariant under p\\, all edges must be the same colors. color, and there are 4 possible
4
xM,l = 16
Every
painted
Same
reason
The
as for
p\\.
that are interchanged must
edges
be the
same
(4 possibilities) and the other edge may also be any of the colors (times 4 possibilities). color
xMJ =
= 16
\\x,
as for
reason
Same
\\i\\.
Then
= 64 +
^2\\Xg\\
120.
4 + 4+16+16+16=
g\342\202\254S3
Thus of orbits)
(number
17.7 Example
and
there
We
repeat number
The
the set be
to
g
are 20
20,
of 24 possiblepainted all edges
color is used on each edge. 4 \342\200\242 3 \342\200\242 2 = 24, and we let X be on X can be considered the group acting = 0 for we see |ZW| = 24 while \\Xg\\
the assumption that of painting the edges Again,
triangles.
are a
different
color,
a different
is then
/ p0. Thus
so there are four
distinguishable
=
of orbits)
(number
-
\342\226\240 120 =
6
distinguishable painted triangles.
17.6 with Example of possible ways
53. Since
-
=
6
\342\200\242 24 =
4,
\342\22
triangles.
17
EXERCISES
Computations In
each
by more
1.
of the
following exercisesuseCorollary methods.
Find
the
number
of orbits
in
{1,2,
2. Find
the
number
of orbits
in
{1,
3.
the
number
of distinguishable
Find
faces
4.
Wooden many
17.2
to work
the problem,eventhough
might be obtained
the answer
elementary
of a
regular
cubes
tetrahedron,
of the
distinguishable
7, 8}under 3,4, 5,6, 7, 8} under tetrahedral dice that 3,4,
2,
rather
5, 6,
than a
the
cyclic
the
subgroup
can
subgroup
be made
of 58.
((1,3,5,6))
of 58 generated
using one, two,
by
three,
(1,
3) and
and four
(2,4,7). dots on the
cube.
same size areto be painted a different color on eachface to blocks can be made if eight colors of paint are available?
make
children's
blocks. How
165
Section 17 Exercises
5.
of the and
are
a pair
6 leaving
one to Find
9 that
identity,
6. Each of the
7.
4 if
Exercise
Answer
all
the
of opposite
corners
eight
corners.
eight
number
available
colors may be repeatedon different faces at will. [Hint: The 24 rotations leave a pair of opposite faces invariant, 8 that leave a pair of opposite edges invariant.]
of a cube is to Find the number
of distinguishable
8. Considersix straight Either a 50-ohm or A
rectangular
colors.How a. no color b. each
the
ways
a. no color is usedmore than once. same color can beusedon any
9.
be tipped with one of distinguishable
edges
of a
of four
colors,
each of
markings possible. of cardboard
square
which may be usedon
(See the
can be
hint
painted
from
in Exercise if six
5.) colors of paint
and
b. the
of eachtype
of a cube consist vertices invariant,
wires
prism
2
lengths is to
available. How
of resistor
many
of equal
100-ohmresistor
number
ft
long
with
distinguishable
of edges. with
ends
be inserted essentially
many
soldered in
the
together to form edges ofa regular of each wire. Assume there
middle
different
faces,
of faces?
at least
six
wirings are
1-ft square ends is to have each of painted prisms are possible if
is to be repeated on different color may be usedon any number
tetrahedron. are
possible? its six faces painted with
one of
six possible
PART
Fields
and
Rings
IV
section
18
Section
18
Rings
Section
19
Integral Domains
and
Fields
Section 20
Fermat'sand
Section 21
The
Theorems
Euler's
22
Rings
Section
23
Factorization
Domain
of an Integral
of Quotients
Field
Section
of Polynomials
of
Section 24
!Noncommutative
Section25
fOrdered
Rings and
a Field
over
Polynomials Examples Fields
and Fields
Rings
thus far has been concernedwith sets on which a single binary operation defined. Our yearsof work with the integers and real numbers show that a study of sets on which two binary operations have been defined should be of great importance. Algebraic structures of this type are introduced in this section. In one sense,this section more intutive than those that precede seems studied are closely it, for the structures related to those we have worked with for many years. However, we will be continuing with our axiomatic approach.So,from another this study is more complicated viewpoint than to deal with. group theory, for we now have two binary operations and more axioms work
our
All
has been
The most called
A
with two
structure
algebraic
general
Properties
As Example 18.2following since grade school.
a ring.
with
18.1 Definition
Basic
and
Definitions
rings
ring
addition
{R. +, and
^\302\276). {R,
set R together
is a \342\226\240}
+} is an
J%2. Multiplication
Sections
with
18.1 indicates,
two
that
+
operations
binary
on R such
the
and
which \342\200\242,
axioms
following
we call
are satisfied:
abelian group. is associative.
c e R, the left distributive law (a +
For all \342\226\240M3.
the right
defined
multiplication,
we shall study is we have all worked
that
operations
binary
Definition
a, b,
24 and 25 are not required
for the
law, a
distributive b)
\342\226\240 c =
(a
\342\226\240
remainder of the
c) +
(b
c) =
\342\226\240
(b +
\342\226\240 hold.
c)
(a
\342\226\240
b) +
(a
\342\226\240 and
c)
\342\226\240
text.
167
168
Part IV 18.2 Example
Fields
and
Rings
aware that axioms
well
are
We
complex
(Z, +,
For example,
(Q, +, \342\200\242>,
of two of rings grew out of the study theory The particular classes of rings, polynomial rings in n variables over the real or complex numbers and the \"integers\"
field. It was introduced
under
addition
(ffi, +, \342\200\242>,
that is
and
and (C, \342\200\242>,
hold in closed under
a ring
for \302\253^83
+,
subset
any
of the
multiplication.
are rings. \342\200\242}
\342\226
Note
\342\226\240 Historical
(Section 22)
^gj, Jl?2, and
is a group
that
numbers
example,
of
the
twentieth
definition
appeared.
an
number
algebraic
David Hilbert (1862-1943)who
the term ring, but it was
latter
of
in connection
that a century The theory of
with the
decade
the second
until
not
first
abstract
fully
invited of
was given a firm axiomatic foundation by Emmy Noether (1882-1935) in her monumental paper \"Ideal Theory in Rings,\" which appeared in 1921. A
chain of this paper is the ascending concept condition for ideals. Noether provedthat in any ring in which every ascendingchain of ideals has a maximal element, every ideal is finitely generated. the Emmy Noether received her doctorate from of Erlangen, Germany, in 1907. Hilbert University major
\342\226\240 b. We
before addition becomes
1915,
but his
to
efforts
position were blockedbecause complained, \"I do not see that
her candidate is an argument against [to the faculty]. After all, we are a not a bathing establishment.\" Noether was, university, however, able to lecture under Hilbert's name.
admission
the political changes accompanying First World War reached Gottingen, she was given in 1923 a paid positionat the For the next decade, she was very influential University. in the development of the basic concepts of modern with other lewish faculty members, algebra. Along however, shewas forced to leave Gottingen in 1933. She spent the final two years of her life at Bryn Mawr College near Philadelphia. the
end
after
of the
It is customary of a
in
Gottingen
the sex of the
Ultimately,
commutative rings
her to
her a paid her sex. Hilbert
secure
in a ring to denote multiplication shall also observe the usual convention in the absence of parentheses, so the
a(b + c) = ab
+
using ab in place juxtaposition, that multiplication is performed left distributive law, for example,
by
ac,
without the parentheses on the right side of the equation. Also, as a convenience analogous to our notation in group theory, we shall somewhat incorrectly refer to a ring R in place of a ring {R, +, \342\200\242}, that no confusion will result. on Z In particular, from now provided and Q, ffi, and C will also be the rings will always be (Z, +, \342\200\242}, in Example 18.2. We on occasion refer to (/?,+) as the additive may group of the ring R. 18.3
Example
Let
of R and
R
be as
any ring and entries. The
multiply
matrices
check that {Mn(R), and the two distributive straightforward
calculations
let M\342\200\236(R) be the collection of all n x n matrices having elements and in R allow us to add operations of addition multiplication in the usual in the appendix. We can quickly fashion, explained
+} is
an
laws
indicate
The associativity of matrix group. in Mn{R) are more tedious to demonstrate,
abelian
that they follow
from
the
same
properties
multiplication
but
in R.
We will
in
operation Example
Let F be under the
any
set
the
of all
define
on F
multiplication
ring;
a/x
we
leave
for the
E. We : E \342\200\224*-
compositionof
Greek
8)(x) =
18.5
Example
f o g
for
abelian
group
= f{x)g{x).
when
cr(p,(x))
multiplication the
almost
and the
letters,
+} is an
f(x) + g(x).
value at x is
function
function
both
functions
f(x)g(x).
usual
with
so no
used
that F is a
this juxtaposition
notation
If discussing permutation multiplication. and function composition in F, we would
However,
in this
be
will
using
we will denote
example chiefly
when
by
multiplying
result.
should
confusion
we
which
homomorphisms,
defined
product
checked
is readily
It
have
function.
composite
exclusively
function f{x)g{x),
polynomial
know that {F,
by
whose
function
composite
notation
the
the
we have a commutative
\342\226\262
the demonstration to Exercise 34.We
to use
we were
use
is a ring. In particular, is not multiplication
2.
(fg){x)
That is,/g is the
169
Fields
addition,
if + We
>
n
functions f
function
usual
Mn{R)
Note that M\342\200\236(\302\243).
rings for
of these
that
know
and Af\342\200\236(Z), Af\342\200\236(Q), Af\342\200\236(R),
rings
18.4
on that we
now
from
assume
Rings and
18
Section
\342\226\262
of Z under addition consisting of cyclic subgroup = n. Since {nr){ns) n{nrs),we see that nL, is closed integer assure under The associative and distributive laws which hold in Z then multiplication. on in the text, we will consider us that {nL, +, \342\200\242} is as ring. From now nL to be this ring.
Recall that in group
theory,
all integer multiples
of the
nZ
is the
\342\226\262
18.6
Example
the cyclic group If we define for a,b e Ln the product ab as the (Z\342\200\236, +}. of the usual product of integers when that divided by n, it can be shown is a ring. We shall feel free to use this For example, in L\\q we have fact. +, \342\200\242} (Z\342\200\236, is multiplication modulo n. We do not check the ring (3)(7) = 1. This operationon Z\342\200\236 axioms here, for they will follow in Section 26 from some of the theory we develop will always there. From now on, Z\342\200\236 be the ring (Z\342\200\236 \342\226\262 ,+,\342\200\242}. Consider
remainder
18.7
Example
\342\226\240 \342\200\242 \342\200\242 x Rn of all ordered R n are rings, we can form the set R\\ x R2 x \342\200\242 \342\200\242. R2, \342\226\240 \342\200\242 \342\200\242 where addition and of n-tuples e rt Rt. Defining r\342\200\236), n-tuples (rj, r2. \342\200\242, multiplication as for we see at once from the axioms in each by components (just groups), ring componentthat the set of all these n -tuples forms a ring under addition and multiplication by \342\200\242 \342\200\242 \342\200\242 The x x x is the direct of the \342\226\262 R\\ R\342\200\236 /?,-. R2 components. ring product rings
If Rx,
Continuing
ring. The
matters
additive
occasion to refer to
inverse
of notation, we shall of an element a
let 0
always
of a
is
ring
be
the
\342\200\224a. We
additive
shall
identity
frequently
of a
have
a sum
a +
a+
having n summands. We shall let this sum is not to be constructedas a multiplication not be in the ring at all. If n < 0, we let n
\342\226\240 a =
(-a)
\342\200\242 \342\226\240 \342\200\242 a
+
be n ofn
+ (-a)
\342\226\240
a,
and
+
always
a
in
\\-
a using the dot. However,n \342\226\240 the ring, for the integer n may
(-a)
170
Part IV
Rings and
for
\\n\\
Fields
summands.
we define
Finally,
Q-a = 0 Z on
left side of the equations and 0 e R on the right side. Actually, the holds also for 0 e R on both sides. The following theorem this proves and various other elementary but important facts. Note the strong use of the distributive laws in the proof of this theorem. Axiom for a ring concerns only and addition, c%j This shows that in order to prove axiom ^%2 concerns only multiplication. that anything to have to use axiom between these two operations, we are going gives a relationship show in Theorem 18.8 is that 0a = 0 for the first thing that we will ,^B3.For example, element a in a ring R. Now this relation and multiplication. involves both addition any The multiplication 0a stares us in the face, and 0 is an additive Thus we will concept. law to prove this. have to come up with an argument that uses a distributive for
0 e
18.8 Theorem
If
the
= 0
0a
equation
ring with additive
R is a
1. 0a 2.
= aO
=
3. (-a)(-b)
=
Proof
For Property
the
e R
a,b
ab.
1, note that
by
axioms
&x
aO =
law for
cancellation
0a
the
and
a(0 + 0) = additive
= (0 +
+ 0a
\302\253>g2, aO
group
=
0 +
0. Likewise,
0a =
a(-b)
since aO= 0 by
For Property 3, note
+ ab
show
distributive
= a(-b + b)
=
aO =
0,
+ ab
= (-a + a)b
=
0b =
0.
that, by that
definition, =
a(\342\200\224b) \342\200\224(ab),
law,
that
= -(a(-b))
(-a)(-b) 2. Again
remember to
1. Likewise,
Property
(-a)b
Property
we have aO =
0)a = 0a = 0 + 0a
In order
by
aO.
(/?,+},
0. This provesProperty 1. to understand the proof of Property 2, we must is the element that when added to ab gives 0. Thus \342\200\224(ab) we must show precisely that + ab = 0. By the left a(\342\200\224b) that
implies
we have
= -(ab),
aO +
Then by
any
0,
= (-a)b
a(-b)
for
then
0,
identity
by Property
2,
-(a(-b)) = -(-(ab)), and
of
and \342\200\224(ab) It is
use
element that when added to \342\200\224(ab) 0. This is ab by definition gives = ab. the of an in a inverse by uniqueness group. Thus, (\342\200\224a)(\342\200\224b)
is the \342\200\224(\342\200\224(ab))
that you understand rules for signs.
important
our usual
the
preceding
proof. The
theorem allows us
\342\231\
to
Section 18
and Isomorphisms
Homomorphisms
18.9Definition
our
into
a ring
For
in group R' should be
work
From
1. 2.
b) =
+
=
: R \302\242)
quite clear
how
map of a ring
a structure-relating
R
+
->\342\226\240 R' is
if the
a homomorphism
two
following
e R:
for alla,b
satisfied
are
it is
theory,
defined.
R', a map
R and
rings
conditions
171
Fields
and
Rings
0(fe), \342\226\240
(p(a)(f>{b).
that 0 is a definition, Condition 1 is the statement relate group {R, +} into (/?', +}. Condition 2 requires that \302\242) is also the multiplicative structures of the rings R and R' in the same way. Since \302\242) a group homomorphism, all the results concerning group homomorphisms are valid is one to one if and only structure of the rings. In particular, \302\247 if its for the additive kernel Ker(0) = {a e R | 0(a) = 0'} is just the subset {0} of R. The homomorphism of the group {R, +} gives rise to a factor group. We expect that a ring \302\242) This is indeed the case. We delay of will discussion give rise to a factor ring. homomorphism this to Section our treatment of factor groups in will 26, where the treatment parallel In
the
preceding
the abelian
homomorphism mapping
Section
18.10 Example
14.
Let F be the ring of all functions mapping E into E defined in Example 18.4. For each : F ->\342\200\242 a e E, we have the evaluation homomorphism \302\247a E, where 0a(/) = /(\302\253)for the in e F. this for We defined 13.4, but we (F, f homomorphism +) Example group did not rest of
text,
with it in group a real
homomorphism property 18.11
Example
We of
solution
such that
ael
equations.
polynomial
solving
theory.
for finding
to finding
precisely with
much
do this
4>a{p)
will
= 0.
We leave
2 for 4>a to Exercise
Much of the the
deal with it in the equation p(x) = 0 amounts remainder of this text deals
be working a great
a polynomial
of the
demonstration
multiplicative
35.
\342\226\262
n is a ring is the remainder of a modulo = n. We know + +
The map
for homomorphism
We
: Z \342\200\224>\342\226\240 where Z\342\200\236 4>{a) \302\242)
each
realize
importance is the just like the other
that in the concept
except
study
of two
of
any sort
of mathematical
systems being structurally this concept is for names. In algebra
an
structure,
identical, always
idea
that is, called
of basic one being
isomorphism.
Part IV
172
The
An to
for groups,
the
our work in
From
on
is satisfied for
argument).
Similarly,
(Z, +} groups, with $(x) = 2x for ieZ. 4>{x)4>{y)= 2x2y = Axy. abelian
As
R'
is a homomorphism
as
just
you
and
the
to
(2Z,
Here
that is
one
\342\226
isomorphic.
that isomorphism gives an equivalence that the multiplicative property of an
expect
check
R (to complete the symmetry map 0_1 : R' \342\200\224*\342\200\224*: R' R\" also a is \\i ring ismorphism, then the
inverse
that if
map
composite
this in
do
: R \\l\302\247
a ring
\342\200\224*R\" (to
complete
the
Exercise 36.
are isomorphic
+}
is not \302\242)
the
under
isomorphism, for
0 :Z
map =
4>{xy)
2xy,
->\342\200\242 Z,
while A
Fields
Questions:
Multiplicative
of elements
to a ring
then
We need to
holds for
We ask
argument).
transitivity
the
check
requirement
multiplicative
we
theory,
of rings. we
R' are
R and
group
collection
any
isomorphism
a ring R
\342\200\224*R' from
: R \302\242)
leads us,
names
definition.
following
R'. The rings
and onto
relation
18.13Example
to
isomorphism one
things being just alike exceptfor
of two
concept
it did
18.12 Definition
Fields
Rings and
of the rings we have mentioned, such as Z,Q, and R, have a multiplicative identity Note also element 1. However,2Z does not have an identity element for multiplication. in the matrix rings in Example 18.3. that multiplication is not commutative described the zero It is evident that {0}, with 0 + 0 = 0 and (0)(0) = 0, gives a ring, ring. Here 0 acts as multiplicative as well as additive identity element. 18.8, By Theorem this is the only case in which 0 could act as a multiplicative for from element, identity 0a = 0, we can then deduce that a = 0. Theorem 3.13 shows that if a ring has a it is unique. a multiplicative We denote element, multiplicative identity identity element in a ring by 1. Many
18.14 Definition
A
ring
the
in which
identity element is a ring
multiplicative
is a commutative
is commutative
multiplication
with
unity;
ring.
A
ring
with
1 is called \"unity.\" In
a ring with
\342\226
1 the
unity
distributive
laws show
that
\342\226\240 \342\200\242 \342\200\242 \342\226\240= 1 + \342\226\240 + 1) (1 + 1 + \342\226\240 + 1) (1 + n summands m summands nm
(1 +
that is,
(n
a
identity element
the multiplicative
\342\226\240\342\226\240 =
l)(m
1)
(nm)
\342\226\240 1. The
next example
1+
\342\200\242 \342\226\240 \342\200\242
+
1),
summands
gives an
of this
application
observation.
18.15
Example
We claim
that
for
additive
group
isomorphism.
r and 5 where gcd(r, s) = the are both cyclic abelian groups
1,
integers
isomorphic.Additively, 1 and (1, 1) respectively.
they
Thus
: Zrs
To check
Zs denned
\342\200\224>\342\226\240 Zr x
the
multiplicative
rings
of order
by
Condition
Zrs
and Zr x Zs rs with
are
generators \342\226\240 \342\226\240 is an 1) = n (1, 1) 2 of Definition 18.9,
Section 18 we use the observation and compute. =
4>{nm)
(1,
product
only if each Rj In a ring R with
if and
1) =
\342\226\240
(nm)
that a direct
Note
this example for
preceding
x
Ri
[n
R2 x
\342\226\240
Fields
unity
(1, 1) in
the
1)] =
\342\200\242
l)][m
(1,
(1,
\342\226\240 \342\200\242 \342\200\242 x R\342\200\236 of rings
or has the set
is commutative
the
and
Rings
173
x Zs,
Zr
ring
0(n)0(m).
\342\226\262
is commutative or has
unity
respectively.
unity,
R* of nonzero if closed under elements, 1/0, a be if will inverses exist. ring multiplication, group multiplicative multiplicative A multiplicative 1 / 0 is an element inverse of an element a in a ring R with unity = 1. Precisely as for groups, a~l e R such that aa~l a multiplicative =a\"la inverse for unity
the
a in R
element
an
0 + 0 = 0 and We are thus led in
18.16
Definition
this
introductory
Let
R
be
field).A
Let us
0 as both additive the existence of multiplicative
to discuss unity.
on rings.
find
the
18.18
Example
Z is
not
in Z.
ri) =
units
in Z14.
skew
strictly
20 will show
for
to be
defined in
done.
is a unit of R if it has a multiplicative unit, then R is a division ring (or skew noncommutative division ring is called
in R
is a A
ring.
of terminology
Of course, 1 and
therefore can
be units,
of 14; that
the
= 13 \342\200\2241
= 11 \342\200\2243
and
are units. Since are also units.
(3)(5)= 1 we
= 9 \342\200\2245
since no
multiple
a common
they
all
have
units
in
are precisely Z\342\200\236
None
of the
6, 7, 8, factor, either 2 or 7, those m e Z\342\200\236 such
of
2, 4,
or
10
can with
that A
\\.
a field,
The only
division
u
a lot
that
ring {0}, where identity element. nonzero elements
\342\226\240
3 and 5
Section
gcd(m,
An element
element of R
multiplicative
shows
18.8 the
field.\"
are units; elements of Z14 remaining be one more than a multiple 14.
We are almost
and
inverses
is unavoidably
There
is a commutative
skew
that
see
inverse for 0 exceptfor
= 0, with
(0)(0)
section
field
Exercise 43). Theorem
exists at all (see
a multiplicative
a ring with unity 1/0. R. If every nonzero
a \"strictly Example
if it
have
with nonzero
a ring
inverse in
18.17
is unique,
be hopelessto
it would
because units
field
2, for example, has no multiplicative are 1 and \342\200\224 1. However, Q and
in Z
is given
in
Section
inverse, ffi
are
fields.
so 2 is not a unit An example of a
24.
\342\226\262
have the natural concepts of a subring of a ring and subfield of a field.A subring induced from the whole ring is a subset of the ring that is a ring under operations a subfield for a subset of a field. In fact, let us say here once and is defined similarly all that if we have a set, together with a certain specified type of algebraic structure We
of a ring;
for
of this set, field, integral domain, vectorspace,and so on), then any subset a natural an algebraic structure induced algebraic structure that yields of the same type, is a substructure. If K and L are both we shall let K < L denote structures, K is a substructure of L and that K < L denote that K < L but K / Exercise 48 of R. gives criteria for a subset S of a ring R to form a subring (group,
together
ring, with
L.
Fields
Rings and
Part IV
174
element, identity the multiplicative identity element -1 is -1 is a in unit but Z, example,
For
fields were implict Although on the
it was
first
published
of rationality\":
R\", R'\",
of the
words
is any
element
or unity not
is a unit,
but
Unity is
unity.
having a
that is,
unity,
and
unit
multiplicative
not every unit
is
/ 1.
-1
functions of the coefficients.\"
integral
that
insisted
be constructible the domain of view
any
must
subject
many steps, as a complete
supplement to
did
not
of those
every
in
finitely
rationality
entity, but merely as a regionin which place various operations on its elements. Richard Dedekind (1831-1916), the inventor of a real number, of the Dedekind cut definition took
considered a field as a completedentity.
In
in his definition edition of Dirichlet's text on of number theory: \"By a field we mean any system which in infinitely many real or complex numbers, that the addition, itself is so closed and complete, and division of any two subtraction, multiplication, numbers always produces a number of the same system.\" Both Kronecker and Dedekind had, however, dealt with their varying ideas of this notion as early as the 1850s in their university lectures. A more abstract definition of a field, similar to the one in the text, was given by Heinrich Weber (1842-1913)in apaper of 1893. Weber's definition, that of Dedekind, specifically includedfields unlike with finitely many elements as well as other fields, such as function of fields, which were not subfields the field of complexnumbers. he published
and
of
\342\226\240 \342\200\242 \342\200\242 one
\342\226\240 \342\226\240 \342\226\240 with
R\", R'\",
Kronecker, however, who mathematical
by Abel
domain
\"The
\342\226\240 \342\226\240 contains \342\200\242)
are rational
which
quantities
work
the early
Leopold Kronecker (1823-1891) with his own work on this subject he called in 1881 a definition of what
in connection
a \"domain
in
of equations
solvability
who
R',
unit
Note
\342\226\240 Historical
quantities
use
our
while a
Thus
unity.
(/?', rationality
confuse
multiplicative
inverse.
Galois,
not to
careful
be
Finally, the
1871,
the
the
following
second
18
EXERCISES
Computations Exercises
In
1.
(12)(16)
in
5. (2,3)(3,5)in the
13,
decide
and give a
ring
the ring
is commutative,
with
the usual
addition
with
the
whether
Z x Z
with
10. 2Z x Z with
usual
and
addition and
addition addition
Z32
in Z26
6. (-3,5)(2,-4)
x Z9
Z5
set,
9.
ring.
Zis
through
8. Z+
given
4. (20)(-8)
(closed) on nZ
in the
product
2. (16)(3)in
In Exercises 7
7.
the
in Z24
3. (11)(-4)
state
6, compute
1 through
the
whether
indicated
ring is not it has unity,
operations
If a
structure.
whether multiplication multiplication
and multiplication and multiplication
by components by components
of addition
Z4
x Zn
and
tell why this is whether it is a field.
formed, and
in
the
multiplication
case.
If a ring
are
defined
is formed,
11.
\\a,b e Z} with
+ b*/2
[a
12.
[a +
13.
The set
b\\fl
a, b e
|
of
all
Q}
the
usual
addition and
multiplication
the
usual
addition and
multiplication
complex numbers ri
imaginary
pure
14 through
In Exercises
with
all units
19, describe
14. Z
15.
17. Q
18. Z x Q
20. Consider the a.
matrix
order
the
Find
b. List
all
units
an example
/ 0',
0(1)
where
and
16. Z5
M\342\200\236 (M). Is
23. Describe all
24. Describeall 25. Describeall 26.
How
ring ring
of a homomorphism 0(1) / 1'.
27. Considerthis whence Is this
28. Find
homomorphisms
of Z x Z
of the X2
X = /3
of the
det(A) is the
determinant
with
unity
1/0
and
or
=
X
x Z
Z
the
zero matrix,
0, the
/3
A
for
Z.
I3 in
=
X2
matrix
of the
not?
x Z.
into
of Z x
equation
\342\200\224 =
correct? If not,
solutions
all
into Z? ring
M3(R).
we have (X
so factoring,
\342\200\224
h)(X
+ 73) =
0
\342\200\224/3.
out the
point
equations2 +x
error,
if possible,
and
give a counterexample
to
the
conclusion.
\342\200\224 6 =
OintheringZubyfactoringthequadraticpolynomial.Compare
27.
Exercise
with
R' are rings
Z.
homomorphisms
either
reasoning
Why
of Z into Z
are there
M where
or why
of Z into
X2 = 73 implies
R and
R \342\200\224> R' where \302\242):
M\342\200\236 (M) into
homomorphisms
solution
it.
/ 0' and
homomorphisms
many
elements in
number of
is, the
that
map det of det a ring homomorphism?
ring
19. Z4
x Z
22. (Linearalgebra) Considerthe A e
multiplication
ring
given
x Z
Z
and
addition
ring M2(Z2).
of the ring, in the ring.
21. If possible,give 1'
the
in
with the usual
reR
for
175
Exercises
Section 18
Concepts
29 and 30, correct the definition of the italicized term without reference to the needed, so that it is in a from acceptablefor publication. 29. A field F is a ring with nonzero unity such that the set of nonzero elementsofF is a group In Exercises
30.
A
in a ring
unit
31. Give an
example
is
an
element
32. Give
an example of a ring direct product, or a subring
33. Mark
each of the a. b. c. d.
following
with
two elements a and unity
b such
that has a
1/0
of 1^.]
true or
false.
Every
field
is also a
Every
ring
has a multiplicative
Every
ring
with unity
at
least
Every
ring
with
at
most
ring.
has unity has
under
correction is
multiplication.
1.
of magnitude
of a ring having
text, if
identity. two units. two units.
subring
that ab with
= 0 but nonzero
neither
a nor
unity 1'
b is zero.
/ 1. [Hint:
Consider
a
Part
176
IV
Fields
and
Rings
of some field
to be a ring but not a subfield, under are not very important. Multiplication in a field is commutative. Thenonzero elements of a field form a group in under the multiplication Addition in every ring is commutative. element in a ring has an additive inverse. Every
e.
for a subset
is possible
It
f. The distributive g.
h. i.
j.
the induced
operations.
for a ring
laws
field.
the
Theory
34.
defined on
the multiplication
that
Show
the
F of
set
functions
in
Example
18.4 satisfies
axioms .M2 and
requirement
for a homomorphism.
,i%3
for a ring.
35.
the evaluation
that
Show
36. Completethe argument on a collection of rings. 37.
Show
39. Let (R, 40. 41.
e Zp.
Show that
Show
44.
An
that the
a.
Show Find
45. (Linear P =
of
the set
that
then (U,
equivalence
relation
is a \342\226\240)
group. [Warning:
ring R ring
of a field
that
Be
ab =
K
the fields
is commutative.
0 for all C are
and
a, b
e R.
not isomorphic.
in the ring Zp we have (a + b)p = ap + bp for (a + b)\" is valid in a commutative ring.]
be the
must
define
we
if
only if R
and
if
of the
unity
whole field, in
for
all
32
to Exercise
contrast
row of A. AT
A(ATA)~]
all
if
is unique.
unity
= a.
a2
of a commutative
elements
idempotent
in the ring 2,(, x
under multiplication.
is closed
ring
Z12.
that for an m x n matrix A, the transpose AT of A is the matrix whose jth that if A is an m x n matrix such that ATA is invertible, then the projection is an idempotent in the ring ofnxn matrices. Show
a of a ring R
element
is idempotent
with
Recall
algebra)
of a commutative
48.
with unity, \342\200\242)
is a \342\226\240)
(R,+,
of a unit a ring
inverse
multiplicative
all idempotents
is the j'th
47.
b in a
and
isomorphic. Show be a prime. Show that usual binomial expansion
in a subfield
element
a of a ring R
element
b.
An
the unity
(R, +,
gives an
rings.
43.
46.
[Hint:
all a
that
Show
Let p that the
that isomorphism
show
are not
3Z
and
exponentiation) Observe
(Freshman
for
2Z
the rings
multiplicative
multiplication.] b)
group.
18.12to
in a ring
units
all
\342\200\224 for
+ b)(a
be anabelian
+) that
Show
a, b 42.
the collection of that U is closed under
a2 \342\200\224 b2 = (a
that
after Definitions
outlined
U is
if
to show
sure
38.
that
Show
18.10 satisfies the
of Example
map 0fl
is
ring, Show that a ring R has no nonzero Show that a subset S of a ring R
0 for
if a\" =
nilpotent
a + b
then
is also
Show
e Z+.
if a
that
and b
are nilpotent
matrix
elements
nilpotent.
if and
element
nilpotent
gives
some n
column
a subring
of
R
if 0
only
if and
is the
only if the
only
following
solution
of x2
= 0 in
R.
hold:
OeS;
(a-fe)e
ab e
49. a. Show b. Show 50. Let R be
that
an intersection
that
an intersection
a ring,
and let a
of of
be a
subrings sub
fixed
fields
5 for
S for
all
ring R is again of a field F is again
of a
element
of R.
a,
fee
all a, b
5; e S.
a subring a subfield
Let Ia = [x e
R
\\ ax
of R. of F. = 0}.
Show
that
Ia is a
subring of
R.
Section 19 IntegralDomains
177
and let a be a fixed element of R. Let Ra be the subring of R that is the intersection of all of R generated 49). The ring Ra is the subring by a. Show that the containing a (seeExercise abelian group {Ra, +} is generated (in the sense of Section 7) by [an \\ n e Z+}. 52. (Chinese Remainder for two congruences) Let r and .s be positive integers such that gcd(r, s) = 1. Theorem the 18.15 to show that for n there exists an x Use isomorphism in Example e Z, such that x = m m, integer = n and x (mod r) (mod s).
Let fibea
51.
ring,
of R
subrings
53. a. State
18.15 n factors. for a direct product with prove the generalization of Example \342\200\242 n Let a,, bt e Z+ for i = 1,2, \342\200\242 and let gcA{bt, bj) the Chinese Remainder Theorem: \342\200\242, \342\200\242 i / j. Then there such that x = a; (mod b{) for i = 1,2, \342\200\242 n. exists ieZ+ \342\200\242, that are binary operations on S such Consider (S, +, \342\200\242), S is a set and + and \342\200\242 where and
b. Prove
54.
(S, +)
is a group,
S* consistsof
group where
is a \342\200\242)
(S*,
a(b Show that
+ c) = (ab)
+ (ac)
{S, +,
division ring.
and
is a \342\200\242)
55. A ring R is a Boolean ring if Boolean ring is commutative. (For
of S
all elements
(a + b)c = (ac)+ (be) [Hint:
Apply
for
except the all a,b,c
the distributive
laws
additive
element,
identity
\342\202\254 S.
to
\\){a +
(1 +
b) to
prove
the commu-
of addition.]
tativity
56.
= 1 for
students
having
subsets of S.Let
binary
a2
= a
for
all
R, so
a e
that
element
every
some knowledge of the laws of set theory) For a on SP{S) be defined by + and \342\200\242 operations
A +
B =
{A\\JB)-{AC\\B) = {x\\x&Aovx
is idempotent.
set S, let
SP(S)
be the
Show
that
every
collection of all
\302\243Bbu\\.xi{AC\\B)}
and
A-B
for A,B&
SP{S).
a. Give
tables
b.
Show
the that
has four elements.] and \342\200\242 for SP{S), where S = [a, b). [Hint: SP{S) is a Boolean ring (see Exercise 55). set S, {^(S), +, \342\200\242)
for +
for any
section 19
=ADB
Integral Domains a careful
While
we
motivation
treatment of polynomials is not make intuitive use of them
of Zero
Divisors
given
shall
in this
until Section section.
22, for purposes
of
and Cancellation
most important algebraic propertiesof our usual number system is that a of two numbers can only be 0 if at least one of the factors is 0. We have used product times in solving equations, perhaps without that we were using this fact many realizing for example, we are askedto solve the equation it. Suppose, One
of
the
x2 The
first
thing
we do
is to factorthe x2
- 5x + 6 = 0.
left
- 5x +
side:
6 = (x
-
2)(x
- 3).
Part
IV
Solution
is 0 if
and
Solve
the
number
only
x2
any
+ 6
5x
= (x
\342\200\224
+ 6
5x
But
in Zi2.
number
in
= (8)(6) = find,
for (6 -
These ideas 19.2 Definition
If a
=
still valid if we think = 0 for all a e aO
(4)(3)= (3)(8) (4)(9)= (9)(4)
(6)(10)=
=
(10)(6)
that we
importance
elements of a
=
(8)(3)
=
(6)(6)
ring
formalize them R such that
in
ab
also
= (6)(8)
= (9)(8)
(8)(9)
as standing
of x
Zj2, but
= 0. 6 and
11, A
a definition.
are
a and b
\342\200\224 0, then
\342\226
shows that in Z12
19.1
gcd with 12 is
1. Our
not
2, 3,
elements
the
the numbers
are exactly
these
that
whose
is,
is 0a
0 (or
Example
of 0. Note that
are of such
two nonzero 0 divisors).
b are
and
divisors of
=
3)
3)
our equation has not only 2 and 3 as solutions, but also - 3) = (9)(8)= 0in Z12. 2)(11 (4)(3) = 0 and (11
in fact, that
2)(6-
3)
2)(x
= (3)(4)= (4)(6) = (6)(4) =
=
The reason is that the resulting numbers
3. Why?
2){a
\342\200\224 \342\200\224 is
not only
Zj2,
2 and
\342\200\224 \342\200\224 of
Z12.
= (6)(2)
(2)(6)
We
= 0 in
\342\200\224
x2
equation
The factorization
for
for* are
values
possible
only
a, the product (a by any if either a \342\200\224 2 = 0 or a \342\200\224 3 = 0.
is replaced
if x
that the
conclude
we
Then
19.1 Example
Fields
and
Rings
in
Zn
8, 9, and
4, 6,
that are
next theorem shows
not
are divisors
10
to 12, example of a prime
relatively this
that
is an
generalsituation. 19.3
Theorem
Proof
In the
ring
relatively
prime
Let m
of 0
the divisors Z\342\200\236,
are preciselythose
that are
elements
nonzero
not
to n.
0,
m ^
e Z\342\200\236, where
the gcd
let
and
of m
and
1. Then
nbed
^
= 0 in
while Z\342\200\236,
-(2)-(^)and
0 as a multiple
(m/d)n
is 0, so m is On the ms =
0,
then
relatively
19.4
Corollary
Proof
gives a divisor other
If p
is a prime, then corollary
Another
hand,
is relatively suppose m e Z\342\200\236 prime to the product ms of m and s as elements
prime to m, boxedProperty
= 0 in
Thus m{n/d)
m nor
neither
1 following
Example
n. If for s in
the
ring
6.9 shows
we have e Z\342\200\236
Z. Since n divides
that
n
is
s,
\342\231\
Z\342\200\236.
Zp
has no
indication
0
19.3.
of 0 divisors is importance of the concept and let a,b,c e R. The cancellation implies b = c, and ba = ca with a ^ 0 implies
of the
Let R a ^
divisors of 0.
from Theorem
is immediate
following theorem. R if ab = ac with
n/d
of 0.
n divides
so s
This
of n.
be
a ring,
\342\231\
shown
b
in
the
hold in = c. These
laws
Section 19
are
Proof
hold in a ring
laws
cancellation
The
course, the
R
only if R has no
if and
a ring in which the cancellation laws hold, and R. We must show that either a or b is 0. If a 0,thenafo a, b \342\202\254 cancellation laws. Similarly, b ^0 implies that a = 0, so by R
Let
in
R,
be
divisors of 0.
suppose ab
^
0 if the
=
= 0 for
there can be no
some
that b
a0 implies
= 0 of
divisors
cancellation laws hold. that
suppose
Conversely,
R has no
divisors of 0, and
that
suppose
ab =
ac
with
0. Then
a ^
ab Since a A
laws hold
cancellation
additive
{R, +) is a group.
since
19.5 Theorem
laws. Of
cancellation
multiplicative
179
Integral Domains
similar
^ 0, and that R
Suppose
in i? can have and
by
at
a unit in R with case that i? is
is a ring
most
19.5
Theorem
=
no divisors
with
x
in
Integral
c)
0.
X2, since R
of 0. Then an for if
i?,
multiplicative inverse a ~1,then
commutative,
in
particular
ax\\
=
fo
ax2 =
and
the
solution
R has
x
\342\200\224 =
c
0, so b
= c. \342\231\246
ax
equation
has no divisors of 0. If
fc,
= b, then
unity
a ^ 0, = a^2, ax\\
with
1^0
and
of ax = b is a ~1b. In
a is the
is a field, it is customary to denote a~lb This quotient by the formal quotient b/a. R is not commutative, for then we do not the inverse a-1 of particular, multiplicative
if R
are equal by commutativity) notation must not be used in the event that know whether denotes a~lb or ba-1. In b/a element a nonzero a of a field may be written ba~l
and
\342\200\224 =
no divisors of 0, we must have b = ca with a ^ 0 implies b = c. ba
one solution x\\
a(b
i? has
since
shows that
argument
\342\200\224 ac =
(they
I/a.
Domains
The integers are really our most familiar number system. In terms of the algebraic we are Z is a with unity and no divisors of 0. commutative discussing, ring properties this is the name the next for that definition to such a structure. Surely gives responsible
19.6 Definition
An
integral
domain
D is a commutative ring
with
unity
1^0
and
ofO.
no divisors \342\226\240
Thus, a polynomial
are from an integral domain, one can solve coefficients of a polynomial in the which in the into linear factors equation polynomial can befactored
if the
usual fashion by setting each factor equal to 0. In our hierarchy of algebraic structures, an integral domain a commutative with and as we shall show. Theorem field, unity ring cancellation laws for multiplication hold in an integral domain.
19.7Example
containing
belongs between a 19.5 shows that the
have seen that Z and Zp for any prime p areintegral is not an integral but Z\342\200\236 domains, domain if n is not prime. A moment of thought shows that the direct product R x S of two nonzero rings R and S is not an integral domain. Just observe that for r e R and s & S both nonzero, we have (r, 0)(0,s) = (0,0). \342\226\262 We
Part
IV
19.8Example Solution
and Fields
Rings
that
Show
need
We
Z2 is
although
only
observe
an
the matrix
domain,
integral
ring
has divisors
M2(%2)
of zero.
that
0
0
0 0 next
Our
is,
19.9Theorem Proof
the
Every
richest)
theorem one we
F is
field
e F, and
Let a,b
an
shows that the have defined.
the
still
most
restrictive
(that
domain.
integral
suppose
field is
of a
structure
that a
^ 0. Then if ab =
-W)
a
)
= 0, we
have
= 0.
(-)0 \\a )
But then
0=
-
(ab) =
a ^ 0 implies that We have shown that ab = 0 with of 0 in F. Of course,F is a commutative ring with
19.10
Figure having
we ask
It
diagram view
exhibit is done
far
the
some
fields we fields of finite
only
by counting.
Counting
b =
0 in F,
unity,
be
for the fields
Q, M, and C. The corollary The proof of this theorem
is one
of the
19.10 Figure
are
most powerful
A collection
of rings.
In Exercise
20
as well.
order!
know
\342\231\24
algebraic structures
concerned.
chiefly
skew
so there are no divisors is proved.
theorem
so our
of containment
which we will two binary operations with this to redraw to include you figure strictly
Thus will
gives a Venn
lb
\\a
a
of the
next theorem
is a personal favorite. in mathematics. techniques
19.11
Theorem
Proof
finite
Every
181
Integral Domains
Section 19
domain is a field.
integral
Let 0, l,ax, be
of a
the elements
all
there exists b
= 1. Now
ab
that
these elements of D are
the cancellationlaws that
of these elements
hold
19.12
Corollary
Proof
is a prime, then
If p
a 7LP is
a ring
we find
are
using
we have
where
the fact that
from
over
matrices
that there
indicates
19.14
Example
R be
for
all a
approximate an
idea
is an
domain
integral
from
and
The 1 +
can be no
very
much
systems, determinants, to
depend involving
try
of the
Cramer's to diagonalize
only on
the
notions
of
bases, only
are
linear algebra
arithmetic magnitude,
make senseusing
relation
... + 1 = 0
summands
of magnitude
notion
natural
we
Mn(Lp),
numbers are usedin
transformations they
ring
undergraduate
or orthonormal
solutions
the
consider
algebra
of magnitude.
in
the
field
Zp.
of a Ring
We might ask whether there is a positive integer n such that n \342\226\240 a \342\200\224 0 \342\200\242 \342\200\242 a means a + a + \342\200\242 e R, where n \342\226\240 + a for n summands, as explainedin
any ring. 18. For
example,
the
integer
m has
If for a ring R a positive integer n exists such positive integer is the characteristic exists,then R is of characteristic 0.
this
such
property
that n of the
for
the ring Zm.
0 for all a e R, then the least R. If no such positive integer ring \342\226\240 a =
\342\226\240
of a characteristic be using the concept chiefly for that the characteristic of an integral domain is either
shall
to show
ring
aj, by divisors, none \342\226\240 are elements al, aa\\, \342\226\240 aan \342\226\240, a = 1, or act; = 1 for some i.
typical
field;
any
Considerations of linear
Characteristic
Let
The
7LP
we the
similarity
p
We
is,
=
at
no 0
field.
p.1=1+
us
0,
\342\231\246
and eigenvectors, and
valid
of a field. properties such as least-squares
19.13 Definition
that
= 1, that
field properties of the real or complex as matrix reduction to solve linear
the
eigenvalues
a matrix
Section
= aaj implies that
integral domain. Also, sinceD has
corollary shows that when of matrices over afield. In
preceding about
course,only
The
a ^
\342\231\246
work. Such notions
fields
D, where
19.11.
The
rule,
a e
inverse.
This corollary follows immediately Theorem
talking
for acn
distinct,
Hence by counting, order, so that either al
Thus a has a multiplicative
that for
\342\200\242 \342\200\242 aan. \342\200\242,
is 0.
\342\226\240 \342\200\242in some \342\226\240, a\342\200\236
1, a\\,
in an
show
to
consider
a\\, aa.\\. We claim that all
need
D. We
domain
finite
e D such
is of Z\342\200\236
characteristic
n,
while
Z,
Q, M,
fields. Exercise29 asks 0 or a prime p.
and C all have characteristic0.
\342\226\262
Fields
Rings and
IV
Part
182
first glance, of a ring seems to be a tough determination of the characteristic job, the ring is obviously of characteristic0. Do we have to examine every element a Definition theorem 19.13? Our final of this section shows ring in accordance with it suffices the ring has unity, to examine only a = 1.
At
unless the
of
if
that
Let R
19.15 Theorem
Proof
a ring with
be
0 for
n
\342\226\240 1 =
If
n \342\200\242 1 ^ 0
some n
for all
integer n, Suppose that
n
then surely
e Z+,
Defimtion
n is a
for all
we cannot
19.13,
R
positive integer such
n-a = a + a + ---+a
n
then R has characteristic is the characteristic of
e Z+,
smallest such integer n
then the
e Z+,
so by
positive
1 ^ 0 If n \342\226\240
unity.
n
have
characteristic
that
n \342\200\242 1 = 0,
= a(l + 1+
0 for all 0.
\342\226\240 a =
has
1) =
\342\200\242 \342\200\242 \342\200\242 +
Then for a(n
a e a
any
e R,
aO =
\342\200\242 =
1)
for
R
0. If R.
some
we have
0,
Our theorem followsdirectly.
\342\231\246
19
\342\226\240 EXERCISES
Computations
1.
2. Solve the 3. Find
all
of the
solutions
5 through
In Exercises
x2
x3 \342\200\224 2x2 \342\200\224 3x \342\200\224 0 in Zn-
equation
3x =
equation
all solutions of
Find
4.
of the
all solutions
Find
2 in
field Z7; in
the
the
equation x2 + 2x + 2 + 2x + 4 = 0 in Z&.
10, find
the characteristic of
5. 2Z
6.
8.
9. Z3 x Z4
x Z3
Z3
11. Let
R
be a
commutative
ring
12. Let
R be a
commutative
13.
Let R be a commutative
14.
Show
that
the matrix
1
Z
field
Z23.
\342\200\224 0 in Z&,
the
given
ring.
7. Z3 x
x Z
10.
with
unity
of characteristic
ring
with
unity
of characteristic
4. Compute and 3. Compute and
ring
with unity
of characteristic
3. Compute
2
2 4
is a
of zero in
divisor
and
Z6
3Z
x Zi5
b)4 for a, b + b)g for a, b
simplify
(a +
simplify
(a
simplify
(a + b)6
for
e R. e R.
a, b e
R.
M2(Z).
Concepts In
15.
If ab
16. If
16, correct the
15 and
Exercises
needed, so that
it is
in a form
= 0, then
n \342\200\242 a = 0
a and b
are divisors
for all elementsa
17. Mark each of the
following
of the
definition
italicized term
without
reference
acceptablefor publication. in
of zero.
a ring
true or
R, then n is
false.
a. nZ has zero divisors if n is not prime. b. Every field is an integral domain. c. Thecharacteristic of nZ isn.
the
characteristic
of R.
to the
text, if correction is
d.
As
e.
The
Z is isomorphicto
a ring,
. f. Every integral domain g. The direct product of h. A divisor of nZ is a
.
ZisasubfieldofQ.
j.
of the
18. Each
19.
. i.
the
commutative
(For
of the
is again
ring
unity
unity,
with
an
domain.
integral
can have
no
inverse.
multiplicative
example of a (it is inside
integral domain. Let F be a field. Give five algebra)
not be an
but
had a semesterof linear
include a subset
characterizations
different
of 0.
are divisors
of M\342\200\236(F) that
19.10 to
Fig.
domains
of Z.
have
circle),
elements A
20. Redraw
integral
domain.
an integral
to
to a certain regions in Fig. 19.10 corresponds type of a ring. Give an cells. For example, a ring in the region numbered 3 must be commutative
who have
students
two
a commutative
subdomain
of the six
in each
> 1.
in
numbered
six
ring
zero in
for all n
nZ
any ring that is isomorphic of characteristic 0 is infinite.
law holds
cancellation
183
Exercises
Section 19
skew fields.
to strictly
corresponding
Proof Synopsis
21. Give
synopsis of
a one-sentence
22. Givea one-sentence synopsis
the
of the
of the
proof
\"if\"
19.5,
of Theorem
part
proof of Theorem 19.11,
Theory
23.
a of a ring
An element
R is
idempotent
if
= a.
a2
Show
ring contains
a division
that
two
exactly
idempotent
elements.
24. 25.
of
an intersection
that
Show
integral domain
of an
subdomains
D is again
of D,
a subdomain
Show that a finite ring R with unity 1/0 and no divisors of 0 is a division a field, although ring. (It is actually In your proof, to show that a / 0 is a unit, commutativity is not easy to prove. See Theorem 24.10.)[Note: inverse\" of a / 0 in R is also a \"right you must show that a \"left multiplicative inverse,\"] multiplicative
26. Let
R
be
be
R
such
a.
a ring that that aba
that
R has
that
bab
c.
Show
that
R has
d.
Show
that R is a
27.
Show that the Show
if D
that
the
a
nonzero
each
\342\202\254 R, there
exists a unique
= b. unity. division
ring. of a
characteristic
that
two elements. Supposefor
no divisors of 0.
Show
b. Show
28.
at least
contains
= a.
is
an
domain,
integral
of
subdomain
then [n
an
e Z} is a
11
D is
domain
integral
\342\226\240 n
to the
equal
of D
subdomain
of D.
characteristic
contained
in every
subdomain
ofD.
29.
Show
of D is mn,
30. This exercise shows as R.
characteristic
in S
be the
of an integral domain 1) D.]
characteristic
consider
(m that
Let S
addition
usual
n
every
=
R
ring R can be enlarged (if if R has characteristic
x Z
by components,
the meaning explained
\342\226\240 r has
be either
0 or a prime
p.
[Hint:
If the
characteristic
\\){n
in
=
Section
(r\\r2 + 18.
n\\
to a ring
necessary) 0, and
and let multiplication
(n, ni)(r2, n2) where
D must
\342\226\240 \342\226\240 in
R x
be defined \342\226\240 +
r2
n2
S
with
unity,
having
has characteristic n.
if R Z\342\200\236
by
\342\226\240
n,n\\n2)
the same
Let addition
Part
IV
Rings and
a. Show
that
S is a
ring.
b.
Show
that
S has
unity.
c.
Show
that S
d.
Show
SECTION
that
20
the
and
Fields
the same characteristic.
R have
map 0 : /?
\342\200\224>\342\200\242 5 given
by 0(r)
FERMAT'SAND
=
0) for
(r,
r e
R
R isomorphically
maps
onto a subring
of S.
THEOREMS
EuLER's
Theorem
Fermat's
We know
are naturally isomorphic, with the coset in Z/nZ of cosets Zn. Furthermore, addition and them in be Z, finding the may performedby choosing any representatives, adding can be made into a ring by coset of nZ containing their sum. It is easy to see that Z/nZ any chosen representatives. multiplying cosets in the same fashion, that is, by multiplying While we will be showing this later in a more general situation, we do this special case is well defined, because the now. We need only show that such coset multiplication of from and the laws will follow distributive immediately associativity multiplication of in those the Z. To this choose chosen end, representatives properties representatives a + rn and b + sn, rather than a and b, from the cosets a + nZ and b + nZ. Then that
as additive
an
which
is also
form a ring The
In
following
for Zp,
particular,
and Z/nZ Z\342\200\236
a e
each
+
(as
the
we must
elements form a group
This gives Theorem
of order
20.2
Corollary
If a
the field
multiplication.
1
p
\342\200\224 1
a group
have
us
at once
(Little Theorem
ap~l
our
18.
Section
under
and
the elements
ap~l
20.1
is well-defined,
under multiplication modulo p. Since the order of any we see that for b ^ 0 and b e Zp, we divides the order of the group, = 1 in Zp. Using the fact that Zp is isomorphic to the ring of cosets of the a + pZ described above,we see at once that for any a e Z not in the coset 0 + pZ, a group
element in have bp~l form
+ rsn)n,
multiplication
1,2,3,---,/7form
+rb
isomorphic to the ring Z\342\200\236. is a special case of Exercise 37in
the nonzero
field,
any
for
(a + rn){b+ sn) = ab element of ab + nZ. Thus
cosets
For
groups,
corresponding to a
a + nZ
- 1,that
e Z,
then
ap
= a
p).
the so-called Little Theoremof
of Fermat)
is, ap~l
= 1 (mod
If a
= 1 (mod
e Z
and
p is a
p) for a
# 0 (mod
(mod p) for
any
prime
p.
Fermat.
prime not p).
dividing
a,
then
p divides
The corollary follows from
Proof
sides reduce to 0 modulo
Theorem 20.1if
20.1 occurs
of Theorem
statement
number aT of the 1. Furthermore, divides aT \342\200\224 p also divides all numbers
a letter
in
this
progression
to note
failed
Fermat
that
p not
that
then both
p),
perhapshe felt
divide a;
the
the
that
result
20.3 Example
Let us compute
the
came
of
study
interest perfect
in
this
numbers.
is a positive (A perfect number integer m that is sum of all of its divisors less than m; for 6=1 Euclid + 2 + 3 is a perfectnumber.) example, if 2\" - 1 is had shown that 2n~x(2n - 1) is perfect prime. The question then was to find methods for whether 2\" \342\200\224 1 was prime. Fermat noted determining \342\200\224 1 was composite if n is composite, and that 2\" from his theorem the result that if n then derived 1 are is prime, the only possible divisors of 2\" \342\200\224 those of the form 2kn + 1. From this result he was 1 was to show, for example, that 237 \342\200\224 able quickly \342\200\242 \342\200\242 = 2 3 37 + 1. divisible by 223
of 8103 when
remainder
from his
the
it was fails in that case.) Fermat did not in the letter or elsewhere of the result and, in fact, never indicate a proof But we can infer from other parts of mentioned it again.
obvious
= 0 (mod
Fermat's
that
correspondence
result
such that p T divides p \342\200\224 1 and of the form aKT \342\200\224 1.
is a least
curious
(mod p). If a
\342\231\246
\342\226\240 \342\226\240 \342\226\240 \342\226\240 there \342\226\240, a', \342\226\240,
a, a2,
progression
geometric
(It is
#0
p.
to Bernard The from Pierre de Fermat (1601-1665) 18 October 1640. Fermat's Frenicle de Bessy,dated was that for any prime version of the theorem p and
conditionthat
a
185
Note
\342\226\240 Historical
any
Fermat's and Euler's Theorems
20
Section
divided
Using Fermat's theorem, we
by 13.
have
gl03
20.4 Example Solution
that
Show
- 1 is
211213
s (812)8(37) s
(1^)(87)e87e
=
(-l)3(-5)EE5(modl3).
not
(25)3(-5)s
210
_l 211.213
10-.1,121 [(21U) 23
Thus
is prime, 2P 20.5
Example
Solution
A
by 11.
divisible
10 = 1 (mod 11),
By Fermat'stheorem,
(-5)7
-1
=
so 23]-l
23]- 1
= [lU21
8- 1 = 7 (mod
11).
\342\200\224 the remainder of 211-213 1 when divided by 11 is 7, not 0. (The number 11,213 1 is a prime number. Primes of the and it has been shown that 21L213 \342\200\224 form
p is
\342\200\224 1
where
Show that for
This seemslike
prime areknown
every integer n, an
incredible
the
as Mersenne
number
n33
result. It means
primes.) \342\200\224 n is
that
\342\226\262
divisible
15 divides
by 15. 233
\342\200\224 333 \342\200\224 433 \342\200\224
2,
3,
4,
etc.
Now 15 by both 3
and
= 3-5, 5 for
and
every
we
shall
use Fermat's
n. Note that
n33
theorem to show
\342\200\224 n =
n(nn
\342\200\224
1).
that
n33
\342\200\224 n is
divisible
Part
IV
Rings and
Fields
n, then
3 divides
If
Fermat's theorem, n2
=
3 divides
surely
1 ee
(n2)16 -
and hence 3 divides n32 \342\200\224 1. If n = 0 (mod 5), then n33 theorem, n4 = 1 (mod 5), so
\342\200\224 n =
-
n32
n33
Thus
\342\200\224 n =
0 (mod
1 =
-
n32
1).
n, then by
divide
not
1
5). If
0 (mod
- 1=
(n4)8
5) for every
1 = 116-
-
18
=
0 (mod
n
#
3),
0 (mod
5),
then
by
Fermat's
s 0 (mod 5).
1
n also.
^
Generalization
Euler's
Euler gave a generalization of Fermat's from our
20.6 Theorem
3 does
\342\200\224 If
n(n
3) so
1 (mod
next
as in
Theorem
The
set
theorem,
His generalization
theorem.
is proved
which
by
will
essentially
using
counting,
at once
follow
the same argument
19.11. of
elements
of nonzero G\342\200\236
that Z\342\200\236
not 0
are
divisors
forms
a group
under
multiplication modulo n.
Proof
under multiplication modulo n. Let a, b \342\202\254 is closed Gn. If G\342\200\236 = 0 implies would exist c ^ 0 in Z\342\200\236 such that (ab)c = 0. Now (ab)c = 0. Since b e G\342\200\236 But then and c ^ 0, we have be ^ 0 by definition of G\342\200\236. that a{bc) that a ^ Gn contrary to assumption. Note that we have shown that a{bc) = 0 would imply for any ring the set of elements that are not divisors ofO is closed under multiplication. other than ring structure has been involved so far. No structure of Z\342\200\236 n is associative, We now show that Gn is a group. Of course,multiplication modulo and 1 e G\342\200\236. It remains to show that for a e G\342\200\236, there is b e G\342\200\236 such that ab = 1. Let First
we
ab
Gn, \302\242.
must
show that
then there
1,a.\\, be the
elements of
The G\342\200\236.
elements
a\\, aa\\, are all
different, for if aat
a divisor of 0, we must either a 1 = 1, or some
\342\226\240 \342\226\240 \342\226\240, ar
= aaj, then
have
a,-
aa,-
must
\342\200\224 =
a,-
\342\226\240 \342\226\240 \342\226\240, aar
and thus is not 0, and since a & G\342\200\236 Therefore we find that by counting, a;-. \342\231\2 inverse. multiplicative
= a (a,- \342\200\224 a,) 0 or a,- =
be 1, so a
has
a
Note that the only property of Z\342\200\236 used in this last theorem, other than the fact that it was a ring with unity, was that it was finite. In both Theorem 19.11 and Theorem 20.6 a we have (in essentially the same construction) employed counting argument. Counting are often simple, but they are among the most powerful toolsof mathematics. arguments Let n be a positive integer. Let
20.7 Example
Let n = 12. The positive integers less than 1,5, 7, and 11, so \302\242)(12)= 4.
or equal
to 12
and
relatively
prime
to 12 are
\342\226\2
Theorem
By
divisors of Euler's
describe
20.8Theorem
If
that
a is
and
= 1 (mod
avin)
prime
relatively
b < n
Using
group
20.6,
consisting
bVln)
Let n =
12. We
in Example a4 =
saw
(49)2= 2, 401=
74 (mod
12),
b can
be viewed elements
a contains
containing
an
of these
multiplication
integer
cosets by
have
= 1 (mod
as an
of the
element
of
multiplicative to n. Thus
prime
Z\342\200\236 relatively
n),
20.7
\302\242)(12)
12). For example,with 74 = 1 (mod 12).Of course, the
Euler's
using
if we
4. Thus
=
that
1 (mod
is
theorem,
to
compute
take
a =
a
integer
any
7, we have 74
easy
it in
Z12.
12 =
1.
way In
=
to compute Z12,
we have
so \342\200\2245
20.6, we
Theorem
Using
We prefer to
20.10Theorem Let m
with
work
a positive
be
(-5)2 = (5)2 =
to ax = b
Application
Proof
1, so
+
12(200)
without
72 =
the
divisible
\342\231\246
prime to 12, then
relatively
7 =
\342\200\224 1 is
av{n)
theorem follows.
and our Example
not
are now
ny
(mod
of the
b^
20.9
nZ
that
fact
representatives is well-defined,we
19.3 and
of order G\342\200\236
a + nL of the
a
But by Theorems
then
to n,
prime
relatively
the coset
to n.
n of
modulo
that Z\342\200\236
We can
phi-function.
n).
n, then
to
prime
relatively
multiplication
an integer
a is
Euler
theorem.
of Fermat's
generalization
is,
: Z+
q>
nonzero elements of
number of -> Z+ is the
the
is
function
(Euler'sTheorem) If by n,
Proof
19.3,
0. This
187
Theorems
and Euler's
Fermat's
20
Section
(mod
can find
all
an equation
integer
let
and
ax = b has a unique
equation
By Theorem20.6,a is sides
both
Multiplying
a unit
of ax
1
Zm
=
A
m)
of a linear congruence ax = b (mod m). the results for congruences.
solutions
in
Zm
a e
and
5 =
and
= b on
the
interpret
prime to
be relatively in Zm.
Zm
solution
in
74
and
a~lb is
left
by
m. For eachb e
a solution
certainly
a~l, we see this
is
the
Zm,
of the equation. only solution. \342\231\246
Interpreting
20.11 Corollary
If
a and
b (mod
m are m)
this theorem
relatively
has as
has
lemma for
in
the
precisely
general
any
at once
integer
one residue
the following
corollary.
congruence ax class modulo m.
b, the
case.
integer and let a, b e Zm. Let d be the gcd of a and m. The equation solution in Zm if and only if d divides b. When d divides b, the equation exactly d solutions in Zm.
Let m ax
as a
then for
intergers,
solutions all integers
Theorem20.10serves 20.12 Theorem
prime
for congruences, we obtain
be
a positive
= b has a
=
Part
IV
Proof
Fields
Rings and
First
= b in Zm unless d divides b. Suppose5 e Zm is so b = as \342\200\224 Since d divides both a and m, we qm. the right-hand side of the equation b = as \342\200\224 and hence divides qm, 5 can exist only if d divides b. that d does divide b. Let
there is no
show
we
Then as
a solution.
see that d divides b. Thus a solution
Supposenow
of ax
solution
\342\200\224 =
b
qm in Z,
a = aid,
=
b
and
bid,
mid.
can be rewritten
b\\) dqmi. We see of m^. Thus the solutions 5 of ax = b in Zm are precisely the elements that, read modulo mi, yield solutions of = fci in \302\2531* Zmi. Now let 5 e Zm, be the unique solution of a\\x = bi in Zmi given by in Zm that reduce to 5 modulo mi Theorem 20.10. The numbers those that are precisely can be computed in Zm as Then
as equation Ms a multiple
the
thatas
\342\200\224 =
b
qm in Z
of m if and
\342\200\224
+ mi, s + 2m 1, s are exactly d solutions of the s, s
Thus
there
20.12
Theorem
gives us
at
multiple
\342\226\240 s + (d + 3m 1, \342\226\240 \342\226\240,
\342\200\224
l)mi.
in Zm.
equation
\342\23
result on
classical
this
once
bi is a
\\fais
only
\342\200\224=
as d{a\\s
\342\200\224
the
of a linear
solutions
congruence.
20.13 Corollary
Let d be the exactly d
of positive integers a and m. The congruence ax = b (mod m) has a if d divides this is the When the solutions are the integers in b. case, only
gcd
if and
solution
our
Actually,
modulo
classes
residue
distinct
of Theorem
proof
m.
20.12 shows a bit
the solutions of ax = b if any solution 5 is found, classes (s + km\\) + (mZ) \342\200\224 1. It also tells us that we
about
more
stated in this corollary; namely, it shows that are precisely all elementsof the residue where mi = m/d and k runs through the integers from 0 to d can find such an 5 by finding a\\ = aid and b\\ = b/d, and solving a\\x To solve this congruence, we may consider a\\ and b\\ to be replaced by modulo mi and solve the equation a\\x = b\\ in Zmi. (mod
then
20.14 Example
Solution
Find
m) than we the solutions
The gcd of
12 and
there are no 20.15
Example Solution
of the
solutions
all
18 is
congruence 6, and
6 is
=
Ylx
27 (mod
by
the preceding
solutions.
Find all solutions of 15
of the
b\\ (mod
mi).
remainders
18).
of 27. Thus
not a divisor
= their
corollary,
\342\226
congruence
3, and 3
15*
does
= 27
(mod 18).
as explained before 27. Proceeding consider the Example everything by congruence 5ih9 (mod 6), which amounts to solving the equation 5x = 3 in Ze- Now the units in Zs are 1 and Thus in this group of units. the solution in Z(, is 5, and 5 is clearly its own inverse a x = (5_1)(3)= the of solutions 15* 27 (mod 18) are the Consequently, in residue the three classes. integers The
gcd
20.14,
and
18 is
we divide
divide
3 and
(5)(3)= 3. 3 +
18Z =
\342\200\242 -33. \342\200\242, {\342\200\242
-15,
9 +
18Z =
\342\200\242 \342\200\242, -27, {\342\200\242
-9,
9, 27,
18Z =
\342\200\242 -21, \342\200\242, {\342\200\242
-3,
15, 33, 51,--
15
+
3, 21,
\342\200\242 39, \342\200\242 \342\200\242},
45,
\342\200\242 \342\200\242 \342\200\242}. \342\200\242},
20.13. Note the d = displayed residueclassesmodulo
in the three
3+
6Z modulo
they came from
6, for
solutions
3
Corollary
illustrating
18 can solution
the
189
Exercises
20
Section
3,9, and 15 in be collected in * = 3 of 5* =
Zi8 the 3
\342\200\242 All the
solutions
one residue
class A
in 1,(,.
20
\342\226\240 EXERCISES
Computations We will
the multiplicative
later that
see
for
a generator
this
for the
group
6.
field.
Fermat's
theorem,
Fermat's theorem the
Compute
217 modulo
the remainder of 3749 when
to find
of 2(2' *
remainder
1 when
+
of values
of
for
n <
Compute
where
p is a prime.
9.
Compute
where
both p
10. Use Euler'sgeneralization 11 through
Exercises
6 (mod
2* =
13. 36*
is
of 347 when it
remainder
the
find
8.
11.
3. divided
You
[Hint:
Z17
to compute the
of
remainder
30.
q are
and
18, describe all
primes.
theorem to
of Fermat's
of the
solutions
22* = 5
14. 45* =
15.
39* =
125 (mod
9)
16.
41* =
125 (mod
17.
155* = 75 (mod
65)
18.
39* =
52 (mod
19.
Let p
20.
Using
Exercise
21. Using 22. Using
Exercise
28 below,
find
the
remainder
of
Exercise
28 below,
find
the
remainder
of
>3.
28 below to
Use Exercise
28 below,
remainder
the
in Examples
did
of (p
24) 9) 130) \342\200\224
2)!
modulo
p.
37. 49! modulo 53. 24! modulo 29.
of 34! modulo
remainder
the
find
find
divided
(mod 15)
15(mod
24)
be a prime
when
71000
given congruence,as we 12.
4)
of
remainder
the
find
15 (mod
Concepts of the
each
by finding
23.
will need
=
23. Mark
this
divided by 7.
it is
divided by 19.
by
Illustrate
is cyclic.
field
18.]
7. Makea table
In
elements of a finite
2. Zn
Using
5. Use
of nonzero
finite
given
1. Z7 4.
group
following
b. ap~l
= 1 (mod s 1 (mod
c.
< n for
a. ap~l
p) for all
all n e
units
f. The
product of two
are the Z\342\200\236
of two of a
integers a integers a
and
primes
such
that
p. a ^
0 (mod p) for
a prime
p.
Z+.
all n e
< n \342\200\224 1 for
g. The product h. The product
for all
p)
d. (p{n) e. The
in
or false.
true
units
Z+.
positive integers in
nonunits
is always Z\342\200\236 in
less than
Z\342\200\236 may be
unit and a nonunit
n and
relatively prime
a unit. a unit.
in Z\342\200\236 is never
a unit.
to
n.
by 24.
20.14
and
20.15.
Part
190
IV
i.
Every
j.
Let d
Give the
group
b (mod p), where gcd integers a and d incongruent solutions. ax =
congruence
has 24.
Fields
and
Rings
be the
exactly
for the
table
multiplication
prime, has a solution. d divides b, then the congruence
p is a m. If
of positive
multiplicative
of units
group
in Zi2.
To which
ax = b
(mod m)
of order
group
4 is
it
isomorphic?
Proof
Synopsis
25. Give a one-sentence synopsis
26. Give
of the
synopsis of the
a one-sentence
proof of Theorem 20.1.
20.8.
of Theorem
proof
Theory
27. Show
that
1 and
Consider the 28.
p
\342\200\224 1 are
x2
equation
the only elements 0.]
Zp that
field
are
own
their
inverse. [Hint:
multiplicative
\342\200\22 1 theorem that states that if p is a prime, then (/? \342\200\224 27, deduce the half of Wilson's 1)! = \342\200\224 1 (mod n), then n is a prime. >1 such that (n half states that if n is an integer (mod p). [The other think what the remainder of (n \342\200\224 Just 1)! would be modulo n if n is not a prime.] Using
Exercise
1)!s
29. UseFermat's 383838
30.
of the
\342\200\224 1 =
Referring
section
to show that theorem = (37)(19)(13)(7)(3)(2).] to Exercise
21
29,
If
an
is such
domain
integral
it is a
then
that
field. However, many
field. This dilemma
is not
the
that
every integral
too serious.
It
divisible by 383838.
integer
n37
\342\200\224 n is
divides
n37
\342\200\224 n for
all positive
[Hint:
n.
integers
Domain
of an Integral
of Quotients
Field
The
larger than 383838
a number
find
positive integer n,
for any
element has a multiplicative inverse, such as the integers domains, Z, do not form a to show that every is the purpose of this section nonzero
integral regarded as being containedin a certain field, afield of quotients domain. This field will be a minimal field containing the integral domain the of integral in a sense that we shall describe. For example, the integers are contained in the field can all be expressed as quotients of integers. Our construction of a Q, whoseelements field of quotients of an integral domain is exactly the same as the construction of the
domain
can be
from the integers, which often in a course in foundations numbers or appears advanced calculus. Tofollow this construction is such a in use exercise the through good of definitions and the concept of isomorphism that we discuss it in some detail, although to write out, or to read, be tedious. We can be motivated at every every last detail would
rational
step by the
way
Q can
from Z.
be formed
The Construction Let
D be
outline
domain that steps we take is as
an integral
of the
1.
Define what
2.
Define the
the binary
we
desire
to enlarge
to
a field
of quotients
follows:
elements operations
of F
are to
be.
of addition
and
multiplication
on
F.
F.
A
coarse
3.
Check
4.
Show
The Field of
21
Section
the field
all
Steps 1,2, and with
proceed
4 are very interesting, the construction.
Step 1 Let D be a
are
that
is, if D
= Z,
be the
pair
x D
of D
subset
element
21.1 Definition
Two
only
product
e D]
as representing a formal the number cut the set D
represent
and suggests that
we
|
a/b,
quotient for
us.
x D down
The pair a bit.
by
{{a,b)\\a,b
\342\202\254 D,b
^0}.
not
still
F,
elements
if
Cartesian
chore. We
going to be our field as is indicated by the fact that, with D = Z, of integers such as (2, 3) and (4, 6) can represent the same rational pairs the same We next define when two elements of S will eventually represent of or, as we shall say, when two elements of S are equivalent.
Now S is number.
and form the
(a, b)
eventually
given
S =
different
pair
will
operations.
largely a mechanical
3 is
Step
= {{a,b)\\a,b
ordered
(2, 3) element of Q
the
(2, 0) representsno Let S
of an
to think
going
and
these
under
integral subdomain.
as an
D
domain,
integral
given
DxD We
F is a field
that
show
to
viewed as containing
can be
F
that
axioms
191
Integral Domain
of an
Quotients
(a, b)
and
(c,
d) in
ad = be.
S are equivalent, denoted
(a, b)
by
~
(c, d),
if
and \342\226\240
definition is reasonable, sincethe criterion for (a,b) ~ (c, d) is an ad = be involving elements in D and concerning the known in multiplication equation of equality of definition D. Note also that for D = Z, the criterion gives us our usual | = since (2)(6) = (3)(4).The rational that we usually with f\302\260r number example, 2-, \302\247 f of as the collection of all quotients of integers that reduce denote by | can be thought that this
Observe
to, or 21.2 Lemma Proof
are equivalent
Therelation
to,
~ between
We must check the
|.
of
elements
three
set
the
S as just
described is an
of an equivalence
properties
equivalence
relation.
relation.
in D is commutative. Reflexive(a, b) ~ {a,b) since ab = ba, for multiplication ~ = If then ad in D is be. Since Symmetric {a,b) (c,d), multiplication ~ = that and we deduce cb da, commutative, (c, d) {a, b). consequently If
Transitive
{a, b)
these relations
and
~ (c, the
fact
d)
asd = d
Now in
the
^0,
and
argument.
and
(c,
d)
~ (r,
that multiplication
sad = sbc = foes
5), then in
=
D
ad
=
foe
and
is commutative,
bdr
cs =
dr. Using we have
= ford.
is an integral domain, so cancellation is valid; this is a crucial step Hence from asd = brd we obtain as = br, so that (a, b) ~ (r, s).
Z)
\342\231\246
192
Part
IV
Fields
Rings and
of Theorem0.22,that ~ gives a partition of S into equivalence To avoid long bars over extended classes. we shall let [(a, b)], rather than expressions, class of (a, b) in S under the relation ~. We now finish Step 1 (a, b), be the equivalence by defining F to be the set of all equivalence classes[(a,b)] for (a, b) e S. now
We
Step 2 Observe that Q give 21.3 Lemma
the
in view
know,
next
The
usual
For [(a, b)] and
serves to define
lemma
[(a, b)] is viewed
= Z and
if D
addition
and
as (a/b)
e Q,
in F.
multiplication
these definitions
applied
to
operations. d)] in
[(c,
F,
the
equations
= [(ad + be, bd)]
+ [(c, d)]
[(a, b)] and
= [(ac,bd)]
[(a,b)][(c,d)]
give well-definedoperations
Proof
that if
first
Observe
of
[(a, b)]
^0. BecauseZ)
and
and multiplication
addition
[(c,
d)] are in
F,
then
on F.
b) and
(a,
(c, d)
are in
S, so fo ^
0
integral domain, bd ^ 0, soboth (ad + be, bd) and (ac, bd) shows are in 5. (Note the crucial use here of the fact that D has no divisors of 0.)This are at least in F. that the right-hand sides of the defining equations It remains for us to show and multiplication that these operations of addition are in S of elementsof well defined. That is, they were defined by means of representatives that if different representatives in S are chosen, the same element F, so we must show of F will result. To this end, suppose that (a\\, b\\) \342\202\254 [(c, d)]. We [(a, b)] and (ci, d{) \342\202\254
and d
must
is an
that
show
[(ad +
+ b\\Ci, b\\di) e
(a\\di
be, bd)]
and
e [(ac,bd)].
(axci,bidx) Now
b\\) (\302\253i,
e [(a,
b)] means
that
~ (a,
b\\)
(a\\,
a\\b = (c\\, d\\)
Similarly,
e [(c, d)] implies
b); that
is,
b\\a.
that
c\\d = d\\c.
To get a \"common denominator\" (a\\,b\\),
by
b\\b.
(c,
d), and
Adding
(c\\,
d\\),
the resulting
the first equations, we obtain +
a\\bd\\d
Using various
axioms for an
integral
(a\\d\\
c\\db\\b
=
the
b\\ad\\d
we see
domain,
+ b\\C\\)bd
member) for the equation by d\\d and the
second
(common
we multiply
=
b\\d\\(ad
following d\\cb\\b.
+ that
+
be),
four
pairs
second
equation
(a, b),
equation
in D:
The Field of
21
Section
193
Integral Domain
of an
Quotients
so
so,
of D,
axioms
using
which implies
a\\bc\\d
= b\\ad\\C,
a\\C\\bd
=
e
important to understand This completesour it. proving
for this
but that
done. Thus working We list the things left to the exercises.
we have
this construction.
1. Proof
Now
in
Addition
F is
it is
true,
Let
through
that must
the
[(0, 1)]is
an
4.
[(\342\200\224a, b)]
is an
5.
Multiplication
in F is
Multiplication
in F
F. If a
for
necessity
be provedand
of
of them.
a couple
prove
[{c,d)] + [{a,b)] bd) ~ {cb + da, db). axioms of D.
= db, by
bd
the
is associative.
3.
b)] e
and
[{ad + be, bd)]. Also to show that {ad + be,
b)] +
Addition
[{a,
lemma
to work a few of these through work through them unless we understand them will contribute to our understanding
element
identity
for
addition
additive inverse for
in F.
[{a,b)] in
F.
associative.
is commutative.
7. The distributive laws hold in F. 8. [(1, 1)]is a multiplicative identity element in F. 9. If [{a, b)] e F is not the additive identity element, [{b, a)] is a multiplicative inverse for [{a,b)].
Proof
\342\231\246
proof.
commutative.
2.
6.
For
obtain
good for us
cannot
we
definition
is
F.
M).
of the last
meaning
[{c, d)] is by definition [{cb + da, db)].We need ad + be = cb + da and since
[{a,
we
2.
Step
is
~ (ac,
b\\d\\)
the
Step 3 is routine, reason
Therest are
in
c\\d = d\\c,
b\\d\\ac,
which completes the
M)],
[(ac,
It is
what
of addition
and
that
Thus (aiCi, fci^i)
Step 3
care
we get
(aiCi,
details.The
be,bd),
e [{ad + be, bd)].This takes a\\b = b\\a multiplying the equations
F, on
in
multiplication
{ad +
+ b\\Cx, b\\d\\)
{axd\\
giving
b\\d\\) ~
+ b\\C\\,
(a\\d\\
= 0, then a\\
=^0
= 0,
so
{a,b) ~ (0, 1).
then
a
^ 0 in
D
and
is
by
This \342\231\246
194
Part
IV
and Fields
Rings
that
is not in F.
= [(0, 1)].But
[(a, b)]
is,
the
additive
Now
[(a,
a)]
b)][(b,
1)] is
[(0,
the
we have a = [(ab, ba)]. But in F,
identity
additive
^
by Part sense to
identity
0, so it makes in D we have
ab =
3. Thus talk about
ba, so
if
[(a,
b)]
[(b,
a)]
= (ba)l,
(ab)l
and
1).
(ab,ba)~(l, Thus
=
[(a,fe)][(fe,a)] and
1)] is
[(1, This
the
Step 4
renamethe
that image
lemma
The next
suggest injection 21.4 Lemma
The
D. To do to show that F can be regarded as containing of F. Then if we is an isomorphism i of D with a subdomain of D, we will be done. of D under i using the names of the elements We use the letter i for this isomorphism to us this gives isomorphism. will on we D into F. the footnote (see inject page 4);
for us
remains
It
show
there
i{a) =
i : D \342\200\224>\342\226\240 F given by
map
\342\23
Step 3.
completes
this, we
[(l, 1)],
8.
by Part
identity
multiplicative
[(a, 1)]is
an
of D
isomorphism
a subring
with
off.
Proof
For a
b in
and
D, we
have
i{a + b)
=
[{a +
b, 1)].
Also,
i(a) + i(b) = [(a.1)] so *'(a
+ fe) = z'(a)
= [(al
[{b. 1)]
+
+
lfe,
1)]
=
[(a + b,
1)].
+ i{b). Furthermore, 1)],
[(ab.
i(ab)=
while
i(a)itf>) = so j'(afc)
=
[(a, l)][(fe,
to
show
that i is
only
one to one. If
[(a, 1)] = 1) ~ (b, 1) giving
al
=
lfo; that
i
an isomorphism
is
of D
with
[(ft,
i(a)
= i(b),
then
1)],
is,
a = Thus
[(ab, 1)],
i(a)i(b).
It remains for us
so (a.
=
1)]
b. of course,
;'[\302\243>], and,
i[D] is
then
a subdomain
off.
\342\23
Since
we have
21.5 Theorem
Any
[(a. now
integral
element
field of
of F
b)] = proved
[(a, 1)][(1,&)] the following
=
[(a.
quotients of
D.)
1)] =
i(a)/i(b)
clearly
holds
in F,
theorem.
D can be enlargedto can be expressed as a quotient
domain
l)]/[(fc.
(or
in) a field F such that elements of D. (Such a field
embedded
of two
every
F is a
of Quotients
The Field
21
Section
of an
195
Domain
Integral
Uniqueness in the
said
We
21.6
Theorem
Let F
be a
The
\342\200\224> L that
e D.
for a
of D are
e
with
D
theorem b ^ 0. The next which is a field of quotients
as a minimal field D must contain will show that every of D, and that any
isomorphic.
of D
let
and
L be
diagram
in
field
any
of F
gives an isomorphism
and mapping
subring
for
evident,
a subfield
of quotients
field
ty : F
map
Proof
of quotients
fields
be regarded in some sense since every field containing
could
F
that
intuitively
contains
D
containing
two
is
a/b for every a,b
all elements
field
beginning This
D.
containing
21.7
Fig.
with
D. Then there exists a of L such that \\jf(a) = a
containing a subfield
may help
the situation
to visualize
you
theorem.
this
An element of F is of the form a /F b where /f denotesthe quotient of a e D by to map a /r b onto a /i b where D regarded as elementsof F. We of course want the quotient of elementsin L. The main job will be to show that such a map denotes
b e /l
is well defined. define
must
We
ir :
F
we start
\342\200\224>\342\226\240 L, and
\\jf{a) = a e F
x
Every
attempt
to
is a
define
quotient
a
/r
a e
for
some
b of
defining
by
two
D. a and
elements
if
(a
/F b) =
^ (a)
forfo
sense. If a the identity
show that this map if is sensible and well-defined. of ir (a /F ^ Owe have i/{b) ^ 0, so our definition b = c /f d in F, then ad = be in D, so ir{ad) /F and
ir{bc)
=
Thus
ir{a) L, so ir
Since
b) as i/r(a) =
\\jr{bc).
on D,
ir{ad) = ir{a)ir{d)
in
0, of
D.
Let us
if(b).
/L
We must first onZ),
b, b ^
ty by
/L
ir(b)
= f{c)
/l f{d)
is well-defined.
The equations ir(xy)
= ir{x)ir{y)
21.7 Figure
ir{b)ir{c).
ir is
the
identity
ir (b) makes But since \\jr is /i
IV
Part
196
Rings and
Fields
and
follow
easily
If \\jf{a
/f
from the definition of ty on F b) = \\j/(c If d), we have
the fact that
from
and
is the
ty
on D.
identity
t(b) = t(c)/L t(d)
t(a)/l so
=
i/{a)i/{d)
Since ty is one to
is the
Every field
Corollary
i/r(a) =
21.9
L
Any two fields
Corollary
a for
a
domain
21.6
element
every
D contains a field of quotients of
subfield
the
^
D.
of
of L is a quotient
ty [F]
in L
\342\231
of quotients
of an
integral
domain D
in Theorem 21.6 that L is a field of quotients in the form a /L b for a,b e D. expressed of Theorem 21.6 and is thus isomorphic to F. be
can
Thus
d.
\342\231
are isomorphic. of D,
Suppose
Proof
soa /p b = c /p
e D.
an integral
containing
In the proof of Theorem of elements of D.
Proof
= be,
that ad
deduce
one.
By definition,
21.8
we then
on D,
identity
i/r(b)i/r(c).
element x of L i/r[F] of the proof
so that every
ThenL is the
field
\342\231
21
\342\226\240 EXERCISES
Computations
1. Describethe
field
F of quotients
of
the
D =
of C. \"Describe\"
D are
the
means
Gaussian
2. Describe(in ofR.
the sense
give the
subdomain
integral {n
+ mi
elements of C that
m e Z}
\\ n,
up the field
make
of quotients
of D
in C.
(The elementsof
integers.) of Exercise
1) the
field
F of
quotients of the
integral
subdomain
D =
{n
+ m-j2
\\n,m&Z}
Concepts
3.
Correct
a form
the
definition
of the
italicized term
of quotients of an integral element of D is a unit in F. A field
without
reference
to the
text,
if
correction
is needed,
so
that
it is
acceptablefor publication. domain
D is a
field
F
in which
D can be embeddedso that
every
nonzero
in
Section
4. Mark
of the
each
following
a. Q is a
b. c. d. e.
or false.
true
field
of quotients
of Z.
R is a field R is a field
of quotients
of Z.
of quotients
of E.
C is a field of If D is a field,
of E.
quotients
of D is isomorphic any field of quotients has no divisors of 0 was used strongly several
then
to
D.
times in the construction D. domain integral quotients of D. g. Every element of an integral domain D is a unit in a field F of quotients D is a unit in a field F of quotients of of an integral domain h. Every nonzero element D' of an integral domain D can be regarded i. A field of quotients F' of a subdomain of some field of quotients of D. j. Every field of quotients of Z is isomorphicto Q. f. The
fact
D
that
F of
5. Show by an
be a field
example
of quotients
197
Exercises
21
of a
field
of the
that
a field
F' of
of a
quotients
proper subdomain
D' of an
integral
D. as a
subfield
D may also
domain
for D.
Theory
6. Prove
7. Prove Part
3 of
Step 3. You Step 3. You
8. Prove Part
4 of
Step 3.
You
may
assume
any preceding
part
of Step
3.
9. Prove Part
5 of
Step 3.
You
may
assume
any preceding
part
of Step
3.
10. Prove Part
6 of
Step 3.
You
may
assume
any preceding
part
of Step
3.
11. Prove Part
7 of
Step 3.
You
may
assume
any preceding
part
of Step
3.
2 of
Part
12. Let R
be
this
assume
any preceding
part
of Step
3.
may
assume
any preceding
part
of Step
3.
and subset of R closed under commutative ring, and let T be a nonempty multiplication the construction in neither 0 nor divisors of 0. Starting with R x T and otherwise following exactly we can show that the ring R can be enlarged to a partial ring of quotients T). Think about Q{R, and see why things still work. In particular, show the minutes or so; lookbackover the construction
a nonzero
containing this
may
section, 15
for
following:
a. Q(R, b. 13.
T) has unity
even
if R
In Q(R,
Prove
T), every nonzero from Exercise 12 that
does
every
can be enlargedto a commutative
14. 15.
not.
of T
element
a that is ring containing an element 30 of Section 19. Compare with Exercise are there in the ring Q(Z^, {1,3})?
commutative
nonzero ring
is a unit.
with unity.
With
reference
to Exercise
12, how
With
reference
to Exercise
12, describe the
ring
g(Z,
to Exercise
12, describe the
ring
<2(3Z,
elements
many
{2n | n
e Z+}), by
describing
a subring
not
a divisor
of 0
of R
to
which
it
of R
to which
it
is isomorphic.
16.
With
reference
{6\"
|
n e
Z+}) by describing a
subring
is isomorphic.
17.
the condition that T have no divisors of zero and just require reference to Exercise 12, suppose we drop that nonempty T not containing The attempt to enlarge R to a commutative 0 be closed under multiplication. of T is a unit must fail if T contains an element element a that is a ring with unity in which every nonzero of 0, for a divisor of 0 cannot where a construction parallel to that in the divisor also be a unit. Try to discover text but starting with R x T first runs into trouble. In particular, for R = Z6 and T = {1, 2, 4},illustrate the With
first difficulty
encountered. [Hint:
It
is in
Step 1.]
Part
198
IV
22
section
Fields
and
Rings
of Polynomials
Rings
in an
Polynomials
Indeterminate
a pretty workable idea of what constitutes a polynomial in x with coefficients R. We can guesshow to add and multiply such polynomials and know what is with meant by the degree of a polynomial. We expectthat the set R [x] of all polynomials in the ring R is itself a ring with the usual operations of polynomial addition coefficients and that R is a subring of R[x]. with and multiplication, However, we will be working different than the in from a school viewpoint approach high algebra slightly polynomials or calculus, and there are a few things that we want to say. rather than a variable. Suppose, for In the first place, we will call x an indeterminate coefficients is One of the in the ring Z[x] is Ix, our of Z. that ring polynomials example which we shall write simply as x. Now x is not 1 or 2 or any of the other elements of Z[x]. Thus from now on we will never write such things as \"x = 1\" or \"x = 2,\" as we have all have
We
a ring
in
in other
done
courses. We callx an
Also,
change.
we
will
write
never
indeterminate
than a variable to emphasize this 4 = 0,\" simply because as \"x2 \342\200\224 to speaking of Z[x], We are accustomed in a lot of time the remainder of spending rather
an expression such
in our ring not the zero polynomial will be and a \"solving polynomial equation,\" refer to it as \"finding a zero of a polynomial.\" our text discussing this, but we will always of algebraic structures not to say in In summary, we try to be careful in our discussion context that they are not equal. one context that things are equal and in another
x2
\342\200\224 4 is
Note
\342\226\240 Historical
use of
x
other letters near
and
the
end
of
The the alphabet to represent an \"indeterminate\" is due to Rene Descartes (1596-1650). Earlier, Francois Viete (1540-1603) had used vowelsfor inis also
Descartes
responsible for
of the factor theorem The Geometry, which
of analytic
geometry; curves
can be
Descartes was
the
first
appeared
as an
(1637).
This
publication his work
appendix to work
also
basic concepts showed how
of the
Descartes
described algebraically. born
to
a wealthy
family
in
of delicate France; since he was always health, he formed the habit of spending his mornings in bed. It was at these times that he accomplished his most productive work. The DiscourseonMethod was Descartes' the proper attempt to show procedures for \"searching for truth in the sciences.\"The first step in this process was to reject as absolutely La Haye,
everything
since
it
was
\"I think,
philosophy:
therefore
I
least
the
was
\"something,\" he conceived his first
thinking of
principle The
am.\"
but, was
doubt;
most
Discourse on Method, however, The Optics, The Geometry, appendices: It was here that Descartes and The Meteorology. provided examples of how he actually applied his method. Among the important ideas Descartes discovered and published in these works were the sine law of refraction of light, the basics of the theory enlighteningparts
quantities.
(Corollary23.3)in
his Discourse on Method contained the first publication geometric
for known
consonants
and
determinates
of which he had that he who necessary
false
arethe
of the
three
of equations, and a geometric explanation of the rainbow. In 1649, Descartes was invited by Queen Christina of Sweden to come to Stockholm to tutor her. Unfortunately, the Queen required him, contrary
to his
hour.
1650.
He
long-established contracted
soon
habits, a lung
to
rise
at an
disease and
early
died
in
knows
If a person
precisely
the
nature
such a polynomial
Section 22
Rings of Polynomials
polynomials,
it is
about
nothing
x with sum
of a polynomial in to be a finite formal
199
an easy task to describe in a ring R. If we just define
not
coefficients
n
= ao
22aix'
+
\342\200\242 \342\200\242 \342\200\242
+ anx\",
+
a\\x
where a,- e R, we get ourselvesinto a bit of trouble. For surely 0 + a\\x and 0 + a\\x but we want to regard them Ox2 are different as formal as the same polynomial. sums, as an infinite formal sum solution to this problem is to define a polynomial practical
+ A
oo
2^ajXl
=
+ '
+ a\\x
flo
''
+ anx\"
' '
+
'
i
(=0
where
22.1 Definition
=
a,-
than
more
R be a
Let
there is no problem of having number of values of i. Now 0 for all but a finite one formal sum representwhat we wish to consider a single polynomial.
ring.
A
polynomial
f{x)
with coefficients
= ao
+
in
R is
an
infinite
sum
formal
oo
2_\\ciix' (=0
+
a.\\x
\342\226\240 \342\226\240 \342\226\240 \342\226\240 + a\342\200\236x\" + \342\226\240 \342\226\240,
at e R and a,- = 0 for all but a finite number of values of i. The a,- are coefficients of f{x). If for somei > 0 it is true that a,- ^ 0, the largest such value of i is the degree J of f{x). If all a,- = 0, then the degree of f{x) is undefined where
To
working with polynomials, let us agree that = 0 for i > n, then we may denote f(x)
simplify
= ao
if f{x)
\342\226\240\342\226\240 has at + \342\226\240 a\342\200\236xn
by
+ Q\\x
a0
+
a\\x
+
+
\342\226\240
\342\226\240 \342\226\240 \342\226\240
+
1- a\342\200\236xn.
lxk in such a sum as xk. For example, Ix as 2 + x. Finally, we shall agree that we may omit altogether from the formal sum any term Ox', or a0 if ao = 0 but not all at = 0. Thus 0, 2, x, and 2 + x2 are polynomials with coefficients in Z. An element of R is a if R
Also, in
Z[x],
has
constant a way
write
the
write
will
a term
2 +
polynomial
polynomial. and
Addition in
0, we
1 ^
unity
we will
of polynomials
multiplication
with
coefficients
in a ring R
are defined
to us. If
familiar
f(x) =
+ axx
a0
h anx\"
H
H
and
g(x) then
for polynomial f(x)
'
= bo
+
biX
+
--- +
+ \342\226\240\342\226\240\342\226\240, b\342\200\236xn
addition, we have
+ g(x)
= co +
C]_x
H
h cnxn
H
where
=
+ b\342\200\236, c\342\200\236 a\342\200\236
The degree of the zero polynomial is sometimes defined to be \342\200\224 1, which is the first integer less than 0, or to be \342\200\224 oo so that the degreeof f(x)g(x) will be the sum of the degrees of f(x) and g(x) if one of is zero. them defined
200
Part IV
Fields
Rings and
and
for
f(x)g(x) =
The set R[x] ring
under
and
if
all polynomials
of
addition
polynomial
R has
1^0.
unity
is an
then
where
h d\342\200\236x\" H
H
With these definitions
commutative. theorem.
Proof
+ dxx
need
and multiplication,
of addition
in an indeterminate
and 1 is
unity
x
for
we have
Applying
abelian
ckx
,,=0
E
in a ring R is a so is R[x],
for
multiplication
E 5-0
computations.
E
.4=0
\\ ('=0
J2 n=0
E(
E
(Ea'fc\302\253-''
\\ i=0
i
aibJck)xs
E^-\342\204\242 m=0
Efc
J^m-j
,;=o
m=0 \\ j=0 oo
./=0
Whew!!
/ \\
/
\302\260\302\260
/ \\i=0
the fourth expression, having computation, just two summation signs, as the value of the triple product f (x)g{x)h(x) of thesepolynomials this associative we view f(g(h(x))) as the fashion, multiplication. (In a similar of the associative composition (f o g o h){x) of three functions f, g, and h.) be
should
under value
this
In
viewed
are similarly proved. (See Exercise26.) to the statement of the theorem show that R [x] is a commutative R is commutative, and a unity 1 ^ 0 in R is also unity for R[x], in view of the in R[x]. of multiplication \342\231
The distributive The ring
is not
ckx \302\243*'\342\226\240*\302\273-'\342\226\240*\"
J2
J=0
R
following
R[x].
group
E
2>,v
so these
then
are
laws
2>*'')
the
is commutative,
is apparent. The associativelaw but slightly cumbersome, straightforward, the law. associative proving to at, bj, ck e R, we obtain axioms ring
That {R[x], +) and the distributive We illustrate by
aib\342\200\236-i
^=0
coefficients
with
If R
multiplication.
also
=
d\342\200\236
number of values of i, not equal J2\"=obian-i if
a finite
but
make
definitions
22.2 Theorem
d0
both c, and dt are 0 for all sense. Note that J2'i=oa'bn-i
that
Observe
we have
multiplication,
polynomial
if
definition
Thus
Q[x] the
comments
Z[x] ring
laws
prior
is the
ring
of polynomials
of polynomials
in x
with
in the indeterminate x with integral and so on. coefficients,
rational
coefficients,
Section22
22.3 Example
In la
[x
have
], we
(x + Still working
1)2= {x+ l)(x+ 1)
7Li\\_x\\, we
in
(x +
is, the
naturalway
in
+
1.
obtain
y are
and
l)x + 1 = x2
(1 +
x2 +
+
(1 +
two inde terminates,
0 = 0.
1) = Ox +
then
can form
we
\342\226\262
the ring
(/?[>])[)>],
that are polynomials in x. Every in y with coefficients polynomials that are polynomials in x can be rewritten in a with coefficients
of
ring
polynomial
=
1) + (x + 1)= (1+ l)x
and*
is a ring
Ifi?
that
201
of Polynomials
Rings
y
in x
as a polynomial
with
that
coefficients
are polynomials
in
y as
illustrated
to (/?[)>])[;t], 20. This indicates that isomorphic (R [*])[>'] is naturally a careful these rings by means of this natural although proof is tedious. We shall identify isomorphism, and shall consider this ring R[x, y] the ring of polynomials in two inde\342\226\240 of in in R. The ring R[xi, \342\226\240 terminates x and y with coefficients \342\226\240, x\342\200\236] polynomials the n indeterminates defined. xt with coefficients in R is similarly We leave as Exercise 24 the proof that if D is an integral domain then so is D[x]. In is not a field, for if F is a field, then is an integral domain. Note that F[x] F[x] particular, x is not a unit in F[x]. That is, there is no polynomial f(x) e F[x]such that xf(x) = 1. of F[x]. Any element F{x) By Theorem21.5,one can construct the field of quotients in F{x) can be represented as a quotient of two polynomials in F[x] with f{x)/g{x) of F[x\\, define F(x\\, \342\226\240\342\226\240\342\226\240, to be the field of quotients ---,.\302\275]. g(x) ^ 0. We similarly x\342\200\236) \342\226\240 is the field of rational functions in n indeterminates This field F{x\\, \342\226\240 over F. \342\226\240, x\342\200\236) These fields play a very important role in algebraic geometry.
by Exercise
Evaluation
The
Homomorphisms
to proceed to show how homomorphisms to as \"solving a polynomial equation.\" a subfield of E, that is, F < E. The next theorem asserts of F[x] into E. These homomorphisms homomorphisms for much of the rest of our work. We
are
have
22.4 Theorem
now ready
referred
always
(The Evaluation E, let a be any defined
for
(f>a
The
+
a\\x maps
homomorphism
Proof
be a
Let F
indeterminate.
an
E and
The
of a field
subfield
map
4>a
: F[x]
\342\200\224*E
by
+ (\302\253o and
Field Theory)
and let x be
of E,
4>a(aa +axx
a,
for
Homomorphisms element
used to study what we F be fields, with F the existence of very important will be the fundamental tools can be
Let
subfield
The dashed
\342\226\240 \342\226\240 \342\226\240
e + anxn) F isomorphically 4>a is evaluation
and mapping
lines indicate
h anxn)
H
+ axa -\\
\\-
anan
F[x] is ahomomorphism of F[x] into E. Also, 4>a{x) = that is, 4>a{a) = aforae f. The by the identity map; at a.
diagram an
= a0
element
in
Fig.
22.5
of the
may help us
set. The theorem
to
situation.
visualize
this
is really
an immediate
IV
Fields
Rings and
[fix]] N
a =
\"\342\226\240 a = (pa{a)
22.5
consequence
of our
well defined,
that
is,
of our
independent
such
since
If
a finite
Ox',
which
f(x) =
\342\226\240\342\226\240\342\226\240 = bo + a\342\200\236xn, g(x)
+ cxx +
+
\302\242(^(/^)
\342\226\240 \342\226\240 \342\226\240
+
crxr,
map 0a is sum
a finite
only
+
b\\x
+
+
qa
H
or deletion
by insertion
h{x) =
\342\226\240\342\226\240\342\226\240
+ bmxm,
and
then
= 4>a{h{x)) =
+ g(x))
The
F[x].
h a\342\200\236x\",
H
sum representing f(x) can be changed does not affect the value of 4>a(f(x))-
+ axx \302\253o
f(x) + g(x)= co
in
representation of f{x) e F[x]as
a0 + axx
of terms
and multiplication
addition
of
definitions
Figure
Co
+
crar,
while
0\302\253(/(^))
+ Mg(x))
Since by definition
=
(a0 +
of polynomial Mf(x)
Turning
to multiplication,
aia + ---+ anan)
+
(b0 +
addition
we have c,-
=
+ g(x)) =
Mf(x)) +
$a(g(x)).
we see that
f(x)g(x)
= d0
bxa +
+ fc,-, we
a,-
\342\226\240 \342\226\240 \342\226\240
+ bmam).
see that
if
+
dxx
+
\342\226\240 \342\226\240 \342\226\240
+ dsxs,
then
t>a{f{x)g{x)) = d0 +
dia + ---+ dsas
while
[
=
+ (\302\253o
axa
+
\342\226\240 \342\226\240 \342\226\240 + a\342\200\236a\") (bo
+ bxa
+
\342\226\240 \342\226\240 \342\226\240
+ bmam).
Sinceby
multiplication dj = X!/=oaibj-i,
of polynomial
definition
= [0a
fa(f(x)g(x)) 4>a is a
Thus
we seethat
(/(*))] [0a(g(* ))]\342\200\242
homomorphism.
of 0a applied to a constant so gives 4>a{a) 0a maps F isomorphically of 0a, we have 4>a{x) =
The
203
Rings of Polynomials
22
Section
definition
very
= a,
a e
polynomial by the identity
map.
F[x], where a
e F.
by definition
Again
\342\231\246
with the identical proof if F point out that this theorem is valid commutative rings with unity rather than fields. However, we shall be they are fields. primarily in the case in which We
merely
and
E
are
interested
It is hard to overemphasize the importance of this simple theorem for us. It is the very It is so simple that it could justifiably foundation for all of our further work in field theory. a little misleading to write be called an observation rather than a theorem. It was perhaps notation makes it look so complicatedthat you out the proof because the polynomial it is a difficult theorem. may be fooled into thinking
22.6
Example
Let F be Q and E be E 0o : QM -+ E. Here 0o(\302\253o+
Thus
22.7
Example
polynomial
every
be Q and : -+ KQM 02 Let F
E
Theorem
in
\342\226\240 \342\226\240 \342\226\240 = ao + a\342\200\236x\")
a\\x +
is mapped onto its
be E in
Thus x2 + x
a\\x
\342\200\242 \342\200\242 \342\200\242
+ anQ\"
homomorphism
=
\302\253o-
term.
consider
and
= a0 + a\342\200\236x\")
H
in the
\342\200\224 6 is
kernel TV
the reason
that 02(;t2
be Q and E be C 0; :Q[x] -+ C.Here
Let F
0,-(ao
and 4>i{x)
= i.
+x
of 02-
\342\226\262
the evaluation
homomorphism
+ a{l
+
H
a\342\200\2362\".
H
1 is
+
2-6
22.4
anxn)
that
and
02(;t
-2)
consider
= a0
+ axi
= i2+1
= 0,
Note that
in the
kernel TV* of
t
= 0.
-6 = (x -2){x + 3),
1-
kix2
+
22
Of course,
- 6) = 0 is
in Theorem
+ a\\x
=
- 6)
+x
x2 +x
so x2
constant
Theorem 22.4
4>2{x2
Example
+ \302\253i0
evaluation
that
Note
22.8
+
the
Here
02(\302\2530+
and
and consider
22.4,
+ 1)
=
2-2 = 0.
the evaluation
-\\
\\-
anin
\342\226\262
homomorphism
204
Part
22.9Example
IV
Fields
Rings and
Let
F be
0JT
: Q[x]
Q
E
E be
let
and
fa(a0 + a^x be provedthat the kernel of
It can Thus
polynomials
We now
of Theorem
between our Rather than speak
the connection
equation. of a
= 0, then
a is a
of this
terms
In
r such
, numbers
let
and
\342\200\242 n. 1, \342\200\242 \342\200\242
all formal
to Q[x] in a
natural
the classical concept of a polynomial equation, we
and
ideas
new
of solving
a be an
+ r
h ana\".
H
f(x).
\342\226
can rephrase
we
definition,
r2
that
zero of
+ axa
a0
- 6=
0 by
the classical problem of finding Q and E = E and finding
F =
letting
all alia
real e E
that
all zeros
is, finding
It
seem
may
that we have
mappings, and we can now to develop for will continue
We continue concerned
0} =
=
[r e
E | r2
merely succeededin
use
done is to
all the
+
r
- 6
the problem
mapping machinery
that
answer, since
same
= 0} = {2,-3}.
a simple
making
phrase
the
we
problem seem quite in
have
the
language
developed
of
and
solution.
its
Goal
Basic
Our
6)
what we have
In fact,
complicated.
= 0,
6 in E. Both problems have of x1 + x \342\200\224
E |$a{x2+ x -
{a e
- 6)
+x
4>a{x2
and
= 0,
Ofon
shows that
element of E. Let f{x) = a0 + E be the evaluation homomorphism F[x], and let 0a : F[x] ->\342\200\242 f(a) denote
of a field E,
Let
22.4.
If f(a)
that
if a,- =
only
polynomial.
a zero
4>a{f{x))=
such
if and
an7tn.
\342\226\240 \342\226\240 \342\226\240 be in + a\342\200\236x\"
+
a\\x
+
\342\226
be a subfield
F
aXTZ
4>M is
refer to finding
Let
\342\226\240 \342\226\240 \342\226\240 = 0 + a\342\200\2367t\"
+
Approach
complete
solving
22.10Definition
= a0
h anx\")
H
+ a\\Tt +
a0
a polynomial
shall
-\\
homomorphism
(j>\342\200\236(x)n.
New
The
evaluation
=
with
way
the
is a one-to-one map. This {0}, and \302\242^ with rational coefficients form a ring isomorphic
it
in
and consider
22.4
Theorem
in
Here
-> E.
to
homomorphisms
to put
attempt
with topics
in
ring
for group
in perspective. Sections 26 and 27 are on factor groups analogous to the material theory. However, our aim in developing these analogous
our
theory
future
work
that are
from our aims in group theory. be quite different In group the concepts of factor groups and homomorphisms to study the structure of a given group and to determine the types of group structures of certain orders that about homomorphisms and factor could exist. We will be talking rings in Section 26
conceptsfor theory
we
used
rings
will
Section22 with an eye to fundamental
which is one of the Let us take a moment to talk about the language of \"solving polynomial
of polynomials,
zeros
finding
in algebra.
problems
mathematical
using
history,
205
of Polynomials
Rings
and most
oldest
this aim
the
in
equations\"
light of to which
we are accustomed. start with the Pythagorean We school of mathematics 525 B.C. The of about that all distances are commensurable: that Pythagoreans worked with the assumption a and b, there should exist a unit of distance u and integers ;; and m is, given distances and b = (m)(u). In terms such that a = {n){u) of numbers, then, thinking of a as being one unit of distance, they maintained that all numbers are integers. This idea of comcan be rephrased accordingto our ideas as an assertion that all numbers mensurability are rational, for if a and b are rational numbers, then each is an integral multiple of the For example, if a = reciprocal of the least common multiple of their denominators. -^ = (35)(^) and b = (76)(^). and b = |f, then a The Pythagoreans knew, the Pythagorean of course, what is now called theorem; of length that b and a hypotenuse c, is, for a right triangle with legs of lengths a and
b2 =
a2 +
c2.
had to grant the existence of a right triangle two also of a hypotenuse having of such a right triangle would, say one unit each. The hypotenuse legs of equal length, consternation and as we know, have to have a length of V2. Imagine then their one of their society\342\200\224according himself\342\200\224 to some stories it was Pythagoras dismay when in our terminology in the following is stated came up with the embarrassing fact that They
theorem.
22.11 Theorem
The polynomial
\342\200\224 2 has
x2
no zeros
numbers.
rational
the
in
Thus y/2 is
a rational
not
number.
Proof
Suppose that
m/n
terms
lowest
e Z is a rational
for m, n
we have canceled
that
factors
any
with gcd(m, n)
both m2
2n2
and
factor of 2n2, m2 has as factorsthe 2n2 must have two as a factor. We have 2 is a
any m, n e Thus
equation, x2
delightful
the
fraction
same
integer,
is
and
since
be
one
2n2
of the
are the
factors of m2.
But
as a square.
repeated twice. Thus m2 must have two factors 2. Then must have 2 as a factor, and consequently ;; has 2 from m2 = 2n2 that both m and n must be divisible by 2, fraction terms. Thus we have 2 ^ (m/n)2 m/n is in lowest
the
Z.
\342\231\246
= 0. We
account
in
\342\200\224 2n ,
Since m2 and must
assume
m/n
2, so n2
deduced
the Pythagoreans
- 2
so that
= 2. We
of m
factors
that
that (m/n)2
= 1. Then
are integers. see that 2
factors
the fact
contradicting for
we
such
to m and n,
common
m
where
number
refer
of this
ran
right
the student Pythagorean
into
the question
to Shanks dilemma
of a solution
[36, Chapter and its
of a 3],
polynomial
for a lively
significance
in
and
mathematics.
totally
Part
206
IV
of polynomial
solution
Babylonians
Fields
Note
\342\226\240 Historical
The goalof
Rings and
mathematics developed
equations
has been a
for nearly 4000 years. The versions of the quadratic
quadratic equations. For example, to solve x2 \342\200\224 x = 870, the Babylonian scribe to take half of 1 (|), square it instructed his students and add that to 870. The squareroot of 870|, (|), 30 is then added to to as the namely 291, give | was the scribes did not discuss, answer. What however, what to do if the square root in this process was not a rational number. Chinese mathematicians, however, from about 200 B.C., discovered a method similar to what is now called Horner's method to solve since they used numerically; equations quadratic a decimal system, they were able in principle to formula
to solve
carry
out
the
rational
distinction between
Chinese, in techniques
In the
to
to as
computation
and could
necessary
places
many
thereforeignore
as
the
and irrational solutions. The their numerical
extended
fact,
equations of higher degree. poet-mathematician methods (1048-1131) developed
polynomial
Arab
world,
the
Persian
Omar Khayyam for solving cubic equations geometrically by of intersection of appropriately finding the point(s) while Sharaf al-Din al-Tusi(died chosen conic sections, 1213)used, in effect, techniques of calculus to or not a cubic equation had a real determine whether positive root. It was the Italian Girolamo Cardano (1501-1576)who first published a procedure for solving cubic equations
algebraically.
In our of a group, we commented on the of motivation of the definition necessity such as x + 2 = 0 might have solutions. having negative numbers, so that equations caused a certain amount of consternation in some The introduction of negative numbers circles. We can visualize 1 apple,2 apples,and even j| apples, but how can philosophical and say that it is \342\200\22417 consideration of the equation we point to anything Finally, apples? of the number i. The very name of an \"imaginary x2 + 1 = 0 led to the introduction number\" this number was regarded. Even today, students many given to i shows how are led by this name to regard i with some degree of suspicion. The negative numbers were introduced to us at such an early stage in our mathematical that we development
accepted them
without
question.
in high school freshman algebra. The first there problem in both freshman was to learn how to add, multiply, and factor polynomials.Then, algebra in the second course in algebra and in high school, considerable emphasiswas placed on solving polynomial equations. These topics are exactly those with which we shall be concerned. The difference is that while in high school, only polynomials with real
We first
met polynomials
number coefficients were considered,we
shall be doing our work with for polynomials coefficients from any field. Once we have developed the machinery of homomorphisms and factor rings in Section with our basic goal: to show that given any polynomial of 26, we will proceed of the polynomial may be from degree > 1, where the coefficients any field, we can find a zero of this polynomial in some field containing the given field. After the machinery is developed in Sections 26 and 27, the achievement of this goal will be very easy, and is really a very elegant piece of mathematics. All this fuss may seem ridiculous, but just think back in history. This is the than 2000 years of mathematical in working with polynomial culmination of more endeavor the After achieving our basic goal, we shall spend the rest of our time studying equations.
of these
nature
this material.
solutions of polynomial equations. We need have no fear in approaching We shall be dealingwith familiar topics of high school algebra. Tins work-
seem much In conclusion,
should
is not
see Artin
than
group theory. the machinery of factor rings
that
in order for us to achieve our
p. 29].
[27,
However, factor rings
we should
that
ideas
more natural we remark
necessary
really
207
Exercises
22
Section
grasp,
goal.
and ring homomorphisms For a direct demonstration,
are fundamental ring homomorphisms once we have goal will follow very easily and
basic
our
and
basic
mastered them.
22
\342\226\240 EXERCISES
Computations 1 through
Exercises
In
4,
sum and
the
find
the
of the
product
5. How
6.
How
7.
9. &[ix4
+ 5x\"
14. x5
+ 3x3 + x2 +
ipa
03 ix231
22.4. Compute
for
in Theorem
Use
Fermat's
2x
fix)
2)iAx2
+ 3)(x7
+ 3x2
homomorphism.
+ 1)]
all zeros in the indicated to try all candidates!]
finite field of
the
polynomial
given
with coefficients
in
+ 2x +
2 in
Z7
in Z5
= x3
+3x111 -2x53 theorem
evaluation
indicated
the
is simply
+ 2x2 + 5 and
gix)
= 3x2
+ 2x
an evaluation homomorphism as
\342\200\224> Z5 be
: Zs[x]
+
+ 2)
-x2+3x
theorem.]
13. x3
where
evaluation homomorphism.
the indicated
22.4. Compute for
Z2
Fermat's
(Include 0.)
Z5 [x]?
10. 05[(x3 [Hint:
15, find way
+ 1 in
15. fix)gix)
ring.
polynomial
(Include 0.)
in Z2M?
+ 3)]
3x2
+ 2x53)
12. x2
Use
in Theorem
= E = Z7
11, F
+ 2x)ixi -
In Exercises 12 through that field. [Hint: One
17.
3
8. ^i2x3
9 through
11. 04(3x106
given
Z5[x].
+ 3)
faix2
Let
of degree <
there of degree< 2 in
= \302\243 = C
8, F
7 and
In Exercises
16.
are
polynomials
many
Exercises
In
are there
polynomials
many
the
in
g{x) = 2x2 -
1. f(x) = Ax Ax + 2 in Z8[x}. = = x + 1 in Z2[x]. 2. f(x) x + 1,g{x) 3. f{x) = 2x2 + 3x + 4, g(x) = 3x2 + 2x + 3 in Z6[x]. 4. f{x) = 2x3 + Ax2 + 3x + 2, gix) = 3x4 + 2x + A in - 5,
given polynomials
in
in
Z7
Theorem
22.4.
Use Fermat's theorem
to
evaluate
+ l).
to
find
all zeros
in Z5 of 2x219 +
3x1A
+ 2x51
+ 3x44.
Concepts In
Exercises
needed, so that
18 and it is
19, correct the in a form
definition
of the
acceptablefor publication.
italicized term
without
reference
to the
text, if correction is
18. A
Fields
Rings and
IV
Part
208
with coefficients
polynomial
R is an
a ring
in
sum
formal
infinite
oo
=
y]atx'
+ a2X2
+ o\\x
flo
+ a\342\200\236xn + \342\226\240\342\226\240\342\226\240
H
i=0
where 19.
Let F is the
=
/? for 2 \342\202\254
a,-
be a field and let f(x) e f[i]. Aze/wo//(x)isana x into a. evaluation homomorphism mapping
20. Considerthe
y) =
evaluation
the
(3x3 +
f(x, y) as it
Write
(QM)[;y]-
21. Consider
e F such
that
= 0, where
0a(/(x))
0a
: F(;c)
-\302\273 F
element
f{x, of
\342\200\242 \342\200\242 \342\200\242.
0, 1,2,
would
if viewed
appear
05 : Q[x]
homomorphism
l);y2 + (x4 -
(x2 -6x+
+
2x)y3
as an
2x)y
+
- 3x2 +
(x4
2)
of (Q[j])M.
element
six elements
\342\200\224>\342\226\240 R. Find
in
the
kernel
=
0, for
of the
05homomorphism
22.
of degree>0 in
Find a polynomial
23. Mark each of the
false.
true or
following
a. Thepolynomial
that is a unit.
Iii[x]
+
(anxn
1-
+
a\\x
flo)
\342\202\254 R[x]
is 0
if
and
only if
a,-
i
=
0, 1,
\342\200\242 \342\200\242 n. \342\200\242,
R[x] is commutative. ring, then c. If D is an integral domain, D[x] is an integral domain. d. If R is a ring containing divisors of 0, then has divisors of 0. R[x] e. If R is a ring and f(x) and g(x) in R[x] are of degrees3 and 4, respectively, then f(x)g(x) may be of degree 8 in R[x]. f. If Ris any ring and f(x) and g(x) in R[x] are of degrees 3 and 4, respectively, then f(x)g(x) is b. If R
is a commutative
then
7.
of degree
always
g. If F is a subfield all g(jc) e F[x].
E and
a
\342\202\254 \302\243 is a
/(x) e F[;c],then
zero of
a is a
zero of
h(x) =
f(x)g(x)for
h. If F is a field, i. If
R
j.
R is a
If
is a
in F. then the units in F[x] are precisely the units R divisor 0 in then x is never a of [x ]. ring, ring, then the zero divisors in R[x] are preciselythe zero
divisors
in R.
Theory
24. Prove
that
25. Let
D be an
if D
is
an
integral
b. Find
the units the units in
c.
the
integral
domain,
domain
and x
a. Describe
Find
units
26.
Prove the
27.
Let F be a field
left
in
ShowthatD
an
an
integral
domain.
indeterminate.
in D[x].
Z[x]. Z7
[x].
distributive
law for
of characteristic D(oq
a.
then D[x] is
: F[x]
+
a.\\x
R is a ring and x is an indeterminate. R[x], where D be the formal polynomial differentiation zero and let
+ a2X2
\342\200\224>\342\226\240 F[x]
isagroup
+
\342\200\242 \342\200\242 \342\200\242
+ anxn)
=
homomorphism
a\\
+ 2
map, so that
\342\226\240 \342\200\242\342\200\242\342\200\242 + +n \342\200\242 \302\2532* a\342\200\236x\"~l.
of {F[x], +) into
itself.
Is Da ring
homomorphism?
Section 23
28.
b. Find
the
kernel
of D.
c.
the
image
of F[x]
Find
Let F be a
a. Define an
D.
under
field E.
of a
subfield
evaluation
homomorphism ->
: F[xu ---,\302\276] 0\302\253,,...,\302\253\342\200\236 stating
b. With c.
E
the analog of Theorem 22.4. = F = Q, compute \302\242-3,2(^1\302\2763+
the concept in the text definition Define
29. Let
R be a
209
Field
over a
of Polynomials
Factorization
ring, and let
of a zero of zero of
of a RR
be the
E
for
\342\202\254 \302\243,
a,-
3^14\302\276).
a polynomial
f
\342\226\240 \342\226\240 \302\243 \342\226\240, xn) F[x\\,
{x\\,
\342\226\240 \342\226\240 in \342\226\240, xn]
a way
analogous
to
the
f(x).
set of all
functions
mapping
R into
= 0(r)
+ ^(r)
R. For
\302\242,^/e
RR, define
the sum \302\242+11/
by
(0+^)(r) the
and
\342\200\242 \342\226\240f by
0
product
(0
for r efi. 30.
f(x) e F[x]such a.
that
the set
that
Show
b. Show that
F[x] don't
31. Refer
function composition.
29, let F
to Exercise
Referring
a. How
0(a)
PF is not necessarily isomorphicto F[x]. ring even have the same number of elements.]
29 and
30 for the
are there in
(Z2Zl, +} and
(Z3 Zl,
following
Z2Zi? in
is a \342\200\242)
ring.
is a polynomial
on F forms a
functions
polynomial
(RR, +,
of FF \302\242)
on
function
exists
F, if there
F.
a e
all
that
Show
An element
field.
= f(a) for
Pf of all
elements
many
be a
= 0(rW)
the
to Exercises
b. Qassify
i;){r)
\342\200\242 is not
that
Note
\342\200\242
of
subring
[Hint:
FF.
Show
that
is a
if F
finite
field,
PF and
questions. Z^z,l
+} by Theorem
11.12,the
Fundamental
Theorem
of
finitely
abelian
generated
groups.
c.
Show
Note
if F is a finite that if . that
field,
then FF
f,(x) = then
fi(cij)
that every
SECTION
23
= 0
for
j on
function
with F g(x),h(x)
that
f(a)
concerned
Suppose
that
\342\202\254 F[x]
PF c
can be controlledby
FF. LetF have
a \\,
as elements
\342\226\240 \342\226\240 \342\226\240, a\342\200\236.
and let
with
f{x)
a e
=
zeros
factors
for
(pa(g(x)h(x))=
of c
e F. Usethis
(pa(g(x))(t>a{h(x))
=
F
to show
of polynomials.
be fields,
g{x)h(x) for
we
have
g(a)h(a).
either g(a) = 0 or h(a) = 0.The of finding a zero of a factor of
Thus if a e E, then f(a) and only if find a zero of f(x) is reducedto the problem is one reason why it is useful to study factorization = 0 if
choice
of polynomials. Let E and so that f(x) = in F[x], the evaluation homomorphism
finding
e F[x]
E. Now
the
over a Field
of Polynomials
we are
< E.
Of course,
- at-i)(x - a\342\200\236), \342\226\240 \342\226\240 \342\226\240 \342\226\240 c(x - a{)\342\226\240 (x (x ai+l) \342\226\240
j^ i, and the value ft (at) F is a polynomial function.]
Factorization Recall
= PF. [Hint:
attempt f(x).
to This
Part IV
Fields
Rings and
The Division Algorithm in F[x] The
the division
been amply 23.1
Theorem
basic tool for our
is the
theorem
following
with
Z given in
for
algorithm
section. Note
in this
work
Theorem 6.3,
the
the
similarity
has
of which
importance
demonstrated.
for F[x]) Let
(Division Algorithm
f{x) = anxn
+
H \302\253o
H a\342\200\236-i*\"_1
and
= bmxm +
g(x)
elements of F[x], with there are unique polynomials
be two
either
where
Proof
Considerthe such
that
we are
= 0 or
r(x) set
-
f{x) done.
the
q(x)
and of
some
be an
let r{x)
Otherwise,
We must
e F[x].
q(x)
F and
e
Cj
0. If
c, ^
t >
of minimal
element
is of
latter
the
that the
c,x' + c,-xx'~lH
is a
(1) can be written
(c,x'
so
Thus
in S,
the
in the
the
contradicting
that
Suppose
co,
(1)
of lower degree), degree of r (x). However,the
t, the
polynomial
form
-
g{x)[q{x)+ (c,/bm)x'-m], that r(x) was selected to m of g(x). the degree
fact
is less than
of r(x)
degree
1-
m.
= r(x)-(ct/bm)x'-mg(x),
+ terms
of degree lowerthan
polynomial
fix) it is
S. Then
m, then
\342\200\224
r(x)
in Eq.
r(x) = 0,
form
the
which
in
an s(x)
exists
s{x) and
+ r(x) degree of r(x) is lessthan
f(x)-q(x)g(x)-(ct/bm)x'-mg(x) and
degree
there
= g(x)q(x)
show
r(x) = with
elements of F and m > 0. Then r(x) in F[x] such that f(x) = g{x)q{x)+ r{x), r(x) is less than the degree m of g(x).
S = {f(x) - g(x)s(x)\\ s(x) e F[x]}. If 0 e S then = q{x) g(x)s(x) = 0, so f(x) = g(x)s(x). Taking
f(x) for
b0
+
both nonzero
and bm a\342\200\236
degree
\342\226\240 \342\226\240 \342\226\240
+
bm~\\xm~l
have minimal
in S.
degree
For uniqueness,if f{x)
= g(x)ql(x)
+ rl(x)
fix)
= g(x)q2(x)
+ r2(x),
and
then
we have
subtracting
- q2(x)]=
g(x)[qi(x) Because either r2(x)
of g{x),this r2(x)
We
hold
ri(x)
= 0
can
compute
\342\200\224
just as
can
so
\342\200\224
rx(x)
= 0
or
only if q\\{x) r\\{x)
=
the
= 0
so
q\\{x)
r^x). \342\200\224
r\\{x)
=
is less than
q2(x).
Then we
the degree must
have
\342\2
r2(x).
the polynomials
we divided polynomials
q2(x)
-
of r2(x)
degree
\342\200\224
r2(x)
in M.[x]
q(x) and in high
r(x)
of Theorem
school.
23.1
by
long
division
Section23
23.2Example
us work with
Let
polynomials
in
by
= x2
g{x)
\342\200\224
+ 3
2x
be easy to follow,
but
to
find
q{x)
-
3x3
- 1 23.1. The long
of Theorem
are in
so, for
Zs[x],
example,
should
division
Ax
\342\200\224 = (\342\200\2243.v> 2x.
\342\200\224
x
x
+ 3 ;r,-4 -
x-2x
2x3 + Ax
+
andr(x)
that we
remember
211
Field
and divide
Z5 [x]
f(x) =x4
over a
of Polynomials
Factorization
x4
3x3
+
2x2
\342\200\224
+ 3x~
- x3
-
2x3
\342\200\224
x3
x2
+ Ax
- 1
Ax
+
+ 2x~
\342\200\224
- 3x2
+ 2x -- 1 x - -4 +
-
3x2
3x
x +
3
Thus = x2
q{x)
We
three
give
23.3 Corollary
Proof
(homomorphism)
for a
that
Suppose
\342\202\254 F[x]
3,
for
&
have
we
F
f{a) =
first
0. By
one
our proof
e F[x]
if
Theorem
and
23.1,
appears in high in terms of the
if x
only
there
exist
\342\200\224 a is
q(x),
= 0 or
r(x)
the
degree
-a)q(x) + r(x), is less
of r(x)
than
we must have
1. Thus
r(x) = c
F, so
/(-*) = our
Applying
evaluation
must
c=
be that
Conversely, if
x
(-^
homomorphism, 0 =
so it
+ 3.
that
such
either c e
r(x) = x
F is a zero of f(x)
a e
element
f(x) = (x where
and
corollaries of Theorem23.1.The case F[x] = Rpc]. We phrase special 22. approach described in Section
the
(Factor Theorem) An a factor of f(x) in F[x]. r(x)
x
important
school algebrafor mapping
\342\200\224 \342\200\224
0. Then \342\200\224 a is a
evaluationhomomorpohism4>a
-
4>a
/(a)
a)q(x) :
F[x]
= 0q(a)
+ c. Theorem
\342\200\224>\342\226\240 F of
22.4, we find
+ c,
= (x \342\200\224 so x \342\200\224 a is a factor of a)q(x), /(x) factor of f{x) in F[x], where a e F, then
to f{x)
= (x \342\200\224 a)q(x),v/ehave
f
f{x). applying
(a) = 0q(a)
= 0.
our \342\231\246
212
Part
IV
23.4Example
Rings and
Fields
in Zs[x], note
again
Working
1 is a
that
(x4 + Thus
by
in Zs[x].
23.3,
Corollary
Let us
the
find
3x3
factorization
eZ5W.
+2*+4)
be ableto
we should
x ~l
zero of
by long
x3 +
Ax2
+
x4 +
3x3
+
3x3 +
x4 +
factor
2x + 4
into
-
(x
(x)
Y)q
division. + 1
Ax
2x +
4
4x3
4x3 --4x2
Thus x4 + 3x3 a zero of x3 +
+
2x +
4x2 + 4x
4 = (x
- 1)03+ 4x2
+
4x2
- 4x
4x +
+
we can divide
1 also,
+
2x
4x2
this
x+
4
x
1
-
1) in
Since
Z5[x].
polynomial
by x
1 is
seen
\342\200\224 1 and
to
be
get
r +4 x~
1
x3
+
+ 1
+ 4x
4x2
+ 4x + 1 -4
4x
0
Since x2 +
4 still
has
1 as a
zero, we
can divide
by x
again
\342\200\224 1 and
get
x + 1
+4 x+ x
-
4
1
0 Thus x4 + The
23.5 Corollary
A
3x3
+
2x +
next corollary
nonzero
polynomial
4 = (x also
should
f(x)
- l)3(x+ 1)in look
e F[x]
Z5[x].
familiar.
of degreen
can
have
at most
n
zeros
in a field
F.
Section 23
Proof
The
shows that if
corollary
preceding
= ix
f(x)
where, of course,the
is
of qiix)
degree
F is a
e
a\\
fix),
then
-ai)qiix), \342\200\224 1. A
n
zero of
213
Field
over a
of Polynomials
Factorization
zero a2
&
then results
of q\\(x)
F
a
in
factorization
we arrive
this process,
Continuing
where qr(x) has no further \342\200\224 ix a,-) can appear on the
b ^ a; for
since F the
i =
1,
final
that
surprising
the result inZ.
If G is a
that
the
of the
ar)qrix), is n,
of fix)
degree
at
preceding equation,
n factors
most
so r
Also, if
< n.
F, then
fib)
= ib-ai)---ib-ar)qrib)^Q,
of 0
and
of b
none
zeros
the
qr (b) are 0 by of fix).
\342\200\224 or
a;
in F
Hence
construction.
\342\231\246
of the multiplicative F* of the structure group than with factorization in F[x]. It may at first seem such a result follows from the division in F[x], but recall that algorithm a subgroup of a cyclic group is cyclic follows from the division algorithm
finite
the
F, rather
field
of the
subgroup
In particular,
ix
is concerned with
corollary
nonzero elements of a
Proof
&
\342\200\242 r < n are all \342\200\242, a,- for i = 1, \342\200\242
Our
23.6 Corollary
b
-
\342\200\242 \342\200\242 \342\200\242
Since side
right-hand
\342\200\242 \342\200\242 r and \342\200\242,
divisors
no
has
- ai)
in F.
zeros
a2)q2ix).
at
= ix
fix)
- ai)ix -
= ix
fix)
group
multiplicative
group
multiplicative
{F*,
of all nonzero
field F,
of a \342\200\242)
elements of a
G is
then
finite
cyclic. is cyclic.
field
By Theorem11.12as
a finite abelian group, G is isomorphic to a directproduct Z,*, x where each is a of a Letus think each of the of dt power prime. Zdr, Z^. as a notation. Let m be the least common cyclic group of order dt in multiplicative multiple ' = so \342\200\242 \342\226\240 of all the d; for i = 1, 2, \342\200\242 If a,- e Z^, then a; -dr. 1, \342\200\242, r; note that m < d\\d2a.m = 1 since d; divides m. Thus for all a e G, we have am = 1, so every elementof
Zd2 x
G is
\342\200\242 \342\200\242 \342\200\242 x
zero of
zeros
the
in
cyclic
5 through
Exercises
the
Our next importance school
\342\226\240 \342\226\240
dr
elements,
23.5, so m
powers group Zm. prime
> d\\d2
d\\,d2,--,dr
8 ask us to find all generators that the multiplicative group
some finite fields. The fact has been appliedin algebraic Irreducible
d\\d2
F by Corollary
the primes involved
isomorphic to
G has
1.
field
the
in
\342\200\224 But
xm
while xm
\342\200\224 1 can
m =
\342\226\240 \342\226\240 \342\226\240 Hence
dr.
are distinct,
and
have at most m \342\226\240 \342\226\240 \342\226\240 so
dr,
d\\d2
the group
G is \342\231\246
of
the
groups
cyclic
of units
of a finite
of units for field
is cyclic
coding.
Polynomials definition
out a type of polynomial in singles The concept is probably already familiar. in a more general setting.
to us. algebra
that
F[x] We
will be of utmost are doing high
really
Part
214
IV
23.7Definition
Rings and
Fields
A nonconstant
polynomial f(x) F[x] if f(x) cannot
polynomial
in
g(x) and is
polynomial
the
F, then
over
irreducible
not
reducible
f{x) is
\342\226
that the preceding Note the concept irreducible.
be irreducible
23.8 Example
lower degreethan
that is
is an irreducible of two polynomials g{x)h{x) of degree f(x). If f(x) e F[x] F or
F.
over
just
both of
in F[x]
h(x)
a nonconstant
e F[x] isirreducibleover be expressedas a product
if
the concept irreducible over F and may be irreducible over F, but may F. We illustrate this. containing
not
not
f(x)
polynomial
a larger
over
viewed
concerns
definition A
field E
2 viewed in Q[x] has no zeros in Q. This shows that that x2 \342\200\224 2 = (ax + b){cx + d) for a,b,c, over Q, for a factorization x2 \342\200\224 \342\200\224 \342\200\224 2 viewed in R[x] is not to 2 in would rise zeros of x2 deQ Q. However, x2 give \342\226 2 factors in R[x] into (x - */2)(x + V2). irreducible over R, because x2 \342\200\224
shows
Theorem
22.11
x2
irreducible
\342\200\224 2 is
It is worthwhile to remember that we F. Thus could have denned of
such that in
polynomial
the
units
factorization
any
in F
an irreducible
=
f{x)
[x] arepreciselythe nonzero elements polynomial f{x) as a nonconstant g(x)h(x) in F[x], either g(x) or h(x)
is a unit. 23.9
Example
Let us show
that
f(x)
= x3
2 factored in Zs[x] least one linear factorof be 0, by Corollary 23.3. 3x +
+ 3x
into f(x)
+ 2 viewed
in
Zs[x]
over Z5. If
is irreducible
of lower degree then there polynomials a for some a e Z5. But of the form x \342\200\224
then
x3
+
exist at
would f{a)
would
= -2, /(2) = 1, and /(0) = 2, /(1) = 1, = \342\200\2242, has no zeros in Z5. Thus f(x) is irreducible over that f(x) /(-2) showing and cubic Z5. This test for irreducibility by finding zeros works nicely for quadratic \342\226 over a finite field with a small number of elements. polynomials
Irreducible
The problem difficult.
23.10 Theorem
We
now
give for
Let f(x) e F[x],and zero in
23.8 and
let
f(x)
of
irreducibility
determining
in Examples
only if it has a Proof
will play a very important role in our work from now on. irreducibleover F may be whether a given f(x)&F[x]is that are useful in certain cases. some criteria for irreducibility
polynomials of determining
One technique illustrated
/(-1)
However,
23.9. We
be of
formalize
degree 2 or
it in a
3. Then
cubic
and
quadratic
was
polynomials
theorem.
f(x)
is reducible over F if
and
F.
is reducible so that f(x) = g(x)h(x), where the degree of g(x) and the degree of or cubic, h(x) are both less than the degree of f{x), then since f(x) is either quadratic either g(x) or h(x) is of degree 1.If, say, g(x) is of degree 1, then except for a possible factor in F, g(x) is of the form x \342\200\224 that f{a) = 0, so a. Then g{a) = 0, which implies f(x) has a zero in F. If f(x)
Conversely, Corollary23.3shows of f(x), so f(x) is reducible.
that
if f(a)
We turn to some conditions for irreducibility most important condition that we shall give is not this theorem here; it involves clearing prove
= 0 for a
e F, then
a is a x \342\200\224
factor
\342\231
over
Q of
contained in denominators
polynomials the
next
in
theorem.
and gets a bit
Q[x].
The
We shall messy.
23.11 Theorem
If f(x) r
e Z[x],then
degreesr Proof 23.12
Corollary
Proof
=
If f(x) Q, then If
x\"
+
zero m
in Z,
a
in
+ a0 is in Z[x] and m must divide
\342\226\240 \342\226\240 \342\226\240
then
Q,
is in
Thus ao/m
Example
with
degrees
of the
polynomials
same
^
a0
{x-
+
m)(xn~l
0, and if
fix) has a zero in
a0\342\200\224
am
Q[x] by Corollary factor in Z[x],
a linear
23.3. so for
ao/m).
divides oq.
so m
Z,
with
has a linear factor x
/(x)
f(x) =
23.13
of lower
polynomials
f(x) has a factorization with
by Theorem 23.11, e Z we must have
somem
two
\342\231\246
+ an-ixn~l
has a zero
then
But
into a product of has such a factorization
is omitted here.
it has a
f(x)
if it
only
215
a Field
s in 7L\\x\\.
and
The proof
factors
f(x)
Q[x] if and
s in
and
of Polynomialsover
Factorization
23
Section
\342\231\246
2 2 over Q, for x2 of the irreducibility of x2 Corollary 23.12 gives us another proof in Q[x] if and only if it has a zero in Q by Theorem 23.10. By factors nontrivially Corollary23.12,it has a zero in Q if and only if it has a zero in Z, and moreover the only are the divisors \302\261 none of these numbers 1 and \302\2612 of 2. A check shows that possibilities
\342\200\224
\342\200\224
2. is a zero of x2 \342\200\224
23.14
Example
\342\226\262
use Theorem23.11to
Let us
that
show
+ 8x +
=x4-2x2
f{x)
l
has a linear factor in Q[x], then it has a Q[x] is irreducible over Q. If fix) and have to a in be divisor Z of 1, that is, Z, by Corollary 23.12, this zero would = 8, and /(-1) = \342\200\2248, But /(1) \302\2611. so such a factorization is impossible. in
viewed
zero in either
If fix)
into
factors
two
quadratic
factors
in
Q[x],
by Theorem
then
23.11, it
a
has
factorization.
ix2 + in Z[x].
of powers
coefficients
Equating
ax + b)ix2
ad + be = 8,
bd=l,
+
ex +d)
of x, we find
ac+
b
+ d
=
that
\342\200\224
2,
have
we must and
a +
c= 0
e Z. From either bd = 1, we see that b = d = lor b = d = = d and b we deduce from ad + be that case, 8, + c) = 8. But dia = since a c a into two 0.Thus factorization + impossible quadratic polynomials impossible and fix) is irreducible over Q. for
a, b,c,d
integers
In any
We
conclude An
irreducibility.
23.15 Theorem
our irreducibility criteria with the famous Eisenstein is useful criterion in Exercise 37. given very
Then fix)
is
an
\342\200\224
this
1.
is
is also \342\226\262
criterion for
additional
(Eisenstein Criterion) in Z[x], and
=
^k 0 (mod
irreducible
Let
p), over
p
e Z but
Q.
be a prime. Supposethat
at = 0
fix)
(mod p) for all i < n,
= anxn with
ao
+
# 0
\342\200\242 \342\200\242 \342\200\242
+ a$
is
(mod p2).
Part
216
IV
Proof
Fields
and
Rings
By Theorem23.11we need lower degreein Z[x]. If =
f{x) is a factorization co = 0 (mod Let m be the
in
are
bo and cq
that
p).
Now
smallest
(brxr +
into
factor
polynomials
of
\342\226\240 \342\226\240 \342\226\240 \342\226\240\342\226\240\342\226\240
+
+
b0)(csxs
+
c0)
cs ^0andr,s < \302\253,thenao # 0 (mod p2) implies modulo p. Supposethat \302\243>o congruent # 0 (mod p) and = brcs. 0 that br,cs a\342\200\236 (mod p) implies ^0 (mod p), sincea\342\200\236 ^ value of k such that q ^ 0 (mod /?).Then ^ 0,
withfor
Z[x], not
does not
show that f(x)
only
to 0
both
r >
m, brcm-r if r < m.
fcmCo if
\342\200\242 \342\200\242 \342\200\242
+
+ b\\Cm-\\
+
\342\200\242are all \342\200\242, cm_i, \342\200\242 co p while congruent to 0 modulo s = n, congruent to 0 modulo p implies that am ^ 0 modulo p,som =n. Consequently, was nontrivial. contradicting our assumption that s < n; that is, that our factorization
that neither
fact
The
cm are
nor
fo0
\342\231\
23.16 Example
2 over of x2 \342\200\224
irreducibility
p
Taking
p = 2, the
if we take
that
Note
the
= 3,
we see by
gives us
criterion
Eisenstein
still
another
proof
of
Q. 23.15
Theorem
-
25x5
that 9x4
-
12
3x2
is irreducible over ( 23.17
Corollary
The polynomial
1
Proof
over
irreducible
by
Again
23.11,
Theorem
22.5
can be f{x)
used for
commutative
: Q[x]
0X+1
-+ Q[x].
It
+XP~2 + .
+x+ I
factorizations
consider
only
proof
in Z[x].
We remarked
evaluation homomorphims use the evaluation homomor-
that
shows
actually
Here we want to rings. is natural for us to denote
by f(x
+ 1) for
6 QM.Let
g{x) = The
we need that its
= xp~l
p
prime
any
Theorem
following
phism
Q for
x-l
coefficient
is divisible 0 < r < p.
by
+ 1) = $\342\200\236(x
of xp~r p
because
(x +
l)p - 1
xp
+
+
(x + 1)- 1 for 0 < r < p is the binomial p divides
pi
but
does
coefficient p\\/[rl(p not divide either r! or
Thus
g{x)
= xP-1
+
+ (PA Xp-2 +
\342\200\242 \342\200\242 \342\200\242
+ p
PX
\342\200\224which
r)!]
(p \342\200\224 r)! when
satisifies the =
$>p(x)
Factorization of Polynomials
23
Section
for the prime criterion were a nontrivial factorization
Eisenstein
h{x)r{x)
+ 1)
$>p(x
is thus
and
of
&p(x)
ever Z.
irreducible in Z[.r]. then
Br: r
= h{x + l)r(x+ 1)
factorization of g(x) inZ[x].Thus
give a nontrivial
would
= g(x)
p
217
a Field
over
must
$p(x)
also be
irreiuc:rle
over Q.
\342\231\246
The
in
Polynomials
can be
F[x]
factored into
in an
polynomial.
cyclotomic
in F[x]
of Factorization
Uniqueness
23.17 is thepth
0>p(x) in Corollary
polynomial
of irreducible
a product
g(x) e F[x] we say
polynomials in
f(x), g{x) divides essentially unique way. = the such that Note if there exists e F[x] g(x)q(x). q(x) F[x] f(x) similarity theorem that follows with boxed Property (1) for Z following 6.9. Example
23.18 Theorem
For
Let p{x) be an
F[x], Proof 23.19
Corollary
Proof 23.20
Theorem
We
irreducible
either
then
the proof
delay
of
Using mathematical induction, If F
is a field,
into
a product
then
nonconstant
every
of irreducible for
unit
and
g{x)
into
h(x) are
both
polynomials
polynomials, nonzero constant) polynomial.
and the
degree.
It remains
for us
for to
show
fix) = are two
Piix)
factorizations of
divides
some
qj(x),
degree ofh(x)
we stop
i =
1, 2,
u\\ ^
0, but
u\\
is in
not
not,
process,
at least one we arrive at
being
then
\342\231\246
in F[x]
unique
f(x)
=
degree of f(x). of them factors a factorization
irreducible
into
thus
is a
qi(x)q2ix)
\342\226\240 \342\226\240 \342\226\240
Then by Corollary is irreducible, q\\{x)
Since
Prix)
23.19,
uiPiix),
unit. Then
substituting
canceling, we get \342\226\240 \342\226\240 \342\226\240
qs(x)
polynomials.
assume q\\(x).
F and
that
Suppose
uniqueness.
let us
Piix)
than the
F[x],
23.18.
factored
irreducible,
less
e
\342\200\242 \342\200\242 r. \342\200\242,
\342\226\240 = \342\226\240 pr{x) pi(x)p2(x) \342\226\240
fix)
s(x) e
= Pl(x)p2(x)---pr(x).
qiix) = where
here. If this
Continuing
is both
be
polynomials in F.
factors
If f(x)
Theorem
F[x] can
irreducible
(that is,
f(x)
where pt{x) is irreducible
f(x) e
polynomial
irreducible,
of lower
this is immediate from
in
\342\231\246
\342\200\242 \342\200\242 \342\226\240rn(x)forri(x)
the
Let f(x) e F[x]be a nonconstant g(x)h(x),wi\\h the degree ofg(x) If
find that
we
i
of the
27. (See Theorem27.27.)
to Section
theorem
this
F[x]
fix
for r(x),
r(x)s(x)
Ifp(x)isirreducibleinF[x]andp(x)dividestheproductr1(x) then p(x) divides rt(x) for at least one i.
except for order and Proof
If p(x) divides divides p(x) s(x). in F[x].
polynomial
divides r(x) or
p(x)
that
= u\\qiix)
\342\226\240 \342\226\240 \342\226\240
qs(x).
u\\P\\{x)
for q\\(x)
and
Fields
Rings and
Part IV
218
By a
similar
= u2p2ix),so
say q2(x)
argument,
Pi(x)-- -Pr(x)= uiu2q?,(x)---qs(x). in this
Continuing
manner, we eventually 1 =
possible if s = r, so that irreducible factors pt{x) and qj{x)
\342\226\240 \342\226\240 \342\226\240
\342\200\242 \342\200\242 \342\200\242
uiu2
qs(x).
urqr+\\{x)
this equation is actually were the same except
is only
This
at
arrive
1 =
the unit
\342\200\242 \342\200\242 \342\200\242 Thus
mm
ur.
possiblyfor
and
order
factors. 23.21
\342\231\246
is (x \342\200\224 of x4 + 3x3 + 2x + 4 in Zs[x] + 1). \\)3(x Example 23.4 shows a factorization These irreducible factorsin Zs[x] are only unique up to units in Zs[x], that is, nonzero \342\200\224 \342\226\262 constants in Z5. For example,(x \342\200\224 + 3). + 1) = (x \342\200\224 l)2(2x 2)(3x \\)3(x
Example
23
\342\226\240 EXERCISES
Computations 4,
1 through
Exercises
In with
= 0
r(x)
1. f{x) =x6 + 3x5 2. f(x) =-xb + 3x5 3. f{x) = x5 - 2x4 5
In Exercises (Review
5.
q(x)
the
- 5
+ 3x
+ 1 in
+
the
3
2x -
that
= g{x)q{x)
f(x)
+ r(x)
in Z7[x]. in Z7[x].
3
Z\342\200\236 [x].
- x + 2 in
all generators of
8, find
through
g(x)
g{x)
= 2x
g(x) = 5x2
and
3x2
+ 2 and
and
= 3x2
algorithm so
division
the
= x2 + 2x -
andg(x)
- 3x
+4x2
as describedby degree of g(x).
and r(x)
+ Ax2-3x+2
=x4 + 5x3 -
4. f(x)
find
or of degree less than
Z\342\200\236 [x].
group of
multiplicative
cyclic
of
units
the given
field.
finite
Corollary 6.16.) 6.
Z5
can be factored into
9. The
polynomial
x4
+ 4
10. The
polynomial
x3
+ 2x2
11. Thepolynomial 12. Is x3
+ 2x + 3
+ 2x
+ 1 can
Ix
an irreducible
5 can
polynomial
factors
linear
be factored
\342\200\224 \342\200\224
+ 3x2
2x3
7.
Z7
8.
Z17
in Zs[x].
into linear factors
be factored
of Zs[x]? Why?
in
factors
linear
into
this
Find
factorization.
Z7[x].
in Zn
it as a
Express
Z23
Find this
[x].
Find
factorization.
this
factorization.
product of irreducible
polynomials
ofZsM.
+ x2 + 2x + 2 an
13. Is 2x3 polynomials
14. Show
that
+ 8x
= x2
fix)
15. Repeat Exercise14with 16. Demonstrate
that
x3 +
17. Demonstrate
that
x4
In Exercises
irreducible
in Z5 [x]?
polynomial
Why?
Express
it as a
product of
irreducible
in Ijs[x].
18 through Q.
ducibility over
21,
- 2 is irreducible
over
Q. Is
+ 6x + 12in place is 8 over Q. 3x2 irreducible 22x2 + 1 is irreducible over Q. gix)
= x2
determine
whether the
polynomial
fix)
irreducible
over
E? Over
C?
of fix).
in
Z[x]
satisfies an Eisenstein
criterion
for
irre-
219
Section23 Exercises
18.
-
x2
20. 4x10
19. 8x3 + 6x2 -9x+2A
12 - 9x3
+
-
2Ax
21. 2x10-
18
of 6x4 + 17x3+ lx2 + x \342\200\224 10 in Q. (This is a tedious of analytic or and calculus and make a graph, geometry best candidates for zeros.)
22. Find
all zeros
use a bit
- 30
\\0x2
+
25x3
school
high
algebra problem.
use Newton's
method
reference
to the
You
might
are the
which
see
to
Concepts In
24, correct the
23 and
Exercises
needed, so
A polynomial
23.
g(x),
24. A
f(x) h{x) e F[x].
nonconstant
25.
Mark
f(x)
factors
is in
of the
each
a. x
over
is irreducible
\342\202\254 F[x]
polynomial
one of the
in F[x],
italicized term
of the
definition
following
the
F if
field
e F[x] is irreducible F.
and only the field
over
over
g.
6 is
In
Exercises
of degree
f(x)
polynomial f(x) of degree such
F
that
F
if
and
if in
only
any
polynomials
of it
factorization
any
numbers p
27 through
30,
such
find
n with
that
1 in
elements
of F.
nonzero
elements
of F.
coefficients
in a field F can have
coefficients
in a field F can have
x +
has at
F[x]
can have
in F[x]
polynomial
with
n
nonzero
n zeros
most
at at
most
in
F.
n zeros in
any
given
< E. of degree
polynomial
Every
all prime
Find
^ g(x)h(x) for
if f(x)
irreducible
polynomial
j. Each 26.
is
correction
\342\200\224
field E
i.
if
Q.
over Q. c. x2 \342\200\224 3 is irreducible over Q. d. x2 + 3 is irreducible over Z7. the units of F[x] are precisely the e. If F is a field, f. If F is a field, the units of F[x] are precisely the h. A
text,
or false.
true
- 2 is irreducible
b. 3x
A
without
acceptablefor publication.
in a form
it is
that
most
at
2 is a factor
all irreducible
a
least one zeroin the field F. finite number of zerosin the + x3, +
of x4
polynomials
of
the
x2
\342\200\224 1
x +
indicated
degree
field F.
in Zp[x]. in the
given
ring.
27.
Degree 2
in
Z2[x]
28.
Degree 3
in
Z2[x]
29.
Degree 2
in
Z3 [x]
30. Degree 3
in
Z3[x]
31.
in Zp[x], where Find the number of irreducible p is a prime. [Hint: quadratic polynomials of reducible polynomials of the form x2 + ax + b, then the number of reducible quadratics, from the total number of quadratics.]
Find
the
number
this
and
subtract
1 /a
is a zero
Proof Synopsis
32. Give 33.
a synopsis
Give a synopsis
of the proof
of Corollary
23.5.
of the
of Corollary
23.6.
proof
Theory
34. Show that 35. If F is a an
+ an_\\x
for p a field
+
prime,
and
a ^
\342\200\242 \342\200\242 \342\200\242 +
a0xn.
the
polynomial
0 is a zero of
xp + f(x)
a =
is not
in
Zp[x]
flo
+ a\\x H
irreducible + anxn
for
any
in F[x],
a e show
Zp. that
of
36. (Remainder 37. Let
a.
->
: Z
am
for a
x
\342\200\224 in
a,
that
5^
: Z[x]
->\342\200\242 Zm[x]
ff^(a0 + is a homomorphism b. Show
if fix)
that
the
H
onto
1-
(b)
r(x)
when
by
given
= (the remainder
amia)
of a
divided
when
by m)
=
anxn)
+
ffm(a0)
am(tfi)*
H
1-
^(aj*\"
7Lm\\x\\.
both have
ffm(/(;c))
less than n, to show that x3 + 11x of degree
polynomials
c. Usepart
of ZLr] e Z[x] and
the remainder
that
by
given
flix
field, and let a e F. Show algorithm, is /(a).
division
the
with
homomorphism
e Z.
Show
F is a
where
accordance
the natural
be
Zm
e F[x]
Let f(x)
Theorem)
is divided by
fix)
Fields
Rings and
Part IV
220
then
degree
\302\253 and
+ 36is irreducible
does not
a^ifix))
in Zm[x]
factor
into two
in Q[x].
is irreducible
fix)
[Hint: Try a prime
in Q[x].
of m
value
that simplifies
coefficients.]
SECTION
24
^NoNCOMMUTATrVE EXAMPLES the only example we of all n x n matrices M\342\200\236iF)
Thus
far,
ring
presented
with
rings and
noncommutative
with
nothing
have
important noncommutative of such rings. examples
rings
of a ring that
is
F.
We
entries
in a field
strictly
skew
shall
be doing
To show that in algebra,
fields.
naturally
very
occurring
commutative
not
there
we
give
is the
almost are other several
of Endomorphisms
Rings
into itself is an endomorphism Since the composition of two into itself is such a we define multiplication of A again homomorphism, homomorphisms on End(A) is associative. by function composition, and thus multiplication we have to describe the value of (0 + xjs) on To define addition, for 0, xjs e End(A), each a e A. Define Let
A be
of A.
Let
any abelian group. A homomorphism set of all endomorphisms of
the
of A
A
be
End(A).
(0 + f){a) = 4>ia)
+
f{a).
Since
(0 +
f)ia + b)
=
$ia
+ b)
+ b)
=
ma)
+
+ fia 0(i>)] +
[Vr(fl)
+
Vr(fl)]
+
[00)
= [^a)
= we see
that
Since A
+
\\[r
is again
is commutative,
(0 + for all a
e
A,
so
associativityof addition
t
This section is
follows
not used in
=
+
+
ir)ia)
i4>
+
irib)]
+ irib)]
ir)ib)
in End(A). we have
f)ia) = 4>ia)
+ xjs \302\242)
i
+
xjs
+ 0
+
ir{a)
and addition
from
the remainder
= ir{a)
of
the text.
in
+
4>ia)
End(A)
=
(i/r +
0)(a)
is commutative.
The
+ [\302\242
= =
If
the additive
e is
for a
e
A
is an
additive
identity
[Vr(a)
+ 9(a)]
if (a)]
[\302\242(^+
0(a)
= e
in End(A).
Finally,
0 e defined \342\200\2240
+
0(a)
+ 0(a)
= (0 + i/O(a) + 0(a) = M + is) + e](a). the homomorphism 0 defined
then
of A,
identity
221
+ [(^ + 0)(a)]
= 0(a)
6)](a)
(^+
Noncommutative Examples
24
Section
by
for
End(A),
by
= -0(a)
(-0)(a)
is inEnd(A),
since
+ b)
(-0)(a
= -0(a = -0(a)
+ ft)
-
=
=
0(fo)
+ 0(fo)]
-[0(a)
+ (-0)(^),
(-0)(a)
Thus (End(A), +} is an abelian group. Note that we have not yet used the fact that our functions
= 0 + (\342\200\2240) 0.
and
are homomorphisms except Thus the set AA of all functions homomorphisms. definition of addition, and, from A into A is an abelian group under exactly the same in AA. of course, a nice associative multiplication function composition again gives are homomorphisms now to However, we do needthe fact that these functions in End(A) to
prove
+
that 0
show
the left
i/r
and
are \342\200\2240
distributive law in axioms for a ring.
again
law, (AA, +, Except for this left distributive and 0 be in End(A), and let a e A. Then
End(A).
all the
satisfies
Let
0,
(0(0 + f)){a)= 0((0+ yr)(a)) Since 0 is a
\342\200\242}
xjs,
=
+ f{a)).
0(0(a)
homomorphism,
+ Vr(fl)) =
0(0(a)
0(0(a)) +
=
0(Vr(a))
+ (0V)(a)
(00)(a)
= (00 + 0iA)(a). Thus
0(0
+ i/r) =
00 +
0i/f.
The
right
law causes no
distributive
trouble,
even
in AA,
and follows from ((Vr
+ 0)0)(a)
=
= Thus
24.1
Theorem
we
have
proved the
The set End(A) homomorphism
addition
of
+
(Vr
= ^(0(a))
0)(0(a)) +
(Vr0)(a)
= (Vr0
(00)(a)
+ 0(0(a)) + 00)(a).
theorem.
following
all endomorphisms and homomorphism
of
an
abelian
group
multiplication
A forms a
(function
ring
under
composition).
to show relevance to this section, we should give an example that Again, showing Since function composition is in general not End(A) need not be commutative. in some seems reasonable to expect. However,End(A) be commutative commutative,this may
cases. Indeed,Exercise15asks
us
to show that
End((Z, +}) is commutative.
Part
222
24.2Example
IV
Fields
Rings and
Consider
the abelian group (Z x Z, +} discussed that two elements of End((Z Z, +}) are
x
verify
\342\200\224 +
\302\242((171, n))
Note that
n, 0)
(m
and
onto the
everything
=
n)
(\\[r4>){m,
of Z
+ n,
\\[f(m
defined
xjs
0) =
(0, n).
\302\253))
x Z,
to
straightforward
by =
ir((m,
factor
first
11. It is
in Section
and
xjs
first
the
collapses
(0, 0).
while
Hence0i/r 24.3
Example
\342\226\262
^\302\242.
and let {F[x], +} be the additive zero, group of the with coefficients in F. For this example, let us denote this this notation. We can considerEnd(F[x]).One group by F[x], to simplify in F[x] of End(F[x]) acts on each polynomial by multiplying it by x. Let this
be a field of characteristic
Let F
F[x]
ring
^
0(0, n) = (n, 0).
n) =
(0i/O(m,
additive element
of polynomials
endomorphism be X, so
Another
+
+ a\\x
X(ao
=
element of End(F[x]) is formal of a sum is the an endomorphism of F[x].)
differentiationis
Y(ao +
+ anxn+
with respect to x. derivatives\" guarantees this endomorphism, so
of the
sum
Let Y
be
\342\200\242 \342\200\242 = a\\ + \342\200\242 + a\342\200\236x\") + \302\2532*2
a\\x
\342\200\242 \342\200\242 \342\200\242
+ \302\2532*3
+
a\\x
differentiation
\"the derivation
formula
+
1- a\342\200\236x\")a$x
+ \302\2532*2
+
+ 2\302\2532*
.
(Thefamiliar that
\342\200\242 \342\200\242 \342\200\242 + na\342\200\236x\"~l.
1, where 1 is unity (the identity map) in in F[x] of End(F[x]). Thus XY ^ YX. Multiplication of polynomials by any element F also gives an elementof End (F[x]). The subring ofEnd(F[x]) generatedby X and Y and is important and multiplications of F is the Weyl algebra in quantum by elements
us
17 asks
Exercise
to
show
\342\200\224 XY =
YX
that
mechanics.
\342\226\262
and
Rings
Group
Let G = {gi\\i ring
nonzero
with
e
Algebras
Group
1} be unity.
any
written
group
Let RG
be the
and let
multiplicatively of all formal
R
be
any commutative
sums.
set
iel for
at
\342\202\254 R and
two elements
gi e G, where all but of RG by
( J2a'Si \\ iel
Observe is again
E/6/0&.
that in
(at RG.
)
/
+
a finite
number
( J2biSi ) \\
is/
0 except for a + b{) \342\200\224 It is immediate that {RG,
the
at
are
0. Define
the
sum
of
= J2(a> + b^Si-
)
finite
of
number
iel
of indices
+} is an abelian
group
i, so with
Sj-\302\243/(a,' +
additive
bfigi
identity
Section 24
of two as follows:
G and R
iel
\\
we formally
J \\
iel
iel
J
223
Examples
is defined by
of RG
elements
Multiplication in
Noncommutative
of the
use
the
multiplications
I
\\gjgk=gi
the sum E,e/fc,gj and rename a over Since at and bt are 0 for all but a finite of nonzero summands number of i, only a finite number thus an element at most a finite number of \342\202\254 R and be viewedas of R. may Again, cijbk are nonzero. Thus multiplication is closed on RG. such sums Y,g.gt=g.ajbk The distributive at once from the definition of addition and the formal laws follow Naively, term a
way we
the
distribute
sum
E,\302\243/a,g,-
by a-jbkgi where g}gk = gt the a, bt contains sum Ss \342\226\240 gt=giuj
used
to define
distributivity
in G.
For
multiplication.
the
= !>\342\226\240
iel
,
J2b'Si)(j2CiSi) iel I \\
iel
!>\302\243<)
/J =
I]
I]( iel
.
J2
I]
I L iel
iel
\\
of multiplication
associativity
ahbjCk\\gi
J2
I
ahbj\\gi
J
I
\\gi,gj=g,
jSi .
'
\\ghgjgk=g,
J2\\ iel
bJCk
\\gjgi.=gi
[J^CigA \\
iel
I
.(\302\247o*)tehs')]tec'4 Thus we have
24.4 Theorem
the following
proved
If G is any group written multiplicatively is a ring. unity, then {RG, +, \342\200\242}
Corresponding 1g
with g,
to
we see that
each {RG,-}
Definition
The ring the
24.6
Example
group
G=
the
{e,a}is
cyclic
of order
Oe +
If we denotethese
and
addition
Oa.
elements
is not a
RG
abelian,
RG defined aboveis the of G over F. algebra
Let us give
R is a
and
commutative ring
g e G, we have an element Ig in RG. If can be considered to contain G naturally
subsystem. Thus, if G is not 24.5
theorem.
over R. If
in the
0,
F
is a field,
then FG is \342\226\240
tables for the multiplication 2. The elements of Z2G are Oe +
(rename)
multiplicative
commutative ring.
of G
ring
group
we identify as a
nonzero
with
le + Oa,
la,
obvious, a,
e,
and
natural
way
and
e +
by
a,
group
le +
algebra
la.
1aG,
where
Rings and Fields
Part IV
224
24.7 +
0
a
e e+a
Table
24.8
Table
a 0 e e+ a a 0 e e+ a e a 0 e+ a a e e+ a 0 e e+ a a 0 get Tables
we
respectively,
24.7
0 a e e+ a
to see that
For example,
24.8.
and
a e 0 0 e a a e e+ a e + a
0 0 0 0 0
e +
0
e+ a
e+ a 0
a)(e +
(e +
a
a) = 0,
we have (le This
shows
example
+ la)
+ \\a)(le
that a group
= (1 +
algebra may
Oe+ 0a.
l)a =
(1 +
+
\\)e
0 divisors.
have
Indeed,
this
the
is usually
case.
\342\226\2
The
Quaternions
We have
not
of Hamilton
given an example of the standard example
Rowan
Sir discovered quaternions searching for a way (vectors in E3). Six
Hamilton
(1805-1865) 1843 while he was number triplets multiply in
to
years earlier he had numbers abstractly as pairs (a, b) developed the complex of real with addition {a, b) + {a'+ numbers = + b') and multiplication (a + a',b (a, b){a'b') \342\200\224 for an bb', ab' + a'b); he was then {aa' looking for 3-vectors that was analogous multiplication and distributive such that the length of the product of the lengths of the vector was the product factors. After many unsuccessful attempts to multiply of the form a + bi + vectors 1, i, j are (where
b') =
cj
mutually
a noncommutative of a strictly
division skew
ring.
let us
field;
The quaternions
describe them.
Note
Historical William
yet are
perpendicular),
he realized
Let
group
while
Canal
Royal
in
on October
Dublin
the set H, for Hamilton, under addition by components, operation
1=
16,
k 1843, that he needed a new \"imaginary symbol\" to be perpendicular to the other three elements. He could not \"resist the impulse ... to cut with a knife on a stone of Brougham the fundamental Bridge\" these defining formulas on page 225for multiplying
quaternions.
The exampleof
were
quaternions
the
skew field.
a strictly
first known many
Though
others
were subsequently discovered, it was eventually noted that none were finite. In 1909 Joseph Henry
Maclagan Wedderburn tor at Princeton
thenaprecep-
(1882-1948), University,
gave
the
Theorem24.10.
walking
four times. This gives the H. We shall let
the
along
be E
x E x E x E. Now
the direct of addition
(1,0,0,0),
j = (0,0,1,0),
i = (0,l,0,0), and
(E
product of E under on H. Letus rename
= \302\243 (0,0,0,1).
x E
first proof
x E addition
certain
of
x E, +} is with elements
a
itself of
furthermore
We
to let
agree
ai =
(ai,0,0,0), = a3j (0, 0, a3,0) In view of our
225
Noncommutative Examples
24
Section
and
we
of addition,
definition
0,0),
(0, a2,
= (0,0,0,
a4k
a4).
have
then
a3, a4) =
(a.\\, a2,
=
a2i
+ a4k.
+ a3j
+ o.2i
a\\
Thus
=
define
To
+ a3j
+ a2i
(ai
+ a4k)
bi) + (a2 +
(a\\ +
(a3 +
+
b2)i
on H, we start
by
la =
for
multiplication
=
a\\
a
b3j + b4k) b3)j + (a4 + b4)k.
b2i +
(b\\ +
+
defining
aei,
and
Note to
jk = i,
= k,
ij
the similarity with if we think
remember
=
ki
ji =
j,
the so-called crossproduct
of the
=
\342\200\224k, kj
\342\200\224j.
These formulas
vectors.
of
ik =
and \342\200\224i,
are easy
sequence
k, i, j, k.
i, j,
Theproduct
from is the next one to the right. The left to right of two adjacent elements to two the of the next one to the from left of elementsis right adjacent negative product it must be to make the left. We then define a product to be what distributive laws hold,
namely, (a\\
+ a3j
+ a2i ~
(aib\\
+ (a\\b3
+
+
+
a4k)(b\\
b2i +
\342\200\224 \342\200\224 \342\200\224
(a\\b4
b3j + + (a\\b2
a2b2
a3b3
a4b4)
\342\200\224
+ a3b\\
+ a4b2)j
+ a2b3
\342\200\224
+ a4b\\)k.
a2b4
a3b2
b4k)
+
+ a3b4
a2b\\
\342\200\224
a4b3)i
are isomorphic to a subring Exercise 19 shows that the quaternions of M2(C), so we see that multiplication is multiplication is associative. Since ij = k and ji = \342\200\224k, not so H is definitely not a field. Turning to the existence of multiplicative commutative, inverses, let a = a\\ + a2i + a3j + a4k, with not all a,- = 0. Computation shows that (a\\ If we
+ a2i
+ a3j
0
0
+ a4k)(a\\
\342\200\224\342\200\224\342\200\224
a2i
a3j
a4k)
=
a^ + a22 +
a3 +
a4~-
let \\a\\
=
ax
0 + a2
0
+ a3 +
0 a4
and
a =
a\\
\342\200\224\342\200\224\342\200\224
a2i
we see that
a
ai 12
/\302\2532
i\342\200\236|2
\\\\a\\2
._
\302\253
Vial2/
._
U. Via12
a3j
a4k,
Part
226
IV
Fields
Rings
and
is a
multiplicative
We consider that
for a.
inverse
have
we
demonstrated
the
following
theorem.
The
Theorem
24.9
Note This multiplication.
G =
that
i4 =
There are no
24.10Theorem
See Artin,
= i3j. of a famous
content
the
theorem of
ring is a field.
division
[24] for proof of Wedderburn's
and Thrall
Nesbitt,
ji
quaternion
proof. finite
Every
and
fields. This is
skew strictly we state without
(Wedderburn'sTheorem)
Proof
j2 = i2
1,
finite
which
Wedderburn,
order 8 under
a group of j, where
i and
by
and multiplication.
addition
under
is \302\261i, {\302\2611, \302\261j, \302\261k]
is generated
group
skew field
a strictly
H form
quaternions
theorem.
\342\231
24
\342\226\240 EXERCISES
Computations in the
3, let
1 through
Exercises
In
group algebra Z5G
G=
{e,a, b}
4 through
In Exercises 4.
3/)(4 +
(i +
(4e + 2a+ 3b)
+ Ob) +
+ 3a
2/ -
7, write
the
for
+ sa+tb
re
,.
1. <2e
group of order 3 with
a cyclic
be
(2e +
2-
element
of
in the
EI
3\302\253 + 0b)(4e
form
5.
k)
to the
group S3 given
(Opo +
in the
+ aii
a\\
+ 2a
group
9. Find the
lyoi
the
element
in Example
+ 0p2
8.7, compute
+
+ 0/ii
lfl2
+
the
l/i3)(lpo
3. (3e+ 3a+ 3b)4
+ 3b)
+ a3j
+
a$k
for
a;
\342\202\254 ffi.
ffkji5
+ 3^-1
7. [(1+30(4/
Referring
e. Write
element
r,s,(eZ5.
6. (i+JT]
8.
identity
form
the
in
product
+ lPi + 0p2
+ 1/ii + 0fl2
+
l(l3)
Z2S3.
algebra
center of the
(H*,
group
where \342\200\242),
H* is
the set of nonzero
quaternions.
Concepts
10. Find two subsets of the
11.
addition
induced
Mark
each of the
a. b.
Every
c.
End(A)
different from C and from and multiplication from EL
EI
has M\342\200\236(F)
d. End(A)
of 0 for any
no divisors
nonzero
a ring
is never a ring
is a
of which
n and
any field F.
of M2CZ2) is a unit.
element
is always
each
false.
true or
following
each other,
with
with
unity
unity
=\302\243 0 for =\302\243 0 for
every abelian
any abelian
group
A.
group
A.
field
isomorphic
to C
under
e. The
subset
f. R(Z, g. The
Iso(A)
for every
End(A)
of End(A), consisting of abelian group A.
+) is isomorphic
+,
to (Z,
G is a
group
of A
isomorphisms
commutative
for every \342\200\242)
of an abelian
RG
ring
group
the
Rings and
Ordered
25
Section
onto
A,
a subring
forms
ring R with unity. ring for any commutative
commutative
227
Fields
ring R
of
with
unity.
quaternions are a field. is a group where HI* (HI*, \342\200\242) No subring of HI is a field.
h. The
i. j.
12.
a.
of the
each
Show A
following
of degree n
polynomial
in a strictly
coefficients
with
quaternions.
an example.
giving
by
set of nonzero
is the
skew
field
have more
may
than
n zeros
in the skew
0 is a
right divisor
field.
b.
A
finite
subgroup of a
multiplicative
not be cyclic.
field need
skew
strictly
Theory
13.
Let
14. Show 15.
element of
be the \302\242)
of 0. Show that
0 and
1.]
Show
that
is also \302\242
that
M2(F)
End
End((Z x Z, +)) given a left divisor of 0.
has at least six units +)) is
((Z,
for
field F.
every
+,
to
x Z2,
exampleshowedthat [Hint: F has
units.
these
Exhibit
to (Z,
isomorphic
naturally
24.2. That
in Example
and that \342\200\242)
at
least
is naturally
End((Z\342\200\236, +))
two elements,
isomorphic
to
(Z\342\200\236,+,-).
16. 17.
= {e},the
19. Thereexists
24.3, K e
(f>{a
for all a,
b,c,d e
a. Find the b. What 8
c.
What
section
matrix
K,
25
show
gives
+ cj
+
=
that
that
0 :
HI
a
\342\200\236 + i
of
HI
=
dk)
an isomorphism
+,
\342\200\242).
1.
show
M2(C) such
+ bi
(Z2
to
is isomorphic
RG -^
defined
M2(C)
with
^
_,
n
R
for
any ring
R.
by
~*~c '
~*~ ^^'
0
0[HI]
K.
equations
other
isomorphic that YX - XY
one element,
group^of
a matrix
is not
+))
Z2,
to Example
Referring
18. If G
x
that End((Z2
Show
should
you
check
thing should you check
^Ordered
to see to
show
that
\302\242) really
that
4>
gives
is a homomorphism? an isomorphism
of
HI
with
0[H]?
Rings and Fields the set E
on any subset of E. (We Definition 0.7.) We regard < as providing an ordering of the real numbers. In this section, we study orderings of that the this section under discussion and fields. We assume have throughout rings rings 1. nonzero unity a is positive, so the order relation < In the real numbers, a < b if and only if b \342\200\224 We use the if we know which real numbers are positive. on E is completelydetermined the notion of orderin a ring. to define idea of labeling certain elements as positive We
are
remind
with the inequality that relations were you
familiar
This section is
not
used in
the remainder
relation
discussedin
of the text.
< on
Section
and
0. See
Part IV
228
25.1Definition
Rings and
An
Fields
ring is a
ordered
R
ring
subset P
with a nonempty
together
these
R satisfying
of
two properties.
Closure
For
Trichotomy
For each
a, b e
all
a e
of P
a + b
both
one
R,
a e Elements
P,
P,
are in
and ab
one of the
and only
a = 0,
P.
holds:
following
\342\200\224a e P.
are called\"positive.\"
\342\226\24
is an ordered ring with set P of positive elementsand S is satisfies the requirements for a set of positive elementsin the of S. Exercise an (See gives ordering 26.)This is the induced ordering ring from the given ordering of R. We observeat once that for each of the rings Z, Q and ffi the set of elements that we satisfies the conditions of closureand trichotomy. have always considered to be positive We will refer to the familiar and the induced ordering on their ordering of these rings now an as the natural We unfamiliar illustration. give subrings ordering. It is
easy to
of R, S, and thus
a subring
25.2
Example
see that P
then
if R
D S
elements. There are two Let R be an ordered ring with set P of positive We two define an ordering of the polynomial describe R[x]. ring possible in of positive elements. A nonzero be written can R[x] polynomial Aiigh,
f(x) =
+
to
ways
sets, Pjow and in the form
\342\200\242 \342\200\242 \342\200\242 + a\342\200\236x\"
that and anx\" are the nonzero terms of lowest and arxr be the set of Let all such ar e P, Ptow highest degree, respectively. f(x) for which e and let Phigh be the set of all such f(x) for which P. The closure and an trichotomy that P\\ovl and Phigh must satisfy to give orderings of R [x] follow at once from requirements of addition and multiplication those same properties for P and the definition in R[x].
where
ar ^
Illustrating
0
arxr + ar+\\Xr+l
natural
in
0, so
and
a\342\200\236 ^
with
Z[x],
ordering
given
Piow,
by
be positive because\342\200\2242 is not positive would be polynomial positive because3 is
would not same
Suppose any
nonzero
also in
now
that P is the set of positive of R. Then either a or
element
P. Thus all squares of nonzero
is positive. By closure,we seethat 1 + in P, so it is never zero.Thus is always
1 +
With
elements of R
given
ordering
\342\200\2242x + 3x4
by
Phigh,
this
in Z.
positive
\342\200\224a is in
elements
f(x) =
the polynomial
in Z.
in
by
closure,
are positive. In
\342\200\242\342\200\242\342\200\242 finite + 1 for
any
ring R. a2 =
ordered
an
P, so
\342\226\262
particular,
number
Let a be is (\342\200\224a)2 1 =
12
of summands
ring has characteristic zero. must be positive, we see that the natural are precisely the ordering of E is the only possible ordering. The positive real numbers and the set could not be enlarged without destroying squares of nonzeroreal numbers 1 + 1+---+1 Because must be positive, the only possible trichotomy. ordering of Z is the natural ordering also. All ordered rings have characteristic zero so we can, by identification consider every ordered ring to contain Z as an ordered (renaming), subring. \342\200\224 If a and b are nonzeroelements of P then either \342\200\224a or a is in P and either b or \342\200\224 either ab or ab is in P. By trichotomy, b is in P. Consequently by closure, ab cannot be zero so an ordered ring can have no zero divisors. We summarize these observations in a theorem and corollary.
Becausesquares
of
nonzero
an ordered
elements
25.3Theorem
R be an
Let
ordered
25.4 Corollary
We
Z to
consider
can
Z from
is
R
there
elements of
of nonzero
All squares are no zero
ring.
characteristic 0, and
Ordered Rings and
25
Section
Fields
are
R
229
R has
positive.
divisors.
be embeddedin any ordered ordering of Z. The only
the natural
ring R,
the
and
induced
ordering
possible
of
ordering
of 5 is
the
natural
ordering.
25.3 shows
be ordered, because that the field C of complex numbers cannot i2 are squares. It also shows that no finite can be ordered ring is zero. because the characteristic of an ordered ring The theorem that follows defines a relation < in an ordered ring, and gives properties of <. The definition of < is motivated that, in the real numbers, a < b by the observation
Theorem
if
and
if b
only
definedin 25.5 Theorem
Let R the
be
\342\200\224 1 =
and
1=12
both
\342\200\224 a is
an ordered
with
on R
by
relation
<
ring defined
Trichotomy
If b <
Isotonicity
If
given
Conversely,
P =
set
Definition
if (b \342\200\224 a) e
only
one of
only
the
b <
c,
then
0<
c and
a, then
been
\"is less
read
than,\" be
P
(1)
a, b, c e
all
R.
holds:
following
b < a. < c.
a
a +
a + b <
could have
ordering
ab
c. < ac
and ba
< ca.
< on a nonzero ring R satisfying these three conditions, the two set of satisfies criteria for a x) positive elementsin and the relation < P defined this P is the given as in Condition (1) with
{x e
25.1,
and
a < b, a = b, and b < c, then
lfa
Transitivity \342\226\240
the
and
One
that
properties.
positive elements. Let <,
P of
< b if
also shows
listed
< has these propertiesfor
relation
\342\202\254 R. The
the
having
set
a
for a,b
The theorem
positive.
of a relation
terms
a relation
R
|
0 <
relation <.
Proof Let R
be an
ordered ring
We prove the
three
Trichotomy
P of
set
with
positive elements, and
let a
< b mean (b
\342\200\224
a)
properties
Let a, b e appliedto
the
R. By b
a,
holds. These
a=
of P in
Definition 25.1
one of
exactly
= 0, (a -
b-a in terms
translate
a
property
trichotomy
\342\200\224
(b-a)&P,
b <
b,
b)
e P
of < to a
respectively.
Transitivity
Leta < closure
e P.
for <.
of
b
and
P under
c. Then (b \342\200\224a) e P and (c-fc) addition, we have
b <
(b-a)
soa < c.
+
(c-b)
= (c-a)&P
eP.By
230
Part
IV
Fields
and
Rings
Let b < c, so (c -
Isotonicity
P so a + b
<
\342\200\224 =
ac
a(c
b)
ba
and
We leave the Exercise 27.) view
In
The
ring.
25.6 Example
Let R
e
b)
(a + c) - (a + b) a > 0, then by closure
\342\200\224=
\342\200\224
aband(c
= (c of P
Then
P.
c. Also if
b)a
ca
\342\200\224
are in
ba
b) e both
P, so ab
< ac
exercise.
(See
< ca.
theorem as an
of the
part
\"conversely\"
easy
equally
\342\231\2
25.5, we will
of Theorem >, <,
notations b >
a +
and
a means a
>
feel free
now
are defined
as
usual
to use the in terms
a b means either b < a or
< b,
in an ordered
< notation
of <
and
a
=.
Namely,
< b,
b =
a.
of R[x] given by to think what the orderings ring. It is illustrative 25.5. Example 25.2 mean in terms of the relation < of Theorem \342\200\224 x is positive sox < a. Also, Taking Piow, we observe, for every a > 0 in R, that a \342\200\224e x = x \342\200\224 0 is positive, so 0 < x. Thus 0 < x < a for every a e R. We have (xl x;) Piow when i < j, so xi < x' if i < j. Our monomials have the ordering be
and
Piow
an ordered
Phigh in
0 <
for
any
a e
positive many
infinitely
We leave a Exercise
R.
positive
\342\200\242 \342\200\242 < -x6
x^ < x4
x2
< x
< a
E, we see that in this ordering of Rpc] that are less than any positive real number! of R[x] given by discussion of < for the ordering R =
Taking
there are
elements
similar
P^h
1.
to
\342\226\2
preceding example is of interest because it exhibits We give a definition explaining this terminology. consider to be a subring of every orderedring. The
Archimedian.
an ordering
Remember
that is that we
not
can
Z
25.7
Definition
An ordering
of a ring
For each given na > b. is an
Archimedian
R with positive
ordering.
this property: a and
b
in
R,
there
exists a positive integer
n
such
that
\342\226\2
but the ordering of R[x] given by Piow ordering of E is Archimedian, in Example 25.6 is not Archimedian because for every positive integer n we have (17 \342\200\224 e Piow, so nx < 17 for all n e Z+. nx) We give two examples and fields that are of interest describing types of ordered rings in more advanced work. The
natural
discussed
25.8 Example
(Formal Power SeriesRings)
Let R be a ring. In Section 22 we defined a polynomial of the a,- are 0. If formal sum X!^o aix> wnere ^ but a finite number we do not require any of the at to be zero, we obtain a formal power series in x with coefficients in the ring R. (The adjective,/orwa/, is customarily used becausewe are not with convergence of series.)Exactly the same the formulas are used to define dealing sum and product of theseseriesas for polynomials in Section 22. Most of us had some in
R[x]
to be a
Ordered Rings
Section 25
231
Fields
and
and multiplying series when we studied calculus.These seriesform a adding which we denote by /?[[*]], and which contains R[x] as a subring. If R is an ordered to R[[x]] exactly as we extended ring, we can extend the ordering
practice ring the
to R[x] using The monomials have
ordering
not?) 25.9
Example
the
set
the
same
Pjow of
positive elements. (We cannot use that we displayed in Example
Phiah-
Why
25.6.
ordering
\342\226\262
with Laurent Series Fields) Continuing the idea of Example 25.8,we let F series of the form Y^n a'x' where and consider formal N may be any integer, zero, or negative, and a,- e F. (Equivalently, we couldconsiderY2u=-oc positive, aix' where all but a finite of the a,- are zero for negative number values of i. In studying for functions of a complex variable, oneencounters of this form called \"Laurent calculus series With the natural addition and multiplication of these series, we actually have a series'') + 0 + (k + Ox2 + \342\226\240\342\22 ^zeW which we denoteby F((x)).Theinverseofxistheseriesx_1 Inverses of elements and quotients can be computed three by series division. We compute \342\226\240 terms of (x 1 +x \342\226\240x2 in R((x)) for illustra+ \342\226\240 +x3 + 2x4 + 3x5 + \342\200\242) -)/(x3
(Formal
be a field
tion. 3x~' + 2x4 +
x6 +
3x5 +
x - 1+
-
x
+
x2
+
+x3
+ 3x +
+ 2
x~l
Ax~l
\342\200\242 \342\200\242 2x + \342\200\242 6x - 9x2 + Ax
F is an
If an
ordered field, we can use the In Exercise 2 we ask
of Pjow in i?[[x]] to define analog to you symbolically order the monomials \342\200\242 \342\200\242 \342\200\242 as we did for R[x] in Example 25.6. Note that
of F((x)).
ordering
\342\200\242 \342\200\242
-x~3,x~2,
=
x_1,x\302\260
l,x,x2,x3,
obvious
F((x)) contains, as a subfield, a field of quotients on this field of quotients. ordered ring and let \302\242) : R identification (renaming), the
R be an
Let clear
that by
of
to provide
R
for a skeptic,
25.10Theorem
Let
R be an
positive
4>{a) <'
4>{b)
We call by\"
25.11Example
<^> from
Example
22.9
ordering \342\226\262
be a ring
-^- R' map
induces an
and thus
F[x],
It is
isomorphism.
to carry over theorem what would have
4>
be used
can
of R'. We state as a as Exercise 25.
intuitively
the ordering to
be
proved
set P of positive elementsand let
ordered
ring
with
subset
the ordering of R' described in the ordering of R. that the
stated
0(ao + is one
of
the proof
leave
The
isomorphism.
set of
an ordering and
\342\226\240\342\226\240\342\226\240
+
to one. Thus
it
evaluation
aix +
provides
the preceding
theorem
0W : Q[x]
homomorphism
\342\226\240 \342\226\240 \342\226\240 = ao + a\342\200\236x\")
an isomorphism
the \"ordering
+
a\\it
of Q[x]
+
-\302\261 E
induced
where
\342\226\240 \342\200\242 \342\200\242 + a\342\200\236n\"
with
0[Q[x]].
We denote
this
232
Part
IV
Rings
and
image
ring
Fields
natural (and only)
by the of Q>!
induced element An Theorem 25.10
the set
the ordering Q>[tt]. If we provide Q[x] with using on Q[7t] induced by (j>\342\200\236 is very and 25.6, the ordering
by
25.2
Examples
of
ordering
In the
ffi.
Piow ordering,
it
of
Piow
different is less
from
that
than
any
\342\226
of a ring isomorphism can be used to exhibit that do
of R
automorphisms
not
is called
itself
onto
R
different
orderings
an
automorphism ordered
of R. R if there ring
P of positive elementsonto
the set
carry
an
of
itself.
exist
We give
an example.
25.12 Example
Exercise11of
This
by Z[\\/2].
ring
However,we claim automorphism.
homomorphism
induced by
clearly one to to
property
e Z}
\342\200\224>\342\226\240 defined Z[V2]
Z[V2]
n*Jl \\m,n
order induced
a natural
has
ring
that
It is
that {m +
18 shows
Section
by
is a ring. ffi
in
+ n*jT)
Let us denotethis *Jl is positive.
which
= m
\342\200\224
ny^is
an
of the Z[V2]. We leavethe verification we see the Because 0(\\/2) = \342\200\224 \\/2, ordering
and onto
one
17.
Exercise
from
order on Z[V2], an element positive! In the natural both positive, or if m is positive and 2n2 < m2, or if n is positive and m2 < 2n2. In Exercise 3, we ask you to give the analogous descriptions A for positive elementsin the ordering of Z[-s/2] induced by \302\242. m +
be one
4> will
where
n~j2 is positive if m
\342\200\224 is
^/2
n are
and
of Examples 25.11and 25.12, the induced orderings,we wonder one. Our final theorem shows that
In view are not natural
25.13 Theorem Let D be an
ordered
integral
of D.
of quotients
field
domain
this
with P
as set of
elements,
positive
and let
F be a
The set
P' = {xe F | x
=
for a,b
a/b
is well-definedand gives an order on F that P' is the only subset of F with this property. Proof
of ffi that exhibit orderings on subrings other than the Q can have an ordering is not possible.
which whether
D and \342\202\254
ab e
the given
induces
P}
order on D.
Furthermore,
suppose that x = a/b = a'/b' for a, b, a', b' e D and show that ab We must that a'b' e P. From a/b = a'/b' we obtain ab' = a'b. = we have Now P and by assumption, ab & P. b2 \342\202\254 (ab)b' a'b2. Multiplying by b, = = and the properties of a ring, we see that Using a(\342\200\224b) {\342\200\224a)b \342\200\224(ab) trichotomy a' and b' are both in P or both not in P. In either case, we have either a'b' e P. We proceed to closure for P'. Let x = a/b and y = c/d be two elements of P', so P and cd e P. Now x + y = (ad + bc)/bd and ab \342\202\254 + bc)bd = (ab)d2 + b2(cd) (ad is in P because squares are also in P and P is closed under addition and multiplication. Thus (x + y) e P'. Also xy = ac/bd is in P' because acbd = (ab)(cd) is a product of two in P. of P and thus elements that for x = a/b, the product For trichotomy, we need only observe ab satisfies just To show
that
P'
is well-defined,
& P.
one of
ab \342\202\254 P, by
trichotomy
We have is
in
P'
ordering
if and
For P',
for P.
that
shown
only
if
al
from F by P'.
ab
thesetranslate
P' does = a is
give in
P,
an
so
ab i P
= 0,
= 0, and* \302\243 P', respectively. a e D, weseethata=a/\\ For ordering the given ordering on D is indeedthe induced into
x
P',x of F.
&
Exercises
Section 25
P\" is a
that
suppose
Finally,
set of positive
of F
elements
satisfying
of F
shows maintains
that
conditions
the
P\" where P. Let* = a/b \342\202\254 a. b e D. Then = P. Thus xeP'so P\" c P'. The law of that we then must have P' = P\".Therefore P' gives the only ordering \342\231\246 of D. order for elements original
of Definition 25.1 and such that P\" n D = = ab must be in P\", so ab e (P\" flfl) xb2
trichotomy
233
25
\342\226\240 EXERCISES
Computations
1. Leti?
of a positive element a of .Rand ordered ring. Describethe ordering as we did in Example 25.6, but using the set Phigh of Example 25.6
be an
in R[x]
2. LetF be an Example
describedin
of F((x))
of m
terms
and
4 through
In Exercises
that
n,
of Z[V2] = {m + n~Jl \\m,n elements of Z[V2] in that ordering.
ordering
all positive
9, let
R[x] have
the
25.2. In Example to increasing
4. a.
-5 + 3x
5. a.
-1
6.
a.
7. a.
8. a.
-3 + 5x2
-2x2 + 5x3 - 3x2
Ax
9. a. x -
3x2
x3 +
b.
+x3
Ax4
- 3x2 + 5x3
(ii), list
the
e Z}
in which
+ lx2 + lx2
-5x
c.
-5
c.
2x x
the labels a, b, c, d, e of by the relation
V2
positive. Describe,
d. x - llx4 d. %x2
3x2
ordering
to increasing
e.
+ x5
e. -3x3
-
e.
5x5
2x
e.
3x
d.
Ax4
e.
x +
25.9. List the
labels
3x2 +
x +
described in Example order of the elements
6x3
as described
-
8x4
e.
d. 5* Ax3
2 + Ax2
lx2
d. 6x3 + 8x4 - Ax2 d. -3x
- 6x3
- 3x2 +
polynomials in an Theorem 25.5.
given < of
-
b. \342\200\224
11.
a. 1
12. a.
c.
^V
J\\,
- x
5-lx
-
, , \342\200\236
x2+3x3
b.'
,_
b.
d.
^V
l+x -2
-
c.
+ Ax
-\342\200\224\342\200\224-
A-3x
c.
x-x2
3x2
a, b, c,
by the relation
1 +2x -\342\200\224\342\200\224
A~3x
\342\200\224 ^V
d.
d.
\342\200\224\342\200\224t
l+x2
9 - 3x2 2 + 6x
e. x3
+ Ax
3 -5x
e.
Ax5
- 2x2 - 2x2
Theorem 25.5.
10. a.
in
ordering
\342\200\224 is
the
as described
c.
c.
elements
given
and
polynomials
c. Ax
+ 2x2
Ax
b. 2
discussed
ii- Phigh
c. -x
10 through 13, let Q((x))have in an order corresponding
In Exercises the
+ 5x3
each case (i) order of the
b. 5 - 3x b. 3x- S,x3 b. -2x + 5x2 b.
in F,
\342\200\242 in the ,\342\226\240\342\226\240
\\,x,x2,x3
by
given
ordering
i. Pi0w as described in order corresponding
= x\302\260
of R[x].
elements
example.
25.12 describedan
3. Example in
let F((x)) be the field of formal \342\200\242 \342\200\242 the ordering of the monomialsx~3, x~2, x~l,
field and
ordered
25.9. Describe
\342\200\242 \342\226\240 x\". \342\226\24 \342\200\242,
x2, x3,
the monomials*,
as setof positive Laurent series with coefficients
~6+2x
-
Ax3
d, e of < of
IV
Part
234
Rings
b' 3~5x
l~X
13 a'
and Fields
l
a'
\"
3+5x
1+x
1
c
Ax
&
~3x + x2
Ax+x2
+ x2
i-x
Concepts
14. It
be shown that
can
containing
Explain
subfield of C that
of a
ordering
15. Mark
the smallest subfield
\\i2{~l+^).
each of the
of R containing ifl is isomorphic to there is no ordering why this shows that, although contains some elements that are not real numbers.
true or
following
the
C
subfield of
smallest
C, there may
for
be
an
false.
a. Thereis only
one ordering possible for the ring Z. R can be ordered in only one way. in only one way. subfield of R can be ordered
b. The field c.
Any
d.
The field
e.
If R
f.
An
g.
An
is
an
Q can be ordered ordered
of a
ordering
ring,
in only
one
then R[x] can
way.
be orderedin
ring i? is Archimedianif is Archimedian
h. If R i. If R
is an is an
ordered ring
j. Every
16. Describean
ordered ring ordered ring of the
ordering
has
if for
R, then
a e
and
fo
foreacha,
of a ring R ordering such that b < na.
e R,
e
each a,b
\342\200\224a cannot
that induces
a way
the
there exists such
R
that 0
n
order
given e Z+
< a,
on R.
such that b
< na.
n e Z+
exists
there
be positive.
a or \342\200\224a is positive. R, then either infinite number of elements. a e
and
an
ring Q[tt],
discussedin
in which jt is greater
25.11,
Example
than
rational
any
number.
Theory
17.
25.12, show
to Example
Referring
the
that
map
-\302\273 R
Z[V2]
where
(p(m
+ n^/l)
=
m\342\200\224 n4l
is a homo-
morphism.
18 through
In Exercises
24, let R
an ordered
be
Theorem 25.5. Prove the Theorem 25.5. For example,you must 0 < b then ab < 0.\ R defined in
on set P of positive elements, and let < be the relation 25.1 and (All the proofs have to be in terms of Definition know that negative times positive is negative, so if a < 0 and
ring
with
statement.
given
not say,
\"We
18. IfaeP,thenO
19.
If
a, b
20.
If a
21.
If a
e P
< b, then < 0 and
0 < b,
R is
a field and
23. If
R is
a field
24. If
R
then ab
the
field and
27. Show
that
if R
requirements
&
P.
< 0.
a and b are positive, and 0 < a < 1,then 1 <
-1 < a < 0,then
25. Prove Theorem25.10of 26. Show
d = 0 or cd either c \342\200\224
\342\200\224 b < \342\200\224a.
22. If
is a
= bd, then
ac
and
the
positive.
l/a. < -1.
l/a
text.
is an ordered ring with for a set of positive
that if < is a relation stated in Theorem 25.5, then
then a/b is
on there
set P
of positive elementsand the ring S, and thus
elements in
a ring R exists a
satisfying the subset PofR
properties satisfying
S is gives
a subring of R, then an ordering of S.
P
of trichotomy, transitivity, and conditions for a set of positive
the
n S
satisfies
isotonicity elements
in
Definition
25.1,
the relation
28. Let
R be
then
29. Let R
a =
and such
the
relation
domain.
Show
that
by
a
b if and only
235
Exercises
25
Section
if
(b
R
and
-
a) e
P is the
same
as
<.
an ordered
integral
that
if
a2n+l
= b2n+l
where a,b
e
n is
a positive
integer,
b.
in two variables with coefficients in R. the ring R[x, y] of polynomials be an ordered ring and consider on and Example 25.2 describestwo ways in which we can order R[x], and for each of these, we can continue two ways, giving four ways of arriving at an ordering of R[x, y]. There are order (/?[;c])[y] in the analogous then (/?[y])[;c]. Show that another four ways of arriving at an ordering of R[x, y] if we first order R[y] and of R[x, y] are different. You might start by considering whether x < y or all eight of these orderings [Hint: < in this in of these and continue x each fashion.] y orderings,
PART
and
Ideals
Factor
Rings
V
section
26
and Factor
Section
26
Homomorphisms
Section
27
Prime and
Section
28
fGr6bner Basesfor
Ideals
and Factor
Homomorphisms
Rings
Ideals
Maximal
Rings
Homomorphisms the
We'defined
since we
concepts to talk
wished
rings.
isomorphic
homomorphism
their additive
26.1 Definition
A
map
0 of
a
of homomorphism and isomorphism about evaluation homomorphisms
We repeat is a structure-relating and
structure
ring
R into
some
their
a ring
map.
easy reference.
for
here
definitions
A homomorphism
multiplicative
in Section
for rings
and
for polynomials
for
rings
Recall
must
relate
18,
about that
a
both
structure.
R' is a homomorphism 0(a + b) = 4>{a)
if +
\302\2420))
and
^ab) for all
elements a
and
bin
= 0(a)0(fc)
R.
In Example 18.10we defined evaluation and Example 18.11 homomorphisms, that the map 0 : Z ^ Z\342\200\236, where is the remainder of m when divided 0(m) We give another n, is a homomorphism. but fundamental simple very example of a
showed
by
homomorphism.
26.2
Example
Let Rl,R2,--,Rn be rings. For each i, the map m : (Projection Homomorphisms) \342\200\242 \342\200\242 \342\200\242 \342\200\242 = x Rn -> Rt defined Rx x R2 x \342\200\242 jt,- (ry, r2, \342\200\242, by r,- is a homomorphism, r\342\200\236) onto the ith component. The two required projection of a homomorphism hold properties Section
28 is not
required for
the remainder
of
the text.
237
238
Part
V
for
Factor
and
Ideals
of
Properties
26.3 Theorem
our
work
individual
are computed
direct product
the
in
multiplication
by
\342\226
component.
Homomorphisms
the exposition of Section13but
through
way
for
homomorphisms.
ring
Let R into a ring R'. If 0 be a homomorphism of a ring = in in then 0' is the additive R', and if a e R, R, 0(0) identity identity = If then is a of R'. Going the then S is a of R, 0[S] \302\242(-a) \342\200\224\302\242(a). subring subring of R. Finally, if R has unity of R', then 0_1 [S'] is a subring other way, if S' is a subring to subrings, and for 0[/?]. 1, then 0(1) is unity Loosely speaking, subrings correspond a to with under with unity rings ring homomorphism. unity correspond rings
(Analogue of Theorem 13.12)
0 is the
Proof
and
in each
and multiplication
addition
We
addition
both
since
iti
Rings
Let 0
additive
be a homomorphism
R into a ring R'. of into {R, +} (/?',+'), group homomorphism is the additive element of R' and that identity a ring
of
Since,
= C
0(0)
also tells us
13.12
Theorem
{S, +}, the elements of 0[5],
group
set
if S
that
+'} gives
(0[S],
is a
and
=
0(\342\200\224a)
\342\200\2240(a).
then, considering the additive (/?', +'}. If 0(si) and 0(\302\276)are two
of R,
subring
a subgroup
of
then
=
0(^)0(\302\276)
Thus
0(^1\302\276) \342\202\254 0[5].
is a
0[5]
Consequently,
in particular, 0 can be 13.12 tells us that
Theorem
as a
viewed
e 0[5],
0(^)0(\302\276)
Going the other way, (0-1 [\302\243'],+} is a subgroup
so 0[5] is
closed
under
multiplication.
i?'.
of
subring
0(^1\302\276)
also shows
13.12
Theorem
that
e 0\"1[5'],so
of (#,+}.Leta,fo
if S' that
is a 0(a)
subring
of
then
R',
e 5'.
e S'and0(fo)
Then
\342\202\254 5', we
is a
and thus
Finally,
of
subring if
see that
e 0_1
afo
4>{a)4>{b).
[5'] so
for
all r
26.4 Definition
Let
26.3
that
0(1)
is unity
R be a homomorphism
0:7?^
kernel Now
into
(/?',
give us at
this
of 0,
denotedby
of rings.
but
not
necessarily
for R'
as we
The subring
{retf|0(r) = O'}
\342\226
Ker(0).
is the same as the kernel of the 0. Theorem 13.15and Corollary results for ring homomorphisms. analogous
Ker(0)
+} given by once
for 0[/?],
9.
0-1[O'] = is the
)0(1),
\342\231
in Exercise
to illustrate
a map
= 0(r
0(l)0(r)
0[/?].
in Theorem
Note
ask you
for
multiplication
e i?,
0(r) = 0(lr) = 0(rl)= so 0(1) is unity
0_1 [S'] is closedunder
R.
unity 1, then
R has
=
group
13.18
of {R, +} homomorphisms
homomorphism
on group
26
Section
26.5
Theorem
26.6 Corollary
and Factor Rings
Homomorphisms
239
: R => R' be a ring (Analogue of Theorem 13.15) Let \302\247 homomorphism, = \302\253 H = Ker(0). Let a e R. Then + // = // +a, where a + 0_1[0(a)] a of the commutative is the coset containing additive group {H, +}.
of
(Analogue map
A
13.18)
Corollary
: R \302\247
homomorphism
ring
=> R' is
a one-to-one
= {0}.
only if Ker(0)
if and
let
and
H = H+a
Factor (Quotient) Rings ready to describe the of Theorem 14.1. analogue
We
26.7
Theorem
are
now
14.1) Let0 : R of H form a ring That is, the sum
of Theorem
(Analogue
Then the
additive
choosing
representatives.
cosets
and
the
of the
product
cosets is defined (a
the
Also,
Proof
Once again,
additive
the
show
that
multiplication that R/H
show
and
multiplication are
computed
+
= 0(afo
and
hi)
for us
in
isomorphism.
We proceed
14.1.
Theorem
of cosets by
choosing representatives a + hi the representatives
consider
+ ah2 + h\\b
= ab
the
+ a/i2 +
0(afc)
+
0(a)O'
4>{ab)
+
0' +
is a
ring,
distributive
by choosing
+
h\\b
<$>{ab) + 0(a/i2)
by choosing
corresponding propertiesin
is well of
a +
+
7/
h\\h2. 7/= 0_1[0(afo)], and 4>{h)= 0' for
^\302\276)
+ 0(/ii&) + + 0>(fo)
0' +
+
0' =
0(/ii/i2)
+ 0'0' (1)
4>{ab).
representatives is well defined.
it remains
laws hold
to show in
R/H.
that
the
Since
representatives, these properties
associative
addition follow
property
for
and at once
from
7v.
14.1 shows that the map /x defined in the statement of Theorem26.4is one to one, onto 0[7v],and satisfies the additive property for a homomorphism.
Theorem defined,
is done
= 4>{a)is an
dies in the coset ab + 7/. Sinceafo = 0(c) 4>{ab).Since 0 is a homomorphism
that
= =
multiplication
+ 7/)
(i(a
by
theory
h2, e 7/
=
To
= (ab)+ H.
this element
0(c)
Thus
H,
b) +
+
Let
+ 7/.
we need only show h e 7/, we obtain
H.
with kernel
homomorphism
by
that multiplication
c = (a + h\\)(b We must
the
with
start
aspects.
let h\\,
end,
of the
part
to check the multiplicative We must first show
defined. To this and b + h2 of b
H) = (a
+ 7/)
+ H)(b
a ring
14. We
whose binary operations are defined by R/H of two cosets is defined by
-> (j>[R]defined
/x : R/H
map
R' be
=>
+ (b +
+ H)
(a
of Section
for rings
analogue
well
Part
240
V
Ideals and
Factor Rings
we have
Multiplicatively,
fi[(a +
This completes 26.8
Example
the
shows
choosing representatives
shows
this
that
It remains
multiplication
Eq. (1)is
due
Ker(0),
Theorem 26.9
which is
below,
Theorem
of
cosets of
if ah
only
the
14.4)
the
H)(b +
ring R.
the
of
subring
of
Multiplication
equation
H)=ab + H
all a,b
\342\202\254 R and
and hb & H for all a,b are also representatives ofthe
& H
ah
+ /i2
&
h
&
H.
R and
all
cosets
a +
/i
Let /ii,
e H.
7/andfo +
h2
\302\243 H
so
//containing
b. Then
(a + Since ah2 +
be a
Let H
H for \342\202\254
Kb
14.4.
of Theorem
analogue
H is well defined by
\342\202\254 H and
Suppose first that that a + /ii andfo
afo
\342\22
Z\342\200\236.
H of a ring R such that only to characterize those subrings is well defined. The coset cosets of H by choosing representatives in Theorem to be well defined in Eq. (1). The success of 26.7 was shown = 4>{hxb) = ${hxh2) = 0'. That H where to the fact that 4>{ah2) is, if h \342\202\254 then for every a,b e 7? we have ah e H and Kb e H. This suggests
(Analogue
a and
\342\23
map
(a +
Proof
isomorphism.
n, is
is isomorphic to
Z/nZ
ring
additive
,if and
H).
additive
multiplicationof
26.9 Theorem
/x is an
+
: Z -> Z\342\200\236 defined by 0(m) = r, where r is the \302\242) Since Ker(0) = nZ, Theorem 26.7 ahomomorphism. on residue classes canbe computed by operations the corresponding operation in Z. The theorem also and performing
of m
remainder
that
demonstration
18.11 shows that the when divided by that Z/nZ is a ring where
Example
H =
H){b + 7/)]= ii{ab + H) = 4>{ab) = 0(a)0(fe)= /i{a+ H)/i{b
and
h\\b
h\\){b
and h\\h2
+
hi)
are all
in
\342\200\224 ab +
H
ah2 +
by hypothesis,
h\\b
+
h\\h2.
we see that
(a
+ h\\){b
+ h2) e
//.
of additive cosets by representatives is well the coset product {a + //)//. Choosing representatives = 0+// =//. Sincewe a e (a + //) and 0 e //, we see that =a0+// (a + //)// also compute can a e (a + //) and /i e //, we see that (a + //)// by choosing any ah & H for any /i e //. A similar argument the with starting product //(fo + //) shows
Conversely,
defined. Let a
that
hb
&
H
suppose
eR
and
for any
/i
that
multiplication
consider
e //.
\342\23
In group the type of substructure are precisely of groups theory, normal subgroups with a well-defined operation on cosets given required to form a factor group by operating with chosen representatives. Theorem26.9 shows that in ring theory, the analogous substructure must be a subring H of a ring R such that aH c H and H for Hb \302\243 \342\202\254 a// = {a/j | /i e //} and //fo = {/jfo | /i e //}. From now on we will alla,b /?, where we started usually denote such a substructure using N by N rather than H. Recall that to mean a normal subgroup in Section 15.
An addive subgroup
Definition
26.10
We see
is an
nL
that
be
Let F
Example
the
of all
ring
function.
unique
algebraic
functions mapping E in F. Is functions
For example, the
of a constant
+
into
+ \302\2532<22
numbers
primes \342\200\242 \342\200\242 \342\200\242
of
+
xp =
ap-i
certain
n >
of the form where a is a
26.13
be
the
of F
subring
function is again
a constant
2 sin
C is
x.
Thus
not
1 (p
that
Solution
Let/ we
just
be as in
/(2)
= 0.
e N
and
find that
observing
the
let
gf e
an
Dedekind
g e
ideal
=
zn
states &
if
L+
out that an \"ideal
not a
number,\" which was uniquely
was
at all,
\"number\"
by
be written
uniquely
example, and let N be Why or why not?
and show that algebraic
as a
the
any
number
product of prime
subring
of all
in the field could
ideal
ideals.
functions f such
in Fl
N. Therefore the
prove
Last Theorem, which has no solutions x,y,z
product of two ideals ring of integers of any
F. Then (fg)(2)
that N is
to
the set of integers it \"divided.\" Richard took advantage of this fact to identify the ideal factorwith this set; he therefore called the set itself an ideal, and proceeded to show that it satisfied the in definition the text. Dedekind was given then able to define the notions of prime ideal and determined
preceding
Is N
y\"
turned
are
prime factors\" and \"ideal of certain congruence relationships; factors\" were then used as the divisors
Let F
able
in fact
2.
the
\"ideal
Example
By use
factorization.
unique
of Fermat's
cases
It
in terms
these\"ideal
=
s{nm)
Why?
the function
was
Kummer
in general
two
Kummer defined
in F7
ideal
to preserve
that x\" +
to the expected results; the such \"unfavorable\" numbers may well be divisible by other \"unfactorable\" numbers. numbers\"
let C
and
E,
with every 2 is
and
necessary
not lead
does
x
of these,
to
wanted
prime) and the a,Kummer had noticed that ordinary integers. of primes as \"unfavorable naive definition product of
and
\342\226\262
Kummer
In particular,
complexroot
numbers\"
into
C an
function
of sin
product
of
to factor
+ a\\a \302\253o
a subring,
it is
know
seZ.
Kummer
Eduard
the
integers.
be able
L since we
ring
\342\226\262
who (1810-1893) an \"ideal concept complex in 1847 in order to preserve the notion in certain rings of factorization
was Ernst It introduced of
R
Note
\342\226\240 Historical
number\"
e
all a,b
for
Nb\302\243N
off.
ideal
an
constant
the
in
all
true that the product
is not
It
and
ideal
e nL for
consisting of all the Solution
the properties
satisfying
\342\226\240
{nm)s = n{ms)
26.12
R
241
ideal.
is an
Example
ring
c N
aN
26.11
of a
N
Homomorphisms and Factor Rings
26
Section
N kernel
is
= f(2)g(2)= 0g(2) = an ideal of F. We could
of the
0,so also
fg e N. Similarly, have proved this
by
E. \342\226\262 evaluation homomorphism \302\2422 '\342\226\240 F \342\200\224>\342\200\242
Part
242
V
Ideals and
that multiplication
we know
Once
distributive
laws for
at once
corollary
this
of
(Analogue
of N
these cosetsfollow
R/N
associative
the
denned on
multiplication
same propertiesin
from the
at once
is well
R.
and
the
We have
26.9.
of Theorem
Let N be an ideal of a ring 14.5) the binary operations dennedby
Corollary
form a ring
representatives law for
choosing
by
subring N of R,
of a
cosets
additive
26.14 Corollary
Factor Rings
R. Then the
additive
cosets
with
{a +
(b + N)
N) +
= (a + b)
+
N
and
(a +
26.15 Definition
The
ring
in the
R/N
N)(b + N)
=
ab +
preceding corollaryis the
factor
N. ring
(or quotient
ring) of
If we
of
use
quotients
Tocomplete
26.16
Theorem
Proof
term
the
of an
Fundamental
ring, be sure
quotient
integral
13 and
(Analogue of Theorem 14.9) Let N , by y(x) = x + N is a ring homomorphism
be an
analogy
it with the
to confuse
of the,
notion
field
21.
Section
Theorem
Homomorphism our
not
domain, discussed in
with Sections
and
R
*
byN.
14, we
give the
analogues of Theorems14.9
14.11.
The additive
part
is done in
ideal of aring R. Then kernel N.
: R
-> R/N given
the multiplicative
question, we
y
with
Theorem 14.9. Turning
to
see that y(xy) 26.17
Theorem
= (xy)
+ N = (x
+ N)(y +
N)
=
\342\231
y(x)y(y).
Let 0 : Theorem; Analogue of Theorem 14.11) with kernel N. Then is a and the ring, map ring homomorphism 0[/?] (i: R/N -> 0[/?]given by /i(x + N) = 4>{x) is an isomorphism. If y : R \342\200\224>\342\200\242 R/Nisthe homomorphism given by y{x) = x + N, then for each x e R, we have
Homomorphism
This follows
once
R
Proof
at
from
Theorems
26.7
and
26.16.
Figure
26.18 is
Fig. 14.10.
the
analogue
of
\342\231
W)
R/N
26.18 Figure
Exercises
Section 26
26.19
26.11 shows that 18.11 shows that
Example Example
Example
homomorphism, and /j,
: Z/nZ
is
an
->
where Z\342\200\236
the factor
form
can
remainder of m
is the
0(m)
Ker(0) = nZ. Theorem26.17then of m modulo + nZ) is the remainder
see that
we
where Z\342\200\236
->
: Z
so we
ideal of Z,
is an
nZ 0
/x(m
nZ.
Z
ring
a
n is
modulo
the map well defined and that
shows is
n
243
\342\226\262
isomorphism.
R gives rise to a factor ring every ring homomorphism with domain ring R/N gives rise to a homomorphism mapping R into R X. An ideal in ring theory is analogous to a normal Both are subgroup in the group theory. needed to form a factor structure. the type of substructure to Theorem 26.3 on properties of homomorphisms. We should now add an addendum and let N be an ideal of R. Then 0[7V]is an ideal Let \302\247 : R -> R' be a homomorphism, In
R/N,
summary,
and
of 0[/?],
factor
every
although
of R', then
0_1
need
it
[W]
not be an
ideal of./?'. Also, if
of R. We
is an ideal
the proof
leave
N' is an of
of either
ideal
0[.R] or
22.
to Exercise
this
26
\342\226\240 EXERCISES
Computations
1. Describeall
of Z x Z into Z x Z. [Hint: Note that ring homomorphisms = and 0)) 0((0, 1)) = 0((0, 1))0((0, 1)).Consider 0((1, 0))0((1,0)) a subring isomorphic to all positive contains integers n such that Z\342\200\236
0((1,
2.
Find
3. Find
N of
ideals
all
Zi2.
In
each
?
case compute
Z^/N;
that
if 0
0((1,
a homomorphism,
then
0)(0, 1)).]
Z2.
a known
is, find
is such
also
ring
the quotient
to which
is
ring
isomorphic. Give addition
4.
and
tables
multiplication
Are 2Z/8Z
for 2Z/8Z.
and
rings?
Z4 isomorphic
Concepts 7, correct the a form acceptable
5 through it is in
Exercises
In
needed, so that 5.
An
6. An
ideal
and nr
TV
of a
ring R is an
with
italicized term
text, if correction is
to the
reference
without
for publication. a ring
additive
R' is
subgroup
a homomorphism of (R,
/?' such 0 : /? ->\342\226\240
+) such that
for
all r
e
R
that Ker(0)
and
all n
e
iV,
= {0}. we have
rn
&
N
e N.
7. Thekernel 8.
of a ring R
isomorphism
of the
definition
/?' is {0(r) = 0' | r e W}. a ring R into a ring 0 mapping Let F be the ring of all functions mapping R into R and having derivatives of all orders. Differentiation gives = /'(*)\342\226\240 Is 5 a homomorphism? Why? a map 5 : F \342\200\224>\342\226\240 F where S(f(x)) the connection between this Give exercise and Example 26.12.
9. Give an unity
of a
homomorphism
of a ring
homomorphism 0 : R
following
true or
example
->\342\226\240 /?' where
W
has
unity
1 and
0(1)
7^
0',
but 0(1)
for R'.
10. Mark each of the
a. Theconcept b.
A ring
c.
A
ring
d. Q is an
false.
of a ring homomorphism is closely connected with the idea of a factor R' carries ideals of R into ideals of R'. homomorphism 0 : R \342\200\224>\342\226\240 is one to one if and only if the kernel is {0}. homomorphism ideal
in R.
ring.
is
not
Part
244
V
Ideals
e.
Every
ideal
f.
Every
subring
Every
quotient
h.
The rings
is a subring of the ring. of every ring is an ideal of the ring. ring of every commutative ring is again a commutative
j. The conceptof an concept of a group. 11.
Let R be a ring. interest? Why?
12.
Give an
is to
ideal
{0} and
that a
factor
to
example
to show
that
a factor
to show
that
a factor
13. Give
14. Give
an example
15.
a subring
16. A
isomorphic. with unity 1 is
example
an
Find
R
that
Observe
show
of the
Z
Z x
ring
the
ring of an integral ring of a ring with
Assume
to prove that a quotient ring of a ring N for all r, 5 & R. The student starts out: rs = sr for allr.se R/N is commutative. Then
a. Why
does
of a
concept
factor
the
normal
rings R/R
is to
subgroup
and
the
of real
R/{0}
field.
be a
may
may have divisors of 0.
domain
of 0
divisors
an integral
be
may
domain.
x Z.
an ideal of Z
is not
that
the
Are
integral domain
of an
ring
ideals of R.
both
N.
if 1 e
only
ring as
of a
concept
are
R
if and
of R
all
is asked
student
ring.
Z4 are
Z/4Z and N in a ring
ideal
An
Rings
in a ring
g. i.
Factor
and
an ideal
R modulo
if and
is commutative
N
only
if
\342\200\224 e
(rs
sr)
the instructor
reading have written?
this
b.
What
should
c.
Prove
the assertion. (Note
the student
expect
\"if and
the
nonsense
R/N. from there on?
only if.\
Theory
17. Let
R =
b\\[2
a subring of R and that \342\226\240 , . 2b-] is\342\226\240 an isomorphism.
[a,
if
that
R', and
R,
composite
a commutative
be
R
(j>p(a) =
21. Let R
ap is a homomorphism R' be
and
unity
22. Let
R -> that
Show
b. Give an
c. Let 23. Let
F
f(x\\,
is
N'
an
rings
R' has
1 and
and
with
a ring
is an
ideal
ring is
R ->
homomorphism.
R' and
\\fr :
p.
-> R' be a ring homomorphism then \302\242(1) is unity for R'.
be
Show
R', where
or maps everything
that
Section
(f>p
{0'}. =\302\243
$[N] need not be or of R'. Show
of cf)[R]
either
an ideal that
Show
that
a factor
25.
Show
that
if R
cj>~1
[N1]
is an
: R \342\200\224>\342\226\240 R given by
Show
that if R has
the
set Ns of all Section 22)
ideal of R.
be
,xn]
ring of a field
is a ring
then the
of R'.
\342\226\240 \342\226\240 x F for n factors. Show of F x F x \342\226\240 that \342\200\242\342\200\242 of have element S as a zero Exercise \342\200\242, (^, every (see ,xn) an) \342\226\240 This is of in F[x\\, \342\226\240 in algebraic \342\226\240, xn]. importance geometry.
24.
0.
onto
ideal of cf)[R].
a field,
ideal
=
13.)
let S be and subset any \342\226\240\342\200\242\342\200\242 \342\200\242\342\226\240 & that
F[xi,-
Show
+ bsl%
homomorphisms,
the map
0[i?]
Z.
of R.
ideal
an
R ->
a, b e
for
yt
->\342\226\240 R\" are
R'
such that
: R let \302\242)
and let N
cj> :
(Use Exercise 49 of
of prime characteristic Frobenius homomorphism).
homomorphism
that
to one
one
either
form
of the
show
unity (the
to show that
example
is a
x 2 matrices
of M2(Z). Then
to a
and if
rings, R\"
0 divisors,
no
R' be
be an
->
ring
a subring
from a field are
R\"
\\jf
function
20. Let
R' is
each homomorphism
that
Show
19. Show
a.
and let R' consist of all 2
e Z}
\\a,b
R is
that
18.
{a +
with
is either
the
unity and N is
trivial an
ideal
(zero)
of R
ring of
such
that
one elementor is isomorphic N
R, then =\302\243
R/N is a ring
28 of
to the with
field. unity.
26. Let
27. Show
28. Let
a commutative
be
R
R' be
R'. Show
a
0 induces
natural
{x e W | ax an ideal of R. again Ia =
that
Show
of a ring
and W
let N
and
rings
that
e R.
let a
and
of ideals
an intersection
that
R and
into
ring
R is
Prime and Maximal
27
Section
be ideals
R and
of
= 0}
is
an
245
Ideals
of W.
ideal
/?', respectively.Let 0 be a homomorphism ->\342\226\240 if 0[A^] c W. (UseExercise /?'/Af'
of
0* : R/N
homomorphism
W
39 of
Section 14.)
29. Let0 be a homomorphism is a unit in R'. 30. An element a of a ring elements in a commutative
31. Referring the
32.
to Exercise
the
ideal
33. Let R 34. Let R some
be e
36. What is the Show
30, find
3. What if N
that
is the
a,
of Z?
unit
be a
that o(;/1
W. Show
in
of all nilpotent
collection
the
that
ring Z12 and
of the
nilradical
is the nilradical
that it
observe
is one of
of Z32?
of a commutative
R, then
ring
R/N
as nilradical
has
ring
and
if every
that
element of N
the nilradical
34, show
that for
examples
by
proper ideals N b. -JN
N
equal
of the
relationship
ideal ~JN of Exercise
of a
the nilradical
34 to
commutative equal
may
e R,
such
that
a\" & N
for
ring R,
N.
of R/N (seeExercise
30)?
Word
your
carefully.
:C^
that 0
given
Afc(R)
by
*(* + for
the
nilradical
w
N}.
to Exercise
a. */N neednot
37.
ideal,
Exercise
in
Exercise
R'. Let
ring
to Exercise N an ideal of R. Referring 30, show of R/N is R/N, then the nilradical of R is R. that the set */N of all a a commutative ring and N an ideal of R. Show of N. Z+, is an ideal of W, the radical
Referring
answer
an
a nonzero
= 0 for some n e Z+. Show the nilradical of R.
if a\"
nilpotent
30, show
{0 +
and
n
in
unity onto
R with
ring R is
a commutative
be
is nilpotent
35.
found
Referring trivial
R is
definition given
to the
of Z12
ideals
of a ring
fc
e K
38. Let fitea Section
of C with
gives an isomorphism ring
24. Let
R,
be the
a. Show b. Show
of endomorphisms
of (R, +)
as described
in
=ax
R. that
Aa is an
that
R' =
c. Prove the SECTION
ring
by
Xa(x)
for* e
\302\273)
0[C] of M2(IR).
subring
unity and let End((W, +)) and let Xa : R ^ Rbe given
with
a e
the
K)=(4
{Xa
analogue
27
endomorphism of (/?,+). of End((/J, | a e tf} is a subring +)). of Cay ley's theorem for R by showing
12 through
Exercises
R/N
some
27.1 Example
As
where
examples
was
for p
of (b) is isomorphic
section
asked us to structural
to R.
and Maximal Ideals
Prime
rings
that R'
shown
a prime.
of
14 of the R and R/N
this
situation,
in Corollary a factor
Thus
preceding have
very
and in
different
provide
which is domain
of factor
properties. We start
the process, provide solutions
19.12, the ring Zp, ring of an integral
examples to those
isomorphic to
may be afield.
Z/pZ,
with
exercises. is a
field A
Part
246
V
27.2 Example
Factor
and
Ideals
The
Z is
Z x
ring
Rings
not
an
for
domain,
integral
(0,1)(1,0) = that
showing
27.3
Example
Z)/N is
ring
= {0,3} of
The subset N
elements, 0 + N,l show that Z(,/N ~
+ N, Z3
(0 +
to Z
isomorphic
Thus a factor is not.
meZ.
where original
are 0divisors.Let
and (1,0)
(0,1)
Z, and (Z x
Z x
of
ring
showed This the
a
that
example
Every
nonzero
and
is easily 2 + N.
Z6
While factor
27.5 Theorem
Proof
examples
factor
ring of
If R
then
Corollary
Proof
an
domain,
integral
of
seen
be
to
of Z(,, and
an ideal
These add and
multiply
has
Z(,/N
a
in such
fashion
three as
to
N)**0,
(2+A0\302\253*2.
(1+A0\302\253*1,
an
that is,
domain,
integral
ifR has zerodivisors, A
The examples preceding better than the original ring. seem worse than that of may
not.
ring R has
at
to R.
of
indicate,
rings
rings of a
and integral
Corollary 27.6,
a field is really not
are R/R,
ring
useful
ring with unity, and N is
an
R
is
and the trivial one which element, and only Just as for a subgroup cases. an ideal N of R such that N ^ R
domains follows
which
ideal R
the improper
two ideals,
These are uninteresting
ideal
nontrivial
rings
least
factor
has
may
be
interest, as the theorem, shows that a
of great
our next
to us.
ideal
of R
containing a unit,
then
N =
R.
LetN be an rN
27.6
is a
be
ideal
+ N] -o-m, even though the
[(m, 0)
the correspondence
under
theseideals,the
a group, a proper N ^ {0}.
above
Now N is an
\342\226
and
is isomorphic
which
e Z}.
\\n
\342\226
ring.
ideal {0}. For of
{(0, n)
correspondence
~ Z^ is integral domain, but Z/6Z factor ring may have a structure that seems indicates that the structure of a factor ring
original
'R/{0},
= the
Z is an
that
Note
N
under
a ring may
This example shows that ifR is not even it is still possible for R/N to be afield.
27 .4 Example
(0,0),
of R, and suppose that u e N for some unit uin R. Then the condition ideal c N for all r e R implies, if we take r = u~l and u e N, that 1 = u~l u is in N. But c rN N for all r e R implies N = R. that r 1 = r is in N for all r & R,so \342\231
A field contains
no
proper
ideals.
nontrivial
Since every nonzero elementof a field is a unit, an ideal of a field F is either {0} or all of F.
it follows
at once from
that
and
Maximal We
now
take
case
in which
with
\342\231
Ideals
up the question
domain. The analogy the
Prime
Theorem 27.5
of when
groups
the factor ring
is
a factor
in Section a field.
ring is a field and when 15 can be stretched a bit
it is an integral further to cover
Prime and Maximal
Section 27
27.7 Definition
A
27.8
Example
of a ring R is an ideal M containing M.
ideal
maximal
idealN of
R such that
from
different
247
Ideals
there is no proper
R properly
Let p
be a prime
about
multiplication
We know
integer.
positive
\342\226\240
that
is isomorphic and Zp
Z/pZ
moment and regarding and consequently Zp is a simple group, for the
Z/pZ
to Zp. as
Forgetting
additive
groups,
normal pZ must be a maximal 15.18. Since Z is an abelian Theorem is a subgroup by group and every subgroup of Z. Since pZ is an normal subgroup, we see that pZ is a maximal proper subgroup that pZ is a maximal of Z. We know that Z/pZ ideal of the ring Z, it follows ideal is
we know
that of Z
isomorphic
27.9
Theorem
(Analogue maximal
Proof
to
the
ring
the next
illustrates
Zp, and
that
a commutative is a field.
R be
Let
R/M
in
suppose
Conversely,
Observe
if
that
of R
ideal
Since Z/nZ the
Corollary Proof
unity.
a
Then M is
R is
with
commutative
such
that R/M that
M
is a c
c N
field.By the R
and
of Section final paragraph is the canonical y homomorphism {(0 + M)} cy[N]c R/M. But
onto R/M, then y[AT| is an ideal of R/M with states that the field contrary to Corollary 27.6,which ideals. Hence if R/M is a field, M is maximal.
27.11
with
ring
R/M.
N is any
Example
This
a commutative ring with if M ^ R, which is unity ring so M with a that a is not e + + (a M) R/M, ^ M, the additive inverse identity element of R/M. Supposea + M has no multiplicative does not in R/M. e Then the set (R/M)(a + M) = {(r+ M){a + + M) | (r M) #/M} of It is an ideal is 1 + M. We easily see that nontrivial contain + M) R/M. (R/M)(a it does not contain because 1 + M. By the because a \302\242 and it is a proper ideal M, \342\200\224>\342\226\240 if : R is of Section the canonical final 26, y homomorphism, then paragraph i?/M M. is ideal of R But this contradicts a + M)] proper properly containing y~l[(R/M)(a our assumption that M is a maximal ideal, soa + M must have a multiplicative inverse in R.
ideal
maximal
unity, then R/M is also a nonzero Let the case if M is maximal.
27.10
Z/pZ is a field.
\342\226\262
of Theorem 15.18) if ideal of R if and only
Suppose M is a
a field. Thus
is actually
Zp
theorem.
maximal
R/M contains no
proper
26, if of
R
this
is
nontrivial \342\231\246
and Z\342\200\236 if n is a prime, we see that is isomorphic to Z\342\200\236 is a field if and only ideals of Z are preciselythe ideals pZ for prime positive integers p. \342\226\262
A commutative
ring
with
unity
is a field if and
only
if it
has no
27.6 shows that a field has no proper nontrivial if a commutative ring R with unity has no ideal and R/{0], which is isomorphic to {0} is a maximal
Corollary
Conversely,
proper nontrivial
ideals.
ideals. proper R,
is a
nontrivial
field
by
ideals, then
Theorem
27.9.
We now turn to the question of characterizing, for a commutative R with unity, ring N ^ R such that R/N is an integral domain. The answer here is rather obvious. The factor ring R/N will be an integral domain if and only if {a + N)(b \342\200\224 A\") = N the
ideals
implies
that
either
a +
N = N
or
b+ N
=
N.
Part
248
V
Factor
and
Ideals
Rings
This is exactly the statement of 0, since the coset N plays that R/N has no divisors the role of 0 in R/N. at representatives, we see that this condition amounts to Looking that ab e N implies that either a e N or b e N. saying
21.12Example
are of the form nZ. For n = 0, we have nZ = {0}, and Z/{0} ~ Z, which and Z\342\200\236 is an integral domain if domain. For n > 0, we have Z/nZ ~ Z\342\200\236 integral if that and n is a the nonzero ideals nZ such is an integral domain Thus Z/nZ only prime. a field, so that pZ are of the form pZ, where p is a prime. Of course, Z/pZ is actually is a maximal ideal of Z. Note that for a product rs of integers to be in pZ, the prime \302\243> in this example makes the use of the must divide either rors. The role of prime integers \342\226\262 in the next definition more reasonable. woidprime of Z
ideals
All
is an
27.13
Definition
An ideal N
^
commutative
in a
R
ring
is a
R
ideal if ab
prime
e
TV
that either
implies
a&Norb&Nfora.b&R. that
Note
27.14 Example
Note that
x {0}
Z
= 0 in
have bd
Z x
{0} is
{0}.Note
\342\226\240
a prime
ideal
and indeed, in
in Z,
integral
any
domain.
ideal of Z x Z, for if (a, b)(c, d) e Z x implies that either T> = 0 so (a, b) e Z to Z, which is an (Z x Z)/(Z x {0})is isomorphic
is a prime
that
{0},then
x {0}ord = 0
Z. This
integral
we so
must
d) e domain. (c,
\342\226\262
remarks preceding Example is illustrated by Example 27.14. \\V,hich Our
27.15 Theorem
Let R an
27.16
Corollary
Proof
a commutative
be
ring
if and
domain
integral
Every maximal ideal in
If M is maximal is a prime ideal
unity
a field, hence an
of the
ideal
an
is a
in R.
Then R/N is
prime ideal. domain,
integral
theorem,
following
and therefore
been presentedregarding maximal and prime ideals is the main ideas well it quite a lot. We should keep of maximal understand the definitions and prime ideals and facts that we have demonstrated.
shall be using
mind. We must know and must remember the following in
For a commutative
ring
R with unity:
is maximal if
1.
An
ideal
M of R
2.
An
ideal
TV
3.
Every maximal idealof
of
M
\342\231\246
that has just and we
important
ring R with
a commutative
R, then R/M is Theorem 27.15. by
a proof
and let TV ^ R be prime ideal in R.
unity,
if TV is a
only
in
material
The very
with
27.12 constitute
R is
prime if R is
and
and
only
only
a prime
if R/M is
if R/N
ideal.
is
an
a field. integral
domain.
Section27 Prime and Maximal
Fields
Prime
We now with
n
=
0.
If
R
is a
the
to show that
proceed
rest, and
unity
ring with unity n > 0, and (-1)
27.17 Theorem
1. Recall
+ (-1)
ring with
e Z is a homomorphism
for n
upon which all
foundations
form Z\342\200\236
rings
fields. Let R be any \342\200\242 \342\200\242 1 we mean 1 + 1 + \342\200\242 that by n \342\226\240 + 1 for n summands for n \342\226\240 1 = 0 for for n < 0, while + --- + (-1) for \\n\\ summands
the map
1, then
unity
Z and
rings
Zp perform a
Q and
that
: Z \302\242)
of Z
into R.
m)
\342\226\240 1 = (ft
for all
service
similar
0(n) =
Proof
249
Ideals
\342\200\224>\342\226\240 R given
by
\342\200\242 1
n
Observethat
The distributive
=
+ m)
0(n
1 +
+
show
in i?
laws
(1 +
(ft
n summands Thus
l)(m
for
that
(m
0(n) +
1)
0(m).
= (1+1+ -+ nm
summands
\342\226\240
1)
summands
> 0. Similar arguments
ft, m
\342\226\240 1 for
(nm)
1)
all n, m
1 + -+1)
m
\342\200\242\342\226\240 =
(n
laws show
1) +
that
1)(1 +
---+
\342\226\240 =
\342\200\242
the distributive
with
we have
e Z,
(ft
\342\200\242\342\226\240 =
\342\226\240 1.
(nm)
1)
l)(m
Thus
(p(nm) = 27.18
Corollary
If R to
Proof
is a ring
If R Z\342\200\236.
0(m) =
: Z \342\200\224>\342\226\240 i? given by
A field
must be
The kernel
to
isomorphic Z.
If the
characteristic of
isomorphic characteristic
ideal
m
\342\200\242 1 for
in Z. All
\342\231\246
F is either of prime
to
an
Z.
of characteristic0 and Proof
\342\231\246
1)
l)(m
with unity and characteristic n > 1, then i? contains a subring a subring isomorphic to has characteristic 0, then R contains
is isomorphic Theorem
\342\200\242\342\200\242 =
m e Z is a homomorphism by ideals in Z are of the form sZ for some 5 e Z. By Theorem19.15we see that if R has characteristic n > 0, then the kernel of
27.19
1 = (n (nm) \342\226\240
contains
a subfield
F is not
0, the
n must
contains a
isomorphic
to
subfield
p, or F
would
isomorphic
isomorphic
to Zp
or
Q.
above corollary shows
be a prime F must contain a subring
Then Z\342\200\236. 0, then
p and
characteristic
F
that
have
to Z.
contains
0 divisors.
In
this
case
a subring
If F is Corollaries
of
PartV
Ideals and
show
21.8
and 21.9
field
of quotients must
contain
must
F
that
be isomorphicto
a field
which
in
Structure
Ideal
subring
and
this
that
to Zp
isomorphic
Q are the
for some prime
fundamental
p
ova
blocks on
building
prime fields.
Q are
and
of this
\342\231\24
every
The fields Zp
of quotients
Q.
field contains either a subfield to Q>. These fields Zp and isomorphic all fields rest.
Thus subfield
27.20 Definition
Factor Rings
\342\226\24
F[x]
that F is a field. We give the next definition this section, we assume in the case for a general commutative ring R with unity, although we are only interested a R = F[x]. that a i? with and e Note for commutative ring R,theset{ra \\r e R] unity is an ideal in R that contains the element a.
Throughout
27.21Definition
If R of a
is a commutative ring with unity and a e R, the ideal {ra \\ r e R] of all multiples is the principal ideal generatedby a and is denoted by {a). An ideal N of R
ideal
principal 27.22
27.23
Example
Example
rest of
the
{a) for
=
N
if
of the ring is a principal ideal.
Z is
The ideal
consists
Every ideal
{x) in
of
some a form
the
is a
e R.
nZ,
\342\226\24
which is
generated by
n,
ideal of Z
so every
\342\226\26
F[x]
of
all
in F[x]
polynomials
having zero
term.
constant
\342\226\26
The
for
algorithm
division
27.24 Theorem Proof
If
F
as
simple
the
important application of the of this theorem is to
a subgroup
that
proof
but very
23.1.) The proof
of a cyclic group
is
division the
cyclic
division
is to
the
in Z.
algorithm
is a field,
another
(See Theorem
F[x].
in F[x]
algorithm
is
theorem
next
every ideal in
is principal.
F[x]
If N = {0},then N = (0}. Suppose that N ^ {0}, and let g(x) If the degree of g(x) is 0, then of N of minimal e F degree. g(x) = and is a unit, so N \342\200\224 is If Theorem soN the of 27.5, F[x] (1) by principal. degree g(x) is >1, let f{x) be any element of N. Then by Theorem 23.1, f(x) \342\200\224 + r(x), g(x)q(x) where r(x) = 0 or (degreer(x)) < (degree Now f(x) e N and g(x) e N imply g(x)). = that f{x) \342\200\224 an a nonzero is in N definition of ideal. Since is g{x)q{x) r{x) by g(x) elementof minimal degree in N, we must have r{x) = 0. Thus f(x) = g(x)q(x) and
Let N be an
be a
N =
ideal
of F[x].
element
nonzero
(g(x)). We
achieving
can
\342\231\24
now characterize
our basic
zero in some field
goal:
E
containing
to
the
F.
ideals of F[x]. This any nonconstant polynomial
maximal
that
show
is a crucial step in f{x) in F[x] has a
27.25
Theorem
Proof
ideal
An
{0}of F[x] is
{p(x)} ^
Prime and
27
Section
if and
maximal
only
if
251
Maximal Ideals
p{x)
over F.
is irreducible
Then {p(x)) ^ F[x], so p(x) \342\202\254 F. {p(x)) ^ {0} is amaximal idealof F[x]. = a be factorization of in Since a maximal is F[x]. p(x) f(x)g(x) p{x) {p{x)) ideal and hence also a prime e {p(x)} implies that f(x) e {p(x). or ideal, (f(x)g(x)) a t hat as e either or has factor. But then can't have we is, g{x) {p(x)}; f{x) g{x) p{x) the degrees of both f{x) and g{x) less than the degree of p{x).This shows that p(x) is
Supposethat
Let
irreducible over F. if
Conversely,
ip(x))
p{x)
q{x)
Example
{p(x)}. Thus {p(x)}C N
Example
is a {x2
supposethat
ideal
N
ideal such
is an
so N
that
= {g(x))for
27.24, F[x]. by that p{x) = g(x)q(x) for some q(x) e F[x].But .Then p(x) e N implies is irreducible, which implies that either g{x) orq(x) is of degree0.If g(x) is of degree a constant in F, then g(x) is a unit in F[x], so {g(x)) = N = F[x]. If nonzero is, = c, where c e F, and g(x) = (l/c)p(x) is in (/?(*)), so is of degree 0, then q{x)
N =
27.26
over F,
is a principal
N
Now
Theorem
e N
someg(x)
0, that
is irreducible
p{x)
cJVc
23.9 shows
Similarly, \342\200\224 2) is a field.
We
to
Application
x3
that
Theorem
field.
shall
c
F[x]
so {p(x)}is maximal.
is impossible,
\342\231\246
+ 2 is irreduciblein Z5[x], so Z5[x]/{x3 + 3x + 2) 2 is irreducible in Q[x], so Q[*]/ 22.11 shows that*2 \342\200\224 such fields in more detail later. \342\226\262 examine + 3x
in F[x]
Factorization
Unique
we stated without Theorem 27.27, which follows. (See proof this theorem, we provedin Section 23 that factorization of in F[x] polynomials into irreducible polynomials is unique, for order of factors and except in F, We delayed the proof units of Theorem 27.27 until now since the machinery we us to give such a simple, four-line proof.This proof fills the gap have enables developed in F [x]. in our proof of unique factorization In
Section
23,
Theorem 23.18.)
21.IH Theorem
Proof
Assuming
Let p(x) be an irreducible F[x], then either p{x) Suppose
p(x) divides
Theorem 27.25.
If p{x) divides
in F[x].
polynomial
divides r{x)
or p{x) divides
r(x)s(x). Then
e {p(x)},
r(x)s(x)
{p{x))imphes
that
which is maximal
Corollary
divides
Hence
27.16.
r(x), or
that
s(x)
by
Preview
We
close
goal.
We
e {p(x)},
from this Basic
Show
that
this
\342\231\246
Basic Goal
of Our
have
e
r(x)s(x)
giving p(x) divides s(x).
A
s{x) e
s(x).
{p(x)} is a prime ideal by either r(x) e {p(x)}, giving p(x)
Therefore,
for r(x),
r(x)s{x)
section with an outline all the ideas for the proof
of
demonstration
the
at hand
in Section
now; perhaps you
can
29 of our basic the details
fill in
outline.
goal: there
Let F exists
be a
a field
field
E
and
containing
let f(x) F
be a nonconstant
and containing
polynomial
a zero
a of f(x).
in F[x].
252
Part V
Factor
and
Ideals
Rings
Outline of the
1.
Let
2.
Let E
3.
Show that no
be an
p{x)
be the
irreducible factor of f{x) in
homomorphism
f(x)
in
x +
coset
the
4>a
are in
27.25 and the
{p{x)) in
27.9.)
coset of
same
consider F to be (isomorphic
that we may
4. Let a be
of F
elements
different
two
F[x].
(See Theorems
F[x]/{p(x)}.
field
deduce
and
Proof
to)
that for the evaluation have 0O(/(*)) = 0. That
Show
E.
\342\200\224>\342\226\240 E, we
: F[x]
F[x]/{p(x)), of E.
a subfield
is a zero
is, a
of
E.
An example of a field constructed accordingto this outline is given in Section 29. and multiplication tables for the field Z2[x]/{x2 + x + 1). We There, we give addition the cosets show there that this field has just four elements,
1),
+x +
0+{x2
\\
x + {x2+
{x2 +x+\\),
+
x + l),
and
(x
+ \\)+{x2+x
a,
rename these four cosets 0, 1, and 29.21 for addition and multiplication
We
are a{\\ and
+
\\).
and obtain Tables a + 1 respectively, 29.20 in this 4-element field.To see how these tables a + a = that we are in a field of characteristic2, so that constructed,remember = \342\200\224 0. Remember a0 also that a is a zero ofx2 +x + l.sothata2 +a + 1 = 0 + 1) \342\200\224 = = a2 \342\200\224aI a + I. consequently and
27
\342\226\240 EXERCISES
Computations
1.
Find
all prime
2. 3.
Find
all prime
ideals and
Find
all prime
ideals and
all prime
ideals and
all
4. Find 5. Find
ideals and
maximal
ideals
of Ze-
all
maximal
ideals
of Zn-
all
maximal
ideals
of Z2
maximal
ideals
of
all
is a field.
all c
e Z3 such
that
Zt,[_x]/(x2
+ c)
6.
Find
all c
e Z3 such
that
Z3M/U3
+ x2 +
7.
Find
all c
e Z3
that
Z3 [x] /(x3
+ ex2+
that
Z5[x]/{x2
+ x
that
Z5M/(.x2
8. Find
all
9. Find
all c
such
Z5 such
c e
e Z5
such
x Z2. Z2 x Z4.
c) is a field.
1)is
a field.
+ c) is a field. + ex + 1) is a field.
Concepts In
Exercises
is needed,
10 through 13, correct the definition of the it is in a form acceptablefor publication.
italicized term
to the
without
reference
other
ideal of R.
text,
if
correction
so that
10.
A maximal
11.
Aprime
ideal of a ring
ideal
R is
of a commutative
an ideal
ring
R
that
is an
is not
contained
ideal of
the
form
in
any
pR
= {pr | r
e
R}
for
some
prime p.
Exercises
Section 27
12. 13.
A
prime
A
principal
subfields.
no proper
is an ideal
unity
with
N
the property
there
that
exists
a e
.V
such
smallest
the
of the
each
has
that
of a commutative ring with ideal that contains a.
ideal
is
that N
14. Mark
is a field
field
253
following
or false.
true
a. Every
prime ideal of every
b.
Every
maximal
c. Q is its
own
d. The prime
of every
ideal prime
field
f.
with zero
commutative
is a
unity
with unity
ring
maximal ideal.
is a prime ideal.
subfield. of C
subfield
e. Every
ring with
commutative
is E.
a subfield
contains
isomorphic
to
a prime
field.
as a subring. divisors may contain one of the prime fields a subfield g. Every field of characteristic zero contains isomorphic to Q. h. Let F be a field. Since F[x] has no divisors of 0, every ideal of F[x] is a prime i. Let F be a field. Every ideal of F[x] is a principal ideal. ideal of F[x] is a maximal ideal. j. Let F be a field. Every principal A
ring
15. Find
a maximal
16.
Find
a prime
17.
Find
a nontrivial
ideal of
ideal.
Z x Z.
Z x Z that is not maximal. proper ideal of Z x Z that is not
ideal of
18. Is Q[x]/(x219. Is Q[x]/(x2-
prime.
field? Why?
5x
+ 6) a
6x
+ 6) a field?
Why?
Proof Synopsis
20. Give 21. Give 22. Give
of Theorem 27.9. of Theorem 27.9.
or two-sentence
synopsis
a one-
or two-sentence
synopsis
of \"if
a one-
or two-sentence
synopsis
of Theorem
27.24.
synopsis
of the
if\" part
a one-
23. Givea one-or
two-sentence
if part
of \"only
part
\"only
of Theorem
27.25.
Theory
24. Let
25. Corollary
26.
Zp and
28.
29.
25, is it possible
Exercise
fields Zp
27. Following
Show
that every ring with unity with unity may simultaneously give an example. If it is impossible,
is possible,
to the
unity.
a ring
that
Continuing
with
ring
tells us
27.18
possible
If it
finite commutative
R be a
the
TLq for
and
7Lq for
of Exercise
idea
p ^q
that
a ring
two different primes
and p
and
26, is
it
possible
q both prime?
with
that
every
prime
ideal in R is a maximal
a subring isomorphic contain two subrings isomorphic prove it.
contains
unity
p and
ql
simultaneously Give an example
to
either
to
ideal.
Z or
some
and Zm Z\342\200\236
contain two subrings
may
or prove it
to contain for an integral domain Give reasonsor an illustration.
Is it Z\342\200\236.
for n ^ml isomorphic
is impossible.
two subrings
isomorphicto
Prove directly from the definitions of maximal and prime idealsthat every maximal ideal of a commutative ring R with unity is a prime ideal. [Hint: M is maximal in R, ab e M, and a g M. Argue that the smallest Suppose a and M must contain 1 as ra + m and multiply ideal {ra + m \\ r e R, m e M) containing 1. Express by b.] Show
that
no proper
N is a nontrivial
maximal ideal ideals.
in
R if and only if R/N is a simple ring, with Theorem 15.18.)
a ring
(Compare
that
is, it
is nontrivial
and
has
254
Part
30. Prove that
and f(x),
field
and let
is an
ideal of
both
be irreducible
F[x]. Show
g(x) e
that
N =
{r{x)f{x) +
if f(x)
and g(x) have
to
of Algebra:
Theorem for
of
is a sort
34. If A
Let f\\{x),
C[x]:
\342\200\242 \342\200\242 \342\200\242, f,(x)
ideals of a ring
35. Let
A and
B is
A +
that
Show
B be
an
ideals of a
ring
36. Let
AS is an
that
A and
S be
ideal
ideals of a
that
A
37. Show
that
for a field
S is
F,
the
in
cannot
and g{x)
/(*)
in
B of A
B =
define sum,
exercises
defined
B is
and
{a + b | a
a e
C
C
in
is a
that
of g(x) is in
zero of
the
all
smallest
r of
ideal
e A,
b e
of A
and
and
quotient
of ideals.
by
B}.
that
S is
product,
A c
A+
defined
B and
B c
A+
B.
by
n
e A, fc,- e S,
n e Z+
b. Show
AS
R.
ring R. The :
a zero
has
every
some power
Then
C[*].
three
y^a,fc; | a,I '=i
commutative
an ideal
Show
The next
= i
A :
then
\342\200\242 \342\200\242 \342\200\242, /,\342\226\240(*).
product AS
R. The
in C[x]
and suppose that
b. Show
f
Show
F[x],
theorems:
polynomial
ideal.
AB
a.
e {f(x)).
only if g(x)
and
if
and N ^
degrees
two
e C[x]
g{x)
sum A +
R, the
A +
a.
different
nonconstant
Every
in a ring.
of ideals
arithmetic
S are
and
divides g(x)
s(x)g(x)| r{x),s{x)e F[x]}
of these
the equivalence
prove
these polynomials is also a zeroof a polynomial of C[x] that contains the r polynomials f\\{x), There
F[x] is maximal.
ideal of
prime
over F.
33.Use Theorem27.24 Nullstellensatz
nontrivial
F[x]. Show that f(x) /(*), g(x) e F[x].Show that
field
Fundamental
Rings
is a field, every proper
if F
31. LetF be a 32. Let F be a
Factor
and
Ideals
V
S = {r
e
tf
A : S
quotient
|
e A
rb
that
c (A
n S).
of A by S is
for all b e
defined
by
S}.
of i?. set
S of all
matrices of the
form
(oo)
for a, b
e F is a
multiplication
38.
Show
section
that
the
28
but not a
ideal
right
on the
right
matrix
ring M2(la) is a
by
any
section with
concerned
zeros of a section
of
book by
This
element
this
that
is, MiJJLj)
has no proper
nontrivial
closed under
ideals.
for Ideals In particular, we are we can of the set of common accomplish our goal in a single
gives a brief introduction to algebraic geometry. the problem of finding a description as simple as finite
of polynomials. In order to be stating a few theorems without We recommend proof. and Loustaunau for the and further [23] proofs study.
number
text, we
Adams
section
of M2(F). That is, show that S is a subring of M2(F), but is not closed under left multiplication. ideal
simple ring;
Bases
^Grobner This
left
will
is not usedin the remainder
of
the text.
the
28
Section
Recall
a field.
be
F
Let
terminants
\342\200\242 \342\200\242 x F for F x \342\200\242
F x
that
the ring
\342\226\240 \342\226\240 \342\226\240is
,xn]
F[X\\,X2,
\342\226\240 \342\226\240 \342\226\240 with ,x\342\200\236
X\\,X2,
255
Ideals
and Ideals
Varieties
Algebraic
Grobner Bases for
in F.
coefficients
We let
of
Cartesian
the
n inde-
in
polynomials
be
F\"
product
For easein writing, we denote an element {a\\, ai, \342\200\242\342\200\242\3 \342\200\242\342\226\240\342\200\24 For ,xn]. Using similar economy, we let F[x] = F[X\\,X2, as in each a g f\", we have an evaluation F just homomorphism 0a: F[x] \342\200\224*\342\226\240 e That is, for /(x) = f(xu x2,\342\226\240 Theorem 22.4. \342\226\240, x\342\200\236) F[x], we define 0a(/(x)) = /(a) \342\200\224 \342\200\242 \342\200\242 The follows from the ci2, \342\200\242, an). proof that 0a is indeed a homomorphism f(al, and distributive properties of the operations in F[x] and F. Just as commutative, associative, = 0. a of F\" is a zero of /(x) e F[x] if /(a) for the one-indeterminate case, an element of
F\" by
In what
a,
bold
in
n factors.
type.
abbreviate follows, we further section we discuss the problem
In this
the
28.1 Definition
polynomials set of all these
Let S be a
use*, y, 28.2Example
F\"
examples, which of Xi,x2, and^3.
z in place
-
2} c
R[x, y].
y
-intercept
2.
ring
I = ideal of
29 the
to Exercise
\"We leave
in a commutative
i? =
polynomials
illustrative
{2x + y a:-intercept 1 and
is an
of the
R. We
with
R
subject
and
of algebraic
The
of a finite
properties
geometric
studying
usually
algebraic
of
geometry. in F\"
V(S)
is
the
set
of all
S.
in
\342\226\240
at most three
involve
variety
indeterminates, we
in E2 is
V(S)
the
line
with \342\226\262
proof
straightforward
unity,
that for r
elements /1,
\342\200\242 \342\200\242, /2, \342\200\242 /r
the set
\342\226\240 \342\226\240 + Crfr {clfl + cih + \342\226\240
denote
f(x) by \"/\342\200\242\" common zeros in F\"
The algebraic variety
of F[x].
subset
finite
S =
Let
F[x]. Finding
common zerosis the
common zeros in In our
\342\200\242 \342\200\242in \342\200\242, /r
/i, /2,
numberof
a polynomial of finding
this ideal by all the /,-
for i =
e R I c,\342\226\240
(/i, /2,---,
fr)- We
are polynomials
in
\342\200\242 \342\200\242, 1, \342\200\242 r}
are
interested
in the
case
F[x] F[x]. We regard the c,- as \"coefficient this I is its ideal the smallest ideal containing construction, polynomials.\" By \342\200\242 it can also be describedas the the polynomials /1, /2, \342\200\242 intersection of all ideals \342\200\242, /r; 28.3 Definition
where
containing
these
Let I be an basis for /
if
Unlike
ideal I =
all
the
Proof
and
r polynomials. in a commutative
ring
R
with
unity.
A subset
{bl, fc2,
(b\\, b2,---,br).
the situation
elements of a basis,or of 28.4 Theorem
c,-
in linear unique
\342\200\242 of \342\200\242 \342\200\242, br}
I is a \342\226\240
of algebra, there is no requirement of an ideal member in
representation
for independence terms of a basis.
\342\200\242e Let /1, /2, \342\200\242 in F\" of the polynomials zeros \342\200\242, F\\x\\. The set of common /r f-t for \342\226\240 \342\226\240 i = 1, 2, \342\226\240, r is the same as the set of common zeros in F\" of all the polynomials in \342\200\242 the entire ideal I = (/1, /2, \342\200\242 \342\200\242, fr)-
Let
/ =
c1/1+c2/2 + --- + crfr
(1)
256
Part
V
Ideals and
be
Factor Rings of I,
any element
the
evaluation
= Cl(a)/i(a)
/(a)
polynomial
an ideal I in We can summarize
every
we let
F[x],
Applying
\342\200\242 \342\200\242 \342\200\242
+
+
the
cr(a)/r(a)
= 0,
\342\200\242 \342\200\242 \342\200\242
cr(a)0
f in I.
becauseeach
V(/) be 28.4
Theorem
+
polynomial
of each fi
a zero
be
will
For
of I.
zero of
also a
a is
in I
+ c2(a)/2(a)
+ c2(a)0+
= ci(a)0 that
\342\226\240 \342\226\240 and fr. \342\226\240,
f2,
fi,
a to Eq.
homomorphism
showing
be a common zero of (1), we obtain
a e Fn
let
and
fi
of all
set
Of course, a zero
of
every
e I.
\342\231
common zeros
of all
elements
as
V({fl,f2,---,fr})=V((fl,f2,---,fr)).
We state without 28.5 Theorem
the Hilbert
proof
(Hilbert BasisTheorem)
Our
ideal
Every
Given a
objective:
Basis Theorem. (See Adams
basis for
basis that better associated algebraicvariety
become a
exhibits
in F[xi,
x2,
Loustaunau
and
\342\226\240 \342\226\240 has \342\226\240, xn]
a
finite
[23].)
basis.
ideal I in F[x], modify it if possible to the structure of I and the geometry of the
an
V(I).
that follows provides a tool for this task. You should notice that the information the division in about gives algorithm that we did not mention Theorem 23.1. We use the same notation here as in Theorem but with x rather 23.1, = g(x)h(x) in F(x), then g(x) and h[x] arecalled\"divisors\" than x. If f(x) or \"factors\" theorem
The
theorem
of/(x).
28.6 Theorem
(Property of in F[x] such are the same
the
Division
that
f(x)
Let f(x), g(x), q(x) and r(x) be polynomials zeros in F\" of f(x) and g(x) + r(x). The common of g(x) and r(x). Also the common divisors in F[x] as the common divisors of g(x)and r(x). members of a basis for an ideal I of F [x], then replacement
Algorithm)
= g(x)q(x) common zeros
as the
of /(x) and g(x) are the same If /(x) and g(x) are two of f(x) by r(x) in the basis still Proof
If a
e Fn is f(x)
equation so
a is
a zero
then
applying
r(b)
= 0 The
a common
zero
= g(x)q(x)
of both b
I.
and r(x),
then
of g(x)
+ r(x), we obtain If b = g(b)g(b)
and g(x).
f(x)
yields
yields a basis for
/(b)
/(a)
applying
= g(a)g(a)
e F[x] is a common + r(b) so0 = 0g(b)
a
both
to
of f(x)
zero +
sides of
the
= 0g(a)+ 0 = 0,
+ r(a) r(b)
and g(x),
and we
see that
as well as g(b). proof
concerning
common
divisors is essentiallythe
same,
and is
left as
Exercise 30.
let B be a basisfor an ideal I, let f(x), g(x), e B and let f(x) = g(x)q(x) + r(x). Let B' be the set obtained by replacing f(x) by r(x) in B, and let /' be the ideal B' as a basis.Let Sbe the set obtained from B by adjoining r(x) to B. Note that having S can also be obtained by adjoining to B'. The equation f(x) = r(x) f(x) Finally,
g(x)q(x)+
that
shows
so we have
e /',
/(x)
r(x) = /(x) \342\200\224 q(x)g(x) for /. Therefore1 =
A
basic
linear
28.7 Example
in linear is finding all common solutions algebra we abandon our practice of never equations. For the moment a nonzero polynomial, and work a typical problem as we do in a
\342\200\224 0\" for
course.
algebra
(Solutionas in
a Linear
Course)
Algebra
x +
the
We multiply
first
3z = z =
+
y
add it
-y +72
the
first
as
solution
linear
system
8 \342\200\2245.
to
the
second,
the new
obtaining
system
= -21
preceding one. For any value z, we can and then determine x from y-value equation corresponding \342\200\224 as we obtain z {(\342\200\224Az 13, lz + 21, z) \\ z e R} parameter, equation. Keeping is a line in Euclidean \342\226\262 set, which (\342\200\22413, 21,0). 3-space through the point
set
same solution
the
has
tri the
notation
of
this
as the
R3
in
from
the
find
of the
3z = 8
-
y
in R3
all solutions
Find
-
y
\342\200\2242 and
by
equation
x+
which
\342\231\246
Illustration
2x + Solution
equation basis
5 is
of linear
number
\"/(x)
writing
and
have
for problem solving
technique
of a finite
1'
Linear
Familiar
A
e I, so we for /.
257
S is a basisfor /'. The 5 c S c /. Thus
/', Thus
B'CK
that r(x) 5' is a basis
shows
Ideals
Bases for
Grobner
28
Section
the
second
in the
the problem
section,
preceding examplecan
be phrased
as follows:
Describe We solved it by finding
+
V({x
a more useful {x +
the
two basis
original
basis,
y-3z-8,
second member,
Notice that the
- 3z -
y
\342\200\224
y +
lz
8, 2x+ y
y
-
-y + 7z + 21, of this
2x
z+
that
it is
r, as
polynomial (/,
g) =
clear that
the
can be phrased given
ideal
variety.
in
{g, r). We method
in terms
basis
Theorem chose
28.6, a very
introduced of applying
into one
that
better
in
a division
5
6z
+ 21
+7^
We
which assures us
+ 21),
an
that
illuminates
algorithm
V({f,
process
the geometry
of the
expression
the polynomial f by g}) = V((g. r)) and
replaced
simple, 1-step problemin Example in a linear algebra coursefor solving
a division
from
namely
process,
\342\200\224 \342\200\224 16
+ lz (\342\200\224y z).
can be obtained
basis
new
y+ -y
the
R3.
in
21}.
y, z)
+ 2y
Thus 2x + y+z + 5 = (x + y \342\200\224 3z \342\200\224 + 8)(2) form f(x, y, z) = g{x, y, z)q(x, y, z) + r(x, y,
+
r (x,
2x+
-3z-8
5))
namely
polynomials as a remainder
x +
z +
+
28.7.
repeatedly
of the
However,
a linear
associated
system to change a algebraic
258
Part
V
Factor
and
Ideals
F[x],
we want to
that
now the
Illustration
Indeterminate
A Single Suppose
Rings
the
find
single so there exists f(x) e F[x]such a single polynomial, and {/(*)}
ring
idealin F[x] is principal, consistsof the zeros of for I as we could desire.We generator
polynomial.
expecttwo but
28.8 Example
will
have
basis
division
I in
ideal
By Theorem 27.24, that I = (/(*)). Thus
every V(I)
V in R consisting
for (f,g)
2. We
3x2
of common zeros of ! 6x \342\200\224 + 3x~ \342\200\224
g(x) = x
and
having polynomials ofas small degree as possible, = g(x)q(x) + r{x)in Theorem 23.1, where r(x) the basis {/, g) by the basis {g, r). replace
f{x) then
x x3 +
is probably
carefully!
example
\342\200\224 5x
algorithm
at most
degree
our
variety
3x~
to find a new
so we use the
in
algebraic
= x4
f(x)
want
an
x.
an example
the basis
Let us describethe We
associated with
indeterminate
as simple a basis give illustrating computation of such a single for I in a case the where basis for I contains more than one f(x) given Because a polynomial in R[x] has only a finite number of zeros in R, we or more randomly selected polynomials in R[x] to have no common zeros,
constructed
we
in R
V(I)
variety
in the
of polynomials
\342\200\224 2
\342\200\224 6x \342\200\224
x3
x4+
-3x2
2
\342\200\224 5x-
-6x2 8x -2x3 + + 3x - 2 -2x3 -6x2 + 12x+ 16
x4 + 3x3
3x2
9x2
Becausezeros
9x2
of
x2 \342\200\224 x \342\200\224 2, and
\342\200\224 9x \342\200\224 18 are
take as
-
as zeros of
the same
9x
- 18
x2
\342\200\224 x \342\200\224 we
2,
let r(x)
=
basis
new
= (x3 +
[g, r}
g(x) by r(x) to obtain dividing {r(x), ri(x)} consisting of polynomials
By
-
3x2
\342\226\240x-2).
r\\(x), we will at most 2.
a remainder
be able
now
to
a basis
find
of degree
x + 2
-
6x
- X --2
x3
+
x3
-
4 3x2
x2 4x2
4x2
- 6x
-
8
-
8
- 2x -
4x - 4x
8
0
Our (x2
new - x
basis [r(x),r\\(x)}
- 2) = {(x-
Theorem
+
now
becomes
1)), and
{x2-x-2}. V = {-1,
we see that
same
divisor\" (abbreviated gcd) of
f{x)
and g{x).
I =
Thus
(/(*), g(x)}=
2}.
tells us that the common divisors of f(x) and as the common divisors of r(x) and ri(x). itself divides 0, so the common divisors of f(x) of course, include r(x) itself. Thus r(x) is called
28.6
example are the
we seethat r(x) of r(x), which,
2)(x
\342\226
g(x)
in the
Because and
g(x)
preceding
0 =
(O)r(x),
are just
a \"greatest
those
common
Section 28
259
Ideals
Bases
Grobner
We tacklethe In view
Grobner Basesfor
of
of finding a nice basisfor an ideal I in illustrations for the linear and single indeterminant
problem
our
F [x] =
F [xi,x2,
\342\200\242 \342\200\242 \342\200\242. x\342\200\236 ].
cases, it seems
replace polynomials in a basis by polynomials of lowerdegree,or containing It is crucial to have fewer indeterminates. a systematic way to accomplish this. Every in linear algebra has had an occasional instructor student who refuses to master matrix and creates zero entries in columns reduction of a matrix in an almost random fashion, rather than the first column and then proceeding to the second, etc. As a first finishing this problem of specifying an order for polynomials in a basis. step in our goal, we tackle \342\200\242 \342\200\242 in F[x] have terms of the form ax^'x\342\204\2422 \342\200\242 a e F. Our polynomials xnm\" where reasonable to
try to
Properties for
an
of Power
Ordering
1. 1 < P for all power products P^l. 2. For any two power products P, and < pJt Pi = Pj, Pj < p, holds. pt 3.
If Pi
< Pj
4.
If
< Pj,
Pi
and
< Pk,
Pj
then PPj
Let us considera powerproduct p =
Xlm'x2m2
exactly
Pj,
Products one of
then P, < Pk.
< PPj
for
in F[x]
to be
product
power
any
an
P.
expression
\342\226\240 \342\200\242 \342\226\240 where all x\342\200\236m\302\273
the m,- >
0 in
Z.
that all x,- are present, perhapssomewith exponent 0. Thus in F[x, y, z), we must write xz2 as xy\302\260z2to be a power product. We want to describe a total ordering < on the set of all power products so that we know just what it means to say that P, < Pj for two We providing us with a notion of relative size for powerproducts. power products, can then try to change an ideal basis in a systematic way to create one with polynomials terms a,P, with as \"small\" power products Pt as possible. We denote by 1 the having of the power products 0, and require that an ordering power product with all exponents has the properties shown in the box. Supposethat such an ordering has been described = PPt where 1 < P. From and that 4 P, ^ Pj and P, divides Pj so that Property Pj in the box, we then have 1P, < P P, = Pj, so P, < Pj. Thus P, divides Pj implies that to show by a counterexample that Pi < Pj.ln Exercise 28, we ask you P, < Pj does not that P, divides Pj. imply It can also be shown that these properties guarantee that any step-by-step process for modifying a finite ideal basis that does not increase the size of any maximal power at least one by something smaller at each step product in a basis elementand replaces Notice
terminate
will
In F[x] by
Property
in a with
1, we
finite
x the
number of steps. only indeterminate, there
must have 1 <
x. Multiplying
is only
one
repeatedly
power by x
using
ordering, for Property
4,
\342\200\242 \342\200\242 < x2, x2 < x3, etc. Property 3 then shows that 1 < x < x2 < x3 < \342\200\242 is the order. Notice that in a basis 28.8, we modified basisby replacing possible Example
we have x only
product
and
polynomials
by polynomials
containing
smaller
power
products.
260
Part
V
Factor
and
Ideals
Rings
There are a number terminates. We present define
orderings for power products in the one, lexicographical order (denoted by
of possible
just
n indeIn lex, we \"lex\-") with
F[x]
(2) s, < tt for the first subscript i, reading from left to right, such that Si ^ t,-. in the order xnym, we have y = x\302\260yx< y], if we write power products = x and xy < xy2.Usinglex, the order of n indeterminates is given by 1 < xn < x1y\302\260 \342\200\242 \342\200\242 < \342\200\242 < X2 < xi. Our reduction in Example 28.7, where we first got rid of all \"big\" x\342\200\236_i the \"smaller\" x's that we could and then y's, corresponded to the lex order z < y < x, all power that is, to writing products in the xm ynzs order. For the two-indeterminate case with lex term order schematically is y < x, the total if
only if in F[x,
and
Thus
1<
<
y
An
y
<
in
xy < xy2
<
x'y < x~y~ <
< x\" <
<
xy
products P inducesan obvious ordering which we will refer to as a term order. From
of terms
of power
ordering
polynomial
x <
<
y
F[x],
ordering of powerproducts,
now
aP of a
on, given
to be
an
written
in F[x] every polynomial f We term has the order. order so that (first) decreasing leading highest the leading term of f and by lp(/) the power product of the leading denote by lt(/) in F[x] such that lp(g) divides lp(/), then we can term. If f and g are polynomials a division of execute cases, f by g, as illustrated in the linear and one-indeterminate = < lp(/)Note that we did not say that to obtain f(x) where g(x)q(x)+ lp(r)
we
consider
of terms,
in
the
r(x)
< lp(g).
lp(r)
28.9 Example
By division,
We
We
see
that
{xy2,
maximum term
one with smaller
Solution
the basis
reduce
an example.
with
illustrate
y2 divides xy2 and
\342\200\224 for
y}
y2
size,
the ideal
I =
y2
(xy2,
the order lex with
assuming
\342\200\224 in
y)
]
y]to
y < x.
compute
xy
y-y
\342\200\224
xy
xy-
xy
y2 does not lp(y2
Because
is not less than for/ is {xy,y2
divide
xy,
\342\200\224 =
y)
we cannot
the
continue
y2. However,
we dohave
< lp(xy2).
lp(xy)
= xy Our new basis
Note that lp(xy)
division.
\342\200\224
\342\226\262
y).
with more than one indeterminate, it is often easier to perform basis a a basis and adding it by multiplying polynomial g(x) by polynomial \342\200\224q{x) to obtain as we matrix in reduction linear r(x), fix) polynomial perform algebra, than out the division display as we did in the preceding writing example. Starting When
dealing
reduction
to a rather with
basis
and \342\200\224x We
can
polynomials
xy2 and
the resulting adding do that in our head,
Referring
any polynomial
again
y, we
+ xy \342\200\224xy2
and write
to Example
f(x, y) =
\342\200\224 y2
28.9,
to xy2,
down it
can reduce
will
the
xy2 the
obtaining
the result follow
a(x, y)(xy) + c2(x,y)(y2
by multiplying replacement
\342\200\224
y2 xy for
y by
xy2.
directly.
from what \342\200\224 in
y)
we state later that
(xy, y2
\342\200\224 either
y),
given
xy or
GrobnerBasesfor
Section 28
y2
divide
will
(See Exercises
lp(/).
Grobnerbasis.
28.10
Definition
31.) This illustrates
the
261
Ideals
defining
property
of a
\342\226\240 \342\200\242 with nonzero polynomials in F\\x\\,x2, \342\200\242, x\342\200\236], term ordering \342\226\240 \342\226\240 = <, is a Grobnerbasis for the ideal I gr) if and only if, for each nonzero {gi,g2, \342\226\240, \342\226\240 f e I, there exists some i where 1 < i < r such that lp(g,) divides lp(/).
A set
{gi, g2,
\342\226\240 \342\200\242 of \342\226\240, gr)
of a Grobner basis from a given basis illustrated the computation 28.7, 28.8, and 28.9, we have not given a specificalgorithm. The method consists of multiplying We refer the reader to Adams and Loustaunau [23]. in in the basis by any polynomial some F[x] and adding the result to another polynomial that reduces the size of power products. In our in the basis in a manner polynomial division of f(x) by g(x) where lp(g) we have treated the case involving illustrations, if divides lp(/), but we can also use the process lp(g) only divides some other power \342\200\224 in a basis are xy \342\200\224 in f. For example,if two elements 1, we can y3 and y2 product \342\200\224 \342\200\224 \342\200\224 28.6 1 by y and add it to xy \342\200\224 xy y3 to xy y. Theorem y3, reducing multiply y2 shows that this is a valid computation. \342\226\240 can fail to be a Grobner basis for You wonder how any basis {gi, g2, \342\226\240 \342\226\240, gr} may \342\226\240 \342\226\240 = element in we an when form I h crgr I, c\\gi + c2g2 + (gi> 82, \342\226\240, gr) because, \342\200\242 \342\200\242 i = r. However, cancellation of for we see that lpfe) is a divisor of lp(c,-g,-) 1, 2, \342\200\242, with an example. power productscan occur in the addition. We illustrate While
for
28.11 Example
an
we have
in Examples
ideal
Considerthe
I =
ideal
\342\200\224
(x2y
2, xy2
shown cannot be reducedfurther.
\342\200\224 in
y)
The
K[x, y).
the
in
polynomials
basis
\342\200\224 \342\200\224 I contains y(x-y \342\200\224 2) x(xy\342\200\224 = xy leading power product xy is not divisible by either of the leading y) 2y, whose \342\200\224 \342\200\224 2, xy2 y) is not a Grobner power products x2y or xy2 of the given basis. Thus {x2y \342\226\262 basis for I, according to Definition 28.10.
in Example 28.11, we realize that a Grobner a smaller leading power product than those in the given basis. Just as we did in be polynomials f and g by as small power products as possibleso that
When we run into a situation basis must contain some polynomial
like
that
with
given basis. Let f and g Example 28.11, we can multiply the resulting two leading power products (lcm) of lp(/) and lp(g), and then subtract in the
cancellationresults.We without
28.12Theorem
A
basis
if,
for
proof a G = all i
{g\\,
^ j,
remainders by
denote
theorem that ,gr}
polynomial
elements
of G,
a Grobner in
the
or add
with
formed in test whether
basis for the
same, the
the
division
least
this
fashion
a basis
ideal
(gi, zero
common
multiple
so g). We state a Grobner basis.
coefficients
suitable
gj) can be reducedto
S(gi,
as
be
will
a polynomial can be used to
\342\226\240 \342\226\240 \342\226\240 is
g2, the
the ideal
However,
from F
by S(f,
is
g2,
\342\226\240 \342\226\240 if \342\226\240, gr)
by repeatedly
and only
dividing
algorithm.
we may prefer to think of reducing As we mentioned before, S(gi, gj) by a sequence of operations consisting of adding (or subtracting) multiples of polynomials in G, rather than writing out division. we can obtain a Grobner basis from a given basis. First, We can now indicate how in the basis as far as possible themselves. Then choose reduce the polynomials among
Part
262
Factor
and
Ideals
V
Rings
polynomials gt and gj in the basis, and form can be reduced to zeroas just described. If and
above,
them. If S(gj, gj) cannot with basis this given S(gt, gj), and
the procedure
repeat
the polynomial S(gt, a different so, choose
with
gj). See if S(gt, pair of polynomials,
be reducedto
gj)
as described
zero
all over,
reducing this basis as possible. By Theorem28.12,when S(gt, gj) for all i ^ j every polynomial can to zero using from the latest be reduced basis, we have arrived at a polynomials with a continuation of Example28.11. Grobner basis. We conclude the
augment
start
as much
28.13
let gi = x2y
2.8.11,
Example
Continuing
Example
28.11, we obtained the gi and g2. We
In Example
reduced to zero using
{x2y {2xy {2xy
\342\200\224
\342\200\224
\342\200\224
\342\200\224
2y}
by adding
(third) (\342\200\224x)
to first
\342\200\224
\342\200\224
\342\200\224
by adding
(third) (\342\200\224y)
to second
y,xy
2, xy2
2, 2y2
y,xy
-
2, 0,
xy
2, 0,
\\x
1}
2}
\\,
(2, i), The
2y] -
(xy
in
a Grobner
In Exercises 1 through
-
4, write
z <
where
xmynzs
if f
=
y
\342\200\224 and \\
| + 2y, which
\342\200\224
2y)
can
2(y
basis, we seethat
\\)the
readily
x
\342\200\224 then
2,
be reduced
V(I) contains
variety
algebraic
only
one
\342\2
of Grobner have
bases
in
is due to the fact that they are to engineering and computer scienceas
applications
applications
mathematics.
+ 10z3-
zmynxs
y
the polynomials
8, write
x <
R[x,
- 3x3
y
y, z] in
decreasing term
2. 3yV 4. 38 -
+ 2x2yz2
2xyV
where
in
the order
order,
using
5yV
- 8z7
the polynomials <
in
R[x,
y, z] in
4xz
4x
+
+ 2yz
-
8xy
+
decreasing term order, using
3yz3 the
z.
1.
6. The polynomial
in Exercise
2.
7. The polynomial
in Exercise
3.
8. The polynomial
in Exercise
4.
ordering,
deglex,
lex for power
< x.
in Exercise
Another
2y},
\342\200\224 \342\200\224
polynomial
5. The
g =
R2.
importance
+ lx2y2z
5x2yz3
In Exercises 5 through products
\342\200\224
28
\342\226\240 EXERCISES
- 7x
(xy
|)
2)
machine computable.They as to
basis. Note that
\342\200\224 \342\200\224\342\200\224=
\342\200\224 and
the Grobner
From
2y) 2y}
-
\342\200\224 = S(f, g) = xf yg to zero by adding j(x
3y
y,xy
(-2) (third) to first by adding by adding (-^) (first) to second to third by adding (-f) (first) by adding (\\) (first) to third
\342\200\224
\342\200\224 x \342\200\224 is
{y
Clearly,
2y]
-y,xy-
\342\200\224
- 2, 0, {Ay \\x
3.
\342\200\224
basis
augmented
2y)
y,xy
2, 2y2
-
{Ay
1. 2xy3z5
2, xy2
R2.
cannot be
2y,
\342\200\224
\342\200\224
{Ay
products
(gi, g2) in
\342\200\224which
\342\200\224
2, xy2
{Ay
well
y,
S(gi, g2) = xy the basis {x2y
reduce
I =
\342\200\224 and
each step.
indicating
point,
2,g2
polynomial now
= xy2
\342\200\224
for power
products
in
F[x]
is defined
as follows:
order
lex for power
Exercises
Section 28
and
if
st ^
tt.
s' < \302\243\"= only if either Xw=i 1 fi > or these two sums are equal 9 through 13 are concerned Exercises with the order deglex.
9. List,
in
z <
where
xmynzs
20 power products
y, z] for
in R[x,
order
the
with power
deglex
the polynomials
13, write
<
terms
of decreasing
order
in
order
the
using
products
xmynzs where z
10. The
polynomial
in Exercise
1.
11. The polynomial
in Exercise
2.
12. The polynomial
in Exercise
3.
13. The polynomial
in Exercise
4.
For Exercises 14through
a single-stepdivision
of i such
that
products
y < x.
10 through
In Exercises
the smallest
order,
increasing
smallestvalue
for the
st < tt
and
263
y
deglex
with
power
< x.
power products in M[x, y, z] have order lexwhere z < y that changes the given ideal basis to one having
17, let
< x. If
reduction
algorithm
possible, perform maximum term
smaller
order.
14. (xy2 -2x,x2y 16. {xyz-3z2,x3
+ 4xy,xy
-
y2)
+ y2z3,x2yz3+4) InExercises18and19,lettheorderofpowerproductsinR[w,x,
basisfor 18.
(w
19. {w
the + x
given
-
- Ax
21.
(x4
22. In
{x5
23.
- 3, 2w
+ 4z 3y
- z
lex
with
+
2x - 5, x3-
{x2y+x
17.
{y2z3 +
y3, y3
-
3, y3z2
- y4)
x
+z,
2z, y2z2 +
3)
y, z]belexwithz
- 2z
y
+
y
+ 4, w
- 2z
- 4x -
x3 +
x2
- x2
- 4x + 4,
- x2
-
+ 5, w
basis for
a Grobner
-
3x
+
l(k +
-
1>
in R[x].
4>
x3
- 3x
+ 2)
+ x - 1) ideal in algebraic
24.
+ 2y-9)
+ l,xy2
8y
- 5) - z
ideal
indicated
the
z
+
3y
23 through 26, find a Grobner basis for the given y < x. If you can, describe the corresponding
-x-2,xy
{x2y
+
- 2x
22, find
- 3x2 - 4x - 4, - 4x3 + 5x2 - 2x,x3
+ x2
+ x
+ 2, 2w
+ x3
Exercises
to be
25.
(xA
+
{xy +
ideal.
20 through
In Exercises
20.
y
15.
26.
1> +y\342\200\224
(x2y
M[x, y + x,
the order
]. Consider
variety in M[x, xy2
{x2y +xy2,
xy
-
of powerproducts
y].
y)
-
x)
Concepts
27. LetF be a field. a.
Mark
each of
the
or false.
true
following
in F[x] has a finite basis. subset of R2 is an algebraic variety. The empty subset of R2 is an algebraic Every finite subset of R2 is an algebraic ideal
Every
b. Every c. d.
e. Every
line
f.
Every
finite collection
g.
A greatest
in R2
h. I have computed i. Any ideal in F[x] The
ideals
division
the
Grdbner
is an
algebraic
number
algorithm bases
variety.
variety.
of lines in R2 common divisor of a finite
computed using
j.
is an algebraic
variety.
before
variety.
of polynomials
repeatedly. I knew what
unique Grobner basis. are equal because they (x2,y2)
they
in M[x]
(one indeterminate)
can
were.
has a
{x, y > and
both
yield the
same algebraicvariety,
namely
{(0,0)},inR2.
28. Let
R[>, y] be ordered
by
lex. Give an
example
to
show
that Pt
< Pj does not
imply
that Pt
divides Pj.
be
264
Part
Factor
and
Ideals
V
Rings
Theory
29. Show
if /i, f2, I for i = \342\202\254
that
crfr\\Ci
30. Show
R with
ring
then
unity,
I =
{ci/i + C2/2H
if /(x) = g(x)q(x) + r(x) in F[x], then the common divisors in F[x] of /(x) and g(x) arethe common divisors in F[x] of g(x) and r(x). \342\200\224 that {xy, y2 \342\200\224 y2 y) is a Grobner basis for (xy, y), as asserted after Example 28.9. that
as the
31. Show 32. Let
elements of a commutative \342\200\242 an ideal of /?. 1, \342\200\242 \342\226\240, r) is
\342\226\240 \342\226\240are \342\226\240, fr
F be
a field.
Show
that
if S
is a nonempty
I(S) = is an
of F\",
F[x] ] /(s) {/(x) \342\202\254
then = 0
for
all
s \342\202\254 5}
ideal of F[x].
33. Referring
to Exercise
32, show
34.
Referring
to Exercise
32, give
35. Referring
to Exercise
32, show
36.
to Exercise
32, give
Referring
subset
S
that
an
example
that
an
c V(7(5)). of a
if AT is an
example
subset S of R2
ideal of
of an
F[x],
ideal N
in
that
such then
R[x,
N
V(I(S))
^ S.
C I(V(N)).
y] such that
I(V(N)) ^
N.
h
same
part
Fields
Extension
VI 29
Introduction
Section
30
Vector Spaces
Section 31
SECTION
29
to Extension
Section
Extensions
Algebraic
Section
32
Geometric
Section
33
Finite
Constructions
Fields
Fields
to Extension
Introduction
Fields
Our BasicGoalAchieved now in a position to achieve our basic goal, which, that stated, is to show loosely nonconstant polynomial has a zero.This will be stated more precisely and proved
We?are
every
29.1 Definition
29.3.
Theorem
in
A field
E is an
some new
We first introduce
extension field of a field
F
if F Fix,
C
for some
terminology
old ideas.
< E.
\342\226\240
y)
Fix)
Fiy)
29.2 Figure
Thus K is an
Q.
As
picture
configuration
Fig. 29.2, 1
Section
extensionfield
study of groups, it extension fields, the larger
in
the
where
there is just
is often referredto,
32 is not required
for the
of will
field
Q, often
and C is an extensionfield of both H and be convenient to use subfield diagrams to
being on
one singlecolumn
without
any
remainder of the
top. We of
illustrate
fields,
this
as at the
precise definition, as a tower
in Fig. left-hand
29.2. A side
of
of fields.
text.
265
Extension Fields
Part VI
266
This great and
for our basic goal! the techniques we now
Now
from
29.3 Theorem
polynomial
elegantly
F be
Let
Then there exists an
extension
a field and let f{x) bea nonconstant field E of F and an a e E such
that
that are irreducible in F[x] into polynomials By Theorem23.20,f{x)has a factorization in a It is clearly Let be an irreducible such factorization. over F. p(x) polynomial \342\200\224 find field of F an element a that 0. to an extension E such sufficient containing p(a) We a in so is a field. Theorem is maximal ideal 27.25, (p(x)} F[x], F[x]/{p(x)) By with a subfield of F[x]/{p(x)) in a natural claim that F can be identified way by use of
Proof
the
i/f : F
map
\342\200\224> F[x]/{p(x))
by
given
i/(a) = F. This
e
fora
map is oneto
forsomea,^e p(x),
a
\342\200\224 =
b
choosing
that
Now
b. We
F]
field of
by
F.
We
show
e F
a,b
map
Thus
i/f.
F[x]/{p(x)}. we shall view
must
by
F[x]/{p(x)) i/f
F
We identify
E=
homo-
is a
with
F [x]/{p(x)) as
our desired extensionfield manufactured E contains a zero of p(x). now
have
(p(x)}
polynomial
F. Thus we in
e (a + (p(x))).Thus of
subfield
of the
a \342\200\224 b is in
and multiplication
choose a
onto a
that
(p(x)} = b +
E
of F.
set
us
+ {p(x)),
a =x
so a e E. Consider Theorem22.4.
implies
addition
defined
of this
that
= i/f(&),thatis,ifa + \342\200\224 b must be a multiple
(a)
i/f
so we may means
a + {p(x))
(p(x)),soa
F one-to-one
It remains for us to Let
a=
maps
|a e
{a + (p(x)} an extension
>1.
representatives,
any
morphism
b)
degree
0, so
for if
one,
\342\200\224 e
F,then(a
is of
which
have
the
If p{x)
evaluation
a\\x +
\342\200\224 +
oq
4>a{p(x)) =
a : F[x] homomorphism anxn, where at e F, then
\342\226\240 \342\226\240 \342\226\240
+
+ ai(x \302\253o
+ (p(x)))
h an(x
H
\342\200\224\342\226\272 E, given
by
we
have
+ {p(x)))n
Note
\342\226\240 Historical
Kronecker is known Leopold on constructibility \"God
for
his insistence
objects. As
of mathematical
made
work of man.\" Thus, struct new \"domains
the integers; all
to be
he wanted of rationality\"
only the existence
of integers
He did not
in starting
and
else is
able to
the
con-
indeterminates.
believe
p{x);
that is,
his new field consisted of
in the original field elements a with the condition that p{a) = 0. of the theorem presented in the text 29.3) dates from the twentieth century,
expressions and
The
rational
new root
his
proof
(Theorem
Kronecker completedhis dissertation
(fields) by using
with the real or combecause as far as he was concerned, numbers, plex those fieldscould not be determined in a construeHence in an 1881 paper,Kronecker tive ereway. ated an extension field by simply to a adjoining a a of an nth field root irreducible given degree polynomial
and
= 0.
f{a)
he noted,
follows quickly
disposal.
(Kronecker'sTheorem)(BasicGoal) in F[x].
result
important
at our
have
at the after,
of
University
he
managed
Berlin.
the
family
in
For many years business,
1845
there-
ultimately
He then returned becoming financially independent. to Berlin, where he was elected to the Academy of Sciences and thus permitted to lectureat the univerOn the retirement of Kummer, he became a prosity. fessor at Berlin, and with Karl Weierstrass (1815-
1897)directed
the
influential
mathematics
seminar.
Section 29
in
But we can compute in F[x]/(p(x)} by choosing = x + (p(x)}.Therefore, of the coseta
= F[x]/(p(x)}.
E
andx is a representative
=
h anx\") (a0 + aix -\\ = p(x) + (p(x))= (p(x))
p(a)
We have found
in F[x]/(p(x)}. therefore
and
representatives,
+ (p(x)} =
0
= F[x]/(p(x)}
such
that
p(a)
= 0,
= 0.
f(a)
\342\231\246
the construction involved
illustrate
We
a in E
element
an
267
Fields
to Extension
Introduction
the
in
proof
of Theorem
29.3 by
two
examples.
29.4
Example
= R, and let f(x) = x2 + 1, which is well known to have no zerosin ideal is irreducible over R by Theorem 23.10. Then (x2 + 1) is a maximal r + (x2 + 1) in R[x]/{x2 r e R with + R[x]/{x2 + 1) is a field. Identifying R as a subfield of E = R[x]/{x2 view + 1).Let Let F
= x
a
in R[x]/{x2
Computing
Example
\342\200\224
(x2
2)(x2
can
The
3),
construct
+ 1. We
shall
+
(x2
M[x]/{x2
identify
construction
As we
said before,
element
for some
over F.
= x4
-
5x2
+
6. This
1))
+ 1)
We
one
the close
C at
with
of
an
factors in
of the
into
Q[x]
seen. We can start a such that a2 - 2 = 0,or such that f32 \342\200\224 3 = 0. element^ we
have
29.4.
\342\226\262
Elements rest of
is devoted to the study an element of an extension study by putting two categories.
most
commence
of
time f(x)
irreducible over Q, as field E of Q containing
Transcendental
and
polynomials. a field F into
f(x)
factors being an extension
an extension fields of Q containing in either case is just as in Example
construct
Algebraic
An
consider
and
\342\200\224 both
- 2 and
x2
with
29.6 Definition
+ 1) +
+
1) = 0.
\342\226\262
= Q,
Let F
we
of x2
can
section.
this
29.5
is a zero
so
1), we
1).
1))2+ (1+ (x2
(x + (x2 +
= (x2
Thus a
+
(x\"
and thus
+ 1), we find
+ 1 =
a2
+
R
in R[x],
this
text
this
a of an extension field nonzero f{x) e F[x]. If a
F is
\302\243 of
a field
is not
algebraic over F,
algebraic over then
a
F
of zeros field
of
E of
if f(a)
\342\200\224 0
is transcendental \342\226\240
Part
268
VI
Extension Fields
29.7 Example Cis an 29.8
Example
It is well
known
do
as we
Just irreducible
is an
i
algebraic
of x2 \342\200\224 2, we
element over Q, being
not
speak
simply
see that V2 is an algebraic a zero of x2 + 1. \342\226
n
of an
transcendental
e are
and
\342\226
of an irreducible polynomial,
rather
but
of an over F, similarly we don't speaksimply F. The illustration element algebraic over following
polynomial
rather
Since V2 is a zero
not easy to prove) that the real numbers base for the natural logarithms.
(but
Heree is the
over Q.
but
Q. Also,
over
element
of Q.
field
extension
of an
algebraic element, shows the reason
for this. 29.9
Example
The real number n
n is algebraic 29.10
Example
It is
V3, so a2 - 1 = V3 zero of x4 \342\200\224 2x2 \342\200\224 2, which
a
a is
Definition
real number
the
= 1 +
a2
To connect
29.11
is transcendental over Q, as we stated \342\200\224 R, for it is a zero of (x n) e R[x].
easy to see that
then so
over
these ideaswith
of C that number is an element
An element
is an
There
those
Vl+V3 and
(a2
algebraic of C that
29.8. However,
\342\226
is algebraicover Q. - 1)2 = 3. Therefore
For if a4
a =
Vl+V3, - 2 = 0,
- 2a2
is in Q[x]. of number
\342\226
theory, we give the
over Q is an algebraic is transcendental over Q.
is
in Example
extensive and elegant theory
of
A transcendental
number.
\342\226
numbers.
algebraic
definition.
following
(See
the
Bibliography.)
next
The
evaluation
gives a
theorem
elements over F
homomorphisms
terms of
useful
(j)a.
Note
of algebraic
characterization
field E
extension
an
in
that
of F.
It
also
illustrates
once more we
and transcendental
the importance
are describingour
of our
concepts
in
mappings.
29.12Theorem Let E be an
of a field F and let a e E. Let <$>a : F[x] ->- E be the of e F and a(x) = a. F[x] into E such that a(a) = afora homomorphism Then a is transcendental over F if and only if 4>a gives an isomorphism of F[x] with a subdomain of E, that is, if and only if 4>a is a one-to-one map. field
extension
evaluation
Proof
The element
a is
F[x],
is true
which
F[x],
over F if and only if f(a) ^ 0 for all nonzero f(x) e if and only if 4>a{f{x)) ^- 0 for all nonzerof{x) e definition) (by if and only if the kernel of a is {0}, that is, if and only if a is a
transcendental
is true
which
one-to-one
map.
\342\231
Irreducible
The
Polynomial
for a
over F
field K of Q. We know that V2 is algebraic overQ,being a zero of is also a zeroof x3 - 2x and of x4 - 3x2 + 2 = (x2 - 2)(x2 - 1). Both these other polynomials of x2 \342\200\224 2. The next having ~J2 as a zero were multiples theorem shows that this is an illustration of a general This theorem plays a situation. central role in our later work. Consider
x2
- 2.Of
the
extension
course,
~Jl
Introduction to
Section 29
29.13 Theorem
Let Ebe an
e E, where that p(a) e such polynomial p{x) F[x] F, andleta
field of
extension
an irreducible
is uniquely determined up to a constant >1 in F[x] having a as a zero. If degree then divides f{x). p(x)
factor
p(x)
Proof
269
Fields
Extension
algebraic over F. Thenthere = 0. This irreduciblepolynomial F and is a polynomial of minimal for e F[x], with f(x) = f{x)
a is
in
f{a) = 0
the evaluation
be
of 4>a is
an ideal
of F[x]
homomorphism
and
by
into
27.24 it must
Theorem
\342\231\246
irreducible.
highest
as the
Let E be an extension monic polynomial irreducible
Example
that
irr(V2,
VI + V3
in
R, a is
over Q
Example23.14),
we
over
see
F,
F
and
will
denoted
with
Eisenstein
polynomial.
a e E be algebraicover F. The unique described in Theorem 29.13 is the be denoted by irr(a, F). The degree of
let
by deg(a,
F).
\342\226\240
p =
2, or
by
of the
application
technique of
that
Q) = x4
irr(VTTV3,
is algebraicof
Thus
and
property
F,
is a monic
of x appearing
to Example x2 - 2. Referring 29.10, we see that for a = which is in 2x2 \342\200\224 2 is of x4 \342\200\224 2x2 \342\200\224 2, Q[x]. Since x4 \342\200\224
Q) =
a zero
(by
the
a over
degree of a
We know irreducible
p(x)
of a field
field for
of Theorem
power
p(x) having
polynomial
is the
in
we can assumethat the coefficient of the 1 29.13 is 1.Sucha polynomial having
in F,
constant
suitable
appearing coefficient of the highest
irr(\302\253,F)
29.15
by a
multiplying of x power
By
Definition
0.
E, given by Theorem 22.4. The be a principal ideal generated by some p{x) e F[x]. Now (p(x)} consists precisely of those elementsof F[x] having a = 0 for f(x) ^ 0, then if f{a) as a zero. Thus, e (p(x)}, so p(x) divides f(x). f(x) is a polynomial of minimal >1 having a as a zero, and any other such Thus p(x) degree polynomial of the same degree as p(x) must be of the form (a)p(x) for some a e F. remains for us to show were a It only that r(x)s(x) p(x) is irreducible. If p(x) \342\200\224 = 0 would imply that factorization of p{x) into polynomials of lower degree, then p(a) = 0, so either r(a) = 0 or s(a) = 0, since \302\243 is a field. This would contradict r(a)s{a) >1 such that p(a) = 0. Thus is the fact that p(x) is of minimal p(x) degree
Let a kernel
29.14
is
degree
4 over
-
2x2
-
2.
Q.
\342\226\262
Just as we must speak of an element a as algebraic over F rather than as simply algebraic, we must speak of the degree of a over F rather than the degree of a. To take a trivial illustration, V2 e R is algebraic of degree 2 over Q but algebraic of degree 1 over R, for irr(V2, R) = x - s/l. The here is due to the machinery of the theory of homomorphisms quick development that we now have at our disposal. and ideal theory Note especially our constant use of the evaluation homomorphisms cf>a.
Extensions
Simple
Let E
be an extensionfield
homomorphism
of F[x]
We consider
two
into E cases.
of a with
field F,
a (a)
and
a e
let
\342\200\224afora
e
E. Let
a
F anda(x)
be
the evaluation
= a,
as in
Theorem
22.4.
Part
270
VI
Extension Fields
CaseI
over F. Thenas in Theorem 29.13, the kernel of and Theorem is a maximal ideal 27.25, F)) by {irr(a, F)) of F[x]. Therefore, \302\243[x]/{irr(o:, is a field and is to the F)) isomorphic in This subfield of E is then the smallest E. image a[\302\243[x]] a[\302\243[x]] F and a. We shall this field by F(a). sub field of E containing denote is algebraic
a
Suppose
4>a is (irr(o;,
Supposea is
Case II
F[a],
is thus denote 29.16 Example
F. Then
over
transcendental
Theorem
by
functions
element
that
this
field
gives
by F(a).
over Q, the field Q(n) is isomorphic Q in the indeterminate x. Thus from transcendental over a field F behavesas though
over
is
a subdomain
Since n is transcendental rational
29.12,
of E. Thus in this case domain that we shall denote integral by a field of quotients of F[a], which contains By Corollary 21.8, \302\243 the smallest subfield of E containing F and a. As in Case I, we
of F[x] with a field but an
an isomorphism is not 4>a [F[x]]
to
the
a structural it were
field Q(x)
of
viewpoint,
an
an indeterminate
over F. 29.17
Definition
\342\226
An extension field a e \302\243.
\302\243 of
29.18Theorem
Let
\302\243 be
the degree
expressed
of irr(a, the
extension be \302\243)
\302\243(a) of
n >
Proof
For
b{
the usual
are
a field
1. Then
every
ha
+
in
bQ
+
the
form
p{a)
0a (/(*))
= f(a), a formal =
\302\243)
be uniquely
bn-lan-1,
p(x)
=
every
element
0a[F[*]]
polynomial x\"
of
+
an-ixn~l
coefficients
in a with -\\
\\-
in
\302\243. Let
a0.
\342\200\224
0, so
an This
= \302\243 \302\243(a)can
\342\226\240 \342\226\240 \342\226\240 +
evaluation homomorphism 4>a,
irr(a, Then
over \302\243. Let
be algebraic
let a /J of
\302\243.
F(a) =
is of
and \302\243, element
form
P = where the
some
F.
a simple in
= \302\243(a)for \302\243 if \302\243
this section. We have now developed so results appear throughout results are starting to pour out of our efficient plant at an alarming gives us insight into the nature of the field \302\243(a)in the case where
much machinery that rate. The next theorem
a is algebraicover
extension of
a simple
\302\243 is
\342\226
important
Many
a field
equation
powers of a
in that
\302\243(a)can
are
less
\342\200\224 \342\200\224 \342\226\240 \342\226\240 \342\226\240 \342\200\224 \342\200\224an_\\an~l a0.
be used to express every monomial than n. For example,
am
for m
>
n
in terms
of
Section29
a\"+1
if /J e
Thus,
=
\342\200\224 \342\200\224 \342\200\224 \342\226\240 \342\200\242 \342\226\240 \342\200\224an-\\an a\342\200\236_2a\"~ a^a
=
aa\"
=
\342\200\224 \342\200\224\342\200\224 \342\200\224 \342\226\240 \342\226\240 \342\226\240 an-2a\"_1 ao) a\342\200\236_i(\342\200\224an-\\an~x
F(a),
bQ
for
in
b'0)
g{a) =
and
F[x]
+
\342\226\240 \342\226\240 \342\226\240
+
(h
we must
\342\226\240 \342\226\240 \342\226\240
+
bn-ian~x
-
b[)x +
29.19 Example
bn-ian-\\
+
b\\a
--- +
b^a\"'1
have
= 0.
g(x)
b'^x\"-1
of minimal
polynomial
of the
uniqueness
give
an impressive
The polynomial
is less
of g(x)
degree
\342\200\224 = b\\
Therefore, bt
= g(x)
\342\226\240\342\226\240\342\226\240 + (b\342\200\236-i
*< =
We
required form
=b'0 +
+
0. Also, the
Sinceirr(a, F) is a nonzero
and the
a^a.
if
bia
(b0 -
is
\342\200\224 \342\200\224 \342\226\240 \342\226\240 \342\226\240
F, then
e
b\\
+
bia +
= b0 +
P
For uniqueness,
in the
be expressed
can
ft
271
Fields
to Extension
Introduction
than
degree
in
the
of irr(a, F). a as a zero,
degree having
F[x]
0, so
*;,
is established.
b\\
\342\231\246
example illustrating
29.18.
Theorem
+ x + 1 in
over Z2 by Theorem 23.10, of p(x). By Theorem29.3,we a zero a of x2 + x + 1. By extension field E of 1o_ containing Theorem29.18,Z2(a)has as elements 0 + 0a, 1 + 0a, 0 + la, and 1 + la, that is, 0, and 1, a, and 1 +a. This gives us a new finite field, of four elements! The addition in For this field are shown Tables and 29.21. to tables for 29.20 example, multiplication = = observe a2 a 1 in we that since then + + + + 0, p(a) a)(l a) Z2(a), compute (1 = x2
p{x)
since neither element know that there is an
0 nor
is irreducible
Z2[x]
1 of 1o_
element
=
a2
a zero
is
\342\200\224 \342\200\224a 1 =
a +
1.
Therefore,
(1 + a)(l +a) = we
Finally,
that
show
1+ct 1+a
\342\226\262
(x~
+
1).
29.21 Table
29.20 Table
0 0 1 a
= l+a + l=a.
Theorem 29.18 to fulfill our promise of Example 29.4 and to the field C of complexnumbers. We saw in + 1) is isomorphic we can view R[x]/{x2 + 1) as an extension field of R. Let
a =x+
+ 0 1 a
= l+a2
can use
R[x]/{x2
Example 29.4 that
+ a2
l+a+a
1 1 0 1+a a
a
1 +a
a
1 +a
0
1 +a
a
1
0
1
a
1
0
1 +a
0 0 0 0 0
a 0 a
1
1 0 1 a
1+a
1
1+a
1
a
+
0
1+a
a
Part
272
VI
Extension
Fields
ThenR(a)
= R[x]/(x2
+ 1)andconsistsofallelementsoffheforma we see that
by Theorem 29.18. But since a2 + 1 = 0, a + ba plays the of (a + bi) e C. Thus role
+ tofor a
~ C.
R(a)
the role
plays
This is
the
a,beM.,
of ;' e
C, and
algebraic
elegant
Cfrom R.
way to construct
29
\342\226\240 EXERCISES
Computations =
5, show
1 through
Exercises
In
the
that
number a
given
/(\302\253)
1. 1 + V2
2. V2 +
4. Vl+^2
5. y/W-~i
that your
6. V3
8, find
6 through
Exercises
In
Q) and
irr(a,
over
Q by
e Q[x]
/(x)
finding
such
that
- V6
over
algebraic
7.
deg(a, Q) for
9 through 16, classify the F, find deg(a, F).
^(1) + a e
given
to Example
a product
18.a.
a = 1
+ i, F
12.a = v^r,
to prove
F
=
over
the given field
F. If a
is
R
= K
= n2, F = Q 16. a = n2,F = Q(tt3) x2 + x + 1has a zero a 14. a
in Z2(a) and thus must factor into a by long [Hint: Divide x2 + x + 1 by x \342\200\224
1.]
x2 + 1 is irreducible + 1 in an extension
the polynomial
that
Show
C.Beprepared
+ f
V2
C as algebraic or transcendental 10.
a e
do so.
text. The polynomial factors in (Z2(a))[x]. Find this factorization. fact that a2 = a +
the
using
algebraic number
given
8.
29.19 of the
of linear
division,
the
77
9. a = i, F = Q 11. a = 4%, F = Q 13. a = VJr. -F = QW 15. a = n2, F = Q(n) Refer
1 + i
3.
V3
over Q if challenged to
irreducible
are
polynomials
In Exercises
17.
e C is algebraic
o.
a be a zero of x2 addition tables for the
b. Let
nine
in Z3[x].
of Z3. As in Example of Z3(a), written in the order 0,1,
elements
field
29.19, give
the
multiplication
and
2, a, 2a, 1 + a, 1+ 2a,2 + a, and
2 +2a.
Concepts In Exercises
19 through
is needed, so that 19.
An element
it is
a of
22,
the definition of the acceptablefor publication.
correct
in a form
field
extension
an
\302\243 of
term
italicized
a field F if
without reference to
algebraic over F if
and
only
if correction
the
text,
if a
is a
zero of
some
polynomial.
20. An
element p of an
polynomial
21.
A
22. A
monic field
subfield
extension
field
\302\243 of a
field F is
over
transcendental
F if and only if
fi
is not
a zero of any
in F[x]. polynomial \302\243 is
a simple
of \302\243 contains
in F[x] is
one having
extension of a.
a subfield
all coefficients F
equal to
if and only
if
there
1. exists
some a
e
\302\243 such
that no
proper
23. Mark
of the
each
a. The
following
c.
d. R e. Q f.
Let
eC
i. Every j. If x stated
have
a. Find a
b. Find 25.
polynomial
nonconstant
polynomial
F
in F[x] has in F[x] has
indeterminate, Q[n]
of R
\302\243 of
of degree n.
over Q
algebraic
without proof
subfield
n.
degree
If
/(a)
= 0
for
If
/(a)
= 0
for nonzero /(x)
that
such
extension
field of
every
extension
field
over
transcendental
Q.
algebraic of degree 3 over F. e2 is algebraicof degree 5 over \302\243.
1is irreducible
over
Z2.
+aia'
flo +a\\a
trying
26.
+ x2 + eight
e R[x], then
1by
F. of F.
some
a be a zero of x3 + x2 + 1 in an extension field of Z2. Show factors in (Z2(a))[x] by actually finding this factorization. [Hint:
the
then
n is
that
R such that
x3 + x2 +
~ Q[x].
e are
n and
a zeroin a zeroin
b. Let
Divide x3
e Q[x],
/(x)
> n.
nonconstant
is an
a subfield
a. Show that
be
/(x))
(degree Every
of
> n.
/(x))
(degree
24. We
is algebraic over F.
of a field F
of Q. an extension field of Z2. a e C be algebraic over Q
g. Let a h.
of R.
extension field
is an is
nonzero
over Q.
transcendental
extension
element
Every
29
or false.
true
n is
number
b. C is a simple
273
Exercises
Section
for
a,-
that Every =
x3 + x2 element
+ 1 factors into three linear of Z2(a) is of the form
0,1.
long division. Show that the quotient possible elements. Then complete the factorization.] a by x \342\200\224
also has
a zeroin
Z2(a)
\302\243 of degree 3 over Z2. Classifythe groups be an extension field of Z2 and let a e \302\243 be algebraic and ((Z2(a))*, \342\200\242) to the Fundamental Theoremoffinitely abelian groups. As usual, according generated is the set of nonzero of Z2(a). elements
Let
27. Let \302\243 be an extension field of a field to as the minimal polynomial referred
F and for
let a e E be algebraic over F. The polynomial irr(a, a over F. Why is this designation appropriate?
by simply
(Z2(a),
+)
(Z2 (a ))*
F) issometimes
Proof Synopsis
28. Give
or three-sentence
a two-
synopsis
of Theorem
29.3.
Theory
29. Let\302\243 be an that
Show
30. Let
\302\243 be
Show
b. Show
that from
e E.
Suppose a
is transcendental
over
F but
algebraic over F(fi).
algebraic over
an extension
n. Prove that 31. a.
F, andleta, fi F(a). of a finite field F,
extension field of
/6 is
field
has qn
there
exists an irreducible
part (a) that
where F has q
elements.Leta
eE
be algebraic
elements.
F(a)
of degree
polynomial
there exists a finite
field
of 27
3 in Z3 [x].
elements.
[Hint:
Use
Exercise
30.]
over F of degree
Part
274
32. Consider
a. Show b.
the
field Zp
prime
for p
that,
in Zp.
Deduce the
Using
part
33. Let \302\243 be of F(a)
Fields
Extension
VI
^ 2, not desired
(a), show
that
of characteristic p
in Zp is a square of an element of Zp. [Hint: 12 =(p conclusion by counting.] exist finite fields of p2 elementsfor every there p in Z+. prime
field of a field in F is also transcendental
is not
and let
F
over
a e E
be transcendental
over
Following
36.
Let F
37.
idea of Exercise
be a finite Let
[Hint:
a e
the
F*
SECTION
of characteristic
field
the set of nonzero zero of some polynomial
F* be
is a
Use Exercises number
31, show
30 and
36 to
show
there
that
p.
that
Show
of F.
elements in Zp[x] that
of finite
every
F. Show
that
1)2
every
1
element
F.
34. Show that {a + b(J/2) + c02f \\ a, b, c e Q} is a subfield of R by using of the field axioms. [Hint: Use Theorem 29.18.] by a formal verification
35.
\342\200\224 =
element
every
an extension
that
^0.
exists a
of 8
field
every element
Apply group the form x\" field is of
the
ideas
of this
section,
rather
than
16 elements; of 25 elements. over the prime field Zp < F. to show that every group (F*, \342\200\242)
elements; of
of F
is algebraic
theory
to the
\342\200\224
1.]
prime-powerorder, that
is,
it has
a prime-power
of elements.
30
Vector
Spaces
vector space, scalars,independent and bases may be familiar. vectors, the scalars we present these ideas where section, may be elements of any field. We use Greek letters like a and p for vectors since, in our application, the vectors will be elements of an extension field \302\243 of a field F. The proofs are all identical with those often given in a first course in linear algebra. If these ideas are familiar, we suggest and 30.22, and then reading Theorem 30.23 30.4,30.8,30.11,30.14, studying Examples If the examples and the theorem and are understood, then do some exercises its proof. and proceed to the next section. notions
The
In
of a
this
and
Definition
Elementary
Properties
of linear algebra. Since linear algebrais not spaces is the cornerstone treatment of vectorspaceswill be brief, designed for study in this text, our the subject of linear to develop only the concepts independence and dimension that we need for our field theory. familiar from calculus. Here we allow The terms vector and scalar are probably and develop the theory by not the real to be elements of scalars field, numbers, any just structures we have studied. as for the other axioms just algebraic The
30.1 Definition
topic
of vector
Let F bea field. V
under
V
by
each
A vector
addition
element
space over F (or F-vector space)consists with an operation of scalarmultiplication on the left, such that for all a, b e F and a,
together
of F
of an
abelian group element of
of each /5 e
V the
following
conditionsare
aa e V.
%_.
a{ba)
= (ab)a.
= (aa)+ (ba). P) = (aa) + (afi).
+ b)a
9%. (a
+
a(a
= a.
la
9%.
The elements of V
F is under
275
Spaces
satisfied:
%.
%.
Vector
30
Section
are
discussion,
elements of F
vectors
and the
we drop
the reference to
F
scalars.
are
to a
refer
and
When
only one field
vector space.
\342\226\240
scalar multiplication for a vector spaceis not a binary operation on one aa of V with each we denned it in Section 2. It associates an element a of V. Thus scalar ordered pair (a, a), consisting of an element a of F and an element for V, the is a function multiplication mapping F x V into V. Both the additive identity for F, the 0-scalar, will be denoted by 0. 0-vector, and the additive identity that
Note
set in
30.2
sense
the
Consider
Example
ordered
the
abelian
\302\273-tuples under
= R x R x +) group (K\342\200\236, addition by components.
for n factors,
\342\226\240 \342\200\242 \342\200\242 x R
which
Define scalarmultiplication
of
consists for
scalars
inRby
ra
Note
\342\226\240 Historical
ideasbehind The spaceoccurred
notion of a vector concrete examples
abstract
the
in many
century and earlier. For example, dealt with complex as pairs of real numbers numbers explicitly and, as noted in Section 24, also dealt with triples and eventually during
the
William
nineteenth Rowan
Hamilton
in his invention of the quadruples of real numbers In these cases, the \"vectors\" turned out quaternions. could both be added and to be objectswhich multiplied by scalars, using \"reasonable\" rules for both of these operations. Other examples of such objects included differential forms (expressions under integral
and algebraic
signs)
dimensional
in
out
working
spaces
slehre of 1844and to give an abstract
in
his
1862,
a detailed
Die Lineale the
definition
to Definition
equivalent
(1858-1932)in Peano's aim
theory
of
n-
Ausdehnungfirst mathematician of a vector space
the
30.1 was Giuseppe Peano of 1888. Geometrico
Calcolo
book,
as
the
calculus. develop a geometric of Peano, such a calculus\"consists
title
indicates,
According a system of
to
to those of algebraic calculus, the objects with which the calculationsare are, instead of numbers, performed geometrical Peano's work had no objects.\" Curiously, immediate effect on the mathematical scene. operations
but
analogous
which
in
Hermann
Although
Peano's
repeated
announced
for
1945) in the dealing with complete
Weyl (1885-1955) essentially definition in his Space-Time-Matter
of a vector space did not mainstream until it was a third time by Stefan Banach (1892of his dissertation 1922 publication what we now call Banach spaces, definition
the mathematical
enter
(1809-1877)
in
his
was to
of 1918,the
integers. Grassmann
Hermann
Although
succeeded
,ran)
(rau
normed
vector
spaces.
Part
276
VI
Fields
Extension
eR and a = over K. The axioms a vectorspaceover
for r
Euclidean
of the 30.3
Example
For any field is ordinary
of F[x] by 3\302\276-for
(a\\,
K\". With these operations,R\" becomes a vector space In particular, l2=Kxlas vector space arereadily checked. can be viewed as all \"vectors whose points are the origin starting in the sense often studied in calculus courses. \342\226\2 \342\226\240 \342\226\240 e \342\226\240, an)
for a R
plane\"
can be viewedas a vector
F, F[x]
of F is
element
an
in F[x]
of polynomials
addition
immediately
F, where
over
scalar
multiplication in F[x]. The
from
the
multiplication
ordinary
space then follow
a vector
space and
that
fact
of vectors
addition
element axioms ^f through F[x] is a ring with aa of an
\342\226\2
unity. 30.4
Example
be an extensionfield of a field F. Then E addition of vectors is the usual addition
Let E F,
where
the
usual
field
from the
multiplication for E.
this example
are assuming
We
we need
everything
is the
that
nothing the
from
vector
about
definition,
as a
regarded
in E
scalar
and
and a e E. The field of scalars is actually important one for us. a
with
Here our
axioms
field It is
vectors.
in E
be
can
e F
vector spaceover
multiplication follow
axioms
a subset
of our
aa is at once
space of
\342\226\2
from previous work and shall prove even though the results may be familiar from spaces
calculus.
30.5Theorem
If
a vector e F and a
V is
all a
Proof
spaceover
The equation 0a
read \"a(O-vector) for a ring and
= 0 is =
an
on
depend heavily
again
in the
equation
0, aO =
0 and
=
=
for \342\200\224(aa)
(\342\200\224a)a a(\342\200\224a)
read \"(O-scalar)a = 0-vector.\"Likewise,aO = 0 is to be The proofs here are very similar to those in Theorem 18.8
to be
0-vector.\"
= (0
(0a)
is
0a =
then
F,
e V.
abelian
laws
distributive
+ 0)a
{V,
group
the
+),
9^ and 9%. Now
= (0a)+ (0a) so by the
cancellation
group
law,
0 =
0a.
Likewise,from = a(0
a0 we
that aO
conclude
= \342\200\224(aa). so (\342\200\224a)a Likewise,
conclude
Linear
30.6 Definition
Let
V be
generate)
aO =
and
a vector
F.
if for
spaceover every
fi
e /5
for some a-} combination of
the
e
F and
a, .
= aa
(-a))a
+
(\342\200\224a)a,
= aa a(a + (\342\200\224a))
+
a(\342\200\224a),
= also. a(\342\200\224a) \342\200\224(aa)
Independence
V
+
from
0 = that
= a0 + a0,
= 0. Then
0 = 0a = (a
we
+ 0)
a,-; e
Bases
The
V, we =
\342\231\2
vectors in a
{a;| i e 1}
of
V span
(or
have +
flja,-,
S, j =
subset S =
1,
a2ah_
+
\342\200\242 \342\200\242 n. A \342\200\242,
h anain
vector YTj=i
ajaij 1Sa linear
\342\226\
30.7Example In the
vector
---,0) + ^(0,1, ---,0)H
= 01(1,0,
(^,02,---,^)
> 0 span F[x] over
xm form
monomials
the
0, ---,1)
for
spanK\",
Also,
277
Spaces
vectors
1, ---,0),---,(0,
---,0),(0,
(1,0, clearly
of Example 30.2,the
W over K
space
Vector
30
Section
the vector
F,
1).
0,---.
\\-an(0,
space of Example 30.3. \342\226\262
30.8
30.9
30.10
Example
Definition
Example
be a field
Let F
space
{1, a,
where
\342\200\242 \342\200\242 \342\200\242, a\"-1},
extension field of F. Let ae\302\243be algebraic over F. Then F and by Theorem in 30.18, it is spanned by the vectors n = deg(a, F). This is the important example for us. \342\226\262
E an
and
F(a) is a vector
over
A vector
since
Example
The
subset S =
in a
vectors
we have
ajaij
J2\"j=i
over F,
the vectors
Thus
could not
degree
that the
Observe
1,
e
vectors are that
\342\226\240 \342\226\240 n such \342\226\240,
vectors
combination
the
spanning
linearly
space
over
V
that
is a
F(a)
for us.
a field
e
F are linearly Fandw
e Z+,
\342\200\242 n. If the vectors are not \342\200\242, j = 1,\342\200\242
over F.
\342\226\240
0>
where
not all
W that are given
in
aj =
Example
0. 30.7 are linearly
in {xm | m > 0} are linearly independent over R. Likewise,the vectors independent vectors of F[x] over F. Note that (1, \342\200\2241), (2, 1), and (\342\200\2243, 2) are linearly dependent
+ (2, 1)+ 3(-3, 2)
7(1, -1) Let
an extension field
E be
n, then
by
Theorem
e
F. In
for 0. Thus
the
=
(0, 0)
= 0.
ofa field F, and let a e E be algebraicover F. If element of F(a) can be uniquely expressed
29.18, every b0
for bi
in
K, smce
over
30.14Example
\342\226\262
independent over F if the only way the of the vectors a, is to have all scalar over F, then there exist dependent
=
ajaij
YTj=\\
is
section.
e S, coefficientsaj
dependent
linearly
1} are linearly
a linear
0. If the
to
are
they
space
a,
= 0 for
if aj
only
a vector
vectors
distinct
F
be linear \342\226\262
idea in this
important of
in {a,- | i
as 0-vector canbe expressed
coefficientsequal aj e F for j =
most
the
{a,| i e 1}
for any = 0 in V
F if,
independent
linearly
of V
subset
finite
vector spaceF[x] over
of arbitrarily polynomials large any finite set of polynomials.
contains
definition
next
independentover
30.13Example
is a
If F < E and a e E is algebraic over the field F, Example 30.8 shows finite-dimensional vector spaceover F. This is the most important example The
30.12Definition
of
dimensional. The
is finite
R\"
that
dimensional,
combinations of elements 30.11
if there
dimensional
finite
\342\226\240
30.7 shows
Example not finite
F is
a field
space V over whose vectors span V.
particular, 0 elements
+b1a
+ ---
= 0 + 0a -\\
1, a,
\342\200\242 \342\200\242 a\"\"1 \342\200\242,
+
deg(o\\ F) = in the form
+ bn.1a\"-1 0a\"_1
are linearly
must be
a unique
such
independent vectors
in
expression over F{a)
278
Part VI
Extension Fields
for F(a)
30.15Definition
If
over F. This
is the
a vector spaceover
V is
for
a basis
definition,
1, a,
example for us. In fact,
form a basis reason we are
\342\200\242 \342\200\242 a\"\"1 \342\200\242,
this
is the
spaces.
\342\226\262
F, the
a field
F if they
over
V
next
the
important
on vector
material
this
doing
also span F(a), so by
F. They
field
the
span
a subset
are linearly
and
V
vectors in
B =
{$ | i e 7}of
V form
independent.
\342\226\240
Dimension to prove about vector spaces are that every finiteThe only other results we wish dimensional vector space has a basis,and that any two bases of a finite-dimensional Both these facts are true without the vector of elements. space have the same number the more that the vector is finite but dimensional, assumption proofs require space case is all we need. and the finite-dimensional knowledgeof set theory than we are assuming, we
First
30.16Lemma
lemma.
a vector space overa field F, and let a e V- If a is a linear combination \342\200\224 \342\200\242 \342\200\242 of vectors in V for i m and each fr is a linear combination 1, \342\200\242, Yj in \342\200\242 \342\200\242 then a is a linear combination of the Yj \342\226\240 \342\200\242, \302\273, j = 1, V be
Let
vectors
for
Proof
an easy
give
/J,-
Let a
= J2?=i aiPi>
and
let
/J,-
Ha\\ r=
30.17 Theorem
In
Proof
Let
V
that
be
is a
finite
l
where n
n
HbiiYi
\\y = l
vector space,
a finite-dimensional
a subset
J2\"j=i bijYj> I
m
01 =
=
/
= ) I]= l
every finite
y
set of
and
at
I
btj are in
of V
F. Then
m
HaibH
\\r = l
)yj>
/
vectors spanning
the
space
contains
basis.
dimensional
over F,
and
let
\342\226\240in V \342\226\240, a\\, \342\226\240 an span V. Let us list starting at the left with i = 1, and discard
vectors
a row. Examine eacha,- in succession, is some linear combination of the preceding a; for i < j. Then continue, with the following and discard the next 0\302\276 that is some linear combination starting o/+i, of its remaining and so on. When we reach an after a finite number of predecessors, of the steps, those a, remaining in our list are such that none is a linear combination preceding a, in this reduced list. Lemma 30.16 shows that any vector that is a linear combination of the original collection of a,- is still a linear combination of our reduced, and possibly smaller, set in which no a, is a linear combination of its predecessors. Thus in the reduced set of a,- again the vectors span V. For the reduced set, suppose that the
at in
the first cij that
aiatl for that
i'i
ar
<
< \302\2432
7^ 0,
\342\200\242 \342\200\242 \342\200\242 <
or we
ir and
that some
could drop
arair
h arair
-\\
aj from
= 0
may assume from the left side of the equation.
0. We =\302\243
Theorem 30.5 Then,
using
Section 30
30.5
Theorem
shows
which
Thus
construction.
so they 30.18
Corollary
Proof
that air is a the vectors
for
is
theorem
= {a\\,
LetS
F.
over
V
the
\342\200\242 \342\200\242 be \342\200\242, ar)
\342\231\246
a finite
By Corollary sequence
a finite
the
work on vector spaces.
of linearly
vectors of a finite-dimensional to a basis for V over F. Furthermore,
independent
be enlarged
F, then
over
basis B =
is a
that span
\342\231\246
of our
afinite set
of vectors
set
proof.
culmination
there
30.18,
basis.
vector space has the
vectorspaceV over a field F. Then S can \342\226\240 is if B = [Pi, \342\226\240 \342\200\242, any basis for V pn) Proof
of its predecessors, contradicting our set both span V and are linearly independent,
reduced
the
in
30.17 completes
Theorem next
a,
a finite-dimensional
By definition, The
linear combination
vector spacehas
A finite-dimensional
space.
30.19Theorem
we obtain
again,
a basis
form
279
Vector Spaces
r < n.
\342\200\242 for \342\226\240, p\342\200\236]V {Pi,\342\226\240
over F.
Consider the
finite
of vectors -
Oil,
--
Pi,
,
\342\200\242 \342\226\240 \342\200\242
, Pn-
B is a basis. Following the technique, since used in Theorem 30.17, each vector that is a linear combination of its remaining predecessors, we arrive at a basis for V. Observe that no a,- is cast out, since from left to right, working the a; are linearly Thus S can be enlargedto a basis for V over F. independent. For the second part of the conclusion, consider the sequence
These vectors span
of discarding
V,
in turn
\342\226\240\342\200\242\342\200\242
Oil, Pi,
These vectors
are not
linearly
since
the
/3,-
form
over F,
independent on
, Pn-
-\\
=t>iPi
because a\\
is a
linear combination
\\-bnpn,
Thus
a basis.
al+(-bl)Pi
+ --- + (-bn)pn = 0.
vectors in the sequence do span V, and if we form a basis by the technique of working from left to right and casting out in turn each vector that is a linear combination at least one pt must be cast out, giving of its remaining a basis predecessors, The
where m <
n
\342\200\224 1.
Applying
the same technique to \302\253!, a2,P{i\\
we
arrive
at a
the
---,/^,
new basis
{ai,a2,/8?\\---,/8,(2)},
sequence
of vectors
280
Part
Fields
Extension
VI
with s
<
where
0 < t
we arrive
\342\200\224 2.
n
Continuing,
{ai,
30.20
<
Let B =
B as an independent
symmetric argument If
Definition
is a
V
of
Let
algebraic over F and F is
n. This is the
An
Application
30.22,
an additional 30.23
Theorem
of field them
be an extensionfield
i^(a)
is an
stated
F
shown
have
We
of
in
the
\342\231\246
a e
E. Example
30.14
of F{a)
dimension
shows
as a vector
that if a is space
over \342\226\262
vector
and let
a e
E be
over F
space
is algebraic
space
over
F,
ideas to field
basis
{1, a,
and deg(/6,
F) <
and
theorem
gives
theory.
over F. If
algebraic
with
30.4,30.8,30.11,30.14,
sentence of this
The last
theorem.
one
into
in Examples
contained
theory
of F,
of F(a)
deg(a, F)
\342\226\240 \342\226\240 \342\200\242, a\"-1}.
= n, then
Furthermore,
deg(a,F).
in the preceding examplesexceptthe very important result of the above theorem. Let fi e F(a), where a is algebraic over the elements
everything
last sentence n. Consider
degree
m. A
Theory
vector
\302\273-dimensional
the
then
nice application of these
every element p
Proof
to Field
Let E
30.19, n <
for us.
example
important
and incorporate
two bases. Then by Theorem B' as a basis, we seethat
of
number
n.
and let
F,
= n,
F)
deg(a,
the results
collect
We
m =
so \302\273,
field of a field
an extension
\302\243 be
and
vectors
same
the
have
vector space overa field F, the number of elements in a basis \342\226\240 of basis, as just shown) is the dimension of V over F.
choice
the
set of
F
V over
space
\342\226\240 \342\226\240 be \342\226\240, fi'J
{fi[,
m <
gives
finite-dimensional
(independent Example
vector
B' =
\342\226\240 \342\200\242 and \342\200\242, /6\342\200\236}
{fii,
regarding
30.22
\342\231\246
bases of a finite-dimensional elements.
Proof
30.21
\342\226\240\342\226\240-,ar,p{{\\---^^),
Any two
Corollary
basis
r
\342\200\224 r. Thus
n
at a
finally
\342\200\242 \342\200\242, /6\". 1,/6,/6 , \342\200\242
These
cannot be
for by
Theorem
elements of F(a) that are linearly independent over F, any basis of F(a) over F would have to contain at least as many elements as are in any set of linearly independent vectors over F. However, the basis \342\200\224 \342\226\240 has just n elements. If /6! = pj, then \342\226\240, {1, a, \342\226\240 an~1} /6! pj = 0, so in any case there
exist b( e
F
n
1 distinct
+
30.19,
such
that
b0 + not
where
such
that
+
blp+b2p2
=
all bt = 0. Then f(x) + bnxn = 0. Therefore, /6 is algebraic /(/6)
---+^=0,
\342\226\240 \342\226\240 \342\226\240
+
b\\x +
over F
and
bo is a nonzero deg(/6,
30
\342\226\240 EXERCISES
Computations
1.
Find
three
In Exercises
bases
2 and
for R2
over R, no two
3, determine
whether the
of which given
set
have a
vector in
of vectors
common.
is a basisfor
R3 over
R.
element
F) is at
of F[x]
most n.
\342\231\246
2. {(1,1,
4 through
Exercises
In
3. {(-1,1,2),
1, 1)} 9, give a basisfor
1), (0,
(1, 0,
0),
vector space
indicated
the
5.
over Q
4. Q(V2)
7. C over 9. Q(v^)
over Q over Q Q(z')
6. Q(^2) 8.
10. According
Theorem
to
irreducible
the element 1 +
30.23,
for 1 +
polynomial
a in
a
of
R over
Q
29.19 is algebraic over Z2. Find
of Example
Z2(a)
-14, 0)}
field.
R
over
R(V2)
1), (10,
-3,
(2,
over the
281
Exercises
Section 30
the
Z2[x].
Concepts
11 through
In Exercises
is needed, so that 11.
12.
the definition of
14, correct
it is
in a form
V over a field The vectors in a subset S of a vector space combination of the vectors in S. uniquely as a linear
The vectors in a zero vector cannot
13. Thedimension 14.
A
15.
Mark
F of
over
for a
basis
a finite-dimensional
F span
a field F
over
vector
are
V if and
space
V over
if each
only
p e
over
independent
linearly
of vectors in
combination
linear
vector space V following
a field F
over
if correction
V
can
be expressed
F
if
only if the
and
S.
a field
F is
the minimum number
of vectors
of two
vectors
b. Thesum
of two
scalars is
c. The
product
of two
d. The
product
of a
e. Every
vector
g. The
0-vector
is a
h. If F < E and i. If F
< E
and
vectors
in V that
span
V
and
are linearly
dependent.
vector.
a vector. scalars is a scalar.
scalar
a vector
and
space has a finite f. The vectors in a basis are linearly may
is a setof
or false.
true
a. The sum
j.
text,
the
V.
of the
each
as a
be expressed
to span
required
a vector spaceV
S of
subset
reference to
term without
italicized
the
acceptablefor publication.
be part
a
e E
a
e E
is a
dependent.
of a basis.
is algebraic over the is algebraicover the
Every vector spacehas
veaor.
basis.
field
F, then
field
F, then
a2 is algebraicover a + a2 is algebraic
F. over
F.
a basis.
The exercises that follow deal with the further of veaor spaces. In many for cases, we are asked to define study vector spaces some conceptthat is analogous to one we have studied for other algebraic structures. These exercises should our ability to recognizeparallel and related situations in algebra. Any of these exercises may assume improve of concepts defined in the preceding exercises. knowledge
16. Let V
a vector
be
a. Define
b. Prove
space over a field
a subspace of the
vector
F.
space
V over
F.
of subspacesof V is again a subspace of V over F. over a field F, and let S = {ctt | i e 1} be a nonempty collection space a. Using Exercise 16(b), define the subspace of V generated by S. b. Prove that the vectors in the subspaceof V generated S are precisely the (finite) by vectors in S. (Compare with Theorem 7.6.)
17. Let
18. Let
V
that
be
V\\,
spaces
an intersection
a vector
\342\200\242 \342\200\242be vector \342\200\242, V\342\200\236 Vt for
i =
1,
spaces over the same field F. Define show that the direct sum is again
\342\200\242 \342\200\242 \342\200\242, n, and
the
direct
a vector
sum
of vectors in V.
linear
combinations
\342\200\242 \342\200\242 \342\200\242 V\\ \302\251 \302\251 Vn of
space over F.
of
the vectors
282
Extension Fields
Part VI
19.
Generalize
30.2 to
Example
for any field
20. Define an
F.
of a
isomorphism
obtain
vector
the
ordered \302\273-tuples of
F\" of
space
of F over
elements
the
field
F,
basis for F\"?
is a
What
vector space V
a field F
over
space V
a vector
with
over the
same
field F.
Theory
21. Prove that for
22.
is a
if V
F
V over
Let F be
any
where fl;j,&i
e F.
a. Show
the \"system
that
only if the
and
(This result is regarded as the
b. From
part
has a = f)
+
012\302\276 +
\342\200\242 \342\200\242 \342\200\242 +
#21-^1
+
^22\302\276 +
\342\200\242 \342\200\242 \342\200\242 +
^ml-Xi
+
O.ml^-2
fundamental
=
b\\,
CllnX-n =
' '
'+
&2>
=
OmnXn
Dm,
\342\226\240 e F that is, there exist Xi, \342\200\242 \342\200\242, X\342\200\236 satisfy all m equations, if Fm is a linear combination of the vectors cij = {a\\j,- \342\226\240\342\226\240 ,amj). of a solution, but should to prove, being practically the definition really be solution theorem for a simultaneous existence of a system of linear equations.)
(a), show that if n
\342\226\240 of \342\200\242 \342\200\242, bm)
\342\200\224 m and
{cij
vector
vector spaces over the are following conditions
V be
and
V if the
into
equations in n unknowns\"
that
solution,\"
Prove that every finite-dimensional space F\" of Exercise 19.
24. Let V
uniquely
a\\nKn
+
\342\200\242 of V is a basis a subset {#, ft, \342\200\242 \342\200\242, ft} a as linear combination of the ft.
F, then
linear
simultaneous
flnXi
(&i,
straightforward
over a field be expressed
= 1,
\\ j
basis for
\342\200\242 \342\200\242 is a \342\200\242,\302\273}
F\",
the system always
then
has a
solution.
unique
23.
vector
V can
the \"system of m
Consider
field.
space
every vector in
only if
and
if
vector
finite-dimensional
V of dimension
space
same
A function a, fi e
field F.
for all
satisfied
0(a + /8)
=
n
over
a field F
is isomorphicto
V is a linear 0 : V -\302\273\342\200\242
V and
a e
the
transformation
vector
of V
F:
+ 008).
0(a)
0(aa) = a(0(a)).
a.
V is completely determined e /Jisabasisfor VoverF,showthatalineartansformation0 : V \342\200\224>0(#) e V. Let set of vectors, not necessarily distinct, of V. {fa | j e 1} be a basis for V, and let {ft' | z e 1} be any Show that there exists exactly one linear transformation V such that 0(ft) = ft'. 0 : V ->\342\200\242 {/6, |z'
If
by
b.
25. Let a.
vectors
the
V be
V and
To what linear
vector spacesover
concept
transformation
b. Define
that
we
have
the
studied for
Let
V
that
it is
the kernel (or nullspace) of 0, and when 0 is an isomorphism of V
27. Let the
a.
b.
vector space over a field a vector space over F.
be a
the
let
and
0 : V
structures
algebraic
\342\200\224>V be
a linear
of groups
and
F,
show with
rings
does the
concept of a
and let
that it
is a subspaceof
V.
V.
S be a subspace
of
V. Define
the
quotient
vector spacesover the same field F, and let V be finite dimensional vector space V over F. Let 0 : V ->- V be a linear transformation. Show that 0[V] is a subspaceof V'. Show that dim(0[V]) = dim(V) dim(Ker(0)). [Hint: Choose a convenient For example, enlarge a basisfor Ker (0) to a basis for V.] Theorem30.19. V and
dimension
transformation.
correspond?
c. Describe 26.
field F,
same
V be
space
over
V'/S, and
F. Let
dim(V) be
of the
basis
for
show
V, using
section 31
283
Algebraic Extensions
31
Section
Algebraic Extensions
FiniteExtensions In
30.23
Theorem
F, then every
over
we
in F[x], elements
31.1 Definition
Definition
extension field ofa field
algebraic
almost
interested
over F. In
exclusively
F and
studying
of F
extensions
in
E is algebraic
a e
of polynomials
zeros
containing
only
over F.
algebraic field
\302\243 of
F is an algebraic extensionof
a field
F
element
if every
in
over F.
is algebraic 31.2
is an
if E
that
element of \302\243(a)is be
shall
extension
An
we saw
If an
extension field
E is
a finite
\342\226\240
\302\243 of
a field
of degree
extension
F is of finite dimension n over F. We shall
n as let [E
a vector spaceover : F] be the degree
then
F,
n of
over F. To
field.
It
E
E \342\226\240
say that a field that just asserts
\302\243 is a \302\243 is a
of a field
F does not mean vector space overF, that
finite extension finite-dimensional
that
is finite
E
is, that
: F] [\302\243
is finite.
We and
use the
often
shall
only
fact
We need
= F. if \302\243
over enlarged to a basis for \302\243 Let us repeat the argument field F must be an algebraic
31.3 Theorem Proof
finite
extension
We must
show
A
field that
for
finite extension
of F,
then,
: F] [\302\243
= 1 if
be by Theorem 30.19, {1} can always only = \302\243(1)= F. : F] = 1 if and only if \302\243 F. Thus [\302\243 that a finite extension \302\243 of a of Theorem 30.23 to show of F. extension that
observe
a field
\302\243 is an
algebraic
extension of
a is \302\243,
algebraic
over
Theorem
\302\243 of
a e
if \302\243 is a
that
\302\243. By
\302\243.
30.19 if
=
[\302\243:\302\243]
n,
then
cannot be linearly
so there
elements,
independent
anan + all
=
0. Then
and
not
and
f(a) = 0. Therefore,
a,-
_f(x) = a
\\-
+ oq =
\342\200\242 \342\226\240 \342\200\242
+ axx
+
anxn
a\\a
over
is algebraic
exist a, e
\302\243 such
that
0,
+ oq is a nonzeropolynomial
in F[x].
\302\243.
\342\231\246
the importance of our next theorem. It plays a role in of the theorem of Lagrangein group theory. While its proof followseasily from our brief work with vector spaces, it is a tool of incredible An elegant application of it in the section that follows shows the impossibility power. of performing certain geometric constructions with a straightedge and a compass. Never a theorem that counts something. underestimate We
field
31.4 Theorem
If
\302\243 is a
is a
cannot
theory
finite
overemphasize analogous to the
finite extension field extension
of
role
of a field
K is
and \302\243,
a finite
extension
and \302\243,
[K
:
=
\302\243]
[K
:
: \302\243]. \302\243][\302\243
field
of
then \302\243,
K
284
Part
VI
Extension Fields
\342\226\240 Basis j
Basis
{a.iPji \342\226\240 Basis {or;]
31.5
Proof
Figure
\342\200\242be = 1, \342\200\242 as a vector a basis for \302\243 \342\200\242, \\ i n] space over F, and let the set \342\200\242 \342\200\242 be a basis for isT as a vector = The theorem will be 1, \342\226\240, m} space over \302\243. [Pj | j proved if we can show that the mn elements atPj form a basis for K, viewed as a vector over F. (See Fig. 31.5.) space we have Let y be any element of K. Sincethe fij form a basis for K over \302\243,
Let
{at
m
=
Y
for bj e
the a,
\302\243. Since
form a basis for
J2bjh F, we
over
E
have
n
bj
=Y^aiiUi
;=i
for
fl,j
so the
F. Then
e
I
j=i
\\
n
\\
remains
that
\"EijCijiatPj)
and
(E\"=1Cjjo;i)
i,j
1=1
/ span ^T over F. for us to show that the mn elements = 0, with c;j e F. Then
m\302\273 vectors It
m
a*/6,
e E.
Since the
are
over F.
independent
independent over E,
/^- are
elements
a, fij
we must
Suppose
have
n
=0
^C-jQ!,
(= 1
for
all ;'.
for
all
i
so J2\"=i
now the a, are independent over F, ;'. Thus the a, Pj not only span K over form a basis for ^ over
But and
Thus they Note remembering that
F.
that if
we proved {a,-
over E, for K over F. Figure
for K in a
moment.
\\i
= fields 31.5
1,
theorem
this
jWjisabasisforFoverFandf/Sj E then the
F < gives
< K,
a diagram for
this
but
cijai also
= 0
implies
that
are independent
c,j
= 0
over F.
\342\231\246
exhibiting a basis.It is worth \342\200\242 \342\200\242is a basis \342\200\242, | j = 1, m] set {a,/8j} of m\302\273 is a basis for products situation. We shall illustrate this further
by actually
\342\200\242 \342\200\242 \342\200\242
F
Algebraic Extensions
Section 31
31.6
Corollary
If
F,- is
\342\200\242 = 1, \342\200\242 r and \342\200\242,
a field for i of Fl,
extension
31.7
Corollary
[Fr
If E
is an extension field of F, a
By Theorem F(^) <
30.23,
E
= 2, while
contains field of F
that
extension 0:2, so
31.4
over
algebraic
a
[F(a) : F] and
by induction.
\342\231\246
e F(a),
F, and p
then
F)
deg(^,
illustrates a type
: F]. We
= [F(^)
F)
deg(0,
have
F <
divides [F(a) : F].
: F]
[F(fi)
of
argument
\342\231\246
one often
makes using
2. Note that is no element of Q(\\/2) that is a zero of x3 \342\200\224 2 is of degree 3 overQ,but 3 does not divide 2. \342\226\262 zero of x3 \342\200\224
extension field of a field over F. By definition, algebraic
necessarily
with
is
FJ.
: Fr_2] ---^:
F^W^
E be an
Let
in
:
eF
31.4
Theorem
there
31.7,
deg(V2, Q)
F
F) =
deg(a,
following example or its corollaries.
By Corollary
a finite
F).
F(a), so by
The
then Fr is
of F;,
extension
straightforward extension of Theorem
is a
proof
Theorem 31.4
31.8 Example
=
The
divides deg(a,
Proo/
a finite
and
[Fr : Ft]
Proof
F,-+1 is
285
F,
of E, not let a1( ao_ be elements is the smallest extension field of
and
F(ai)
can be characterized as the smallest both o^ and ai_. We could equally have started = (F(q!2))(q!i).We denote this field by F^, a2). Similarly, for
a\\.
E
in
(F{oli))(0:2)
(F(at))(0:2)
Similarly,
containing
\342\200\242 \342\226\240 \342\200\242 all the a, for an) is the smallest extension field of F in F containing \342\200\242 \342\200\242 \342\200\242 from = n. i 1, \342\200\242, We obtain the field F(ai, \342\200\242 the field F by adjoining to F the \342\200\242, an) elements a, in F. Exercise 49 of Section 18 shows that, analogous to an intersection of subgroups of a group, of an intersection of subfields of a field E is again a subfield \342\200\242 can be characterized as the intersection E. Thus F(o!i, \342\200\242 of all subfields of E \342\200\242, an) \342\200\242 F and all the a,for i = 1, \342\200\242 n. \342\200\242, containing
a,
31.9
Example
e F,
F(ui,
Consider Q(V2). the
a zero that
of x4 \342\200\224 10x2 +
{1, V3}
is
Theorem 31.4
Thus and
(see
demonstrated
method
the
easily
Q(\\/2)
over
that -Jl
discover
Q. Using
+ V5 is
in Example 23.14, we can show irr(V2 + V5, Q) = x4 - 10x2 + 1,
: Q]
the comment following V3) over Q.
the
theorem)
then shows
{1, V2,
that
V5, V6}
is \342\226\262
Q(V2,
real cube root of 2 and 2l/2 be the positive square root of 2. Then because deg(21/2,Q) = 2 and 2 is not a divisor of 3 = deg(21/3, Q). = 2. Hence {1,21/3,22/3} : Q(21/3)] is a basis for Q(21/3) over Q [Q(21/3,21/2) is a basis for Q(21/3,21/2) over Q(21/3). 31.4 Furthermore, {1,21/2} by Theorem
Let 21//3 be 21'2
1. By
is a basis for
{1, V2}
29.10, we can
is irreduciblein Q[x]. Thus = 4. Thus (V2 + V5) i Q(V2), so V5 i Q(V2).Consequently, a basis for Q(V2, V5) = (Q(V5))(V5) over Q(V2). The proof of V3)
(see
a basisfor Example
that
in Example
this polynomial
so [Q(V2 +
31.10
Theorem30.23shows
demonstrated
technique
the
i Q(21''3)
the
comment
following
the theorem),
{1,2^,2^,2^,2^,27^} is abasis for Now
21/6
Q(21/2,21/3)
is a
over
Q. Because 21'6 = 2(2^), we is irreducible over Q, by
zero of x6 \342\200\224 2, which
have
21'6
Eisenstein's
e Q(21/2,
21/3).
criterion,
with
Part
286
VI
Fields
Extension
/7 =
2. Thus 1/6)
and
31.4
Theorem
by
6 = [Q(21/2,21/3): Q] = = [Q(21/2, we
Therefore,
: [Q(21/2,21/3)
21/3) : Q(21/6)](6).
have
must
[Q(21/2, 21/3) :Q(21/6)]= so
21/3) = Q(21/6),
Q(21/2,
the
31.11
Theorem
even
extension,
a simple
E be an
ai,
\342\200\242in \342\200\242 \342\200\242, an
Suppose that E = F(ai, algebraic overF, so each is algebraicover F, and \342\200\242 n. Corollary \342\200\242, j = 2, \342\200\242
:
[F(ofi)
F]
Continuing
arrive
at
an
a field
F
form F(ai,
\342\200\242 \342\200\242 in \342\200\242, an)
the case
all
that
E is an algebraic extensionof F, each a;, is over algebraic every extensionfield of F in E. Thus F(a!) \342\200\242 \342\200\242 is \342\200\242 for \342\200\242, \342\200\242, F(al5 algebraic over F(at, \342\200\242 general, aj_i) o^)
\342\200\242 \342\200\242 Since \342\200\242, an).
a,- is
in 31.6 applied to
that
E is
a
the
of finite
sequence
is a
E
that
finite
done.
are
we
extensions
\342\200\242 \342\200\242 = \342\200\242, an)
E
of F.
extension
finite
Conversely, suppose = F, and F = F(l)
then
\342\200\242 \342\200\242 of \342\200\242, an)
1.
n >
\342\200\242 F, F(a{), F(au a2),\342\200\242 \342\200\242, Ffe,
then shows
\342\226\262
of elements algebraic extension ofa field F. Then there exist a finite number \342\200\242 if and if E is a finite-dimensional E such that E = F(a1( \342\200\242 \342\200\242, an) only extension of F. over F, that is, if and only if E is a finite
Let
space
for an extensionF(ai,
of the
F
1,
Theorem 31.3.
preceding
though
Let us characterize extensionsof a, are algebraic over F.
vector
Proof
comment
the
by
shows that it is possible
31.10
Example
to be actually
: Q]
Q(21/6)][Q(21/6)
> 1. If F(o!i) = F, we this process, we see from such that
are
^ F,
o^
let
ao_
[E : F]
^ F.
where ai
e F,
that since
31.4
,an) =
F(<*i
let
if not,
done;
Theorem
of F. If
extension
algebraic
If E
e E,
where
[E : F] is
= 1, Then
0\302\276 F(ai). \302\242.
finite,
we must
E.
Closures Algebraically ClosedFieldsand Algebraic We
31.12 Theorem
not yet
have
overF,
then
observed
so area
+ p,
afi, a
the
following
Let F be an
field
of F.
extension
is a subfield Leta, Theorem
of
p e
F,
=
the algebraic
extension of a field
- /^anda//},if
and is also included in
FE
Proof
is an
if E
that
^
fi
F and a, fi e E are algebraic O.ThisfollowsfromTheorem31.3
theorem.
Then
{a e
E
\\ a
is algebraic
closure of F in
Fe.ThenTheorem31.il shows 31.3 every element of F(a, /J) is
that
over F}
E.
F (a,
algebraic
p) is a finite over F,
that
extension is,
F(a,
and by c Fg. Thus P) of F,
Section 31
Fe contains
a +
also contains
\342\200\224 and
a
afi,
fi,
fi,
287
Extensions
Algebraic
a/fi for ft
^
0, so
Fe is a subfield
of
E. 31.13
Corollary
Proof
\342\231\246
of all
The set
numbers
algebraic
forms a
field.
corollary is immediate from Theorem is the algebraic closureof Q in C.
of this
Proof
numbers
because
31.12,
the set of all
algebraic \342\231\246
numbers have the property It is well known that the complex that every nonconstant Theorem of polynomial in C[x] has a zero in C. This is known as the Fundamental in Theorem 31.18. We now give a proof of this theorem is given Algebra. An analytic to other fields. definition generalizing this important concept
31.14Definition
F is
A field
inF.
algebraically closed if
nonconstant
every
polynomial
in F[x] has
a zero \342\226\240
of F in an extension field F canbe the algebraic closure field E without is the For of closed. closure algebraic Q in Q(x), but example,Q being algebraically 1 has no closed because x2 zero in is not + Q Q. algebraically of a field being algebraically closedcan the concept The next theorem shows that of polynomials also be defined in terms of factorization over the field. that a
Note
F
31.15Theorem
field
A
factors
Proof
\342\200\224
Then
(x \342\200\224 a)(x
\342\200\224
a)g(x).
and
only
if every
nonconstant polynomial
in
F[x]
be a nonconstant in F[x] polynomial \342\200\224 a x is so a factor of 23.3, f(x) = By Corollary f(x), if g(x) is nonconstant, it has a zero b e F, and we have f(x) = we a factorization of into linear in F[x] get Continuing, f(x)
Let F be algebraically Theri f(x) has a zero (x
if
F is algebraically closed in F[x] into linear factors.
b)h(x).
closed,
let f(x)
and
e F.
a
factors.
suppose
Conversely,
F
31.16
Corollary
An algebraically
Let
\302\243 be
x
a,
closed field F F
with
an algebraic
by Theorem F = E.
\342\200\224
have
that every nonconstant b is a linear factor
\342\200\224
polynomial
of f(x), then
of F[x] is a b/a
has a factorization zero of f(x).
closed.
is algebraically
extensions E Proof
If ax
factors.
linear
into
Thus \342\231\246
has
no
proper
algebraic
extensions,
that
is,
no algebraic
< E.
extension of 31.15, since F
F, so F < E. Then if is algebraically closed.
E, we have irr(a, F) = Thus a e F, and we must
a e
\342\231\246
that just as there exists an algebraically show closed extension an algebraic extension F K, for any field F there exists similarly of F, with the property that F is algebraicallyclosed.Naively, to find F we proceed as If a polynomial follows. f(x) in F[x] has a no zero in F, then adjoin a zero a of such an to thus the field F(a). Theorem 29.3,Kronecker'stheorem, is strongly F, f(x) obtaining used here, of course.If F(a) is still not algebraically closed,then continue the process is that, contrary to the situation for the algebraic closure C of K, further. The trouble infinite number we may have to do this a (possibly of times. It can be shown (see large) 33 and 36) that Q is isomorphic to the field of all algebraic numbers, and that Exercises In a
C of the
moment real
we shall
numbers
Part
288
VI
Extension Fields
we cannot
number of algebraicnumbers. We shall machinery, Zorn 's lemma, in order to be able to handle such a situation. This machinery is a bit complex, so we are putting the proof under a separateheading. The existence theorem for F is very important, and we state it here so that we will know this fact, even if we do not study the proof. 31.17 Theorem
It is
well
closure,
that
extension F
an algebraic
is,
is
that
the
f(z)
polynomial
e C[z] have no
is an
numbers
zero in
Then
C.
gives
l//(z)
an
entire
function;
C, lim^oo \302\242 \\f(z)\\ = oo, so lim^^co analytic everywhere. = Thus 1// must be bounded in the plane. Hence by Liouville's theorem of function and thus / is constant. Therefore, a nonconstant theory, 1 // is constant, \342\231\246 in C[z] must have a zero in C, so C is algebraically closed.
1// is
complex
polynomial
of an
Existence
the
of
if f
Also
0.
|l//(z)|
Proof
of complex
field.
closed
that is,
The field C
of Algebra)
Theorem
(Fundamental algebraically
algebraically closed field.We recall an analytic proof for of a complex variable. Thereare algebraic functions
C is an
a coursein are much longer.
has had
but they
proofs,
Let
a finite
adjoining
by
set-theoretic
algebraic
that
known
who
student
the
Proof
F has an
Q
closed.
algebraically
31.18Theorem
discuss
first
field
Every
Q from some
obtain
have to
Algebraic Closure
that every field has an algebraic extensionthat is algebraically closed. the Axiom should have the opportunity to see someproof involving the time finish This is a for a Choice natural such by of they college. place proof. We of the Axiom of Choice. To state shall use an equivalent form, Zorn's lemma, Zorn's we to a definition. have set-theoretic lemma, give shall
We
prove
Mathematics
31.19
Definition
A partial
students
ordering of
of elements
of S
a
1.
a
7^
Ifa
In
a partially
a e
S (reflexive law).
b <
a, then
either
have
a = b (antisymmetric a < c
ordered set, not
m of
collection
given Z and
by
c.
Q+,
a partially
T of
are comparable, bound for a
T
element
The
conditions
law).
law).
(transitive
or b <
a < b
\342\226\240
need be
two elements
every
ordered pairs
a. As
usual,
comparable; that is, b denotes a < b
a <
for but
b.
upper
31.20 Example
and
we need not
A subset in
all
< defined for certain are satisfied:
a relation
by
following
and b < c, then
3.
e S,
a,b
such that the
for
2. lfa
S is given
a set
set S
is a
a < b or
b <
subset
a partially of all
chain if
ordered
that is, either A
ordered
of partially
ordered
subsets of
For example, if the Z c Q+ nor neither
set S
set S if
is maximal
a set forms
a (or if
Q+ c
Z.
a
a partially
< u is no
there
whole set is K, we have
two
every
both).
An
for all a e s e S such
ordered Z
elements
c Q.
u e
element A.
a and b S is an
Finally, that
m <
the relation set under Note, however, that
an
s. < for \342\226\262
31
Section
31.21 Zorn'sLemma
If S
is a
S has
at
set such that
ordered
partially
every chain
an upper
S has
in
The lemma is equivalent There is no question ofproving Zorn's lemma. Choice. Thus we are really taking Zorn's lemma here as an axiom for a proof to the literature for a statement of the Axiom of Choice and
Zorn's lemma.
Axiom of
to the
Refer our set theory. of its equivalence to
[47].)
Edgerton
(See
is often
Zorn's lemma
then
in S,
bound
element.
one maximal
least
289
Extensions
Algebraic
we want to show the existence when of a largest kind. If a field F has an algebraic extension F that is be a maximal algebraic extensionof F, for closed, then F will certainly algebraically since F is algebraicallyclosed,it can have no proper algebraic extensions. The Given a field F, we shall idea of our proof of Theorem31.17is very simple. contain of F that is so large that it must first a class of algebraicextensions describe
or
of some
structure
maximal
F.
We algebraic extension of any conceivable and show subfield ordering, on this class,
(up to isomorphism) ordering, the ordinary
of Zorn'slemma extension F of
useful
then
a partial
define
that the hypotheses a maximal algebraic
By Zorn's lemma, there will exist this class. We shall then argue that, as a maximal element, closed. extension F can have no proper algebraic extensions, so it must be algebraically it uses in many texts. We like it because Our proof differs a bit from the one found into it throws no algebra other than that derived from Theorems 29.3 and 31.4. Thus lemma. of both Kronecker's theorem and Zorn's strength sharp relief the tremendous we are writing out The proof looks long, but only because every little step. To the the construction of the proof from the information in the mathematician, professional matter. This proof was suggested to the author during preceding paragraph is a routine his graduate student Norman student, days by a fellow graduate Shapiro, who also had
a strong
of
Axiom
the 1870
explicitly by
that such
for
of
a least Axiom
of
the set
for it.
preference
the
although used implicitly s, was first stated
in 1904
the result
theorem,
well-ordering
every
respect
asserted
Choice
S of
with
all
subsets
function, a
to <.
that, given
any
function
f
set
always
M and
in
deduction.\" A
: S \342\200\224> M such
e M' for
that
few
his collection
years later he included
of axiomsfor
set
theory,
this
that
to
equivalent his
the
the
what
into
theorem
well-ordering
of Choice),
Axiom
was more
lemma
he realized
Although
to
natural
the well-ordering theorem somehowa \"transcendental\" principal. Other mathematicians soon
appeared
is
consistently essentialpart
in
use
was
with his reasoning. The lemma 1939 in the first volume of Nicolas
in that
of the
de FAnalyse.
work and
quickly
mathematician's
it
(itself
agreed in
Structures Fondamentales
a collection
that
he claimed
Bourbaki's Elements de Mathematique:
in
axiom
1935.
in
algebra because
exists
M' in S. Zermelo noted, every \"this fact, logical principal cannot... be reduced to a still simpler one, but it is applied in mathematical without hesitation everywhere
that f(M')
(1906-1993) was
equivalent to
Zermelo's
of M, there
in 1930
modified
slightly
Zermelo-Fraenkel set theory, the axiom as used abasis of that system generally today theory. Zorn Zorn's lemma was introduced by Max
now called with
connection
in
was
which
exists an order-relation< nonempty subset B of A contains
element
a \"choice\"
1880
set A, there
any
that
Choice,
s and
Zermelo
Ernst
his proof
in this
Note
\342\226\240 Historical
The in
satisfied.
are
F
Les
It was became
toolbox.
used an
290
Part VI
Extension Fields
31.22 Restated
Theorem 31.17 Proof
It
can
in set
A =
of Theorem
31.17,which
we
restate
here.
algebraic closure F. that
theory
Suppose we form
our proof
out
carry
F has an
field
Every
be shown
elements.
to
now ready
are
We
any set, there
given
exists a set with
strictly
more
a set
e F[x\\,i = 0,---, (degree
[cofj | f
f)}
an elementfor every possiblezero of any f(x) e F[x]. LetQ bea set with strictly Q by Q U F if necessary, we can assume F c \302\2431. elements than A. Replacing extension of F and that, as sets, consist of all possible fields that are algebraic Consider field of F, If E is any extension One such algebraic extension is F itself. elementsof \302\2431. F and deg(y, F) = n, then and if y e E is a zero e F[x] for y \302\242. y by f(x) renaming of F(y) elements a^+aiy + \342\226\240\342\226\240\342\226\240 co for co e \302\2431 and co \302\243 + an_iyn~l F, and renaming distinct elements of Q as the at range over F, we can considerour renamed \302\243(y) by = 0. The set Q to be an algebraic extension field F(co)of F, with F(co) c ^ and f(co) elements to form F(co), sinceQ has more than enough elements to provide has enough n in any subset of F[x]. n different zeros for each element of each degree a set All algebraic extension fields Ej of F, with Ej c Q, form that has
more
S = that
ordered
is partially
itself. The preceding closed, there will *
a,
=
LetT /6 e
paragraphs be
many
{Ejk} be
W. Then
under
shows in
S, and
there exist Ej1, Ej2
e
e /}
sub field inclusion that if F is far away
usual
our
fields Ej
a chain in
{Ej | j
<. One element from
being
of
S is
F
algebraically
S.
let W = U^\302\243^. We now make W into a field. Let T is a chain, Since 5, with a e Ej1 and p e \302\243y2.
and \302\243,-2 is a subfield of the other, say E^ < E-n. Then a, p e Ej2, \302\243,-, the field operations of Ej2 to define the sum of a and p in W as (a + p) e \302\243^ the product as (ap) e \302\243j2. These are well defined in W; they and, likewise, operations of Ej2, since if a, p e Ej3 also, for Ej3 in 7\\ then one are independent of our choice of the other, since T is a chain. of the fields Ej2 and Ej3 is a subfield Thus we have addition and defined on of W. multiplication operations All the field axioms for W under these now follow from the fact that operations these operations were definedin terms of addition and multiplication in fields. Thus, for 1 F as in a if 1, a e \302\243,-,, e serves since for e W, W, identity example, multiplicative then we have la = a in \302\243^., so la = a in W, by definition of multiplication in W. As to check the distributive laws, let a, p, y e W. Since T is a chain, further illustration, all three elements a, p, and y, and in this field the we can find one field in T containing laws for a, p, and y hold. Thus they hold in W. Therefore, we can view W distributive as a field, and by construction, \302\243Jt< W for every Ejk e 7\\ If we can show that W is algebraic over F, then W e S will be an upper bound for T. But if a e W, then a e \302\243,-, for some Ejx in 7\\ so a is algebraic over F. HenceW is an algebraic extension of F and is an upper bound for T. The hypotheses of Zorn's lemma are thus fulfilled, so there is a maximal element F of S. We claim that F is algebraically closed. Let f(x) e F[x],where F. f(x) \302\243 that f(x) has no zero in F. Since Q has many more elements than F has, we Suppose
one of the and
we use
fields
291
Section31 Exercises where and form a field F(co) c \302\2432, with \302\2431, a^F, first paragraph of this proof. Let /3 be in F(co). Then
co e
take
can
in the
saw
a zero
by
as we
of /(*),
Theorem
30.23,
f)
is
of a polynomial
a zero
=
g(x)
+ 0\302\276
ot\\x +
\342\226\240 \342\200\242 \342\226\240 + a\342\200\236x\"
31.11 the field Thenby Theorem \342\200\242 we is over F(a0, F(ao, \342\226\240 \342\200\242, an), p algebraic \342\226\240 \342\200\242 \342\226\240 \342\226\240 Theorem also see that F(ao, is a finite extension over 31.4 \342\226\240, a\342\200\236, \342\200\242, F(ao, /6) an). \342\200\242 then shows that F(ao, \342\200\242 extension of F, so by Theorem \342\200\242, an, j3) is a finite 31.3, ft is the choice of F as over F. Hence F(
in
with a,
F[x],
\342\200\242 \342\200\242 is \342\200\242, an)
in
maximal
The
Sinceit proof
e F,
S. Thus
mechanics may
and
hence
be the
out in
detail.
the
degree
over F.
at algebraic
a finite extension
of F,
since
and
f(x) must have had a zero in
F,
of the preceding proofare routine first proof that we have ever seen
so F
is algebraically closed.
to the
professional mathematician. Zorn's we wrote the lemma,
using
\342\231\246
31
\342\226\240 EXERCISES
Computations 1 through
Exercises
In
13,
find
and a
basis for the
Be prepared
extension.
field
given
to
your
justify
answers.
1. Q(V2) over Q 3. Q(V2, V3, vT8) over
5. Q(V2,^2)over 7. Q(V2V3) over 9. Q(v^, $6,4/24) 11. Q(V2 + V3) 13. Q(V2,V6 +
2. Q(V2,
S) over Q + V3) over Q Q(V2, ^5) over Q
6. Q(V2
8.
Q over Q
over
over Q
4. Q(X/2,
Q
Q
VW)
V3)
Q(V3)
over Q(V3
+
10. Q(V2,
V6)
over
Q(V3)
12. Q(V2,
S)
over
Q(^
+ V3)
Vf5)
Concepts
14 through the definition of the 17, correct is needed, so that it is in a form acceptablefor publication. In Exercises
14.
extension of a field
An algebraic
F is
a field Fiai,
term without
italicized
\342\226\240 where ao, \342\226\240 \342\226\240, an)
reference
each a, is
to
a zeroof
the
text, if
some
correction
polynomial
in
F[x].
15.
A
16. The A
field
Show
algebraic F is
by
an
a field F is one that can be obtained by adjoining Fe of a field F in an extension field E of F is the
field of
closure
algebraic are
that
17. 18.
extension
finite
field
finite
number
consisting
of elements
to F.
of all elements
of E
over F.
algebraically closed if and only if every field example that for a proper extension
be algebraically
a
closed.
polynomial \302\243 of
a field
has
F,
a zero in the
F.
algebraic
closure
of F in E
need not
Extension Fields
Part VI
292
19.
of the
each
Mark
a. If a
following \302\243 is a
field
or false.
true
finite extension
of a field
finite field.
\302\243 is a
then
F,
b. Every finite extension of a field is an algebraic extension. c. Every algebraic extension of a field is a finite extension. d. Thetop field of a finite tower of finite extensions of fieldsis a finite extension e. Q is its own algebraic closure in R, that is Q is algebraically closed in R. f. C is algebraically closed in C(x), where x is an indeterminate. g. C(x)
is algebraically
i.
An
Proof
closure,
algebraic
closed
algebraically
since C
numbers.
all algebraic
contains
already
0.
of F,
field
extension
bottom field.
indeterminate.
an
be of characteristic
field must
closed
algebraically \302\243 is an
j. If
no
x is
where
closed,
C(x) has
h. The field
of the
then
\302\243 is an
algebraic extension
of F.
Synopsis
20. Give a one-sentencesynopsis of the proof of Theorem 31.3. 21. Give a one- or two-sentence synopsis of the proof of Theorem 31.4.
Theory
22. Let
23. Show F
bi) e
(a +
if
that
\302\243 is a
2A. Prove
that
25.
degree
F(a)
for
b ^
and
finite extension
E =
indeed,
and,
C where a, b e R
0.
of a field a e
every
E
Show
F and
that
C =
[E :
F] is a prime
R(a +
bi). then
number,
simple extension
\302\243 is a
of
in F.
not
over (
x1 \342\200\224 3 is irreducible
field extensions can we obtain adjoining to a field F a square root of an element by successively in this new field, and a square in F, then square root of somenonsquare so on? Argue from this that a zero of x14 \342\200\224 3x2 + 12 over Q can never be expressed as a rational function of square roots of rational of square roots, and so on, of elements of Q. functions 26. Let \302\243 be a finite extension field of F. Let D be an integral domain such that F c D c E. Show that D is a field. What
of
F not
27. Prove in 28.
29.
30.
that Q(V3
detail
+ V7)
=
V7).
Q(V3,
0, then Q(^/a + Vb) Generalizing Exercise 27, show that if Ja + 4b \302\261 [Hint: Compute (a - b)/{Ja+\302\253Jb).] Let \302\243 be a finite extension of a field F, and let p(x) e F[x] be irreducible a divisor of [E : F].Show that p(x) has no zeros in E.
Let
\302\243 be
31. Show 32. Let
F, and
over
if F,
that
over F,
and
\302\243 be
an
E,
\302\243 be
K are
and
F. Leta
F(a) =
over
an algebraically
\302\243 \342\202\254 be
of odd
algebraic
F and
over
degree over F. Show
for
all a
and
&
have degree that that
in Q.
is not
a2 is
algebraic of
if E
is algebraic
F(a2).
fields
algebraic over E. extension field of a field K is
E is transcendental 33. Let
field of
an extension
odd degree
= Q(Va, Vb)
< E
with
F
(You
must
F.
< K,
then
not assume
that every a
Prove
K is
the
algebraic over are
extensions
e E that
is not
F if and
only
finite.)
in the
algebraic closureFE
of
F in
FE.
closed
algebraically
closed.
algebraically
closed field.)
(Applying
extension
field
this exercise
to
of a field C and
F.
Q, we
Show
that
see that
the algebraic field of all
the
closure FE of F in E is is an algebraic numbers
Section32 GeometricConstructions
293
F and contains all zeros in F of every fix) e F[x]. then \302\243 is an algebraic extension of a field E algebraically closedfield. is algebraically closed. (Actually, no finite field of characteristic 2 that no finite field of odd characteristic Show field F, some polynomial x1 \342\200\224 is algebraically closed show that for such a finite a. either.) [Hint: By counting, for some a e F, has no zero in F. See Exercise32, Section 29.]
Show that if
34.
is an
35.
algebraic closure of Q in C is not a finite extension of R is either R itself or is isomorphicto C. 37. Argue that every finite extension field in 38. Use Zorn's lemma to show that every proper ideal of a ring R with unity is contained
36.
Prove that, as
SECTION 32
asserted in
the
text, the
of Q.
some
maximal ideal.
tGEOMETRICCONSTRUCTIONS to give an application demonstrating the power of this section we digress briefly to Theorem 31.4. For a more detailed study of geometric constructions, you are referred Courant and Robbins [44, ChapterIII]. in what types of figures can be constructed with a compass and We are interested We shall discuss the in the sense of classicalEuclidean a straightedge geometry. plane of trisecting certain angles and other classical questions. impossibility In
Numbers
Constructible
a single line segment that we shall define to be Let us imagine that we are given only one unit in length. A real number a is constructible if we can construct a line segment of length |a | in a finite number of steps from this given segment of unit length by using and a compass. a straightedge The
rules
of the
at the moment, the to the points (0,0)
game are pretty
endpoints
We suppose that we are given line segment, letus supposethat Euclidean plane. We are allowedto strict.
of our unit
and (1,0) in
the
just they draw
two points correspond a line only
already located. Thus we can start and drawing the line through (0, 0) and (1, 0). We are allowed by using the straightedge we have already found. Let us to open our compass only to a distance between points open our compass to the distance between (0,0) and (1, 0). We can then place the point of the compass at (1, 0) and draw a circle of radius 1, which passes through the point located a third point, in this way, we can have (2, 0). Thus we now (2, 0). Continuing and so on. Now open the compass locate points (3,0), (4,0), (\342\200\224 the distance 1,0), (\342\200\2242,0), from (0,0) to (0,2), put the point at (1,0), and draw a circle of radius 2. Do the same with two new points, where these circlesintersect, the point at (\342\200\2241, 0). We have now found and we can put our on them to draw what we think of as the y-axis. Then straightedge from (0, 0) to (1, 0), we draw a circle with center opening our compass to the distance at (0, 0) and locate the intersects the y-axis. Continuing (0, 1) where the circle point in this fashion, we can locate all points with integer coordinates in any rectangle (x,y) that it is (0, 0). Without containing the point going into more detail, it can be shown to erect a perpendicular to a given line at a known point possible, among other things, with
our
straightedge
1 This chapter is
through two
not usedin
points
that
the remainder of the text.
we have
294
Part VI
Extension Fields
on
32.1 Theorem
Proof
the
line,
and find a line
Our first
result is
If a
/3 are
and
the
a known
passing through
constructible
point and
parallel to a given
line.
theorem.
following
real numbers,
so
then
are a
+ /J, a
\342\200\224
/J, afi,
and a/fi,
if
that a and fi are constructible, so there are line segments of lengths \\a | and to us. For a, fi > 0, extend a line segment of length a with the straightedge. at one end of the original Start of length a, and lay off on the extension the length segment a + /J; a \342\200\224 with the This constructs a line fi is similiarly /J compass. segment of length an obvious breakdown into constructible (see Fig. 32.2).If a and fi are not both positive, to their signs shows that a + fi and a \342\200\224 cases fi are still constructible. according The construction of afi is indicated in Fig. 32.3. We shall let O A be the line segment from the point O to the point A, and shall let | OA| be the length of this line segment. If OA is of length |a|, construct a line I through O not containing OA. (Perhaps, if O is at (0, 0) and A is at (a, 0), you use the line through (0, 0) and (4, 2).) Then find the PA and 1 and OP is of length |/J|. Draw points P and B on I such that OP is of length to PA and intersecting OA extended at Q. By similar construct /' through P, parallel We
are
given
| fi | available
triangles,
we have
1
so O
Q is of length
|a/J
III
_
l\302\253l
\\OQ\\'
|.
if fi ^ 0. Let O A be of length \\a |, Finally, Fig. 32.4 showsthat ct/fi is constructible and construct I through O not containing P on I such that OB OA. Then find P and 1. Draw BA and construct /' through is, of length to P, parallel | fi \\ and OP is of length OA at Q. Again BA, and intersecting by similar triangles, we have
1061 _ 1
so OQ
is of length
l\302\253l
III'
\342\231\2
\\a/p\\. a
a
fj
\342\200\242
\342\200\242
QT +
*
\342\200\242 *
<*
/S
32.2Figure
32.3
Figure
- 0
\342\200\242
0
Geometric Constructions
32
Section
295
32.4Figure 32.5
Corollary
Proof
The set
of all constractiblereal
Proofof
this
the field
Thus
since
numbers,
From now our
Regarding
on,
given
constractiblereal
F of all
Q is
we proceed segment
with
1 as
We can construct
analytically.
as an intersectionof points
2.
as an radius
and a
in the
be found
two
lines,
number.
in one
each of
of the
which
can locateby
three
following passes
#2) in the
(qi,
point
plane that we
through
plane using
ways:
two known
line
that
two points having through rational coordinates
passes
has
rational
and whose
is rational.
of lines
Equations
point
circle whosecenter
as an intersection of two whose radii are rational.
3.
rational
coordinates,
of a
intersection
coordinates
we can locate any
Any further can
rational
having
any
1
an x-axis,
on
unit
rational.
and a straightedge
a compass
1.
the basic
coordinates
both
of rational
Q, the field
contains
numbers
0
of length
\342\231\246
of R.
subfield
smallest
the
numbers.
of real
field
from Theorem32.1.
is immediate
corollary
F of the
a subfield
forms
numbers
of the
and circles
centers have
whose
circles
in 1, 2,
discussed
type
ax + by
+
+ dx
+
c =
0
+
f
rational
and
coordinates
3 are
of the
and
form
and
x2
where a, b, c, d, e, and with
f
+ y2
are all in
Q. Since
ey
= 0, 3 the
Case
in
intersection of
equations
x2 + y1 +
dix + eiy
+
fi
x2 +
d2x +
+
/2=0
= 0
and
is
the
same
as the
y2
intersection of the x2 +
y2
+
first +
dix
eo_y
circle
+ ay
having equation
+ /1
= 0,
two
circles
296
Part VI
Fields
Extension
line (the common chord)having
and the
{di
- d2)x +
equation
(ei -
+
e2)y
/i
-
/2 = 0,
we seethat
Case 3 can be reduced to Case 2. For Case of two 1, a simultaneous solution linear equations with rational coefficients can only lead to rational values of x and y, us no new points. However, finding a simultaneous solution of a linear equation giving with rational coefficients and a quadratic with rational coefficients, as in Case 2, equation
leads, upon
the
32.6 Theorem
a
smallest
>
can obtain and
Proof
square
when solved by the of numbers that are not
equation,
roots
argument,
field
containing
consists constractible real numbers precisely from Q by taking square roots of positive numbers a finite number of field operations.
F of
field
The
Such an
to a quadratic equation. have solutions involving
field axioms. Q except nothing was really usedinvolving constructed so far, the argument those real numbers for some a e H, the \"next new number\" constructed lies in a field H(^/a) 0. We have provedhalf of our next theorem.
preceding
If H is the shows that where
may
in Q.
squares In
substitution,
formula,
quadratic
applying
of all real numbers that we a finite number of times
shown that F can contain have no numbers from Q by except those we obtain and applying a finite number of square roots of positive numbers taking a finite number of field operations. However, if a > 0 is constractible, then Fig. 32.7 shows that ^/a is constractible. Let OA have length a, and find P on OA extendedso that OP has with PA as diameter. Erect of PA and draw a semicircle length 1. Find the midpoint a perpendicular to PA at O, intersecting the semicircle at Q. Then the triangles OPQ and O QA are similar, so We
1061
=
\\OA\\
and | O Q |2 = numbers
are
Theorem
la = a. Thus
O Q
is of length
\\OP\\
\\OQ\\' ^/a.
Therefore
square roots
of constractible
constractible.
32.1 showed that
field
operations
are possible
a
1
32.7 Figure
by
construction.
\342\231\24
GeometricConstructions
Section 32
32.8
Corollary
If
Proof
y such
\342\226\240 \342\226\240= \342\226\240, an
a\\,
In
that
of the
existence
then
Q, \302\243
y
Q(a!,
\342\226\240 is \342\226\240 \342\226\240, at)
a, is immediate =
2\"
from
The We
32.9
Theorem
Proof
32.4,
which
the proof.
\342\231\246
Constructions
of Certain
of certain geometricconstructions.
show the impossibility
now
can
32.6. Then
Theorem
\342\226\240\342\226\240-,\302\253\342\200\236) :Q(y)][Q(y):QL
completes
Impossibility
> 0.
[Q(ai,---,a\342\200\236):Q]
= [Q(ai, by Theorem
there is a finite of real numbers sequence \342\226\240 an extension of Q(ai, \342\226\240 of degree 2. \342\226\240, a,_i)
2r for some integer r
Q] =
[Q(y):
particular,
The
and
constructible
is
y
the cube is impossible,that is, given a side of a cube,it is to construct with a straightedge and a compassthe side of a cube volume of the original cube.
the cube have a side of length 1, and hence a volume of given of 2, and hence a side of length sought would have to have a volume x3 \342\200\224 2 over Q, so zero of irreducible
Let
shows
32.8
Corollary
for someinteger Theorem
not always possible that has double the
Doubling
[Q(^2) : Q] =
32.10
297
that to
r, but no
double this
a
with
construct
1, we would
to have
need
such r exists.
being
is a
3 = 2r \342\231\246
the circle is
Squaring
cube
But ifl
ifl.
3.
of volume
cube
1. The
impossible;that is, given straightedge and a compassa square
a circle,
is
it
not
always
area equal
having
to
the
possible to area
of the
given circle.
Proof Let the
circle
given
a square of side*Jtt also.
32.11
Theorem
Proof
Trisecting the with a straightedge
Figure 32.12 length
|
cos
an area of n. We would need to construct over Q, so *Jtt is transcendental over Q \342\231\246
angle is impossible;that and
is,
the
9 can
angle
be constructed. Now
be trisected.Note cos W
exists an angle
there
be trisected
cannot
that
a compass.
indicatesthat
9 | can
that it cannot
have a radius of 1,and hence . But n is transcendental
be constructedif
60\302\260 is
a constructible
and
only
angle,
that
= cos(2# +
9)
= cos29cos9 \342\200\224 sin 29 = (2 cos29
\342\200\224
= (2cos2
-
= 4cos3#
9
1) cos
9
sin 9
\342\200\224 2
sin 9
cos 9 sin 9
1) cos 9 - 2cos9(1-
-3cos#.
cos2
9)
and
if a we
segment of shall
show
Extension Fields
Part VI
298
6
cos
32.12 Figure
[We realize that
today have
students
many
not
the trigonometric
seen
1 repeats Exercise 40 of Section1 3 cos 6 from Euler's formula.] 4 cos36 \342\200\224
Exercise
used.
=
cos 36
Let 6 =
3cos6 =
cos
cos 36 we see that
so 20\302\260, 36,
that
=
\\,
let a
and
4a3
and
asks
you
= cos 20\302\260. From
identities we just
to prove the identity
the
identity
4 cos3 6 -
-. 2
-3a=
zero of 8x3 \342\200\224 6x \342\200\224 1. This polynomial is irreducible in Q[x], since, by Theorem 23.11, it is enough to show that it does not factor in Z[x]. But a factorization in Z[x] would entail a linear factor of the form (8x \302\261 1), (2x \302\261 1), or (x \302\261 1), (4x \302\261 1). We can quickly check that none of the numbers and is a zero of \302\2611 \302\261|,\302\261|,\302\261|, 8x3 - 6x - 1.Thus is a
a
Thus
[Q(cO
so by Corollary32.8,a is
not
mathematicians as far Greekcentury
B.C.
to trisect
circle. Although
prove that
such
problems
century
who
the
60\302\260 cannot
be trisected.
\342\231\2
made
tools, including a detailed
with
equations
p prime whoseroots /7-gon.He showed
the
of the form that
the
early
study
solution
his
cube,
anyone
they
of
conic
constructibility
of cyclotomic
form xp
\342\200\224 1 =
the
second.
who
proved Corollary 32.8 and also demonstrated 32.9 and 32.11. The proof of Theorem 32.10, on the other hand, a proof that n is requires
Theorems
0 with
of a regular all such equations
the vertices
although
sections.
nineteenth
1 is not a power if p \342\200\224 must involve roots higher
In fact, Gauss assertedthat to find a geometric attempted construction for a /7-gon where p \342\200\224 1 is not a power of 2 would \"spend his time uselessly.\" Interestingly, Gauss did not prove the assertion that such were That was constructions impossible. accomplished in 1837 by Pierre Wantzel (1814-1848), who in fact than
and
solutions
the
Carl Gaussin
in connection equations,
Hence
are solvableusing radicals, of 2, then the solutions
and
never able to
were
were impossible, to these
constructions
other
using
It was
doublethe
they
to construct
manage
using straightedge
the angle,
square the did
back as the fourth without success to
tried
had
constructions
geometric
compass
= 3,
Note
\342\226\240 Historical
find
constractible.
: Q]
transcendental,
Ferdinand
a result finally achieved Lindemann (1852-1939).
in
1882
by
Section32 Exercises
that
Note
is constractible,
for n >
the regular \302\273-gonis constractible which is the case if and only
if a
3 if and
if the
only
line segment of length
299
angle 2n/n
cos(2n/n)
is
constractible.
32
EXERCISES
Computations
1. Prove
the
identity cos
trigonometric
39 = 4 cos3
3 cos 9 \342\200\224
9 from
the
Euler
e'6
formula,
: cos.9
+ isin.9.
Concepts
2. Mark
of the
each
following
true
or false.
a. It is impossibleto double constructible
b.
cube of
any
by compass
edge
g.
and
fact
(up to
It is impossible to double every cube of constructible edge by compass and
conclusion
c. It
h.
constructions.
is impossible to radius
constructible
square
any circle
by straightedge
of and
. i.
compass constructions. angle can betrisected by and compass constructions. straightedge number is of degree e. Every constructible r > 0. 2r over Q for some integer of f. We have shown that every real number r > 0 degree 2r over Q for someinteger is constructible. No constructible
that
into
integer
constructions. straightedge
straightedge
d.
The
of a positive of primes is unique
factorization
a product
order) was used strongly of Theorems 32.9 and arguments
Counting
are
the
Using
4. Show 5.
proof
of Theorem
algebraically to Fig.
Referring
that
32.13, where
thereforethat
OAP is
similar to triangle
r is
that
possible to AQ
that
APQ.
the regular an angle
construct
bisects
10-gon is constructible pentagon is also). [Hint:
the regular the regular
show
it is
32.11, show
and
straightedge
j. Wecan find numbers
starting length
a compass.
of all constructible number of stepsby with a given segment of unit and using a straightedge and a the totality
in
a finite
compass.
9-gon is not of
constructible.
30\302\260.
angle OAP, (and Triangle
Show algebraically
exceedingly
tools. powerful mathematical We can find any constructible number in a finite number of steps by starting with a and using a given segment of unit length
Theory
3.
at the
32.11.
that
constructible.]
32.13 Figure
300
Part
In Exercises
VI
6 through
6. Theregular 7. The regular
20-gon 30-gon
8. The angle
72\302\260 can
9. The regular
15-gon
10. Supposeyou who
section
that
the
is true.
statement
is constructible. be trisected. can be
constructed.
to explain
course in what you would
33
to show
needed
where
is constructible.
wanted
down
of Exercise5
the results
9 use
had a
never
Write
Fields
Extension
roughly
abstract
in
three or four sentences,for a high school plane geometry teacher how it can be shown of 60\302\260 that it is impossible to trisect an angle
just
algebra,
say.
Fields
Finite
fields. We shall of all finite of this section is to determine the structure purpose one finite field (up show that for every prime p and positive integer n, there is exactly referred to as the Galois field to isomorphism) of order pn. This field GF(/?\") is usually of order p\".We shall be using quite a bit of our material on cyclic groups.The proofs are simple and elegant.
The
The Structure of a FiniteField
be 33.1Theorem Let \302\243
Proof
has
qn
Let
{a\\,
a finite
extension of degreen
\342\226\240 \342\226\240 be \342\226\240, an}
be
can
for bi e
If
\302\243 is
bt
order.
field F. If F
q elements,
has
a finite
field of
b\\ax
over F. By Exercise21of
be any of the q elements of the a,- is qn.
characteristicp, then
Section
E
contains
of F,
the
total
number
We
of such
exactly
pn elements for
some
n.
now turn will
30,
\342\23
field isomorphic is a finite extension of a prime to the field 7LP, Every finite field \302\243 The corollary follows at once from Theorem33.1. p is the characteristic of \302\243.
theorem
then E
h bnan
-\\
may
combinations
integer
positive
Proof
F. Sinceeach linear
distinct
Corollary
a finite
over
as a vector space a basis for \302\243 in the form written uniquely
P=
33.2
have prime-power
must
fields
finite
elements.
p e E
every
that all
show
now
We
to the
show us
study how
of the any
of multiplicative structure field can be formed from
finite
a finite the
where
field. The following
prime
subfield.
\342\23
33.3
Theorem
Let
Proof
of
a field
\302\243 be
The setE* of
in
zeros
7Lp
of the
algebraic closure7LP
polynomial
\342\200\224 x in
xp\"
forms a multiplicative
of E
elements
nonzero
in an
contained
elements
pn
are preciselythe
of E
301
Finite Fields
Section 33
of 7LP. The
elements
Zp[x].
of order
group
pn
\342\200\224 1 under
the field multiplication. For a e E*, the order of a in this group divides the order pn \342\200\224 1 = a. Therefore, of the group. Thus for a e E*,wehaveap\"~1 element every \342\200\224 x can have at most pn zeros, we see that E contains in E is a zero of xp\" x. Since xp\" \342\200\224
= ljSoa^'
precisely the 33.4
Definition
An element unity if a\"
33.6
Corollary
of a
Proof
actually
been
we now
state
The
is an \302\253th root
am ^ 1 for
0<
elements of
if
of unity < n.
m
a finite
a\"
is a primitive
1. It
=
\302\253th root
of \342\226\240
elements are all
of pn
field
(pn
\342\200\224 roots
l)th
in Corollary 23.6, we showed that the multiplicative group of nonzero finite finite field is cyclic.This is a very important fact about fields; it has in For of this section, to the sake completeness applied algebraic coding. and illustrate with an example. it here as a theorem, give a corollary,
Let a
elements
of nonzero \342\200\242)
of a finite
field
F is
cyclic.
23.6.
extension
finite
{F*,
group
multiplicative
See Corollary A
\342\231\246
7LP.
that
Recall
Proof
1 and
x
unity.
elements
33.5 Theorem
a of a field =
\342\200\224 in
xp\"
the nonzero
Thus
of
of
zeros
\342\231\246
a finite
\302\243 of
bea generator
field F
the cyclic
for
is a simple
group E*
of F.
extension
of nonzero
of E.
elements
Then E
= F(a) \342\231\246
33.7
Example
Consider
the
By Theorem 33.5 force and ignorance. We
field
finite
generator of Zn* by must be an element
Zn.
brute
24 =42
22 =4,
Thus
22 nor
neither
is a primitive By the 10th
10th theory
roots of
unity
of order dividing
of Zn*
25 is
root
of
1, but, unity
of cyclic in
Zn,
=5,
groups, all the
= 10 =
1, so 2 is a generator
the
form
of Z,,*, that
generators
2\", where
n is
relatively
elements are
21 = 2,
23
to =
find
a
10, 2
-1. of Zu*,
that is,
2
We were lucky.
in Zn.
are of
Let us try 2. Since \\Zn*\\ by trying is, either 2, 5, or 10. Now is cyclic. \342\226\240)
= (2)(5)
25
210 =
of course,
start
that
10,
and
(Zu*,
=
8,
27 =
7,
29 =
6.
is, all
the
prime
to 10.
primitive
These
Part
302
VI
Fields
Extension
Note
ft Historical Carl
had shown
F. Gauss
Although residues modulo a prime
several papers
set of the field
the
that
p satisfied
the basic
was Evariste Galois (1811-1832) who with what he called \"incommensurable to the congruence F{x) = 0 (mod solutions\" p), where F(x) is an nth degree irreducible polynomial modulo p. He noted in a paper written in 1830
He
brilliance
became influential.
properties, it first
dealt
politics
active
however,
was,
following
the
was
he
before
essentially established
20
Galoistheory.
French
revolutionary
in
of 1830. In the life
revolution
July
May 1831, he was arrested
and
ideas of
for
threatening
of King Louis-Philippe. he was acquitted, Though he was rearrested for participating, armed, heavily in a republican demonstration on Bastille one that should consider the roots of this Day of that one that Two months after his release from of imaginary congruenceas \"a variety year. prison symbols\" the he was killed in a \"the as one uses Galois can use in calculations duel, V\342\200\224T. March, following just victim of an infamous and her two dupes\"; of F{x) = 0 then showed that if a is any solution coquette \342\226\240 \342\226\240 \342\200\242 the a letter to a friend + flo + aia + aial + previous night he had written (mod p), the expression in the theory of some of his work takes on precisely pn different values. fln-ia\"_1 clarifying equations and requesting that it be studied by other he proved results equivalent to Theorems 33.3 Finally, Not until 1846, however, were his major mathematicians. and 33.5 of the text. it is from that date that his work Galois' life was brief and tragic. He showed published; papers in
mathematics
The
2,
on,
early
5th roots
primitive that
turn
now
power
pr,r
If F
is a field
Because
4,
24 =
of
unity
in Zn
5,
=
26
is 25
of prime
F is
9,
= 10 =
characteristic
p
with
the gcd of m
and
10 is
28 =
3.
\342\200\224 1.
\342\226
field
of order
closure
pr for every prime
F, then
xp\"
\342\200\224 x
has
p\"
\342\231
algebraically closed,xp\"
Sincewe have
finite
algebraic
in F.
it suffices to factors x a, the factorization.
show
that
\342\200\224 x
over that field into a product of linear of these factors occursmore than once in
factors
none
an algebraic theory introduced of derivatives, this elegant to us, so we proceedby long division. Observe that 0 is a zero of \342\200\224 x of multiplicity 1. Suppose a ^ 0 is a zero of xp\" \342\200\224 x, and hence is a zero of \342\200\224 \342\200\224 a is a factor of f(x) in F[x], =xp\"~l l.Thenx and by long division, we find
technique is not
f(x)
2m, where
of GF(p\
\342\200\224 so
xp\"
form
the
to the question of the existence of a > 0. We need the following lemma.
zeros
distinct
Proof
are of
in Zn
22 =
root
square
Existence
The
33.8 Lemma
of unity
is,
The primitive
We
publishing
available
not
303
Fields
Finite
33
Section
that
\342\200\224
(x
= xp\"~2
Now g(x) has
+
\342\200\224 1
+
a1
ap\"~3x \342\226\240\342\226\240\342\226\240+ +
each summand
and in g(a),
summands,
pn
a)
+ a2xp\"-4
axp\"~3
a?\"-1 \342\200\236\302\273 ,
is
1
a
a
Thus
g(a) =
a field of characteristicp. Therefore,
are in
we
since
--.a
[(pn-1)-1)-=a
0, so
^
g{a)
a is a zero
of
of /(*)
multiplicity 1. 33.9
Lemma
Proof
is a field
If F
all
and
\342\231\246
of
F.
p e
binomial
py = ap
(a +
33.10
Theorem
Proof
A
field
finite
Let 7LP
(-l)pnap\"
in
Proof
+ pp.
suppose (ap\"~'
+
we
that pp\"~r
elements existsfor
(1/a)^\" = 1/a.
Thus
of pn elements,
field
l\\aP~2p2
)p
+
(a +
have =
\342\226\240\342\226\240\342\226\240 + Qapp~l fip
ap\"
every
+
=
fi)p\"
ap\"
+
.Then
Pp\"
\342\231\246
pp\".
power pn.
prime
= 2then-1 = 1.
= -landifp
Now 0 and 1 are zerosof xp\" K is a subfield of Zp containing
\342\200\224 x. For
since Lemma33.8showed
that
a ^ 0, ap\" Zp. Therefore, \342\200\224 x has
xp\"
p\"
finite
have
is a
=
a
K is
distinct \342\231\246
q
then for
field,
of degree
Let F there
ap
=
\342\226\240
of all closure of Xp, and let K be the subset of Z,p consisting 33.9 shows that (a + P) e K, and the Zp. Let a, p e K. Lemma = a we obtain = ctp\" pp\" = ap shows that e f.Froma''\" ap (\342\200\224a)p\"
that
is any
F[x]
F
have
we
/3)p,
in Zp.
zeros
If F
a, p e
\342\200\224 x in
xp\"
desired
the
Corollary
=
so \342\200\224a e K. (\342\200\224a)p\" \342\200\224a
implies
33.11
+
=
Thus
+
(ap )p\" = = (-l)^\"a.If/7isanoddprime,then(-l)^
equation
for all
pp\"
an algebraic
be
zeros of
to (a
theorem
l)ap^1 + Pp Qap~lp + Qap-2P2+
ap
of pn
GF(/?\")
+
+ {p-
=
py\"'1 y
[(a +
ap\"
1]
on n,
induction
=
py\"
+ P)p\" =
(a
+ (p- lK-1^ + (P{P~
+ ---
Proceeding by
then
\342\231\246
the
Applying
(a +
p,
n.
integers
positive
Let a,
characteristic
prime
= pr
field K
every positive integer n,
is an
there
irreducible polynomial
n. elements, where p
< F
containing
is the
7LP (up to
of F.
characteristic
isomorphism)
and
By Theorem
consisting
precisely
33.10, of the
Part
304
VI
Fields
Extension
zeros of xp\" \342\200\224 x. We want to show F < K. Every element of F isa zero of xp' \342\200\224 x, by = Theorem 33.3. Now prs to the exponents prpr^s~v>. Applying this equation repeatedly and the fact that for a e F we have ap' = a, we see that for a e F, using rn
= ap
aF
Thus F < K. Then that K is simple
33.12
Theorem
Proof
be of
must
irr(/6,
F)
Let p
be a prime
= ap r(n~2)
=
r \342\226\240 \342\226\240 \342\226\240 = aF
= a.
F] = n.
shows that we must have [K : 33.6 so K = F(fi) for some Corollary 33.1
Theorem
over F
rfr-l)
in
We
fi e
have
seen
K. Therefore,
degree n. e Z+. If
let n
and
\342\231
E and
E'
of order p\",
are fields
then
E ~
\302\243\".
is a By Corollary 33.6, \302\243 prime field, up to isomorphism. an irreducible of so there exists n, polynomial f(x) degree ~ the elements of \302\243 are zeros that E Because Zp[x]/{f(x)}. of xp\" \342\200\224 of xp\" \342\200\224 x in Zp[x]. Because \302\243' also consists of x, we see that /(*) is a factor \342\200\224 in Zp[x]. Thus, zeros of xp\" x, we see that E' also contains zeros of irreducible f(x)
Both
\302\243 and
Z^, as
\302\243' have
simple extension of Zp degree n in Zp[x] such
because
of
contains
\302\243\" also
exactly
pn elements,
E' is also
isomorphic
to Zp[x]/{f(x)).
\342\231
Mathematical that
used
in
coding.
algebraic
In an article
Monthly 11 (1970):249-258,Norman a finite field of order up to three errors using
in
constructs
Levinson
correct
can
have been
fields
Finite
the
American
a linear
code
of elements.
(A
16.
\342\226\240 EXERCISES'\"33
Computations
In Exercises 1 through
calculator
whether there
3, determine
field
having
the
the given
number
3. 68,921
2. 3127
1. 4096
4. Find
exists a finite
be useful.)
may
number
of primitive
8th roots of unity
in GF(9).
of unity of unity
5. Find the
number
of primitive
18th roots
6. Find the
number
of primitive
15th roots
7. Find the
number
of primitive
10th roots of
unity
in GF(19). in GF(31). in GF(23).
Concepts
8. Mark each of the
following
a. Thenonzero b. The
of every
elements
elements of every
zeros in C of There exists a finite
c. The d.
false.
true or
e. Thereexistsa finite f. Thereexistsa finite
(x28
finite
field
finite
\342\200\224 e
1)
form
field
Q[x] form
elements. 125 elements. 36 elements.
field
of 60
field
of
field
of
form
a cyclic
a cyclic group
a cyclicgroup
group
under under
under
multiplication.
addition. multiplication.
Exercises
Section 33
g.
The
h.
There
i. The
a primitive
exists an irreducible nonzero elements
j. If F is
of F
i is
number
complex
a finite
then every field, of F. automorphism
is an
of degree
polynomial
of Q form
root of
4th
a cyclic
unity.
58 in
group Q*
isomorphism
305
Ta\\x\\.
field
under
F onto
mapping
multiplication.
a subfield of
an
closure
algebraic
F
Theory
9. Letla Using
10. Show
be an algebraic the results of
that
every
closure of la, this
section,
irreducible
and
show
let a, /J e la be zerosof that 7q(oi) = la(P).
polynomial
in
is a
Zp[x]
x2 + 1 and
x3 +
divisor of xp\"
\342\200\224 x for
of x3
+ x
+ 1,respectively.
some n.
11. Let F be a finite the cyclic
12.
Show
13.
Show that
14.
Let p
that
F is a generator field of p\" elements the prime subfield containing Zp. Show that if a e of nonzero elements of F, then deg(a, Zp) = n. group {F*, \342\226\240> a finite field of p\" elements of pm elements for each divisor moin. has exactly one subfield \342\200\224 x is
the product
xp\"
be an
odd
of all
irreducible
monic
in
polynomials
Z,p
[x]
of a
degree d dividing
p), the
x1 =
congruence
an equivalent
statement
a (mod /7) in
the
has
a solution in
finite field
Zp,
of cyclic groups.] part
n.
prime.
a ^ 0 (mod a. Show that for a e Z, where if a{p~l)l2 = 1 (mod p). [Hint: Formulate
b. Using
of
(a), determine
whether or not
the
polynomial
x2
\342\200\224
6 is
irreducible
in Zn[x].
and
Z if and
use
only
the theory
Advanced
\"m^ii
Section
34
Isomorphism
Section
35
Series
Section 36
Theorems
of Groups
Sylow Theorems
Section37 Applications of the Section38 Free Abelian Groups Section39 Free Groups Section40 Group Presentations
section
34
are
Sylow Theory
Theorems
Isomorphism There
Theory
Group
concerning isomorphic factor groups that The first of these is Theorem theory.
theorems
several
isomorphism theorems of group restate for easy reference.The
in Fig.
is diagrammed
theorem
are
known
as the we
14.11,
which
with
kernel
34.1.
//
s \\l s
//
/
/
/
(isomorphism)
g/k'
34.1Figure 34.2
Theorem
(First
: G -\302\273G/K
yK
lemma
The
34.3 Lemma
Let iV
that follows
two isomorphism
other
be
be the
-> 0[G] such that
fi : G/K
of the
Let <\\> : G -* G' be a homomorphism canonical homomorphism. Thereisa unique for each x e G. (*) = KYk{x))
Theorem)
Isomorphism
and let
a normal
homomorphism. to the set of
subgroups
of great aid in
our
proof
and intuitive
understanding
theorems.
subgroup of
Then the normal
will be
K,
isomorphism
map $
a group from
of G/N
G the
and
set of
given
by
let y : G ->- G/N be the canonical normal subgroups of G containing onto. (L) = y[L] is one to one and
N
307
308
Part VII
Proof
Advanced GroupTheory 15.16 shows that if L is a normal subgroup of G containing N, then (L) = of G/N. Because N < L, for each x e L the entire coset y[L] is a normal subgroup xN in G is contained in L. Thus by Theorem 13.15, y _1 [(L)] = L. Consequently, if L = H, then and M are normal subgroups of G, both JV, and if 4>{L) = 4>{M) containing L = y~x [H] = M. Therefore is one to one. If H is a normal of G/N, of G by then y\"x[H] is a normal subgroup subgroup Theorem
e H and y[y_1[H]]= if. This
Theorem 15.16. BecauseN
(y-1[H])=
we see that A7 c y_1[H]. Then the set of normal subgroups
= N,
y~x[{N}]
that
shows
is onto
4>
of G/N.
\342\23
and
If H
A7\"
are
of a
subgroups
G,
group
we let
then
{hn\\he H,n e N}.
HN =
and A7 as the intersection of all subgroups of G that Of course is the smallest HN. contain subgroup of G containing of both and since such G containing H H v N is also the smallest N, any subgroup we not a of G. must contain HN. In general,HN need be However, subgroup subgroup have the following lemma. the join if thus if HN;
define
We
34.4Lemma
If
a normal
A7 is
We show h\\,
e if
h2
- some
is a
HN
that
and
e A7. \302\2733
G, and if
is also normal
if H
Furthermore,
if
of
of
subgroup
NH. Proof
v A7 v A7
subgroup of
e A7'. \302\2732 \302\273i,
Since
We
34.5 Theorem
A7 is
Proof
that
suppose
H is
HN
which
if
a normal
are now
ready for
= A7
also normal e HN,
the
(SecondIsomorphismTheorem) subgroup
then
{h\\n\\){h2n2) = h\\(n\\h2)n2 under the induced operationin
= (ghg~x)(gng~x)
of G.
G,
G, from
and \302\273 e A7, we have (Aw)-1 = n~xh~x subgroup. Thus (hn)~x e if A7, so if subgroup, so A7if = H v N = HN. Now
in
Then (HN)/N
second
<
G.
in
so if A7
G. A
Let
if
for some
argument
let
e H,
is indeed
be a
~ H/(H n N).
A
ee is in
e A7, \302\2734
similar
normal
Let
at once. n\\h2 = h2n3
follows have
= (hih2)(n3n2)
e =
Clearly
G.
in
HN we
h\\{h2nz)n2
G, and
isomorphism
=
A7
subgroup,
=
h~xriA
is normal v
= HN =
then H v N
of G,
subgroup
Then
so if A7 is closed
ghng~x
H is any
if A7. For
since shows
\302\273 e A7,
A7
is a
that
and g
A7
for
e if A7, h e H normal if
is a
e G. Then
in G.
\342\23
theorem.
subgroup
of
G and
let N be
a normal
canonical homomorphism and let H < G. Then y[H] is a Theorem 13.12. Now the action of y on just the elements of H G/N by subgroup to H) provides us with a homomorphism (calledy restricted mapping H onto y[H], and the kernel of this restriction is clearlythe set of elements of N that are also in H, that HON. Theorem 34.2 then shows that there is an isomorphism is, the intersection Let
y
: G \342\200\224> G/N
be the
of
Mi:if/(ifnA0^y[if]. HN of M2
to HN also providesa homomorphism On the other hand, y restricted mapping onto y[H], because y(\302\273)is the identity N of G/N for all n e N. The kernel to HN is N. Theorem then provides us with an isomorphism 34.2 y restricted -+ y[H]. : (HN)/N
Section 34
isomorphicto each
More
isomorphism.
= LetG Z x Z x
n N)
H/(H
Proof
~ Z.
of G/K.
normal
two
The
third
N
and
x
= h(H n
^~\\h)
an
tf).
\342\231\246
Then clearly ~ Z and we (HN)/N
= {0}
{0}. We have
H. Then G/H
~
x Z x Z.
=
HN
have
also
and
Let H and
K
K be
then H/^T these groups.
< H,
concerns
theorem
is a normal
of a group
normal subgroups
G with
(G/K)/(H/K).
be
-+ (G/K)/(H/K)
onto (G/K)/(H/K),
of G
subgroups
isomorphism
Theorem)
Isomorphism
K <
(/. : G
will be
\342\226\262
(Third
Let
{0},
N =
H n
and ^T are
H
If subgroup
34.7 Theorem
and
=
n{-\\iJL2{{hn)N))
H = ZxZx {0}x Z
ZxZxZ, Z
are
explicitly,
4>{{hn)N) = Example
309
and H/(H f! N) are both isomorphic to y[H], they (HN)/N -> H/(H n N) where = fif1^ other. Indeed, : (HN)/N
Because
34.6
IsomorphismTheorems
for
and
(/.(afe)
(a)
by
given
e G. Clearly
for a
(aK)(H/K)
is
a,beG,
= [(aK)(bK)](H/K)
= [(ab)K](H/K)
= [(aK)(H/K)][(bK)(H/K)] so
is a \302\247
The kernel
homomorphism.
These x are just
the
of
elements
consistsof
those
x e
Then Theorem
H.
G such
34.2
that
(x) that
shows
= H/K. G/H
A nice
as being
\342\200\224 \342\231\246
(G/K)/(H/K).
way of viewing
34.7 is
Theorem
factored via a normal
subgroup
to regard the G, K <
K of
Yh =
canonical
H < G, to
map
yh
'\342\226\240 G
G/H
give
Yh/kYk,
in Fig. 34.8. Another way of visualizing to a natural isomorphism, as illustrated this in Fig. 34.9, where each group theorem is to use the subgroup is a normal diagram subgroup of G and is contained in the one aboveit. The larger the normal subgroup, the of G collapsed smaller the factor group. Thus we can think by H, that is, G/H, as being smaller than G collapsed by K. Theorem 34.7 states that we can collapse G all the way down to G/H in two steps. First, collapse to G/K, and then, using H/K, collapse this to (G/K)/(H/K). The overall result is the same (up to isomorphism) as collapsingG up
byH. yH \342\226\240G/H
Natural
H
isomorphism
>- (G/K)/(H/K)
G/K
K
yH/K
34.8
Figure
34.9
Figure
Advanced GroupTheory
Part VII
310
34.10
K = 6Z < Consider = Z/6Z has elements
Example
Then G/H
Z.
3 + 6Z,
4
6Z, 2 + 6Z, and 4 + 6Z He in 2Z/6Z. is isomorphic to Z2 also. Alternatively, imder ffe isomorphism to the cyclic
and
elements
corresponds
2Z/6Z
~ Z6/(2)~ (Z/6Z)/(2Z/6Z)
Z2
~
= Z/2Z ~ Z2.
Now
Thus
6Z. has
(Z/6Z)/(2Z/6Z)
~ Ze, and Z/6Z
see that
we
G/K
5+
and
6Z,
+
six cosets,
these
two
G =
<
2 + 6Z,
1 + 6Z,
6Z, Of
H = 2Z
(2) of
subgroup
Z6. Thus
\342\226
Z/2Z.
34
\342\226\240 EXERCISES
Computations In
using
the three
isomorphism theorems, it is the fact that the groups just
isomorphism : Z12 1. Let \302\242)
-* Z3 be
a. Find the kernel
b. List the c. Give the 2. Let # : Zrg List the
c Find the d. Give
3. In the
4.
the
actual
the
between Zn/K
the homomorphism
-\302\273\342\226\240 Z12 be
elements cosets
in
and
Z3 given =
where
coset. by the
map /j,
a. List
b. c. d. e.
the
List the List the List the Give the
6. Repeat
G =
34.2.
10.
\302\242(1)
the elements
showing
Z\\%/K,
in
each
-
and
(4)
N =
and
coset.
[Z18] given
by the map
[i describedin
34.2.
Theorem
(6).
in HN
groups)
and m.HC[N.
in HN/N,
Z24, let
H = (4) and
K =
the
elements
in each
in G/K,
the
elements
in each
cosets
in
the
elements
in
cosets
in (G/K)/(H/K), between
correspondence
5 for
the
group
in each coset. showing the elements G/H and (G/K)/(H/K) describedin
G = Z36
with
proof
of Theorem
34.5.
coset. coset. each coset.
in G/H,
cosets
showing showing H/K, showing
the
(8).
cosets
Exercise
in Theorem
described
(which we might write H + N for these additive in each coset. showing the elements c. List the cosets in H/(H R N), showing the elements in each coset. d. Give the correspondence between HN/N and H/(H n N) described in Exercise 3 for the group Z36 with H = (6) and N = (9). Repeat group
the
of \302\242.
let H
Z24,
b. List
5. In the
by
for this.
us training
[Zi8].
group
a. List the the
given
correspondence
exercisesgive
2.
\302\242(1)
in each
elements
the
showing
the correspondencebetweenZ^/K
group
isomorphic. =
know
The first six
Kofty.
correspondence
cosets
to
necessary
are
such that
homomorphism
in Zn/K,
cosets
a. Find the kernel K b.
often
and not
H =
(9) and
the
proof
of Theorem
34.7.
K = (18).
Theory
7.
Show normal
directly from the in G, then H n
definition N
is normal
of a
normal
in H.
subgroup
that
if H
and
N
are
subgroups
of a
group
G,
and N
is
Seriesof
Section 35
8. Let
a.
9.
H, K, and
B and
that
Show
b. To
what
Let K
and L be normal
SECTION
factor
C are normal
of A,
subgroups
< L. Let A
< K
H
and B
=
B =
G/H,
and
C =
G/JT
~ L
and G/L
G,
which
K/H,
L/H.
< C.
G is (A/B)/{C/B) isomorphic?
group of
35
of G with
normal subgroups
L be
311
Groups
of G
subgroups
L =
Kw
with
G, and
= {e}.
KC\\L
Show that
~
JT.
of Groups
Series
Subnormal and Normal Series
are not Theorem
too
ease of 35.1
Definition ~\342\200\224
important
11.12.
Many
for of our
of our
groups because
gives insight groups. They strong
from abelian groups, however, for
be taken
will
illustrations
group
and nonabelian
abelian
abelian
generated
finitely
of a
a series
of
notion
The results hold for both
of G.
structure
the
into
with the
is concerned
section
This
computation.
A subnormal of subgroups = {e}andH\342\200\236 of normal
(or subinvariant) of G such that G. A normal
< H;+i
H,-
(or invariant)
of G such that
subgroups
group G is a finite
of a
series
H
Ho, Hi,
sequence
and H; is a normal subgroup of G is a finite series sequence Ho =
< H,+i,
{e},and
Hn
\342\200\242 \342\200\242 \342\200\242, H\342\200\236
of H;+i with
Ho, Hi, = G.
Ho =
\342\200\242 \342\200\242 \342\200\242, H\342\200\236 \342\226\240
and normal series coincide, for abelian groups the notions of subnormal is normal. A normal series is always but the converse subnormal, need not be true. We defined a subnormal series before a normal series, sincethe concept of a subnormal series is more for our work. important that
Note
since every
35.2 Example
Two
subgroup
of normal
examples
series of
Z
are
addition
under
{0} < 8Z
< 4Z < Z
and
{0} <
35.3
Example
the
Consider
group
D4
< {/\302\276} is a
35.4
Definition
A subnormal a group G
35.5
Example
<
Z.
(normal) series{Kj}is if
{H,}
c
Mi. M2} <
Pi,
8.12.
Table
It is
each H,-
of a is one
72Z< 24Z
not
a normal
series since
subnormal (normal) series{H,} of the Kj. \342\226\240
The series
{0} <
DA
\342\226\262
a refinement
{Kj}, that is, if
8.10. The series
in Example
square
(A)>
A
normal
is not
of
(A). Mi)
the
series, as wecouldcheckusing in D4.
subnormal
{po, fMi)
of
of symmetries
9Z <
<
8Z
< 4Z
<
Z
Part
VII
Advanced
Group Theory
is a refinement
series
of the
{0} < 72Z new
Two
24Z, have
both
inserted.
been
in studying the structure and subnormal normal
interest
Of
dennedfor 35.6Definition
4Z and
terms,
< 8Z < Z.
of G
\342\226
are
factor
the
series, since H
groups normal
is
Hi+i/H;. in Hi+1 in
These are either case.
G are isomorphic if subnormal (normal) series {H} and {Kj} of the same group between the collections of factor groups {Hi+\\/Hi} there is a one-to-onecorrespondence
Two
and {Kj+\\/Kj}such
that
are isomorphic.
subnormal (normal) seriesmust
two isomorphic
Clearly,
factor groups
corresponding
have
the same
\342\226
number of
groups.
35.7Example
two
The
series of
Z^,
{0}< (5) <
Z15
and
(3) < Z15,
{0} <
areisomorphic. to
(5)/{0},
The
Both
Z15/ (5)
and (3)/{0}are
to Z5,
isomorphic
and Z15/(3) is isomorphic
\342\226
ortoZ3.
Theorem
Schreier
We proceedto prove that two subnormal series ofa group G have isomorphic refinements. result in the theory of series.The proof This is a fundamental is not too difficult. However, students lost in the and then we know from experience that some tend to feel proof, get that cannot understand the theorem. We now give an illustration of the theorem they before
35.8
Example
Let us
we proceed to
try
to find
its
proof.
isomorphic
refinements of the {0} <
8Z <
series
4Z <
Z
and
{0} < given
in Example
35.2. Consider the {0} <
of
{0} <
8Z <
4Z < Z
and
9Z <
Z
refinement
72Z < 8Z
< 4Z <
Z
the refinement
{0} < 72Z
< 18Z< 9Z <
Z
of
< Z. In
< 9Z
{0}
Z4, Z2,
both
Zg,
cases
have
the refinements
or Z. Theorder in
and 72Z
which
Series
35
Section
four
the factor
factor
313
of Groups
groups
isomorphic
groups occur is different
to
to be
sure.
\342\226\262
with a rather called the butterfly
start
We
sometimes
has a butterfly
technical
lemma since
lemma,
developed by Zassenhaus. Fig. 35.9, which accompanies
This
lemma
is
the lemma,
shape.
of a group G, and let H* be a normal K be subgroups of H subgroup of K. Applying the first statement in Lemma 34.4 to H* and K* be a normal subgroup P K) is a group. Similar and H P K as subgroups of H, we seethat H*(H arguments show that H*(H n A\"*), .\302\243*(# n K), and \302\243*(#* n K) are also groups. It is not hard of H P K (see Exercise22).The same to show that H* P K is a normal subgroup to H* K and 34.4 P H Pi K* as of Pi K Lemma H applied subgroups argument using n K*) is a group. Thus we have the diagram of subgroups shows that L = (if* f! ^T)(jFf relations in Fig. indicated shown 35.9. It is not hard to verify the inclusion by the
H and
Let
diagram.
Since both
H
of H PK. We middle line in
relationships,
subgroup
relations
are
L =
normal
(H* P
subgroups
K)(H P K*)
of H is
P K,
a normal
second
subgroup the heavy
denoted
35.9.
kernel PK)^(HP K)/L, and show that
: H*(H
H*(H
the
this particular normal subgroup relationship by We claim the other two heavy lines also indicate normal and that the three factor groups given by the three normal we shall define a homomorphism are all isomorphic. To show this,
have Fig.
subgroup
and H* PK 34.4 shows that
P K*
in Lemma
statement
H
K
H*CiK
HnK*
35.9 Figure
Part VII
314
Advanced
n K*) ~ (H n \302\243)/L. A similar result for line in 35.9 then follows by symmetry. groups right-hand heavy Fig. Let : denned as follows. be For h e H* and x e H*(H n^)->(fffl K)/L H n ^T, let (/;x) = xL. We show is well-defined and a homomorphism. Let /zl5 /i2 e = H* and xi, x2 e H n isT. If /zun = /;2x2, then e H*D(H DK) = /z2\342\200\2241/z1x2x~l H* 0 K c. L,sox\\L = x2L. Thus is well denned. Since H* is normal in H, there is
in H*(H n ~~\"
Theory
Group
h-i
K),
that
and
n K)/H*(H
H*(H
on the
the
such that
H*
in
=
x\\h2
Then
h3xi.
((/ilXi)(/!2X2))=
=
0((^1^3)(-^1\302\276))
0 is
Thus
(xlx2)-^
= 0(/iiXi) \342\200\242 (/z2x2).
= (xiL)(x2L)
a homomorphism.
onto (if n \302\243)/L. Finally if h e H* and x e if n K, then (/.(Ha:) = xL = L if and only if x e L, or if and only if hx e H*L = H*(H*n K)(Hn \302\243*)= n A\"*). Thus Ker() = H\"(H n \302\243*). H*(H We have proved the following lemma. 0 is
Obviously
35.10
Lemma
Lemma) Let H and K be of H and K, respectively.
(Zassenhaus normal
subgroups
1.
H*{H n K*) is a normal
2.
is a \302\243*(H* n \302\243)
3.
H*(H n K)/H*(H n
Theorem
subgroup of \302\243*(H n
normal
~ A\"*)
\302\243*(H
(H
let H* and
be
isT*
K).
n
n K)/[(H*
G and
\302\243).
n
n K)/K*(H*
Two subnormal (normal)
Theorem)
(Schreier
n
of H*{H
subgroup
~
35.11
of a group
subgroups Then
\302\243)
n \302\243*)].
A\(H")
series of
G have
a group
isomorphic
refinements.
Proof
Let G bea group
and
let
{e} = Ho
< Hi
< H2
<
\342\200\242 \342\200\242 \342\200\242 = G < H\342\200\236
< K2
<
\342\200\242 \342\200\242 \342\200\242 <
(1)
and
{e} =
be two
subnormal
series
H = H(H+i n This
for
inserts
each
m\342\200\224\\ not
i where
0
< \302\2430
for G. K0)
*i
Km =
For i where 0 <
< H(H+i
n Ki)
distinct necessarily - 1 and let Hy
i
<
H0,o
<
H0,i
< H0>2<
< Hi,i < < #2,1 < <
Hlj2 H2j2
the chain n Km)
\342\200\242 \342\200\242 \342\200\242 <
H(H+i
groups between
= H(H+i n
H
^),
and then
\342\200\242 \342\200\242 \342\200\242 <
< Hi,0
<
\342\200\242 \342\200\242 \342\200\242 <
< H2j0
<
\342\200\242 \342\200\242 \342\200\242 <
< H3j0
H0,m_i Hi,m_i H2>m_i
H+i. we
of groups
= Hi+1. If we
obtain
do
this
the chain
...
< Hi-1,1 < Hi-1,2 < =
(2)
1,
of groups
{e}=
G
\342\200\224 form
G.
- - -
<
Hi-l.m-l
<
Hn-\\ym
(3)
Series of
Section 35
315
Groups
1 not necessarily distinct groups, and ft;o = H for each i. chain (3) is a subnormal that is, each group is normal chain, in the following This chain refines the series (1). group. = Kj(Kj+i n H;) for 0 < j < m \342\200\224 In a symmetric fashion, we set 1 and 0 < Kjj chain i < n. This gives a subnormal
This chain
By the
(3) contains nm
Zassenhaus
+
lemma,
{e} =
K0io
<
5 #0,2
^0,1
< ^1,1 5 < ^2,1 <
<
^
< ^1,0
\342\200\242 \342\200\242 \342\200\242 <
< ^2,0
Ko,n-i
^1,2< ^2,2
\342\200\242 \342\200\242 \342\200\242 <
^1,n-1
''
'
\342\200\224 K-iyn-l
~S.
^3,0
< .. .
< Km-l,l
<
-STm_!
2
<
\342\200\242 \342\200\242 \342\200\242 <
^m-l,n-l
< ^m-l,n
= G. mn + 1 not necessarily refines the series (2). we have the Zassenhaus lemma 35.10, (4) contains
chain
This
(4) distinct
groups,
and Kj0
= Kj for
each
j. This chain
By
n Kj+i)/Hi(Hi+i n
H(Hi+l
~
Kj(Kj+i \302\243,\342\226\240)
n Hi+i)/Kj(KJ+i
n ft),
or
Htj+i/Hij ~ Kji{+l/Kj,i <
(5)
of relation (5) give a oneisomorphisms the chains factor between subnormal (3) groups correspondence = = that and while and (4). To verify this note ft correspondence, ft+i, H,i0 ftI>m of array Kjfi = Kj and K^n = Kj+\\. Each chain in (3) and (4) contains a rectangular from rise to a factor group. The factor groups the mn symbols <. Each < gives arising rth row of <'s in chain (3) correspondto the factor groups arising from the rth column in (3) and (4), we obtain of <'s in chain (4). Deleting repeated groups from the chains series of distinct groups that are isomorphic refinements of chains (1) and subnormal the theorem for subnormal series. (2). This establishes in G, we merely observe that For normal where all ft and Kj are normal series, in G, so the same proof all the groups formed are also normal and above Kjj fty of and follows at once from the second assertion in This applies. normality ftj Kjj of normal subgroups of a group Lemma 34.4 and from the fact that intersections yield for
0 <
i
<
n
\342\200\224 1 and
normal
The
We now 35.12
Definition
0 < 7
aw \342\200\224 1. The
of isomorphic
to-one
subgroups.
Jordan-Holder come
to the
A subnormal series
\342\231\246
Theorem real meat of the
{ft} of a group are simple.A normal series ft+i/ft, factor groups ft+i/ft- are simple.
theory.
G is {ft}
a compositionseriesif all of G is a principalor chief
the factor series
groups the
if all
\342\226\240
Part
VII
Advanced
Group Theory
Note that coincide.
seriesfor 35.13 Example
We
abelian and principal series groups the concepts of composition series is a composition every normal series is subnormal, every principal group, abelian or not. for
since
Also,
any
composition (and also no
Z has no
that
claim
{0} = Ho <
Hk
principal)
series.
<
Z H\342\200\236
\342\226\240 \342\226\240 \342\226\240 <
Hn-i
For if
=
is a subnormal series, Hi must But then Hi/Ho is be of the form rZ for somer e Z+. isomorphic to r Z, which is infinite cyclic with many nontrivial proper normal subgroups, for example, 2rZ. Thus Z has no composition (and also no principal) series. \342\226\2 35.14
Example
The series
{e} <
An
<
Sn
a compositionseries(and also a principal series) of S\342\200\236, because is An/{e] is simple for n > 5, and Sn/An is isomorphic to Z2, which is which isomorphic to A\342\200\236, series (and also simple. Likewise, the two series given in Example 35.7 are composition as shown in that example. This illustrates principal series) of Zl5.They are isomorphic, our main theorem, which will be stated shortly. \342\226\2 for n > 5 is
Theorem 15.18, H,-+i/H is simple if and only if H is a maximal of H;+i. Thus for a compositionseries,each H must be a maximal of H;+i- To form a composition series of a group G, we just hunt
that by
Observe normal
subgroup
normal
subgroup a maximal normal
for
normal Hn-2 of G, then for a maximal subgroup and so on. If this process in a finite terminates number of H\342\200\236_i, of steps, we have a series. Note that by Theorem 15.18, a composition series cannot have any composition
refinement.
further
subgroup
35.15 Theorem
To form
H\342\200\236_i of
normal
in
G,
and
Hn-\\
subgroup
G, then
so on.
a principal for a maximal
The main
(Jordan-HolderTheorem) Any
theorem
two
series, we normal
is as
to hunt
have
subgroup
for a maximal that of H\342\200\236_i
normal
Hn-i
is also
follows.
composition
(principal) series of a group
G are
isomorphic.
Proof
Let {H;} and Theorem 15.18
already be
the
group
integer
factors.
be
two composition
(principal) series of G. By factor groups are already
into
Theorem
Thus {H,}
refinement.
35.11,
simple, and
{K;}
must
isomorphic. group, we should regarda composition into simple factor groups, analogous primes. In both cases, the factorization
a finite
For of
{Ki}
refinements. But since all isomorphic shows that neither series has any further
have
they
\342\231\2
series
to the
as a type of factorization factorization of a positive
is unique,
up
to
the
order
of the
Section35 Seriesof
Note
\342\226\240 Historical
of what became the Jordanappearance theorem occurred in 1869 in a the work of Galois by the brilliant French first
the
on commentary
years.
of its
Jordan
Camille appearance
associated Jordan
The context
(1838-1922).
study of permutation the roots of polynomial is the
with asserted that even
groups
the
though
of
sequence
\342\226\240 of the group of the normal subgroups G, I, J,- \342\226\240 not is nevertheless equation necessarily unique, series the sequence of indices of this composition
is unique. monumental 1870 Equations.
Jordan gave a This
latter
35.16 Theorem
Proof
work,
is unique up to who (1859-1937), abstract
determined
the
for
many
theorem, that the in a composition series
the
groups order, was due to
Otto Holder important role in
a very
played
of group theory definition of a group
development
Among his other abstract definition
and Algebraic though restricted to
of
part
factor
of
completely
on Substitutions
Treatise
The Holder
the
his
in
proof
treatise
standard
sequence
equations.
remained
call permutation groups, on group theory
now
we
what
This Holder algebraist
317
Groups
he
contributions,
of
a \"factor
of all
structure
finite
the
once
been
had
gave
given. the first
group\" and of square-free
groups
order.
G has a composition (principal) series, and if N is a proper then there exists a composition series containing N. (principal) If
normal
of G,
subgroup
The series
{e}<
N
<
G
a normal series. Since G has a composition series {H}, then by is a refinement series isomorphic of {e} < N < G to a subnormal to a refinement of {H;}. But as a composition series,{Hi}can have no further refinement. all of whose factor groups are Thus {e} < N < G can be refined to a subnormal series simple, that is, to a composition series. A similar argument holds if we start with a \342\231\246 principal series {Kj} of G.
is both
Theorem
35.17
Example
a subnormal
and
35.11 there
(and alsoa principal)
A composition
{(0, 0)} < The next
35.18 Definition
A
group
G is
is basic
By
to the
can be expressed in
solvable if
it
Hi+i/Hi are abelian.
have
has
x
x Z9 (1) <
((0,
containing
(1) x
(1) =
characterization of those
terms
Z4
1)) is x
polynomial
A
Z9.
equations
of radicals.
a composition
series {Hi} such
that
all
factor
groups \342\226\240
the Jordan-Holder
series{H}must
of Z4
((0,3)) < ((0,1))< (2)
definition
whose solutions
series
theorem, we see that for a solvable group, every composition factor groups H+i/H;.
abelian
Part VII
35.19 Example
Advanced
Group
The group
S3 is
Theory
because
solvable,
the
series
composition
{e}
has factor groups isomorphicto Z3 and Z2, which are solvable, for since A 5 is simple, the series
abelian.The
is
>%
group
not
{e}
is a composition closely
solvable by
We
be shown but a
radicals,
series for a
Recall from Section15 that
groups.
the
Z(G) = {z e and
that
Z(G)
easy to find in
the
row
the
G,
is a normal subgroup center. An element
opposite
the column
a a at
under
the
at
is
in general
is.
4
a
top
and
center =
can be
that
G
group
G \\zg
Z{G) of
a group
gz for all
g e
of
G. If we
will
be in
have
\342\226\
using
centers
of
defined by
G], for a
of G if
and
finite
group
only if the
order as
same
the
in
formed G is
table
the
the center
left are given table.
extreme
the very
Now let G be a group, we can form the factor
is not
Series
subnormal
one
mention
polynomial equation of degree<
Central
Ascending
a polynomial equation of degree5
the fact that
with
connected
The
and As/{e}, which is isomorphicto A5, is not abelian. This group is not solvable. This fact to be the smallest group that
series,
60 can
order
A 5 of
the
G, it
is
elements
elements
in
of the
let Z{G)
be the center and find the
of G. center
Since Z(G) is normal in of this factor Z(G/Z(G))
group G/Z(G) if y : G -+ G/Z(G) is the canonical is normal in G/Z(G), Since Z(G/Z(G)) i s Theorem then a normal 15.16, y~l[Z{G/ map, by subgroup Zi(G) of G. We and find its take can then form the factor group of it to get Z2{G), center, (yi)_1 G/Z\\{G) group.
Z{G))}
and
35.20 Definition
so on.
The series
{e}< Z{G)< Z^G) < Z2(G) in the
described
preceding discussion is the
ascending
<\342\226\240\342\226\240\342\226\240
series
central
of the group
G.
\342\226\
35.21
Example
The center
of S3 is just
the
identity
{po}. Thus
(A)} <
<
central
ascending
(A)} <
series of S3 is
\342\200\242 \342\200\242 \342\200\242 \342\200\242
of the square in Example 8.10 is {po,p2}. D4 of symmetries we said that this (Do you group would give us nice examplesof many we Since is discussed?) things D4/{po,/\302\276} of order 4 and hence abelian,its center is all central series of D4 is of Da/{po, Pi}-Thus the ascending
The
center
of the
(A)}
the
remember
group
that
(A)}
<
(A3, Pi}
<
DA
<
DA <
DA
<
\342\226\240 \342\226\240 \342\226\240.
\342\226\
Exercises
35
Section
319
35
\342\226\240 EXERCISES
Computations 5, give
1 through
Exercises
In
1. {0}<
10Z< Z
2.
{0} <
60Z <
20Z < Z
3.
{0} <
(3) <
and
4.
{0} <
5.
{(0,
25Z <
{0} <
and
Z24
isomorphicrefinements {0} <
245Z <
{0} <
(8) <
Z24
(10Z)
7. Find
all
series
8.
all composition
9. Find
all
(12) < Z72 and
show that
and
of Z4g and
series of
all composition
11.
the
center
of S3
the
center
of
the
ascending
central series of S3 x
Z4.
ascending
central series of S3 x
D4.
12. Find 13. Find 14. Find
the
are
Z x (20Z) < Z
x Z
isomorphic.
x Z5.
Z5
10. Find Find
they
x (80Z) <
0)} < Z
they are isomorphic.
that
show
{(0,
of S3 x Z2.
series
composition
49Z < Z
x Z < Z x Z
series of Zgo
composition
(24) <
{0} <
and
6. Find all composition
Find
Z
and
(18) < (3) < Z72 0)} < (60Z) x Z <
two series.
of the
series of
x Z7.
x Z5
Z2
x Z4. S3 x D4.
Concepts In
15.
16, correct the
15 and
Exercises
needed, so that
it is
A composition
of the
definition
in a form
acceptable series of a group G is
for
of G
16.
A solvable
17.
Mark
each
that
Hi
is one
of the
following subnormal
Every
principal
finite
composition
e. Every
abelian
f. Every
finite
group
<
Hx
maximal
is also
H2 <
normal
series
\342\226\240 \342\226\240 \342\226\240 < #\342\200\236_! < Hn
of abelian
j.
Every
up to
groups.
is a
principal
series.
series. compositionseries. and only if it has a compositionserieswith
has
group
exactly
one composition
has a
order.
group
\342\200\242 n \342\200\224 1. 0,1, 2, \342\200\242 \342\200\242,
compositionseries.
is a
series
is solvable if
finite
for i =
subnormal.
h. S7 is a solvable group. i. The Jordan-Holder theorem has some similarity which states that every positive integer greater uniquely
of Hi+\\
subgroup
= G
is also normal.
series
group
<
sequence
a composition
series
d. Every
text, if correction is
or false.
true
series
normal
Every
g. A
is a
that has
group
a. Every
b. c.
such
to the
reference
without
publication.
a
{e} = H0
of subgroups
italicized term
of prime
order is solvable.
with
than
simple
the Fundamental 1 can
be factored
factor
groups.
Theorem of Arithmetic, a product of primes
into
320
Part VII
Z36. Refer
G be
the
to
S3. Is S3 x
S3 x
Is the group
20. Let
Theory
Group
compositionseriesof D4 of symmetries
18. Find a 19.
Advanced
of the
solvable?
S3
8.10 solvable? 35.11. Let the subnormal series
square in Example
proof of Theorem
(12)<
{0}< and let
{0}< Find chains
(4)
the
in
(3) and
(4)
rectangular
the
Z36
< Z36.
(18>
and exhibit the isomorphicfactor array shown in the text.
Exercise 20 for
21. Repeat
<
(2) be
series
subnormal
the
(3)
(1) be
the
Z24 with
group
series
subnormal
(12) <
{0} <
(4)
in the
as described
groups
proof. Write
chains
(3) and
(1)
< Z24
and (2)
{0}<
(3) <
<
(6)
Z24.
Theory
22. Let
H, and
H*,
23. Show that
K be subgroups
with H*
of G
n K
H*
that
is normal
in H
n K.
if
H0 = is a
Show
in H.
normal
(normal) series for
subnormal
{e}< HkH2<---
Hi+i/Hi is of finite
G, and if
a group
G
order
then G is of
si+u
order
finite
SiS2---Sn.
24. Show that an infinite abelian group can have no compositionseries.[Hint: fact that an infinite abelian subgroup.] group always has a proper normal 25.
Show that a
26. Show
direct
finite
K of
a subgroup
that
compositionseriesfor seriesfor K. Observe
of solvable
product
G.
that the
27. Let
#0
G, and form a By
=
34.5,
{
suppose
<
with
Hi < that N
H
= K
n
Ht
Show
group.
G.
the latter
[Hint:
group is isomorphic
34.7. But
the
Hi <
{e} < i =
0,
Hn = Gbea a composition
\342\226\240\342\226\240\342\226\240 <
\342\200\242 \342\200\242 n form \342\200\242,
Ht_u
H,)]/[Hi-i],
and
that
Ht-i(K
series
for a
that the
distinct
HtN is a
group
< i$.] group G. Let N n #;)
be a normal subgroup of \342\200\242 n also Hq, HtN for i = 0, \342\200\242 \342\200\242, 34.4. Show that H-iN is normal in /\302\276N.
groups
by Lemma
34.5
Ht/Hi-i
~
fli/[ff, n (ffj-itf)],
to
[Hi/Hi-d/W
by Theorem
- [Hi.i(Kn
G be a composition
Hn =
is a simple
N =
and
(HiN)/(Ht-iN) and
[Hint: Let H0 = K n Ht for
among
groups
Hi)/(K n fli_i)
\342\226\240 \342\226\240 \342\226\240 <
compositionseriesfor
Theorem
with
that
(K n by Theorem
solvable.
G is
distinct
23, together
is solvable.
groups
a solvable group
Show
Exercise
Use
is simple.]
n (fli_iA0)/fli_i],
among
28.
Let
a group,
G be
\342\200\242 i = 0, \342\200\242 \342\200\242, n, form
\342\226\240 \342\226\240 < Hn = G be a composition series for G. Let N be a normal Hq = {e} < Hi < \342\226\240 the canonical the be Show that distinct y : G \342\200\224> G/N map. groups among y[Ht] for a composition seriesfor G/N. [Hint: Observe that the map
let
and
subgroup of G, and
321
Sylow Theorems
Section 36
let
f : HiN-+yiHMyiH,-!] defined
by
= yQiin)y[Hi.i]
f{hiii)
is a homomorphism
kernel
with
By Theorem 34.2.
#,_iN.
y(Ht)/y[Hi-i\\ ~
29. Prove
34.5, as shown
via Theorem
Proceed that
series for
homomorphic in the
compositionseries SECTION 36
Exercise 27.]
image of a solvablegroup is solvable. [Hint: Apply Exercise28 to get image. The hints for Exercises27 and 28 then show how the factor
a homomorphic the
hint for
in the
(^-^)/(\302\276.^).
a composition
of this
groups
look.]
image
Sylow Theorems for finitely abelian fundamental theorem generated groups (Theorem 11.12) gives all finite abelian groups. The study of finite nonabelian us completeinformation about give us some important groups is much more complicated.The Sylow theorems The
them.
about
information
We know A4,
which
no
subgroup
has order
that
and
the
set described
in
instances
set
the
d divides
6. Thus
the
\"converse
Sylow Section
is the
it is
useful
give us This 16. time, the
nonabelian
they show subgroup
of
that
order
if d
d.
is
will
groups.
on a of a group from the group; in some of cosets of a subgroup,
of action
application
itself
group itself, sometimes it
a collectionof
Namely,
contain a
finite
another
set
abelian,
15.6 that
also give some information to each other. We relationship
their
in studying
theorems
G is
theorems
Sylow and
divide | G \\. If
showed in Example
a nonabelian group G may have of the theorem of Lagrange\"
converse.
G does
then
\\G\\,
of such subgroups
theorems are very
that these
sometimes
\\G\\. We
of order \\G\\\\
dividing
not a power of a prime.)The
the number
Proofs of
and
order d
G must
group
The Sylow theorems give a weak
6 is
concerning
subgroup of a finite of every orderdividing
of a
12, has no subgroup
some
of
power of a prime
(Note see
order
subgroups
not hold.
does
is a
the
exist
there
then
is formed a collection
subgroups.
p-Groups
Section 17 gave applicationsof Burnside's a finite G-set. Most of our results in this number of elementsin a finite G-set. Let X be a finite G-set. Recall that
Gx = {gxI g contain
e G}.
one element
Suppose
that
there
from each orbit
formula section
x el, orbits in
for
are r in X.
that counted the number of orbits in flow from an equation that counts the the
Now every
of x in X under G is \342\226\240 \342\226\240 \342\200\242 G, and let {x\\,X2, ,xr}
orbit
X under
element of X is in
precisely
one
Part VII
322
Advanced GroupTheory so
orbit,
\\X\\
There may Xq
one-element
the
rewrite
s
of
union
where
orbits,
we may
orbits in X. Let XG = {x e X \\ gx = x for all g e G}.Thus the one-element orbits in X. Let us suppose there are = s, and 0 < s
one-element
be
is precisely
(i)
\\Gxi
J2
(1) as
Eq.
\\Xr
\\X\\
(2)
\\Gx>
J2 i=s+l
of
Most
as in
The proof
of proof.
Eq. (2). We shall to R. J. Nunke for
flow from of this section will Hungerford [10], where creditis given
results
the
theory
Sylow
of Theorem
36.3
line
the
J. H.
is credited there to
theorem)
(Cauchy's
develop
McKay.
Theorem
which follows,
36.1,
numerical conclusion.It In the rest of the chapter,
is not
theorem,
a counting
quite
p. The
modulo
counts
theorem seemsto
but
powerful. on it, and with older
if we choose the correct action set, the correct group what we want seems to fall right into our lap! Compared are extremely pretty and elegant. proofs, the arguments this section, p will always be a prime integer. Throughout Theorem
apply
36.1 Theorem Proof
36.2
Definition
Let G
be
In
notation
the
36.1,
of Eq.
Consequently
p divides
divisible
by p,
Let p the
of order pn
a group
(2), we
|Gx,-1 for
so |X|
and
=
let X
that
know
s + 1<
\\XG\\
(mod
be a finite |Gx,-|
Then
G-set.
divides
i < r. Equation
\\G\\
(2)
by
then
p. A
\\XG I (mod
p).
Theorem 16.16. shows that |X| \342\200\224 \\XG\\ is
\342\231
p).
be a prime. A
prime
=
\\X\\
have a
it does
be amazingly
G is a /7-group if every elementin group G is a /^-subgroup of G if subgroup of a group
G has the
order a
power of a
is itself
subgroup
/7-group.
\342\226
in this section is to show that a finite G has a subgroup group order theorem, \\G\\. As a first step, we prove prime-power dividing Cauchy's if p divides \\G\\, then that G has a subgroup of order p. Our
36.3 Theorem
Proof
Let p be a prime. Let G bea finite group element of order p and, consequently, a subgroup
(Cauchy's Theorem) Then
We
G has an form
of
goal
the set X
that the product
of all p-tuples
of the X =
coordinates
\342\200\242, {gi, gi,- \342\200\242 gp) of
in G
is e. That
{(gu g2,---, gP) \\gi
eG
elements
and let p divide of order p.
of G having
the
\\G\\.
property
is,
and gig2
every
which says
\342\200\242\342\200\242\342\200\242gP=e}.
We
in X, we may let g\\, gi,- \342\226\240 be any \342\226\240, gp-i \342\226\240 \342\226\240 as (gig2\342\226\240 determined Thus |X| = G, uniquely gp gp-{}~1. since p divides \\G\\, we see that p divides \\X\\. \342\200\242in be the cycle (1,2, 3, \342\200\242 \342\200\242, p) Sp. We let a act on X by
elements of and
\\G\\p~l
|X|. In
p divides
claim
a
Let
323
Sylow Theorems
36
Section
a /7-tuple
forming
is then
and
=
\302\260(gU#2,---,
ga(2),
(ga(l),
\302\243/>)
\342\200\242 \342\200\242 \342\200\242
, ga(p))
= (#2,
\342\200\242 \342\200\242 , gp, gl). g3, \342\200\242
\342\200\242 \342\200\242 \342\200\242 \342\200\242 that gi = (g2g3 \342\200\242 \342\200\242, (g2, g3, implies gP, gi) e X, for gi(g2,g3 \342\200\242\342\200\242\342\200\242gP)=e \342\226\240 \342\226\240 \342\226\240 \342\200\224 also. a on and we consider so e Thus acts the X, (g2g3 subgroup (a) gP)gi gp)~l, of Sp to act on X by iteration in the natural way. = we Now |{cr)| = may apply Theorem 36.1, and we know that |X| \\X^\\
Note
that
p, so
p divides \\X(a)\\ also. Let us examine X(ay \342\226\240 \342\226\240 = by {a), if and only if gi = g2 = \342\200\242 Now(gi, g2, \342\200\242 \342\200\242Since in at least one element We know divides e). p |, \\X^) X^), namely (e, e, \342\200\242, gp. a e G,a there must be at least p elements in X^. Hence there existssomeelement ^= e, \342\226\240 and hence ap = e, soa has order p. Of course, {a) is a e X^ such that (a,a, \342\226\240\342\226\240 ,a) order \342\231\246 of G of p. subgroup p). Since p
(mod
divides |X|, it
\342\200\242 \342\200\242 is \342\200\242, gp)
36.4
Corollary
Proof
be a finite
Let G
We leave the The
Let G be a group, a G-set by letting then
g acting
write
H
having
The subgroup
subgroup
\342\231\246
as a
G#
subgroups of
under
is the
discussed
just
conjugation,
G# is
Let H be
the
largest
subgroup
normalizer
of H in
G and
will
be
denoted
N[H] \342\226\240
the
-> H given onto
H,
a /^-subgroup
that
lemma
G, then = ^2#_1,then/!i
map ig then map H
.5^ into
make
normal subgroup.
the proof of of a group : H
G. We
e 5^so H < G and g e G, gHg~l. (To avoid confusion, we will subgroup seen to be a GH = {g e G | gHg~l = H] is easily is a normal subgroup Since G# consists of ofGH.
on.
ifghig~l
of p.
is, if H
That
conjugation.
of G
now
a power
14.
of all
collection
the
on H yields the conjugate this action as gH.) Now
of
In
36.6 Lemma
be on S? by
let 5^
and
G act
G (Exercise 11), and H of G that leave H invariant
from
only if | G | is
if and
corollary to Exercise
all elements
subgroup
36.5 Definition
a /7-group
Theorems
Sylow
never
of this
proof
a, and hence by
G is
Then
group.
be that
must
left fixed
g
=
h2
=
by
^(\302\276)
so gHg~l
of a
follows, we will use the fact that if H is a finite if ghg'1 e H for all h e H. To seethis, note that cancellation in the group G. Thus the conjugation by is one to one. Because ghg'1 |H| is finite, ig must
e N[H]
finite
= H
and
group
(W[H] :H)
g
e iV[H].
G. Then
= (G :H)(modp).
Part
324
Advanced
VII
Group Theory
Note
\342\226\240 Historical
Sylow theorems are due The mathematician Peter Ludvig (1832-1918), paper
in
1872.
permutation
a group
had
re-provedthe even though
yet
in fact
that
of
definition
every group
Proof Let i^be
spent most
Sylow
teacher
in University the
immediately
1898.
equation
of a
prime p
of his professionallife
in
Halden,
a
as
and was
Norway, Christiana
at
of his
years
the mathematical works Niels Henrik Abel.
of editing
project
countryman
cosets of H
solving
any
order a power
to a position He devoted eight
appointed
only
can be
that
radicals.
high school
(Cayley's
group
Sylow himself
8.16]).
[Theorem
is solvableby
of
question of
showed
and
Galois group has
whose
given). Georg Frobenius for abstract groups in 1887,
consideredas a permutation theorem
in terms
to the
theorems
equations
algebraic
been
theorems
he noted
Sylow a brief
Mejdell
who published them in Sylow stated the theorems (since the abstract groups not
appliedthe
the Norwegian
to
life to of his
so that H act on i^by left translation, Note that \\5?\\ = (G : H). are fixed under action by all that is, those left cosets that 3\302\247H, = fr(xH)ifandonTyifH = x^frxH.orifandonlyifx-1^ elements e H. ofH.NowxH = h(xH) = x^x\"1)\"1 eHforalU e H, for all h e tf if and only if x\"1^ ThusxH or if and only if x_1 e N[H] (seethe comment before the lemma), or if and only if in N[H]. x e N[H]. Thus the left cosets in S%n are thosecontained The number of such the
of left
set
h(xH) =
(hx)H. Then Let us determine
cosetsis (N[H]: H), so * Since H is a /7-group, = then us that |^| tells
in
and let
G,
an H-set.
^becomes
=
\\S&H\\
it has
(N[H]
: H).
order a
power of p by
Corollary
that is,
: H)
p),
(mod
\\3ZH\\
that
(G
=
36.4.
36.1
Theorem
(N[H] : H)
p).
(mod
\342\231\246
36.7
Corollary
Proof
be a /^-subgroup
Let H It
Lemma
from
follows
from 1. Thus
of a
H
^
36.6
that
p
Theorem
(N[H]
1.
G
2.
Every
contains
order
We
must
which
Sylow theorems, which prime power dividing \\G\\.
of the any
subgroup
group
of order p'
a subgroup
pl+l for 1 know
: H),
^ H.
then N[H]
: H),
(G
be different
then
\342\231\246
Sylow Theorem) Let G bea finite p does not divide m. Then
(First where
Proof
divides
divides
N[H].
We are now ready for the first of prime-power subgroups of G for
36.8
G. If p
finite group
G of order p'
H of
G contains
a
subgroup
each
for is
and let
i where
a normal
of order
\\G\\
=
pnm
1 < i
subgroup
p by Cauchy's
asserts
the existence
where n > 1
and
of
theorem
that the existence of (Theorem36.3). argument a subgroup of order pl for i < n implies the existence of a subgroup of order a < Let H be of order Since i we see divides \302\273, p'+1. p'. (G : H). subgroup p we know : Lemma then H is a normal divides Since 36.6, (N[H] By p H). We
use
an induction
and
show
Section
Sylow Theorems
36
325
form N[H]/H, and we see that p divides the factor group has a subgroup N[H]/H K which is of order p.lfy: N[H] -> N[H]/H is the canonical then y~l[K] = {x e N[H]| y(x) of e K} is a subgroup homomorphism, hence and contains H and is of order of G. This subgroup N[H] p'+l. We repeat
where in the
36.9
Definition
A Sylow contained
the
in part 1 and note Since H is normal
construction
|y_1[\302\243]
possibly
theorem,
Cauchy's
By
\\N[H]/H\\.
2.
we can
of N[H],
subgroup
| =
p'+1.
smaller
a Sylow
every
/7-subgroup,
this fashion;
that
36.10Theorem (SecondSylow Then Pi
Proof
and
is,
=
pnm
precisely
conjugate
gPg~l
every
\342\226\240
shows 36.8. The theorem If of order pn. P is subgroups
as in Theorem those
of P
is also
/^-subgroup
Sylow
a Sylow
can be
The
/^-subgroup.
obtainedfrom
P
be Sylow /^-subgroups of a finite of G. subgroups
group
J^be
let one
of the subgroups
the collection
\342\231\246
subgroups.
The few
final
illustrations
(Third Sylow
Sylow theorem are given after
congruent
on the number of Sylow /^-subgroups. A and many more are given in the next section.
gives information the
If G
Theorem)
Sylow /7-subgroups is Proof
G.
act on left cosetsof the other, and use Theorem 36.1. ofleftcosets of Pi, and let P2 acton Jz^by >'(xPi) = (yx)Pifor = (G : Pi) = |^|(mod/?),and|^| Then a e ^is 36.1, P2-set.ByTheorem P2. y |^,| = all is net divisible by p, so |J2fPn | ^ 0. Let xPi e Then for xPi y e P2, yxPi 5\302\247Pl. so x~lyxP\\ = Pi for all y e P2. Thus x~lyx e Pi for all y e P2, so x_1P2x < Pi. = |P2|, we must have Pi = x_1P2x, soPi and P2 are indeed conjugate Since |Pi|
Herewe will Let
36.11 Theorem
in
are conjugate.
Pi and P2
Let
are conjugate
is, a /^-subgroup
of G, that
/^-subgroup
any two Sylow /^-subgroups
Theorem) Pi
\\G\\ are
second Sylow theorem statesthat
it is
\342\231\246
/^-subgroup P of a group G is a maximal in no larger /^-subgroup.
G be a finite where group, the Sylow /^-subgroups of G
< N[H] of coursenormal
< y_1[K]
H
y~l[K].
group
Let
that
that
in N[H],
theorem,
is a finite to
and p
group
1 modulo
p and
divides
divides
\\G\\,
then
the number
of
\\G\\.
be one Sylow /^-subgroup of G. Let 5^be the set of all Sylow /^-subgroups and let on <5^by conjugation, Theorem so thatx e P carries T e .9*mtoxTx~l. 36.1, By = = Let then us find If T e xFx\"1 T for all x e P. Thus \\^\\ |^| (mod p). 5?P. 5?P, P < N[T]. Of course T < N[T] also.SinceP and r are both Sylow of /7-subgroups are also of But then in are G, they N[T] Sylow /^-subgroups N[T]. by they conjugate Theorem 36.10. Since T isa normal of N[T], it is its only conjugate in N[T]. subgroup Thus T = P. Then 5^ = {P}. Since \\S?\\ = \\5?P\\ = 1 (mod /7), we see the number of Let P P act
Sylow /7-subgroups
Now let there
is congruent on
J?^by
to
1 modulo
p.
Since all Sylow /^-subgroups are conjugate, = G. If P e <5fthen of P | = (G : G/>) by \\\302\243*\\ |orbit the normalizer of P.) But (G : G/>) is a divisor of \\G\\,
conjugation.
one orbit in Sunder 16.16. (G/> is, in fact, the number of Sylow p-subgroups is only
Theorem so
G act
divides \\G\\.
\342\231\246
Advanced Group
VII
Part
326
36.12Example
The
Theory of S3 have order 2. The
2-subgroups
Sylow
of order
subgroups
2 in
S3 in
Example
8.7
are
Note that of S3. We
are three
there can
subgroups
readily
36.13
= (Po,M3}
order
use the
Sylow
claim
to
congruent
5 and congruent
divides 15,we seethat inner
of G.
G has
ig of
automorphism
G
5. Hencewe must
of order
M2}
are all conjugate. no
that
\342\226
of order 15 is simple.Let G have 5. By Theorem 36.8G has at
group
of order
subgroup
maps P onto a subgroup
= gxg~l
ig(x)
= P for all g e G, so P 37.10 will show that (Example
gPg~l
again
gPg~l,
a normal
is
must
G
subgroup actually be
be cyclic.)
\342\226
that Example 36.13 gives some inkling a theorem that counts something,
trust
We
of
of Theorem
the power modulo
even
underestimate
Never
order
Theorem
by
simple.
must
therefore
with
have
Therefore,G is not
abelian and
6, the
36.11 the number of such subgroups is divides 15. Since 1,6,and 11 are the only positive numbers these to 1 modulo 5, and since among only the number 1 P order 5. each g e G,the one of But for exactly subgroup
of order 5, and
15 that are
than
G has
= (Po,
ipiKpo.Mi}]
they
to show a normal
theorems
1 modulo
and that
illustrating
that
one subgroup
least
less
PjxpJ1,
15. We
Let us
Example
=
iPj(x)
(mod 2). Also, 3 divides
3=1
that
and
check that
^[{Po.Mi}] where
{P0,M3}-
{P0,M2},
Ml}>
{p<),
36.11.
p.
36
\342\226\240 EXERCISES
Computations 1 through
Exercises
In
1.
2. 3.
A
Sylow
A
Sylow
A
of a
group
3-subgroup
of a
group
5.
A
255 =
of order
12 has order
of order
54 has order
Sylow 2-subgroups.(Useonly
or
either
(3)(5)(17)must
Sylow
5-subgroups.
all Sylow
3-subgroups
two
6. Find
have
blanks.
the
information
given
36.11.)
of order
group
Find
24 must
of order
group
in the
fill
3-subgroup
in Theorem
4.
4,
Sylow
2-subgroups
(Use only of S4 and
have
the
demonstrate
of S4 and
either
information
show
that
or Sylow 3-subgroups given in Theorem 36.11.)
that they they
are
all
and
or
conjugate.
are conjugate.
Concepts 7 through it is in
Exercises
In
needed,so that 7.
Let p
8. The
be a prime.
normalizer
itself.
of the italicized term 9, correct the definition a form acceptablefor publication. A p-group
N[H]
of a
is a subgroup
with
group
H
of a
the property
without
that every
group G is the
set
reference
to the
text, if correctionis
elementhas order p.
of all
inner
automorphisms
that
carry
H onto
Section 37
9.
Let
a group whose order is divisible by a prime p. The Sylow p-subgroup that P has some power of p as its order. property
G be
G with the Mark each of the P of
10.
of the
Applications
true or
following
327
Sylow Theory
of a
group is the
subgroup
largest
false.
a. Any two Sylow /^-subgroups of a finite group are conjugate. b. Theorem 36.11shows that a group of order 15has only one Sylow 5-subgroup. of a finite group has order a power of p. c. Every /^-subgroup Sylow
d. Every
/^-subgroup
e.
finite abelian
Every
of every
finite
is a
group
has
group
Sylow
/^-subgroup.
one Sylow
exactly
each prime
for
/^-subgroup
p
order
the
dividing
of G.
f. Thenormalizer
g. If H is h.
A
in G of
a subgroup
Sylow
H of G is always
always a normal G is normal group
of G, then of a finite /7-subgroup
of N[H].
subgroup
and only if it is the
G if
in
G.
subgroup of
a normal
H is
a subgroup
Sylow
only
/^-subgroup
of G.
i. If G is an
j.
A
abelian
order
pn
Show
that Gh
of prime-power
group
a subgroup
and H is
group
no Sylow
has
= H.
then N[H]
of G,
/^-subgroup.
Theory
11. Let
H be
12. Let
G bea finite
a subgroup
Show that every
13.
group
14. Prove Corollary 15. Let G be a finite
primes p
and
so G
subgroup,
45 has
= {g \302\243 G \\ gHg~l
p divide
q ^
is not
a normal
and let
group
[Hint:
Argue
Show
there
a prime P is
that
and let
group
of G.
p be
the
dividing
a prime p divide g e G such
/^-subgroup
\\G\\. Let
P be a < P.
that
(35)3has a normal
gHg~l
every
group of order
18.
Show
that
there
are no
simple groups
of order
255 =
19. Show that there 20. LetG bea finite
are no
simple groups
of order
prm, where
group.
Regard
G as
a G-setwhere
G acts
that Section Gg is the center Z(G) of G. (See Theorem 36.1to show that the center of a finite
Show that a finite group and Hi < Hi+\\ for 0 < i < n.
a prime.
p'
22. Let G be a finite /7-subgroup
section
group
and let
P be
of order [Hint:
a normal
See
pn
on itself
by
use
and
of G
/^-subgroup
Sylow
Theorem
that
36.10.]
and let H
be any
125.
(3)(5)(17). p is a prime,
a. Show
=
Sylow
Show
of G.
/^-subgroup
Sylow
of N[N[P]],
of order
subgroup
a
P be
Sylow
only
exists
Let
\\G\\.
that
1\302\276]
precisely oneproper
9.
of order
subgroup
Show
21. Let p be
of G.
simple.
17.
b. Use
that if G has
Prove
\\G\\.
is a subgroup
= H}
36.4.
16. Let G bea finite /7-subgroup
G.
of order
group
= N[P].
N[N[P]]
and let
a normal
it is
/7-subgroup,
of a group
a positive
r is
integer,
and
m <
p.
conjugation.
15.) nontrivial contains
Exercise
/^-subgroup
is nontrivial.
/7-group normal
20 and get of G.
subgroups an
Show
that
H for 0 5 i < n Section 35.]
such
that
from
idea P
is contained
in every Sylow
of G.
37
Applications
of the Sylow
Theory
In this section we give several applicationsof the Sylow theorems. It is intriguing to see how easily certain facts about orders can be deduced. we of However, groups particular should realize that we are working with of finite order and only groups really making
Part VII
Advanced GroupTheory dent in order of a group be of some use
of determining the structure of all finite groups. few factors, then the techniques illustrated in this section in determining the structure of the group. This will be demonstrated may in Section further all 40, where we shall show how it is sometimes possibleto describe even when some of the groups are not orders, groups (up to isomorphism) of certain abelian. However, if the order of a finite group is highly composite, that is, has a large is in general much harder. number of factors,the problem a small
only
If the
Applications 37.1
Theorem
Proof
group
Every If
normal
of order
{e} =
H0 < Hi
of it.
G has
< r. Then Hr =
\342\226\240 \342\226\240 \342\226\240 <
G
are of order
p,
a subgroup
H of order
and hence
abelian
and
is solvable.
the
now
\342\231\
class equation.The
used the
theorems
Sylow
We
< H2 <
the factor groups
where
avoidedexplicit
form
general
of
proofs
Section 36
pl+l for 1 < i
H+i
series,
36.8 that
from Theorem
is immediate
Thus, G
cyclic.
older
The
only a
to p-Groups and the ClassEquation of prime-power order (that is, every finite /7-group) is solvable.
is a composition actually
problem
general
has
order pr, it in a subgroup
G has
p'
the
of proof in (2) there is a will be familiar line
class equation, although classic class equation so you
mention
of the
develop
the
Eq.
with it.
Let X us
be
G-set where G is a finite
a finite
\\X\\
=
\\XG\\ +
where
x, is
where
X
in the
element
an
= G
and
action
the
i\\h
of G
{x e
=
XG
on G is by
the
of G. If
center
where
which
we let
is \302\273,
the number
that\302\273,-divides
is a
Equation
(2) is
It is readily
\\G\\
divisor of
conjugate class Example
the
now
case of
special
e G
so g
conjugation,
the in
= x for
G | gxg~l
= gx for
c = |Z(G)|and \\G\\
Note
(1)
+1
Consider
in X.
orbit
= {x eG\\xg
37.3
of Section36 tells
Eq. (1),
carriesx e X
=
G
Then
into gxg~l.
Definition
Eq. (2)
\\GXi\\
J2 1=.5
37.2
Then
group.
that
=c
all
g e
= \\Gxt | \302\273,\342\200\242
+ nc+l
of elements in the c + 1 < i < r
for
g e
all
+
ith
G]
G} =
in Eq.
Z(G),
(1), then
we
obtain
\342\226\240\342\226\240\342\226\240 + \302\273,
orbit of
sincein
Eq.
G
under
(1) we
(2)
conjugation
know |Gx,-1=
by itself.
(G : GXi),
\\G\\.
class
equation
of G.
Each
orbit
in
G under
conjugation
G.
checked that
by
G is
a
\342\226\
for
S3 of
(A)},
Example 8.7, the {/01./\302\276}.
conjugate
{Ml>M2,M3}-
classes
are
37
Section
The class
2+3.
6=1 +
37.4
Theorem Proof
in Section
exists somea
turn now
We
Let G H
Proof
v K
G is
a ^
for
\\G\\
c. Now
nontrivial.
< i < Z(G), so c
c + 1
e e
r, so p divides > 1.Therefore
hkh~lk~l two
the
KC\\H Let
that
will be
used
=
to
H
subgroups H x K.
K
such
and
Proof
p
some
of the
that H
f! K
in
theorems
= {e}and
K and h e H. Consider the commutator Since H and K are normal of G, subgroups with parentheses show that is in both K and H. Since hkh~lk~l groupings = {e},we seethat hkh~lk~l = e, so M = ifc/z. G be defined by 0(/z, fc) = M. Then
that hk
showing
kh
for
k e
= h(kh~lk~l).
\342\200\224
Qikh~l)k~l
k')) =
4>(hti,
kk')
= hkh'k'
=
= hh'kk' 4>{h,
k)4>{h',
k'),
is a homomorphism. h and If (j>(h, k) \342\200\224 both e, then hk \342\200\224 e, so h = k~l, and = = = /; \302\243 e, so Ker() is one to one. {(e, e)} and \302\242) H v K, and H v By Lemma 34.4,we know that HK \342\200\224 Thus 4> is onto G, and H x K \342\200\224 G.
Theorem
and \302\273,,
p, and there \342\231\246
to a lemma on directproducts
4>{{h,k)(h',
37.6
each c >
e.
be a group containing normal = G. ThenG is isomorphic
start
We
so
Exercise
follow.
that
Lemma
divides
e Z(G)where
a theorem that
we prove
equation,
/7-group
p divides
Therefore
\\G\\.
class
the
prove.
nontrivial
each nt
(2) for G,
In Eq.
use of
36 askedus to
The center of a finite
divides
37.5
of the
illustration
For
20(b)
329
Theory
S3 is
of
equation
of the Sylow
Applications
4>
For a
prime number
/7, every
group G
of order /72
&
are
K=
in H
G
by
n
isT.
Thus
hypothesis. \342\231\246
is abelian.
except e must be of order p. Let a be such an exhaust G. Also let b e G cyclic subgroup {a) of order p does not = & \302\242 Then with n since an c in n with element would (a). {a) (\302\276) {e}, {a) (\302\276) c/e = both and to construction. From Theorem (a) generate (b), giving (a) (b), contrary in some in all of G. 36.8, (a) is normal subgroup of order p2 of G, that is, normal Likewise (b) is normal in G. Now (a) v (\302\276) is a subgroup of G properly containing of (a) and of order dividing {a} v {b) must be all of G. Thus the hypotheses p2. Hence Lemma 37.5 are satisfied,and G is isomorphic to {a) x (b) and therefore \342\231\246 abelian. If G
is
not
element.
cyclic,
Further
We have
then
every element
Then the
Applications
turn
seen
to a discussion of whether there exist simple groups of certain orders.We that every group order is is simple of prime simple. We also asserted that A\342\200\236
now
Part
VII
Advanced Group Theory
n >
for
5
that
and
A 5 is the smallest
simple
conjecture of Burnside that every of even order.It was a triumph when this
a famous
be 37.7 Theorem
If p and q are distinct subgroup of order q
to 1 modulo
congruent
Proof
with
a single
has
If q
this
is
not
36.11 tell us that G has a Sylow ^-subgroup and that the number is to 1 modulo must of such subgroups congruent q and divides pq, and therefore 1.Thus < there is one the is the divide p. Since p number only Sylow q, only possibility Q of G. This group Q must be normal in G, for under an inner automorphism ^-subgroup G is not simple. it would be carried into a group of the same order, hence itself.Thus and the number these divides pq is a P of of G, Likewise, there Sylow /^-subgroup either 1 or be to 1 modulo This number must If and is congruent q. q is not congruent p. normal in is G. Let us assume that 1 the number must be 1 and P to modulo p, then is of order in e 1 Since element other than Q q and every element in (mod p). every q ^k. = Also P must be a subgroup f! P of P other than e is of order we have {e}. gv Q p, and
36.8
Theorems
G properly containing 37.5 is isomorphic to We
37.8 Lemma
was
is not
p
primes and
of prime order.There was of order must simple group nonprime proved by Thompson and Feit [21]. that
group
finite
If
H
need
Q x P or
Z?
finite
subgroups
=
G and pq. Hence Qv P G is abelian and cyclic.
dividing
x
\"Lp.
Thus
lemma for some of the
another
K are
and
Q and of order
of a
group G,
\\HK\\
=
arguments
counting
by
Lemma
\342\23
that follow.
then
(|ff|)(|*|) \\HC\\K\\
Proof
Recall HK
= = {hk\\h e H,k e K}. Let \\H\\=r, \\K\\ r5elements.H0wever.it is possiblefor hiki to ^T; that is, there may be somecollapsing.If h\\k\\
that HK has at most
and &1;k2
e
x = x
Now
\342\200\224
(h2ylhi
x e (H n
shows
that x
e H,
and
x
\\H
Pi K\\
= t.
equal h2k2,forhi, = h2k2, then let
h2
Now
e H
k2(kiTl
=
\302\2432(^1)-1
shows
that
x e
K. Hence
A\,") and
h2=h\\x~l
On the
{h2ylhx =
s, and
and
^2=-^1-
and &3 = y&i, then clearly y e (H n isT) we let /z3 =h\\y~l e ^T. e H^T canberepresented e Thus each elementM /!3&3 Hand&3 in the form /!,\342\226\240&,-, for A, e H and &,- e ^T, as many times as there are elements of H 0 K, is that t times. the in HK \342\23 number of elements is, Therefore, rs/t. =
other
hand,
if for
/; 1&1, with A3
so do not underestimate Lemma it. The 37.8 is another result that counts something, lemma will be used in the following A finite group G cannot have H way: subgroups and K that are too large with intersections that are too small, or the order of HK would a group of order 24 have to exceed the order of G, which is impossible.For example, cannot have two subgroups of orders 12 and 8 with an intersection of order 2.
Section37
of the
Applications
331
Sylow Theory
Theremainder
of this section consists of severalexamplesillustrating of techniques that all groups of certain orders are abelian or that they have nontrivial proper normal subgroups, that is, that they are not simple. We will use one fact we mentioned H of index 2 in a finite before only in the exercises. A subgroup group G is always we see that there are only the left cosets H itself and the coset normal, for by counting, are the same. Thus even,' right consisting of all elements in G not in H. The right cosets coset is a left coset, and H is normal in G.
proving
37.9 Example
No
a
group
must
order
37.10 Example
where 36.8 such p is a prime. For by Theorem pr for r > 1 is simple, of order pr, which a subgroup of order pr~l normal in a subgroup it has a normal subgroup 16 is not simple; of be all of G. Thus a group of order A 8. G contains
are
abelian of order 15 is cyclic(hence and 5 is not congruent because 15 \342\200\224 (5)(3),
group
Every
This is
37.11 Example
of order
group
and not to
simple, since 15 is not a prime). 3. By Theorem 37.7we
1 modulo
A
done.
20 is simple, for such a group G contains Sylow 5-sub groups in number is to 1 modulo 5 and a divisor of 20, hence only 1. This Sylow 5-subgroup A then normal, since all conjugatesof it must be itself.
No group
of order
congruent
37.12 Example
No
group
of order
for
some
prime p
30 is simple.We dividing
30,
seen that if there is only one Sylow /^-subgroup for the done. By Theorem36.11the possibilities or 6, and those for Sylow 3-subgroupsare 1 or 10.
have
we are
number of Sylow 5-subgroupsare 1 of any two is a subgroup of But if G has six Sylow 5-subgroups,then the intersection each of order dividing 5, and hence just {e}.Thus each contains 4 elements of order 5 of order 5. Similarly, 24 elements that are in none of the others. Hence G must contain 3. The two types if G has 10 Sylow 3-subgroups, it has at least 20 elements of order of Sylow subgroups together would require at least 44 elements in G. Thus there is a A either of order 5 or of order 3. normal subgroup 37.13
Example
No group a normal
of order48 is simple. subgroup
or three Sylow normal in G, by
16 or
2-subgroups of order
show that a group G of order 48 has Theorem 36.11 G has either one of order 16, it is there is only one subgroup
we shall
Indeed,
of either order
16.
order
If
8. By
familiar argument. Suppose that there are three subgroups of order 16,and let H and K be two of them. Then H C\\K must be of order 8, for if H D K were of order < 4, then by Lemma 37.8 have at least (16)(16)/4 = 64 elements, the fact that G has HK would contradicting in both H and K (being of index 48 elements. 2, or Therefore, H f! K is normal only H and K and must of H D K contains both by Theorem 36.8). Hencethe normalizer a now
have order a multiple normal in G.
37.14 Example
>1
of 16
and a
divisor
of 48,
therefore
48. Thus H
OK
must
be
A
of order 36 is simple.Sucha group G has either one or four of order subgroups group 9. If there is only one such subgroup, it is normal in G. If there are four such subgroups, let H and K be two of them. As in Example 37.13, H OK must have at least 3 elements, or HK would have to have 81 elements, which is impossible. Thus the normalizer of > C\\K has as order a 1 of and a divisor of H of 9 36; hence the order must multiple No
Advanced GroupTheory
Part VII
332
in
normal
37.15
36. If the
18 or
either
be
G. If the
order
is 18,
order is 36, then
the normalizer is then f! K is normal in G.
of
index
H
2 and
therefore is
\342\226\262
is abelian (hence cyclic by the Fundamental Every group of order 255 \342\200\224 (3)(5)(17) Theorem 11.12 and not simple, since 255 is not a prime). By Theorem 36.11 such a 15 and is abelian G has only one subgroup H of order 17. Then G/H has order group
Example
15.20, we seethat
subgroup C of G is order 1 or 17. Theorem 36.11 3 and either 1 or 51 subgroups of also shows that G has either 1 or 85 subgroups of order 5. However, 170 elements of order 3, and 51 order 85 subgroups of order 3 would require 5 5 in G; both together would order would 204 elements of order of subgroups require K having in is a subgroup then 375 elements which is Hence there G, require impossible. or order in G. Then G/K has either order (5)(17) either order 3 or order 5 and normal < is Thus C K and and in either case Theorem 37.7 shows abelian. that (3)(17), G/K has order either 3, 5, or 1. SinceC < H showed that C has order 17 or 1, we conclude ~ G is abelian.The Fundamental and Theorem that C has order 1.HenceC = {e}, G/ C H.
in
contained
By Theorem Thus as a subgroup
37.10.
Example
by
11.12 then shows
that
G is
of
H,
the
commutator
C has either
\342\226\262
cyclic.
37
EXERCISES
Computations
1. Let D4 a.
be the the
Find
gxoup of
of D4
decomposition
the class equation
b. Write
2. By arguments
to
similar
group of order not You need
simple.
symmetriesof the for
conjugate
in Example
8.10.
classes.
D4.
used in the examples of this section, convince yourself that every nontrivial and less than 60 contains a nontrivial proper normal and hence is not subgroup out the details. (The hardest cases were discussed in the examples.)
those
a prime
not
into
square
write
Concepts
3. Mark
each
of the
a. Every
b. Every
following group
true
of order
or false.
159 is
cyclic.
102 has a nontrivial normal subgroup. proper c. Every solvable group is of prime-powerorder. d. Every group of prime-power order is solvable. e. It would become quite tedious to show that no group of nonprime 60 and 168is order between simple by the methods illustrated in the text. f. No group of order 21 is simple. g. Every group of 125 elements has at least 5 elements that commute with every element in the group. h. Every group of order 42 has a normal subgroup of order 7. i. Every group of order 42 has a normal subgroup of order 8. where j. The only simple groups are the groups Zp and A\342\200\236 p is a prime and n ^ 4. group
of order
Free Abelian
Section 38
333
Groups
Theory and cyclic. (5)(7)(47)is abelian no group of order 96 is simple. no group of order 160is simple. a subgroup of order 15.[Hint: every group of order30 contains
4. Prove that 5. Prove
that
6.
Prove
that
7.
Show
that and go
37.12,
of order
group
every
factor
the
to
8. This exercise determines
a.
b. Argue
from (a) that
c.
from
Argue
of
order
ai,
classes of
conjugate
(b) that
Sn
for
every
and x is any a cyclein\302\243\342\200\236
\342\226\240 \342\226\240is \342\226\240, am)
of
two cycles in Sn
any
(a) and
in
Example
of s
a product
same
the
element
length
1.
of Sn then x ax~l
\342\226\240 \342\226\240, ={xa\\, xa-i, \342\226\240 xam).
are conjugate.
in Sn of
cycles
disjoint
integer n >
lengths
rt
for
i =
1,2,
\342\226\240 \342\226\240 s is \342\226\240,
conjugate
rt in Sn. disjoint cycles of lengths is pin), where p(n) is the number of ways, neglecting of conjugate classes in S\342\200\236 number is the the summands, that n can be expressed as a sum of positive integers. The number pin)
to every other d. Show that the the
the
a =(ai,
thatif
Show
sentence
the last
Use
group.]
of s
product
number of partitions of n. for n = 1, 2, 3,4,5, 6,7. e. Compute pin)
9.
Find
the
10.
Find the
11.
Show
that
class
that
of
if n
Use Exercise in
classes
p is a prime
the
Exercise
Use
8.]
of the
consisting
subgroup
\342\200\224
8.]
the number of different [Hint: Use Exercise 8.]
is also S\342\200\236
number.
of Sn is
center
for S4. [Hint:
equation
,%\342\226\240 [Hint:
conjugate
p\", where
> 2, the
class
the
& and
for
equation
the number
isomorphism) of order
12. Show
classes and
conjugate
abelian
only.
permutation
identity
(up to
groups
[Hint: Use
Exercise 8.]
section 38
Free AbelianGroups In
this
results
section
concerning
Theorem
of free abelian groups we introduce the concept a demonstration them. The section concludes with
generated abelian groups
of finitely
and
some
prove
of the
Fundamental
(Theorem11.12).
Free Abelian Groups We
should
the notions of a generating 7. In this section
review
group, as given in Section and use additive notations 0 for
the
set
as follows:
+ for
identity,
na = a
+a +
the operation, \\-
a
n summands -na
(-0) + . + (-0)
(-0) + n
Oo
0 for
group G and a finitely generated deal exclusively with abelian groups
for a
we shall
the
first
e Z+
forw
ando
e G.
summands 0 in
Z and
We shall continue to use the symbol to direct sum notation.
x
for
the
second
direct
in G.
product
of groups
rather
than
change
Part
334
VII
Advanced Group
that
Notice
=
(\302\273, m)
38.1
Theorem
that
each
That
is,
Theory
1) for
a subset of
any
x Z. This generating
form
The following
xZ
has the
set
in the
expressed
group G.
abelian
a nonzero
group Z
the
for
set
generating
in Z
(\302\273, m)
x Z can be uniquely \302\273 and m in Z are unique.
of Z the coefficients element
Let X be
1)} is a
0), (0,
{(1, + m(0,
\302\273(1, 0)
\302\273(1, 0)
since
property 1).
+ m(0,
on X
conditions
are
equivalent.
1. Each nonzero summands) in
Proof
X
generates
x,-
e X if
G, and
and
SupposeCondition X,forifx,- = Oandxj the
0
a =
form
the
7^
m
7^ 0. Then
1 is
+ \302\2731x1
only
terms
+
nrxr
We
+
+Xj,
X
have
we
1, X
= 0 for
e Z \302\273,-
/ (0). It
a can be written another such expression in G, we see
two
follows
G, and
generates
+ \302\2731X1
+
H \302\2732*2
1-
+
H \302\2732*2
1\" nrXr,
(n\\X\\
l)Xi
terms
assume
from 1 that 0 ^ the uniqueness of = H + nrxr \302\2732*2 some nrxr = 0 with
we can assume
the uniqueness
all
a
assumption in
Since
e G.
X
generates
a has coefficients in and are of the
\342\226\240 \342\226\240 \342\226\240 + \302\273rxr.Suppose
some
zero
elements
the same
involve
they
and distinct
nrxr)
2 implies Condition 1. Let form a = nxx\\ + \302\2732X2 + of elements of X. By using
in the
we can
expressions,
of Z and
contradict
would
which
that
+ (\302\273!
that Condition
show
now
0 in
0.
of writing xx 7^ 0, contradicting gives two ways 1. Thus Condition 1 implies Condition 2.
Condition
\302\273; 7^
H + n\\X\\ + \302\2732*2 Suppose and renumbering, zero coefficients
with
=
the
\342\226\240 \342\226\240 \342\200\242
{0},
= x,
xi = xi
which
to order
(up
for
nrxr
+
H \302\2732X2
+ \302\2732*2
G/
true. Since 7^ 0,thenxj
dropping
0; by
+
n\\X\\
\342\226\240 \342\226\240 = nr = = \302\2732 = \342\226\240 if \302\2731
From Condition expressionforxj. \342\226\240 \342\226\240 = nr = 0. = \302\2732 = \342\226\240 if \302\273i
nt
G can be expressed uniquely
X.
x,- in
distinct
2.
a in
element
in
X
form
n\\X\\
a =
m\\X\\ +
+
\342\226\240 \342\226\240 \342\226\240 \302\273rxr
+ \302\2732X2 OT2X2
+
\342\226\240 \342\226\240 \342\226\240
mrxr.
we obtain
Subtracting,
0= so
a =
\342\200\224 = \302\273,-m,-
\342\200\224 (\302\273i m,\\)x\\
0 by Condition
2,
\342\200\224 (\302\2732 OT2)x2
+
\342\226\240 \342\226\240 \342\200\224 \342\226\240 + (\302\273r mr)xr,
for i =
= m, \302\273,-
and
+
\342\226\240 r. Thus 1, 2, \342\226\240 \342\226\240,
the coefficients
are unique. 38.2
Definition
An abelian Theorem
38.3
Example
\342\231\2
The group free direct
is a
Zx
free abelian group,and
Z is
free
abelian
group Z x Z of the group products
abelian
set
a generating
having
group
38.1
x Z is
and {(1, {(1,
Z with
0,
itself
X
X is
satisfying
the
a basis for
the
conditions group.
described
in
\342\226\2
0), (0, 1)}is a basis. Similarly, a basis for the 0), (0, 1, 0), (0, 0, 1)},and so on. Thus finite are free abeliangroups. \342\226\2
38.4Example
The
Condition
a 7^
abelian
a free
Suppose and
free abelian, for 2.
is not Z\342\200\236
group
contradict
a has a unique
0, then
a = (Note that in the
of the
that some
that
specified
the
the
in
coefficients
each
for a
finite
basis
of the
form
nrxr
+
X =
included
a, we
1 of
Condition
in
which
n j^ 0,
would
\342\226\240 \342\226\240 \342\226\240
{xi,X2,
G
a e
,xr}.If
e Z. \302\273,-
all elements x,- of our finite basis Theorem 38.1 where the basis
we must
for a
expression zero, whereas
preceding are \302\273,
and Z\342\200\236,
for
\342\226\240 \342\226\240 \342\226\240
+ \302\2732*2
expression
Thus
be infinite.
may
+
preceding expressionfor
X, as opposedto
a
has
expression
n\\X\\
335
Groups
\342\226\262
G
group
every x e
0 for
=
nx
Free Abelian
38
Section
the possibility
allow
Theorem38.1,we
in Condition 1 of
\302\273,\342\200\242 7^ 0.)
We define : G
Zx Zx
^
\342\226\240\342\226\240\342\226\240 x Z
r factors by
4>{a)
=
\342\226\240 \342\226\240 and \302\2732, \342\226\240, (\302\2731. \302\273r)
is an \302\242)
isomorphism. result as a theorem.
38.5Theorem
If
G is
a nonzero
free abeliangroup r factors.
It is a fact
number
Proof
\342\226\240 \342\226\240 It is \342\226\240, (0, 0, 0). to the exercises
The
is really
proof
basis in
in a
basis,
lovely; it
of the
terms
(see Exercise9) and
G is
G contain
group
prove
of elements
straightforward to check that
of r elements, then
a basis
with
any two bases of a free abelian this only if G has a finite
that
elements. We shall basis of G is infinite.
Theorem
details
\342\226\240 \342\226\240 Z x \342\226\240 x Z for
Zx
38.6
=
\302\242(0)
We leave the
although an easy
gives
size of a factor
to
number
of
also true
G
have
a basis
factors. Let 2G =
{x\\,xo_,
{2g\\ g
characterization of the
G/2G
~
(Z x Z2
X
we
have
\342\226\240 \342\226\240 x Z)/(2Z Z x \342\226\240
Z2
X
let
by
{yi,
of
Y.
{yx, >'2,
x
\342\226\240 \342\226\240 \342\226\240 x
2Z)
Z2
in any finite basis X is of elements. infinite basis. Let Y be any basis for G, Y. Let H be the subgroup of G generated of G generated by the remaining elements
of elements
same
number
distinct elements in let K be the subgroup checked that G ~ H x K, so
\342\226\240 \342\226\240 be \342\226\240, ys)
\342\226\240 \342\226\240 and \342\226\240, ys],
yi, It is readily
x 2Z
\342\226\240 \342\226\240 \342\226\240 X
for r factors. Thus |G/2G| = 2r, so the number log2 |G/2G|. Thus any two finite bases have the It remains to show that G cannot also have an and
finite,
\342\226\240 \342\226\240 x Z for r isomorphic to Z x Z x \342\226\240 is readily checked that 2G is a subgroup of G. Since
factors,
~
G is
G is
\342\226\240 \342\226\240 Then \342\226\240, xr}.
e G}. It
G~ZxZx---xZforr
if every
group.
with a finite basis. Then every basis of Let G 7^ {0} be a free abelian group and all bases of G have the same number of elements. Let
the
isomorphic
same
the
it is
state
G/2G ~ (H x K)/(2H
x
2K)
~
(H/2H)
x (K/2K).
Part VII
Advanced
Since s < r. 38.7
Definition
Then Y
>
\\G/2G\\
infinite
set,
of G is the of elements.)
\342\226
11.12)by
where both
\342\226\240 \342\226\240 Z x \342\226\240 x Z)/(dxZ
ds. The
divides
x
x d2Z
x {0} x
dsZ
have n factors,
that any
showing
finitely
form
of the
\342\226\240 \342\226\240 \342\226\240 x
\"denominator\"
and
\"numerator\"
\342\226\240 \342\226\240 divides 6\302\276 which \342\226\240,
basis for G.
in a
Theorem
Fundamental
(Z x
that
see
\342\231
of elements
number
prove the Fundamental Theorem (Theorem abelian group is isomorphic to a factor group
shall
generated
= 2r, we
we have |G/2G| could take s > r.
2s. Since for we
the rank
group,
the same number
the
of
Proof
see
be an
cannot
have
bases
We
2s, we
=
\\H/2H\\
is a free abelian
If G (All
Theory
Group
\342\226\240 \342\226\240 \342\226\240 x
{0}),
and
d2, which 11.12 will
divides
d\\
prime-power decompositionof
Theorem
then follow.
To show that
38.8
Theorem
to such a
is isomorphic
G
factor group,
\342\226\240 \342\226\240 x Z x \342\226\240 x Z onto G with \342\226\240 \342\226\240 \342\226\240 x {0}. The result will then follow
of Z
homomorphism
form
Z
dx
x d2Z
x
dsZ x {0} x follow give the
by Theorem 14.11. Thetheorems purpose in these introductory paragraphs
to
what
details of the argument. Our let us see where we are going as we read
Let G be a finitely
group with generating
abelian
generated
n
be defined
by
onto homomorphism
G.
From
the
h2,\302\2420^1,
meaning
\342\226\240 = \342\226\240, hn)
for
of /!,a,
\342\226\240 \342\226\240 \342\226\240, + K) \302\242[011,
{kx,
ht
=
Since
{a\\, We
\342\226\240 \342\226\240 \342\226\240, a\342\200\236) generates
now
prove a
+
+
Then
hnan.
G, we seeat
once
is
a
that
0(/zi +h,---,hn+kn) (hx
+
ki)ax
=
H
(hxax
h hnan)
\342\226\240 \342\226\240 \342\226\240 + \302\24201, ,kn)
h (hnan
+ knan)
+ (kxax
-\\
\302\242(^,
that
makes
\\-
knan)
\342\226\240 \342\226\240 \342\226\240
the homomorphism is
\"replacement property\"
kn)an
V(hn+
-\\
+ kxai) -\\
G, clearly
\342\226\240 Let set {ax, a2, \342\226\240 \342\226\240, an).
\342\226\240 \342\226\240 \342\226\240
= (Mi
=
is
factors
and a,- e
\342\226\240 \342\226\240 = \342\226\240, kn)]
that
G
h\\a\\ + h2a2
e Z
is a
\342\226\240 \342\226\240 \342\226\2 x
follows.
<$>;Z*Z*---*Z->
Proof
show that there
will
we
of the
kernel
,hn).
onto
G.
\342\231
it possible for us to
adjust
a
basis.
38.9 Theorem
If
X
=
{x\\,
\342\226\240 \342\226\240 is \342\226\240, xr}
a basis for
a free abelian
group
G and t
e Z,
set Y
is also
a basis for
G.
=
{x\\,
\342\226\240 \342\226\240 \342\226\240, Xj-x,
Xj
+ txi,
Xj, Xj+X,
\342\226\240 \342\226\240 \342\226\240, xr}
then
for
i j^
j, the
Section38 FreeAbelian Proof
Since
=
Xj
+
+ (1)(xj
(-t)xt
see that
we
txt),
Xj
from
be recovered
can
337
Groups
Y.
thus
which
G. Suppose
also generates
1-
\302\253i*i-\\
+
+ rijiXj
tij_xxj-i
+
txf)
-\\
nj+iXj+l
= 0.
nrxr
\\-
Then
and\302\273; +
0, 38.10
Example
a basis,
X is
since
and
= 0, it
rijt
4(1. 0) +
(2, 0) in
{(3,
+
-\\
nr = 0. From n;- = 0 \342\226\240 \342\226\240 \342\226\240 \342\226\240 \342\226\240 = ~ ~ \342\226\240 ~ nr = nt rij \342\231\246 is a basis.
=
\342\226\240 \342\226\240 \342\226\240 =
nj
\342\226\240 \342\226\240 \342\226\240 =
\342\226\240 \342\226\240 \342\226\240 =:
=
0also,so\302\273!
= 0.
nrxr
Y
0), (4, 1)} for (4, 1) = For example, we cannot express is {(1,
basis a basis.
n2(0, 1),for n, n2 e Z. Here (3.0) = (1,0)+ 2(1,0),and a basis elementwas added to itself, rather than to a different basis element.
form
of
a multiple
=
rijt
(0, 1)}. Another 0). (0, 1)} is not
0),
{(1,
(0, 1).However,
the
n,+
that\302\273, =
Theorem 38.1 is satisfied.Thus
Z is
Zx
A basis for
\342\226\240 \342\226\240 \342\226\240 =
=
nx
h rijXj
-\\
(\302\273,+ rijt)Xi
follows
2 of
Condition
and
+
-\\
n\\X\\
0) +
^(3,
\342\226\262
abelian group
A free
exist bases
38.11Theorem
Proof
G of finite
free abelian with
is also
K
then
of G and
related to
K nicely
bases. We show that if of G. Equally important,
may have many not exceeding that
rank
rank
K <
G,
there
each other.
G be a nonzero freeabelian of of finite rank n. and let K bea nonzero subgroup group G. Thenar is free abelianof rank s < n. Furthermore, there exists a basis[x\\, X2,---,xn} \342\226\240where \342\200\242 for
of
n. Suppose
at most
rank
a basisof
K has
that
show
We
in K can be expressed
=
some
in Yt. By
elements
there is wx
d\\
division
0 and
*i
=
d\\
we
algorithm,
Now let xi for
in
elements
\\-k\342\200\236y\342\200\236,
all bases Y for G, select one Yx elements of K are written of
elements
the
renumbering
yi
+
=
is the
write
=
rfiCVi
q2yi
= kj +
G. From
-\\
d\\qj
qiyi + our
e
+ k2yi
diyx
minimal
Eq. (1) and \342\226\240 \342\226\240 = rn = 0. Thus r2 = \342\226\240 dxx\\ basis
free abelian
if necessary,
Yi
the yields in terms of the
that
we can
assume
K such that
e
>
K is
nonzero
all nonzero
1\302\276 | as
wx
where
H
Among
minimal such nonzero value basis
All
form
the
is nonzero.
\\kt \\
that
show
G.
\342\226\240 \342\226\240 is \342\226\240, yn]
{y\\,
hyi where
form, which will a basis for
described
the
Y
attainable coefficient as just described. + rj where 0 < r,- < d\\ for ;' = 2,
-\\
qn)'nchoice
K.
h knyn
-\\
h qnJn)
+ riy'2
By Theorem of
Yx
for
-\\
\\-
38.9 [xi, y2,
minimal
the
Using
\342\226\240 \342\226\240 \342\226\240. n. Then
(1)
rnyn. \342\226\240 \342\226\240 is \342\226\240, >'\302\253}
coefficient
d\\,
also
we
see
a that
Part VII
338
Advanced GroupTheory consider bases for in the form
now
We
be expressed
G of the
K, we
e
d\\X\\
can
see that
minimality of d\\ to
multiple of
is a
h\\
element
,yn}.
of K
can
vKyn-
a suitable
subtract
\342\226\240 \342\226\240 \342\226\240Each
{xi,y2,
+k2y2-\\
h\\xx
Since
form
of
multiple
d\\,
and
d\\X\\
see we actually
we
using the
then
have
k2y2
+
\342\226\240 \342\226\240 \342\226\240
+
knyn
in K. Among all such bases {xi,y2,---,yn}, we chooseone Y2 that leads to some k{ 7^ 0 of minimal magnitude. (It is possible all kt are always zero. In this case, K is generated the elements of Y2 we can assume that there by d\\X\\ and we are done.) By renumbering is w2 e K such that W2 =
d2y2
h Kyn
-\\
< 0 and
as in the preceding paragraph, d2 is minimal as just described. Exactly \342\226\240 to a basis \342\226\240 \342\226\240 for \342\226\240 = our basis from Y2 x2, yi, \342\226\240, [x\\, y2, \342\226\240, [x\\, yn} yn} modify = < 0 < we see that G where d\\X\\ e K and d2x2 e K. Writing for r d2 d\\, d\\q +r is a basis for G, and d\\X\\ + d2x2 = d\\{xi + qx2) + rx2 is in {x\\ + qx2, x2,y3,---, yn] K. By our minimal choice of dx, we see r = 0, so d\\ divides d2. \342\226\240 for G and examine We now consider all bases of the form {xx, x2, y3, \342\226\240 \342\226\240, yn] is clear. The The elementsof K of the form \302\2433 \\-knyn. j3 + pattern process continues \342\226\240 where \342\226\240 \342\226\240 a basis {x1;x2, \342\226\240 the until we obtain \342\226\240, xs, ys+i, \342\226\240, yn) only element of K of all is that are zero. We then let xs+\\ = the form \\- knyn zero, is, kt ks+iys+i + \342\226\240 \342\226\240 \342\226\240 = a basis for G of the form described in the statement of and obtain ,xn yn ys+\\, \342\231\ Theorem 38.11.
where d2
we
38.12
Theorem
can
Every
Proof
Zm2 x
x
Zmi
where m,- divides
group is
abelian
generated
finitely
m,+i for i =
1,
isomorphic to a group
\342\226\240 \342\226\240 \342\226\240 x
Zmr x Z
x Z x
of the
\342\226\240 \342\226\240 \342\226\240 x
Z,
\342\226\240 \342\226\240 r \342\200\224 1. \342\226\240,
purposes of this proof, it will be convenient to use as notations Z/1Z {0}. Let G be finitely generated by n elements. Let F = ZxZx---xZfor\302\273
For the 1\\
=
the homomorphism
Consider
factors.
kernel of this
Then
homomorphism.
: F
\342\226\240 is a basis for K and \342\226\240, [dtxi, \342\226\240 dsxs] Theorem 14.11, G is isomorphic to F/K. But
F/K
It is
(Z x Z
~
Z^ x Z^
possible
1, m2 be the We
have
11.12).
Z)/(diZ
x
Zd, x Z x
1, in which
product. Similarly,
dt, and
demonstrated
so on, and
d{ divides
x
~
38.8, and let K be the \342\226\240 form {x\\, \342\226\240 \342\226\240, xn], \342\200\224 \342\226\240 \342\226\240 = s 1. By 1, \342\226\240, di+\\ for i F
\342\226\240 \342\226\240 \342\226\240 x
dsZ
of the
x {0} x
\342\226\240 \342\226\240 \342\226\240 x
{0})
\342\226\240 \342\226\240 \342\226\240 x Z.
case
=
Z^,
d2 may our
= Z/Z
Theorem
a basis for
x d2Z
\342\226\240 \342\226\240 \342\226\240 x
d\\ =
that
next
\342\226\240 \342\226\240 \342\226\240 x
{0} and
be 1, and
theorem
follows
can be dropped (up We let mi be the at once.
so on.
to first
\342\231\2
Fundamental Theorem (Theorem exists sincewe can break the groups decomposition The only remaining of Theorem 11.12 concernsthe part
the
course,
a prime-power
prime-power
factors.
Of
into
x
from this
isomorphism)
Zm.
~
\342\200\224> G of
there is
where
dt >
form
toughest
part
of the
339
Exercises
38
Section
uniqueness of the Betti number, of the torsion coefficients, and of the prime powers. The Betti number appears as the rank of the free abelian group G/ T, where T is the torsion of G. This rank is invariant shows the uniqueness of subgroup by Theorem 38.6 which The uniqueness of the torsion the Betti number. and of prime powersis a coefficients bit to show. We give some exercises more difficult that indicate their uniqueness (see
Exercises 14through
22).
38
\342\226\240 EXERCISES
Computations
1.
a basis {{a\\, ai, a?,),{b\\, (Many answers are possible.)
Find
2. Is {(2,1),
1)} a
(3,
basis for Z x Z? Prove your
1), (4, 1)}a basis
3. Is
(c\\, C2,C3)}for
^),
bi,
{(2, conditions on a,b,c,d (e, /) in R, and see when the
4. Find
e Z for
{(a,
x and
y lie
all
with
at
all
0,
7^
b-x
and
7^ 0,
all
c,-
7^ 0.
assertion.
d)} to be a basisfor
b), (c, in
x Z
x Z
assertion.
Z? Prove your
Z x
for
Z
Z. [Hint:
Z x
Solve x(a, b) + y(c,
d)
=
Z.]
Concepts
In Exercises 5 and 6, correct the definition of the so that it is in a form acceptablefor publication. 5.
The rank
6.
A
for a
basis
distinct
7.
Show
of a freeabelian
by
nonzero and nt
abelian
e X
xt
the
only
in a
of elements
number
is a
G set generating \342\226\240 \342\226\240 = \302\2732 = \342\226\240 = nm if \302\2731
group
e Z
that it is
example
G is
group
possible for a proper
reference
term without
italicized
of a
that
nxx\\
free abelian
text, if
correction is needed,
for G.
set
generating
X c G such = 0.
subgroup
the
to
+
of finite
group
h nmxm
H \302\2732*2
rank r
= 0 for
also to
have
rankr.
8. Mark
each of the a.
true or
following
abelian
free
Every
group is
b. Every
finitely
c. There
exists a freeabelian
d. A in
e. If f. If
generated
some
generating
generates
X
is a
g. Every
nonzero
of every
group
abelian
positive
group is free abelian if
abelian
group
a freeabelian
free abelian
h. Every free abelian i. If K is a nonzero j. If K is a nonzero
abelian group is
a freeabelian
group.
rank.
integer
number equals
its Betti
group of
G and
subgroup
of a
subgroup
of a
c G, then
C.Y c
an
infinite
number
at least
2 has
group
group has rank
X C.Y
G and X
finitely finitely
an
G,
Y generates then
Y is
the that
proof
a free
of Theorem
abelian
group
number
of elements
38.5 (Seethe
contains
two
no nonzero
G.
a basis for
number
of bases.
generated
free abelian
then
K is
generated
free
then
G/K
sentences
G.
of bases.
infinite
group, abelian group,
Theory
9. Complete 10. Show
the
set.
a free
basis for
free.
torsion
torsion-free
generated
finitely
X
false.
preceding
elements of finite
the theorem). order.
free abelian. is free
abelian.
Show
11.
12. Show
Advanced Group Theory
VII
Part
340
free abelian groups, groups of finite rank
G' are
that
if
that
free abelian
G and
nonzero elements of finite Show
13.
and r/s could be contained
Exercises 14 through decomposition of the 14. Let p bea fixed a subgroup Tp 15.
abelian
generated
finitely
to showing groups.]
a free abelian group. set satisfying Condition
[Hint:
2 of
that no two Show Thorem 38.1.]
with showing the uniqueness of the prime powers J of a finitely abelian group. generated
19 deal
numbers
rational
distinct
the prime-power
in
appearing
n/m
subgroup
that the
Show
prime.
elementsof T
as order
having
some power of p, together
with
form
zero,
of T.
16. LetGbe any
that G[n] = abelian group and let n be any positive integer. Show = e}.) G. (In multiplicative notation, G[n] = {x G\\x\" ~ > 1 and prime p. 17. Referring to Exercise 16, show that Zpr[p\\ Zp for any r
{x e
e
of
Exercise 17, show
18. Using
groups containing no
prime-power decompositionof T, the subgroup Tp in the preceding exercise is isomorphic reduces our problem some power of the prime p. [This product of those cyclic factors of order have different into of cyclic that the group Tp cannot decompositions products essentially
direct
the
the
are precisely
in any
that
Show to
in a
torsion
G' is free abelian.
order.
addition is not
Q under
that
G x
then
G | nx
= 0}
is a
subgroup
that
X
(Zpr,
~ \342\226\240 \342\226\240 X \342\226\240 X Zpn\342\200\236)[p] Zp
Zpr2
X Zp X
\342\226\240 \342\226\240 \342\200\242 X
Zp
m factors
provided each rt 19. Let G bea finitely
1.
~
Zpm that m
show
to
ZPH
group and
abelian
generated
\342\226\240 \342\226\240 \342\226\240 x
x
ZpQ
need
>
x
Zpn
=n
rt =
and
defined in Exercise 14. SupposeTp ~ Zpr, x subgroup \342\226\240 \342\226\240 \342\226\240 \342\226\240 \342\226\240 < < < rm and I < Si < S2 < \342\226\240 < sn. We 1 r\\ ri_ < \342\226\240 \342\200\242 n of of the to the demonstration 1, \342\226\240,complete uniqueness
Tp \342\226\240 \342\226\240 \342\226\240 x where Zps\342\200\236,
st for i
=
the
prime-powerdecomposition.
a. Use two
r\\
by
part
(a) of
of the
and
20. Indicatehow 21.
the
Argue
>
wi;
eTp],
Sj, which will and show
numbers
the proof.
complete that
20 that
a finitely
divides
abelian
generated
mi+i for i = that r = s and 20
Exercises
mr
and ns
can
both
|r|, primes dividing p(,], Then prime-powerdecomposition.
\342\226\240 \342\226\240 be \342\226\240, p,h'
mr-\\ and then r = s.
that
ns-\\,
[Hint: Suppose
would then have that this is factors. Then argue
this
of nonzero
subgroup
group. Suppose
\342\226\240 r \342\200\224 1, \342\226\240 \342\226\240, 1, and
for k = through 22. mk
= nk
T~
showing
mr
be
characterized
the highest powers
as follows.
\342\226\240 = 1, \342\226\240 s \342\200\22 \342\226\240, 1, the uniqueness
decomposition. (Observe
Let p\\,
of these primes
' \342\226\240 \342\226\240 = p^ p*2 \342\226\240 \342\226\240 p, they are equal, and continue
Zm|
rij+i for n \342\226\240 1, \342\226\240 \342\226\240, r, demonstrating divides \302\273j
from a torsion-coefficient decomposition can be obtained exercises show the prime powers obtained are unique.)
from Exercise
Characterize
\\x
=
a prime-power
preceding
r \342\200\224 1, and
r,-
different
involving
wish to show This is done in
and let
22.
Show
= {pr'x
pr'Tp
1. We
coefficients.
torsion
that
n i
< j.
all i
for
st
this exercise.]
subgroup of
\342\226\240 \342\226\240 x Z\342\200\2362 x \342\226\240 x Z\342\200\236,, where Z\342\200\236,
and wi i > 1
n =m.
decompositions
prime-power
the torsion
be
that
s\\. Suppose rt = Consider the subgroup
impossible
Let T
show
=
< Sj.
Tj
18 to
Exercise
b. Show
\342\226\240 \342\226\240be \342\226\240, p,
appearing
the distinct
in the
(unique)
= ns
to show
mr-i =
ns-i for
i =
1,
\342\226\24 \342\226\ \342\22
Section39 FreeGroups
section
39
Groups
Free In
of free groups and
and
as well. In
topology
of groups is
presentations
theory that is of great interest and readable discussion and Fox [46, Chapters3 Crowell
of group
fact,
excellent
an
in
found
A be
any (not of the
and
is a syllable and the
introduce
Let A
=
necessarily finite) a, as letters in
a finite
string word 1, which
empty
w
of elements
set the
Any written
symbol of the
alphabet.
of syllables
I. We
i e
for
fl,-
in
think
form
is a
juxtaposition
of
A as an
e Z word. We also
a,\" with n
no syllables.
has
}. Then
[au
a02a3,
a\\a3~
are all
Words
Reduced
and
alphabet
39.1 Example
in
4].
Words Let
but
in algebra
only
40 we discussa portion
and Section
section
this
not
341
words,
we
if
the convention
follow
,
a^a^a^
a^a^
and
a3~
of understanding
a,-1 is
that
the same
as at. \342\226\262
two natural types of modifications first type consists of replacing
are
There
contractions. The
(ype consists of replacing
-pjjg seconci
a.m+n_
it out
is, dropping
of the
word.
every word can be changed contractions are possible.Note the usual
39.2Example
The reducedform
that
of the
these
number
one for
elementary exponents.
word^3^-1^]2^]-7
the elementary a word by of at\302\260 in a word by 1, that of elementary contractions,
certain
occurrence
occurrence
By means of a finite to a reduced word,
of integer
manipulations
an
of an
words, of
no
which
contractions
of Example
a\342\204\242 a\" in
more
formally
elementary amount
39.1 isa22a3a^~5.
to
\342\226\262
It should be said here once and for all that we are going to glossover several points that some books spendpagesproving, induction arguments broken usually by complicated down into many cases. For example,suppose we are given a word and wish to find its
reduced form. Theremay first. How do we know
be a
variety of
the reduced
that
elementary contractions that could be performed word we end up with is the same no matter in The student
probably say this is author tends to agree effort proving herewith the student. Proofs of this sort he regards as tedious, and they have never made him more comfortableabout the situation. However, the author is the first to acknowledge that he is not a great mathematician. In deferenceto the fact that many mathematicians feel that these things do need considerable discussion,we shall mark an occasion when we just state such factsby the phrase, \"It would seem obvious that,\" keeping the quotation marks. what
order
we perform
obvious. Someauthors
the
elementary
spend
contractions?
considerable
this.
will
The
Free Groups Let
F[A]
the
set of all
into a
reduced form
group
of
reduced words formed from our alphabet A be F[A]. We now make in a natural way. For vt>i and w2 in F[A], define vt>i \342\226\240 w2 to be the the word obtained by the juxtaposition w i w2 of the two words.
342
Part VII
39.3 Example
Advanced
Theory
Group
If ?
-5
3
and
VV2
then
\342\226\240 =
wi
W2
=
-9 0-3
-9
9
\"fll
\302\2533^2 >
\342\200\2243\342\200\2249
3
03^2
^2^1
of multiplication on F[A] is well denned operation \"It would seem obvious 1 an element. associative. identity empty word acts as that\" a reduced if form the word obtained the word w e F[A], we by first writing given a\" then the of w in the and second each a;~n, by replacing by syllables opposite order word w~l is a reduced word and also, resulting
39.4 Definition
obvious
seem
would
\"It
this
that\"
The
and
The group
described
just
F[A]
w
\342\226\240 = w~
is
the
free
\342\226\240 w =
w~
1. by A.
generated
group
\342\226\
at Theorem 7.6 and the definition preceding it to see that the present use is consistent with the earlier use. which we will abbreviate set {a,- \\i el] Starting with a group G and a generating by ask if G is free on {a;},that is, if G is essentially the free group {a,}, we might generated by {a,}. We define preciselywhat this is to mean. Look
of
39.5 Definition
term
the
back
generated
If G is a group a map : G of generators
39.6
Example
A = {a,-}of generators, such that (a,) = at,
a set
with
\342\200\224> F[A]
G. A
group
is free
if
it is
and G
then
free on
if G
is isomorphic to on A, and the
is free
some nonempty
set
example of a free group that has occurred before is Z, generator. Note that every free group is infinite.
The only
Theorem
are
a,-
free
A.
\342\226\
on one
is free
which
\342\226\2
Refer to the literature for proofs of the next three theorems. We will these results. They are stated to inform us of these interesting facts. simply
39.7
under
F[A]
If a group G is free on A and also on B, then the sets A and B have the of elements; that is, any two sets of free generators of a free group
not
same
have
be using
number the
same
cardinality.
39.8 Definition
If G is free on Actually,
39.9 Theorem 39.10 Theorem
Two
A
free
nontrivial
the number
A,
theorem is
the next
groups
of elements in
are isomorphic
proper
subgroup
evident
quite
if
and
only
of a free group
A is
from
the rank
39.7.
Theorem
if they
is free.
have
of the free group G.
the
same
rank.
\342\226\
39.11Example
Let
y}] be
F[{x,
free
the
on {x,
group
Let >\342\200\242}.
yk =
for k >
0. The
yk for
k >
generate.
This illustrates
subgroup
may
be much
39.12Theorem
although
of Free
results
The
be generated by A = {at in G', not necessarily any elements G' such that (a,) = : G \342\200\224> G
Let
subgroup of
for the
F[{x, y}] that
they
the rank
of the
a subgroup of a free group is free, the rank of the whole group!
\342\226\262
Groups
concerned primarily with simple and elegant.
section will be here are
in this
xkyx~k
generators
greater than
Homomorphisms
Our work free group.
0 are free
that
343
Free Groups
39
Section
homomorphisms
denned
on a
If at' for i e I are 1} and let G' be any group. then there is at most one homomorphism a,-'. If G is free on A, then there is exactly one such |
i e
distinct,
homomorphism.
Proof
Let
be
for any
G we
for somefinite
G'
that
such
of the
product
generators at, where
since is
the
we
a homomorphism,
^)=n*\302\253o=n(vrj
Thus a homomorphism set. This shows Now
in
G,
that
G is
:G
f
=
Now by Theorem
a/.
7.6,
free on A;
a-h appearing have
in the
product
need
must
j
by its values on elementsof a generating = ai \342\226\240 such that (\302\253<) one homomorphism that is, G = F[A]. For
is completely there is at most
suppose
define
\302\242(^)
have
distinct. Then
not be
from G into
a homomorphism x e
determined
->- G' by fa/)\"'-
tK*)=n y
since F [A] consists are equal.Sincethe the same as those involving
map is well defined, formal products in F[A] The
in
G'
are formally for
-ifr(x)-i(r(y)
any elements
x
and
y in
precisely
G, so
rales
i/f
of reduced
words; no two
different
for computation
exponents is indeed
involving exponents in G, it is clear that ir(xy) = a homomorphism. \342\231\246
should have proved the first part of this theorem earlier,rather than it to the theorem states that a exercises. Note that the having relegated homomorphism of a group is completely determined if we know its value on each element of a generating set. This was Exercise 46 of Section13.Inparticular, a homomorphism of a cyclic group is completely determined by its value on any single generator of the group. Perhaps
39.13 Theorem
Every
group
we
G' is
a homomorphic image of a free
group
G.
Part VII
344
Proof
Advanced GroupTheory = {a,-'| i
LetG'
as G'. Let G into
G
G' such
is
approach,
by
do
be the
\342\226\240 Clearly
Groups the
confuse
not
group
than
free group
with
is not
one
the
notion
abelian.
In
of the
generator that has a basis, group as an abelian group described in Theorem 38.1. Thereis another
abelian
properties free
of a
notion
group on more
denned a free
free
e 1} be a set with the same number of elements there exists a homomorphism i/c mapping the image of G under \342\231\2 xjr is all of G'.
39.12
Abelian
A free
generating set satisfying via free groups, to
Let F[A]
{a, |i
Theorem
at Free
group.
=
A
= ai
V(fl>)
section, we
preceding is, a
Then
that we
important
a free abelian that
that
Look
Another It
e /},andlet
= F[A].
abelian
groups. We
on the
set
generating
now
describe
shall
A. We
this approach. write F in place
of
that F is not abelian if A contains more than one element. of F. Then is an abelian commutator F/C group, and it is not subgroup a If hard basis e aC is renamed a, we to show that F/C is free abelianwith {aC \\ A}. a with basis A. This indicates how a free abelian can view as free abelian F/C group a can be constructed. free abelian set as basis Every group can be group having given in this fashion, That if G is free abelian with basis to constructed is, up isomorphism. of modulo its commutator form the free form the factor X, F[X], F[X] group group and we have a group to G. subgroup, isomorphic
F[A] for the Let C be the
moment.
Note
Theorems 39.7, 39.9, and 39.10 hold for free abelian groups as well as for free version of Theorem 39.10 was proved for the finite rank groups. In fact, the abelian case in Theorem it is true that for 38.11. In contrast to Example 39.11for free groups, a free abelian group the rank is at most the rank of a subgroup of the entire group.
Theorem 38.11also showed
this
for the
finite
case.
rank
39
EXERCISES
Computations
1. Find
the
reduced
form and
inverse
the
of the
reduced form of eachof
2.
Compute forming
3.
4.
5.
the products a basis for
given
(a) and (b) of Exercise1 in the case that {a, b, Find the inverse of theseproducts. group.
a free abelian
How many different
homomorphisms are there
a. Z4?
b. Z6? homomorphisms are there of b. Z6?
How many different
homomorphisms are there
a.
many
Z4?
of a
free group of rank
a free
of a
group of rank
free abelian
group
different
homomorphisms
are there b. Z6?
c.
S3?
c.
S3?
2 onto
of rank
2 into c.
of a free
abelian
group
c} is a set of
2 into
b. Z6?
a. Z4? 6. How
words.
following
in parts
a. Z4? How many different
the
b. a2a-3b3a4c4c2a-1
a. a2b-lb3a3c-lc4b-2
of rank
S31
2 onto c. S3?
generators
Section39 Exercises
345
Concepts
7 and
In Exercises
so that 7.
one of the
9. Take
10.
of a free
rank
The
the definition of the
your
reaction
Mark
each
reference
term without
italicized
correction is needed.
text, if
the
to
acceptablefor publication.
word is one in which also no appearances of
A reduced
letter and
8.
8, correct
in a form
it is
is the
group
in that of the
elementsin
number of
section
in this
instances
are no appearancesin with exponent 0.
there
of two
juxtaposition
syllables
the
having
same
a syllable
the phrase
which
in
a set of
for
generators
group.
seem obvious
would
\"It
the
used
was
that\"
and discuss
instance.
following
or false.
true
free group is a freegroup. of free abelian every Every proper subgroup group is a freegroup. a a free is free c. A homomorphic image of group group. d. Every free abelian group has a basis. a. Every proper
of a
subgroup
b.
e. The
abelian
free
groups of
f. No free group is free. g. No free abelian group h. No free abelian group
i. j.
the
are precisely
rank
finite
abelian
generated
finitely
groups.
is free.
Any
two
free groups
of rank > 1 is free. are isomorphic.
Any
two
free abelian
groups of the
rank are
same
isomorphic.
Theory
11. Let for
G be
a
generated
finitely
G\\f{b\\,---,bn)
abelian
G and
generates
a.
Show
that
{2, 3}
b.
Show
that
both
with
torsion,
group J2\"i=i
with
0. A
identity
= 0
mi^t
and
if
set
finite
each
only if
is not a basis for Z4. Find a basis for Z4. {1} and {2, 3} are bases for Z^. (This shows the number of elements in a basis may vary;
is,
\342\226\240\342\226\240 where ,b\342\200\236),
bt e 0, where mt e
=
wijfo,-
for a
that that
{b\\,-
abelian
generated
finitely
it need
not be an
G, is a basis
Z.
invariant
group G
of the
group
G.)
c. Isa basis d. Show In present-day
a free
abelian
as we
group
that
every
finite abelian
group
Describe algebraicproperties
a basis
used
{b\\,
38 a basisin
the
\342\226\240\342\226\240\342\226\240 ,&\342\200\236},wherethe order
technique
entities with these
Prove that
3.
any
two
algebraic
this three exercisesillustrate give away their identities, first two exercisesasks us to give
The next we
do not
this algebraic
that
properties characterizethe entity. Show that at least one such entity
of these
has
expositions of algebra, a frequently is the following: algebraic entity
2.
So that
defined it in Section
in which it is used in
sense
(particularly
by
divides
offo,-
the
disciples
the order
of bi+\\.
of N. Bourbaki) for
a new
introducing
1.
for
exercise?
this
entity is
to
possess.
properties are isomorphic,
is, that
that
these
exists.
technique
we use the
usual
for three fictitious
algebraicentities,
names
name for
the
for them in entity.
each
the first
of which two
we have met before. The last part
exercises.
Advanced Group
VII
Part
Theory
G be any group. An abelian group G* is a blip group of G if there exists a fixed homomorphism of G G' can be factored as \\j/ = #, where 6 onto G* such that each homomorphism \\j/ of G into an abelian group is a homomorphism of G* into G' (see Fig. 39.14). Let
a.
that any two blip groups Show of G are isomorphic. [Hint: Let Gi* and G2* be two of G. blip groups : G -> Gi* and : G -> G2* can be factored via the other Then each of the fixed homomorphisms \302\2422 \302\2421 = 62
39.14 Figure Let S be each (see
any
set.
G
A group
group G' and map Fig. 39.15).
39.15 with
together
/ : S -> G' there
Figure
function g : S -> G constitutes unique homomorphism / of
a fixed exists
a
a blop G into
G'
group on S if for that / = /g
such
with a. Let Sbea fixed set. Show that if both G], together G2, g2 : S \342\200\224> gi : S -> G\\, and G2, together with that g\\ and g2 are one-to-one and Show are blop groups on S, then G\\ and G2 are isomorphic. [Hint: maps Then proceed in a way analogous to that given by the that g\\S and g2Sgenerate Gi and G2, respectively. hint
Exercise
for
12.]
b. Let 5 be a set.Show c. What
concept
Characterize
section
that
that we
a free abelian group
40
group on S exists.You may use any theorems of the before correspondsto this idea of a blop group
a blop
have introduced by
properties
in a
similar
fashion
to that
used
in
Exercise
text. on
5? 13.
Group Presentations
Definition Following
identity
satisfy.
to these
40.1 Example
We
literature on
group. The idea of
generatorsfor to
in this section we let 1 be the is to form a group by giving a set of presentation or and certain relations that we want the generators group equations the group to be as free as it possibly can be on the generators, subject
of the
most
of a
the
want
group
presentations,
a group
relations.
G has generators x and y and is free except for the relation Suppose xy = yx, which we may express as xyx~ y -1 _ 1. Note that the condition xy = yx is exactly what is needed to make G abelian,even though xyx ^-1 is just one of the many possible of F[{x, y}].Thus G is free abelian on two generators and is isomorphic to commutators of F[{x, y}] F[{x, subgroup y}] modulo its commutator subgroup. This commutator normal is the smallest subgroup containing xyx~xy~x, since any normal subgroup
commutator
l
ly
xyx
containing
gives rise to
subgroup
a factor group
that
and
is abelian
thus
347
contains
the
15.20.
Theorem
by
Group Presentations
40
Section
\342\226\262
Let F[A] be a free group like F[A] as it can be, subject to in a form in which certain equations that we want satisfied. Any equation can be written = side is 1. Thus to be 1 for i e I, where the we can consider the rt right-hand equations If we require that n = 1, then we will have to have r\\ e F[A]. The preceding and suppose that
illustrates the general a new group as
example
we
e F[A]
and
equal 1. Thus any
finite
x
e Z.
n
Also any of the
much
= 1
x(rin)x-1
for any
situation.
to form
want
of elements
product
equal to 1 will
again
have
to
form
product
j where the r,- need checked that the set
not be distinct, of all these finite
group. It is readily F[A]. Thus any as much as possiblelike F[A],subject to the requirements rt = 1, alsohas looking group = looks like F[A] (remember that we multiply cosets r 1 for every r e R. But F[A]/R 1. by choosing representatives), exceptthat R has been collapsed to form the identity We can view this group we are after is (at least isomorphicto) F[A]/R. Hence the group which we will abbreviate as described by the generating set A and the set {r, \\i el], have
will
products
to equal 1
is a
normal
in
the
new
R of
subgroup
{/\342\226\240>\342\226\240}.
ft
Note
Historical
idea of The appears
presentation already Cayley's 1859 paper, \"On
a group
in Arthur
of Groups as Dependingon the Symbolic Equation 9\" = 1. Third Part.\" In this article, Cayley of the five groups of enumeration gives a complete order 8, both by listing all the elements of each Theory
by giving for each a third example is example, his and
octic group; Cayleynotes a4 =
two elements
the
generated by
1,p2 = 1,af$
generally
that
a,
the
ft with
if and
1878,
and noted
what
is
that
this
a,
= f$a3.
/3
the
relations
He also shows
more
is generated =
1, afi
by
= fias
(mod m) (see Exercise13). to the theory of groups returned Cayley a central problem in that theory is the
that
40.2 Definition
a set and
Let
A be
the
An isomorphism r,\342\200\242.
let
{r,-}
c F (j)
of
similar to those discussed in Sections 36, all simple and Holder determined 40, 37, groups of order up to 200 and characterized all the groups of orders p3, pq1, pqr, and p4, where p, q, r are distinct numbers. he Furthermore, prime developed techniques
the techniques for determining of a group if one is G, given the
group is
with
of order mn relations am = 1, ft\" a group
the
called
here
if s\" = 1
only In
For
presentation.
of all groups of a given n. In the order Holder Otto several published early 1890s, papers attempting to solve Cayley'sproblem.Using determination
the
normal
H and
subgroup
G/H.
Interestingly,
group
was still
typically
began
fairly
his papers
structure
the structure since new with
of
of an abstract time, Holder definition of a group
the notion at this the
also emphasized that isomorphic the same object. one and essentially
and
[A]. Let R F[A]/R
be the
onto a
structures of a the factor group
possible
groups are
least normal subgroup of F [A] containing G is a of G. The sets presentation group
348
Part
VII
Advanced Group Theory set A is the set of generatorsfor the {r,} give a group presentation.The r e R is a consequenceof {r,-}. An equation and each r, is a relator. Each is one in which both A and {r,} are finite relation. A finite presentation
and
A
presentation = 1 is a r\\ sets.
\342\226\
40.1, {x, y] may seem complicated, but it really is not. In Example is our set of generators and xyx~ly~l is the only relator. The equation xyx~ly~l = 1, = yx, is a relation.This was an example of a finite or xy presentation. has If a group presentation generators xj and relators r,-, we shall use the notations definition
This
or
(Xj : n)
to denote presentation
the
{xj
group
We may refer
presentation.
group
{a}
ab
group
with
{a6},
presentation
defined
by one
generator a,
consider the group = ba, that is, the group
Now and
as the
F[{xj}]/R
{r,-} =
and
= 1).
(a : a6
This group
1)
with
presentation
A= the
=
Presentations
Considerthe
that is,
to
:7-,-
: r,-).
Isomorphic
40.3 Example
(xj
defined
with
a6 = 1, is isomorphicto \"L6. relation generators a and b, with a2 = 1, b3 = 1,
the
with
by two
presentation
(a, b : a',
b , aba~lb~ ).
a2 = 1 gives a-1 = a. Also b3 = 1 gives b~x =b2. Thus every element a of of a and b. The relation can be written as nonnegative product powers group = 1, that is, ab = ba, allows aba~lb~l us to write first all the factors a and involving then the factors b. Hence element of the to is some ambn. involving every group equal But then a2 = 1 and b3 = 1 show that there are just six distinct elements, The in
condition
this
1, b, b~, Therefore
Fundamental The
this
presentation
also gives
Theorem 11.12, it
preceding When isomorphicgroups.
must
a
again
a, ab, group
ab'. of
order
be cyclic
example illustrates that this happens, we have
different
and
6 that is isomorphic
presentations
abelian,
and
by
to Z6.
the
\342\226\2
may give
isomorphic presentations.To
determine
two presentations are isomorphicmay be very hard. It has been shown (see Rabin [22]) that a number of such problems connected with this theory are not generally and well-defined in all solvable; that is, there is no routine way of discovering a solution cases. These unsolvable problemsinclude the problem of deciding whether two presentations are isomorphic, whether a group free, abelian, given by a presentation is finite, word problem of determining whether a given word w is a or trivial, and the famous of a given set of relations {r,}. consequence whether
Section 40
that
40.4
Example
The importance of this material every group has a presentation.
Let us
show
is indicated
GroupPresentations
by our
349
Theorem 39.13,which
guarantees
that
(x, y
:
y2x
=
y, yx2y
= x)
is a presentation
of the trivial group of one element. We need only show that x and y = 1 and = 1 can be are consequences of the relators or that x and yx2yx~x, y2xy~x y both techniques. deduced from y2x = y and yx2y = x. We illustrate As a consequenceof y2xy~x, we get yx upon conjugation by y~x. From yx we deduce x_1>'_1,and then (x~xy~x)(yx2yx~x) gives xyx~x. Conjugating by x_1. xyx-1 we get y. From y we get y~l, and y~x{yx) is x. = y we deduce yx = 1 upon with relations instead of relators, from Working y2x = 1 into yx2y = x, that is, by y~x on the left. Then substituting yx multiplication = x, we get xy = x. Then by x~x on the left, we have y = 1. (yx)(xy) multiplying Substituting this in yx = 1, we get x = 1. but it somehow seems more natural to amount to the same work, Both techniques \342\226\262 most of us to work with relations.
Applications
We conclude
40.5 Example
this chapter with
two
applications.
Let us determine all groups of order 10 up to isomorphism. We know from the 11.12 that every abelian group of order 10 is isomorphic to Zjo- Suppose Fundamental Theorem that G is nonabelian of order 10. By Sylow G contains a normal subgroup H theory, H must be cyclic. Let a be a generator of order 5, and of H. Then G/H is of order 2 H. and thus isomorphic to Z2. If b e G and b \302\243 must then b2 Since have e H,we every element of H except 1 has order 5, if fo2 were not equal to 1, then b2 would have order our 5, sob would have order 10. This would mean that G would be cyclic,contradicting = that G is not b2 since H is a normal of abelian. Thus 1. assumption Finally, subgroup = so in bab~l Since & is an e H. G,bHb~x H, particular, conjugation by automorphism of H. bab~lmust be another element of H of order 5, hence bab~l equals a, a2, a3, or = a would G would be abelian, sincea and b a4. But bab~l ba = ab, and then give Thus the G are: G. for of generate possibilities presentations
1. {a,b:a5 = l.b2 = l,ba = a2b), 2. (a,b :a5= l.b2= \\,ba = a3b), 3. (a,b:a5 = l,b2 = l,ba = a4b). Note that all three of thesepresentations can give groups of order at the last relation ba = a'b enables us to express every product of a's and form asb'. Then a5 = 1 and b2 = 1 show that the set
S=
a2b\302\260. a3b\302\260, a4b\302\260, a\302\260b\\alb\\ {a\302\260b\302\260,alb0,
includes all elements of G.
a2bl,
most b's
a3b\\a4b1}
10, since in G in the
350
Part VII
Advanced
Theory
Group
It is
not
clear
yet
cases a group
of order
that all these elements in S are distinct, so 10. For example, the group presentation
: a5 =
(a, b
gives a group
a =
b'a = (bb)a =
= (a'b)(ab) Thus a2
=
this
in
a =
group,
1, together
=
=
b(ba)
=
1, which, a = that
1, means
we
have
in all three
= a~b)
\\,ba
have
we
= (ba)(ab)
b(a-b)
= a'{ba)b= a'{a'b)b = a4b~
a4, so a3
a3
with
=
b~
associative law,
using the
in which,
1,
that
a5
= a4 =
together
with
1. Hence
every elementin
1, yields the
a2 = group
1.But with
presentation
=
(a,b :a5 is equal to
1 or
either
b; that is,
this
=
\\,b2
= a2b)
l,ba
is isomorphic
group
(bb)a =
b(ba)
= 1, b~
=
to Z2. A
similar
study
of
for
(a,
shows that This
a =
leaves
a4 again, just
b : a5
so this
also
{a, b : a5 =
as a candidate
a group
yields
1,
b~
=
= a b)
1, ba
isomorphic to
1, ba
TLo_.
= a4b)
order 10.In
this that case, it can be shown does a nonabelian G of give presentation group order 10. How can we show that all elements in S represent distinct elements of G? The way is to observe that we know that there is at least one nonabelian easy group of order 10,the dihedral group D5. Since G is the only remaining candidate, we must have G \342\200\224 attack is as follows.Let us try to make S into a group by defining D5. Another to be of s + w(4f) when divided axby, where x is the remainder (asb')(allbv) by 5, and the of is remainder t when divided in the sense of the v division + 2, y by algorithm ba = a4b as a guide in defining the (Theorem 6.3). In other words, we use the relation two of elements of We S. see that a\302\260b\302\260 acts as and that product (asb')(aubv) identity, can determine t and s successively by letting a\"bv,we given for
of S
all elements
group of
a nonabelian
are distinct,
so
this
t
and
=
\342\200\224
v (mod
2)
then
s =
-w(4f)(mod
5),
is a left inverse for aubv. We will then have a group structure on S if associative law 13 to holds. Exercise asks us out the only carry straight-forward for the associative law and to discover a condition for S to be a group under computation such a definition of multiplication. The criterion of the exercise in this case amounts to asbr,
giving
and
the
which
if the
valid
congruence 42 = 1
(mod 5).
Section40 we do get
Thus
a group
10. Note
of order
GroupPresentations
that
#
1 (mod
5)
32 #
1 (mod
5),
22
351
and
so Exercise13also
that
shows
{a, b
: a5 =
(a,b
:a5=l,b2
1,
1, ba
=
b~
= a'b)
and
40.6
Example
do not
give
Let us
determine all groups of order
= a3b)
10.
of order
groups
= l,ba
\342\226\262
8 up
We know
to isomorphism.
three
the
abelian
ones:
Z8,
Z2 x Z2
x Z4,
Z2
x Z2.
and relations, we shall give presentationsof the nonabelian groups. Usinggenerators Let G be nonabelian of order 8. Since G is nonabelian, it has no elements of order 8, so each element but the identity is of order either 2 or 4. If every elementwere of order = 1. Then since a2 = 1 have (ab)2 = 1, that 2, then for a,b e G, we would is, abab
and
b2
=
1 also,
have
we would
ba = a'bab' = a{ab)'b contrary to order 4.
our
Let {a}bea subgroup of G. Hencea and
b are
G is
that
assumption
not
for G and
generators
ab,
G must have
Thus
abelian.
4. If b
of order
of G
=
the cosets (a) and 1. Since {a) is normal
\302\243 (a),
=
a4
an
b(a) in G
element
of
exhaust
all
(by Sylow
of index 2), G/(a) is isomorphic to Z2 and we have b2 e (a). If b2 = a or b2 = a3, then b would be of order 8. Henceb2 = 1 or b2 = a2. Finally, since is we have bab~x e and since is a to normal, {a) conjugate b(a)b~x subgroup (a) (a), and hence isomorphicto (a), we see that bab~x must be an element of order 4. Thus bab~x = a or bab~x = a3. If bab~x were equal to a, then ba would equal ab, which would make G abelian. Hence bab~x = a3,soba= a3b. Thus we have two possibilities for G, namely, or because
theory,
it is
G,
: (a,
b : a4
= 1, b2
: a4
= 1, b~
b~x
is b in
a3b)
=\\,ba=
and
Go : {a,b Note
that
a~x
= a3, and
the relation
ba = a3b, enable
in Examples elements in
40.3 and each group
40.5.
that
us
Since
to express
a4 = 1
= a', ba
G\\
and
= a b). with G2. These facts, along in as G,- in the form amb\", = 1 or b2 = a2, the possible
b3 in
every element and
either
b2
ab,
a'b,
are
1,
a,
a',
a ,
b,
a b.
Part VII
352
Advanced
Theory
Group
Thus
G\\
from
Exercise
order
8 also.
and
8. That G\\ similar to that used argument order at most
have
each
Gi_
13. An
is a
group of
order 8 can be seen
13 shows
in Exercise
that
has
Go_
Since ba = a3b ^ ab, we see that both Gx and G2 are nonabelian.That the two follows from the fact that a computation shows that G\\ has groups are not isomorphic the other in two of order a and elements a3. On hand, 4, Gi_ all elements only namely, of the tables for these groups but 1 and a2 are of order 4. We leave the computations to Exercise 3. To illustrate we wish to compute (a2b)(a3b).Using ba = a3b suppose we get
repeatedly,
(a2b)(a3b) =
a2{ba)a2b=
=
a5{ba)ab
= aub2.
a\\ba)b
Then for Gj, we have anb2 = if we
but
are
in
an=a\\
we get
G2,
anb2 = an = a_ The
group
of symmetries the
G\\ is
{1,
group
multiplicative
discussed in
octic
the
and is
group
square. The
of the
Gi_ is
group
isomorphic to our the quaternion
old
the group
friend,
group;
of quaternions. \342\200\2241, i, \342\200\224i, k, \342\200\224k] j, \342\200\224j,
it
D4 to
is isomorphic
were
Quaternions
24.
Section
\342\2
40
\342\226\240 EXERCISES
Computations
1. Give a presentation
of Z4
2.
of S3 involving
a presentation
Give
3. Give
for both
tables
the
involving one three
quaternion
generators.
b : a4
In
both
cases,
= 1, b2
=
1, ba
write the
elements in
a4 =
order
the
1, b2
1, a,
=
a2, ba
= a3b).
a2, a3,b, ab, a2b,
a3b.
We know that these presentations product. give groups the rest are forced so that each row and each column products
every
enough
\342\200\224 a3b)
group
{a, b :
compute
generators.
the octic group (a,
and the
involving three
two generators;
involving
generator;
(Note
that we do
of order 8, and of the
not
to
have
once
we have
table has
each element
computed exactly
once.)
4. Determine all
groups
Exercise13,part 5. Determine Exercise
of
order
14 up
to
isomorphism.
all groups of order 21 up to isomorphism. 13, part (b). It may seem that there are two
are isomorphic]
[Hint:
Follow the
outline
of
outline
of Example
Example
40.5 and
use
40.5
use
(b).]
[Hint: Follow presentations
the
giving
nonabelian
groups.
Show
and
that
they
Exercises
Section 40
353
Concepts
In Exercises
so that 6.
it is
6 and 7, correctthe definition of the italicized term in a form acceptablefor publication.
of the
A consequence
7. Two
group
of the first
presentation
a. b. c. d.
first
the
true or
to
is needed,
correction
if
powers.
one-to-one correspondenceof the that yields, by renaming generators, a with those of the second.
presentation
group
has a
group
has many different
Every
group
has two
is a
generators
one-to-one
Every
presentation.
group
has a
e. Every
group
with
f.
cyclic group
Every
raised
text,
false.
Every
Every
of relators
product
if and only if there of the second generators
the
with
following
finite
any
isomorphic
relators of
of the
each of the
8. Mark
are
presentations
correspondence
relators is
set of
to the
reference
without
presentations. presentations that are not isomorphic.
finite
presentation.
a finite has
is of finite
presentation
order.
with just one
a presentation
generator.
of the relator. is a consequence g. Every conjugate of a relator h. Two presentations with the same number of generators are always isomorphic. the set of consequences of the relators i. In a presentation of an abelian group, contains of the free group on the generators. subgroup j. Every presentation of a free group has 1 as the only relator.
the commutator
Theory
9. Use 10.
Show, using
section
of this
methods
the
15. (See alsoExample
and
Exercise
11.
of order
a group that
Show
b : a3
6. Show
that
= 1, b2
that
there
are no
nonabelian
groups
of order
=
1, ba
= a2b)
=
1, ba
= a'b)
nonabelian.
it is
the presentation
(a,b : a3 =
of Exercise
10
isomorphicto 12.
show
Exercise 13,that (a,
gives
(b), to
13, part
37.10).
We showed
in
(up to
gives
isomorphism)
the
1,
fc\"
group of order
nonabelian
only
6,
and
hence
gives
a group
S3.
A4 has no subgroup be isomorphicto either
15.6 that
Example
subgroup of A4 would orders of elements.
have to
of order
Zg or
S3. Show
6. The preceding exercise shows that this is impossibleby again
that
such a
considering
13. Let
S = {a'V\\0
is, S
integer,
consists of
and define
all
formal
products
a!b-i starting
and ending a\302\260b\302\260
with
with am~]bn~].
Let r bea positive
multiplication on S by = axV,
(asb')(aubv)
where x is the remainder in the sense of the division
of s
+ u(r')
algorithm
when
divided
by m,
(Theorem 6.3).
and
y
is the
remainder
of
t
+ v
when divided
by
n,
Part VII
354
Advanced
a. Show
that a necessary this multiplication is
b. Deducefrom
part
Theory
Group
and sufficient condition r\" =
that
(a) that
1 (mod
the group
for
a group
14. Show
that
if n
group (up to
r2,
pq,
isomorphism)
modulo
multiplication
1, r,
of order
\342\200\224with
mn
if and
=
r\"
a(H).
p and q
primes and n. Recallthat
The solutions
the presentations
{a, b : aq
b'
: aq =
\342\200\224
1, bp
1 (mod
m). (See
the
Historical
to be a group
under
for
Note
on page
xxx.)
one nonabelian p), then there is exactly elements of Z? form a cyclic group Z* under xp = 1 (mod q) form a cyclic subgroup of Z* with elements = d, so = = = b : bp ba we bab~l aq have 1, arb), {a, 1, \342\200\224 \342\200\242 \342\200\242 is isomorphicto \342\200\242, 1, this presentation j = 1, p
q > p 1 nonzero the q \342\200\224 of
In the group with presentation since b} generates (b) Thus, (a,
for S
= arb)
and q = 1 (mod
\342\226\240 \342\226\240 rp~l. \342\226\240,
Wab~i =
so all
only if
= l,ba
= l,bn
of order
q. [Hint:
and
presentation
(a,b:am gives
law to hold
associative
the
m).
1, (b]')p =
= 1, ba
=
a{rJ)b)
1,(bj)a = a{r'\\bj)), are
isomorphic]
part
TVfcY-fcrfclrfccn/i
in in
rirnnnc
Groups
Topology1
VIM
41
Simplicial
Section
42
Section
43
Computations of Homology More Homology Computations
Section 44
section
41
Complexes and
Section
Groups
and Applications
Algebra
Homological
Complexes
Simplicial
Groups
Homology
and Homology Groups
Motivation two points are of an idea of when function. Two such sets, or topological one onto the the same if there is a one-to-one function are structurally mapping spaces, this means and its inverse are continuous. Naively, other such that both this function torn or and otherwise deformed, without that one space canbe stretched, twisted, being as the same structure cut, to look just like the other. Thus a big sphere is topologically of a square, a small sphere, the boundary of a circlethe same structure as the boundary and so on. Two spaces that are structurally the same in this sense are homeomorphic. is to topology as the student recognizes that the concept of homeomorphism Hopefully the {where sets have the same algebraic structure) is to algebra. of isomorphism concept and sufficient conditions, The main problem of topology is to find useful, necessary conditions other than just the definition, for two spaces to be homeomorphic. Sufficient a dozen, but some are hard to come by in general. Necessary conditions are a dime has associated with it various kinds of and useful. A \"nice\" are very important space concerns
Topology
for which we have
sets
close together to be
able to
homology groups, namely topy groups. If two spaces
are isomorphic
to the
groups
continuousmapping
1
Part VIII
cohomology groups, are homeomorphic,
groups, homotopy it can be shown
groups,
and cohomo-
of one groups associated with the other. Thus a necessary is that their groups be isomorphic.Some of homeomorphic
be
may reflect very interesting of one space into another
is not required
enough
a continuous
that
the groups
corresponding
condition for spacesto these
define
for the
remainder of the
properties
gives
spaces. Moreover, a homomorphisms from the
of the
rise to
groups
text.
355
356
Part VIII
Groups in of one
Topology
into
the
groups
properties of the
of the
other. These group
may
homomorphisms
reflect interesting
mapping.
student could make
neither head nor tail out of the preceding paragraphs, above paragraphs were just intended as motivation for what of this section to describe some groups, homology groups, purpose some that are associated with certain subset of the spaces, in our work, usually simple R3. familiar Euclidean 3-space
If the
he need not worry. follows. It is the
The
Notions
Preliminary
in Euclidean 3-space R3 for n = we introduce the idea of an oriented n-simplex is adirected 0,1,2, and 3. An oriented 0-simplex is just apoint P. An oriented 1-simplex the points Pi and P2 and viewed as traveled in the direction line segment Pi P2 joining = -P2PiWe will agree, however, that An from Pi to P2. Thus PiP2 + P2P\\PiP2 as in Fig. 41.1, together with a oriented 2-simplexis a triangular P1P2P3, region around the triangle, in Fig. 41.1 prescribed order of movement e.g., indicated by the arrow astheorder Pi P2P3.The order Pi P2P3 is clearly the same order as P2P3P1 and P3PiP2, but the opposite order from P3. We will agree that Pi P3 P2, P3 P2Pi and P2Pi
First
=
P1P2P3
Note
that Pt Pj
Pk
P2P3P1 =
is equal
= -P1P3P2= -P3P2P1 ~P2PiP3-
P3P1P2 =
to Pi
P2P3 if
) I)
0 is
an
and is equal permutation, be said for an oriented 1-simplex
even
could
\302\273-simplex
is an \302\273-dimensional
The definition
of an
oriented
is given by an ordered sequence Fig. 41.2. We agreethat PiP2P3Pi
to
if the permutation is odd. The same = Note also that for n an oriented 2, PiP2. 0,1, \342\200\224
PiP2P3
object.
should now
3-simplex
Pi P2P3Pa of four =
be clear:An of a
vertices depending
\302\261PtPjPrPs,
oriented
solid
3-simplex as in
tetrahedron,
on whether
the
permutation
/1 \\i
2
3
4\\
J
r
s)
is even or odd. Similar definitions hold for n > 3, but that we can visualize. These simplexesare oriented, with the order of the vertices that we are concerned where the vertices are located. All our simplexes will adjective from now on.
41.1 Figure
we
shall
stop here
or have an as well be
as
oriented,
dimensions
with
orientation, with
the
and we
actual shall
meaning points drop
the
Section 41
and SimplicialComplexes
Groups
Homology
357
41.2 Figure of an n-simplex for n = 0, 1, 2, 3. The going to define the boundary the of a 0-simplex P to be the empty We define boundary boundary The notation is this time \"0.\" which we denote by simplex, are now
We
is intuitive.
term
= 0.\"
'%(P)
The boundary of
a 1-simplex
P\\ P2
is defined by = P2-Pi,
di{PiP2)
that is,
the
difference of the end 2-simplex is defined by
formal
of a
boundary
a2(Pi
and
point
= P2P3-
p2p3)
the beginning
Pi Ps +
point. Likewise,
the
P1P2,
that we obtain by it is the formal sum of terms the sign to be + the 2-simplex P1P2P3 and taking is omitted, and + if the third is omitted. if the first term is omitted, - if the second what we naturally to going around to Fig. 41.1, we see that this corresponds Referring arrow. Note also in the direction indicated by the orientation would call the boundary that the equation 3i (Pi P2) = Pi \342\200\224 Pi can be remembered in the same way. Thus we are of a 3-simplex: led to the following definition of the boundary which
we
can remember
dropping each P, in
by saying
succession
hiPiPiPzPA) = Similar
definitions
the boundary
hold
of a simplex
that
from
P2P3P4
for the definition is a face of
~
+ P1P2P4
P1P3P4 of
for n 3\342\200\236
> 3.
the simplex.Thus,
~ PiPiPz-
Each individual of summand P4 is a face of Pi P2 P3 P4,
P2 P3
is aface of PiP2P3P4. aface. However, P1P4P3= -P1P3P4 into simplexes, Suppose that you have a subset of R3 that is divided up \"nicely\" S of the tetrahedron in Fig. 41.2, which is split as, for example, the surface up into four we have some Thus on the surface of the tetrahedron, 2-simplexes nicely fitted together. of the tetrahedron; some 1-simplexes,or the edges of the 0-simplexes, or the vertices, and some 2-simplexes, or the triangles of the tetrahedron. In general, for a tetrahedron; spaceto be divided up \"nicely\" into simplexes, we require that the following be true: but
P1P3P4
is not
1. Each point 2. Each point
of the
space belongs to
at
of the
space belongs to
only
least
one simplex.
a finite
number
of simplexes.
358
Part VIII
in
Groups
Topology
41.3 Figure
3. Two different in common (up to orientation) simplexes either have no points or one is (exceptpossiblyfor orientation) a face of the other or a face of a face set of points in common is (exceptpossiblyfor of the other, etc., or the a a face, or face of a face, etc., of each simplex. orientation) Condition
into
describe some groups associated with definition with the case of the surface
now
us
like those shown in Fig. 41.3. A space divided up to these requirements is a simplicial complex.
configurations
according
Cycles, and Boundaries
Chains, Let
excludes
(3)
simplexes
each
illustrate
a simplicial
complex
our tetrahedron free abelian group S of
X. We in Fig.
shall
41.2. by the
The group C\342\200\236 \302\253-chains of A\" is the (X) of (oriented) generated (oriented) \302\273-simplexes of X. Thus every element of Cn(X)is a finite sum of the form addition of J2t yn-iOi, where the ct,- are \302\273-simplexes of X and m,- e Z. We accomplish chains by taking the algebraic sum of the coefficients of each occurrence in the chains a fixed \302\251f
41.4 Example
For the
for
simplex.
m.i e
our
S of
surface
Z. As
every
mlP2P3P4
+ m2P1P3P4
illustration
of addition,
an
-
(3P2P3P4
tetrahedron,
5PiP2P3) +
element
of C\\ (S)
element
mlPlP2 and
an element
is of the
+ m2PlP3
of C0(S) is
note
(6P2P3P4-
which
that
4P1P3P4)
- 4P1P3P4 ~
9P2P3P4
5PiP2P3.
+ m3P^P4+ m4P2P3
of the
+
m5P2P4
+ m6P3P4,
form
+ m3P3+m4P4.
\342\226\2
= 1, 2, 3.Let us define = is anw-simplex, d\342\200\236(a) e Cn-\\{X)forn C-i(X) of one element, and then we will also have 3o(cr) e C_i(-X).Since group is free abelian, and since we can specify a homomorphism of such a group by its values on generators, we seethat 3\342\200\236 gives aunique boundary homomorphism, we denote again by \"3\342\200\236,\" Cn(X) into Cn-i(X) for n = 0,1, 2, 3. mapping
Now
{0}, the Cn(X)
form
form
m\\Pi+m2P2
giving
of the
+ m3PlP2P4+ m4P1P2P3
=
An
of C2(S) is
if a
trivial
Section 41
41.5 Example
We
SimplicialComplexes
=
^OTiff;
]TV,a\342\200\236(a,).
J
example,
- 4PiP3 +
di&PiPi
=
5P2P4)
-
-
= Pi is reminded again
student
The
are of
great
that
PO
4P3 +
with 0. The elements of the kernel boundary that is, the group of \302\253-cycles, is kernel of 3\342\200\236,
Example
If z
= Pi P2 +
+ P3P1,
Ps
P2
3i(z) = Thus z is
(Pi ~ Pi)
+
\342\226\262
The
two
things of those notation
usual
\"Z\342\200\236(X).\"
+
P2)
(Pi ~
Pi) +
= PiP2 +
2(P3 -
Pi) +
~
(Ps
let c
if we
-
(P2
Pi)
then
a 1-cycle.However, Bi(c) =
-
5(P4
5P4.
are n-cycles.
\302\273-chains
for the
53i(P2P4)
- Pi) +
4(P3
-
2P2
+
43^/\302\276)
-
time you have a homomorphism, consists of 3\342\200\236 image. The kernel
any
and the
kernel
the
interest,
-
33i(PiP2)
= 3(P2
41.6
359
Groups
Homology
have
3\302\253 I
For
and
2P2P3
+
- P3) =
(Pi
=
Ps)
P3 Pi,
0.
then
-P2 + P3
^
0.
Thusc^Zi(X). z =
that
Note
cycle,
\342\226\262
The
Example
under
image
\342\200\224 (\302\273 l)-chains
41.7
P1P1 +
that
+
P1P3
Pi, P2, of (n
the group 3\342\200\236,
to Example 41.6, we seethat
Referring
in
1-chain
then P3
Ci(X),
-
or
circuit,
consists
1)-boundaries, This
group
is denoted
exactly by
of those
\"P\342\200\236_i(X).\"
if
is a
P2
one
P3.
\342\200\224
+ 2P2P3
PiP2
is a
and
of \302\273-chains.
are boundaries
41.6 correspondsto
of Example
P3P1
with vertices
a triangle
around
+ P3Pi Note
0-boundary.
that
P3
-
P2
bounds
PiP3\342\226\262
Let if a
group is
\"0\" rather
41.8 Example
us now the
trivial
group
than \"{0}.\"We shall
Let us compute tetrahedron
Z\342\200\236 (X) and
compute
for of Fig.
First, for
the
2-simplex,we have
n =
a more complicatedexample.In topology, of the 0, one usually denotes it just identity
Bn (X) for
consisting
0, 1, 2 the
by
this convention.
follow
groups
Zn(S)
and Bn(S) for
the
surface
S of
the
41.2. easier C3(5)
cases, since = 0, so
the
B2(S) =
highest
dimensional
= 0. d3[C3(S)]
simplex
for the
surface is a
in
Groups
Topology
Also, since
C_i(5) = 0 by
our
we see
definition,
= C0(S).
Z0(S)
Thus Zq(S) is free abelian the image of a group under
of the and
Pi, P2, P3, and is generated by
generators,
a homomorphism
group. Thus, since C\\(S) is generatedby we see that B0(S) is generated by
original P3P4,
Pi
~
Pi, P3
B0(S) is not ~ (Pi Pi)- It is
However, ~
Pi)
four
on
that
~ Pi,
P1P3,
P1P2,
~ Pi, Pa
~ Pi, Pa
~
Ps-
= (P3 \342\200 P2 generators. For example,P3 \342\200\224 \342\200\224 \342\200\224 and Pi, P3 Pi, B0(S) is free abelian on P2
that
t\302\260 see
easY
P3
Pu
images
on these
abelian
free
~
Pa
easily seen that of generators P1P4, P1P3, PiPa,
It is
P4. the
P4-.P1-
Now let us go after
of integral
multiples
that
is the
end
point
the tougher group Z\\ (S). An element of edges P, Pj. It is clear that 3j (c) =
beginning point of a total of exactly r edges. Thus
all
claim
+ p4p3
Z3
=
PiPi+P2Pa
=
P1P3
+ p3pu + PaPi,
+ P3P1
+ P1P1
end point. These edgesare irij. Then
as an
Pi
z be
in
Z
z + OT2Z4
\342\200\224
OT4Z2
P3, and
of edges
in
say Pt; Pt
P4.
f^AZl
the
serve
the same
cycle, counting
number of times
r. Thus Zi(5) the individual
m2Z4
as a
and an end point
beginning
we see that
multiplicity,
z+
for some integer Since the zt are
-
vertex,
P] P3, and
not
does
P4 must
in the
miZA
+
P\\ P2,
2-simplexes. We
contain the edges PiP2 or P1P4.Thus the only edge \342\200\224 is rriAZi cycle z + MzZa possibly P1P3, but this edge a nonzero a nonzero multiple coefficient as it would contribute of the fact that a cycle has boundary 0. Thus contradicting boundary, consists of the edges of the 2-simplex Since in a 1-cycle each P2P3P4.
a cycle, but again having Pi as a vertex could not appear with the vertex Pi to the is
P2,
vertex edges of c is also the
of the individual These are exactly the boundaries 1-cycles. and choose a particular that the z,- generate Zi(5). Let z e Zi(S),
'let us work on edges having Let the coefficient of Pt Pj
of
P3Pa
P]p4
ZA are
+
=
formal sum
if each
only
+ PaPi,
P2P3
Z2
is a
(S)
and
of r
multiplicity)
(counting
z, =
c of C\\
0 if
m4Zi =
-
is generated by boundaries of the
the
rzi zt, actually
2-simplexes,
by
any
as we
three
of the
observed, we
z,.
see
that
Zi(S)
The
student
should
Finally,
we
= Pi(S).
what this computation means in terms of Fig. 41.2. Now C2(S) is generatedby the simplexes P2P3P4. has coefficient P2P3P4 has coefficient n and P3P1P4
see geometrically describe
Z2(S).
If P3P1P4, P1P2P4, and P2P1P3. r2 in a 2-cycle, then the common
edge P3
P4
has
coefficient
r\\
\342\200\224 in
r2
its boundary.
Section 41
Thus we
r2, and
r\\ =
have
must
SimplicialComplexesand
2-simplexes appears with
Z2(S) is
that is,
41.2. Note that the triangle clockwise, when
that
around
going
the student
Again,
cyclic.
of Fig.
in terms
geometrically to
infinite
+ P1P2P4 +
four
of the
argument,
coefficient.
+ P3P1P4
P2P3P4
in a cycle each one Thus Zi(S) is generatedby
a similar
by
same
the
361
Groups
Homology
Ps,
PiP\\
this computation of each summand corresponds from the outside of the tetrahedron.
should
interpret
orientation viewed
\342\226\262
d2
= 0
and Homology
We
now
come
41.9Theorem
Let the
X be
n
1,
=
3. That
1, 2,
Since a homomorphism
more
briefly,
0. For n
into
41.10
Corollary Proof
n
boundary
operator
For n =
make
3 will
0, 1,2, and
we have
l,and2,
0,
c
(c) for some 3\342\200\236+i
G
= 1 this
0. For n 8\342\200\236_i(3\342\200\236(er))
the
is enough
is obvious,
2,
=
0.
PlPl) Pi) + {Pi-
for the
exercise
student
Pi)
in the
definition
of
the
2).
\342\231\246
of
subgroup
Bn(X) =
C-n-\\-i
everything
We use
values on generators,it =
=
is a
3, Bn(X)
maps
0. 3\342\200\236_1(3\342\200\236(c))
\342\200\224
excellent
an
(see Exercise
\342\200\224
Forw b =
=
3. Then
=0,1,2, =
have
by its
Bi(P2P3 -P1P3 + - (P3 (P3 p2)
81(82(/^2/\302\276)) =
Xforw
shall
= 0.\"
determined we have
a,
\302\273-simplex
we C\342\200\236(X)
\"32
is completely
to check that for an since 30 maps everything
c e
mathematics. We
into Cn-2(X)
Cn(X)
8\342\200\236_i3\342\200\236 mapping
for each
is,
= 0,\" or, \"3\342\200\236_i3\342\200\236
The case
equations in all of for all n > 0.
important it holds
but
complex, andletC\342\200\236(X)bethe\302\273-chainsof
homomorphism
composite
notation
most
and 3,
\342\200\224 2,
a simplicial
intoOforw
Proof
of the
to one
it only for
state
Groups
(X).
Z\342\200\236(X).
b e
Thenif 3\342\200\236+1[C\342\200\236+i(X)].
Bn(X), we must
have
Thus =
=
0, 8\342\200\236(fc) 8\342\200\236(3\342\200\236+i(c))
e Zn(X). For n \342\200\224 3, since B3 (X) = 0. sob
41.11
Definition
The factor
group
we are
not
with
concerned
simplexes
of dimension greater thai:
5. \342\231\246
Hn(X)
=
Zn{X)/Bn(X)
is
the
^-dimensional
homology
group
of X. \342\226\240
41.12
Example
Let us in Fig.
calculate//,, (S) for
n =
0, 1,
2, and 3 and
where
S is
the surface of the
tetrahedron
41.2.
We found are both
0,
and
Zn(S)
and
Bn(S) in Example
41.8. Now
hence
H3(S)
= 0.
C3(S)
= 0,
so Z3(S) and
B3(S)
Part VIII
362
in
Groups
Topology
Also, Z2(S) is that is,
infinite
saw
and we
cyclic
that
H2(S) that
saw
We
\342\200\224
Z\\ (S)
element, that
Thus H2(S) is
= 0.
infinite
cyclic,
~ Z.
factor group
so the
Bx (S),
B2(S)
(S) is
Zt (S)/B1
trivial
the
group
of one
is,
= 0.
Hi(S)
abelian on Pi, P2, P3, and P4, while B0(S) was generated by P3 ^2, ^4 - ^2, and P4 - P3. We claim that every coset of the form r Pi .Let z e Zo(S), and suppose of Zo(S)/Bq(S) contains exactly oneelement that the coefficient of P2 in z is s2, of P3 is 53, and of P4 is S4. Then was free
Zq(S)
Finally,
Pi, P4 -
Pi, P3-
P2-
zfor somer, so z r Pi. If the coset the only
Clearly,
is,
then
any coset e [rPi
r'Pi
a boundary
is
rP\\. We may
form
the
that
that is
of Pi
multiple
S3(P3~ Pi) + s4(P4
Pi) +
e [rPi + #0(^)1, also contains r'Pi,
then
H0(S)
-
Pi)] =
rPi
an element does contain of the form + B0(S)], so (r' - r)Pi isin\302\2430(S). zero, so r = r' and the coset contains the rP\\ as representatives of choose
Hq(S). Thus Hq(S) is infinite
in computing
cosets
-
[^2(^2
exactly one elementof the
Pi,
cyclic,
that is,
~ Z.
A
and computations probably seem very complicated to the student. we but admit that are a bit to write down. However, they very messy are typical for homology theory, the arguments used in these calculations i.e., if you can we can make them understand Furthermore, them, you will understand all our others. the the The next at of section will be devoted to looking geometrically, picture space. definitions
These
The ideasare
natural,
further computations of homology
groups
of certain
simple
but
spaces.
important
41
\342\226\240 EXERCISES
Suggested Exercises
1. Assume a.
c =
that
P4
- 4P3 P4P6 +
3. Describe
b. Is
Q(X), Zt(X), Bt(X),
The line segment
and the
joining
5. Describe Ct(X), Z;(X), Bt(X), Mark
P2 P4
+ Pi P(,P4 is
a 2-chainof
c. Is 82(c)
c a 2-cycle? it
is 0,
for the
completing
space
a certain
the proof of Theorem
consisting
of just
simplicial
complex X.
a 1-cycle? 41.9.
the 0-simplex
P.
(This
a
is really
problem.)
4. Describe
6.
3P3
82(83^1P2^3P4)) and show that Ct(P), Zj(P), Bi(P), and Hi(P)
2. Compute
{Note:
P3
82(c).
Compute
trivial
2Pi
of the
each
following
a. Every
true
is a
boundary
b.
Every
c.
is always C\342\200\236(X)
cycle
and
is a
Hi(X) for two points Hi(X)
for the
or false.
cycle.
boundary. a free abelian
the space X consisting of two distinct 0-simplexes, P is wo^part of the space.)
group.
space X
consisting
of the
1-simplex PiP2-
and
P'.
42
Section
d. B\342\200\236(X) is always
a free abelian
of Homology
Computations
363
Groups
group.
abelian group. f. Hn(X) is always abelian. of a 3-simplex is a 2-simplex. g. Theboundary h. The boundary of a 2-simplex is a 1-chain. i. The boundary of a 3-cycle is a 2-chain. e.
is always Z\342\200\236(X)
j.
If
a free
then
=
Z\342\200\236(X) Bn(X),
is the H\342\200\236(X)
trivial group
to
naturally
of oneelement.
Exercises
More
7. Define
the
1,2, and
so as
concepts
following
generalize
the definitions
in
the
text given
for dimensions 0.
3.
oriented n- simplex b. The boundary of an oriented n- simplex c. A face of an oriented \302\273-simplex a. An
8.
of Exercise7, what
the idea
Continuing
plexes of dimension 9.
c 10.
the
Following
of Exercises 7
ideas
n may
\342\202\254 where C\342\200\236(X),
for B\342\200\236(X)
easy
a question
answer
to
way
a simplicial
complex
X
perhaps
you
asking
containing
to define some sim-
3?
than
greater
be an
would
and
-+ C\342\200\236-i(X), : C\342\200\236(X) 3\342\200\236 Zn(X), C\342\200\236(X),
be greater
and
8, prove
that 32
= 0 in
general,
i.e., that
3,,-1(3,,(0)) =
0 for
every
3.
than
of a is the (n + 1)an (oriented) \302\273-simplex er of X, the coboundary 5(,!)(er) is taken over all (n + l)-simplexes r that have a as a face. That is, the simplexes a as a summand have of 3,!+i(r). Orientation is important r appearing in the sum are precisely those that Let X be the simplicial here. Thus /\302\276 is a facg of P1P2,but Pi is not. However, Pi is a faceof P2P1. complex of Fig. 41.2. consisting of the solid tetrahedron
Let X be a simplicial chain
a.
J2
T>
complex.For
the sum
where
and 5(0)(P4).
5(0)(Pi)
Compute
b. Compute 5(1)(P3P2).
c. Compute 11.
Following
5(2)(P3P2P4).
the idea of Exercise10,let
the same as the a. b. 12.
Define Show
Following
5(n)
the
Following
-+ C
: C{n\\X)
0, that is, that ideas of Exercises10 and ideas
and show
that
of Exercises
Compute H^(S) for the SECTiON 42
a simplicial
complex,and
a way
analogous to
the
S(n+]\\S(\"\\c)) = 0 for eachc
of X, the
X be
let
the group
be
of n-cochains
C(\"\\X)
C\342\200\236(X).
that S2 =
n-coboundaries
13.
group
surface
11, define
B(,!)(X)
10, 11, and S of
the
Computations
the group
way
we defined
-+ 3,, : C\342\200\236(X)
C\342\200\236_i(Z).
e C(n\\X).
Z(n)(X)of n-cocycles
of X,
the group
B(n){X)of
< Z
tetrahedron
define
the n-dimensional cohomology 41.2.
group
H(n){X)
of X.
of Fig.
of Homology Groups
Triangulations calculate homology groups for the surface of a sphere. The first thing the surface of a sphere is not a simplicial you probably say, if you are alert, is that and a since this is curved is a surface plane surface. Remember that complex, triangle two spaces are topologically the same if one can be obtainedfrom the other by bending, Suppose
you
wish to will
364
Part VIII
in
Groups
Topology
twisting, and and to be filled
so on. Imagine
our
air. If the
the tetrahedron, to
3-simplex,
rubber surface is
flexible,like the
have a rubber
surface
of a
balloon, it will deform itself into a sphere and the four faces of the tetrahedron will promptly drawn on the surface of the sphere. then as \"triangles\" This illustrates what is appear of a space. The term triangulation need not refer to a division meant by a triangulation into only, but is also used for a division into \302\273-simplexes for any n > 0. If 2-simplexes a spaceis divided up into pieces in such a way that near each point the space can be deformed to looklike a part of some Euclidean space R\" and the pieces into which the this deformation as part of a simplicial then the complex, space was divided appear after groups of original division of the space is a triangulation of the space. The homology the space are then denned formally just as in the last section. Invariance
with
rubber
Properties
the proofs of important invariance properties of homology groups, that are a lot of but for us to First, machinery, easy explain roughly. require quite are defined in terms of a of a but the homology space triangulation, actually groups groups no matter how the space is triangulated. For they are the same (i.e., isomorphic) example,a square region can be triangulated in many ways, two of which are shown in are the same no matter which is used to groups triangulation Fig. 42.1. The homology
There are
two
very
which
compute them.
is not
This
obvious!
/
\\
.... ..._v.. ,-'
\\
\\
/
\\
42.1 Figure For another
groups of again, not
42.2 Example
The
second invariance property, if one triangulated space is homeomorphic to can be deformed into the other without being torn or cut), the homology are the same (i.e., isomorphic) the two in each dimension n. This is, spaces We shall use both of these facts without proof. obvious.
the
(e.g.,
homology
our tetrahedron
groups in
of the
Example
surface of a sphereare
41.12,
since the
two
the
spaces
same
as those for
are homeomorphic.
the
surface
of
\342\22
Two are the spheres and the cells. Let us important types of spaces in topology introduce them and the usual notations. The \302\253-sphere Sn is the set of all points a distance of 1 unit from the origin in (n + l)-dimensional EuclideanspaceR\"+1. Thus the 2-sphere S2 is what is usually called the surface of a spherein R3, Sl is the rim of a circle,and S\302\2 is two points. Of course, the choice of 1 for the distance from the origin is not important. A 2-sphere of radius 10 is homeomorphic to one of radius 1 and homeomorphic to the matter. The \302\253-cell or \302\253-ball En is the set of all points surface of an ellipsoid for that the origin. in W a distance < 1 from Thus E3 is what think of as a solid you usually El E2 and line is a is a circular sphere, region, segment.
Section42 42.3 Example
of Homology
Computations
365
Groups
and
TheaboveremarksandthecomputationsofExample41.12showthatH2(S2)
are both
isomorphic to Z,
Connected
and
and
HqCS2)
= 0.
Hi(52)
Contractible
\342\226\262
Spaces
of Hq(X) for a spaceX with a triangulation. A space in it can be joined by a path (a concept that we will not in the space. If a space is not connected, then it is split totally up into a pieces, each of which is connected but no two of which can be joined by a
Thereis a very
nice
is connected if
interpretation
two
any
points
define) lying number of path in the space. Thesepiecesare
42.4Theorem
If
X is
a space
to Z x Z x
triangulated
Z, and
\342\226\240 \342\226\240 \342\226\240 x
connected
the
of the space.
components
into a finite number of simplexes,then the Betti number m of factors Z is the
Hq(X)
number
is isomorphic
of connected
components of X.
Proof
Now
is the
Cq(X)
free abelian group
of X.
triangulation
by the
generated
Also, B0(X) is generatedby p. _
where
Pti P,,
is an edge as P,,
in
component
the
p. p.
p. p.
-
+
of the
expressions
P,
in
the
form
p.
Fix Pil. Any P,-, by a finite
triangulation.
can be joined to
of X
of vertices
number
finite
p.
the same connected
Pir in
vertex sequence
p.
of edges. Then Pir
that
showing
with
=
Ph +
P,r e [Pil
select one vertex
followsat Example
Ptl)
-
(P3
We have
-
\342\226\240\342\226\240\342\226\240
+
(Pir
P,;.,),
+ B0(X)]. Clearly,if P,s is not in the same connectedcomponent + B0(X)], since no edge joins the two components. Thus, if we each connected component, each cosetof Hq(X) contains exactly is an integral multiple of one of the selected vertices. The theorem
from
once.
at
once
\342\231\246
that
H0(Sn)
for n
>
0, since Sn
is connected for
n
>
~
(see Chapter41,Exercise
4).
>
Z x
Z
Also,
H0(En)
for n
~ Z
0. However,
Ho(S\302\260)
~ Z
1.
A space but
Ph) +
then P\342\200\236, Pis \302\242. [Pil
one representative that
42.5
(P/:
always
\342\226\262
is contractible kept
within
the
if space
it
be compressed to a point it originally occupied. We
can
without just
state
being torn or cut, the next theorem.
Part VIII
366
42.6 Theorem
in
Groups
If Xis
a contractible
for n >
42.7 Example
Topology
space
triangulated
into a finite
number
of simplexes,
fhenH\342\200\236(X)
=
0
1.
not It is not too easy to prove this fact. The student contractible. to take it as self-evident that probably be willing you cannot compress the \"surface of a sphere\" to a point within the original without it, keeping it always tearing space S2 that it occupied. It is not fair to compress it all to the \"center of the sphere.\" We saw that H2(S2) ^ 0 but is isomorphic to Z. E3 as our solid we consider H2(E3), where we can regard Suppose,however, to E3. The surface S of this tetrahedron tetrahedron of Fig. 41.2, for it is homomorphic are for S (or is homomorphic to S2. The simplexeshere for E3 are the same as they for the whole 3-simplex we examined in Examples 41.8 and 41.12, S2), which except It
is a
a
fact that S2 is
however,
will,
now
that
=
Z2(E3)
and H2(E3)
generator of Z2(S),and in E3,
a. Viewed
= 0. SinceE3
In general, E\"
was
of Z2(E3),
hence
of B2(E3),so this is consistent
an element
contractible,
A for n >
is contractible
1, so we have
Theorem
by
42.6,
0
0.
i >
Further
, We
is 83(0-),
is obviously
Hi(En) = for
this
42.6.
Theorem
with
of
boundary
B2(E3)
that a
Remember
appears.
exactly the entire
Computations seen a
have
illustrate,
the
nice
1-cycles
for
interpretation
in a
by edges of the triangulation. 2-spheres or other closed2-dimensional Hi(X) roughly
to counting
simply because they
appear
of the 2-simplexes)
space.
to counting
Theorem 42.4. As
triangulated space are generated
formed
amounts
Ho (X) in The
2-cycles
can be
in the
surfaces
the
examples
preceding
by closed
curves of of
thought
space. Forming
the
space
as generated factor
the
by
group
= Zl(X)/Bl(X)
the closed curves that
appear
boundary of a 2-dimensional Similarly, forming H2(X) = as the
the closed 2-dimensionalsurfaces
in the
in the
space
piece
are
that
(i .e.,
amounts Z2(X)/B2(X)
space that cannot
not there
a collection of
be \"filled
roughly in solid\"
the space, i.e., are not boundaries of some collection of 3-simplexes. for Thus on the surface of the 2-sphere a 2-dimensional bounds every closed curve drawn piece of the sphere, so Hi(52) = 0. However, the only possible closed 2-dimensional surface, S2itself, cannot be \"filled in solid\" within the whole space S2 itself, so H2(S2) is free abelian on one generator. within
Hi (S2),
42.8
Example
one would expect Hi (S1)to be free abelian on one since the circle itself is not the boundary of a 2-dimensional isomorphic of S1. there of Sl. We You is no 2-dimensional see, part part compute and see whether this is indeed so. A triangulation of Sl is given in Fig. 42.9. Now C^S1) is generated by Pi P2, P2P3, and P3 Pi. If a 1-chain is a cycle so that its boundary is zero, then it must contain Pi P2 According generator,
i.e.,
to the
reasoning
above,
to Z,
Section 42
of Homology
Computations
367
Groups
42.9Figure and
P2Ps
the same number of times, otherwiseits boundary would holds for any two edges. Thus P2. A similar argument
multiple
of
by PiP2
+ P2P3+
we seethat
P3P1.
Since
is free
Hi(51)
= d2[C2(S1)]
#i(S')
It can
be proved
= 0for0
Hi(Sn)
To conform
that
= 0, there being no
abelian on one generator, that HiiS1) for
n >
contain Z\\{Sl)
a nonzero
is generated
2-simplexes,
is,
~ Z.
0, Hn{Sn)
and
A Ho(Sn)
are isomorphic to Z,
while
are
the homology groups of a plane annular us'compute region X between two concentric circles. A triangulation is indicatedin Fig. 42.11. Of course, since X is connected,
it
a coset
to topological
of Bn(X) in
a \"homology Z\342\200\236(X),
is,
homologous.
42.10Example
Let
follows
that
H0(X)
~ Z.
42.11 Figure
368
Part
VIII
Groups
If z without
in
Topology
is any Pi
homologous
and if Pt P2 has
1 -cycle,
coefficientr
P2 homologous to z. By continuing to z containing no edge on the
triangles, we canadjust further no edge Q, P, either. But then
this inner
in
z, then
argument, circle of the
r 32(Pi P2 Q i) is a cycle z \342\200\224 we find that there is a 1-cycle annulus.
Using
the \"outside\"
by multiples of d2 (Qi Pi Qj), and we arrive at z' containing if Q5 Pi appears in z' with nonzero coefficient, Pi appears
coefficient in di(z'), contradicting the fact that z' is a cycle.Similarly, = can for i to a cycle made occur 4. Thus z is homologous 1,2,3, edge <2iP(+i on the outer circle. a familiar such a edges only By cycle must be of the argument, nonzero
with
n(QiQ2
It is
clear
then
+ Q2Q3
+ Q3Q4+
Q4Q5
+
no up
of
form
QsQi).
that
~ Z.
Hi(X)
to the outer circle. Of course,we could well. pushed equally For H2(X),note that Z2(X) = 0, since every 2-simplexhas in its boundary an edge on either the inner or the outer circle of the annulus that appears in no other 2-simplex. Theboundary of any nonzero 2-chain must then contain some nonzero multiples of these We
showed
that
it to
have
we could the
inner
\"push\"
any
1-cycle
circle
edges.Hence H2(X)
42.12 Example
= 0.
A
homology groups of the torus surface X which looks like the To visualize a triangulation of the toras, imagine Fig. 42.13. that marked a, then cut it all around the circle marked b, and flatten you cut it on the circle it out as in Fig. 42.14. Then draw the triangles. To recover the toras from Fig. 42.14, join the left edge b to the right edge b in such a way that the arrows are going in the same direction. This givesa cylinder with circle a at each end. Thenbend the cylinder around and join the circles around the circles. a, again keeping the arrows going the same way Since the toras is connected, Hq{X) \342\200\224 Z. For Hi(X), let z be a 1-cycle.By changing of the boundary of z by a multiple We
shall
surface
the
the compute of a doughnut,
triangle
numbered
of triangle boundary of triangle the side /
as in
1 in 1. Then 2, you
Fig. 42.14, you can get a homologous cycle not containing of the by changing this new 1-cycle by a suitable multiple can further eliminate the side | of 2. Continuing, we can then
42.13
Figure
Computations of Homology
42
Section
369
Groups
42.15Figure
42.14 Figure
of 12,/of 7, | of 8,/of 9, | of 10,/of 11,\342\200\224 to z, can then 17. The resulting homologous cycle, the edges shown in Fig. 42.15. But such a cycle could contain not contain, with only nonzero coefficient, any of the edges we have numbered in Fig. 42.15, or it would not 0. Thus z is homologousto a 1-cycle having edges only on the circle a have boundary or the circle b (refer to Fig. 42.13).By a now hopefully familiar argument, every edge on circlea must appear the same number of times, and the same is also true for edges on circleb; however, an edge on circle b need not the same number of times as appear an edge on a. Furthermore, if a 2-chainis to have a boundary just containing a appears oriented counterclockwise must appear with the same coefficient andi?, all the triangles so that the inner edges will cancel out. The boundary of such a 2-chainis 0. Thus every class (coset) contains one and only one element homology eliminate/
13, | of
of 3, | of 4,/of 14,1 of 15,| of 16,
5,\342\200\224of6,/of
and I of
ra +
where r by the
s are
and
circles
two
integers. Hence H\\{X) a and b. Therefore,
sb,
is free
abelian
on
two
generators,
represented
H^X) ~ Z x Z. for
Finally, with
numbered these
H2(X),
3, also with
not to
triangles
The same holdstrue counterclockwise
multiple
orientation
of the
a 2-cycle must orientation
counterclockwise
counterclockwise
be in
for
the
the triangle
orientation,
formal sum of all
numbered 2
of times
as
in order
for the
it contains
of Fig. 42.14 the
triangle
common edge I of
These orientations are illustrated in Fig. 42.16. triangles, and thus every triangle with the same number of times in a 2-cycle. Clearly, any the 2-simplexes, all with counterclockwise orientation, to Z. Also, B2{X) = 0, there being cyclic, isomorphic
boundary.
any two adjacent must appear the
2-cycle. Thus Z2{X)is infinite no 3-simplexes, so
is a
contain
the same number
H2(X)
~ Z.
Part VIII
370
Topology
42
\342\226\240 EXERCISES
In these
in
Groups
exercises,you
need
not write
out in
detail
your
computations
or
arguments.
Computations
1. Compute
the
2. Compute
the homology
3. Compute
homology
groups
of the
space
groups
of the
space consisting of
consisting
of two two
tangent 1-spheres,
of the
space consisting
of a
2-sphere
with
4. Computethe homology groups a great circle of the 2-sphere. 5. Compute the homology groups
of the
space consisting
of a
2-sphere
with
of a
circle touching
6. Compute
7.
Mark
the
homology
touch
not
the
of the
each
groups
homology
following
true
of the
space consisting of the surface consisting
of a
eight.
2-spheres.
tangent
groups the 2-sphere.
does
i.e., a figure
2-sphere
an annular
with
ring (as in
an annular
ring whose
a 2-sphereat a handle
Fig.
one
42.11)
inner
circle
that is
point.
(see Fig.
42.17).
or false.
a. Homeomorphic simplicial complexeshave isomorphic groups. homology b. If two simplicial have isomorphic homology groups, then the spaces are homeomorphic. complexes c. Sn is homeomorphic to En. d. Hn(X) is trivial for n > 0 if X is a connected space with a finite triangulation. e. Hn(X) is trivial for n > 0 if X is a contractible space with a finite triangulation. f. Hn(Sn) = 0 for n> 0. g. Hn(En)
= 0 for
h.
i. j. 8. Compute
the
n>
0.
A
torus
is homeomorphic
A
torus
is homeomorphic
A
torus
is homeomorphic
homology
groups of the
to S2. to E2. to a sphere with
a handle
consisting
of two
space
on it (see Fig. torus
surfaces
42.17).
having
no points
in
;m
W?
42.17Figure
42.18
Figure
common.
9. Compute the stack
10.
the
Compute
circle of the 11. Compute
the
12. Compute
the
6 and
section
of the
groups
homology
two inner
More HomologyComputations
43
Section
space
stacked toras
of two
consisting
and
371
Applications
surfaces, stacked as one would
tubes. groups of
homology 2-sphere,
the
consisting
space
i.e., a balloon
wearing an inner of the surface consisting
homology
groups
homology
groups of
the
surface
of a toras
to a
tangent
2-sphere at
all
of a great
points
tube.
consisting
of a
2-sphere
handles
with
two
with
n handles
of a 2-sphere
(see Fig.
42.18).
(generalizing
Exercises
11.
43
More
Homology
Computations and Applications
One-SidedSurfaces been free abelian, so that there far all the homology groups we have found have to be the case for the order. This can be shown no nonzero elements of finite always like S2, which has no boundary) that groups of a closed surface (i.e.,a surface homology has two sides. Our next example is of a one-sided closed surface, the Klein bottle. Here will have a nontrivial torsion subgroup the the 1-dimensional homology group reflecting Thus
were
twist in the
43.1 Example
surface.
of the Klein bottle X. Figure 43.2 representsthe as 42.14 bottle Klein Fig. representsthe torus cut apart. The only difference on the top and bottom edgea of the rectangle are in opposite directions is that the arrows this time. To recover a Klein bottle from Fig. 43.2, again bend the rectangle joining the b so that the directions of the arrows match This fabeled that edges gives a cylinder up. is shown somewhat deformed, with the bottom end pushed a little way up inside the in Fig. 43.3. Such deformations are legitimate in topology. Now the circles a cylinder, Let
us
calculate
the homology groups
cut apart, just
so that the arrows be joined go around You must imagine that you are in R4, so that the side without intersecting and \"through\" must
thought,
you can see that
this
resulting
the same you the
surface
43.2
Figure
way.
can bend side,
This
as shown
really has
cannot
the neck of the only
in Fig. one
be done bottle
43.4. With
in
R3.
around a little
side. That is, if
you
Part
VIII
Groups
in
Topology
43.4 Figure
43.3 Figure
at
start
any
We the
can
concept
groups
as we did for
\"one
As
Klein
42.12, by
bottle
much as
we calculated
splitting Fig. 43.2 into
triangles
Of course,
torus.
~ Z,
H0(X)
sinceX is connected. dividing Fig. 43.2 into
the whole
painting
up
bottle.
of the
the homology groups of the torus in Example the
you will wind
side,\"
of a Klein
inside
an
of
calculate
homology
exactly
begin to paint
place and
thing. There is no
we found
for
the
if we
torus,
1-cycle is
every
triangles,
ra +
triangulate
the
Klein
homologous to a cycleof
the
bottle
by
form
sb
for r and s integers.If a 2-chain is to have a boundary containing just a and b, again, all the triangles oriented counterclockwise must with the same coefficient so that appear the inner edges will cancel each other. In the case of the torus, the boundary of such a 2-chain was 0. Here,however, it is k(2a), where k is the number of times each triangle Thus Hi(X) is an abelian group with appears. the homology classes of a and generators b and the relations a + b = b + a and 2a = 0. Therefore, ~
Hi(X)
a group 2-chains
Z2 x
coefficient 2 and Betti that there are no 2-cyclesthis
torsion
with
shows
A torsion
example
surface
coefficient with
of the
does
boundary.
Mobius strip.
not
have
Mostly for
to be the
1. Our
number
time,
H2{X) =
onesided
Z, above
regarding
0.
present
sake
argument
so
in some homology of completeness, we
A
group of
give this
a
standard
Section43 43.5 Example
LetX be the
MoreHomology
and
Computations
373
Applications
can form by taking a rectangle of paperand joining twist so that the arrows match up, as indicatedin Mobius strip is a surface with a boundary, and the boundary is Fig. made of and It one closed curve to a I is clear that the /'. circle) (homomorphic up just Mobius strip, like the Klein bottle, has just one side,in the sense that if you were asked Mobius
to color only Of
which we
strip,
two ends marked a 43.6. Note that the
the
with
a half
side of it, you would since X is connected,
one
course,
wind
up coloring
H0(X) ~
the whole
thing.
Z.
in succession suitable multiples of the triangles 1-cycle. By subtracting 2, | of triangle 3, 3, and 4 in Fig. 43.6, we can eliminate edges/of triangle on 4. to a and Thus is of z' z /, /', and cycle having edges only homologous \\ triangle the same number of times.But if c is a a, and as before, both edges on /' must appear as shown in Fig. 43.6, we oriented 2-chain consisting of the formal sum of the triangles seethat 32(c) consists of the edges on I and /' plus la. Since both edges on /' must appear a suitable multiple of 82(c),we seethat in z' the same number of times, by subtracting to a cycle with edges just lying on I and a. By a familiar argument, z is homologous of times in this new all these edges properly oriented must appear the same number sum is a generator for Hi (X). class containing their formal cycle, and thus the homology
z be any numbered 2, Let
Therefore,
#i(X)~Z. at Q and goes around starting point. If z\" were a 2-cycle, it would have to contain with the indicated 43.6 an equal number r of times = 0, so be r(2a + 1 + 1')^0. Thus Z2(X)
This generating
a,
and
arrives
cycle starts
the
across it
then cuts
strip,
P via
at
at its
H2(X) =
the
1, 2,
triangles
orientation. But
of
Fig.
would
\342\226\262
A
I 1
^
O
82(2\")
0.
\342\226\240\342\226\240
^G-
3, and 4
then
\342\200\242C>\302\253...
^
X\" \" ^
/
-'
-J
43.6Figure The
Euler
Characteristic
Let us turn from the computation of homology groups to a few interesting facts and (or triangulated space) consisting complex applications. Let X be a finite simplicial 3 and less. Let \302\2730 of simplexesof dimension be the total number of vertices in the of 2-simplexes, and \302\2733 the number n\\ the number of edges,m_ the number triangulation,
374
Part
VIII
Groups
in
Topology
The number
of 3-simplexes.
3
=
\302\273o -\302\273i +\302\2732\342\200\224\302\2733 (-1)'\302\273; 5^
1=0
shown to be the same no matter how the characteristic x (X) of the space.
can be
Euler theorem. is the
43.7 Theorem
Let X be a finite be the
Euler
of the
triangulated. This
state
the
number
fascinating
following
space) of dimension
(or triangulated
complex
simplicial
characteristic
X is
space
We just
space X, and let fij
the Betti
be
<3. Let x (X) number of Hj (X).Then
3
=
X(X)
\302\243(-1)^,-.
y=o
holds alsofor X of dimension greater Euler characteristic to dimension
This theorem
of the 43.8 Example
Considerthe
tetrahedron
solid
3, with
than
of the
definition
greater
Fig. 41.2. Here\302\2730=4,
E3 of
the obvious than 3.
=
\302\273i 6,
n2 =
extension
4, and\302\2733 =
1,
so
X(E3) we saw that H3(E3) and A) = 1, so
that
Remember
, fo
= p2
=
+ 4-1 = 1. = H2(E3)= Ht(E3) =
=4-6
0 and
H0(E3)
~ Z. Thus
fa = 0
3
the
For
and
= \302\2733 0,
surface
S2 of
the
fa =
H3(S2) = fa
=
in Fig.
tetrahedron
41.2, we have
n0
= 4,
n\\
=
6,
m_
=
4,
so
X(S2) Also,
= x(E3).
fa = l
J2(-iy
0 and
Hi(S2) = 0, and fa = fa = 1,so
=4-6
+4
and
H2(S2)
=
2.
H0(S2) are
both
isomorphic
to Z.
Thus
3
= x(S2).
\302\243(-lV'i8;=2
Here Thus
Hi(51) fa
Sx in
for
Finally,
= fa
and
= Fig. 42.9, \302\273o
#0(^1)
= l and fa
are
both
3,
=
\302\273i
3, and
X(S1) =
3-3 = 0.
isomorphic
to Z,
= fa= 0, giving
=
=
\302\2732 \302\2733
and #3(^1)
0, so
= H^S1)= 0.
3
^(-1)^-=0
= /(51).
A
43
Section
f mapping a spaceX into a space Y gives rise to a homomorphism into H\342\200\236(Y) for n > 0. The demonstration of the existence of this Hn(X) takes more machinery than we wish to develop here, but let us attempt
function
continuous
/*\342\200\236 mapping
homomorphism
to describe how is true: If z
in
e
Y
these
can be
homomorphisms
That
regarded as the result be an obvious should naively way,
then f(z)
represents
the
=
Bn(X))
class
homology
illustrate
Considerthe
and
this
unit
to show just
attempt
Any
point
f : S1-> S1be
has coordinates
in Sl given
+ Bn(Y).
and f(z) is
in Hn(X)
class under
an
n-cycle
of the homology /\342\200\236\342\200\236
in Y,
class
here by /(z).
mean
we
what
+
y2
=
l}
(cos9, sin9), as
indicated
in Fig.
43.10. Let
by
f is continuous.
this function
= (cos39,sin
sin 9))
/((cos0, Obvieusly,
it down
setting
Y, then
\302\273-cycle in
circle
S1={(x,y)\\x2 in R2.
The following
cases.
up z and
of picking
f(z)
homology
image
certain
z-
containing
43.9 Example
represents a
if z
is,
computed in
if f(z),
and Z\342\200\236(X),
in the
fm(z +
Let us
375
of Spaces
Mappings A
Computations and Applications
More Homology
Now
f should
39).
induce
/,i : HiiS1) -+ H^S1). Here
is isomorphic
Hi(Sx)
P2P3 + P3P1,as seenin the circle,
of the
circle,
then
f
that
maps
to Z
and
42.8.
Example
each of
has
the
arcs
the homology classof z = P1P2 + Now if Pi, P2, and P3 are evenly about spaced onto the whole perimeter P\\Pz, P2P3, and P3P\\ as generator
is,
f(PiPi)
= f(P2P3) =
f(PiPi) =
P\\Pi
+
P2P3
y
9, sin 61)
Il\342\200\224x
43.10
Figure
+ P3Pi-
376
Part VIII
Groups in
Topology
Thus
+ B^S1))
U(z
that is, the fact
f*i maps a generator that
43.9
Example
of
\\PXP2
=
3z
Hi(51)
+
+
P2P3
+ P3P0
+ B^S1)
Bi(51),
onto three
times itself. This
\342\226\2
illustrates our previous assertionthat the homomorphisms with a continuous f may mirror important mapping
of
associated
homologygroups
reflects
obviously
itself three times.
S1 around
winds
f
=
of
properties
the mapping.
concepts to indicate a proofof the famous Brouwer Fixed-Point that a continuous map f of En into itself has a fixedpoint, means for E2, a this is some x E\" that f(x) = x. Let us see what there e such i.e., of rubber stretched out on a table to that you have a thin sheet circular region.Imagine of the rubber circle on the disk. Mark with a pencil the outside form a circular boundary without Then stretch, compress, bend, twist, and fold the rubber in any fashion table. When on the table. but it within the circle you finish, keep always penciled tearing it, some point on the rubber will be over exactlythe same point on the table at which it first started. we
Finally,
use these
Theorem. This theorem
states
y
=
x/
43.11 Figure
The proof we outline is \342\200\224> : E1 we E1, f
for any n > 1. For n = 1, looking at the graph of states find that the theorem simply that any continuous the left and right sides of a square must cross the diagonal somewhere, joining the construction of our proof indicatedin Fig. 43.11. The student should visualize good
a function path
as with
E3
the
illustrating
43.12 Theorem
(Brouwer
forw
Proof
boundary
having
Fixed-Point
S2
S1. The proof contains
E2 having boundary the case of E2.
and
for
construction
Theorem).
A
continuous
map
f of En
a figure
into itself has a fixed
point
> 1.
The casen assume that
=
1 was
considered
f has no
fixed
point
above. Let and
shall
f be a map
of
En into
derive a contradiction.
En for
n
>
1. We
shall
Section 43
Exercises
377
i=y
43.13
If fix)
us extend g(x) =
y,
since
Now
e
can consider the line segment from to x. Let fix) direction from fix) to x until it goes through the Sn~l of En Sn~l with point y. This defines for us a function g : En \342\200\224> as illustrated in Fig. 43.13. Note that for y on the boundary, we have g(y) = y. f is continuous, it is pretty obvious that g is also continuous. (A continuous 7^
x for line
this
boundary
all x
Figure
E\",
we
in the
segment at some
one that maps points that function is roughly are sufficiently close together into points that are close together. If x\\ and x2 are sufficiently close together, then f{x\\) and fixi) are sufficiently close and x\\ is so close so that the line segment joining fix\\) together
to the line is a continuous
joining
segment
mapping
of
f(x2) andx2 that y\\ = g(x\\) is close to y2 = gix2).) Then g En into Sn~l, and thus induces a homomorphism \342\200\224> Hn^iiSn
Hn~iiE\") *(n \342\200\2241)
Now we
said that
=
H\342\200\236^\\iEn)
and n =
i=2
forw
\342\200\224 (\302\273 l)-cycle
an
Sn~1with y e
> 1, since E\"
and we checked it have g*(n-i)(0)= 0.But is the whole complex Hn^\\iEn) = Sn~1, since ^(y) = y for all
is contractible,
we must
ahomomorphism,
the homology class 0 of of simplexes, and #(571-1)
representing orientation
proper
0, for n
3. Sinceg*(\302\253-uis
).
^^.Thus
^(B-i)(0) = 5\"-1+Bfl_i(5\"-1). is a
which
generator
find the
We
a contradiction. ^ 0 of Hn^CS\"^1),
preceding proof very
\342\231\246
aesthetically,
satisfying
and hope
you
agree.
43
\342\226\240 EXERCISES
Suggested Exercises
1. Verify the
2.
by direct Euler
Illustrate
a. The
calculation
characteristic Theorem annular
b. Thetorus c. The Klein
43.7,
both
as we
did
of Example
region of Example bottle
that
triangulations
of the
square
region
X in
Fig. 42.1 give
x(X). in Example
42.10
42.12
of Example
43.1
43.8, for
each of the
following
spaces.
the
same value
for
Part VIII
378
3. Will every
in
Groups
continuous
Topology
of a
map
square
continuous map of a spaceconsisting
4. 5.
groups of
Compute
the
Compute
the homology
6. Mark
homology
each of the a.
Every
b. A
following
into itself have a fixed point? Why or disjoint 2-cells into itself have a fixed point?
consisting
space
space consisting
true or
false.
group
of a
homology
into
Hn(X)
c.
All homology
d.
All
homology
e. All O-dimensional f. If a spaceX has
of a
2-sphere touching
of two
Klein
contractible space is the
map from a simplicial
continuous
phism of
the
of the
groups
region
of two
complex
X
into
bottles
trivial
group
a simplicial
points
of one
Why
or why
bottle at one
a Klein
with no
Will
not?
why
every
not?
point.
in common.
element.
complex
Y
induces
a homomor-
H\342\200\236(Y).
groups are abelian. groups are free abelian. homology groups are free abelian. \302\273-simplexes
but
none
of dimension
greater
than
n and
Hn(X) ^
0, then
is H\342\200\236(X
free abelian. of an g. The boundary boundary of an
h. The
i. The The
j.
\302\273-boundaries \302\273-dimensional
\302\273-chain is
an (n
\302\273-chain is
an
form
l)-chain.
\342\200\224 (\302\273 l)-cycle.
a subgroup
homology
\342\200\224
group
of the \302\273-cycles. of a simplicial complexis always
a subgroup
of the
group
of
n- chains.
More Exercises
7.
Find
8.
We
Euler
the can
characteristic
of a 2-spherewith
n handles
(see Section
42, Exercise12).
the topological real projective plane X, using a so that Fig. 43.14, by joining the semicircles and the directions of the arrows match up. This cannot be done in opposite points come together 3-space R3. Onemust go to R4. Triangulate this space X, starting with the form exhibited in Fig. its homology groups. compute
form
diametrically
Euclidean
43.14, and 9. Thecircular
shown in Fig. 43.14 can be deformed to appear as a 2-sphere a hole in it, as with topologically 43.15. We form the real projective plane from this configuration by sewing up the hole in such a on the rim of the hole are sewn together. This cannot way be done in R3. diametrically opposite points this idea, a 2- sphere with sewn up by bringing together Extending q holes in it, which are then diametrically opposite points on the rims of the holes, gives a 2-spherewith q cross caps. Find the homology groups of a 2-sphere with the space as the disk in Fig. 43.14 but with q - 1 view q cross caps. (To see a triangulation, holes in it to be sewn up as described above. Then triangulate this disk with these holes.) shown
disk
in Fig. that only
43.14Figure
43.15
Figure
44
Section
379
Homological Algebra
43.16Figure It can
Comment:
be
that
shown
homeomorphicto a 2-spherewith to a
2-sphere
genus
of the
surface.
Every
point
homeomorphic
10.
P on
0 cross
q >
with
a regular
X can
torus
nice closed surface, namely a closed 2-manifold, is and is if the surface is two sided, one sided. The number h or q, as the case may be, is the
every sufficiently number h > 0
some
caps if
it is
be describedby
of handles
means
of two
angles 6 and
0,
as shown in
Fig. 43.16.
P. For each of the mappings with That is, we can associate coordinates (6, / of the torus X onto itself given into H\342\200\236(X) for n = 0, 1, and the induced map /*\342\200\236 describe of H\342\200\236(X) 2, by finding the images of the below, for H\342\200\236 in Example 42.12. Interpret these group homomorphisms geometrically as we (X) described generators
0)
did
43.9.
in Example
a. / : X
b. / : X c. / : X 11. With
X
given by
/((0, 0))
-+
X
given
by
/((0,
0)) =
-+
X
given
by
/((0,
0)) =
reference
of a variety for n H\342\200\236(b) a. / :
10, the torus X can be mapped onto For each such below, maps. / : X -+ b given map = 0, 1, and 2, by describing the image of generators
-+
b
given
b. / : X
-+
b
given by
Exercise 11, but
Repeat
13.
Consider onto
the
maps
section
(0. 20) (20, 20)
to Exercise
X
12.
= (26,0)
-+
the map / of the point of b directly
-+ : H\342\200\236(X) /,\342\200\236
44
f((6, 0))
bottle
Klein opposite
Homological
is homeomorphicto
the homomorphism 10. Hn(X) as in Exercise
of
51)by
-+ /H : H\342\200\236(X)
(6, 0)
torus
X into
itself, inducing
maps/,,,
: H\342\200\236(X) -+ H\342\200\236(X).
in Fig. 43.2 given by mapping a point Q of the rectangle in Fig. 43.2 to) it. Note that b is topologically a 1-sphere. Computethe induced of Hn(X). 1, and 2, by describing images of generators
(closest
for n =
Hn(b)
b (which
describe
= (26,0)
/as a map of the
the map
view
0)) =
f((6,
by
its circle
0,
Algebra
Chain Complexesand Mappings The
subject
of algebraic if you see,
algebra. groups Ck(X) and You
maps
Cn(X) -X
topology was responsible for a surge in have a simplicial complex X, then you 3^, as indicated in the diagram H
C\342\200\236-i(X)
-.-^+
d(X) ^
C0(X)
a new
naturally
^+
0,
direction
in
get chain
Part VIII
380
Groupsin
Topology
dk-idk =
with
consider any
0.
of abelian
sequence
the purely
abstract
then
You
groups
1. Sothat
Ak
algebraic portion
and
this
of
homomorphisms
dk
:
Ak
situation and ^ Ak_i such
= 0, always have to require k > 1 in 3^-^^ infinite\" of for all k Z. it is convenient to consider \"doubly e Often, sequences groups Ak = 0 for k < 0 and of such sequences k > n in applications. The study and maps of Ak such is a topic of homological algebra. sequences
that
44.1Definition
dk-i
A chain
0 for
=
dk
k >
(A, 9) is a doubly
complex
A =
of abelian
groups
such that
dk : Ak
setting 44.2
Theorem
Proof
If
and
Ak-\\
A0, A_!, A_2,
A is a
the
then
in
notation
We can
9).
[dk
of homomorphisms
e Z}
\\ k
\342\226\24
we shall be sloppy
theory,
group
now imitate
a completely
in
and
algebraic
of Section 41.
and definitions
chain complex,
=
with a collection 9 = 0.
\342\200\242\342\226\240}
dk-A
similar to our the chain complex (A,
constructions
our
sequence
A2, Au {\342\226\240\342\226\240\342\226\240,
Ak, together
->
infinite
a convenience
As
\"A\" denote
let
do not
you
under
image
dk
is a
of
subgroup
the
of dk-i.
kernel
Consider 3n
3*
Now
dk_idk
once
that
Ak-2-
Ak-
Ak
= 0. This tells us at A is a chain complex. That is, dk-i[dk[Ak]] in is we the kernel of dk-i, which what wished to prove. S^A*] is contained =
0, since
\342\231\246
44.3 Definition
is a chain
If A
complex, then
the
image Bk{A) = dk+i[Ak+l] is the is the &th homology Zk(A)/Bk(A) stated
We
mapping
of the
in the last
f from X
groups
homology
Y, the important
into
kernel group
Let
us
for
arises in the
44.4 Theorem
to the
purely
algebraic
of Hk(X)
situation
fk of that
into
Ck(X)
Hk(Y).
Y,
a continuous
This mapping
triangulations of X
For suitable
way.
following
X and
complexes
simplicial
mapping f gives rise to a homomorphism that it commutes with dk, property
turn
=
\342\226\24
a homomorphism
^kfk =
homology
the
group of A.
section that
Y induces
Zk(A) of dk is the group of A>cycles,and of A:-boundaries. The factor group Hk(A)
into
Ck(Y),
which
and
has the
is,
fk-Aand see
how
induces
this
a map
of
the
groups.
(Fundamental Lemma) Let A and A' with collections 9 be chain complexes, and suppose that there is a collection in the diagram Ak -* A'k as indicated 3^2 .
\342\200\224> Ak+i
|/*+l A'
k+l
3*+ A \342\200\224
Ak
[h A'
3* A \342\200\224> A*.!
\342\200\224
[fkk-l A',k-l
3<-
9'
and
f
of
of homomorphisms
homomorphisms
fk :
Section 44
that every
furthermore,
Suppose,
square is commutative, =
fk-A all k.
for
Proof
Let z e
Then
= /*-i(3*(z))
3i(/*(z)) e
381
is,
d'kfk
->
Hk(A').
Now
Zk(A).
so fk(z)
that
homomorphism f*k : Hk{A)
a natural
induces
fk
HomologicalAlgebra
Let us
Zjt(A').
/*, :
to define
attempt
H,(A) ->
= /,(z)
+ ft(A))
/*,(z
= /*-i(0)
= 0, by
Hk(A')
+ Bk(A').
(1)
defined, i.e., independent of our - z) Then zi e (z + Bk(A)). (zi Bk(A). Suppose representative there exists c e A,+i such that z\\ \342\200\224 z = dk+i(c). But then must first show
We
that
is well
ftk
of z +
this last
term is
choice
e
of a
\302\243,(A),
so
-z) = /*(3*+i(c))= 8'k+1(fk+1(c)), = Bk(A'). Hence d'k+l[A'k+1]
fk(z) = fk(zi
-
fkizi) and
that
of
element
an
fk(zi) e (/,(2)+ 5,(A')). two
Thus
of
representatives
the
same
representatives of just one coset in Hk(A) -> Hk(A') is well defined by Now we compute ftk by taking
in Hk(A)
coset
=
Hk(A')
are
Zk(A)/Bk(A)
mapped
This shows
= Zk(A')/Bk(A').
that
into :
ftk
(1).
equation
representatives of cosets,and we define the of the original of a factor operation group by applying the group group group operation at once from the fact that the action of fk on Zk(A) to representatives of cosets. It follows into into is a homomorphism of Zk(A) that /,, is a homomorphism of Hk(A) Zk(A') fk of
\342\231\246
flt(A'). If the collections of maps /, 9, and 9' have the property, given then 9. the squares are commutative, / commutes with but After another definition, we shall give a seemingly trivial illustration
44.5 Definition
A
chain
is a
(A', 9') is a
complex
subgroup
and
Ak
effect on elementsof 44.6
Example
Theorem
very
44.4,
that
important
44.4.
of Theorem
of
in
the
subcomplex of a chain every c e
= dk(c) for d'k(c) subgroup A'k of
complex A'k,
that
is,
(A, 9), if, for all and dk have the dk
k, A'k same \342\226\240
Ak.
be a chain complex,and let A' be a subcomplex of A. Let;' be the collection of injection mappings Ak given by ik(c) = c for c e A^. It is obvious that;' ik : A'k ->\342\200\242 -> Hk(A). One commutes with 9. Thus we have induced homomorphisms itk : Hk(A') an must be of into Hk(A). that itk might naturally isomorphic mapping Hk(A') suspect This need not be so! For example, let us consider the 2-sphere S2 as a subcomplex of the 3-cell E3. This gives rise to i2 : C2(S2) ->- 02(^3) and induces Let A
i* : But
we have
seen
that
isomorphic mapping.
H2(S2)
H2(S2)-\342\226\272 H2(\302\2433).
\342\200\224 Z, while
H2(E3)
= 0. Thus
^2 cannot
possibly
be an \342\226\262
in Topology
Groups
Relative
Homology
Supposethat
A'
of the
is a subcomplex
this arises is
which
of a simplicial complexX. We just as
the
in
chain A. The topological situation from complex of a simplicial subcomplex Y (in the obvious sense) then naturally consider Ck(Y) a subgroup of Ck(X),
consideration
the
can
where we have A'k
situation
algebraic
of Ak.
a subgroup
Clearly, we would
have
< Cu-^Y).
dk[Ck(Y)]
Let us deal now
with
situation and remember that
the algebraic
at
situation
topological
it can
If A' is a subcomplex of the chain A, we can form complex of factor groups Ak/A'k. We claim that A/A' again gives rise to natural way, and we must exhibit a collection 3 of homomorphisms
that
= 0.
dk-idk
The
of
definition
dk(c +
for c
e
phism,
to show three
Ak.
We
have
and
that
3^_i 3^ =
to
First,
so dk(cl
that
show
dk
\342\200\224
c)
This
the
collection
a chain
attempt is obvious, namely,
\\ to
A'k) =
things:
+
dk(c)
complex
A/A' in a
a;_!
well denned,
dk is
that
define
that
it is a
homomor-
0. is well
denned,
let
c+
be in
also
c\\
A'k.
Then
(c\\
\342\200\224
c)
e A'k,
Thus
e A'k_l.
+
3,(c1)e(3,(c) also.
our
-* (AuMU)
3* : (Ak/A'k)
such
be applied to
time.
any
shows that dk is well
A^_1)
denned.
The equation 3t((Cl
+ Ai)
+ (c2 +
Ai))
=
3t((ci
=
3*(ci
=
= that
shows
\\ is a
+ c2) + c2)
+
Ai) A'k_Y
+ a;_! 3t(c2 + A'k)
+ dk(c2))
{Uci) +
3,(Cl
+
A'k) +
homomorphism.
Finally,
3t_i(3t(c
+ A'k))
= 3n(4(c) =
so
S^-iSjt
+
3^(3^))+Ai_2
Ai_j)
= 0
+
Ai_2,
= 0.
The precedingarguments
routine computations to the homological are routine to you. We gave them in to keep track of dimension, i.e., to keep track great of subscripts.Actually, in homological algebra usually the expert does not write most of these indices,but he always knows precisely with which group he is working. We could keep track of exactlywhich were under gave all the indices so that you groups consideration. Let us summarize work in a theorem. the above
algebraist,
are
typical
and multiplication detail. One has to be a little careful just
as addition
of integers
Section 44
44.7
Theorem
subcomplex of the
If A' is a Ak/A'k,
chain
\\{c + for
The
collection
the
of homomorphisms =
A'k)
+
\\{c)
of factor groups
A/A'
Bk denned
383
by
A\\_x
Ak, is a chain complex.
c e
SinceA/A' 44.8 Definition
A, then
complex
the collection 3
with
together
HomologicalAlgebra
is a
group Hk{A/A') is
homology
we can then
chain complex,
the
relative
&th
the homology
form
homology
groups Hk{A/A').
group of A
modulo
A'. \342\226\240
In our
topological to the
conform
shall
group
from
arising
of
the chains
are
Y
situation where
is a
Y
of a simplicial complexX, we and denote the kth relative homology
subcomplex
of topologists
notation
usual
the subcomplex thus \"set equal
C(Y) of the chain to 0.\" Geometrically,
C(X) by complex this corresponds
\"Hk(X, Y)\" to shrinking
All Y
to a point.
44.9 Example
Let X be the
of the edges (excluding the inside) of the of the the P2P3. We edge subcomplex consisting triangle to a point collapses the rim of the have seen that H\\(X) ~ H\\(Sl) ~ Z. Shrinking P2P3 in Fig. 44.11. The result is still the same as S1. Thus, topologically triangle, as shown we would expect again to have Hi(X, Y) ~ Z. and P3Pi. Since P2P3 e Ci(Y), we seethat Generators for Ci{X) are PiP2, P2P3, of Ci(X)/C\\(Y) are generators complex
simplicial
44.10,
in Fig.
and
PiP2 + To
find
Zi(X,
Y) we
consisting
Y be
let
Ci(Y)
and
compute
MnPiPi +
+
mP3Pl
d(Y))
= 8i(nPiP2) =
n(P2
= sinceP2, is (PlP2
P3
+
n)Pi
+ C0(Y),
-
m{Pl
-
C0(Y)
P3) +
C0(Y)
Thus for a cycle, we must have m = n, so a generatorof Since BX{X, Y) = 0, we seethat indeed Ci(Y). Y)
Hi(X,
Since Pl +
generates
C0(Y)
3^/\302\276^
dimension0
(m
+ 8i(mP3Pi) +
- Pi)
e Co(Y).
+ P3Pi) +
we seethat
+ Ci(Y).
P3Pl
Hq(X,
for connected
44.10
+
Ci(Y))
Z0(X,
= (Pi
Figure
Y)
~ Z.
Y) and
-
P2)
+
0. This is characteristicof simplicial complexes.
Y) =
Z\\{X,
Cq(Y) relative
= Pi+
C0(Y),
homology
groups of \342\226\262
44.11 Figure
Part VIII
384
44.12
Example
in
Groups
Topology
consider S1 as a
Let us
Remember Fig. 44.13).
of E2
boundary)
disk, so S1can
indeed
be
the edge of a circular piece of cloth and then circle comes in to one point. The resulting space
around
of the while
= 0,
H2(E2)
since E2 is a
contractible
and
a point
drawing
H2(.E2,
compute
of as its
thought
of S1to
the shrinking
demonstrate
can
You
(the
subcomplex
is a circular
E2
that
boundary
by putting a drawstring so that the rim string
the
is then a closed bag or we would expect
space,
S1). (see
S2. Thus,
H2(E2,Sl) ~ Z. For
of computation, Sl as the
purposes
region of Fig. 44.10 and 44.13 Figure
we can rim of the
regard E2 topologically Then
triangle.
as
the
C2(E2, S1) is
triangular
generated by
+ C2(51),and
P1P2P3
C2(S1))=
di(PiP2P3 +
(P2P3 -
=
But (P2P3 -
+
P1P3
C2(Sl)is
C,(Sl) +
P1P3
P1P2)
+ C!(Sl).
e CY{Sl), so we have
PiP2)
+ C2(51))=0.
%(pip2/,3
Hence P1P2P3 +
+
d2(P1P2P3)
an
of Z2(E2,
element
S1). Since
B2(E2,S1)= 0, we
that
see
H2(E2,
S1)
~ Z,
as we expected. Exact
The
\342\226\
Sequence
Homology
of a Pair
the exact homology sequence of a pair and give an application. We out all the details of the computations. The computations are routine and straightforward. We shall give all the necessary and shall let the student definitions, supply the details in the exercises.
We
44.14 Lemma
describe
now
not
shall
Let A'
be
morphisms
carry
a subcomplex jk : Ak ->
of a chain complexA. (Ak/A'k). Then
Let
j be
the collection of natural
homo-
jk-A = dkjk, that
Proof
is, j
We leave
commutes
this
easy
with
3.
computation
to the
exercises (see Exercise12).
\342\231\2
Section 44
44.15 Theorem
The map
44.14 induces a natural
Lemma
jk of
Proof
from Lemma
is immediate
This
A' be
Let
z + Bk(A/A') arrive at c
44.14 and Theorem44.4
a subcomplex of the z e Zk(A/A'), and
for
from
implies that
fr
by
dk(c)
two
homomorphism
-* Hk(A/A').
: Hk(A)
M
385
Algebra
Homological
\342\231\246
chain
complex
A. Let
in turn
z =
for A'k
c+
e Hk(A/A').
h
c e
some
Ak.
Then h = that we (Note
choices of representatives.)Now dk(z) = 0, which This, together with dk-idk = 0, gives us dk(c) e Zk-i(A').
successive
e A'kl.
Define
d,k :
-*
Hk(A/A')
fln(A')
by a**(ft)
= a*(c)
+ Bt_!(A').
of it as follows. Start with an looks very complicated. Think a modulo relative element of Hk(A/A'). Now such an element is represented by \302\243-cycle is in A'k_v A'. To say it is a relative fc-cycle modulo A' is to say that its boundary is a boundary of something in Ak, this boundary is in A'k_l and Since its boundary at a in A'k_l. Thus starting with h e Hk(A/A'), we have arrived must be a (k \342\200\224 l)-cycle
(k 44.16
Lemma
Proof
of 3^
definition
This
\342\200\224
The map 3^ : Hk(A/A') is a homomorphism
itk be
Let
which
->\342\200\242 H^_i(A'),
of
Hk(A/A')
this proof to the
We leave
class in
a homology
representing
l)-cycle
the map
exercises
of Example
into
44.17Lemma Proof
The
groups
You
need
this for
in diagram
we have
just
defined,
(see Exercise 44.6.
defined, and
We now
13).
\342\231\246
can construct
fln(A)
the
only check that a sequence of two the exercises (see Exercise14).
the
following
diagram.
Hk(A/A')
Hk(A)
(1), together with
is well
#\302\243_i(A').
Hk(A')
Hk-i(A') ^
JEfc_i(A').
^
fl*_i(A/A')
given
maps, form
consecutive
a chain complex,we could We have been aiming at
maps
(1) a chain complex. always gives
0. We
leave \342\231\246
the homology the answer to which is actually are 0. You may quite easy. All the homology groups of this chain complex think that such a chain complexis uninteresting. Far from it. Such a chain complexeven has a special name. Since
diagram
groups of this
44.18 Definition
A
sequence
sequenceif have
that
chain
(1) gives complex.
under
3^
ask for
a chain complex is an exact homomorphisms 3^ forming groups of the chain complex are 0, that is, if for all k we is equal to the kernel of 3^_i.
of groups Ak and all the homology
the image
(horrors!)
this question,
Part VIII
386
in
Groups
Topology
Exact sequences are of great importance properties of them in the exercises. 44.19
Theorem
Proof 44.20
Definition
The groups We
and
this proof
leave
The exact
of the
maps
to
sequence in
(1) form an
diagram
(see Exercise
the exact
(1) is
diagram
some
give
elementary
\342\22
chain complex in
exercises
the
We shall
in topology.
exact sequence.
15).
\342\231
homologysequenceof
the
pair
(A, A').
\342\22
44.21
Example
Let us
now
that
proof
stated without the
proof that and
for En
result
Theorem 44.19 to topology. We have stated without ~ Z, but that Hk(Sn) = 0 for k 0, n. We have also ^ = 0 for k ^ 0, since En is contractible. Let us assume
an application of ~ Z and H0(Sn) H\342\200\236(S\") give
We can view
S\"
Hk{En)
as a
this the
from
derive
now
of the
subcomplex
to an
is topologically
equivalent
its boundary.
Let us form the
+ (\302\273
exact
and Sn is
l)-simplex,
of
sequence
En+1. For
example, En+l to topologically equivalent
the
pair
{En+l,
=?
Sn). We have
\342\200\242\342\200\242
^^__
^^
=o
Sn.
complex
simplicial
homology
,
for
result
(2)
gives Hk(En+l) = 0 for k > 1. We En+1 as an (\302\273 have indicatedthis on diagram (2). Viewing and Sn as its + l)-simplex < for we see that Ck(En+l) k
for 1 <
k
<
n. The
fact
that
En+l
is contractible
P1P2
\342\226\240 \342\226\240 \342\226\240+
Cn+i(S\.
Pn+2
For 1 < k < n, the exact sequence in the = 0, we see that Hk(Sn) = 0, for from Hk(E\"+l) I,*) =
(kernel
But from Hk+i(E\"+l, S\") = = (image i*k) 3^+1), so Hk(Sn)
The following
chain
0, we =
of reasoning
= 0.
d^+i)
(image
have
Z. Refer
to diagram
= 0, we have
=
=
j*\342\200\236+i) 0
5.
By exactness, (image
(kernel =
(kernel 8\302\273\342\200\236+i)
(2) above.
7*\342\200\236+i)0.
(image
by exactness,
that
~
Since Hn(En+l)
that
From exactness,(kernel
3*\342\200\236+i)Z.
4.
us
Hk(Sn).
< k < n. leads to Hn(Sn) ~
= Hence (kernel 9*\342\200\236+i) (image isomorphic mapping.
3. Therefore(image
that
diagram (2) tells
0 for 1
= 0,we 1. SinceHn+i(En+l) 2.
see
row of
last
=
;*\342\200\236) Hn{Sn).
so Hn(S\") ;*\342\200\236),
~ Z.
is,
is an 3H,\342\200\236+i
Section 44
we
Thus
see that
SinceS\"
Hn{Sn)
is connected,
~ Z and = 0 for Hk{Sn) ~ Z. This fact Ho(Sn)
1 < k could
Exercises
387
< n. also be
deduced from the
exact
sequence -H
Hi(\302\243\"+1,S\")
H0(Sn)
-\302\276 H0(\302\243\"+1)
-\302\276
H0(En+1,Sn).
=0
~Z
=0
44
\342\226\240 EXERCISES
Suggested Exercises
1. Let
S be
A and
is exact.
2. LetA,
additive
C be
B, and
is exact.Show that a. j maps B onto
3. Let A,
the
that
sequence
B.
additive
the sequence
and suppose that
groups
C
isomorphism of A
b. i is an
c. C is
A ~
that
Show
and suppose
groups,
B
into
isomorphic B/ B, C, and D be additive groups i [A]
to
and
let
aXbXcXd
be
exact
an
sequence.
a. i is onto
4.
maps
c.
is a
a.
h and
More
are equivalent:
map
j
sequence of additive
groups,
both
onto 0
map
everything
isomorphism of C onto B and k is one to one c. g is onto i
conditions
if
is an exact
b.
three
onto 0
of B
all
one-to-one
that
Show
following
B
b. j k
Show that the
is an
then the
following are equivalent:
D
Exercises
5. Theorem44.4and 6. In a computation for the torus
homology
be the
X
groups
homology
Theorem
44.7
are closely
connectedwith
Exercise
39 of
Section 14. Show
the
connection.
Examples 44.9 and 44.12 of the text, find the relative homology groups Hn(X, a) in Figs. 42.13 and Fig. 42.14.(Sincewe can regard these relative a, as shown as the homology groups of the space obtained from X by shrinking a to a point, these should groups of the pinched torus.)
to analogous with subcomplex
Part VIII
388
7. For
the
complex by direct
simplicial
(X, a)
and
in
Groups
verify
8. RepeatExercise6 with X of the pinched Klein bottle.) 9. For
Topology
X and subcomplexa of Exercise 6, form the exact homology computation that this sequence is exact. the Klein bottle of Fig. 43.2and Fig. 43.4. (This should give
the simplicial complex X and subcomplex and verify by direct computation that
(X, a)
10.
Find
the relative
consisting of the 11. For the (X,
a of Exercise 8, form is exact. sequence X is the
Y), where
homology groups Hn(X, circles. two boundary
the
verify
exact
the
exact
the
homology 42.11
annularregionofFig.
X and subcomplexY of Exercise 10, form complex is exact. this sequence that by direct computation
simplicial
Y) and
this
and
homology
of the
sequence
homology
groups
sequence of the Y
is the
pair
pair
subcomplex
sequence of the
pair
12. ProveLemma 44.14
13. Prove
Lemma
14. Prove
Lemma 44.17
15. Prove
Theorem
44.19
a. Show b. Show
c. Show
(image y'rf)
d.
f. Show
Let (A, and
gk
by means of
the
(image i*k)
C (kernel
j*k).
(kernel
\302\243 (image
i*k).
C (kernel
8*k).
y'rt)
(kernel d*k) C
Show
e. Show
16.
44.16
(image
(image 3rt) C (kernel (kernel itk-i) c (image 9> and
(A',
9'> be
\342\226\240 such Ak ->\342\200\242
A^
chain
y'^). i**_i). 9**).
/ Dk
and
this g, that
condition
is, if /
\"
by / and
g commute
: Ak ->-
/*(c)
(One abbreviates
and let
complexes,
that both
collection D of Homomorphisms
between / and into Hk(A').
steps.
following
g are
A^+]
S*(c) \342\200\224 =
g
=
and of homomorphisms fk : Ak -> A'k g be collections between / and g is a with 9. An algebraic homotopy such that for all c e Ak, we have
/
d'k+](Dk(c)) +
3D +
homotopic,
Dd.) then
D,_,(9,(c)). that
Show f*k
and
grt
if there
are the
exists
an
algebraic
same homomorphism
homotopy of Hk(A)
part
Factorization
IX
45
Section
Section 46
section
45
Factorization
Euclidean
Domains
Gaussian
47
Section
Unique
and
Integers
Z is
domain
integral
Norms
Multiplicative
Domains
Unique Factorization The
Domains
standard
our
example
of an
in which
domain
integral
there is
(irreducibles). Section 23showedthat for a field F, F [x] is also such an integral domain with unique factorization. In order to discussanalogous in an arbitrary integral ideas we shall give several definitions, some of which domain, are repetitions of earlierones.It is nice to have them all in one place for reference.
45.1 Definition
Let R b =
Definition
a commutative
be
ac,
a
then
ring
divides
b (or
An element
u
of a
commutative
a factor
a, b e
R. If
c e
exists
there
by a
b), denoted
of
relation
equivalence
The only
in
units
ring with unity R is a unit Two elements a, b e R
in R.
Exercise 27 asks us to
Example
and let
unity
a is
such
R
Definition
\"a
that
show
this criterion for
u divides 1, that is, if u in R if a = bu,
of R if are
associates
a and b to be associatesis an
on R.
\342\226\240
Z are 1
and
-1.
the
Thus
only
of 26
associates
in
Z
are
-26. 45.4
that
a\\b as
We read
\\ b.
\342\226\240
has a multiplicative inverse where u is a unit in R.
45.3
with
not divide b.\"
does
45.2
into primes
factorization
unique
26 and A
A nonzero if in every Note
a unit
u,
element p factorization
that then
is not
that
p
an associate any
a unit of
= ab in
factorization
D has
an the
integral property
of an irreduciblep is again of c provides a factorization
D is
domain
that either
an
an irreducible, of
irreducible
a or b is a unit. for
if
p
of D \342\226\240
uc for
p.
389
Part IX
390
Factorization
Note
\342\226\240 Historical
question of unique The integral domain other
was
first
the attempted in connection with Lame Gabriel (1795-1870)of Fermat's proof by Last Theorem, the conjecture that xn + yn = z\"has no nontrivial solutions for n > 2. It is not integral hard to show the conjecture is true if it can that be proved for all odd primes p. At a meeting of the Paris Academy on March 1, 1847, Lame announced that he had proved the theorem and presented a sketch of the proof. Lame'sidea was first to factor xp + yp over the complex numbers as
in public
raised
was
by no
x +
aky
Theorem
ay)(x + a2y)
y)(x +
(x +
a is a primitive
proposed
to
root of
pth
that if the
show
are relatively
unity.
factors in
if xp
this
expression zp, then each
=
to an
lead
would
positive integers,
an
descending
impossibility
Frey
the
truth
Lame
After however, doubts
on
sequence that would prove
gave
the purported proof, noting each of the relatively
was a
/7th
power
depended
uniquely
45.5
power
because
that prime
factors
a
pth
on the result that any integer can factored into a product of primes. It
Definition
An
D is
domain
integral
2.
If
p\\
number
\342\226\240 \342\226\240 \342\226\240 and
pr
irreducibles,
associates.
tragic
idea and
this
what
Conjecture
of Princeton
would
on
this
of lectures
University
problem at
1993
he
which
in
imply
it was
who,
the
finally after
for seven years,
Cambridge
University
announced
350-year-old problemwas now
a unique factorization
following conditions are satisfied: 1. Every element of D that
of a finite
and
Taniyama's
a proof of of the Tamyama-Shimura enough Conjecture to derive Fermat's Last Theorem. Unfortunately, a gap in the proof was soon discovered, and Wiles went back to work. It took him more than a year, but with the assistance of his student Richard Taylor, he finally was able to fill the gap. The result was published in the Annals of Mathematics in May 1995, and this
the
their product was
elliptic
Last Theorem.But
working
a series
seemingly curves
became known as the In 1984, Gerhard Conjecture. and in 1986 Ken Ribet proved that
of Fermat's
in June
finished his announcement, cast serious (1809-1882)
that
conclusion
asserted
two
clarified
Shimura
Tamyama-Shimura
secretly
of
Liouville
Joseph
31,
formulated
Andrew Wiles
theorem.
the
be
infinite
+
at age
Tamyama-Shimura
prime yp of the p factors must be a pth power. He could then demonstrate that this Fermat equation would be true for a triple x', y', z', e#ch than the number smaller in the number original triple. This corresponding and
relationship fields of mathematics, forms. A few years after
eventually
Shimura
Goro
and
between
a curious
\342\226\240\342\226\240\342\226\240{x+ ap~xy)
next
Yutaka Taniyama
1950s,
disparate modular
He
and by techniques of algebraic to Lame and Kummer. In the
unknown
death where
that
this unique
proved,
geometry
late
=
yp
had
was
noticed
xp +
means clear
of the form \"integers\" factorization property. Lame to overcome Liouville's Although attempted the matter was settled on May 24, when objections, a Liouville letter from Ernst Kummer produced in 1844 he had already proved that unique noting that factorization failed in the domain Z[a], where a is a 23rd root of unity. It was not until that Fermat's Last 1994
in an
factorization
than the integers
is neither
0 nor
solved.
domain (abbreviated
a unit
can
be factored
UFD)
if the
into a product
of irreducibles.
q\\ then r
\342\226\240 \342\226\240 \342\226\240 are
qs
two factorizations of the same element of D into pt and qt are qj can be renumbered so that
= s and the
Section45 Unique Factorization 45.6 Example
Theorem 23.20shows
we have in Z
made
use of
2 and
that
Recall
After just An D
fact,
we know that Z
Also
we have
although
is a UFD:
never provedit. For
example,
in these
two
the principal
one more definition
domain D integral is a principal ideal.
= (-2)(-3)(2)(2). (2)(2)(3)(2) as are
\342\200\2242 are associates,
factors
irreducible
45.7 Definition
this
we have
24 = Here
F[x] isa UFD.
field F,
for a
that
frequent
391
Domains
3 and \342\200\2243. Thus
ideal (a) of we
are
associates, the
for order and
except
of 24
factorizations
same.
the
\342\226\262
of all multiples of the element a. what we wish to achieve in this section.
consists
D
can describe
is a principal ideal domain
PID)
(abbreviated
if every
ideal in \342\226\240
ideal is of the form \302\273Z, that Z is a PID becauseevery We know generated by some is a PID. Our purpose in that if F is a field, then F[x] integer n. Theorem 27.24 shows theorems: is to prove two exceedingly this section important
1. 2.
45.17) Every PID is a UFD.(Theorem is a UFD. (Theorem 45.29) If D is a UFD, then D[x]
F is a field (by Theorem 23.20),illustrates both that F[x] is a UFD,where since F has no nonzero elements For by Theorem 27.24, F[x] is a PID.Also, definition for a UFD. Thus Theorem 45.29would that are not units, F satisfies our give a for that is another UFD,except the fact that we shall actually use Theorem F[x] proof of a Theorem 45.29. In the following section we shall study 23.20 in proving properties domains. class of the Euclidean certain special UFDs, Let us proceedto prove the two theorems. The
fact
theorems.
Every
PID Is a UFD
leading up to Theorem 23.20 and its proof indicate the way for our proof of steps handled the Theorem 45.17. Much of the material will be repetitive. We inefficiently of in Theorem it and was the case since was 23.20, F[x] special separately easy only The
we needed for
in general. D is a UFD, it is necessary domain to show that both integral 1 and 2 of the definition of a UFD are satisfied.For our special Conditions case of Condition 1 was very easy and resulted from an argument 23.20, F[x] in Theorem of degree > 0 into a product of two nonconstant that in a factorization of a polynomial of factor was less than the degree of the original the each degree polynomial. polynomials, case
To prove that
Thus
we couldn't
our
field
theory
an
keep on factoring
45.8 Definition
If all
{A, | i e 1} x such that
without
indefinitely
is, polynomials of degree0. For the general turn to this problem. We shall is so. We now
case of need
running
a PID, it
one more
is a collectionof sets, then the union U,e/A, x e A,- for at least one i e I.
into
is harder
unit
factors,
to show
that
that this
set-theoretic concept.
of the
sets
A, is
the set of \342\226\240
Part
392
IX
45.9 Lemma
Factorization
a commutative ring and let Ni c Then N = U; A7; is an ideal of R.
R be
Let
in R.
Proof
Let
Then there are A7,- c A^ or A7; c
e N.
a,b
either
Now
\342\200\242 \342\200\242 \342\200\242 be an
c
N2
ascending
chain of idealsA7,-
ideals A7;
and a e A7; and Nj in the chain, with us assume that A7,- c A7;, so both a and
A7,-; let
b e
Nj. in
& are
a = 0, A7;. This implies that a\302\261b and a& are in Nj,so a\302\261b and a& are in N. Taking we see that b e N implies \342\200\224b e A7, and 0 e N since0 e A7,-. Thus A7 is a subring of Z). since A7, is an ideal, For a e N and d e D, we must have a e A7,- for some A7,-. Then da = ad is in Nt. Therefore, da e U^A7;, that is, da e A7. Hence N is an ideal. \342\231\2
45.10
Lemma
Proof
\342\200\242 \342\200\242 is an Chain Condition for a PID) Let D be a PID. If Ni c A^2 c \342\200\242 = Ns for r that chain of then there exists a such ideals A7,A7,-, positive integer ascending in all ^ > r. Equivalently, proper) every strictly ascending chain of ideals(all inclusions a PID is of finite length. We express this by saying that the ascending chain condition holds for ideals in a PID. (ACC)
(Ascending
By Lemma is
r
N = U,- N is an ideal of D. Now as an ideal in D, e D. Since N = U^A7,, we must have c e Nr, for
we know that
45.9,
a PID, N = (c) for e Z+. For .s > r, we
c
some
=
A7,
A7, for
>
.s
The equivalence In what
follows,
c
{a} c =
(a)
For the
only by
(b) if
Proof
Let
D be
Let
a
e D,
factor. If a neither
only
if a
where a
a\\ nor
element
is neither
b\\
&
(a)
associates
c (flj) and
for
elements
only
if a
a and b
of a domain
D,
c (b) if
and
& divides
a. Using
ad for some and c are units
b = d
the
neither 0 nor a unit
a unit.
We first
done. If a is not
show
first
c,d e D.
is
true
if and
we see property, But then a = adc
that and
and & are associates. a UFD for an integral domain
so a of
definition
e (&), which
this
in D
that a
is a product
has
an irreducible,
at
least
then a
of irreducibles. one irreducible
=
a\\b\\,
where
unit. Now
follows from
b\\ would
\342\231\2
and
(a) C for
(c).
associates.
1 of
that is 0 nor
a, are
that
be and dc. Thus
we are
irreducible, is a
D,
so
Condition
prove
is an
and
that (a)
note
a PID. Every
to remember that
if b divides
some d e and only if a = we obtain 1 =
We can now that is a PID.
45.11Theorem
be useful
only
property,
A7 =
is immediate.
ACC
the
bd for
=
canceling,
and
(b) if and
first
\\ia
= {a}
(&) if
c
A7,
r.
with
will
it
some
have
Thus
which
be a
unit,
(fli),
a\\b\\, and if (a) = (^), then a and fli would be to construction. then, contrary Continuing this procedure
a =
a\\, we arrive
with
now
starting
at
Unique Factorization Domains
45
Section
a strictly
(a) C (fli) By
Lemma 45.10, this
the ACC in
irreducible. Thus a
C
{a2) C
ideals
\342\200\242\342\200\242\342\200\242.
with some
terminates
chain
(ar),
and
ar
then be
must
factor ar.
an irreducible
has
chain of
ascending
393
a that is neither 0 nor a unit in D. By what we have just proved, for an element or a = pic\\ for pi an irreducible and c\\ not a unit. either a is irreducible By an argument similar to the one just made, in the latter case we can conclude that {a) c (cj). If c\\ is for an irreducible not irreducible, then c\\ = /\302\276\302\276 p2 with c2 not a unit. Continuing, we chain of ideals get a strictly ascending
{a) C
(ci)C
(c2)
C---.
This chain must terminate, by the ACC in Lemma irreducible. Then a = p\\pi \342\226\240\342\226\240 -pr
45.12 Lemma
The
results
(Generalization of p is
Proof
completes
to Condition
qr
is an
that
\342\231\246
are of
some
An ideal
27.25)
(p)
in themselves.
interest
in
a PID
is maximal if
and
if
only
irreducible.
an
Let {p)
be a maximal
ideal
(a) = (p). then we must
that
Suppose
{a)
Theorem
cr =
some
of a UFD. Let us of Condition 1 of the definition are parallel to those leading to Theorem23.20.
our demonstration 2. Our arguments here we encounter along the way
This turn
45.10, with
=\302\243 (p),
are associates,
irreducibleof
of D,
= ab in D. Then (p) c p be a unit. If be associates, so b must then since is maximal. But a and 1 D, {p} if p = ab, either a or b must be a unit. Hence p is an
a PID.
Then a and have (a) =
so a isa unit.
Thus,
(a).
Supposethat
p would
(1)=
D.
suppose that p is an irreducible in D. Then if (p) c [a), we must have is a unit, then (a) = {1) = D. If a is not a unit, then b must be a unit, bu = 1. Then /?w = a&w = a, so (a) c and so there exists u e D such that we have = = = c D that either and Thus or (a) (p). {a} {/?) ^ D or p {/?) (a) implies (a) Conversely,
p = ab.
Now
if a
{p),
would
45.13
Lemma
45.14
Corollary
Proof
a unit.
(Generalization either
Proof
be
p \\ a or
(p),
Hence {/?)is a maximal
of Theorem b. p |
ideal.
\342\231\246
27.27) In a PID, if
an
irreducible
p divides
ab,
then
LetZ)beaPIDandsupposethatforanirreducible/7inZ)wehave/7\\ab.Then(ab) e (p). Since maximal ideal in D is a prime ideal every by Corollary 27.16, (a&) e {/?) implies that either a e (p) or b e (p), giving either p \\ a or p \\ b. \342\231\246 is an irreducible in p | a,- for at least one i.
If p
Proof induction.
of this
corollary
a PID
and p
divides
is immediate from
Lemma
the
product
45.13
a^ao_
if we
\342\226\240 \342\226\240 \342\226\240 for
an
a-t
e D,
then
use mathematical \342\231\246
394
Part
IX
45.15 Definition
Factorization
A
p | ab implies
p of
element
nonunit
nonzero
p \\ a or
either
p
an
D is
domain
integral
a prime if,
all a,
for
b e
D,
\\ b.
\342\226\24
45.13 focused our attention on the defining property of a prime. In Exercises Lemma 25 and 26, weaskyou to show that a prime in an integral domain is always an irreducible and that in a UFD an irreducible is also a prime. Thus the concepts of prime and irreducible coincidein a UFD. Example 45.16 will exhibit an integral domain containing some that are not primes, so the concepts do not coincide in every domain. irreducibles
45.16 Example
F
Let
be a
field
and
let
the subdomain F[x3, xy, y3]
D be
in D,
y3 are irreducibles
show
The
defining
property of Condition 2
Theorem45.17by 45.17 Theorem
Proof
we see that
xy divides x3y3 but not x3 or y3, that neither x3 nor y3 is a prime.
of factorization,
45.11
Theorem
a prime
Theorem
shows
not a
xy is
x3, xy,
and
of
the uniqueness
Every PID
23.20)
that if D is
prime. Similar
arguments
\342\226\26
what is is precisely of a UFD. definition
the
in
demonstrating
(Generalization of
y]. Then
F[x,
= (xy)(xy)(xy).
(x3)(y3)
Since
of
but
a PID, then
needed to establishuniqueness We
now
factorization
the complete in a PID.
proof of
is a UFD.
each
a e
D, where a
is neither
a
0 nor
unit, has a factorization a =
It remains
irreducibles.
into
for us to
pr
pip2---
show
Let
uniqueness.
a =
qiq2 \342\226\240\342\226\240\342\226\240qs
Then we have p\\ such factorization into irreducibles. that for some 45.14. implies p\\ \\ qj j by Corollary By changing we can assume that j = 1 so p\\ \\ q\\. Then q\\ = p\\u\\, necessary, irreducible, u\\ is a unit, so p\\ and q\\ are associates. We have then be
another
the
qs),
order and
of the
since
q}-
if
p\\ is an
= P\\u\\qi---qs,
P\\Pi---Pr
so by
\342\226\240 \342\226\240 \342\226\240 which
\\ {q\\q2 the
law in D,
cancellation
Pi--- Pr =uiq2---qsthis
Continuing
process,
starting with p2 1 =
Since
the qj Example
45.17is
false.
Ulu2
are irreducibles,we must 45.31 That
at the
end of this
is, a UFD need not
and
so on,
we
finally
arrive
at
\342\226\240\342\226\240\342\226\240Uyqr+i ---qs.
r =
have section be
s.
will
a PID.
\342\231\246
show that the
converse to Theorem
Section 45
Many algebra texts start have assumed that you were
Unique Factorization
395
Domains
the following corollary of Theorem 45.17. We with this corollary and used it freely in our other
proving
by
familiar
work.
45.18Corollary Proof
seen
that all ideals
We
have
and
Theorem It is
worth noting
We
proved
Theorem
we shall
We
the
that
examine this
start
of the
= form \302\273Z
(n) for n
a UFD.
e Z. Thus Z is a PID. \342\231\246
the
is a PIDby more
parallel
D[x]
Is
is a PIDwas
that Z
proof
6.6 by using
Is a UFD, then
now
Z are
in
Z is
domain
integral
applies.
27.24, that F[x]
Theorem
If D
45.17
The
of Arithmetic)
Theorem
(Fundamental
division
algorithm for F[x].
the division
using
way back in Corollary for Z exactly as we proved,
really
algorithm
In
Section
6.7. in
46,
closely. a UFD
the proof of Theorem 45.29, our is as follows. Let D
main
second
bea UFD.
result for
this
section.
The
a field of quotients F is a UFD by Theorem and we shall show that we can recover 23.20, a factorization for f(x) in F[x]. It will be necessary to e D[x] from its factorization of course. This approach, which in F[x] with those in D[x], comparethe irreducibles we prefer as more intuitive some more efficient modern ones, is essentially than due to idea of the argument of D. Then F[x]
We
can form
Gauss.
45.19 Definition
\342\226\240be nonzero Let D bea UFD and let a\\, ao_, \342\200\242 d of D is elements of D. An element \342\200\242, an \342\200\242 = a greatest common of d divisor all of the a, if \\ a; for i n 1, \342\200\242 \342\200\242, (abbreviated gcd) and any other d' e D that divides all the at also divides d. \342\226\240
we called d \"a\" gcd rather than \"the\" gcd because gcd's are only \342\200\242 units. Suppose that d and d' are two gcd's of a,- for i = 1, \342\200\242 n. Then \342\226\240, d | d' and d' \\ d by our definition. Thus d = q'd' and d' = qd for some q, q' e D, so Id = q'qd. By cancellation in D, we see that q'q = 1 so q and q' are indeed units. The technique in the example that follows shows that gcd's exist in a UFD. In this
defined
45.20Example
definition,
up to
we obtain 420 = 22 \342\200\242 gcd of 420, -168, and 252 in the UFD Z. Factoring, \342\200\242 \342\226\240 \342\200\242 = 23 \342\200\242 = and 32 252 22 7. choose one of these We 7, numbers, (-3) of its of each irreducible factors to say 420, and find the highest power (up associates) that divides all the numbers, 420, -168 and 252 in our case. We take as gcd the product of these highest powers of irreducibles.For our example, these powers of irreducible factors of 420 are 22, 31, 5\302\260, and 71 so we take as gcd d = 4 \342\200\242 3 \342\226\240 1 \342\200\242 7 = 84. The only other \342\226\262 units. gcd of these numbers in Z is -84, because 1 and -1 are the only Let
us
find a
3 \342\226\240 5 \342\226\240 7, -168
Execution element
of a
Section 46 will gcd's
without
of the
UFD
into
exhibit
factoring
technique
in
a product
a technique,
in a
Example
45.20
of irreducibles.
depends
on being able to
This can be a tough
the Euclidean Algorithm,
class of UFD'sthat
includes
that
Z and
will
factor an
in Z. allow us to find job,
even
F[x] for a field F.
Part IX
396
45.21
Definition
45.22
Example
Factorization
be a UFD.A
Let D
nonconstant
in D[x]
is primitive
if 1 is
In Z[x],
4x2 + 3x +
2 is
polynomial
f{x)
=
a gcd
of the
that every
Observe
45.23
Lemma
in
Proof
\342\200\242 \342\200\242 n. \342\200\242,
0,1,
+ 2 is not,
+ 6x
since
\342\226\
2, a
nonunit
in
Z,
2.
is a
\342\226\
nonconstant irreducible in
must
D[x]
be a primitive
polynomial.
nonconstant f{x) e D[x] we have f{x) = (c)g(x), where and g(x) is primitive. The elementc is unique e D, D[x], up to a unit factor is unique up to a unit factor in D. D and is the content of f(x). Also g(x)
If D is
c
+ anxn
i =
at for
but 4x2
primitive,
common divisor of 4, 6, and
\342\200\242 \342\200\242 \342\200\242
&\\x +
+
flo
a UFD,then
for
g{x)e
e D[x]be given
Let fix) flo, fli,
where
nonconstant
;' =
at for
the
for some qt e D. By the in D divides all of the
= cq; irreducible
is a
f(x)
a gcd of
c be
\342\200\242 \342\200\242 Let \342\200\242, an.
at
distributive
For uniqueness,
if
also
go, q\\,
ford e
= (d)h(x)
f(x)
of c
each irreducible factor
D, h(x) e D[x],andh(x)
d and
divide
must
i, we have = f{x) (c)g(x), where no \342\226\240 \342\200\242of \342\200\242, qn g(x). Thus g{x) is a for each
we have
law,
coefficients
coefficients
with
polynomial
\342\200\242 n. Then 0, 1,\342\200\242 \342\200\242,
polynomial.
primitive
then
every
conversely. By
primitive, (c)g(x)
setting
=
= (v)h(x) for of c into d, we arrive at (u)g(x) (d)h(x) and canceling irreduciblefactors a unit u e D. But then v must be a unit of D or we would be able to cancel irreducible factorsof v into u. Thus u and v are both units, so c is unique up to a unit factor. From
f(x) = (c)g(x),we
polynomial g(x) is also unique
the primitive
that
see
up
factor. 45.24
Example
\342\231\
InZ[x],
4x2 + where
45.25
Lemma
Proof
to a unit
2x3
6x -
=
8
+ 3x
(2)(2x2
- 4),
\342\200\224 4 is
+ 3x
primitive.
(Gauss's Lemma) is again primitive.
\342\226\
If D is a UFD,then
a product
of two
primitive
polynomials
in D[x]
Let =
f{x)
flo
+
d\\X +
h fl\342\200\236x\"
and
be
primitive
doesnot
divide
in D[x], and all a; and
ar be the first (that is, /7 does of xr+s in /z(x)
-r+s
coefficient not divide =
let
(flo^+i H
= f(x)g(x).
h(x)
p doesnot of f(x)
ar). is f{x)g(x)
+ ---
= b0+biX
g{x)
not
h ar^bs+i)
bmxm
be an irreducible in D. Then p since f(x) and g(x) are primitive. Let by p; that is, p \\ at for i < r, but /?f ar The coefficient | bj for ;' < s, but p{^. Let p
all bj,
divide
Similarly,
+
divisible let
p
+ arbs
+ (ar+1^_! H
h ar+A)-
Now
Unique FactorizationDomains
45
Section
for i < r implies that
p | a,
p | (a0br+s
and also
p
s implies
for 7 <
\\ bj
h ar-ibs+i),
H
that
45.26
Corollary
Proof
This shows
cr+i.
f(x)g(x)
divisible
not
If D
is a UFD,then
This
corollary
F[x] is a UFD.
a finite
p doesnot
given
of
product
45.25 by
F be
let
said earlier,
Let D be
a UFD
and let
/(*))
> 0. If
fix) is an
Also,
if
is primitive
fix)
F be a field
\342\231\246
a field of
\342\231\246
of
quotients
that
show
Theorem 23.20,
D. By
is a
D[x]
UFD
by
D[x]. The next lemma F[x]. This is the last important one in
into
of
is again primitive.
in D[x]
induction.
we shall
of /(x) e D[x]back nonconstant irreducibles of D[x]to those 45.27 Lemma
and consequently p doesnot there is some coefficient of
p e D,
polynomials
primitive
in F[x]
factorization
ar&s,
Thus f(x)g(x) is primitive.
UFD and
we
divide
an irreducible
from Lemma
be a As
that
by p.
follows
let D
Now
ar or bs,so
not divide
does
/7
divide
h ar+sb0).
H
p I (ar+A_i
But
39\"
carrying
a
relates
the
step.
of D. Let _f(x) e \302\243>[x],where (degree then _f(x) is also an irreducible in F[x]. in F[x], then _f(x) is irreduciblein irreducible
of quotients
in D[x],
irreducible
in D[x]
and
D[x].
Proof
Suppose that
F[x],
that a
nonconstant
factors
e D[x]
f(x)
r(x),
r{x) and
of lower
polynomials
degree
in
is,
f(x) = for
into
s(x)
e F[x].
s{x) is of the
Then since F form
a/b
is
r{x)s{x)
a field
of quotients
for some
a, b e
id) f{x)
= riix)syix)
D. By
clearing
of D,
each
coefficient
denominators,
in
we can
get
for
siix)
of ri(x) and siix) are the degrees D, and ri(x), ^i(x) e Dfx], where the degrees = ic)gix),r\\ix) = ic\\)r2ix), and six), respectively.By Lemma and 45.23, fix) = ic2)siix) for primitive polynomials gix), r2ix), and s2ix), and c, cuc2 e Z).
d e
of r(x) Then
idc)gix) and c\\c2
by Lemma = dcu for
= iclc1)r2ix)s1ix),
45.25, r2(x)s2(*) is primitive. some unit u in D. But then idc)gix)
By the
uniqueness
= idcu)r2ix)s2ix),
part
of Lemma
45.23,
398
Part IX
Factorization
so
= (c)g(x)
f{x) We
that if fix) factors nontrivially of the same degrees in D[x].
shown
have
into
= icu)r2ix)s2ix).
polynomials
then fix) factors nontrivially if fix) e D[x] is irreducible
in F[x],
Thus
be irreduciblein F[x]. Dfx], A nonconstant fix) e D[x] that is primitive irreducible in Dfx], since D[x] c F[x].
in
it must
shows that if D is
45.27
Lemma irreducibles
the
irreducible
in D,
together with in F[x], where F is a field
in D[x] and
is also
F[x]
\342\231\246
a UFD, the
in D[x]
irreducibles
nonconstant
the
irreducible in
are precisely
polynomials
primitive
are
that
of Dfx].
of quotients
is very important in its own right. This is indicated by the The precedinglemma was our Theorem 23.11. (We admit that it corollary, a special case of which of a lemma a theorem. does not seem very sensible to call a specialcaseof a corollary on the context in which it appears.) The label assigned to a result dependssomewhat
following
45.28Corollary
If D
is a UFD and
into
a product
a factorization Proof
It was
into in
shown
same degreesin
45.29 Theorem
Proof
If D
of the
of Lemma
degree
in
(see the
Dfx]
since
holds
Dfx]
is a
then a nonconstant
degrees
same degreesr
45.27 that
s
r and
main
then
if
fix)
in
fix)
Dfx].
into a product
factors
it has
into first
factors has
e Dfx]
if and only if it
F[x]
s in
and
a factorization next to last sentence of the Dfx] c F[x]. F[x],
prepared to prove our
is a UFD,then
of D,
polynomials of lower
the proof
The converse
are now
field of quotients
polynomials
of lower
polynomials
We
is a
F
of two
of the
polynomials
paragraph
of the
of two
proof).
\342\231\246
theorem.
UFD.
Let fix) e Dfx], where fix) is neither 0 nor a unit. If fix) > 0. Let since D is a UFD. Suppose that (degree fix))
is of
degree 0, we are
done,
\342\226\240 \342\226\240 \342\226\240 fix) = gi(x)g2(x) gr(x)
factorization of fix) in Dfx] having the greatest number r of factors of positive of such factors because r cannot exceed the (There is such a greatestnumber Now factor each giix) in the form g,(x) = c,- ht(x) where c,- is the content degree of/(x).) of giix) and /z,(x) is a primitive polynomial. Each of the ht(x) is irreducible, because if it could be factored, none of the factors could lie in D, hence all would have positive of giix), and then to a factorization of degree leading to a correspondingfactorzation with more than r factors of positive degree, contradicting our choice of r. Thus we fix)
be a
degree.
now
have
= Cihiix)c2h2ix)
fix)
the ht(x)
where
we obtain The
are irreduciblein
a factorization
factorization
UFD; see the
comment
of fix)
of fix)
e
following
\342\226\240 \342\226\240 \342\226\240
crhrix)
now factor the c, into irreducibles of irreducibles in Dfx]. Dfx], where fix) has degree 0, is unique since Lemma 45.27. If fix) has degreegreater than Dfx].
into
If we
in D,
a product
D is
a
0, we
Exercises
Section 45
of any factorization is, the factors in
view
can
into units
By Theorem in F. But as that
45.30
have
We
that
example
45.31
Let
Example
F be
a unit
a field
Then N is
an
Another
up to
unique
in
by
see (by
x and
let
and
the set N
but not
ideal,
of a
example
y
of all
be
factors
polynomials
that
D[x]
associates.
\342\231\246
\342\226\240 \342\226\240 is a \342\226\240, x\342\200\236]
UFD.
\342\226\240 \342\226\240 is a \342\226\240, x\342\200\236]
F[x\\,
UFD.
it easy for us to
makes
F[x,y] is not
a PID is
not
is
that
that is, associates. is the content of f(x).
Then F[x, y] is a UFD by and y in F[x, y] having constant
a principal ideal. Thus
UFD
the
= F[xux2]. 45.29, so is (F[xi])[x2]
induction)
inx
in
45.23, this
factors,
f(x)
then F[xi,
indeterminates.
F[x] 45.27.
Lemma
45.23. Thus all irreducibles in
order and
By Theorem
a UFD.
procedure, we
unit
of
by Lemma
factor
are
up to
D[x] factorization
45.30 seen that a PID is a UFD.Corollary shows that not every UFDis a PID.
Consider
45.30.
polynomials
\342\200\242are \342\200\242 \342\200\242, x\342\200\236indeterminates,
x\\,
in this
Continuing
in
the
in D in
irreducibles
By Theorem23.20,F[xi]is
Proof
in F[x]
>0
unique up to in the factorization
is a field and
If F
Corollary
irreducible
are unique, except for possible constant in D[x], each polynomial of degree appearing is primitive. By the uniqueness part of Lemma
irreducible
an
is again
appearing
as a factorization
in D[x]
irreducibles
into
and
23.20, these polynomials
The product of the which
D)
of f(x) in D[x] these polynomials are unique
factorization shows
f(x)
(that
399
\342\231\246
give an
Corollary
0.
term
a PID.
Z[x], as
shown
indicated
domain.
\342\226\262
Exercise
in
12,
Section 46.
45
\342\226\240 EXERCISES
Computations Exercises
hi
8, determine
1 through
whether
the
irreducible of the
is an
element
1. 5inZ
2. -17inZ
3.
4. 2x
in Z
14
5. 2x
- 10in
7. 2x-
6. 2x
Z[x]
8. 2x -
10inQ[x]
9.
If possible,
10.
Factor the domain
give four the
11 through
In Exercises
associates
different 4x2
polynomial
Z[x]; of
integral 13,
hi
Exercises the
14 through
indicated
14. 18x2-
of 2x
+ 8 into domain Q[x];
\342\200\224 4x
find all
gcd's of the
of
irreducibles domain
integral
given
Q[x]
10
inZn[x]
as an element
\342\200\224 7 viewed
a product of the
Z[x]
elements
of Z[x];of
viewing Zn[x].
17, express the
given
polynomial
it
as
16. 2x2 - 3x+ 6 in
Z[x]
of
the
integral
13.2178,396,792,594
as the product
15. 18x2 17.
of Zi i [x].
an element
of its
content
with a primitive
UFD.
12x+48inZ[x]
Q[x];
of Z.
12. 784,-1960,448
11. 234,3250,1690 in
- 3 in - 3 in
2x2
-
12x+48
- 3x + 6 in
inQ[x] Z7[x]
polynomial
Factorization
IX
Part
400
Concepts In
18 through it is in
Exercises
is needed, so that
a
Two elements
18.
of the italicized term 20, correct the denmtion a form acceptablefor publication.
an integral
b in
domain
D are
integral domain D is an
irreducible
and
in D if
associates
to the
reference
without
and
only
if their
text, if correction
quotient a/b
a
D is
in
unit.
19.
An
20.
An
21.
Mark
of an
element
and
of D if
integral domain D is a prime
of an
element
of D
if and
smallerelements of D. of the
each
following
a product
into
of two
a product
into
or false.
true
a. Every
field
is a
UFD.
b.
field
is a
PID.
Every
cannot be factored
cannot be factored
if it
only
if it
only
elements of D.
of two
c. Every PID is a UFD. d. Every UFD is a PID. e. Z[x] is a UFD. f.
g. If D is a PID,
then
h. If D is a UFD,
22. Let
in
two irreducibles
Any
i.
In
any
UFD,
j.
A
UFD
has
D[x]
then
are associates.
UFD
any
PID. a UFD.
is a
D[x]
is
for an irreducible no divisors of 0. if p | a
p, then
p itself appears in
a UFD. Describe the irreducibles in D[x] in terms F[x], where F is a field of quotients of D. D be
23. Lemma 45.26states
is also
an irreducible
in D
irreducibles
and
the
if D is a UFD with a field of quotients F, then a nonconstant irreducible f(x) of F[x]. Show by an example that a g(x) e D[x] that is an irreducible of F[x]
that
a commutative
of D[x]. in this sectionwas restrictedto integral ring with unity, consider factorizations
in particular
(1,0).
in
irreducibles
of D[x] neednot
irreducible
be an
24. All
of the
of a.
factorization
every
work
our
domains. into
Taking the same irreducibles in Z x
section but for Consider
in this
definition
Z. What
can
happen?
Theory
25. Prove
that
if p
is a prime
26. Prove that
if p
is an irreducible
27.
For a for
28. Let
in R
unit
D be an
of D. Show
29. Let
a UFD,
D,
p is
then
then p is
integral that the
be a
UFD.
Exercise 37, Section 18 showed nonunits of D excluding
domain.
set D*
an irreducible.
a prime.
R with unity show that the relation ring is an equivalence relation on R.
the multiplication D
in
commutative
u a
under
integral domain
in an
\342\200\224 U of
a ~ b if that 0 is
a is an
associate
of b (that
group where U is multiplication.
= bu
Is this
units set a group
again
a primitive
is a \342\226\240>
(U,
closed under
is, if a
the
set of
of D?
Show
divisor of
a nonconstant
that
a primitive
in D[x]
polynomial
is
polynomial.
30.
Show
31. Factor
that x3
in a
PID, every ideal is contained
\342\200\224 into
y3
There are severalother
ideals in
a ring.
in Q[x, y]
irreducibles concepts
The following
often three
in a and
considered
exercises
maximal ideal. [Hint:
prove that
concern
that each
of
the
Use Lemma45.10.]
factors
similar in character to some of these concepts.
are
is irreducible. the
ascending
chain condition on
Section46 EuclideanDomains
chain condition (ACC) for ideals holds in R if every strictly increasing in R is of finite length. The maximum condition (MC) for ideals in holds an ideal not properly set S of ideals in R contains in any other ideal of contained nonempty the set S. Thefinite basis condition holds in R if for each idealN in R, there is a finite for ideals set (FBC) \342\200\242 c N such that N is the intersection of all ideals of R containing BN. The set By is a finite Bn = {bi, \342\200\242 \342\226\240, bn} generating set for N. Show that for every ring R, the conditions ACC, MC, and FBC are equivalent. Let R be any ring. The descending chain condition (DCC) for idealsholds in R if every strictly decreasing \342\226\240 \342\200\242 of ideals in R is of finite The minimum condition (mC) for ideals Ni d N2 D N3 d \342\200\242 sequence length. in R if given any set S of ideals of R, there holds is an ideal of S that does not properly contain any other ideal in the set S. Show DCC and mC are equivalent. that for every ring, the conditions Give an example of a ring in which ACC holds but DCC does not hold. (See Exercises32 and 33.)
32. Let R
be
sequence
33.
34.
401
any
The ascending
ring.
Ni c N2 R if every
section
46
C N3
\342\200\242 \342\200\242 \342\200\242 of ideals
c
Domains
Euclidean We
several times on the importance of division algorithms. Our first them was the division was for Z in Section 6. This algorithm algorithm theorem that a subgroup used to prove the important of a cyclic group is shows at once that Z is a PID. The is, has a single generator. Of course,this
remarked
have
with
contact
immediately
cyclic,that
23.1 and was used in a completely appeared in Theorem a modern F[x] is a PID. Now technique of mathematics is to related situations and to try to bring them under one roofby abstracting take someclearly the important ideas common to them. The of this definition is an illustration following division
for F[x]
algorithm
way to
analogous
technique, as is this existence of a fairly 46.1
Definition
that
show
A Euclidean norm on an
of D into
the
division
For
2.
For all a,b
such
integers
see
can integral
function
Disa that
we
in an
what
algorithm
domain
integral
nonnegative
1.
Let us
text!
whole general
the
An integral
domain D
a nor
neither
is a Euclidean
domain
conditions
b is 0, v{a)
D such <
that
a =
bq +r,
v(ab).
exists a
if there
elements are satisfied:
the nonzero
mapping
r in
and
the
with
domain.
following
all a, b e D with b ^ 0, there exist q where either r = 0 or v(r) < v(b).
e D, where
v
by starting
develop
Euclideannorm
on D. \342\226\240
The
of Condition 1 is clear from our enable us to characterize the
importance
Condition 2 is that
46.2Example
The for
it will
domain integral n 7^ 0 in Z is a
Example
If F
is a field, then
(degreef{x))for Theorem 23.1, and is the sum of their
of a
The importance of Euclidean domain D.
Z is a Euclideandomain, for the function v defined by Euclideannorm on Z. Condition 1 holds by the division
for Z. Condition 2 follows 46.3
discussion. units
from
is a
F[x] f{x)
\\ab\\
=
\\a\\\\b\\
and
\\a\\
>
1 for
a ^
0 in
Z.
v(n) =
\\n\\
algorithm \342\226\262
v defined by v{f{x)) = Euclidean domain, for the function and f(x) ^ 0 is aEuclideannorm. 1 holds by Condition 2 holds since the degree of the product of two polynomials
\342\202\254 F[x],
Condition degrees.
\342\226\262
Part
402
IX
Factorization
Of course,we should opening
46.4 Theorem
Proof
some
give
ones that motivated the we anticipate remarks,
familiar
Every Euclidean
is a
domain
the
of Euclidean domains other than these do this in Section 47. In view of the
examples
We shall
definition.
theorem.
following
PID.
domain with a Euclidean norm v, and let N be an ideal in D. be a Euclidean = {0},then N = (0) and N is principal.Supposethat N ^ {0}. Then there exists b 7^ 0 in N. Let us choose b such that v{b) is minimal among all v(\302\273)for n e N. We there exist claim that N = (b). Let a e N. Then by Condition 1 for a Euclidean domain, and r in D such that q
Let D If N
a = bq =
whereeitherr
< v(b).Nov/r
Oorv(r)
is an ideal. Thus v(r) < v(b) is a was any element of N, Since 46.5
Corollary
Proof
\342\200\224
e iV,sofhatr
bqanda,b
impossible
by our
we seethat
iV =
choice of b.Hencer
\342\202\254 iV since
\342\231\246
(b).
Theorem
46.4,
domain is a PID and
a Euclidean
by
Theorem
45.17,
a PID
UFD.
is a
\342\231\246
we should
Finally,
mention
PID is a Euclidean
not every
while
that
domain.
a Euclidean
domain is a PIDby
of PIDs
Examples
that
are
Theorem
not Euclidean
46.4,
are
not
however.
found,
easily
iV
= 0,soa = bq.
domain is a UFD.
A Euclidean By
= a
+ r,
Domains
in Euclidean
Arithmetic
some properties of Euclidean domains related to their that the arithmetic structure of a Euclidean domain v on the domain. A Euclidean is not affected in any way by a Euclidean norm norm is tool for possibly throwing some light on this arithmetic structure of the merely a useful structure of a domain D is completely domain. The arithmetic determined by the set D on D. and the two binary operations + and \342\226\240 domain with a Euclidean norm v. We can use Condition 2 of Let D be a Euclidean a Euclidean the units of D. norm to characterize We
shall
multiplicative
46.6 Theorem
Proof
now
investigate
We emphasize
structure.
For a Euclidean with a Euclidean norm v, v(l) is minimal domain nonzero a \342\202\254 if and only if v{u) = v(l). D, and u e D is a unit Condition
2 for
v tells
us
at
once
that for
a j^0,
v(l) < v(lfl) On the other
hand,
if u
is a
unit
in D,
=
v(a).
then
v{u) < v{uu~l)
= v(l).
Thus
v{u)
for a unit
u in
D.
= v(l)
among
all v(a)
for
EuclideanDomains
Section 46
that a nonzero there exist q and r in
suppose
Conversely,
division
algorithm,
1 =
either r =
where
nonzerod e 46.7
0 or
D,
v{r)
v{u) is
such +
uq
that
= v(l).
v(w)
Then
by
r, =
v(w)
v(l)
is minimal
= 0 and
1 =
of v(n) for nonzero 1. Of course, 1 and
over all v{d) for
uq, so
w
unit.
is a
1 1, and 1 and \342\200\224 \342\200\2241 are exactly the units \342\226\262
For F[x]
Example
\342\231\246
Z is \302\273 \342\202\254
=
ofZ. 46.8
the
that
impossible. Hence r
For Z with v(\302\273)= \\n\\, the minimum are the only elements of Z with v{n)
Example
e
But since
< v(u).
v(r) <
such
D is
u
D
403
with
(degree
/(*)) for f(x)
f(x) e F[x]is 0.The of F,
elements
nonzero
=
v(f(x))
for all nonzero
these
and
nonzero
^ 0, the
the
are precisely
of
units
of v(f(x))
value
minimum
of degree
polynomials
0 are exactly
the \342\226\262
F[x].
that everything we provehere holds in every Euclidean domain, in in we can show that 45.20, F[x]. As indicated Example any a and b UFD have a one a and b into in a but irreducibles, gcd and actually compute by factoring such factorizations can be very tough to find. However, if a UFD is actually Euclidean, and we know an easily Euclidean norm, there is an easy constructive way to computed We
emphasize particular in Z and
find gcd's,
as the
theorem
next
Note
\342\226\240 Historical
Euclidean algorithm
Euclid's Book VII,
in
appears
The Elements as propositions1 and
2 of
where it is used as here to find the greatest common divisor of two integers. Euclid uses it again in Book X (propositions 2 and 3) to find the greatest of two
measure
common
determine
magnitudes (if
two
whether
it
and to
exists)
are
magnitudes
incommensurable.
The
and
mathematician
solve the integers,
indeterminate
Brahmagupta
\"reciprocallydivide\"
seventh-century Indian Brahmagupta. To rx + c = sy in equation uses Euclid's procedure to
r by
s
until
By then
substitution
procedure
based
algorithm
for
his equation.
Brahme-
astronomer
remainder.
using,
the
he reaches in effect,
on the various
he produces a finding the smallest
remainders,
the
of the
nonzero
and
in
(Correct Astronomical System (628)
Brahma)
again
appears
algorithm
sphutasiddhanta
of
shows.
of the
solution
Chinese Remainder Shushu jiuzhang
Sections) (1247). Qin's goal a method for solving the system of congruences N = r, (mod mi). As part of that method he neededto solve congruences of the form Nx = 1 (mod m), where iV and m are relatively The solution to a congruenceof this form prime. to display
is again
found
the
differentfrom
by
sources.
a substitution
Indian
remainders from to iV and m. It is
independentlyin
to
Qin in his
algebraist algorithm
in Nine
(Mathematical Treatise
was
Euclidean
quotients
Chinese Euclidean
so-called in the
problem published
element in
final
a
straightforward
positive solution
The thirteenth-century Jiushao also used the
one,
using
the
Euclidean
not
known
procedure, the
quotients
algorithm whether the
and Chinese algorithms, was discovered itself, algorithm these cultures or was learnedfrom the
Indian
and applied
common the
Greek
404
Part IX
46.9 Theorem
Factorization
Let D
Algorithm)
(Euclidean let
a and
that
is,
b be nonzeroelements
be a Euclidean D. Let
of
a =
-\342\200\224\342\200\224
where either
=
r\\
v(r{) <
0 or
v(b). If
bqi+ri, 0, let
r\\ ^
b = riq2 either
where
r2 =
0 or v(r2)
<
In
v(7^).
r;_i
with a Euclidean norm v, and Condition 1 for a Euclideannorm,
domain
as in
be
r\\
+
that
such
be
r2
r2,
let ri+i
general,
be such that
+ ri+u
=rtqi+i
\342\226\240 \342\226\240 \342\226\240 must terminate < v(?-,). Then the sequence where either rl+i = 0 or v(ri+1) r,, r2, with some rs = 0. If r\\ = 0, then b is a gcd of a and />. If n j^O and rs is the first r,- = 0, then a gcd of a and /> is rs_i. such that if d is a gcd of a and />, then there exist A. and /x in \302\243> Furthermore, cf
/Voo/
=
la
+ ///>.
Since v(r,) < If cf |
v(r,_!)and we
of steps
number
= 0, then
n
a =
is a
v(r,)
that after
it follows
nonnegative integer, rs = 0.
some finite
at some
arrive
must
is a gcd of
bq\\, and b
a and
/>. Suppose
n ^
0. Then
set as
the set of common
if cf | a
and
b, we have cf |
if di
However,
SQd\\ri.
\\ r\\
and d\\
(a
\342\200\224
bqi),
then
\\ b,
+ ri),
di | (bqi
sodi \\a.
Thus
the set
of common divi
sors
and b is the
of a
same
argument, if r2 7^ 0, the set of common divisors of b and this process, r\\ is the same set as the set of common divisors of r\\ and r2 \342\226\240 Continuing the set of common divisors of a and b is the same set as the set of we see finally that common divisors of rs_2 and rs_i, where rs is the first r, equal to 0. Thus a gcd of rs_2 and rs_i is also a gcd of a and b. But the equation divisors
of b and
r\\.
By
a similar
=
rs-2
qsrs-i
qsrs-i
+rs=
shows that terms
a gcd of rs_2 and is rs_!. rs_i It remains to show we can express that of the construction just given, if d =
d = rs-i, then, form l,r;_i +
backward
working /Xjr,_2
for
some
a gcd d
b,
then
of
d =
and
/>
0a +
lb
a
as and
d = Xa we
+ ///>.
are done.
In
If
through our equations, we can expresseach r,- in the D. To illustrate Xt, /x,- \342\202\254 using the first step, from the
equation
rs-3
=
qs-\\rs-i
+ rs-i
we obtain d =
rs-i
= rs_3
- qs-irs_2.
(1)
We then
rs_2 in terms of and rs_4. Eventually,
express
of rs_3
terms
d
=
=^\302\243l+M3Ii
=
+
X3b
can be expressed in the form d = d' = ud for some unit u, so d' = {Xu)a
46.10
Example
Let us
illustrate
gcd
22,471
the
of
(/x3
express d
in
- ^3^2)^1
+
If d'
[ib.
is
any
other
gcd of
a
and
b, then \342\231\246
{^u)b.
that is
of anything
that
(1) to
Eq.
it can
be implemented on
a computer. Of
labeled an \"algorithm.\"
norm | | on Z by computing a Euclidean algorithm for the Euclidean the division 3,266. We just apply algorithm over and over again, and 46.9 remainder is a gcd. We label the numbers obtained as in Theorem The computations are easily the statement and proof of the theorem.
the
and
nonzero
last
to further
Xa
+
Theorem 46.9is that
thing about
nice
course,we anticipate
+
in
bqi)
which
The
+ ^n = X3b
- nq2)
X3q2)(a-
-
(/23
we
>s(b
substitute
rs_4 and will have
and
rs_3
405
Domains
Euclidean
46
Section
illustrate
checked. a =
b=
22,471= (3,266)6 3,266
+
n=
2,875
= (2,875)1+391
r2
+ 138 2,875 = (391)7
391
(138)2
138 =
(115)1+23
=
'
115=
Thus r5
= 23 is a gcd
is important,
for
=
r6
138 115
r5 =
0
+
(23)5
r4
3,266 2,875
= 391
r3 =
+ 115
22,471
=
23 0
and 3,266. it is very difficult
of
22,471
sometimes
We found to
find
a gcd without factoring! This a factorization of an integer into \342\226\262
primes.
46.11
Example
the definition of a Euclidean norm says about r being \"positive.\" In computing a gcd in Z by the Euclidean algorithm as in Example 46.10, it is surely to our interest to make \\rt \\ as small as possible division. be more efficient to write Thus, repeating Example 46.10,it would
Note that the division nothing for in
I
I,
each
1 in
Condition
algorithm
a = 22,471 b
3,266= (391)8 391 = 138
We can
of r, and
- 391
= (3,266)7
22,471
= (23)6
change the are the \342\200\224r,
r2
23
r3 =
+ 0
sign
=
138
+
(138)3 -
=
n =
of r,
same.
r4
=
3,266
-391 138
-23 0
from negative to positive when
we
wish
since the
divisors \342\226\262
Factorization
IX
Part
406
46
\342\226\240 EXERCISES
Computations 1. The function 2. The
function
v for v for
of f(x))
term
4. The function
for
v for Q
f(x)
given by
7. Find a
8.
gcd of 49,349 the
Following
Z in the form 9. Find
and
n2 for
nonzero
v
is a
of f(x)) for /(x)
absolute value of the
= (the
norm for
Euclidean
the
domain.
integral
given
e Z
n
= (degree
\342\202\254 Z[x],
f{x)
^ 0
of the
coefficient
highest degreenonzero
= a2 for nonzero
v{a)
= 50
the
find values
actually
you get 23
for
the
in
in Example
computation
so substituting,
(391)8,
23
the gcd
express
actually
line of
a e
Q for nonzero a e Q
v(a)
last
-
before that, 138 = 3,266 work your way back up to
=
v(f(x))
46.11, the
function
e Z[x]
Q given by
to Example referring the From next to [#('\302\273\302\243.\342\226\240
given
v(f(x))
Z[x] given by nonzero
the
v(\302\273)
Z[x] given by
v for
function
whether
Z given by
v for
3. Thefunction
5. The 6. By
5, state
1 through
Exercises
In
X
and
A(22,471) + //(3,266) for X, From 46.11, 23 = (138)3-
form
391.
- 391,and = [3,266- (391)8]3
so on.
Z. /x \342\202\254 the
line
That is,
//.]
15,555 in Z.
idea of Exercise 6 and referring A(49,349)+ //(15,555)for A,/x
to Exercise
7, express the
positive
gcd of
49,349
and
15,555
in
\342\202\254 Z.
a gcd of -3x9
xw
- llx7
+3x*
+ llx6- llx5+19x4-13x3+ 8x2-9x+3
and
- 3x5 +
x6
3x4 -
- 5x +
9x3
+ 5x2
find
the gcd
2
inQM.
10.
how
Describe
the Euclidean
Algorithm can be usedto
of n
members
(\302\276, m.,
\342\226\240 \342\226\240 an of \342\200\242,
a Euclidean
domain.
11. Using
devised in Exercise
method
your
10, find
the
gcd of
2178, 396,792, and
726.
Concepts
12. Let us
consider Z[x].
a. Is Z[x] b.
Show
c. Is
a UFD? that
{a +
Why?
xf(x) | a
Z[x] a PID?(Consider
part
domain? Z[x] a Euclidean Mark each of the following true
d. Is
13.
\342\202\254 2Z, f(x)
a.
Every
Euclidean
\342\202\254 Z[x]}
is an
ideal
in
Z[x].
(b).) Why?
or false.
domain
is a
PID.
b. Every PID is a Euclidean domain. c. Every Euclidean domain is a UFD. d. Every UFD is a Euclidean domain. e. A gcd of 2 and 3 in Q is |. f. TheEuclidean algorithm gives a constructive method for finding g. If y
is a
Euclidean
norm on a Euclidean
domain
D, then v(l)
a gcd
of two integers.
< v{a) for all nonzero
a \342\202\254 D.
h. If visa
i.
on
norm
v(l) then
407
Norms
1.
for all nonzero a e D,a ^ for all nonzero nonunits
< v(a)
< v(a)
v(l)
D.
ae
j. For 14.
Euclidean domain D, then a Euclidean domain D,
on a
norm
Euclidean
a Euclidean
v is
If
and Multiplicative
Gaussian Integers
47
Section
field
any
is a Euclidean
F, F[x]
Does the choice of a particular of D in any way? Explain.
domain.
norm
Euclidean
v on
a Euclidean
D influence
domain
the
structure
arithmetic
Theory
15. Let D be a Euclidean then
and let
domain
v
be
a Euclidean
norm on D.
that
Show
and
if a
&
are
in D,
associates
= v(b).
v(
16. Let D be a Euclidean and let v be a Euclidean norm on D. Show that for nonzero a,b e D, one domain y (a) < y (afo) if and only if b is not a unit of D. [ffiwf: from Exercise 15 that v(a) < v(ab) implies Argue show that v{a) = v{ab) implies of D. Using the Euclidean b is not a unit (a) = (ab). Conclude algorithm, < if b is not a unit, then v(a) v(ab).]
17. Prove or disprove the D | y(a) > y(l)} U {0}
18. Show 19.
Let
that y
a. Showthatifs is a
c.
that
there
Euclidean
domain
D,
{a e
then
e Z+, A : D* exists a Euclidean
D.
> 0,then?j
= v(a) + s for nonzero D a \342\202\254 ?j (a) of D. Z given by A(
+ v(l) ->
-> Z defined by nonzero elements
: D*
usual, D* is the
set of
be a UFD.An
c in D is a least common a and element (abbreviated 1cm) of two elements multiple and if c divides element of D that is divisible by both a and b. Show that every two every and b of a Euclidean D have an 1cm in D. [Hint: Show that all common multiples, nonzero domain in the obvious sense, of both a and b form an ideal of D.] D if
in
21. Use if
b |c elements a \\ c,
last statement
the and
a
only
and s,
if r
Using the last has a solution
statement in
Z if
in Theorem
46.9 to
as integers
viewed
in Theorem
a and
n
show that two nonzero elements r,seZ the group generate in the domain Z, are relatively that is, have a gcdof prime,
46.9,
are relatively
show
that
for nonzero
a,b,n
Following congruence
to
find
section
a
\342\202\254 Z, the
congruence
ax =b
(Z, +) 1. (mod
n)
prime.
23. GeneralizeExercise22by showing that for nonzero a, b,n e Z, the in Z if and only if the positive gcd of a and n in Z divides &. Interpret 24.
on
D. \342\202\254
Let D
b
22.
that s D. As
for t
norm
domain.
on
norm
that
Show
a Euclidean
norm on a Euclidean domain
Z such \342\202\254
b. Show a 20.
Euclidean
Euclidean
that
ideal of D.
is an
field is
every
a Euclidean
be
If v is a
statement:
following
has that
ax congruence this result in
= b (modn) has a solution the ring Z\342\200\236.
the idea of Exercises6 and 23, outline a constructive method for finding a solution in Z of the ax = b (mod \302\273) for nonzero a solution. Use this method a, b, n \342\202\254 Z, if the congruence does have solution of the congruence22* = 18(mod 42).
47
Gaussian
Integers
and Multiplicative
Norms
GaussianIntegers We
47.1
Definition
should
give an
example of a Euclidean
domain
different
A Gaussian integer isa complex a + bi, where a, b number a = a + bi, the norm N(a) of a is a2 + b2.
from Z \342\202\254 Z. For
and
F[x].
a Gaussian integer \342\226\240
Factorization
IX
Part
408
We shall
let
Z[i]. Note
his Disquisitiones Arithmeticae, Gauss In in detail the theory of quadratic residues, the theory of solutions to the congruence showing of the
to
attempting
of quartic was
much more
Gauss's investigations
and
the
ordinary
that there
47.2
are four
Lemma
is,
x2 =
p
p
q)
(mod
Gauss realized that it the Gaussian
consider
the ordinary integers. of the Gaussian
paper published
Z[i] is
a e
rational
a Euclidean
on
norm
that is, all
integers,
In Z[i],
the
in
1832
them
between
analogies
the
generalized
prime
noting
as in
as the
product
units.
He
real
other
two
integers,
a
neither
able to determine which that
a real
be 2 or of the form 4\302\273 + 1. \342\200\224 = 2 and real + (1 0(1 ;') prime primes to 1 modulo 4 like 13 = (2 + 3 0(2 \342\200\224 30 can only
which
prime,
of
was then
are prime: A Gaussian integers integer if its norm is is prime if and only
Gaussian
The real
as the product
factor
47.1,
prime integer by denning Gaussian integer to be onethat cannot be
of them not
and \342\200\224i,
he
of a
notion
the
expressed
is
Definition
primes of
the
in
Exercise
10.
of two 4\302\273 +
Gaussian
3 like
7
and
Real primes. 11 are still
integers. See
of Gaussian
domain
the
prime
form
among
properties
following
denning
and 1, \342\200\2241,;',
namely
integers,
the norm
congruent
integers
in
For example, after integers. units elements) (invertible
Gaussian
the
primes. results to theories
his to
are containedin a long which he proved various
that
q are
and
p
however, natural
than
rather
where
generalize
residues,
integers
= congruences x2
(mod p)
studied
famous quadratic the relationship between
the
proved
theorem
x2 = q
and
In
and
solutions
the
leads
and
Note
\342\226\240 Historical
(mod q)
Z[i]
by v(a) = N{a) for nonzero Gaussian integers include all the
the
that
The following lemma gives to a demonstration that
integers.
of Z.
elements
reciprocity
Gaussian
norm function N on
of the
v denned
function
the
set of all
be the
Z[i]
somebasicproperties
of the
norm function
for all a, p
hold
N
e Z[i]:
1. N(a) > 0.
Proof
47.3
Lemma
Proof
If
2.
N(a)
3.
N(aP)
we let a =
= =
a\\
We computations.
leave
Z[i] is
integral
an
0 if and
obvious that of 0. Let a, p e It is
only if a
N(a)N(P).
+ a2i and p =b\\ +bii, the proof of theseproperties
ap
= 0
that
either
implies
are all
results
these
as an
exercise
straightforward (see Exercise 11).
Z[i]
Z[i].
commutative ring with unity. We show Using Lemma 47.2, if ap = 0 then
is a
= N(aP)
there
that
are no
Of
divisors
= 0.
= N(0)
implies that N(a) = 0 or N(P) = 0. By Lemma a = 0 or p = 0. Thus has no divisors of Z[i]
47.2
again,
0, so Z[i] is an
domain.
it
\342\231\24
domain.
N(a)N(P) Thus
= 0.
this integral
\342\231\24
since Z[i] is a subring obvious that Z[i] has no
course,
is really
of
C, where
0 divisors.We
C is gave
the
field
of complex
the argument
numbers, of Lemma 47.3to
the use of the going outside of Z[i] in
illustrate
47.4 Theorem
Proof
v given The function Thus Z[i] is a Euclidean
Integers and
Gaussian
47
Section
our
the norm
3 of
property
multiplicative
MultiplicativeNorms
avoid
and to
N
function
409
argument.
= N(a) for
by v(a)
Z[i] is a Euclidean
a e
nonzero
on Z[i].
norm
domain.
for all Note that for fi = b^ + b2i + 0, N(bx + b2i) = fci2 + b22, so N(P) > 1. Then < This a, fi 7^ 0 in ZD'], N(a) N(a)N(P) N(ap). proves Condition 2 for a Euclidean norm in Definition 46.1. It remains to prove the division Condition 1, for N. Let a, p e Z[i], with algorithm, = = We must find er and p in Z[;] such that a fli + a2i and p bi + b2i, where ^/0. = = 0 < or N(p) for a iVX/S) = &i2 + b2. Let a//} fia + p, where either p
=
= r + si
Let qi
r,jeQ.
#2 be
and
q2i
of a, we seethat
by construction
- a
-
AM
)
\\r
-
q\\
\\
<
and
\\
=
N((r
+ si)
- (q, + q2i))
=
N((r
- qi)
+ (s-
q2)i)
- g2| <
|s
\\.Therefore
<
we obtain
Thus
N(a - pa) = N[
N(p) =
so we do indeedhave
47.5Example
integers
= q1 +
Let a
s, respectively.
in Z as close as possible to the rational numbers r and \342\200\224 If = = a we are done. and p 0, Otherwise, pa. p
N(p)
---
fi\\
a
N{P)N[--o\\
as desired.
< N(j3)
= 1, of Section 46 to Z[i]. In particular, now apply all our results since N(l) = = 1. From of Z[i] are exactly the a = a\\ + a2i with the a\\2 + a2 N(a) that the only possibilities are a\\ = \302\2611 with fact that a\\ and a2 are integers, it follows Thus the units of Z[i] are \302\2611 and One can also \302\261i. a2 = 0, or fli = 0 with a2 = \302\2611. to of use the Euclidean a two nonzero elements. We leave compute gcd Algorithm We
can
the
units
such
to
computations
5 is no
longer
the
irreducible
an
exercises.
note
Finally,
in Z[i],
for 5 =
that
while
(1 +2()(1
5 is
an
\342\200\224 and
2i),
in Z,
irreducible
neither
1+
2i nor
1 \342\200\224 2i is a unit.
\342\226\262
Norms
Multiplicative
Let us
point
out
again
duciblesand units domain.
However,
suitably
defined
is
we
to determine
an
integral
not affected
in
any
domain way
arithmetic
concepts ofirre-
on this section show, be defined
the arithmetic structure where for a domain
of D.
the
a
This
number theory, of algebraic norms of the domain, each doing its part in helping of the domain. In a domain of algebraic structure integers, for each irreducible (up to associates), and each such norm
in algebraic many different
we have essentially onenorm concerning gives information
D, the arithmetic
by a norm that may our work thus far in
preceding sectionand norm may be of help in determining
consider the
that for
as the
illustrated
strikingly
integers
are
the behavior
in the
integral
domain
of the
irreducible to
Part IX
410
Factorization
This is
it corresponds.
which
in an algebraic Let us study integral 3 of N on Zp] given
elements
2 47.6 Definition
and
Let D be an
1. N(a) = 0 if
2. N(aP) =
Proof
If D is an for every an
element
Let
D be
integral
A multiplicative norm N on D is a function conditions are satisfied: following
for
domain
with a
n in D,
alia,
that
1. Also, if
=
N(l)
1 =
SinceN(w) be
the
p,
N
prime =
so either |N(a)| unit of D. Thus 47.8
Example
On Zp'], the in the sense
= 1 or \\N(fi)\\
irreducible
is an
7r
1. By
in
Z.
definition.
then
Then
1. of norm
elements
Then
\302\2611. Let
n
D \342\202\254
= a/6, we have
iin
\\N(a)N(P)\\,
assumption,
this
that
means
either a
or
fi
is a
of D.
\342\231\246
by N(a + bi) = a2 We saw that the function
N defined
function
of our
=
= 1
\\N(u)\\
a unit in D,
= N(u)N(u~l).
implies that 1^(^)1 = of D are exactly the
p=\\N(n)\\
1 and
1 is
then \302\243>,
= N(uu~l)
p is a
\342\200\224
is an irreducible of D.
norm N.
units where
=
\\N(a)\\
e Z,
p
N(l)
= N(1)N(1)
((1)(1))
unit in
is a
u
N(l)
integer, this
is an
. Now suppose that such that 1^(^)1 =
for a prime
a multiplicative
N(l) = shows
a such that
every
= p
\\N(n)\\
N, then
norm
multiplicative
domain with
an integral
D
mapping
D.
e
fi
furthermore,
with
Properties
= 0.
if a
only
D. If,
satisfying
the
that
N(a)N(P)
u in
unit
norm
multiplicative
them.
47.2.
in Lemma
and
a
that have
of
properties
studying
associated with
of mappings
means
by
importance of
of the
example
domains
domain.
integral
into the integers Z such
an
structure
+
b2 gives
a
multiplicative
norm
by v(a) = N(a) for are precisely the elements a
v given
a Euclidean norm on Zp], so the units N(\\) = 1. Thus the second part of Theorem 47.7 appliesin Zp]. We saw in Example 47.5 that 5 is not an irreducible in Zp], for 5 = (1 + 2()(1 -2i). Since = 12 +22 = 5 and 5 is a prime in Z, we see from Theorem 47.7 N(l N(l -2i) 1 + 2i and 1 - 2i are both that in Zp]. irreducibles As an application of mutiplicative norms, we shall now give another example of an domain that is not a UFD. We saw one examplein Example 45.16. The following integral a
nonzero
is \342\202\254 Z[i] =
ofZp]withAf(a)
+2i) -
is the 47.9 Example
standard
illustration.
LetZ[V^5] = {a + addition,
domain.
subtraction,
Define
N on
ib
\342\202\254 Z}. As
and multiplication,
Z[V^5]
a subset of the and containing
numbers
0 and
Z[V~5] is an integral
by
N(a
+ b-J^5)
= a-
closed under
complex
+ 5b2.
1,
Section47 (Here 7=5 = i V5.)
and Multiplicative
Integers
= 0 if
N(a)
Clearly,
is a
= N(a)N(P)
N(afi)
Gaussian
and
if
only
that
computation
straightforward
411
Norms
= 0.
a = a + b
That
exercises
12). Let us
find all candidates for units in Z[-v/\342\200\2245]by finding all N(a) = 1. If a = a + fc-s/=5, and N(a) = 1, we must have a2 + 5b2 a and b. This is only possible if b = 0 and a = \302\2611.Hence \302\2611 Since \302\2611 the units in are the only candidates for units. are units, they are then precisely Z[V=5]. we have 21 = (3)(7)and also Now in Z[V=5],
Exercise
(see
a
elements
in Z|V=5] with = 1 for integers
21 = If
3,7,1 +
can show that
we
then
will
know
Suppose
2V=5)(1 -
(1 +
and 2\\[\342\200\2245,
1
2V-5).
\342\200\224 are 2\\f\342\200\2245
all irreducibles
be a UFD, sinceneither
that Z[V=5] cannot that 3 = aft. Then 9 =
AT
3 nor
7 is
in Z[v^ + \302\261(1
we \342\226\2405],
2\\ /=5).
= N{u)N{P)
(3)
9. If
= 1, then a is a unit. If a = that we must have N(a) = 1, 3, or iV(a) = then a2 and for no choice of a and b is a + b-sf\342\200\2245, N(a) + 5b2, integers N(a) = 3. = = = 3 then so is a unit. Thus from we can conclude that If N(u) 9, 1, N(/J) fi a/3, an in is a unit. 3 is irreducible A similar either a or /6 Therefore, Z[V~5]. argument in Z[-v/\342\200\2245]. shows that 7 is also an irreducible = If 1 + 2-s/\342\200\2245 y<5, we have
shows
21 =
+
Af(l
27^5)
= N(y)N(S).
so N(y) = 1,3, 7, or 21. We have seen that there is no element of Z[-v/\342\200\2245]of norm 3 or'7. This either N(y) = 1,and y is a unit, or N(y) = 21,so N(8) = 1, and S is a 1 + 2^f-5 is an irreducible in Z[V~5]. unit. A parallel argument shows that Therefore, 1 \342\200\224 is also 2-v/\342\200\2245 In
summary,
an irreducible we have shown
in
= {a
Z[V^5]
but not
domain
w fl\302\273 integral
21
+ ibS
\\a,b e Z}
a UFD.In particular,
21 = of
Z[V\342\200\2245]-
that
3
\342\226\240 7 =
(1 +
27^5)(1
there
are two
different factorizations
- 2V^5)
These irreducibles cannot be primes,for the property of a prime of factorization (see the proof of Theorem 45.17). \342\226\262 uniqueness
into irreducibles.
enablesus to
prove
with a classical application,determining We conclude which primes p in Z are equal to a sum of squares of two integers in Z. For example, 2=l2 + l2,5 = l2 + 22, and 13 = 22 + 32 are sums of squares. Since we have now answered this question for the even prime number, 2, we can restrict to odd primes. ourselves only
47.10Theorem
(Fermat'sp = a2 integers
Proof
First,
a and
b in Z if
that p
suppose
p is an odd p = 1 (mod
+b2
number. 4).
This
Theorem)
and
only
if
Let p be an odd prime = 1 (mod 4). p
in Z.
Then p
= a2
b2
for
= a2 + b2. Now a and b cannot both be even or both be oddsince = 2r and b = 2s + 1, then a2 + b2 = 4r2 + 4(s2 +s) + l,so takes care of one direction for this \"if and only if\" theorem.
If a
Factorization
IX
Part
412
For the
other
divisor of p \342\200\224 1, we that p),
n2 has so p
we assume that finite field Zp
direction,
elementsof
of nonzero
see
multiplicative
divides
n2
+
p
=
1 (mod
4).
Now
the
is cyclic, and has order that Zp contains an element n of multiplicative \342\200\224 order 1 in 2, so n2 = Zp. Thus in Z, we the
1 in
multiplicative \342\200\224 1. Since p
group 4 is a
order 4. It follows have n2 = -1 (mod
Z.
Z[i], we see that p divides n2 + 1 = in + i){n \342\200\224 i). Viewing p in then /7 would have to divide n + i or n \342\200\224 i. If p divides that is irreducible Z[i]; Suppose p some e Z. Equating coefficients n + i, then n + i = p(a + bi) for a,b of;', we obtain \342\200\224 n i would lead to an 1 = pb, which is impossible. divides Similarly, p impossible = our that is irreducible in must be false. \342\200\2241 Thus Z[i] p assumption equation pb. in Z[i], we have p = (a + bi)(c+ di) where neither a + Sincep is not irreducible bi nor c + di is a unit. Taking norms, we have p2 = (a2 + b2)(c2 + d2) where neither we have a2 + b2 = 1 nor c2 + d2 = 1. Consequently, p = a2 +b2, which completes that we see this is the factorization of p, our proof. [Since a2 + b2 = (a + bi)(a \342\200\224 bi), \342\231\246 that is, c + di = a \342\200\224 bi.] and
you to determine which
10 asks
Exercise
1 in
n2 +
primes
remain
pinZ
irreducible
in
Z[;].
47
\342\226\240 EXERCISES
Computations In
1 through
Exercises
of a
factor
\342\202\254 Z[i ] must
4, factor the have norm
Gaussian
> 1 and
to consider as possibleirreducible quotient is again in Z[i].]
2.
1. 5
5.
6 does
that
Show
factors
into a product of irreduciblesinZp][Hint: Since an irreducible a finite there are number of Gaussian integers a + bi N(a), dividing only a by each of them in C, and see for which of a given a. Divide ones the integer
not factor
(up
uniquely
4. 6 -
4 + 3f
3.
7
to associates)
into irreducibles
in
Z[V^-5].
li Exhibit
two different
factorizations.
6. Considera
=
[Hint: Use the
7. Use
7
+ 2i
and p
= 3-
Ai
in Z[i].
a =
pa +
Find a
p
and
p in
with
N(p)
proof of Theorem 47.4. algorithm in Z[f] to find a gcd of 8 + 6i and
construction
a Euclidean
Zp] such
that
< N(fi).
in the
5 \342\200\224 15; in
Z[i]. [Hint:
Use the
construction
in
of Theorem47.4.]
the proof
Concepts
8. Mark
of the
following a. Z[i] is a PID.
each
b. Z[i]
true
or false.
domain. c. Every integer in Z is a Gaussianinteger. d. Every complex number is a Gaussianinteger. e. A Euclidean algorithm holds in Z[i]. f. A multiplicative norm on an integral domain is a
domain.
Euclidean
is sometimes
an
aid
in finding
irreducibles of
the
g. If
N is
h. If F
norm on an integral N defined
a multiplicative
is a field,
the function
then
D, then \\N(u)\\ = (degree N(f(x))
=
domain by
413
Exercises
47
Section
1 for
every
D.
u of
unit
isa multiplicative
of fix))
norm
on F[x].
i. If F is a field, is a multiplicative j. Zt-y^-5] is an
9. Let
norm integral
an integral domain ofD.Letjr be such that
D be
unit
norm N
a multiplicative
with
is minimal
\\N(n)\\
among
all
of 2
=
defined by N(f(x)) on F[x] according to our domain but not a UFD.
the function
then
^ 0
f(x)
and
= 0
N(0)
definition.
\\N(a)\\ = 1 for a e D if and only if a is a > 1 for ft e D. Show that it is an irreducible
that
such
\\N(J})\\
ofD.
10. a. b. 11.
equal to the odd prime p
Show
that
2 is
Show
that
an
of a
product
irreducible
Z is
in
12. Prove that N of Example 47.9 is multiplicative, 13. LetD be an integral domain with a multiplicative Show
of D.
that
{a)
be a
a.
Show
that
nonzero
c.
is a
Z[i]/{a)
b. Show that if
n is
to part
Referring
an
n
47.10.)
[Hint:
ring.
N such that | N(a)
D has a factorization of 16 +
li
\\
1 for
Z[i]-
e ZlV^]. and only
a
D if \342\202\254
if
a
a is
in D.
irreducibles
into
10 \342\200\224 5i in
and
=
a, 0
Use
[ffiwf:
the construction
in
Use the
division
algorithm.]
then Z[i]/(7r) is a field. and characteristic of eachof
of Z[f],
irreducible the
order
that
is,
not
the
following
fields.
+ i>
ii.Z[i]/(l
square free,
N(a)N(fi) for
ffl.Z[i]/(l+2i>
divisible by the square
of
any
prime
integer.
Let ZtV^J]
= {a +
\302\243 Z}.
a. Show that
the norm
N,
defined
N(u) = 1for a
b.
Show
that
c.
Show
that every nonzero Use part (b).]
[ffi'wf:
finite
N(aP) =
is, that
norm
a gcd
find
j
\342\202\254 Z+ be
\\a,b
ib^/n
(mod 4). (UseTheorem
in Z[i].
ideal
(b), find
i.Z[f]/(3> 16. Let
Z[i] to
principal
that
nonunit of
nonzero
every
14. Usea Euclidean in algorithm the proof of Theorem 47.4.]
15. Let
in Z[f ].
47.2.
Prove Lemma
unit
the square of an irreducible in Z[; ] if and only if p = 3
and
unit
by N(a)
\342\202\254 Z[V^\302\253]
if and
nb2
for a
only if a is not a unit
\342\202\254 that is Z[V\342\200\224\302\253]
a
17. RepeatExercise16forZ[V\302\253]
= a2 +
=
{a +
bsfn
\\a,b
\342\202\254 Z}, with
= a
a unit has
+
i
b^fn,
is a multiplicative
norm on Z[V\342\200\224\302\253].
of Z[V\342\200\224\302\253].
a factorization
N defined by
N(a)
into irreducibles = a2
\342\200\224for
nb2
in
a =
Z[V\342\200\224\302\253].
a+
bs/n
inZ[V\302\253].
18.
Show by in
the
domain
a construction
integral
is Euclidean.
Euclidean.)
analogous
domain 2^-/-2]
See Hardy
to that
for v(a) and
given in
the
= N(a) for
Wright
[29] for
proof nonzero
of Theorem 47.4 that a in this domain
a discussionof
which
the
division
algorithm holds
(see Exercise16).(Thus
domains
Z[~Jn] and
Z[V^\302\253]
this are
PART
Galois
X
Theory
Section
48
Automorphisms
Section
49
The
Section 50
48
of Fields Theorem
Extension
Isomorphism
Splitting
Section51 Section 52
section
and
Automorphisms
Fields
Extensions
Separable
Totally Inseparable
Extensions
Section
53
Galois
Section
54
Illustrations of Galois Theory
Theory
Section 55
Cyclotomic
Section 56
Insolvability of
Extensions the
Quintic
of Fields
Automorphisms
FieldTheory The Conjugation Isomorphisms of Algebraic Let
F be
a field,
of F that is of a particular closures
and
algebraically F is not
of F
F be
let
an algebraic closureof
closed. Such
that
F,
a field F exists,by
critical, since, as we shall show a map leaving F under
are isomorphic
is, an algebraic extension 31.17. Our selection
Theorem
in Section
fixed.From
49,
any
two
on in
now
algebraic
our
work,
algebraic over afield F algebraic F of F. closure are contained in one fixed under consideration algebraic that we are engaged in the study of zeros of polynomials. In the Remember in F[x] amounts to studying of Section 31, studying zeros of polynomials terminology of algebraic extensions of F and of elements the structure algebraic over F. We shall a and /J have the of F with extension show that if E is an algebraic a, (5 e E, then = this fact We if if shall and same irr(/J, F). irr(o;, F) phrase properties only algebraic We achieve all along in field theory. as we have been doing in terms of mappings, there exists an isomorphism -fa^ this by showing that if irr(o;, F) = irr(/J, F), then of F(u) onto F(J5) that maps each element of F onto itself and maps a onto fj. The
we
next
shall
assume
theorem
fundamental
momorphisms
that all
this isomorphism for the study of algebraic
exhibits tools 4>a
of Theorem
isomorphisms.Before
stating
and
extensions
ira
all elements
These
extensions;
isomorphisms
they supplant
will become
our
evaluation
ho-
the
make their last contribution in defining these some more this let us introduce theorem, proving
22.4, which and
terminology.
Section 52 is not required
for the
remainder of the
text.
415
Part X
416
48.1 Definition
48.2
Example
Galois
and
Automorphisms
Theory
Let E be an algebraic extension F if irr(o:, F) = irr(/J, F), that over F.
just
conforms
that
by conjugate and b ^ 0,
R. If a, b e K zeros of x2 \342\200\224 lax +a2
bi and
classic idea of
with the
defined
understand
are conjugate over a \342\200\224 bi are both
that a +
numbers
if we
numbers
complex
numbers
\342\226\
of conjugateelements
The concept conjugate
of a field F. Two elementsa, (3 e E are conjugate over a and (3 are zerosof the same irreducible polynomial
is, if
complex numbers we mean the
complex
conjugate
+ b2,
in
is irreducible
which
R[x]. 48.3
Theorem
\342\226\2
Isomorphisms) Let F be a The map ij/a_p : F(a) ->-
(The Conjugation
F
with
= n.
F)
deg(a,
is an
e F
for c,
= 1- c\342\200\236_!q;\"_1)
+ cta-\\
fa,p(c0
co +
onto F(/J) if
of F(a)
isomorphism
and let a and fi F(fi) defined by
field,
c^
-\\
and
\\-
algebraic
over
cn_ifin~l
and
if a
only
be
are
/3
conjugate
over F.
Proof
Supposethat isomorphism.
: F(a)
^ra^ Let
irr(a,
faj(ao + By
Conversely, supposeirr(a, F) = irr(/J,
By Theorem
and
F[x]
maps
F(a)
to
\\-
+
anp\"
48.4 where
As the
4>a
:
\\F\\x\\[
composition of two For (cq + cia
onto F(f}). faj(co
F[x]
+ Cld
= (f/ifa
+
\342\200\224> F(a),
Then
is an
theorem +anan
= 0. that
irr(/J, F)
= ^p.a shows
gives \302\242^
that
irr(o;, F)
monic,
=
homomor-
evaluation
the
0, so
\342\226\240 \342\226\240\342\226\240 =
have the same there is a natural
Similarly,
Let \\j/a^
kernel (p(x)). isomorphism
rise to
an
are elements under corresponding isomorphisms, -fa^ is again an isomorphism \342\226\240 \342\226\240 \342\226\240 + + cn-\\an~l) e F(a), we have lines
dashed
the
= p(x). both
= F(a).
are
polynomials
\342\200\224> F(f3)
onto F(f3).
F[x]/{p(x))
in Fig.
mappings.
onto
F)
: \302\242^F[x]
corresponding
F[x]/{p(x))
mapping
diagrammed
and
\342\200\224> F(a)
26.17,
i/ra mapping isomorphism -ifrp
of the a\\a
+ axp -\\
F) divides irr(/J, F). Therefore,since both so a and fi are conjugate over F. F),
phisms
the
= a0
+
in the statement of Theorem29.13this implies A similar the isomorphism F). argument using (i/a,p)~l
irr(o:,
irr(a, irr(/J,
\342\226\240 \342\226\240 \342\226\240 +fl\342\200\236x'!.Thenflo
h anan)
-\\
a\\ot
+
+a\\x
assertion
last
the
divides
flo
in the statement
as defined
\342\200\224> F(fi)
F) =
=
These mappings
i/p(ifa)~l-
indicate
+ cn_1an-1)
^(CQ+C^H
48.4 Figure
\\-Cn_iCCn
l)
Section 48
= ^((co
= 00 + Thus
is the
iraj
+
map denned
+ {p(x)))
cn-xxn-1)
of the
statement
the
in
theorem.
Theorem 48.3 is the cornerstone Theorem of Section49 and of
following
important
+
010+ ---+^-^-1.
corollary of Extension Isomorphism
The
\342\226\240 \342\226\240 \342\226\240
+
cxx
417
of Fields
Automorphisms
\342\231\246
of our
proof of
of the
most
the
rest of our
work.
48.5
Corollary
be algebraic of F such that if
a field F. Every over isomorphism ir mapping F(a) onto a subfield (a) = aforaeF maps a onto a conjugate fi of a over F. Conversely, exists each conjugate one isomorphism i/aj of F /3 of a over F, there exactly each a e F onto itself. a subfield of F mapping a onto /3 and mapping
Let a
(a)
for onto
Proof
Let
of F(a)
an isomorphism
be
ij/
Let
F) = ao
irr(a,
+
a\\x
- - -
+
+
that
if {a)
= a for
a e
F.
Then
anxn.
= 0,
\342\200\242 \342\200\242 \342\200\242
a\\ot +
+
flo
of F such
a subfield
onto
+
anan
so
0= and
fi
=
if (a)
if
+ ci\\a
(ao
= ao
\342\200\242 \342\226\240 \342\226\240
+
+
anan)
+ aiir(a)
+
\342\226\240 \342\226\240 \342\226\240
+
anir(a)n,
is a conjugateof a. for each conjugate
/3 of a over F, the conjugation isomorphism iia^ with the desired properties. That is the only of Theorem 48.3 is an isomorphism iraj such isomorphism follows from the fact that an isomorphism of F(a) is completely \342\231\246 determined by its values on elements of F and its value on a.
Conversely,
48.6 CoroUary
corollary of Theorem48.3,we
a second
As
Let
e R[x]. If
f(x)
0 also.Loosely,
prove
a familiar
result.
= 0 for
f(a +bi)
complex
can
zeros
(a + hi) e C, where a, beR, then hi) = f(a of polynomials with real coefficients occur in conjugate
pairs.
Proof
have
We
seen
that C
= R(i). Now R) =
irr(i,
so
i
and
where
\342\200\224i are conjugate
ifi-t(a
over R.
hi is + hi) = a \342\200\224
f(a
+ hi)
=
flo
irr(-i,
R) =
x2
1,
+
By Theorem48.3,the
+
a\\{a +
hi) +
\342\200\242 \342\200\242 \342\226\240
+
map
conjugation
an isomorphism. Thus, if for an(a
at
ifi-i
:C
\342\200\224> C
e R,
+ hi)n
= 0,
then
0 =
i/i_i(f(a
+ bi))
= =
that
is,
f(a
\342\200\224=
bi)
0 also.
a0
f(a
+
ai(a
\342\200\224
bi)
+
- - \342\226\240
+
an(a
\342\200\224
bi)n
- bi), \342\231\246
Part
418
48.7
X
Example
and Galois Theory
Automorphisms
Consider are \342\200\224V2
Q(V2) over Q. The zeros of irr(V2, Q) = x2 - 2 are over Q. According to Theorem48.3the conjugate
Q(V2) defined
isomorphism of Q(V2) onto
As
in the
illustrated
48.8 Definition 48.9
Definition
An
Example
If ct is an fixed by ct
isomorphism of a field if a {a) = a. A collection
e F is left
a, b, c,
of \302\243 onto It is studying
map
by every
ct
:
E
field.
of the an
then
nontrivial
that follows.
work
the
in
element
isomorphisms of E leaves ct e S. If {ct}leaves F
=
of E; it
automorphisms form
concerning
group
extension. Thus much
of
a +
is
\342\226\240
a of
a subfield fixed,
then
\302\243 is
left
E
F of
ct leaves
structure
of an fixed
by
- cV3 -
b4l
the
We E as (Q(V2))(V3).
our purpose to study the structure the automorphisms of E that leave
theorems
importance
some field,
E defined
-*
automorphism
if we view
itself
S of
have a
a field may
example,
automorphism
+ cV3 + dV6)
+ bV2
Q is an
d e
show that these results
three
fixed
onto
E
and
of utmost
\342\226\240
a (a for
field onto itself is an
= Q(\\/2, V5).The
Let E
Fields
preceding corollary Such maps will be
of a
fixed if each a F fixed. 48.10
\342\226\262
onto itself.
isomorphism
= a \342\200\224 bV2
+ bv2)
itself.
Fixed
and
Automorphisms
isomorphism
_V5 :
i/c^
by
V'VJ -V2(fl is an
so s/l and Q(V2) ->-
and -s/l,
V2, map
that
a leaves
^
i//-^
Q(\\/2) fixed.
of a algebraic extension\302\243 of F. We shall
each element
\342\226\262
field F
by
presently
in a natural We can then apply the way. information about the structure of our field
a group to
get
work is now being proved, but the ideas contained These theorems are therefore of
our preceding
are readily that follows.
isomorphism
conjugation see
dV6
brought
in them
together.
form the
The
next
foundation
importance to us. They ideas contained in them that are important. A big step in mathematics does not always consist of proving a hard theorem, but may consist of noticing how certain known mathematics may relate to new Here we are bringing situations. of zeros of polynomials. Be theory into our study group the concepts sure to understand involved. Unlikely as it may seem, they are the key to the solution of our final in this text. goal for everything
really
to
amount
observations,
Final Goal (to be (degree every quintic starting
with
elements
more
rather than theorems;
precisely
stated later):
5) polynomial f(x) in the field containing
it
great
is the
To show
that
in can be expressed
the
coefficients
not terms
of f(x).
all zeros of of radicals
Section 48
Note
\342\226\240 Historical
was Richard Dedekindwho It idea of an
of the
of group
had been
through
first
in 1894.
field,\"
theory to
the
he called
what
Artin
The earlier
fields in great detail. Artin emphasized is now called Galois theory the goal of what should not be to determine solvability conditions the for algebraic equations, but to explore
groups and
of
the
polynomials. Dedekind extended of the entire field and proved several of the theorems of this section. Heinrich Weber continued Dedekind's Though on fields in his algebra approach to groups acting in other text of 1895, this method was not pursued texts near the turn of the century. It was not until the abstract approach to 1920s, after Emmy Noether's of certain
roots this
idea
to mappings
If {crt\\i which i
{oy \\i
e 1} is a collection el] gives the
This first
e I.
of our
fixed elementsof 48.11
Theorem
Proof
a and
extensions and groups his approach in
field
between relationship
detailed
Artin
automorphisms.
1942. In fact,
and
1938
basedon
Artin's
of Galois
development
of a field E, the are those a e E left
information
almost all
contains
theorems
of this
remainder
the
that
fixed
can
be
text is
theory.
about for o,by every said about these of E
elements
of automorphisms
three
of a lecture
in given in 1926; his method was first published B. L. Van der Waerden's Modern Algebra text of himself in lecture notes in 1930 and later by Artin
least
E.
Let {o,-| i e 1} be a collection a e lijeft fixed by every o,If CT,(fl)=
that Emil this relationship of
at Gottingen,
developed
(1898-1962)
that
of equations
theory
of permutations
groups
algebra becameinfluential
the
developed
of a field,
automorphism
a \"permutation application
419
of Fields
Automorphisms
o,(&)
of i
for
= b for all
Gi(a
forms a
i
e I,
\302\261 b)
E. Then the
of a field
automorphisms
el
set
E{m] of all
E.
of
subfield
then
= ai{a)
= a
\302\261 Oiib)
\302\261b
and
Oiiab)
for all i
e I.
Also, if b
e
I. Since
the
o,-
are
48.12
48.13
Definition
Example
('
e I.
Hence 0, 1
we have
automorphisms,
e E^)
The field
E^} of Theorem
automorphismo,
we shall
Consider
the automorphism
we
48.11
refer to E^
= a/b
= Oiia)/
0-,-(0)=0 for all
ab
^ 0, then cfiia/b)
for all i
= CT,(fl)CTj(fo)=
Thus
and
E{ai]
i//-^ _^
is a
=
subfield of
fixed field of {o,| i field of a.
is the
as the
0-,-(1)
1
E. e /}.
\342\231\246
For a
single
fixed of
Q(\\/2)
given
\342\226\240
in Example
48.7. For a, b
e Q,
have
^yi and a
\342\200\224
by/2
= a
+ by/2 if and
_^ia only
+ b4l) if b =
=a
0. Thus
\342\200\224
by/2,
the
fixed
field of
i//^ -^. ^sQ-
A
Part
420
X
and Galois Theory
Automorphisms
E,
that
ax is again
permutation
of a field E of E. If a
an automorphism is, a permutation
that
Note
E onto
an
automorphism
Proof
The set of
all
since,
This
is how
Theorem
be afield,and
Let E <
F fixed
For a,
so
e
x
function
composition.
is a subgroup
of the
that the field
immediately
fixed
EG(e/f) of all
implies
e G(E/F).
er_1
automorphisms of \302\243. element of G(E/F),
by every
fixed
E left
of
elements
Also, if
G(E/F).
that
follows
it
G(E/F)
by
F.
contains
\342\231\2
The group leaving F
Do not but
G(E/F) of the preceding the fixed, or, more briefly, of E/F
think
as meaning
rather
The ideascontained example. We urge you
group
defined
preceding three theorems this example carefully.
in the to study
a, b
shows
31.9
that [Q(V2,
-V2^a
= a
+ bvl)
\342\200\224Ji^-Ji, -V5
V5) : Q] = -^.
ij/^
4. If
we
\302\260f
\342\200\224
&\\/2
an of Q(\\/2, V3) having Q(\\/3) automorphism the automorphism i//^ _^ of Q(\\/2, V5) having of two automorphisms is an automorphism, product
have
field. Sincethe
following
\342\226\2
e Q(\\/3) is we
in the
E
of some
space
are illustrated
isomorphism
conjugation
of
by VV2
Similarly,
of automorphisms
in the notation as denoting a quotient G(E/F) that E is an extensionfield of the field F.
the field Q(V2, V5). Example Q(V2, V3) as (Q(V3))(V2), the
Theorem48.3
is the group of E over F.
theorem
Consider view
^V2
i is in
e G(E/F)
a
group of all
is left
a,
automorphism
identity
a~l{a) so
of F
element
every
= a (a) =
= a(x(a)) the
e F,
a
for
Since
sort,
Of course, then a =
e G(E/F).
ax
for
of
e F, we have
and a
G(E/F)
Thus G(E/F)
Example
its entrance.
makes
theory
EG(e/f).
a (a) = a
48.17
group under
of homo-
composition
general,
group
of all automorphisms let F be a subfieldof E. Then the setG(E/F) forms a subgroup of the group of all automorphisms of E. Furthermore,
(ax)(a)
Definition
in
of the
then
of automorphisms of E is defined and is thus composition, by function ( : E \342\200\224>E given permutation multiplication). The identity permutation the by i(a) = a for a e E is an automorphism of E. If a is an automorphism, then a _1 is alsoan automorphism. a subgroup Thus all automorphisms of E form permutation 8.5. \342\231\2 of SE, the group of all permutations of E given by Theorem
F
48.16
of E,
(it is
E leaving Proof
\302\243 is a
one-to-onemapping
automorphisms
Multiplication associative
48.15
of a field
automorphisms
particular a
x are
of E,
morphisms again yieldsa homomorphism. 48.14 Theorem
is in and
wmcn
moves both
;=
ff2
=
03 =
the
V2
and
identity
VV3,_V3>
V5,
that is,
automorphism,
and
f ^2,-^73,-V3.
leaves
neither
as fixed field. Q(\\/2) as fixed we number
can
consider Let
fixed.
48
Section
48.18Table (
The
C\\
er2
i
G\\
er2
er3
C\\
C\\
i
CT3
er2
er2
er2
a-i
L
C\\
CT3
er3
er2
G\\
i
l
of all automorphisms of Q(a/2, a/3) has must contain Q, since every automorphism
group
This fixed
field
421
of Fields
Automorphisms
field, by Theorem
a fixed
of a
48.11.
field leaves 1 and
hence
Since prime subfield fixed. A basis for Q(a/2, a/3) over Q is {1, = \342\200\224 and <72(a/3) = \342\200\224a/3, we see that Q is exactlythe fixed o\\ (a/2) a/2, o\\ (V6) = \342\200\224a/6 field of {t,
a/2,a/3,a/6}.
the
=
CT1CT3
=
^,-^2(1/72,-,/2^,/3.-73)
^/V
CT2-
V3
isomorphic to
the Klein We can show that G is the full 4-group. r of Q(\\/2, a/3) maps a/2 onto because every automorphism either 48.5. Similarly, x maps \302\261a/2, by a/3 onto either \302\261a/3. But since Corollary of Q(a/2, a/3) for Q(a/2, {1, a/2, a/3, a/2a/3}is a basis a/3) over Q, an automorphism and 0\302\276 i,ai,cr2, give all leaving Q fixed is determined by its values on a/2 and a/3 . Now, possible combinations of values on a/2 and a/3, and hence are all possible automorphisms
The
G is
group
a/3)/Q),
G(Q(a/2,
group
ofQ(V2,V3).
Note but
accident,
Frobenius
The
Let F several
one
Proof
w
\302\2730 \342\226\262
by
abstract
For an
of all automorphisms the group a generating element, and it may cyclic group there is no way of distinguishing that
definition
of F have
as being more important than any other. However, for the cyclic field there is a canonical (natural) automorphisms of a finite generator, This fact is of automorphism (classically, the Frobenius substitution). in some advanced work in algebra. The next theorem exhibits importance
Frobenius
automorphism.
Let F bea finite field of characteristic p. Then the map ap : F apioia e F is an automorphism, the Frobenius automorphism, Let a,b
7to
generator
considerable
48.19 Theorem
group has
elements.
generating
of all group the Frobenius this
shall show later
We
field.
a cyclic
Now
[Q(a/2, a/3) : Q] = 4. as we shall see later.
4, and situation,
Automorphism
be a finite
is cyclic. any
that G(Q(a/2, V3)/Q) has order rather an instance of a general
e F.
Taking
n =
1 in
Lemma 33.9, we seethat
(a
->- F
defined
of F. Also,
+ b)p =
ap +
by
=
F{a
ap(a) ~ }
bp. Thus
7Lp.
we
have
crp(a
+b)
= (a+ b)p
= ap + bp
=
ap{a)
+ op(b).
Of course,
ap(ab)= (ab)p so op is at least op is {0}, and Thus op is an
a homomorphism. a one-to-one op is
=
If ap(a)
map. Finally,
automorphism of F.
apbp
=
ap{a)ap{b),
= 0, then
ap
since
F
and a = 0, so the is finite, op is onto, by = 0,
kernel
counting.
of
X
Part
422
and Galois
Automorphisms
Theory
is
field Zp must be contained(up to isomorphism) in F, since F of theorem p. For c e Zp, we have op(c) = cp = c, by Fermat's (see Thus the polynomial xp \342\200\224 the elements of Zp. By x has p zeros in F, namely Corollary20.2). of degree n over a field can have at most n zeros in the field. Corollary polynomial The
prime
characteristic
23.5,a
elements fixed under
Since the
^P = in
Freshmen an
+bn.
exponent
m
Here we
the zeros in
are precisely
ap
still sometimes make
see that
this
is actually
valid
all
conjugates
in C
F of
see that
x,
\342\231\2
of
error
the
saying
that {a
{a + b)p = ap + bp
exponentiation,
freshman a field
in
\342\200\224 we
F^v
college
p,
of xp
F
+
\342\200\2
b)n
with
characteristic p.
48
EXERCISES
Computations In
8, find
1 through
Exercises
of the
1. V2overQ
2. V2overR
3.
4. V2-V3overQ
5.
3
V2overQ
+
+
V2
i over
6.
Q
7. y/l + V2 over Q In Exercises 9 through notation of Theorem
14,
the
we consider
48.3, we have
the
field
following
shorter
9.
r2(V3)
11. (r3r2)(V2 +
3
13.
15. Referring
to
automorphism or set of 16.
the
the
-*
(Q(V2,
V5))(-V3),
-*
(Q(V2,
V3))(-V5).
^75
-V5- Compute
= ^75.-V5' T3
=
that
order
>/3))(V5)
^V3 -V3'
^
T5
=
find the
^(72
12.
(r5r3)
21, refer to
the
automorphisms
for Exercises
each of the automorphisms of an element of a group.)
fields
fixed
following
in E
the
for Exercises
element
of E.
^ \" 3^
'
^2V3-4l) -
V3
+
(t2t5)(V^0))]
= Q(V2,S). c.
\302\243{\342\200\236,}
directions
indicated
the
E):
+ 75)
10.
14. r3[r5(V2
b.
directions
V2 over
: (Q(V5, V5))(V3)
T3
19. T5T2 22. Referto a. Show
+
V5))(-V2),
48.17,
16 through
8. y/l
(Q(V3,
\302\243{(71.(73}
Exercises
i over
Q(\\/2, V3, \\/5). It can be shown that [E : Q] = 8. In are here automorphisms of conjugation isomorphisms (which
>/45)
Example
+
E =
V5)
+
(r52r3r2)(V2
a.
let t2
notation,
R
\\/2
-*
(\302\253^2> ^VS.-VS \342\226\240
For
field.
given
^/2.-72 : (Q(>/3.V5))(V2) ^/3.-73
In
over the
number
given
9
through
and
14
\302\243(,7,.(73}
find the
fixed
field
of the
of E. 18.
17.
t32
20.
T3T3T2
9 through t2,
21. {T2,T3,T5}
14 for
t3 and
{t2, t3}
this exercise.
T5is of order 2 in
G(\302\243/Q).
(Remember
what
is meant
by
b.
Find H of G(\302\243/Q) generated the subgroup by the elements n, T3, and T5, and give the There are eight elements.] Just as was done in Example H of part (b) is the full group 48.17, argue that the group
c.
423
Exercises
48
Section
table.
group
[Hint:
G(\302\243/Q).
Concepts
23
In Exercises
needed, so that 23.
and
Two elements, a both zeros of the
an algebraic extension f(x) in F[x]. polynomial
25. Thefields
Q(V2)
and Q(3
a.
a conjugate
/3 ^
Find
to
b. Referring
26.
+ V2) are the
a of a
over
extension
\302\243 of
\302\243 and
\302\242^F[x]
->
\302\243 have
Let a
over
conjugate
are
a field F
:
of course.
same,
are
a field F
\302\243 of
to
the
text, if
F if
and F if
over
conjugate
correction is
only
and
if they
only
are
if the
kernel.
the same
= 3 + a/2.
Q.
the conjugation
compare
automorphism
\302\260f Q(V2)
^75.-72
with
the conjugation
ijra.p.
of the
value
the
18 of
in Exercise
29.
Describe the value of the Frobenius automorphism 29.19. Find the fixed field of 02in Example
27. Describe 28.
(a),
part
automorphism
an algebraic -> : F[x]
/6, of
and
evaluation homomorphisms
without reference
term
italicized
/6, of
and
same
24. Two elements, a
of the for acceptable publication. the definition
correct
24,
in a form
it is
Frobenius
Section 29. Find
Let F
be a field
of characteristic
for a
e F need
not
Mark
each
of the
be an
fixed field
0. Givean
p ^
automorphism
following
the
in
on each
on 0\302\276
automorphism
the
ct2
of
element of the
finite
field of
four elementsgiven
of the
finite
field of
nine elementsgiven
each element
0\302\276.
example
to show that
is infinite.
case that F
->
the map ap : F What
may go
\302\243 given
by a,,
(a) = ap
wrong?
or false.
true
a onto /6. automorphism of \302\243 mapping b. For a, /6 algebraic over a field F, there is always an isomorphism of F(a) onto \302\243(/6). and conjugate over a field F, there is always an isomorphism of F(a) onto \302\243(/6). c. For a, fi algebraic of every field \302\243 d. Every automorphism fixed every element of the prime subfield of \302\243. leaves e. Every automorphism of every field \302\243 fixed an infinite number of elements of \302\243. leaves f. Every automorphism of every field \302\243 fixed at least two elements of \302\243. leaves of every field \302\243 0 leaves fixed an infinite number of elements of characteristic g. Every automorphism
a. For
all
a,
is always an
there /6 e \302\243,
of \302\243.
h.
j.
of a field
All automorphisms
i. The
set of all
For fields
\302\243 form
of a field
elements
< \302\243 \302\243
\302\243 left
a group
under function composition. a single automorphism
fixed by
of \302\243 forms
a subfield
of
< G{K/F).
G(K/E)
Proof Synopsis
30. Give
a one-sentence
31. Give a one-sentence
synopsis of synopsis
the
of the
\"if\" part \"only
if
of Theorem
48.3.
part of Theorem
48.3.
Theory
32. Let of
a be
algebraic of degreen over F. Show from Corollary F fixed. a subfield of F and leaving
\302\243(a)onto
48.5 that
there
are at
most n
different
isomorphisms
\302\243.
Part
424
X
\342\226\240 be 33. ~LetF(ai,\342\226\240 \342\226\240, an) is completelydetermined
34. Let
\302\243 be
an
that
a
Show
an ex tension
values er(o!;). a field F, and let a be an automorphism extension of algebraic of the set of all zeros of irr(a, F) that induces a permutation
Es
saw
an algebraic
in Corollary
23.17
the
that
is irreducible
over Q for every
a.
f, f2,
c.
prime
xp = xp~l x \342\200\224 1 \342\200\224 1
p. Let f
+ xp~2
be a zeroof
+ x
H
\302\242,,0),
and
the
H
subgroup
described
48.3
F. Is
over
over
\302\243?
\302\243 be
the
and let x be an their values on x.
of a field
automorphism
squares of elements of \302\243. An automorphism of the field
c. If a is an d. Theonly section
automorphism
of
automorphism
of R
49
of F(a)
over
M.
combinations
linear
isomorphisms for
indeterminate
sequence of
following
which
isomorphism
a field,
by describing
39. Prove a. An
conjugation
there a similar
e E.
of
of
\302\243 such
then G(\302\243/\302\243),
+ 1 the field Q(f).
consider
are distinct
and consider
Let a
collectionof automorphisms
generates
of f,
elements
b.
leaving F fixed.
in E.
basis for Q(f) over Q, elements of G(Q(f )/Q).
\342\200\242 \342\226\240 \342\200\242, fp_1
is a
Let
if S
that
be a
e 1}
{ct,- \\i
of E are
that they are all )/Q) is abelian
37. Theorem
leaving
,an)
zeros of \302\242,,0), and conclude Deduce from Corollary 48.5 and part (a) of this exercisethat G(Q(f Show that Show that the fixed field of G(Q(f )/Q) is Q. [Hint: that
Show
F fixed
\342\226\240 \342\226\240 \342\226\240
F(ai,
polynomial
cyclotomic (X) =
0
b.
a of
automorphism
any
by the n
We
that
38.
Show that
of F.
field
Theory
extension of a field F. Let S = element of F fixed. Show leaves each every at = Eh-
35. Let \302\243 be
36.
Galois
and
Automorphisms
case
the
where
in \302\243(/6)
with
\302\243. Determine
f2,
a and
the case that
fi
a
the
and
all automorphisms
fi
of
\302\242,,0).
I. of order p \342\200\224
\342\200\242 \342\200\242 \342\200\242, fp_1
were
zeros
are left
fixed by
all
conjugate algebraic are both transcendental
of F(x)
leaving
\302\243 fixed,
theorems. \302\243 carries
R of and
is the
real
a <
elements
that are squares of
numbers
carries
b, where a, b
identity
The Isomorphism
positive
e R,
then
elements in
\302\243 onto
elements
that are
numbers onto positive numbers. < o(b).
o(a)
automorphism.
Extension Theorem
The ExtensionTheorem and the next, we shall automorphisms of fields.In this section and the number of automorphisms of a field E. existence \302\243 an F want that is extension of and that we to find some algebraic Suppose of E. know from Theorem 48.3 that if e E are over We a, /3 automorphisms conjugate Of E then there is an of onto e F, course, a, /3 F(a) \302\243(/6). isomorphism i/^^ implies < E and F{fi) < E. It is natural to wonder whether the domain of definition both F(a) of Vv./S can be enlarged from F(a) to a larger field, perhaps all of E, and whether this A an E. of this situation lead to of is might perhaps automorphism mapping diagram than of the domain of definition of shown in Fig. 49.1. Rather \"enlarging speak -fa^,\" it Let
be
us continue
concerned
studying with both
the
Extension Theorem
The Isomorphism
49
Section
E
:
>-E
F{a)
'-
\342\226\272 Ftf)
E
\342\200\224
49.1 Figure
all of E.
we are
that
Remember
Theorem shows that that is,
the subfield
maps E into
are very
with
to
try
fi
F'[x].
It
Sincea
a
situation
we need Zorn'slemma
49.3
Theorem
an
=
ao +
a\\x
+
of
E,
extension
this
\342\200\242 \342\200\242 \342\200\242
+
anxn,
+
is
an
cr(ai)x
isomorphism,
-\\
h cr(an)xn.
we know
that
is irreducible
q{x)
can be mapped isomorphically /J. (Thisis not quite Theorem renamed by the isomorphism a.)
onto
F'(fi)
48.3, but
it
in by a
is close
= E, we are If F{a) element in E not in F{a) and continue the of an algebraic much like that in the construction closure very the trouble is that, in general, where E is not a finite extension, to be repeated a (possiblylarge) infinite number of times, so to handle it. For this reason, we postpone the general proof of
to the end
of this
find
another
section.
Extension Theorem) Let E be an algebraic extension of F F'. F' be an onto a field Let closure isomorphism algebraic
(Isomorphism
a be
50. Thus
the conjugation isomorphisms -fa^ will guarantee theorems at for fields. Extension least, mappings, many
that F(a) and mapping a onto
reasonable
F of a field F. Again the process may have 49.3
gives an automorphism
in Section
our
ff(flo)
map extending to it; a few elementshave been done.If F(a) ^ E, we have to
Theorem
under
The Isomorphism Extension be extended to an
F.
F' of
e F'[x].
seems
It is
of F
mapping of
eE
a
process.
extension
is a
look
q(x) Here q(x)
algebraic
which
situations. particularly in algebraic and topological E situation. that is an at this general algebraic Suppose a field F and that we have an isomorphism a of F onto a field F'. Let F' closure of F'. We would like to extend a to an isomorphism r of E onto F'. This situation is shown in Fig. 49.2. Naively, we pick a but not in extend a to F(q). If
a zero in
be
t,\"
algebraic closureF of
p(x) = in(a, F) let
Figure
to a map
question we shall study
is a
itself,
take a more
extensionof be an algebraic a subfield of
all
that
^ f
in mathematics,
important
Let us
i//^
can indeed always mapping i/^ this extension of F. Whether
of lots of isomorphism
existence
map
always assuming
used in conjunction
theorem,
F and
a
E onto
isomorphism of
the
in a fixed
contained
are
consideration
49.2
speak of \"extending
is customary to
*-t\\E\\
Z
F
425
of a field F. Let of F'. Then a
426
Part X
can be for all
extendedto a e F.
We
If
E
< F
an
an
algebraic
isomorphism
to an isomorphism of E
Proof
Proofof
this
we replace As
49.5Corollary
Proof
corollary
F
by
another
of F is unique,
E onto a
x of
isomorphism
corollary the existence ifa,p, as discussedat the
is
conjugation
Theory
as a
give
isomorphisms
49.4 Corollary
Galois
and
Automorphisms
F(a).
of an
extension
of F
-fa^ :
F{a) -> F(f3),
onto
F' by
E are conjugate
by
the
identity
F onto
onto
of a
the
extended
of the
statement
a
theorem
\342\231\2
that
F is isomorphic
closure
an algebraic
to
under
F'
an
to an
49.6). x~l : x[F] -> F
need
We
onto
can
be
F, we
\342\231\2
49.6
Index
can be
49.3, the map of F. Since x~l is already subfield
Theorem
by
conjugation
over F, then
of F onto F can be extended that leaves F fixed (see Fig.
isomorphism of F'
a subfield mapping that is onto F'. But show x only to an isomorphism of F' extended must have x[F] = F'.
The
in the
as we promised earlier, corollary, we can show, an to up isomorphism leaving F fixed.
49.3,
a{a)
F.
F and F' be two of F. Then closures algebraic isomorphism leaving each element of F fixed.
isomorphism t
our
givenbyTheorem48.3,
from Theorem49.3if
F{fi), and F'
Let
By Theorem
one of
\342\200\224
section.
fi e
a,
such that x{a)
F'
of F.
a subfield
is immediate
and
of
extension of
of this
start
subfield
Figure
Field Extension
discussed the question of existence, we turn now to the question of how many. For a finite extension \302\243 of a field F, we would like to count how many there isomorphisms are of E onto a subfield of F that leave F fixed.We shall show that there are only a finite in G(E/F) is such an isomorphism, number of isomorphisms. Sinceevery automorphism a count of these isomorphisms will include all these automorphisms. Example 48.17 that G(Q(V2, and that 4 = [Q(V2, 7\302\276: Q]. While showed \\/3)/Q) has four elements, such an equality is not always true, it is true in a very important The next theorem case. Having
Section 49
takes the
49.7
Theorem
first
big
need, but
shall
we
Let
\302\243 be
F',
and
a finite F'
let
The IsomorphismExtension
in proving this. We state the theorem not make the proof any harder.
step
The
extension of
a field
Let
F.
algebraic closureof
be an
ct be an isomorphism the number of
Then
F'.
and
independent
determined
by the
finite,
in Fig. 49.8 may help us to follow This diagram is constructed in the following
F[ and F2' are
where
isomorphism of F[
algebraicclosuresof
onto
F'T
F', two fields E
that
way. Consider
and
F[
two
F'2, respectively.
by Theorem 49.3
Then
ct to
of F',
we are
an
and
ct.
and
F;
to
about
isomorphisms
17 0nt0 Z7' ct2 : F \342\200\224> F2,
0nt0 E\" Z7 : F \342\200\224> F,,
Ct!
onto a field
of F
extensionsof
construction
the
diagram
make.
general terms than
in more
does
it
isomorphism r of E onto a subfield of F' is That is, the number of extensionsis completely it is intrinsic to them.
Proof
427
Theorem
and
Now
o^ct,-1
49.5
Corollary
is an
there is
an
isomorphism
X:Y[^Y2
extending
this
isomorphism
each ti : E -> F[ that at E
and
going
first
:
aia^1
F[\342\200\224>F'T Referring
T2(a) = for a
e E.
ti
defining
by
Clearly t2
ct2 . The
extends
is,
by chasing
between
n :E
the other
-> F[ and
t2
way
: E
Written
could have
with t2
started
and recovered
F'2
the is one
diagram, shows to one. In view
that
of this
the
correspondence one-to-one
number of x extending a is independent of F', F' and ct. number of mappings ct is finite follows from the fact that since E is extending \342\200\242 \342\226\240 \342\226\240 \342\226\240in extension of F, E = F{a\\, \342\200\242, an E, by Theorem 31.11. a\342\200\236)forsomeo!i, \342\226\240,
correspondence,
the
That the
a finite
algebraically,
= (X~1r2)(a),
around
->
r2
49.8, corresponding to : E -> F2, by starting
(A.Ti)(a)
fact that we
Ti(a) that
to Fig.
a\\ we obtain an isomorphism to the left, then up, and then to the right. extends
f;
t{[E]
Extends
Fi
l U2a\\
-<-
>\"
T,(or) = (Arj)(a)
a2 ^1^-
49.8 Figure
r2[E]
-+ f;
Part
428
X
and Galois Theory
Automorphisms
a
are only
There
of possible
number
finite
= ai0
irr(a;, F) where
a^ e
F,
then
[cr(ai0)+ 49.9 Definition
49.10
Corollary
Proof
a(an)x -\\
Let \302\243 be a finite extension of a field of F leaving F fixed is the index < E < K, where : {K E}{E : F}.
If F
It
of F
F
leaving
fixed
F.
-\\
\\-
of E
each
subfield of F.
of e F'[x].
of isomorphisms of E onto
a subfield
over F.
\342\226\240
of the field
field
extension
that
in F', for if
x{at)
images
aimixmi,
in F'
zeros
The number
has {isT :
the
h cr(aimi)xm']
{E : F}
finite
49.7
Theorem
from
follows
subfield
is a
K
aax
of the
be one
must
xicct)
+
candiates for
F,
then
{E : F} isomorphisms extensions to an isomorphism E} of the
: F}
=
\302\243 onto
a
{K
of
t,
a
of K onto
\342\231\246
The preceding corollary was really the main thing we were after. Note that it counts even if it is only called something, something. Never underestimate a result that counts a \"corollary.\" field of characteristics We shall show in Section 51 that unless F is an infinite p ^ 0, we always have [E : F] = {E : F] for every finite extension field E of F. For the case : F} extensions E = Fia), the {F(a) of the identity map i : F ->- F to maps of F(a) of F are given onto a subfield by the conjugation isomorphisms ijratp for each conjugate : F} = n. if irr(a, distinct zeros in F, we have {\302\243 fi in F of a over F. Thus F) has \302\273
We shall show later that unless F distinct zeros of irr(a, F) is deg(a,
49.11Example
F =
Consider
that {E
: Q} =
is
We
restate
the
the Isomorphism
a be an isomorphism can be extendedto for a e F. Consider isomorphism of
pairs L onto
a
0, the
number
of
that : Q}
example
= 2,
shows
so
= (2)(2).
: Q}
: Q(V2)}{Q(V2)
in
Theorem Extension Theorem 49.3.
Let E be an algebraic extensionof a field F. Let onto a field F'. Let F' be an algebraic closureof F'. Then a x of E onto a subfield of F' such that x{a) = a{a) isomorphism Theorem
of F an
(L, X), where
all
p ^
\342\226\262
Extension
IsomorphismExtension
Proof
of characteristic
: F].
49.10
Corollary
of
Proof
and
= [F(a)
Q(\\/2, V3) overQ,as in Example 48.17. Our work [E : Q] = 4. Also, {E : Q(\\/2)}= 2, and {Q(V2)
4 = {E : Q}= {E This illustrates
infinite
F)
subfield
of
F'
L is a field
such
such that X{a)
that
F <
L < E and a e F. The
{a) for
\342\200\224 a
A is set
an 5 of
such
Section 49
pairs (L,
this relation <
Let T
since (F,
is nonempty,
X)
< (L2,
(Li,Xi)
The IsomorphismExtension
X2), if Li
doesgive
<
and
L2
a partial
a) is such a pair.
a e
Li. It
checked
is readily
that
of S.
ordering
1} be a chain
ordering on S by
a partial
Define
= X2(a) for
Xi(a)
429
Theorem
claim that H = {Ji\342\202\254l H is a subfield of \302\243. Let a,b e H, where a e Hi and & e H2; then either Hi < H2 or H2 < Hi, since T is a chain. If, say, Hi < H2, then a,b, e H2, so a \302\261 b, ab, and a/& for b 7^ 0 are all F c H c E. Thus in H2 and hence in H. Since for each ie/,FCfljC\302\243,we have H is a subfield of E. c e H. Then c e H, for some i e I, and let Define X : H -> F' as follows.Let =
I;' e
{(Ht,Xi)
of
S. We
A.(c) =
The map
X is
oi(H2,X2)
is an a,
isomorphism of H onto
b e
H,
Hi andc e H2,
because if c e T is a chain. since Ai),
well defined
< (Hi,
Xt(c).
In
either
of F'. If
a subfield
(Hi, Xi) <
either
then
case, Xi(c)
a, b e
H then
X-^b)
= X(a)
=
X(a)X(b).
= X2(c).We there
is an
(H2, X2)
claim
that X
H such
that
and
+
A(a
=
\302\243)
+ b)
=
Xi{a)
= X:(ab)
=
Xi{a)Xi{b)
Xi(a
+
+
X(b).
Similarly,
X(ab)
= 0, then a e Hi for some i implies that X;(a) = 0, so a = 0. Therefore, X is of H and X that an isomorphism. Thus (H, X) e S, and it is clear from our definitions (H, X) is an upper bound for T. have shown that every chain of S has an upper bound in S, so the hypotheses We exists a maximal element (K, r) of S. Let of Zorn's lemma are satisfied.Hence there if K + E, let a e E but a \302\243 K. Now a is algebraic x{K) = K', where K' < F7.Now over K. Also, let p(x) = irr(a, ^T). Let -fa be the canonical over F, so is algebraic If X(a)
a
isomorphism
fa
to the
corresponding
-*
: *[*]/(*)>
*(\302\253),
: K[x] evaluation homomorphism \302\247a
/?(x) =
+
flo
+ \302\253i*
\342\200\224> ^T(a).
If
\342\200\242 \342\226\240 \342\226\240
+
anxn,
consider
#(*) = in i\302\243'[x]. Since
is a zero
a'
x is an
of q(x)
T(fl0)
+
T(fli)x
h T(fl\342\200\236)x\"
H
isomorphism, q(x) is irreduciblein
in F'.
Let : K'[x]/(q(x)} Vr\342\200\236-
-*
*'(\302\253')
isT'[x].
Since
K' <
F', there
X
Part
430
and Galois Theory
Automorphisms
\342\226\240
K(a)-*-
\342\226\240
K\\x]/ip{x))
49.12
isomorphism analogous to
be the
f :
be the
Fig.49.12.)
the
Then
K
x on
composition
isomorphism of that
contradicts
which
K(u) (K,
and
K'[x]/(q(x)) x +
mapping
(q(x)}.(See
(p(x)} onto x +
of maps
fa'ff-1 is an
let
Finally,
ijra.
K'{a')
K'[x]/(p(x))
Figure
K[x]/{p(x)) -+
extending
isomorphism
\342\226\240
-*-
: Kia) -+
x) is maximal.
Clearly, (K, x) <
of F'.
a subneld
onto
K'ia')
(K(a),fa'xf~x),
have had K
we must
Therefore
= E.
\342\231\246
49
\342\226\240 EXERCISES
Computations
Let E =
V5,
Q(\\/2,
1. i
It can
V5).
of mapping of a subfield Describe the extensions Q.
E, give
^15)
-+ Q(V2,
2. o- : Q(V2,
715)
-+ Q(V2,
3-
: Q(^30)
It is
a fact,
we can
which
72, as
4. Describe
all
5. Describe
all
6. Describeall onto
7. Let
is the
usual,
i
where
1 through [E : Q] = 8. In Exercises 3, for the given isomorphic of the mapping to an isomorphic mapping of E onto a subfield of the
is the
V5, V5} for E over Q.
set {-Jl,
generating
identity map = V2
cr(V2)
that the
cubing,
by
verify
a2 =
^2,
real cube root
extensions
of the
identity identity
of the
and
-715
zeros of x3
-l + iV3 ' 2
72
of 2. Usethis
and \"\"\"
of Q
to
an
map
of Q
to
an
iAv5,_V3
a3 \"J
isomorphism isomorphism of
72 \"
in Exercises
information
map
automorphism
Q are
\342\200\224 2 in
Q(75)
-1-(73 2
4
6.
through
mapping
Q(72)
onto a subfield
mapping
Q(72,
73)
to an isomorphism
onto
mapping
of Q.
a subfield Q(i,
of Q.
the automorphism
the fixed field b. Describe all extensions a. Describe
values on
715)
of the
a subfield a be
that
-/15), where
extensions
extensions
extensions
-\342\226\272 Q(V30)
= \302\253i
where
shown
all
by giving
: Q(V2,
^V30,-V30
be
of Q(jt) that
maps
n onto
\342\200\224 it.
of a. of a
to
an
isomorphism
mapping
the field Q(/jt)
onto
a subfield
of Q(jt).
75,
of Q. 7\302\276
Section 50
431
Fields
Splitting
Concepts
8. Mark each of the
true or
following
a. Let F(a) be
false.
extension of a field F. Then every isomorphism of F onto a subfield of F isomorphism of F(a) onto a subfield of F. b. Let F{a)be any simple algebraic extension of a field F. Then every isomorphism of F onto a subfield of F has an extension to an isomorphism of F{a) onto a subfield of F. c. An isomorphism of F has the same number of extensions of F onto a subfield to each simple
has
simple to an
any
extension
an
algebraicextension
of F.
d. Algebraic
closuresof isomorphic
e. f.
closures
g.
Algebraic Any The
closure
algebraic
of a
index
finite
h. The index behaves i. Our remarks prior
with respect to finite towers of statement of Theorem49.3essentially E over F. for a finite extension 49.5 shows that C is isomorphicto Q.
theorem
j.
fields are always isomorphic. are not isomorphic are never isomorphic. of Q(V2) is isomorphicto any algebraic closure of Q(\\/l7). E over a field F is finite. extension
of fields that
Corollary
multiplicatively to
the first
of fields.
extensions
finite
a proof
constitute
of
this
Theory
9. Let K be an algebraically K is algebraic over a [K]
10. Let F
\302\243 be fixed
closed is an
field. Show
automorphism
an algebraic extension of a field be extended to an automorphism
that
F.
can
11. Prove that
if
is \302\243
an algebraic
a of K onto a subfield of isomorphism Theorem is, is an onto map. [Hint: Apply
every
of K, that
extension of
that every
Show
isomorphismof E onto
a subfield
itself
such
that
49.3 to a~1.] of F leaving
of F.
a field
F,
then
two algebraic
closures F
and
\302\243 and of \302\243
\302\243,
are isomorphic.
respectively,
12. Prove that the algebraic closure of Q(v^7) in C is isomorphic and x is an indeterminate. field of algebraicnumbers
to
any
algebraic
closure of
Q(x), where
Q is
the
is a finite extension of a field 13. Prove that if \302\243 The remarks preceding F, then {\302\243:\302\243}<[\302\243: \302\243]. [Hint: showed this for a simple algebraicextension of F. Use the fact that a finite F(a) Example 49.11essentially is a tower of simple with the multiplicative extension of the index and degree.] extensions, together properties
section
50
Fields
Splitting
interested chiefly
of a field E, rather than mere of a field that form automorphisms for some of a field F, every isomorphic field \302\243 of F leaving F fixed is actually an automorphism of E. is an algebraic extension of a field E. If a e E and /3 e F is a conjugate Suppose \302\243 over F, then there is a conjugation isomorphism
are
We
going
to be
isomorphic mappings of E onto a group. We wonder whether of E onto a subfield mapping of
a
fa,p
By Corollary49.4,i(raj
of F. Now
if fi ^E,
Thus, if an algebraic
in automorphisms
a subfield
of E. It extension
: F(a)
is the
-+ F(P).
can be extended of E onto a subfield to an isomorphic mapping such an isomorphic mapping of E can't be an automorphism of E. onto extensionE ofafield F is such that all its isomorphic mappings
Part
432
X
a subfield of F leaving
all conjugates
and the ideas
These
are actually automorphisms ofE, then for every a e E, to come be in E also. This observationseemed very that we used a lot of power, the existence of the conjugation namely
out
point
isomorphisms
Let F bea field
suggest
with
subfield
F is a splitting
K <
F
Isomorphism Extension Theorem49.3. definition. the formulation of the following closure F. Let {/,-(x)| i e 1} be a collection ofpolynomials F if E is the smallest field of {/, (x) | i e over splitting F and all the zeros in F of each of the for i e I. A field fix) in over F if it is the splitting field of someset of polynomials m
F is
We
see
1}
the
field
F[x}.
50.2 Example
must
algebraic
A field E < of F containing
in F[x].
Theory
F fixed
a over
of
easily. We
50.1 Definition
Galois
and
Automorphisms
V5] is a splitting
that Q[\\/2,
- 2, x2
of {x2
field
-
also of {x4
3} and
-
5x2 +
6}.
\342\226\2
field of {/(x)} f(x) e F[x], we shall often refer to the splitting F. Note that the field of of over f(x) {f (x)| i e 1} splitting splitting of all subfields of F containing F and all zeros in F of over F in F is the intersection a splitting field surely does exist. such each f(x) for i e I. Thus E < F with fields over F are preciselythose fields the We now show that splitting that all of E onto a subfield of F F fixed are leaving property isomorphic mappings of E. This will a of the next theorem. be Once we are more, corollary automorphisms a concept in terms of mappings. Remember,we are always characterizing assuming F under consideration that all algebraic extensions of a field are in one fixed algebraic closure F of F.
one polynomial
For
50.3
Theorem
over F
as the
A field
E, where
of F
field
F
leaving
F <
E < F,is
maps E
fixed
field over
a splitting
and thus
itself
onto
F if and only
induces
an
if every
automorphism
automorphism of E leaving
F fixed.
Proof
Let \302\243 be F leaving
field over F
a splitting
F fixed.Let {a,-
F of
in
{f(x)
be the work shows that \\ j
e J}
i e I. Now our previous all expressionsof the form
= a0+
g(aj) where nj is finite
products
the
degree
of irr(o,-,
of elements
of
the
F)
\\i
for
a fixed
axaj -\\
and
form
/}, and let a of all zeros
e
collection
ak
h
e F.
g{Uj)
ajt
field
the
Consider
automorphism
F of
F(aj)
all the has as
of
f(x) for elements
,
flnj-ia/
for all j
be an in
the
e J.
set The
S of all finite sums of S is a subset of
E
set
and multiplication and containing inverse 0, 1, and the additive \342\226\240 \342\226\240 \342\226\240 c of each element. Sinceeach element of S is in some we see that S 5, F{aj], ,ajr) also contains the multiplicative inverse of each nonzero element. Thus S is a subfield of E containing all a;- for j e J. By definition of the splitting field E of {f(x) \\ i e /}, we see that we must have S = E. All this work was just to show that {a; \\ j e J] generates E over F, in the sense of taking finite and finite products. Knowing sums we this, see immediately that the value of a on any element of E is completely determined by closed
the
under
values
addition
o{aj).
But
by
Corollary
48.5,
o{aj)
must also
be a zero
of
irr(o;;,
F). By
Splitting Fields
Section 50
Theorem
irr(ay, F) divides the E onto a subfield of E
29.13, a
Thus
maps
F. Sincefor
a~l of
automorphism
for which
f(x)
e E also.
= 0, so a(aj)
fi(aj)
is
the same
However,
isomorphically.
B e
433
the
of
true
E,
P=a(a-\\p)),
we see that
E onto
E,
an automorphism of F leaving every automorphism Suppose,conversely, E. an in F[x] of Let be irreducible g(x) automorphism polynomial a
maps
induces
thus
and
of E.
If
zero of g(x) in F, then by Theorem 48.3, of onto F(fi) leaving F fixed.By Theorem F(a) iraip x of F onto a subfield of F. But then isomorphism fi
is any
:
t_1
x[F]-\302\261
a zero
having
is a
there
conjugation can be i//^
49.3,
induces an a in E.
fixed
F
that
isomorphism
extended to an
F
the image of of automorphism an automorphism of E, so x (a) = fi F leaving F fixed. Then by assumption, x induces in F[x] is in E. We have shown that if g(x) is an irreducible polynomial having one zero is the set of all irreducible in E, then all zeros of g(x) in F are in E. Hence if {ft(x)} \342\231\246 E is the splitting of {gk(x)}. in F[x] having a zero in \302\243, then field polynomials can be
extended to
t_1 is already
50.4
50.5
Definition
Example
The -
Proof
we see that
F is
zero in
\302\243 splits
If
x must
a
\302\243 is a
splitting
automorphismof
\302\243. The
in
in
factors
over
field
splitting
F.
a field
of
- 5x2 + 6 polynomial xA + V2)(x + V2)(* V3)(x
< If \302\243
F onto a subfield have been onto F,
in
A
so x
f(x)
polynomial
F. Since
of
mapping
isomorphism
Let E be an extensionfield factors into a product of linear
(x
50.6 Corollary
an
of F,
all
is an
e F[x]
splits
it
\302\243 if
in
\342\226\240
\302\243[x].
in
splits
Q[x]
the
field
Q[V2, VJ]
into
A
V3).
irreducible
then \302\243,
every
then \302\243,
every automorphism of Theorem 50.3
polynomial
in
a
\302\243[x]having
\302\243.
field over F
in
half of
second
the
proof
of F inducesan showed preciselythat
\302\243 is
in \302\243[x] the splitting field over F of the set {&(x)} of all irreduciblepolynomials a zero in \302\243. an irreducible polynomial f(x) of \302\243[x]having Thus a zero in \302\243 has having all its zeros in F in \302\243. its factorization into linear factors in \302\243[x],given by Therefore, Theorem 31.15, actually \342\231\246 takes place in \302\243[x],so f(x) splits in \302\243.
also
50.7
Corollary
< \302\243 \302\243 is a splitting field over F, then every isomorphic F fixed is actually an automorphism of F and leaving of field of finite degree over F, then
If
=
: \302\243} {\302\243
Proof
Every isomorphism a
to an automorphism second
of the proof x restricted to
half
Theorem50.3, \302\243 over
mapping
x of
F,
\302\243 onto
that \302\243,
50.3.
is a,
particular,
\302\243 onto
if
a subfield
\302\243 is a
splitting
\302\243 is a
splitting
extended
F
fixed
can be
the
onto
argument
field over F,
then
of
the
by
is an automorphism of \302\243. Thus for a splitting field of \302\243 onto a subfield of F leaving F fixed is an
every isomorphic mapping
automorphism of \302\243.
of F leaving 49.3, together with If
of
\\G{E/F)\\.
a subfield
F, by Theorem
of Theorem
mapping \302\243. In
Part
434
X
The
of
50.8
Example
and Galois Theory
Automorphisms
{E : F] = F, since
equation
finite
degree
mappings
of E onto
{E : F] was
Observe
that Q(\\/2,
a subfleld
F fixed.
{x- -2,x2 over
the
that
\342\231\2
-3} ;,
mappings
leaving Q fixed. (Actually, subfleld fixed, we see that
Q(\\/2, V5) the prime
leave
must
48.17 showed
Example
Q.
automorphisms of
E
of
field
splitting
for a splitting field number of different isomorphic
immediately
as the
defined
F leaving
of
S) is the
follows
then
\\G{E/F)\\
over
and
a2,
a\\,
all the
are 0\302\276
of a field every automorphism are the only automorphisms of
since these
Q(V2,V3).)Then
\342\226\2
to determine conditions
wish
We
E of
extensions
finite
that this
always
equation
F is a finite 0 or when of characteristic p 7^
50.9
Example
Let 1/2 S0T) a linear over
Q
be the < K
only
We
that
can
verify
the
other
Q(^2, i V3). (This Further
study of
16,21,and
of x3 is not
this
the
following
section
\342\200\224
over Q of
x3
Thus the the same field as Q(^, in C.
example
=
6 overQ.
i, y/3), the
2 field which
exercises
E of x3 \342\200\224 2 over Q is is of de gree 12 over Q.) (see
Exercises
7, 8, 9, A
23).
50
\302\245 EXERCISES
Computations In
1 through
Exercises
1. x2 + 4. x3
-
6,
find
the
degree
\342\200\224 2
= 6.
of degree
splitting
is left to
(2)(3)
\342\200\224 2 is
^2-1-^
and
\342\200\224 2
interesting
: Q]
: Q(^2)][Q(^2)
2
for into
Then
the splitting field by cubing that
zeros
in
2 does not split in Q(4/2), 2, as usual. Now x3 of x3 - 2 isreal. Thus x3 - 2 factors in (
r2~1+iV~3
are
We shall show
topic.
root of
factor x \342\200\224 1/2 and is therefore of degree2 over
shown
next is a \302\243
one zero
: Q] = [E [\302\243
We have
= [E:F]
splitting field over a field F of characteristic field equation need not be true when F is an infinite
when This
field.
= {E:F}
0.
cube
real
and
F. Thisis our holds
which
under
\\G(E/F)\\ for
V3)/Q)| = 4.
50.7.
Corollary
illustrating
= |G(Q(V2,
: Q}
73)
{Q(V2,
over Q
3
2.
x4
3
5.
x3
of the - 1 - 1
splitting
field over
Q of the
given
in Q[x].
polynomial
3. (x2 6. (x2 -
2)(x2 2)(x3
-
3)
-
2)
Exercises
Section 50
50.9 for
to Example
Refer
7. What
is the
order
8.
What
is the
order of
9.
What
is the
order of
10. Let
use the
Exhibit
G(Q(-^2,iV3)/Q)? G(Q(^2, i V5)/Q(^2))?
+ 1overZ2.
+x2 + 1 splits
that x3
Show
1,
+ x2 +
of x3
zeros
more
two
of Section
results
9.
of G(Q(-s/2)/Q)?
of x3 +x2
a zero
a be
in Z2(a).
Exercises7 through
435
in
to a,
addition
[Hint: There are
in Z2(a).
among these
elements
eight
elements.
eight
Alternatively,
33.]
Concepts In
11.
Let F
if E 12.
12, correct the
11 and
Exercises
needed,so that
it is
< E
in a form
< F where all the
contains
F is an algebraic closure ofa field F. The field E is a splitting in F[x] that has a zeroin E. zeros in F of every polynomial
in an
/(x) in F[x] splits of lower degree. polynomials
13. Let
be a polynomial in F[x] /(x) bounds can be put on [E : F]?
a. Let a, fi
e E,
leaving
F fixed
extension
field
of degree
n. Let
where E < F is and mapping a
is a splitting R is a splitting
b. R
field
over Q.
c. d. e. f.
field
over R.
field
over R.
C is a splitting Q(i) is a splitting Q(7r) is a splitting
E < F be
the
factors in F[x]
if it
field of
splitting
F
and
if
only
a product
into
fix) over F
F
in
of
What
a splitting
field over F. Then and only if irr(o;, F)
/J if
onto
there
an automorphism
exists
= irr(/J,
of E
F).
over Q.
field
splitting
and only
E of F if
over
field
false.
true or
following
g. For every
text, if correctionis
to the
reference
without
acceptablefor publication.
A polynomial
14. Mark each of the
italicized term
of the
definition
field
over Q(7r2).
field
E over
F,
where
E
< F,
every isomorphicmapping
field
E over
F,
where
E
< F,
every isomorphismmapping
E
< F, every of E.
of
E is an
of E. automorphism
h. Forevery
splitting
of F is an automorphism of E. i. Forevery splitting field E over F, where of F and leaving F fixed is an automorphism F is closure F of a field j. Every algebraic 15. Show
by
an example
that Corollary
16. a. Is |G(F/F)|multiplicative
for
Why or why b. Is |G(F/F)|
not?
[Hint:
Why
finite towers
or why
for
of
finite
if the
finite
7 through
towers of
field
over
that is,
F <
for
a subfield
E onto
a subfield
F.
word irreducible is
extensions,
\\G{K/E)\\\\G{E/F)\\
Use Exercises
multiplicative
bottom field?
50.6 is no longer true
=
\\G{E/F)\\
a splitting
isomorphismmapping
E onto
deleted.
is
E<
K
<
F?
9.] finite
extensions,
each
of which is
a splitting
field
over the
not?
Theory
17.
Show
that
polynomial
if a in
finite
F[x].
extension
E of
a
field
F
is a
splitting
field
over
F, then
E is a
splitting
field
of one
Part
436
18.
Show
19. Show
X
that
if
that
for
: F]
[E
= 2,
E < F,
F <
E is
then
elements.
20.
Show
Q(^2)
21.
Referring to Example 50.9,show
the
has only
a splitting
E isa splitting
each of its that
and Galois Theory
Automorphisms
a. Show
determined
c. Show that natural way 23.
Let E
be
of a
an automorphism by the permutation
that
Show
E
all
contains
over F in F for
conjugates
~
fV3)/Q(rV3)) E over F
field
splitting
+>.
of a polynomial
e F[x]
/(x)
permutes
the
zeros
in E.
of /(x)
b.
only if
and
that
an automorphism
that
if
automorphism.
identity
G(Q(^2, 22.
F
over
field
F.
over
field
the
of a
E over F
field
of f{x) in
E
(a).
then G(E/F)
e F[x],
f(x)
is completely
e F[x]
/(x)
polynomial
in part
given
a polynomial
field over F of is a splitting certain group of permutations.
if E
canbe viewed
in a
as a
field of
splitting
a. What is
of a splitting of the zeros
as
\342\200\224 2
over Q,
in
50.9.
Example
[Hint: Use Corollary
of G(F/Q)?
order
the
x3
and Corollary
50.7
49.4 applied
the
to
tower
Q <
Q(iV3)
b. Show
that
= S3, the
G(E/Q)
on three letters.
group
symmetric
[Hint:
Use
Exercise
22, together
with
part
(a).]
24. Show
that
a prime
for
Corollary 23.17.] F and F' be two
25. Let
F of /(x) in
section
F is
51
p,
the
algebraic
splitting
field
closures
of a
to the
isomorphic
over Q field
over F
\342\200\224 1
is of
degree p
- 1 over
Q.
[Hint:
Refer to
/(x) e
F[x]. Show that the splitting field E over of f(x) in F'. [fffrtf; Use Corollary 49.5.]
Separable Extensions
Multiplicityof Zeros of
consideration
under
Our
v if v
next
aim
e F[x].
is
Polynomial
are now always assuming are contained in one fixed is to determine, for a finite
{E : F] = [E : F].The zeros of polynomials.
Let/(x)
a
that we
Remember
51.1 Definition
and let
F,
field E'
splitting
of xp
the
key
to answering
An element a
greatest
integer
of
F such
such that (x
extensions of a field closure F of F.
that all algebraic algebraic
E of F, under extension what conditions this question is to considerthe multiplicity
that/(a) \342\200\224is
a)v
= 0 is a zero of/(x) a factor of f(x) in F[x].
of
F
of
multiplicity
\342\226\2
the multiplicities of the zeros of one given irreducible The next theorem shows that polynomial over a field are all the same. The ease with which we can prove this theorem of the power of our is a further indication and of our whole conjugation isomorphisms of zeros of polynomials by means of mappings. to the study approach
51.2 Theorem
Let f{x) be
irreducible
in F[x].
Then all zeros of f{x) in
Fhave
the same
multiplicity.
Section 51
Proof
Let a
and
f$
t : F
be
xx leaves
48.3, there is a conjugation
a
xx : F[x]
natural
isomorphism
rx((x
A symmetric
of a.
Corollary
-
multiplicity of ft
(x
tyaj
-
to an
isomorphism
xx{x) =
with
\342\200\224> F[x],
F fixed.
leaves
x.
However,
P)v,
is greater
in f(x)
the reverse
gives
argument
=
a)v)
be extended
can
or equal
than
inequality, so
the
to the
multiplicity
of a
multiplicity
equals
of p.
that
51.3
that the
shows
which
e F[x]and
fixed, since f(x)
f(x)
437
F. Then by Theorem By Corollary 49.4, \\f/aj
in
x induces
\342\200\224> F. Then
Now
of f(x)
zeros
: F(a)^-+F(P).
isomorphism \\jraj
SeparableExtensions
\342\231\246
is irreducible in
If f(x)
then
F[x],
has a factorization
f(x)
in F[x]
form
of the
i
the
where
Proof
The
F.
a e
F and
in
from Theorem 51.2.
is immediate
corollary
zeros of f(x)
the distinct
are
a,-
\342\231\246
we should probably show by an example that the phenomenon of can occur. We shall a zero of multiplicity greater than 1 of an irreducible polynomial over an infinite of field show later in this section that it can only occur for a polynomial this
At
point,
characteristic
51.4 Example
Let\302\243 =
p ^=0.
Zp(y),
\302\243. (See
where
y
is anindeterminate.LeU
51.5.) Now E
Fig.
= F{y)is algebraic
over
= yp, and let F bethe F, for y is a zero of
of
subfieldZp(/)
(xp
- tin F) must divide xp [Actually, By Theorem 29.13, irr(y, irr(y, We a proof of this to the exercises (see Exercise 10).] SinceF{y) leave F, we must have the degree of irr(y, F) > 2. But note that
F[x].
e F[x]. \342\200\224t)
F)
= xp
is not
- t.
equal to
=
F(y) \342\200\236(y)
xp
sinceE has F=
Zp(t)
=
Zp(yP)
a zero of y
of
51.5 Figure
p (see of multiplicity
=
xp
-
yp
=
(x
Theorem 48.19 and > 1. Actually, xp
-
y)p,
the
following
comment).
irr(y, F), so the
\342\200\224 t =
Thus
y
is
multiplicity \342\226\262
here
on we rely that for
on
heavily
Theorem
49.7 and
its
corollary.
Theorem
48.3 and
of F there is one extension a simple algebraicextension F(a) ( zero of irr(a, F) and F F distinct into for isomorphism mapping every the only extensions of;. Thus {F(a) : F] is the number of distinct zeros of
show
identity are
these
irr(a, F). In
If
of our
view
recognize 51.6 Theorem
F)
irr(y,
corollary
of the that
characteristic
t
is p. From
its
-
\302\243 is a
the
work
potential
finite extension
with
of a
the
of F,
then
of Lagrange
theorem
theorem like
{E
this
: F]
next
divides
and
one.
[E :
F].
Theorem
31.4,
we should
Part X
438
Proof
and
Automorphisms
Galois
By Theorem31.11,if
irr(ai, F(ai,
v,-, by
multiplicity
E
\342\200\242 \342\200\242 \342\200\242, &7-1))
Theory
is finite have
a,-
= F(ai, of \302\273,distinct zeros
over F, one
as
E
then
\342\200\242 \342\200\242 where \342\200\242, a\342\200\236),
that are all
a,
of a
Let
e F. common
Theorem 51.2. Then
[F(au \342\200\242\342\200\242\342\200\242,\302\253;): f(\302\253i, ---,0:,-1)]
=
=
\302\273;V;
{F(ai,
\342\200\242 \342\200\242 : \342\200\242, a,-)
F(ai,
\342\200\242 \342\200\242\342\200\242, a,-_i)}v,-.
31.4 and Corollary 49.10,
By Theorem
[E:F] = Y\\niVi, and
=
{E:F}
{E : F}
Therefore,
A
F].
\342\231\246
Extensions
Separable
51.7 Definition
divides [F, :
Y\\ni.
E
extension
finite
F is separable over F if F(a) e F[>] is separableover
a of
element
a separable extension of
of F is
is a
/(*)
polynomial
F
F
if {F
: F}
= [E : F]. An
separable extension of F. An if every zero of fix) in F
irreducible
is separable
over F.
51.8 Example
\342\226\240
field
The
*
E =
= 4
{E : Q}
things a little
To make
a field
extensionof
F to finite
over
separable
easier, extensions
since
we
restricted our definition F. For the corresponding
we have E of
saw
in Example
of a
50.8
that
separable
definition
for
infinite
12.
see Exercise
extensions,
We
Q[\\/2, V3] is = [F : Q].
of distinct zeros of irr(o;, is the number Also, the F). the is the same as of each irr(o;, multiplicity F) multiplicity conjugate of a over 51.2. Thus a is separable over F if and only ifin(a, F, by Theorem F) has all zeros 1. This at that an us once tells irreducible of multiplicity f{x) e F[x] is polynomial F and over 1. has all zeros if separable of multiplicity only if f(x) that
know
of a
51.9
Theorem
If K K
Proof
is a finite
is separable
{F(a)
: F}
in
of E
extension
over F
if
and
and E is a finite extension of F, that is, F < E < K, then if K is over E and E is separable over F. only separable
Now
[K
: F]
= [K
: E][E : F],
{K
: F}
= {K
: E}{E : F}.
and
separable over F, so that [K : F] = {K : F}, we must have [K : E] = case the index divides the degree, [E : F] = {E : F}, since in each by Theorem 51.6. Thus, if K is separable over F, then K is separable over E and E is separable over F. Then
if
{K : E]
K is and
Section51 Separable Extensions For the
converse, note [K
51.10
Corollary
If
\302\243 is
[K
: E][E : F] =
extended in the obvious way, extension top field is a separable
The
separableextension F,
over F, and let
\302\243 is separable
F
a e
extension of
F,
since a,- is
F(a{) <
separableover
a-, is
F,
in(at, F(au
We now an infinite
to
turn
field
of polynomials.
of
the task
that
proving
of characteristic
the
a
if each
in
\302\243 is
\302\243 is
a finite
\342\200\242 \342\200\242 \342\200\242, a,-i),
because
1. Thus
multiplicity
over F
by
Theorem
F(ai, 51.9,
\342\226\240\342\226\240 \342\226\240, a{) is
extended \342\231\246
F[x], that must
Forr =
polynomial is, all at e F.
show
that a,
fail
to be
separable over F only if F is introduce formal derivatives
is to
lemma
instead.
we shall, for one, Formal derivatives are developed
22. of F, and let
fix) = xn monic
a can
One method
elegant technique, and also a useful
following
closure
algebraic
p ^=0.
is an
this
While
of brevity, use Exercises 15 through sake
Let \302\243bean
We
only
Fields
Perfect
Proof
and
\342\200\242 \342\200\242 \342\200\242, 0^))
irr(a,-, F), so that a, is a zero of q{x) of so \302\243 is separable separable over F(ai, \342\226\240\342\226\240\342\226\240,a,_i), by induction.
any
only
F. Since
over
separable over F(ai,
divides
be
one if
\342\226\240 = F(a.\\, \342\226\240 <\342\200\242\342\200\242\342\200\242<\302\243 \342\226\240, an).
ai)
F(a.\\,
q(x) =
51.11 Lemma
tower
finite
any
< E,
< F{a)
shows that F{a) is separableover F. that every a e E is separable \342\200\242such that there exist \342\200\242, o^, \342\200\242 an
F <
in
\342\231\246
E. Then
conversely,
Suppose,
the
: F}.
51.9
Theorem
Now
and
that
it.
under
over F if
is separable
E
then
: F] imply
by induction, to of the bottom
one immediately
of the
= {K
: F]
: E}{E
{K
= {E
[E : F]
over F.
Supposethat
and
: \302\243} and
= {K
: E]
can be
extension of
a finite
separable
Proof
= [K
: F]
51.9
Theorem
of finite extensions. if each field is a
that
439
+ an-\\xn~l
in F[x]. If
e F,
and
we
(f(x))m e
proceed,
+ axx
H
F[x] and
+ a0 m \342\200\242 1 =\302\243 0 in
by induction on
r, to
show
F, then
that
1, (f(x))m= xmn
+
(m
\342\226\240
l)an_lXmn-1
+\342\226\240\342\226\240\342\226\240+ aQm.
an-r
f(x) e
e F.
Part
440
X
and Galois Theory
Automorphisms
Since
e F[x],
(f(x))m
we have, in
particular,
(m
Thus
since m
e F, a\342\200\236_i
As
induction
(m where
^+i(fl,j_i,fl\342\200\236_2,
a\342\200\236_4. By
so
the induction
of the
for r
e F a\342\200\236_r
\342\226\240 = 1, 2, \342\226\240 &. Then \342\226\240,
the
form
\342\226\240 + l)a\342\200\236_(4+1)
\342\226\240\342\226\240 an-k), a+i(fl\342\200\236_i, a\342\200\236-2,\342\226\240,
\342\200\242 \342\200\242 is a \342\226\240, a\342\200\236-k)
\342\226\240 \342\200\242 formal polynomial expression in a\342\200\236_i, 0,,-2, \342\226\240. \342\226\240\342\226\240 we just stated, gk+i{an-i, an-2, \302\243 \342\226\240, F, an-k)
hypothesis
F, since m
e a\342\200\236_(jt+i)
that
suppose
in (f(x))m is
of xmn-{k+l)
coefficient
F.
F.
7^ 0 in
\342\226\240 1
hypothesis,
\342\226\240 e l)a\342\200\236_i
that
F.
7^ 0 in
\342\226\240 1
\342\231\246
a position to handle fields F of characteristic zero and to show that this amounts to E of F, we have {F : F} = [F : F]. By definition, zero is a separable extension. proving that every finite extension of a field of characteristic we give a definition. First, are now in
We
for a
51.12 Definition 51.13
Theorem
Proof
A
extension
finite
if every
is perfect
field
Every field
of characteristiczero
be a finite
Let E
extension
finite
f(x)
= irr(a,
irr^a,
F),
extension
F) factors in say, a = at.
and,
is a
separable extension.
\342\226\240
is perfect.
of a
field F
into
F[x]
of characteristic
Yli(x
zero,
~ ai)v>wnere the
let a e E. Then the distinct zeros of
and
a,- are
Thus
\342\200\242
/(*)=(rft*-a'>)
v
since
and
\342\226\240 1
7^ 0
for a field F
of
0, we
characteristic
eF[x]
(j\\(x-a,)\\ by
zero, we
proof is a bit
Proof
Every
field
finite
need
dj)v,
for
the
case of
of
a
having
F. finite
a
e E. By \342\231\246
field,
although
the
is perfect. field
of F. Let a e E. p, and let E be a finite extension in F into a is separableover F. Now f(x) = irr(a, F) factors the a,- are the distinct zeros of f(x), and, say, a = a\\. Let v = p'e, of characteristic
to show that
\342\200\224 Y\\j(x
through
degree in F[x] F for all a
over
harder.
Let F bea finite We
will also get us
54.11
Lemma
Theorem
then
51.10,
Corollary
51.14
Since f(x) is irreducibleand of minimal see that v = 1. Therefore,a is separable this means that E is a separableextension
51.11.
Lemma
as a
must have
where
Separable Extensions
Section 51
where
p
does
not divide
e. Then
f{x) = Y\\(x
-
=
aty
in
and
F[x],
by Lemma
f(x) = irr(a, F) is
Theorem 48.19 and
the
f{x)
if we
Thus,
~
degree over F
remark
following -
= Y\\(x i
is m
ai)p'
Y\\-,(x
having
it show
=
a-y'
1 j^Q in F. Since F[x] smcee \342\226\240 as a zero, we must have e = 1.
a
f] (xp'
Hence
-
g(x) e
have
must
a?). F[x].
Now
Consider F(ap ) = F(ap'). Then
zeros af . over F with distinct ctp' = (x \342\200\224 we see that a is over F. Sincexp' \342\200\224 a)p', field As a finite-dimensional vector space over a finite field.
that
then
i
as g(xp), we
regard f(x)
'
i
54.11,
of minimal
- a,)A
(T\\{x ^
i
is
441
the
only
F, F(ap
g(x) F(ap')
zero
of xp'
is separable is separable \342\200\224 ap' in
F.
) must be again a finite
the map F(ap') op : F(ap') -\342\226\272
by op(a) = ap for a e F(ap') is an Consequently, (ap)' is also an automorphism
given
automorphism of F{ap) by of F(ap
),
48.19.
Theorem
and
= ap'.
(ap)'(a)
Since an automorphism of F(ap') is an onto map, there is p e F(ap') such that = ap'. But then fip' = ctp', and we saw that a was the only zero of xp' \342\200\224 ap', (ap)'(p) so we must have p = a. Sincep e F(ap ), we have F(a) = F(ap ). Since F(ap ) was a is separable separable over F, we now see that F(a) is separable over F. Therefore, over F and t = 0. We have shown that for a e \302\243, a is separable over F. Thenby Corollary 51.10, E \342\231\246 is a separableextension of F. was to show that fields of characteristic0 completed our aim, which that is, these fields are perfect. have only separable finite extensions, extensions E of such perfectfields F, we then have [E : F] = {E : F}.
We
have
fields
finite
finite
and
For
The Primitive ElementTheorem The
51.15
Theorem
following
theorem
(Primitive Element
is a
If F is
theory.
be a finite separable extension of a field Theorem) Let \302\243
there exists a e E such That is, a finite separable
Proof
classic of field
that
F =
F(a). (Such
extension of a field
a finite field, then E is also finite. Let nonzero elementsof E under multiplication. element in this case. so a is a primitive
an
is a
a primitive simple extension. element
a is
a generator for the cylic (See Theorem 33.5.) Clearly,
a be
F.
Then
element.)
group E* of E = F(a),
Part X
442
Galois
and
Automorphisms
Theory
in the case that E = is general straightforward. Let distinct distinct zeros y = \342\226\240, Pi,- \342\226\240 pn, and let irr(y, F) have \342\226\240 \342\226\240 in is a separable extension F, where all zeros are of multiplicity 1, since \302\243 \342\226\240, Ym F. Since is infinite, we can find a e F such that
We
Yi,
that
assume
now
F(p, y). The in(P, F) have
induction
F is
prove
from this to
our theorem case
the
F
of
a + all i
for
and j,
= p +ay
a
and
infinite,
argument zeros p =
That
with
j^l.
Pi
+ ayj,
j^
(Pi
- P)/(Y ~
is, a(y -
7^ Pt
Yj)
all i
and all j
7^
a = p
P- Letting
+ ay, we have
so
^
a-ayj for
Yj) \342\200\224
1. Let
f(x) h(x)
pt
= in(p, =
F), and consider f(a - ax) e (F(a))[x].
= f(P) = 0. However, h(y) h(yj) 7^ 0 for j 7^ 1 by construction, since the pt and g(x) = irr(y, F) have a common factor the only zeros of f(x). Hence h(x) be linear, since y is the only common in (F(a))[x], must namely irr(y, F(a)), which and therefore is in F(a). Hence zero of g(x) and h(x). Thus y \342\202\254 ay F(a), p = a \342\200\224 Now
were
F(P,y) = F(a).
51.16
A
Corollary
Proof
of a
extension
finite
\342\231\24
This corollaryfollows We
see that
is a finite
the
field of characteristiczero at once
\"bad case\"
possible
only
of an
extension
from Theorems
infinite
is a
51.13 and
51.15.
where a finite
of characteristic
field
simple extension.
p
\342\231\24
may not
extension 7^
be simple
0.
51
\342\226\240 EXERCISES
Computations 1 through a is indeed in the given field. field is Q(a). Show that 4, find a such that the given your that of can direct the for the extension indeed be expressedas formal Q computation given generators by in Q. in your a with coefficients polynomials Exercises
In
Verify
1. Q(V2, ill)
2. Q(^2,Hi)
3.
4. Q(f, i/l)
V3)
Q(V2,
Concepts Exercises
In
so that 5.
it is
Let v
5 and 6, correctthe definition of the italicized term in a form acceptablefor publication.
F be
Z+ if \342\202\254
6. Let F
be
multiplicity
an
an
closure
algebraic
and only
if
(x
without
reference
to the
text, if correctionis needed,
of a field F. The multiplicity of a zero a e F of a polynomial a that is a factor of f(x) in F[x]. is the highest power of x \342\200\224
f(x)
\342\202\254 F[x]
is
\342\200\224
a)v
closure algebraic 1 of irr(a, F).
of a
field
F.
An element
a
in
F is
separable
over F
if
and
only if a
is a zeroof
Section 51
7.
an example of an /(x) Explain how this is consistent Give
8. Mark each of the
a.
Theorem
with
of every of every
Every
finite
extension
b. Every
finite
extension
c. Every
field
d. Every
polynomial
e. Every
polynomial of degreen
f.
irreducible
polynomial
Every
algebraically
closed
Every
field
Every
C
of
are all
443
2.
multiplicity
false.
true or
following
has no zeros in Q but whose zeros in 51.13, which shows that Q is perfect.
that
\342\202\254 Q[x]
Exercises
field
F is
finite
field
separable over F. F is separable over F.
0 is perfect. degree n over every
of characteristic of
F always
field
has n distinct
zeros in
F.
zeros in F. perfect field F always has n distinct of degree n over every perfect field F always has n distinct over
every
zeros
inF.
g.
h.
i. If E is
a finite
j.
a
E is
If
field is perfect.
algebraic extension E that is perfect. separable splitting field extension of F, then field extension of F, then |G(\302\243/F)| splitting
F has an
finite
= [E
\\G(E/F)\\
: F].
[E : F].
divides
Theory
9.
that if a, fi e F are both separableover F, then [Hint: Use Theorem51.9and its corollary.]
a \302\261 fi, afi,
Show
10. Show
that
{1, y, conclude
51.4, Example
11. Prove that
\342\226\240 \342\226\240 \342\226\240, yp~l}
by a
is a
basis for Zp(y) over Zp{yp),
degree argument
algebraic extension
if \302\243 is an
yS
# 0,
are all
y is an indeterminate. Referring t = yp. over Zp{t), where then E is perfect.
to
irreducible
of a perfect
field
F,
12. A
e E, F(a)
of a field F is a separableextension of F if for every a infinite) algebraic extension \302\243 (possibly is a (possibly infinite) is a separable extension of F, in the sense defined in the text. Show that if \302\243 extension of E, then K is a separable extension extension of F and K is a (possibly infinite) separable
13. Let
an algebraic
\302\243 be
14. Let \302\243 be a a.
of
a subfield
forms
finite field of
that the
Show
extension of a field
the \302\243,
b. Deducefrom
ap has
automorphism
(a) that G(E/ZP) is
15.
finite
Exercises
15 through
Let F be
any
/(x)
is
the
splitting
separable
field
=
i
field extension
set of all
elements
Use Exercise
\302\243. [Hint:
in
\302\243 that
of
F.
over F
are separable
9.]
and let
order n.
/(x) =
a$
+ a\\x H
n with
: F} {\302\243 \302\243 over
formal derivatives
22 introduce
=
generator ap.
[Hint:
that
Remember
: F] [\302\243
F.] in
F[x].
+ a,x'
h anxn
H
be in F[x].
The derivative
/'(x)
of
polynomial
fix) where
in
cyclic of order
|G(\302\243/F)|
for a
F
separable
order p\".
Frobenius
part
that the
Show
F.
closure of
separable
F.
over
separable
where
\342\200\224 t is
xp
that
and a/fi, if
its usual
\342\200\242 1 has
meaning
= for
ai-\\
i e Z+
+ (i \342\200\242 l)fl,x!_I
H
+ (n \342\226\240 l)fl\342\200\236xn_I,
and 1 \342\202\254 F. These are formal derivatives; no
\"limits\"
are
involved
here.
a. Prove that the map D : F[x] ->\342\200\242 F[x] given by D(/(x)) = /'(x) is a homomorphism 0. b. Find the kernel of D in the case that F is of characteristic c. Find the kernel of D in the case that F is of characteristicp / 0.
of
(F[x],
+>.
X
Part
444
16. Continuing
of Exercise
ideas
the
b. D(f(x)g(x))= /(x)g'(x) preceding
of
in the
g{x) e
F[x].[Hint:
v. Show that /(x) of multiplicity to Exercise 16 the of factorization (c) of
exercise
(a) of this
part
and
and only if a is
1 if
>
v
= (x \342\200\224 a)ug(x)
/(x)
also a
of /(x)
F[x].]
ring
Show from Exercise 17that every irreducible polynomial over a field the fact that irr(a, F) is the minimal for a over F.] polynomial
18.
Use
degree of /(x)g(x).] \342\202\254 [Hint: Use part (b).] F[x]. /(x) on the
induction
for all
a e F be a zero Apply parts (b) and
[//inf.-
/'(x).
proceed by
\\)f(x)m-lf'(x)
e F.
and a
for all /(x),
/'(x)g(x)
\342\200\242
Theory
that:
\342\202\254 F[x]
and let
\342\202\254 F[x],
/(x)
zero
+
exercise and
= (m c. D((/(x))m) 17. Let
15, shows
= aD(f(x)) for all /(x)
a. D(af(x)) the
and Galois
Automorphisms
F of
0 is separable. [Hint:
characteristic
Use
19. ShowfromExercisel7thatanirreduciblepolynomial^(x)overafieldFofcharacteristicp if and only if
each
20. GeneraUze Exercise 17,showing factor in F[x] have no common
21.
that
\342\202\254 F[x]
/(x)
and
gcdof
> 1 if
of degree>0.
and only
has no zero of multiplicity harder than in Exercise 20, show that /(x) \342\202\254 F[x] in F[x]. [Hint: Use Theorem46.9to factor have no common nonconstant in F[x] also.] and /'(x) in F[x], it is a gcd of these polynomials
/(x)
without
actually
section
52
for determining
procedure computational the zeros of /(x). finding
shows
section
that a
separable extensionK being separable as one
if
> 1 if
[E
is a >1,
\302\243 of
extension
followed
F can be split into of K to E that
a field
by a further
extension
inseparable
extensions
two
stages: a from
is as far
imagine. of totally
in
a fashion
parallel
to
our
extensions.
: F] = 1 < inseparable
F.
\342\226\240
know
We
that {F(a)
totally
inseparable over F
multiplicity
>1.
: F] is the if
and
if irr(a,
Zp(y)
indeterminate.
(Counterpart
of F,
number
only
Referring to Example 51.4,we seethat is an
Theorem
has a zero of multiplicity
E of a field F is a totally of F if {E extension inseparable An element a of F is totally F if over inseparable F(a) is totally
: F].
over
Example
\342\202\254 F[x]
and only if
that if 1
show
21.]
extension
finite
A
finite
can
whether/(x)
Exercise
and /'(x)
/(x)
Extensions
of F,
We develop our theory of separable development
52.1 Definition
Use
[Hint:
Inseparable
^Totally This
52.3
zero of multiplicity
has no
separable
p.
/'(x)
22. Describea feasible
52.2
by
a bit
Working
/(x)
/ 0 is not
q{x) is divisible
term of
of each
exponent
and
inseparable
This section
F
of distinct
zeros of
F) has
one
is totally
only
irr(a, F). Thus
zero
that is
is
a
of
inseparable over Zp(yp), where y \342\226\262
of Theorem 51.9) If K is a finite extension of E, E is a finite extension < K, then JT is totally inseparable over F if and only if K is totally over E and E is totally over F. inseparable < E
is not
used in
the remainder
of
the text.
Proof
E <
F <
Since
E] > 1 and
[K :
have
we
JT,
{.ST: F]
F. Then
inseparableover
=
is totally
: F},
: E}{E {\302\243
< [K : F]
= 1
and
: F].
< [F
= 1
: F}
{F
inseparable over F, and F is totally inseparable over F. is totally inseparable over F and F is totally inseparable over
is totally
K
445
have
{K : E] Thus
> 1. Suppose K
: F]
[E
= 1,and
: F} {\302\243
so we must
Totally Inseparable Extensions
52
Section
if K
Conversely,
F,
then
> 1.Thus
: F]
and [K
52.4 Corollary
is a totally
inseparable
inseparable over F,
is totally
F
F < = F,
F(a)
F(or) < totally
F,
we are
done, by
then
Theorem
inseparable
over
or
in
F,
or
^
F < since
because
a, is totally
F(or) <
F,
inseparable
totally
F.
of or totally
since F is totally
shows that
or \302\243 F. Then
with
inseparable over over
inseparable
for is F,a every or e F, with a \302\243 \342\200\242such that F, there exist o^, \342\200\242 \342\200\242, an
that
over
F(ai) < F(ai, a2) over
inseparable
q{x) = irr(o!;, F(ai,
of q{x) and is \342\200\242 \342\200\242, or,-_i), F(ori, \342\200\242
a e
let
and
definition
the
52.3
Conversely, suppose
F.
If
<
= F(ai,
\342\200\242 \342\200\242 \342\200\242 < F
F, a,- is totally
\342\200\242 \342\200\242 \342\200\242, a,-_i))
divides
We now Theorem
multiplicity and F is totally
is
that
a,-
totally
F, by Theorem
over
inseparable
over
so
\342\226\240 \342\226\240 is \342\226\240, or,-)
totally inseparable
\342\200\242 \342\200\242, a;_i), F(ai, \342\200\242 is the only zero
inseparable over 52.3, extendedby \342\231\246
closelyparalleled
our
work
in Section
51 that
could
we
have
Closures
come to our
main
Let F have characteristicp is totally
<
\342\200\242 \342\200\242 \342\200\242, a\342\200\236).
inseparable
irr(o!;, F)
>1. Thus F{a\\,
of
far we have so Thus handled these ideas together.
Separable
F
F, F(or)
induction.
52.5
then
of F,
extension
F, is
only
it.
under
is a finite
finite
and
F.
over F. SinceF is finite
Now
If F
51.10) if each
of
tower
proper
bottom one if
one immediately
of the
extension
inseparable
finite
of the
extension
F.
Suppose that
If
1, \342\231\246
to any
induction,
by
(Counterpart ofthe Corollary of Theorem F is totally over F if and only inseparable over
Proof
field is a totally
The top
if each field
extended
can be
52.3
Theorem extensions.
=
(1)(1)
inseparable over F.
is totally
K
=
E}{E : F)
= {K:
: F}
{K
for including
reason
^ 0,
inseparable over F
if
and
and
let F
only if
this
be a finite
material.
extension
of F.
there is some integer t
>
Then 1 such
or
e
or
F, \302\242.
that ap'
e F.
F,
446
Part X
Galois
and
Automorphisms
Theory
Furthermore, there is a unique separable over F, and either Proof
e E,a
Let a
F,be \302\242
and,
>1,
multiplicity
K of
extension
F,
E is totally
K or
E =
F
with
< K
< E, such
over
inseparable
is
K
that
K.
over F. Then irr(a, F) has just one zero a of totally inseparable be of the as shown in the proof of Theorem 51.14, irr(a, F) must
form
xp' -ap'.
Henceap e F for
t >
some
Conversely, if ap'
1.
e F for
-
(x^'
as its
a^) e
For
second
the
that
part
=
kr(ai,F)
a
F, \302\243
F) divides (x so a is totally >1,
irr(a,
zero is of multiplicity of the theorem, let E
and this
zero
only
F[x], showing
e E and
then
= (x -a/,
-o/
x^' and
1, where a
t >
some
a)p'.
Thus
irr(a,
inseparable
\342\226\240 Then = F(ai, \342\226\240 \342\226\240, an).
a
F) has over F.
if
-a/),
Y\\(xp,i
i
= at, let
with an
=
We have
aup'.
/\302\276
F(pn,
f,(x) =
\342\200\242 < \342\200\242 \342\200\242, Pni)
fi2i,
-
E, and
is azero
/}n
of
p,j),
Y\\(x
3 to the power p is an isomorphism ap of E onto of the is power p' isomorphic mapping raising (ap)' of E onto a of E. Thus since the 0\302\276 are all distinct for a fixed i, so are the fa for a fixed i. subfield over F, because it is a zero of a polynomial Therefore,/^ is separable fi(x) in F[x] of multiplicity 1. Then with zeros
e F[x], Now
where fi(x)
of E,
a subfield
since
raising
to the
K= over F,
is separable
some /?''
7^
1, then
over .ST,by
inseparable
inseparable over K, by follows
It
in E 52.6
Definition
are
that
the
by
K ^
E,
the the
of Corollary
proof and
/7th roots
/j,-1
of this
theorem.
of Corollary
51.10. If all p': K, showing Hence E =
of Theorem52.5is
of elementsthat
are
the an
the
extension
not already
K(ai,
a,
K
=
\302\2421(1^
\342\226\240 \342\226\240 is \342\226\240, an)
E. If totally
totally
52.4.
separable
precise
= 1, then each
that
Corollaries 51.10 and 52.4 that the field K over F. Thus K is unique. separable
theorem shows preceding of characteristic p. Such
of a field
is in
from
The unique field K The
=
a/\"
first part proof
\342\200\242, F(pu, fin,- \342\200\242 Pnl)
structure can be
pth powers.
consists of all
closure
a \342\231\246
of F in
E.
of totally inseparable
obtained
elements
by
repeatedly
\342\226\240
extensions adjoining
that Theorem 52.5 is true for infinite algebraic extensions E of F. The first assertion of the theorem is valid for the case of infinite extensions also. in the field For the second part, since a \302\261 p, ap, and a//3, for p / 0, are all contained K of E, the separable F(a, fi), all elementsof E separable over F form a subfield over K, since closure of F in E. It follows that ana e E,a \302\243 K,is totally inseparable a and all coefficients of irr(a, K) are in a finite extension of F, and then Theorem 52.5 can be applied. We
proof
remark
of the
Section52 Exercises
447
52
\342\226\240 EXERCISES
Concepts
1. Lety 2. Let y
z be
and
indeterminates,
z be
and
let u
and
indeterminates,
= yn
let
and
andv
=
u = y12
and
the separableclosureof
z18. Describe v
=
Describe
y2zis.
Z?,(u,
z).
v)inZ3(y,
the separable closureof
v) in
Zi(u,
My,z)-
3. 4.
Referring
to Exercise
1, describe
the
Referring
to Exercise
2, describe
the
5. Mark
of the
each
following
a. No proper
totally
Exercise 6) of Zi(u, v)
inseparable
closure (see
inseparable
closure of Z^{u, v)
in
Z?,(y,
in Z^(y,
z).
(See Exercise 6.)
z).
or false.
true
of an
algebraicextension
is totally
b. If F(a)
totally
over
inseparable
infinite field of characteristic F of characteristic
p
/ 0 is ever a separable e F for some t
p
/ 0, then
a^
extension.
0.
>
y, Z5(y) is separableover Z5(y5). d. For an indeterminate y, Z5(y) is separable over Z5(y10). e. For an indeterminate y, Z5(y) is totally inseparable over Z5(y10). f. If F is a field and a is algebraic over F, then a is either separable or totally over F. inseparable E a F a in E. If an extension of field then has closure is F, g. separable algebraic extension of a field F, then E is totally over the separable closure h. If E is an algebraic inseparable
c. For
an
indeterminate
ofFinF.
i. If
E is
an algebraic
of a
extension
inseparable over the separable over j. If a is totally inseparable
field F and E is not of F in E.
extension of
a separable
F, then
E is
totally
closure
F, then a is
the
only
zero of
irr(a, F).
Theory
6. Show
that
if E
7.
Show
power
8. Let and
that
is
over
inseparable
an
extension
algebraic
F forms
a subfield
of characteristic element of F.
a field F
of some E be
a
only
if Fp
finite
subfield of E. Consider
p
of a field F
extension
= F.
[Hint:
the
The
of a field F, then the union of F with the totally inseparable closureof
the
map
diagram
/ 0 is perfect if
and
of characteristic p.
if Fp
only
In
the
= F,
notation
set of
F in
of E,
that
all
elements
of E
totally
E.
is, every
element of
F is a pth
of Exercise 7, show that Ep = E if a e E is an isomorphism onto a
E defined by ap{a) = ap for ap: E ->\342\200\242 in Fig. 52.7, and make degree arguments.]
Part
448
section
X
53
Galois
and
Automorphisms
Theory
Galois Theory
Resume The Galois
theory
< F, a e E, and
has p as a zero F(j8)
also.
there is
Then
an
field
and
group
We shall
Starting
theory.
start
entire
the
recalling
by
text. with
the main
mind.
in
a conjugate
fi be
let
subject matter of
of a
over
is, irr(a, F) F(a) onto
that
F,
\\jraj mapping
isomorphism
and maps a onto p. E, then an automorphism of a over F. conjugate
F fixed
leaves
that
have well
and should
developed
< E
Let F
of the
elegance
interplay of work has been aimed at this goal.
results we have 1.
in
a beautiful
gives
48, our
Section
the climax
is perhaps
section
This
< E < F and a e a of F that leaves F fixed must map a onto some a If F < E, the collection of all automorphisms of E leaving F fixed forms E of the of all of left For S set elements subset G(E/F). G(E/F), group any F < Eg(e/f)fixed Es- Also, by all elements of S is a field < < A field E F, is a splitting field over F if and only if every F, F isomorphism of E onto a subfield of F leaving F fixed is an automorphism of E. If E is a finite and a splitting field over F, then extension = {E : F}. |G(F/F)| If F
2.
3. 4.
5.
: F} divides F, then {E : F} = [E : F].Also, separable if irr(a, F) has all zeros of multiplicity 1 for only If
E is
a
then {F
of F,
extension
finite
over
*
6. If F isa finite
is separable
and
e F.
every a
and is a separablesplitting
of F
extension
|G(F/F)| = {F:F} =
is also over F if
If F
: F].
[F F
over
field
F, then
[F:F].
NormalExtensions We
are
going
of K onto
to be
a subfield
interested in of
F fixed is
of results
4 and
5, these are
K of
F isa finite
an
of K
every
isomorphism
and such
that
: F}.
extensions
finite
the
F such that
automorphism
= {K
[K:F]
In view
K of
extensions
finite
F leaving
of F
that
are
separable
splitting
fields over F.
53.1Definition
extension F.
A finite
Suppose result before,
4, every we let
that
of F if K is
Let K
be
K
a finite
Then ^T
a separablesplitting \342\226\240
^T
is a
finite normal
automorphism of F G(K/F) be the group
one more result,
53.2 Theorem
extension
normal
over
field
we
shall
be ready
extension of F, where
leaving
of all to
illustrate
F
fixed
induces
K < F, as usual. Then by an automorphism of K. As F fixed. After of K leaving
automorphisms the main theorem.
normal extension of F, and let E be an extension of F, where F < E < finite normal extension of E, and G{K/E) is precisely the subgroup
is a
of G{K/F) consisting of F if Proof
is
If K the
and
automorphisms induce the same G(K/F) if they are in the same left coset of
only
the
51.9 shows
Theorem
is a normal
of a
field
over E
field
of this
isomorphism in
two
Moreover,
of E onto
a subfield
G(K/F).
then K is polynomials in F[x], of E[x]. viewed elements as polynomials over F. Thus K over E, since K is separable 1} of
e
\\i
This establishes our
of E.
\302\243 fixed.
G(K/E)
449
Theory
set of
same
is separable
K
that
extension
set {f(x)
leave
that
x in
and
splitting
splitting
all those
of
automorphisms a
Galois
53
Section
contention.
first
Now every element of G(K/E) is an automorphism of K leaving F fixed, sinceit can be viewed as a subset even leaves the possibly larger field E fixed.Thus G{K/E) of Since function is a group under also, we see that G(K/F). G(K/E) composition <
G(K/F).
and only if
x~la e
G(K/E)
for
fM
e
a and x in
for
Finally,
then for
G(K/E),
a e
o-(a) since
= a
^{a)
e
for a
a
G(K/F),
G(K/E) or
only if a
E, we
have
= (tfj,)(a)
e
so \302\243,
leaves
t-1
fM
e
if
G(K/E)
But if a
G(K/E).
= x[i
x(a), e E, then
= t(a) for all a = a
= x~lo is thus
and ^
\302\243 fixed,
same left coset of
for
= r(fj,(a)) =
(t_1ct)(q;)
for all a
= rfi
if er (a)
\302\243. Conversely,
in the
x are
and
if and
in
\342\231\246
G(K/E).
theorem shows that there is a one-to-one correspondence between preceding of G(K/E) in G(K/F) and isomorphisms of E onto a subfield of JT leaving of E F fixed. Note that we cannot say that these left cosets correspondto automorphisms The
left
qosets
sinceE may
is a normal extension splitting field over F. Of course, if \302\243 would be automorphisms of E over F. We might guess this will happen if and only if G(K/E) of G(K/F), and this that is a normal subgroup is indeedthe case. That is, the two different uses of the word normal are really closely related. Thus if \302\243 the left cosets of G(K/E) in G(K/F) is a normal extension of F, then can be viewed as elementsof the factor which is then a group group G(K/F)/G(K/E), of automorphisms on \302\243 and leaving F fixed. We shall show that this factor group acting is isomorphic to G(E/F).
over F,
of F,
then
The
Main
The field
the
these
F,
be a
Theorem Theorem
Main
not
isomorphisms
there
intermediate
of Galois
is a one-to-one fields
where \302\243,
Theory states
that
for
correspondencebetween F <
< \302\243
normal extension K
a finite
of G(K/F) subgroups correspondence associates with We can also go the other way and start the
K. This
field E the subgroup G(K/E). H of G{K/F) and associatewith H its fixed field this with an example, then state the theorem and discuss its proof. intermediate
Kh-
a subgroup
53.3
Example
Let K = Q(\\/2,V3). Now K that there are four automorphisms on the basis {1, \\/2, V3, values
is a
normal extension
V6}
of Q, and
leaving Q fixed.We for K over Q.
of K
We
Example
recall
them
shall
48.17
of a and
each with
illustrate
showed
by giving
their
450
Part X
Galois
and
Automorphisms
Theory
l, (Tj, (T2, \302\24273}
I, \"3i
\"i)
<<.
=
Q(V2,V3)
Jfw
= \302\256(V2) Kll:rj2J
= K, {l,
Q(V6) =
\302\243,,,^,
(7,, (72,O-j}
(b) 53.4
;: The
identity
Figure
(a)
diagram,
Group
(b)
Field
diagram.
map
Maps
\\/2 onto
V6 \342\200\224-J2,
onto
and \342\200\224V6,
leaves
the others fixed
0\302\276'. Maps
\\/3 onto
V6 \342\200\224
onto
and \342\200\224V6,
leaves
the others fixed
: Maps 0\302\276
\\/2 onto
onto
\342\200\224and
leaves the others fixed
cti
:
We saw
that
>/3,
to the Klein 4-group. The complete is isomorphic {;, a\\, a2, 0\302\276} with with each subgroup off the corresponding intermediate paired
subgroups, it leaves fixed, is
list field
of that
as follows: \342\200\242** {1,0-1,02,03} Q,
{1, ay}
\302\253* Q(V3),
{(,a2}^Q(V2),
{1, a3}
^ Q(V6),
{1}<* All subgroups of the abelian intermediate fields are normal Note that if one subgroup
group
{;,
extensions is containedin
a\\,
Q(V2,
a2,
of Q.
y/3).
0-3} are
Isn't
another,
normal subgroups,
that
elegant?
then
the
larger
of the
and
all
the
two subgroups
corresponds to the smaller of the two corresponding fixed fields. The larger the subgroup, the smaller the fixed field, that is, the fewer elements left that is, the more automorphisms, In Fig. 5 3.4 we give the corresponding fixed. and intermediate diagrams for the subgroups to the fields near the bottom. fields. Note again that the groups near the top correspond or turned upside down. Sincehere each That is, one diagram lookslike the other inverted down, this is not a good example for us diagram actually looks like itself turned upside
GaloisTheory
Section 53
53.5
Definition
to use
to
do
look
not
is a finite
over
F.
now state the of the main
shall
the
proof
53.6Theorem (Main Theorem of Galois
with
54.6 to
see
that
diagrams
\342\226\262
of a field
extension
normal
Turn ahead to Fig.
principle.
inversions. F,
then
Galois group
is the
G(K/F)
of K \342\226\240
We complete
their own
like
If K
inversion
this
illustrate
451
of G(K/F)
Galois
set of all
the
Then
F
where
and finally,
example,
G(K/F).
The
of a field
extension
normal
< E
< K, let X(E)
map of the
one-to-one
is a
X
subgroups of
of all
set
be
F,
the subgroup
such intermediate hold for X:
properties
following
= G(K/E).
1.
X(E)
2.
E = Kg(k/e)= Ki(E)-
3.
For H
4.
[K : E] =
< G(K/F),
of X(E)
\\X(E)\\
in
G(K/F).
=
X(EH)
and
:
[E
H. = (G(K/F)
F]
: X(E)),
the
of left
number
cosets
G{K/F).
a normal
E is
5.
another
be a finite
Let K
For a field E,
E fixed.
leaving
then give
theorem,
Theory)
G(K/F).
group
fieldsE onto
main
theorem.
extension
of F if
When X(E) is
a normal
and
if X(E)
only
is a normal
subgroup
of
subgroup
then
of G(K/F),
G{E/F) ~ G(K/F)/G{K/E).
6.
The
of G(K/F)
of subgroups F.
diagram
fields of .ST
on the Proof We theorem. Let us seejust how much 1 is just the definition Property Property 2, Theorem 48.15 shows Observations
of K leaving E
really already proved we have left to prove. of
X found
K is
E. Since \302\243
XiEj),
Now is an
Property onto map.
included our property Property statement
of
the
theorem.
of this For
Kg(k/e)-
a normal
< E,
in
3 is
[K
: E]
=
4 follows
in Theorem
53.2.
and
also
tells
us that
X
is one
to one, for if
2, we have =
Kx(Ei)
=
Kx(E2)
=
Ei_.
to showing that job. This amounts exactly we have H < X(Kh), for H surely is automorphisms leaving Kh fixed. Herewe will be using strongly {K : E}. from [K : E] = {K : E},[E : F] = {E : F}, and the last
going
to be
Of course,for
set of all
the
2
of Property
disposes
then by Property E\\
X
in the statement
part
of E, by using a conjugation extension the Isomorphism Extension Theorem, we can find an automorphism fixed and mapping a onto a different zero of irr(a, F). This implies that a
= Kq(k/e)-This
X{E\\)
a substantial
that
Kg(k/e)
=
of intermediate
diagram
and
isomorphism
so E
inverted
have
E <
Let a e K, where
is the
over
H
our main
< G(K/F),
Part
452
X
and
Automorphisms
We shall have
Galois
Theory
that the
to show
word normal correspondfor
of the
senses
two
Property 5.
already disposed of Property
We have
and 5
to be
remain
6 in
53.3.
Example
Thus
3
Properties
only
proved.
The Main Theorem of GaloisTheory is a strong tool in the study of zeros of If irreducible factor of f(x) is separable over polynomials. e F[x] is such that every F, f(x) field K of over F a normal of F. then the splitting is extension The Galois group f(x) F. The structure of this group is the group of the polynomial f(x) over G(K/F) of will be strikingly information the zeros This f(x). may give considerable regarding in Section 56 we achieve our illustrated when final goal.
Galois Groups over K be
Let
of F
(a
a finite
of K
order
of a
extension
is perfect).
field
finite
Finite Fields
is prn. Then
finite field F. We Suppose that the
we have
that
seen
K is
K is a separable
seen that
have
of F
order the
is pr
splitting
and
[K
:
field
of
xp'\"
extension
= n,
F]
so
the
\342\200\224 x
F.
over
Hence K is a normal of F. extension for a e K, apr (a) = Now one automorphism of K that leaves F fixed is oy, where ap'. Note that (apr )'(a) = ap\". Since a polynomial of degree pri can have at most pr' leave all prn zeros in a field, we see that the smallest power of oy that could possibly elements of K fixed is the wth power. That is, the order of the element oy in G{K/F) is = at least n. Therefore, since \\G(K/F)\\ is cyclic [K : F] = n, it must be that G(K/F) and generated by opr. We summarize these arguments in a theorem. 53.7 Theorem
Let K
be
is cyclic We
a finite
extension of degreen
of ordern, use
is generated
and
this
theorem
and
let K
to give
of a
finite field
F of pr
by oy, where for a
of the
illustration
another
Then
elements.
e K, opr{a)= apr. Main Theorem
G(K/F)
of Galois
Theory.
53.8
Example
Let F cyclic
,
group
(Z12,
= GF(p12), so [K
+). The
diagrams
for
: F] =
12.
Then
the subgroups
to the
is isomorphic
G(K/F)
and for
the
fields
intermediate
are given in Fig. 53.9. Again, each diagram is not only the inversion of the other, but of itself. Examples where the diagrams also looks like the inversion do unfortunately, not look like their own are inversion given in next Section 54. We describe the cyclic
(ap) <*/>
= G(K/F)
K= K{, =
K3)
GF(p6)
G\302\245(p^)
<^6>
<^4>
GF(p^)= K{l,
K^
= GF(p2) F =
(a)
ZP=GF(p) = (b)
53.9 Figure
(a) Group
diagram,
= K,<
(b) Field
diagram.
W
GF(p3) =
K{ap)
Z(
>>
Section 53
that
saw
3
Properties
for example,
generators,
giving
453
Theorem Completed
of the Main
Proof We
= {ap)by
of G(K/F)
subgroups
GaloisTheory
5 are
and
all that remain to
be provedin
of
Theorem
Main
the
Galois Theory.
Proof
H < X(KH) < H
3, we must show
to Property
Turning
G(K/F). Thus
subgroup
aproper
that
we
what
G(K/F), X(KH) = H. We know that is that it is impossible to have really must show H <
for
ofX(KH)- We shall
that
suppose
H < X(KH)
H <
Then
K =
extension,
separable
KH{a) for some
by Theorem
=
n
have
As a finite
a contradiction. 51.15. Let
derive
shall
and
a e K,
\\H\\
KH] =
[K:
G(K/KH) implies that
<
:
[K
= n.
KH]
Let
{K : KH}
=
\\G(K/KH)\\.
= n. Thus we
< \\G(K/KH)\\ of H be
|H|
elements
the
crly
\342\226\240 \342\226\240 \342\226\240, o\\H\\,
to
have
would
and
consider
the
polynomial \\H\\
=
f{x)
Y\\(x
/=i
Then
f{x)
e
degree
cr\\H\\(a).
a e
H, since if
\\H\\
n. Now the coefficients of each a,{a). For example,the coefficient these coefficients are invariant
<
in the
\342\200\224 \342\200\224 \342\226\240 \342\226\240 \342\226\240
a2{a) Oi
is of
expressions
symmetric
Thus
of x1^1 each
under
is
f(x) are
\342\200\224 \342\200\224a\\{a)
isomorphism
\342\226\240 \342\226\240 \342\226\240, GO\\H\\
\342\226\240 the sequence o\\, \342\226\240 \342\226\240, o\\h\\, except coefficients in Kh, and since some ay is
is again
we
of x in
power
H, then CTCTl,
Therefore,
- atipi)).
for order, H i, we see that
being a group. some
Hence
07(0:) is a,
f{x)
so f(a)
has
= 0.
have
would
deg(a, KH)
<
<
\\H\\
n =
[K :
KH] = [KH(a): KH].
3. Thus we have proved Property 5. Every over F, extension E of F, F < E < K, is separable if \302\243 F if and only field over F. over is a splitting by Theorem 51.9. Thus E is normal a subfield of Extension Theorem, every isomorphism of E onto By the Isomorphism F leaving F fixed can be extended to an automorphism of K, since K is normal over F. Thus the automorphisms of G(K/F) induce all possible isomorphisms of E onto a of F leaving F fixed.By Theorem subfield field this shows that \302\243 is a splitting 50.3, over F, and hence is normal over F, if and only if for all a e G(K/F) and a e E, This
is impossible.
We turn
to
Property
a (a) By
t e
Property
2, E
is
the
fixed
field
e E.
of G(K/E),
so a (a) e
G(K/E)
x(a(a)) = a (a).
E
if
and
only
if for all
X
Part
454
and Galois
Automorphisms
in turn
This
holds if and
Theory
if
only
= a
(p~lTo){u)
e
for all a t
and
a \302\243,
and r e G(K/E). But leaves every elementof
e G(K/F),
o~lxo
e G(K/E),
(ct_1tct) e the condition is precisely It remains for us to show
This
G(K/F)/G(K/E). 0 : G{K/F)
when
-* G{K/F)given
\302\243 is
er e
is a homomorphism.
G{K/F) of
automorphism
of
G(K/F).
extension of F, G(E/F) of \302\243 induced automorphism by
F).
e 0\302\276
Thus
~ er
The map
G(E/F).
xe for some r e G(K/F). Thus 0 Isomorphism by the Fundamental is a natural this isomorphism Furthermore,
CT\302\243
extendedto
can be
it is
is,
subgroup
the Isomorphism
By
F fixed
\302\243 leaving
normal
by
0(0\") = for
G{K/F)
a normal
let 0\302\276 be the extension of
\302\243 is
that
a e
that is,
G(K/E). be a
G(K/F)
that
e G(K/F), a normal
a
For
(we are assuming
that
that for all
means
this
\302\243 fixed,
is onto
some
G(E/F).
Theorem,
Therefore,
Extension Theorem, every of K; that automorphism
The kernel of 0 is G(K/E). ~ G(K/F)/G(K/E).
G(E/F)
one.
\342\231\246
53
\342\226\240 EXERCISES
Computations
The field through
K
8,
V3, V5) is a finite numerical compute the indicated = Q(V2,
of Q. It can be shown that [K : Q] = The notation is that of Theorem 53.6.
extension
normal
quantity.
1. {K : Q}
2.
3.
|1(Q)|
4. |1(Q(V2,
5.
|A(Q(V6))|
6.
7. |1(Q(V2
+ V6))|
9.
the
Describe
10. Give the
11. Let K
be the
b. To c.
what
Using of K did
and for
12. Describe the 13. Describe the
the
of G(K/Q)
Exercises
1
\\G(K/Q)\\
V3))|
|1(Q(V30))|
1)
Q.
\342\202\254 Q[x]
over Q.
group G(GF(729)/GF(9)). (Refer
to Example
50.9.)
by giving their values on
answer to part
of G(K/Q),
the back
(a) in
indicating
of the
of the
group
of the
and
5x2 + 6) polynomial (x4 \342\200\224
polynomial (x3
\342\202\254 Q[x]
\342\200\224 Q[x] 1) \342\202\254
text,
iy/3.
(By Example
50.9, K =
give the diagrams for the subfields fields and subgroups, as we
intermediate
corresponding
53.4. group
^2
seen beforeis G(K/Q)isomorphic? in the
given
subgroups
-
of the
of x3 - 2 over
elements
we have
group
in Fig.
a generator
splitting field
the notation
(x4
polynomial
and describe
a. Describethe six Q(^2, i V3).)
In
8. |A(*)| of the
group
order
8.
over Q.
over Q.
Exercises
Section 53
455
Concepts
14. Give
of two
an example
~
a. Two different
d. If e. If
and
K\\
Ko_
not
are
53.6,
of F,
extension
normal
same
may
have the
< E
< L
< K, then
then
K is
a normal
group
F
if
L of
and
field.
fixed
X(E)
< X(L).
extension
of
E, where F
< E <
isomorphic Galoisgroups,
a field F have
K.
then
[E :
[L: F],
\302\243 is a
extension
f. If
Galois
of a
of Theorem
normal extensions E
finite
two
F] =
that
false.
subgroups
the notation c. If K is a finite
same field F such
of the
Ki_
G(K2/F).
true or
b. In
K\\ and
extensions
normal
finite
isomorphic fields but G(KJF) 15. Mark each of the following
finite normal extension of of F. finite normal
\302\243 is any
F
H is
and
a normal
simple extensionof a field
then the
F,
then EH is
of G(E/F),
subgroup
Galois
is a
G(E/F)
group
a normal simple
group.
group is simple. group of a finite
g.
No Galois
h.
The
Galois
i.
An
extension
E of
F is
j.
An
extension
E
F
is not
of F
of a
extension
degree 2 over a field of degree 2 over a field
field
finite
is abelian.
always a normal is always a normal
of F.
extension
of F if
extension
the
characteristic
2.
Theory
16. A
normal
finite
extension
is abelian over F abelian over F.
17. Let
K be
a
finite
and
K of a field
\302\243 is a
F is abelian over normal extension of F, where
of a field
extension
normal
F. Prove that
F
if
F
for
is an
G(K/F)
< E
every
< K, a e
abelian
then
K, the
.ST
group.
is abelian
of a
norm
that if
Show
over E
over F,
and
given
creG(X/F) and the
trace of a
over F, given
by
TrK/F(a)
=
Y^
CT(a)>
creG(X/F)
are elements
18. Consider
K
of F. = Q(\\/2,
V3).
Referring
to Exercise
17, compute
each of the
a. NK/Q(j2)
b. NK/Q(j2
c. NK/Q(V6)
d. t
e.TrK/Q(j2)
g19. Let
h.
7>x/q(\\/6) .ST
be
a normal
extension of F, and irr(a,
Referring
to Exercise
a. NK/F(a)
17, show
= (-l)\"fl0,
let F)
K = = x\" +
following +
(see Example
V3)
NK/Q(2)
TrK/Q(^2 +V3) TrK/Q(2)
F(a). Let H fl\342\200\236_ix\"_1
h ^x
+
fl0-
that
b.
TrK/F(a)
=
-a\342\200\236_i.
53.3).
K
F is
by
X
Part
456
20.
21.
Let fix) the order
of
Let f(x)
\342\202\254 F[x]
Show
the
polynomial of degreen of f(x) over F divides
be a
\342\202\254 F[x]
group
the
that
and Galois Theory
Automorphisms
polynomial such that every over F can be viewed
be a
that each
such
over F.
is separable
factor
factor of f(x) is a separable natural way as a group of permutations
irreducible
of fix)
group
irreducible
Show
that
n!.
in a
over F. zeros of
polynomial of the
/(x)inF
22. Let
F be a field not divide n.
a.
that
Show
b. Show
by this
23.
A
a. Show
F
b.
extension
normal
finite
in F,
unity
where the characteristic
of F is either
0 or
does
of F.
extension
Every
[ffwif:
er \342\202\254 G(F(i;)/F)
some fr
f onto
maps
and
is completely
r.]
K of
a field
F is
cyclic over
E is
a normal
is cyclic over F and cyclic over F. if K is cyclic over F, then
if K
that
F if G(K/F) is a cyclic
where F < E
of F,
extension
group.
< K, then
F is
cyclic over
K is
and
that
Show
K be
a. For a
a
extension
normal
finite
exactly one
exists
there
of [K : F].
each divisor d
24. Let
value
of
wth root
is abelian.
G(F(t;)/F)
determined
a primitive
is a normal
F(f)
that
f be
let
and
e K, show
field
F,
F < F
< K, of degreed over
= irr(a,
F) if and
F for
of F.
that
/(*)=
(*-
f]
U\342\202\254G(X/F)
is in
b.
25. The and
F[>].
join
G(K/(E 26.
F V
L. That
extension
(a), show
to part
Referring
is, F
of a V
L of
L))
With reference
V
two
L is
field
F,
that
extension
f(x)
fields
the intersection and let F and L
is a
power of irr(a, F),
F and
of all
L of F
subfields
be extensions
in
and
only
if K
= F(a).
of F containing both F L. Let K be a finite normal containing F contained in K, as shown in Fig. 53.10. Describe F is
the smallest
of F of
f(x)
both
of G(K/E) and G(K/L). situation in Exercise 25, describeG{K/(En L)}in
subfield
E and
in terms to
the
K
53.9
Figure
terms
of G(K/E)
and G{K/L).
section
54
Illustrations of GaloisTheory
54
Section
457
of Galois Theory
Illustrations
Symmetric Functions a field,
be
F
Let
let
and
y1;
\342\226\240 \342\226\240be \342\226\240, yn
There are
indeterminates.
some natural
F fixed, namely, those defined by permutations F(yi, \342\226\240 \342\226\240 To be more \342\226\240 that a be a permutation of {1, \342\226\240 let \342\226\240. \342\226\240, is, a {yi. >'\302\253}\342\226\240 \302\273}, explicit, \342\226\240 \342\200\242 \342\200\242 \342\200\224> Then a gives rise to a natural map a F(y\\, \342\226\240\342\226\240 \342\226\240, y\342\200\236)F(y\\, ,y\342\200\236) given by \342\226\240 \342\226\240 \342\226\240, y\342\200\236) leaving
of
automorphisms
of
e Sn.
:
---,)^)
_//(yi;
\342\226\240\342\226\240\342\226\240^y\302\273)\\ /(Mi), _
\342\226\240\342\226\240 gCMn. \342\226\240\342\226\240\342\226\240!y\302\253)/ -.^(^)
\\g(yi, \342\226\240 \342\226\240 \342\226\240, >'\342\200\236), g(ji,
for /(yi, immediate
a is an
that
left > y\302\253) F(yi>' \342\226\240'
in the
symmetric
54.1
Definition
An element it is left fixed
of F(j\\,
fixed
for
by
field
aZZ
a,
F \342\226\240\342\226\240-,>'\302\253) leaving
all a
are those \342\202\254 \302\243\342\200\236,
function
a symmetric
of yi,
all the
of
group
0. It is fixed. The elements of rational functions that are \342\226\240 \342\226\240 \342\226\240, yn) ^
g(yi,
\342\200\242 \342\200\242 \342\226\240, y\342\200\236.
\342\226\240 \342\200\242 is \342\226\240, yn)
F{y\\,
isomorphic to
naturally
field of
automorphism
by all permutations
the
\342\226\240 \342\226\240 with \342\226\240, y\342\200\236],
F[yi,
indeterminates y\\,
of the
Sn be
Let
\342\226\240 \342\226\240 \342\226\240, yn) e
\342\226\240 \342\226\240 in the \342\226\240, y\342\200\236,
in
sense just
yi,
\342\226\240 \342\200\242over \342\200\242, yn
F, if
explained.
automorphisms a for a e Sn. Observe \342\226\240 which the subfield of F(yi, \342\226\240 \342\226\240, y\342\200\236)
\342\226\240
Sn is
that
K be Let S\342\200\236.
fixed
is the
the polynomial
Consider S\342\200\236.
n
-
yi);
f(x)Y\\(x
/=i
this
be f(x)
Thus yi,
of a,
fixed
is left
the
by
\342\226\240 \342\226\240 is a \342\226\240, y\342\200\236))[x]
\342\202\254 (F(y1,
f(x)
polynomial
the extension
in
each
of
coefficients
a~x for
map
_f(x)
\342\226\240 \342\226\240 As \342\226\240, illustration, y\342\200\236.
general
way, to (F(y\\,
natural
the
a
polynomial
\342\226\240 \342\226\240 where \342\226\240, y\342\200\236))[x],
of degree n. Let a~x a~x(x) =
x.
Now
\342\202\254 that is, S\342\200\236;
n
n
i=i
i=i
are in K; they are note that the constant
elementary symmetric of f(x)
term
in the
functions
is
(-i)\"yiy2---y\302\253,
the
coefficient
functions
in yi,
The
of x\"^1 is
+ yi \342\200\224(y\\
+
\342\200\242 \342\200\242 \342\200\242 and + y\342\200\236),
first elementary
symmetric
function
second Consider
symmetric
These are symmetric
\342\226\240is in y\\, \342\226\240 \342\226\240, y\342\200\236
si =yi+yi-\\ the
so on.
\342\226\240\342\226\240\342\226\240,y\342\200\236.
is ^2 =
+
yi>'2
the field functions
yiy3
E =
in y\\,
F(s\\,
+
hy\342\200\236,
\342\226\240 \342\226\240 \342\226\240 = + y\342\200\236_iy\342\200\236,andsoon, andthe\302\273this5\342\200\236 \342\226\240 \342\226\240 Of \342\226\240, sn).
\342\226\240 \342\226\240over \342\226\240, y\342\200\236
F. But
course,
F(yi,
E <
K, where
K is
yiy2
finite normal
\342\226\240 is a \342\226\240 \342\226\240, y\342\200\236)
\342\226\240 \342\226\240 -y\342\200\236.
of all extension
the field
458
Part X
and
Automorphisms
of E,
the
namely,
Galois
Theory
field of
splitting
n
=
/(*)
~ Ji)
Y\\(* (= 1
the degree of f(x)
E. Since
over
is n,
[F(yu 13, Section
Exercise
(see
at once
have
we
:\302\243]<\302\253! \342\226\240\342\226\240\342\226\240,?\342\200\236)
50). However,since K \\S~n\\
=
is the
=
\\Sn\\
fixed field
of S\342\200\236 and
n\\,
also
we have
<
nl
\342\226\240 \342\226\240 : K]. \342\226\240, [F(yu y\342\200\236) \342\226\240\342\226\240\342\226\240,yn):K}<
{F(yu
Therefore, < \302\253!
[F(yu
\342\226\240 \342\226\240 \342\226\240 K] < \342\226\240, yn)
\342\226\240 \342\226\240 : \342\226\240, yn)
[F(yu
<
E]
\302\253!,
so = \302\243. \302\243 \342\226\240 \342\226\240 over E is therefore S\342\200\236. The fact that K = E shows ,yn) group of F(ji, \342\226\240 a function can be a s rational function of the elementary expressed symmetric \342\226\240We these in a theorem. functions si, \342\226\240 summarize results s\342\200\236. \342\226\240,
The
full Galois
that
every
symmetric
54.2 Theorem
Let si, Then
\342\226\240 \342\226\240 \342\226\240, sn be
every symmetric
In
\342\226\240 \342\226\240over \342\226\240, y\342\200\236 F
Also, F(y\\, \342\226\240 \342\226\240 and \342\226\240, s\342\200\236), the Galois group
F(si,
finite
of yi,
function
in the
symmetric functions
functions.
elementary symmetric
of
the elementary
\342\226\240 \342\226\240 is a \342\226\240, y\342\200\236)
of this
indeterminates
is a rational finite normal
extension
function
\342\200\242 \342\200\242 y1; \342\200\242 ,yn.
of the
extension of degreenl
is naturally
isomorphic to
of Cayley's Theorem 8.16,it can be deduced from Theorem 54.2 that can occur as a Galoisgroup (up to isomorphism). (See Exercise 11.) group view
S\342\200\236.
any
Examples us
Let
54.3 Example
normal our promised example of a finite extension diagram does not look like its own inversion.
give
whose
Considerthe
in
a Galois
group
four
criterion, zeros
The
C.
number,
-
in C
=
splitting field K of x4 < K, so Q(a) ^ Q(a)
\342\200\224 2
K.
we see that Now
of x4
of x4 \342\200\224 2 over
\342\200\224 2 is irreducible over Q, by Q. Now x4 - 2. Then with p = 2. Let a -^/2 be the real positive zero of x4 \342\200\224 2 in C are a, \342\200\224a, i is the usual zero of x2 + 1 ia, where ia, and
field
splitting
Eisenstein's the
having
subgroup
Q(a, i)
{1, a,
a2,
= K. Letting E a3} is a basis for {1, a,
over
Q thus contains
(ia)/a =
i. Sincea isa real
i) contains all zeros of x4 = Q(a), we have the diagram in Fig. 54.4. E over Q, and {1, i] is a basis for K over E.
However,
a2, a3,
since
i, ia,
Q(a,
ia2, ia3}
\342\200\224
2,
Thus
is a basis
K over
for
Q. Since [K
automorphisms of K
find eight
: Q] =
459
Theory
have | G(K/Q)\\ = 8, sowe need to that any such automorphism a
must
we
8,
We know
Q fixed.
leaving
of Galois
Illustrations
54
Section
values on elements of the basis {1, a, a1, a3, i, ia, ict2, determined be by a(a) and a(i). But o{a) must always a conjugate of a over Q, that is, one of the four 2. Likewise, of irr(a, Q) = x4 \342\200\224 zeros Q) = x2 + 1.Thus the four possibilities for a (a), combined CT(i)mustbeazeroofirr(!, all eight automorphisms. We describe with the two possibilities for a(i), must give in Table 54.5. For example, /)3(0:)= \342\200\224 ia and fo(i) = i, while these po is the identity
is completely
values are in
and these
ia3},
by its
determined
automorphism.
turn
Now
= Mi (Pi
(MiPiXa)
=
=
(\302\253))
MiO'\302\2600
MiO')Mi(\302\2600
=
-ia,
and, similarly,
(M1P1XO= so
=
fj/1pi
A similar
&2-
(PiMiX\302\2600
Thus
= Si,
pifii
be
so p1fii
=
^ fnpi
to one of the two from Table Computing
that
(P1M1XO =
and
ia and
is not
G(K/Q)
nonabelian
isomorphic
Example40.6.
shows
computation
-i. ~i-
abelian. Therefore, G(K/Q) must of order 8 describedin 2, p1 is of order 4, fi1 is of order
groups
54.5, we see that
= Si. Thus G(K/Q) is isomorphic to (Pi. Mi) generates G(K/Q), and pifj/1 = [iip3 for the elements the group Gi of Example40.6,the octic group. We chose our notation of Q(K/Q) so that its group table would coincide with the table for the octic group in Table is that given in Fig. 8.13. We 8.12. The diagram of subgroups H, of G(K/Q) fields repeatit here in Fig. 54.6 and also give the corresponding diagram of intermediate between of the illustrates Q and K. This finally nicely that one diagram is the inversion other.
The
illustrate.
fixed
by
KHi,
we have
leaves
a fixed
Q
and is
of
determination
Let's
{po,
to
find
and a
left fixed
by
fixed fields
the
To find Kh2, Since p\\, P2, /\302\276}.
we merely have
sometimes
Kh,
to find
all pj leave i fixed, Q(i) an extension of Q of degree4 left
is a zero {po,
Q) = x4
of irr(a,
ii{\\. By
Galois
theory,
the one-to-one
requires
an extension
is the fixed
field
by
a bit
of Q of we are and /\302\276
of
ingenuity.
degree
2 left
after. To find ^. Since ^
see that Q(a) is of degree4 over it is the only such field. Herewe are one by the Galois theory. If we find
\342\200\224 we
2,
correspondence given field that fits the bill, it is the one we are after. Finding Kh7 requires more ingenuity. for any /3 e K we see that Since + S1 (/}) is left fixed by H7 = {po, 5j} is a group, po(fi) = a, we see that + 81(a) = a + ia is left fixed by H7. We Po and Si. Taking /} po(a) and 5i are the only a + ia fixed.Thus can checkand see that /\302\276 leaving automorphisms using
strongly
54.5
Table Po
a -*
a
i \342\200\224> i
P\\
ia i
P2
P3
Mi
\342\200\224 \342\200\224a ia a i
i
Si
H2
ia
\342\200\224 \342\200\224a ia
&2
\342\200\224 \342\200\224 \342\200\224i i i \342\200\224i
460
Part
X
Automorphisms
and Galois Theory
CKK/Q)
Hi =
H4
=
H5 =
{p0,fJ,l]
Q(V2)=Jfw
{p0,/j,2}
WJ2)=ZH
H6
=
H7 = {f>o,S
{p0,p2]
+ \302\256(V2
0(^2,0=\302\276
{Po,P2,Si,S2
iV2)=KHi
Hg
=
(a>
^2
Q(V2-1^2)=\302\276
Gt/f/Q)
(b)
54.6 Figure
by
the
one-to-one
(a) Group
(b) Field
diagram,
diagram.
we must have
correspondence,
Q(a + ia) = Q(^2
+
= KHl.
ijl)
Suppose we wish over only
for every conjugate of y to find irr(a + ia, Q).If y = a + ia, then Thus we of K into that need there exists an Q, conjugate. automorphism mapping y to find the other zeros the various different values a(y) for a e G(K/Q) compute
a of G(K/Q) of irr(y, Q). By Theorem53.2,elements giving these different of of G(K/Q(y)) set of the left cosets a representatives by taking is in G(K/Q). A set of representatives for theseleft cosets
be found
{/00, The
conjugates
of
y
=
a +
irr(y, Q)
ia are thus
a
+ ia,
ia -
\342\200\224 \342\200\224 \342\226\240[(x (\342\200\224aia))(x
=
(x2 - 2iax x4
+
4a4
can
= {/%,Si}
Pi./02./\302\276}.
= [(* - (a + ia))(x =
values
2a2)(x2
= x4
+ 8.
\342\200\224 \342\200\224 ia, and a, \342\200\224a
(ia
\342\200\224 ia
+ a.
Hence
- a))]
\342\200\224 + a))] (\342\200\224ia
+
2iax
- 2a2) A
461
Theory
field of a quartic (4th degree) of degree 8 (Example54.3)and of degree 24 of a field F that is a splitting (Theorem 54.2, with n = 4). The degree of an extension field of a quartic F must always divide 4! = 24.The splitting field of (x \342\200\224 over 2)4 over \342\200\224 field of (x2 of degree 1, and the splitting Q is Q, an extension 2)2 over Q is Q(\\/2), an an extension of degree 4 for the extension of degree 2. Our last examplewill give field of a quartic. splitting
54.7 Example
seen
have
We
over
polynomial
the
Consider
and
examples
a field F
is
splitting
field of x4
not
that
we can show
23.11,
zeros of x4
1 are
+
(1
+ 1
x4
that
(See Exercise
in Z[x].
factor
that the
1 shows
shows
A computation
Theorem
it does
that
Section
in
splitting
+ 1 overQ.By
arguing
number
on complex \302\261 (\342\200\2241 0A/2-
the
of F
extension
an
over Q, by
is irreducible work
in which
of Galois
Illustrations
54
Section
1.) The
\302\261 i)/y/2
if
1+i
then
V2 Thus
the
G(K/Q)
and
+ 1 over Q is Q(a),and [K and field diagrams. Sincethere
K of x4
field
splitting
the group
give
mapping a onto eachconjugate we see determined by o{a), Since5
of a, that
the
and since an four elements
(ajak)(a) = and
a8 =
multiplication
1, we
see that
for all j,
G(K/Q) Fig. 54.9.
To find
be
must
the
automorphism
of G(K/Q)
aj(ak) = (c^)*
=
group
isomorphic
a of
: Q]
Table
oi
o-i
05
er7
a
a3
a5
a1
of K Q(a) is completely
are defined
by
ff3(a)
=
a +
a3 =
+ 07(a)
= a
+ a7 =
iy/2.
V2
Table
54.8.
<*jk
it is
ffi(a)
Let us compute
automorphisms
to the group {1, 3, 5, 7} Theorem. of 20.6. Sincea2 = o\\, Gs to the Klein 4-group. The diagrams
Similarly,
a -*
exist
= 4.
is isomorphic
ffi(a) +
54.8
: Q]
only necessary to find an element of K not = 2. Clearly o\\(a) + 03(a)is left fixed [Jr{
K{ai^3-j,
{cti, ct3}, since since {cti , ct3}
G(K/Q)
is
8. This
modulo
V2
V2
under
the
identity,
are
given
in
Q left fixed by both o\\ and ct3, by in
X
Part
462
and Galois Theory
Automorphisms
G(K/Q)
CTi.
07}
Q(V2) =
Jf{apa7}
G(K/Q)
(b)
54.9 Figure
is left fixed by
This
{a\\,07}.
(a) Group
0 \342\202\254 Q. But
by a similar
=
Computations (requiring 2.
text.
the
Verify
each field
Theorem
more than
irreducible
the intermediate
that
Verify
3. For
x4 + 1 is
that
Show
= Q(i) is
field
the
finding
Esi
for
is left
0:0:5
=
fixed
by
both
o^ and
o5> and
\342\200\224i.
after.
we are
A
54
\342\226\240 EXERCISES
1.
Q(-i)
diagram.
+ a5 = 0,
01(0:)05(0:)
argument,
01(0:)05(0:) Thus
use in
= a
+ 05(0:)
ffi(a) and
is of no
technique
(b) Field
diagram,
in Q[x],
fields
given
amount
of theory)
as we
asserted in
in the
field diagram
Example in
Fig.
54.7. 54.6 are
correct (Some are verified
the field
and give
its
diagram irreducible
in
Fig.
54.6,
polynomial
find a
primitive
element
generating
the field over
over Q.
4. Let \302\243 be a primitive 5th root of unity in C. a. Show that Q(f) is the splitting field of x5 - 1 over Q. of \302\243. b. Show that every automorphism of K = Q(f) maps f onto some power \302\243r of G(K/Q). c. Using part (b), describe the elements d. Give the group and field diagrams for Q(f) over Q, computing the intermediate fields as we did Examples54.3
in the
rest.) in
51.15
usual
the
and 54.7.
in
Q (see
Exercises
Section 54
5. Describethe
polynomial (x5 -
of the
group
over
\342\202\254 (Q(\302\243))M
2)
Q(\302\243)>where
is \302\243
463
5th root
a primitive
of unity. 6.
Exercise 4 for
Repeat
easiest way
7. In the
the
describe
possible,
of unity group of the
7th root
a primitive \302\243
(X8
in C. polynomial
- 1) \342\202\254 Q[x]
over Q.
8.
the
Find
and exhibit the words, do the complete over Q
each of the
9. Express
symmetric a. yi2 + b. 10.
C of
in
+
1
functions in
symmetric
1
1
as a rational
function
of the
elementary
yi
yi
C of
in
the polynomial
(x3 -4x2 the polynomial
b. ai2,
over Q
y2, y3
y\\,
1
yi
y3
y\\
ai +
In other
^3-
ji , j2,
Let Q!i,0\302\276. a3 be the zeors
a.
polynomial
fields.
+ yz2
\342\200\224
Find
of the
job.
following
functions
yi2
yi
4x2 - 1) \342\202\254 the polynomial (x4 \342\200\224 the group Q[x]. Compute of G(K/Q) and the intermediate between the subgroups correspondence
field K
splitting
as zeros
having
+ 6x-
precisely the
2)
eQ[x].
following:
+ a3 0\302\276 a32 0\302\2762.
Theory
11. Show
that
every
of some field 12.
group is isomorphic
finite
Galois
some
to
group G(K/F)
for some finite
K
extension
normal
F.
Let /(x) \342\202\254 be a monic polynomial F[x] K < Fbe the splitting field of /(x) over
n having
of degree
F,
and
all
its
that /(x)
suppose
-
f](X
factors separable
irreducible
factors
in
over F.
Let
into
K[x]
Of,-).
/=1
Let
=
A(/)
Y\\(a,
ccj);
i
(A(/))2
product
a. Show
that
b.
that
Show
= 0
A(/) (A(/))2
if
13.
of {a,-1i
An element of C all
algebraic
is an
integers
has as a factor
only if /(x)
and
the
square
of some
irreducible
in F[x].
polynomial
F. \342\202\254
c. G{K/F) may be viewed as a subgroup Show that G(K/F), when viewed permutations
of /(x).
is the discriminant
=
algebraic
forms a
this
if and \302\273},
1,---,
integer subring
is the S\342\200\236
of S\342\200\236, where in
fashion,
only
if it is
of C.
if
A(/)
a zeroof
is a
group of all
subgroup
of
permutations
of {a/
the group A\342\200\236,
\\i
=
1,
formed by
\342\226\240\342\226\240\342 all
even
F. \342\202\254 some
monic
polynomial
in Z[x].
Show
that
the set
of
Part
464
X
55
section
Galois
Automorphisms
and
Cyclotomic
Extensions
The GaloisGroup
a
of
This section
deals with
roots of unity.
The splitting field
Definition
55.1
with the
concerned
primarily
of a\"
that F
Suppose
of Lemma
proof then g(a)
of a
of a field F obtained by F was covered in Section
fields field
finite
1 over
field,
F is
the
and consider
33.8, we seethat
is a
if a
(xn
Gauss consideredcyclotomic the final chapter of his Disquisitiones of 1801. In that chapter, he gave a
in polynomials
constructive the
Gauss's
method, for
example
for actually determining in the case where p is prime. which became an important in the development of the general
procedure
of $p(x)
roots
Galois
solve a seriesof auxiliary equations, of p \342\200\224 the factor 1, with degree a prime coefficients of eachin turn being determined by the roots of the previous Gauss, of course, equation. the roots of ^>p(x) were all powers of knew that one of them, say roots
$i,
which
find
=
/7\342\200\2241
sets of
certain
taking
by
determined
He \302\243.
were
xn
\342\200\224 1 is
a separable
thus
and
a
case
the
anda3
=
these
three
+ \302\243b + \302\24310.Infact, ? + \302\24313 values are the roots of the cubic equation x3 + x2 \342\200\224 6x \342\200\224 7. Gauss then found a second cubic with coefficients equation, involving the a's, + \302\243ia + \302\2434,
xm
the
of these
where p
= 19
Gauss needed to 3x3x2), of degree 3 and one of degree 2
18 =
two equations
as his auxiliaries.It turned out that the first one the three roots, ax = t, + \302\2438 + t,1 + \302\24318 + = + + + + + + a2 f11 \302\24312, ? \302\24316 \302\24314 \302\24317 \302\2433 \302\2435,
had
sums of
roots
desired
the
equations. For example,in (and
characteristic
field of
By long division, as in the \342\200\224 \342\200\224 g(x) = (xn 1)/(x a), of F does not divide n.
the auxiliary
each of
equations
the
\342\200\224 1 and
roots were sums of two of the powers of \302\243, whose coefficients finally a quadratic equation, involved the roots of the previous which equation, had \302\243 as one of its roots. Gaussthen asserted (withouta complete proof) that each auxiliary equation can in turn be reduced to an equation of the form
was to
theory,
\342\200\224 FW. 1) \342\202\254
zero of xn
\342\226\240
Note
\342\226\240 Historical
Arithmeticae
be
shall
of F.
extension
\302\253th cyclotomic
= (n \342\226\240 l)(l/a) 7^ 0, provided that under this the splitting condition, Therefore, normal extension of F.
Carl
so we
33,
some
to F
adjoining
case where F is infinite.
\342\200\224
is any
Extension
Cyclotomic
extension
case
The
Theory
whose and
\342\200\224 which
A,
That
is, he
group
clearly
showed that
in this
can be the
solvedby
solvability
case, the
cyclic group the cyclotomic equation
implied
that
in terms
of radicals.(SeeSection
radicals.
of the of order was
Galois p
\342\200\224
1,
solvable
56.)
from now on that this is the case, and let K be the splitting field of xn \342\200\224 1 \342\200\224 1 has n distinct zeros in K, and by Corollary 23.6, theseform over F. Then xn a cyclic the field multiplication. We saw in Corollary 6.16 that a cyclic group of order n under of order n has
to Theorem wth
roots
of
20.8.For unity.
our
situation
here, these
generators
are exactly
the
primitive
55
Section
55.2 Definition
The
465
Extensions
Cyclotomic
polynomial
=
&n(x)
Y[(X-Cli) i=
are the
the at over F. where
F, is
in
unity
the
nth
polynomial
cyclotomic
\342\226\240
of
an automorphism see
Since
wth roots
of
wth roots
primitive
l
of unity, we as extended in
the
that over
Q,
Galois
the
that
must permute the G(K/F) fixed under every element of
group
is left 3>\342\200\236(x)
primitive G(K/F)
\342\202\254 In particular, for F[x]. way to K[x]. Thus $\342\200\236(x) \342\202\254 is a divisor of x\" \342\200\224 1. Thus over Q, we must F = Q, $\342\200\236(x) and $\342\200\236(x) Q[x], actually that i>p(x) is irreducible over Q, \342\202\254 seen have 3>\342\200\236(x) Z[x], by Theorem 23.11. We have need not be irreducible in the case of the fields Zp, it in Corollary 23.17. While $\342\200\236(x)
regarded
can
be shown
Let us complex
numbers.
numbers
in
is irreducible. $\342\200\236(x)
Let i be
cos
usual
the
. . 2n\\n = I 1-1sin \342\200\224
2n
n
n
so
(cos(27r/\302\273)
root of
i s'm(2n/n))m
+
a zero
unity,
with
the
complex
cos 2n
+
2n
sin
i
= 1,
)
is an wth root = 1 is n. Thus
i sin(Zn/n)
+
cos(27r/\302\273)
of
to subfields work
that
1 shows
Section
to characteristic 0, in particular zero of x2 + 1. Our complex
our discussion
limit
now
natural
of
least
The
unity.
+ i
cos(2n/n)
m such is a primitive
integer
sm(2n/n)
that
nth
of \342\202\254 Q[x]. <&\342\200\236(*)
55.3
Example
A
8th
primitive
root
of
is
in C
unity
ts =
cos
=
2n
2n h i
8
71 \342\200\224 i
cos
JT \342\200\224
.
sin
+
4
4
1
_~
1
of
unity
theory in Q
of cyclic
are
groups,
and \302\243, \302\2433, \302\2435,
in
=
Compare this
Let
with
Example
us still
is irreducible
restrict
over Q.
from
by Corollary
particular
'
V2
6.16 all
the
primitive
8th roots
so \302\2437,
<&\342\200\236(*)(x
We can compute, directly
1+ i
_ ~
V2
V2
By the
\342\200\224
sin
8
- r3)(x-
- ?)(*
this expression,
-
r5)(x
$\302\247(*)
=
r7)-
x4 + 1
(see Exercise
1).
54.7. our
\342\226\262
to F
work
= Q,
and
let
us assume,
Let t,
=
2n cos n
h i
sin
2n \342\200\224,
n
without proof, that
3>\342\200\236(x)
466
PartX
Automorphisms and GaloisTheory
Note primitive nth root of unity. of order all of n multiplicative group consisting that is, all the generators of this roots of unity,
that
so
t,
is a
prime to n. The field Let K = Q(\302\243).If t,m is another primitive over Q, there is an automorphism conjugate m relatively
and
similar
the
that
is a \302\243
All
the
the
cyclic
the form \302\243m for group, the whole splitting field of x\"
Q(\302\243)is
root
nth
xm in
of
since
then
unity,
G(K/)mapping
nth
primitive 1 <
are of
and \302\243
m <
\342\200\224 1
over
Q.
rr be
t,m. Let root of unity
nth
n
are \302\243m
onto \302\243
a primitive
in G(K/)corresponding to
automorphism
of
generator
of unity.
roots
nth
\302\243r.
Then
Galois group
that the
shows
This
consisting of elements group has
and is
Special cases of this For example,a of Example in
55.4 Theorem
that
of the
consisting of the
55.5 Example
Example
*8oo 55.6 Corollary
as the
same
'
of the
Theorem 20.6 n. This
modulo
these results
a theorem.
in
and is
elements
arguments
prime to n
less than n and relatively
integers
positive
we made
and
exercises.
text and
in the
unity,
extension of Q has
n. is
it
splitting field of x8
+ 1 (see Example 55.3 and
The Galoisgroup
of
root
We summarize
this theorem, for
illustrates
54.7
x4 + 1 is the
8th
here.
nth cyclotomic
modulo
multiplication
a primitive
54.7 is
to the isomorphic
group
Gn of
group
multiplication
several times
have appeared
material
The Galoisgroup under
?\"\"\342\200\242
abelian.
to those given
identical
example
to the
to n under
prime
=
(?m)r
is isomorphic
G(K/)
of Z\342\200\236 relatively
elements
=
= (rm(0)r
= rm(n
(rmrr)(0
Exercise
pth cyclotomic
to see
easy
the
that
field of
splitting
Q. This follows
\342\200\224 1 over
from the
fact
that
1).
extension of Q for
cyclic of order
p is
a prime
p-1.
Proof
By Theorem
55.4, the
Galois
1 elements, p \342\200\224
and is
to p prime of nonzero \342\226\240) (Z^*,
under
relatively
this
group
of the
group
isomorphic to the of the
group
p. This field under Zp
field
of Q has
integers less than is exactly the multiplicative
of positive
modulo
multiplication
elements
extension
cyclotomic
pth
By
multiplication.
p and
Corollary
group
23.6,
is cyclic.
Constructible
We conclude with with a compass
\342\231\246
Polygons an and
constructible if and
deterrriining We saw
application a straightedge.
only
if cos(27r/n)
t,
=
which in
n-gons
regular
Section
32 that
is a constructible real number. In cos
h i
sin
In \342\200\224.
n
n
Then 1
2n
t,
n
\342\200\224 = cos
. 2n
i sin
\342\200\224.
n
are
constructible
the regular n-gon is Now
let
Cyclotomic Extensions
55
Section
467
for
In . In \\ ( II cos 1-1 sin \342\200\224
In cos
n J
n
But
2n
\\ \342\200\224 I =
i sin
In
9
h sin\"
n
n )
n
\\
9
cos\"
2n \342\200\224 = 1.
n
then
In
1
32.8 shows Corollary an extension of Q of
Thus generates
If K
If a
is
the
that
- =
+
t,
2 cos \342\200\224. \302\273-gonis
regular
degreea power field of xn - 1 over splitting and CT(0 = T. *en
the
\342\202\254 G(JT/Q)
+
.(f
only if
constructible
+ l/\302\243 \302\243
of 2.
[K :
then
Q,
Q] =
55.4.
Theorem
+i
I)-r
cos
2?rr\\
277T
sin
h i
n
\302\273
= 2 cos
2nr
(
+
cos
2nr
2nr
n
\\
J
i sm
n
.
n
<
< r
Butforl
the only
Thus
of G(K/Q)
leaving
r,
+ Q(\302\243
=
with
r(\302\243)
=
\302\243*_1
1
the regular n-gon is constructible power of 2. It can be shown by elementary arguments
the
where
p-,
are
the distinct
If
first
=
be a power of 2, then and must be one more
q
an
odd
odd
every
the
subgroup
also
is a
theory that if
then
prime dividing
n
must
of 2. Thus we must
a power 2m
and therefore
- 1)\342\226\240 ~ \342\226\240 \342\226\240 1). (p,
V
+
(1) appear only each
to
the
have
1
\342\200\2241 is a zero of xq + 1 for q an odd prime, x + 1 divides xq + 1 for Thus, if m = qu, where q is an odd prime, then 2m + 1 = {2u)q + 1 is 1. Therefore, for pt = 2m + 1 to be prime, it must be that m is divisible
m. Since prime.
divisible by 2\" + 2 only, so /?, has to
by
n,
/>/<\"
than
that
theory,
\342\226\240\342\226\240\342\226\240/>,\",
dividing
Pi = for some
in number
\342\226\240 \342\226\240 \342\226\240
2v~lp^-1
is to
power
if
only
2>\"
odd primes
This shows Galois
\342\200\224 1.
Hence
n =
1/^.
of order 2, so by
fixed is
I/O
= n
2cos(27r/\302\273)onlyinthecasethatr onto itself are the identity t, + l/\302\243
elements of G(K/Q)carrying
the automorphism
automorphismand
=
\302\273,wehave2cos(27rr/\302\273)
have the
form
/,,-=2^
+ 1,
Part X
468
Galois
and
Automorphisms
Theory
that these numbers + 1 were prime for all conjectured = k. Euler showed that while k and 4 give the primes 3, 2, 3, 0,1, nonnegative integers 5, 17,257,and 65537, for k = 5, the integer 2{V) + 1 is divisible by 641. It has been 2(2' + 1 are composite.The case k = 20 is shown that for 5 < k < 19, all the numbers still as far as we know. For at least 60 values of k greater than unsolved 20, including whether the number k = 9448, it has been shown that 22 + 1is composite.It is unknown
a Fermat
prime. Fermat
primes is finite We have thus shown
or infinite.
of Fermat
thosewhere
might be constractible are whose dividing primes squares do not divide n. that for be constractible regular p a prime greater /?-gons might
In particular, the only than 2 are those where
55.7
Example
The divides
p is a Fermat
since 7 is
constractible,
a Fermat
not
a Fermat
3 is
while
Similarly,
prime.
prime,
its
the
square
A
18.
of unity
cos(27r/\302\273)
that all these
Let
are candidates
\302\273-gons that
regular
constractible.
be the
\302\243 again
primitive
wth
We saw above that
+ i sin{2n/n).
2cos
In
\342\200\224 =
n and
that
prime.
It is a fact that we now demonstrate for being constractible areindeedactually root
\302\273-gons
Fermat
are
constractible, for
is not
18-gon
regular
the only regular n
is not
7-gon
regular
that
odd primes
the
1 + \302\243
-, t,
that
f +
+ Q(\302\243
is
I/O
?
a power 2s of 2. of K = Q(\302\243)left the subfield
that cp{n) is
now
Suppose
element of G(K/Q) and r(\302\243)= Hj of order V of G(Q(0/Q) for W
= Ho
E be
Let
+ Q(\302\243
l/\302\243).We
where
=
fixed
{i,t}, by H\\ l/\302\243. By Sylow theory, there exist additional \342\226\240\342\226\240\342\226\240 ,s such that j =0,2,3, Hs =
< Hk
\342\226\240 \342\226\240 \342\226\240 <
saw i
above that
is the
identity
subgroups
G(Q(n/Q).
By Galois theory, Kh, <
and[*/,,_, : KH] then
ccj is a
formula,\"
we
=
zero of
KHs_t
2.Notethat(\302\243+
some (a,x2 + bjX
<
1/f) +
+ 7
\342\226\240\342\226\240\342\226\240<
KH]
=QU
\342\202\254 K,soQ(\302\243
Cj)
\342\202\254 KH [x].
+
1/f)
< R.IfKg^
By the
familiar
= KH,(cij), \"quadratic
have
KH]_l=KHj^bJ^-4ajcj).
Since we saw in Section numbers can be achieved
33 that
construction of squareroots of positive constractible and a compass, we seethat every element in
by a straightedge
Exercises
Section 55
is constmctible. Hence the -1/\302\243), in particular cos(2n/n),
55.8Theorem
The
\302\273-gonis
regular
the odd
with a compass and a straightedge Fermat primes whose squaresdo not
n are
primes dividing
The regular 60-gonis constmctible,
55.9 Example
60 =
since
(22)(3)(5)
if and
if all
only
n.
divide
3 and
and
\302\273-gons where
regular
constmctible
469
5 are
both Fermat
primes.
\342\226\262
55
\342\226\240 EXERCISES
Computations
1.
1.
to Example 55.3, complete the indicated showing that $%(x) = x4 + [Suggestion: computation, = -1 to simplify = the coefficients.] the product in terms of \302\243, and then use the fact that \302\2438 1 and f4
Referring
Compute
2. Classify
3. Using 4.
formula
the
b.
\302\242(60)
Give the first
compass.
3 for which
>
a
the splitting
of x12
field
the regular
constmctible
\302\273-gonis 1\302\260, 2\302\260, 3C,
the
with a straightedge 360-gon,
regular
indicated
and
Fundamental
for a
a straightedge
and so
on.]
- 1 over Q.
e G(K/Q),a1 is 11.12 of
Theorem
finitely
identity automorphism. abelian groups. generated
$s(*) over Z3. are there in the splitting
7.
Find
$3(x)
over Z2. Find
8.
How
many
elements
Classify G(K/Q) according to
the
field of x6
\342\200\224 1 over
Z3?
Concepts
9.
Mark
of the
each
a.
following
or false.
true
over every
is irreducible <3>\342\200\236(x)
b. Every zero c. The group
of
\302\242,, (x)
\342\202\254 Q[x]
over Q
d.
of
\302\242,, (x)
\342\202\254 Q[x]
over
The
group
e. TheGalois f.
The regular
g. Theregular
in C
group
All
integers
All
Fermat
\302\242,, (x)
of the
is a primitive
splitting
17-gon the
of the primes
is constmctible regular
p-gon
wth
root
of unity.
has order n. Q is abelian. of <&\342\200\236(x) over Q
field
25-gon is constmctiblewith
h. For a prime p,
i. j.
of
0.
of characteristic
field
with
a straightedge a straightedge
is constmctible
form 2(2 ' + 1 for are numbers of the
nonnegative form
if
and
has order
and a compass. and a compass. only if
integers
2(2*) + 1 for
p is a Fermat prime.
k are Fermat nonnegative
value:
a compass.
[K:Q].
b. Show that
of
Theorem
\302\242)(8100)
on, constmctiblewith
and so
to constructing
amounts
1\302\260 angle
c.
\302\2425(1000)
degree, that is,
of integral
Constructing
[Hint:
K be
a. Find
30 values of n
smallest angle
5. Find the
6. Let
over Q according to the Fundamental Q[x] polynomial (x20 - 1) \342\202\254 abelian groups. [Hint: Use Theorem 55.4.] of n, as given in Eq. (1), compute the for
group
generated
finitely
a.
the
primes.
integers
k.
the
and a
Part
470
X
Galois
and
Automorphisms
Theory
Theory
10. Show
if F
that
is a field
not dividing n,
of characteristic
=
x\"-l
then
f]^(x) d\\n
in F[x],
11. Find 12. Find 13.
the
where the
$i2(x)
dofn.
over Q <3E>\342\200\236(x)
for n
[Hint: Use Exercises10and
in Q[x].
\342\200\224
1, 2,
n,m e Z+ be relatively
that the
Show
prime.
- 1)over
-
of
in C
3, 4, 5,
and
6. [Hint:
Use Exercise
10.]
11.]
for odd integers Show that in Q[x], <$>2n(x) = $\302\253(\342\200\224x) odd, what is the order of \342\200\224 f ?]
14. Let
> 1. \302\273
[Hint: If f is a primitive
wfh root
of
Q is
field
splitting
in C
xnm
\342\200\224 1 over
of
for
unity
as the
same
the
n
Q. (x\" l)(xm splitting over Q is isomorphic to the direct Let n, m e Z+ be relatively prime. Show that the group of (x\"m - 1) \342\202\254 Q[x] - 1) \342\202\254 Galois and of (xm \342\200\224 over Q. [Hint: Using 1) \342\202\254 Q[x] theory, show Q[x] product of the groups of (x\" 1. Then use of the group of xnm \342\200\224 1 and x\" \342\200\224 1 can both be regarded as subgroups that the groups of xm \342\200\224 Exercises 50 and 51 of Section 11.] field
15.
polynomial
cyclotomic
all divisors
is over
product
SECTION 56
QuiNTIC
Problem
The We
THE
OF
INSOLVABILITY
has
coefficients
real
wi|h
with the fact
familiar
are
for f(x)
e F[x], where
Exercise
4 asks us
to
that
F
f(x) = polynomial in C. as zeros 4ac)/2a
a quadratic
\302\261 ~Jb2 {\342\200\224b
ax2 +bx + c,a ^0,
\342\200\224
this
Actually,
+
show
is true
is any field of characteristic^= 2 and the zeros are this. Thus, for example, (x2 + 2x has its 3) e Q[x]
in
F.
zeros
cubic polynomial over Q can in of The answer is yes, and indeed, even the also be terms radicals. always expressed 4 over Q can be expressed in terms of radicals. After zeros of a polynomial of degree formula\" for zeros of a 5th degree mathematicians had tried for years to find the \"radical it was a triumph when Abel proved that a quintic need not be solvable by polynomial, in
Q(V\342\200\2242). You
whether
the
zeros
of a
first job will be to describe precisely what we have developed is used in the forthcoming
Our
radicals.
the algebra
this
means.
A large
of
amount
discussion.
by Radicals
Extensions
56.1Definition
wonder
may
a field F is an extension of F by radicals if there are elements \342\226\240 \342\226\240 \342\226\240 a\"1 e F and \342\226\240. such that K = F{ax, \342\226\240 nx, nr \342\226\240, a\\,---,ar ar), positive integers \342\226\240 \342\226\240 < 1 and a\"' \342\202\254 for < i r. A \342\202\254 is solvable \342\226\240, a,-i) F[x] F(al5 f(x) polynomial by field E of f(x) overF is contained in an extension of F radicals over F if the splitting An
K of
extension
e K
by
radicals. A
\342\226\240
f(x)
polynomial
every zero
of f(x) by
multiplication,
that the quintic that no quintic
e F(x)
using
is thus
a finite
solvable
sequence of the
by radicals operations
over F
if
can
we
of addition,
obtain
subtraction,
division,
of F. Now
to
say
is not
0, is
to
say
is
with elements and taking \302\273,th roots, starting solvable in the classic case, that is, characteristic as the following shows. solvable, example
not
first
in the
in terms
Ars Magna
the initial discovery due to Scipione
also After
formula for solving of radicals was in 1545
of Girolamo
Cardano,
method is
of the
del Ferro
Ferrari,
in
although also
by radicals,
in Cardano's work. appeared had attempted mathematicians
solve quintics
by
Louis Lagrange
similar who
it was Josephfirst attempted a
methods,
in 1770
of the general principles underlying for polynomials of degree3 and 4, and showed why these methods fail for those of higher in the former His basic insight was that degree. cases there were rational functions of the roots that took on two and three values, respectively, under all detailed
the
analysis
solutions
56.2
Example
degree.
Thefirst of
the
to claim to have a proof of the quintic was Paolo equation in his algebra text of 1799. His
mathematician
insolvability
Ruffini
to
of
could
functions
rational
part
many
the roots, hence these be written as roots of less than that of the original. equations of degree No such functions were evident in equations of higher
possible permutations
Tartaglia. discovered a
equations
quartic
solving
Niccolo
and
Lodovico
student,
method for which
of a
publication
equations
Cardano's
471
Quintic
Note
\342\226\240 Historical
The cubic
Insolvability of the
56
Section
(1765-1822)
proof was along the lines suggested by Lagrange, in that he in effect determined all of the subgroups of S5 and showed how these acted on subgroups of the roots of the equation. rational functions
Abel
who,
versions of the in
settling
this
gaps
his
in
all of
Ruffini's gaps
centuries-old
various
Niels Henrik
It was
proof.
a complete
and 1826, published
1824
closing
proof,
several
were
there
Unfortunately,
published
and
finally
question.
1 is solvable by radicals over Q. The The polynomial x5 \342\200\224 1 field K of x5 \342\200\224 splitting = is generated over Q by a primitive of unity. 5th root \302\243 Then \302\2435 1, and K = Q(\302\243)2 is solvable by radicals over Q, for its splitting x5 \342\200\224 field over Q is generated Similarly, \342\200\224 2. \342\226\262 4/2 and t,, where 4/2 is the real zero of x5 by To say that the quintic is some polynomial of degree 5 We shall show this. We assume
characteristic0. of the
in the classic case means that there exists that is not solvable coefficients by radicals. this section that all mentioned have throughout fields
insolvable with
real
argument is as follows,and
it is
worthwhile
to
to remember
The
outline
1.
show that a polynomial f{x) e F[x]is solvable We shall by radicals over F Galois {ifand) only if its splitting field E over F has a solvable group. Recall that a solvable group is one having a composition serieswith abelian While this theorem goesboth ways, we shall not prove the \"if part. quotients. We shall show that there is a subfield F of the real numbers and a polynomial ~ \342\202\254 5 E with F a over such that f{x) F[x] of degree Ss, splitting field G(E/F) the symmetric group on 5 letters.Recall that a composition series for S5 is {;} < A 5 < S5. Since A 5 is not abelian, we will be done.
try
it.
2.
The following
56.3Lemma
Let
F be
over F,
lemma doesmost
a field of
then
G{K/F)
of our
characteristic0, and is a
work for let
solvable group.
a e
Step 1. F. If K is
the
splitting
field
of
x\"
-
a
472
Part X
Proof
and
Automorphisms
Suppose first unity
Theory
all the
F contains
that
are exactlythe
the generators
wth
By Corollary 23.6 the wth roots of generator of the subgroup. (Actually, Then roots of unity.) the wth roots of unity of unity.
roots
Let \302\243 be a subgroup of {F*, \342\226\240).
a cyclic
form
Galois
wth
primitive
are
l,?,?2,\"-,?\"-1. If
a zero of (xn
F is
e
/6
then all zeros
e F[x], \342\200\224a)
of xn
\342\200\224 a
are
/8,^/8,^/8,---,^/8.
Since
JT
=
an automorphism a in G(K/F) is determined by the value a(P) of the and z(f}) = t,'1$, where r e G(K/F), then p. Now if (7(,8) = \302\243'\302\243
F(/J),
automorphism
on
= T((T(/8))
(T(T)(/8) since
= ?'T(/8)=
= T(f>)
f'V/8,
Similarly,
('eF,
((TT)(/8) =
f^''/8.
solvable. err = tct, and G(K/F) is abelianand therefore Let \302\243 Now suppose that F does not contain a primitive wth root of unity. be a generator in F. Let fi again of the cyclic group of wth roots of unity under multiplication be a zero = field K of xn - a, \302\243 of x\" - a. Since p and \302\243\302\243 are both in the splitting is in (^)/)6 .ST.Let F' = F(\302\243),so we have F < F' < Now F' is a normal extension of F, since F' is the splitting field of x\" \342\200\224 1. Since F' = F(\302\243),an automorphism is r\\ in G(F'/F) \342\200\224 determined by r}(X), and we must have r](X) \342\200\224 1 t,1 for some ;, since all zeros of xn Thus
K.
of
are powers
=
If /ti(\302\243) \302\243.
^
for
e
/U
Oif?)(?) = M(/?(n)
then
G(F'/F),
= M(r')= M(0'= (?J)'
=
?y,
and, similarly,
(^)(0 Thus
is abelian.
G(F'/F)
By
the
Main
M < is
a normal
that to
G{K/F') G(F'/F),
subgroups
G{K/F') < G{K/F)
series of groups. The first part of the proof shows and Galois theory tells us that G(K/F)/G(K/F') is isomorphic which is abelian. Exercise 6 shows that if a group has a subnormal series of with abelian quotient then any refinement of this series also has abelian groups,
is abelian,
Thus
a composition
series of
groups, so G(K/F) is solvable. The
56.4 Theorem
Proof
theorem
following
Let F be a field extension
group
of
that
G(L/F)
characteristic
K is
extension
contained
is solvable.
G(K/F) must
have
abelian
quotient \342\231\246
will complete
and K is an
of F
We first show
the
of Galois Theory,
Theorem
series and hence a subnormal
groups.
quotient
= ^-
Part
1 of
our program.
zero, and let F < E < K < F, where E is of F by radicals. Then G(E/F) isa solvable
a normal group.
L of F by radicals in a finite normal extension and that \342\226\240 \342\226\240 Since K is an extension \342\226\240, by radicals, K = F(a1, ctr)
56
Section
\342\226\240 \342\226\240 \342\226\240, at_{)
e F{ax,
a\"'
where
1 <
for
splitting field Lx of
fi(x) = x\"1
Lemma 56.3 shows
that
<
i
Insolvability
r and
is a
G(Li/F)
a\"1
e
F. To
F. Then Lx
\342\200\224 a\"1 over
form L,
is a normal
solvable group. Now
473
of the Quintic
a\"2
we first
extension
form
the
of F,
and
and we
e Li
the
form
polynomial
we see that under action by any a in G(L\\/F), let We the field of over Then is a splitting be L2 L2 L\\. fi{x) f2(x) splitting field is a normal over F also and extension of F by radicals. We can form L2 from via repeated field of x\"2 \342\200\224 56.3, passing to a splitting Li steps as in Lemma cr(a2)\"2 we at each step. By Lemma 56.3and Exercise see that the Galois 7, group over F of of each new extension thus formed continues to be solvable. We continue this process fields in At we form the field of over F this manner: forming splitting stage i, splitting this
Since
is
polynomial
invariant
e F[x].
the
polynomial
a\342\202\254G(L,_!/F)
L = Lr that is a normal extension of F by radicals, that K < L. solvable group. We see from construction G(L/F) we need only note thatby we Toconclude, Theorem have 53.6, G(E/F) \342\200\224 G(L/F)/ Thus is and hence a factor a G(E/F) G(L/E). group, homomorphic image, of G(L/F). SinceG(L/F)is solvable, Exercise 29 of Section 35 shows that G(E/F) is solvable. over
We
L,_i.
and we
see
a field
obtain
finally
is a
that
\342\231\246
The It
of the
Insolvability
f{x) e F[x]
of
isomorphic Let
so on,
to show
for us
remains
to
yi e until
is a
there
that
the
5 such that
degree
subfield F
of the E of
field
splitting
real
and a polynomial
numbers
f(x) over F
a Galois
has
group
S5. M.
we
be
get
over Q,
transcendental y5 e that such M.
counting argument this fashion are independent and
Quintic
y2
e M. be
over
transcendental
Q(yi,
transcendental ---,)\302\274).
It
can
over Q(ji), be shown
and by
numbers
let 5
f{x)
=
- yi).
Y\\(x
1=1
Thus e E[x]. Now f{x) the elementary symmetry s\\
the
y\\ +
^2 = yiy2
in the yi
+
+yiyi
s5 =
yiy2y3y4y5-
H
yiy3
are, except possiblyfor y;, namely
of f(x)
coefficients
functions
-
a
exist. Transcendentals found in = E elements Let transcendental over Q. Q(yi ,---,)\302\276). real
transcendental
1-)\302\276.
+
+ yiys
yiy*
+
yzy*
+ yiys + +
yiyz
yzys + y^ys,
sign,
among
Part
474
:
X
of x'
The coefficient
Q(>>i,...,ys)
and Galois Theory
Automorphisms
Fig. 56.5). minates over Q, for Q(s,,...,
each
same polynomial
is the
j5)
e F[x] (see F = QCsi, s%,---, ss)',then f(x) Since the as indeterfield F of behave over fix). y, splitting on five a induces an a e 5s, the symmetric letters, group \342\200\224 = = a a Since for e and Q by a(a) a(y,) y,) yCT(,'). Tl?=1(x is
f(x)
a of \302\243 defined
automorphism ^
in
is the
E
Then
as
\302\261j5_,-.Let
\342\200\224
n^=1(x
we have
yCT(,)),
- si
aist) for
;, so
each
a leaves F fixed, and
a e
hence
^
G(E/F).
5!, so
order
has 5\302\276
Now
|G(\302\243/F)|>5!.
56.5
Since the
Figure
F, we
Thus
polynomial of
degree5
\\G(E/F)\\<
51.
F has
over
degree
at
5! over
most
see that
=
\\G(E/F)\\
we summarize
Let yx,
the automorphisms a make S$, so G(E/F) is not solvable.
5!, and
~
G{E/F)
Therefore,
56.6 Theorem
of a
field
splitting
the full Galois This completes
up
group our
G(E/F).
and
outline,
a theorem.
in
\342\226\240 \342\226\240be \342\226\240, ys
transcendental real numbers
independent
over
Q. The
polynomial
5
=
fix)
-
]\\ix
y\302\253)
i=i
solvable
is not
by
over
radicals
in yi,
function
symmetric
F =
Qisi,
st is
\342\226\240 where \342\200\242 \342\226\240, s5),
the
i'th
elementary
\342\226\240 \342\226\240 \342\226\240, ys \342\226\240
shows that (final goal) a that a generalization of thesearguments n > 5. n be radicals need not solvable for degree by of degree 5 in Q[x] that In conclusion, we comment that there exist polynomials are not solvable by radicals over Q. A demonstration of this is left to the exercises (see
evident
It is
polynomial
of
Exercise8).
56
\342\226\240 EXERCISES
Concepts
1. Canthe Z2?
2. Is every
field K of x2 + x + 1 over extension of Z2 by radicals?
polynomial
in
F[x]
of characteristic each of the following
F, if F is 3. Mark
Z2 be
splitting
Is K an
a. LetF be a field splitting b. Let F splitting
of the 0? Why
true of
field
why
+
bx6
+ ex4
+ dx2
a
/ 0,
Z2 of
solvableby
an element
radicals
in
over
not?
contained
0. A polynomial extension of
in an
0. A polynomial a solvableGaloisgroup
of characteristic
in Fhas
+ e, where
a square root to
false.
of characteristic
field in F is
be a field
form axs or
obtained by adjoining
in F[x]
F
by
in F[x] over
F.
is solvable
by
radicals
if and only
by
radicals
if and only if
if
its
radicals.
is solvable
its
c
475
Exercises
56
Section
x11 \342\200\224 5 over Q has a solvable Galoisgroup. are d. The numbers n and over Q. ^/n independent transcendental numbers e. The Galois group finite of a extension of a finite field is solvable. . f. No quintic is solvable by radicals over any field. polynomial 0 is solvable by radicals. over a field of characteristic g. Every 4th degree polynomial h. The zerosof a cubic polynomial over a field F of characteristic 0 can always be attained by means of a finite sequence of operations of addition, subtraction, division, and taking multiplication, square The
field of
splitting
roots
starting
The zerosof
in F.
elements
with
polynomial over a field a finite sequence of operations of addition, in F. roots, starting with elements series of groups of subnormal j. The theory
. i.
a cubic
F of
play
0 can never be attained division, and
characteristic
subtraction,
multiplication,
role
an important
in
by means taking
of Galois
applications
of
square
theory.
Theory
4.
field, and let f(x) = ax2 +bx + c be in F[x], where a ^ 0. Show \342\200\224 the 2, splitting field of fix) over F is F(^Jb2 4ac). [Hint: Complete formula.\"] work, to derive the \"quadratic be a
F
Let
is not school
5. Show
if F
that
is a
different from
of characteristic
field
fix)
where a
6. Show
^ 0, then for a
that
thus
quotients,
7. 8.
Show
that
series
with
This
a.
finite
group,
completing
for a
finite
abelian
exercise
is solvable by
fix)
group,
quotients,
exhibits
= ax4
if the
characteristic of F as in your high
just
square,
2 and bx2
+ c,
radicals over F. of a
every refinement the proof of Lemma a subnormal
with
the
of degree5
series
subnormal
56.3.
series
thus completing
a polynomial
+
that
the
[Hint:
with
Use Theorem
solvable
proof
in Q[x]
abelian quotients
quotient groups of Theorem 56.4. [Hint:
that is
not
solvable
also has abelian
34.7.] can be refined Use
by radicals
to a
Theorem
composition 34.7.]
over Q.
a cycle of length 5 and a transposition r, then H = S5. [Hint: subgroup H of S5 contains and of 9.12. SeeExercise Section S5 39, 9.] every transposition apply Corollary b. Show that if fix) is an irreducible polynomial in Q[x] of degree 5 having two complex and three exactly real zeros in C, then the group of fix) over Q is isomorphic to S5. [Hint: Use Sylow to show that theory the group has an element of order 5. Usethe fact that fix) has exactly two complex zeros to show that the of order 2. Then apply group has an element part (a).] = 5x4 5 is c. The polynomial 2xD irreducible in Q[x], by the Eisenstein with p = 5. Use + criterion, fix) the techniques of calculus to find relative maxima and minima and to \"graph the polynomial function /\" well must have exactly three to see that fix) real zeros in C. Concludefrom enough part (b) and Theorem 56.4 that is not solvable by radicals over Q. fix) Show
that
if a
Show
that
H contains
Matrix
Appendix:
Algebra
here. Matrices appear summary of matrix algebra chapters of the text and also are involved in several exercises. A matrix is a rectangular of numbers. For example, the array
We
a brief
give
in
in some
examples
array
2-14
3 is
having two rows and three so Matrix (1) is matrix,
a matrix
is an m x
Entries
let
in
n
a matrix be
Mmxn(W)
may be any type the set of all m x
notation is abbreviated to Two matrices having
canbe addedin Al Example
In Mi^iT),
the
will
for
m rows
n,
the
n columns
and
matrix
of n
way: we add entries
obvious
use uppercase A +
multiplication,
in
positions.
corresponding
4
-1
1
2
-6
3
letters B = AB,
to
B+
denote A
and
by
If A,
matrices. that
is denned only is, if A is an
of rows of B. That some integer s. We start
to the number matrix
having m =
is square. or rational, real, number\342\200\224integer, complex. We matrices with real number entries. If m = n, the If
we have
easily seen that Matrix
matrix
a 2 x 3 matrix.
can similarly consider M\342\200\236(Z), etc. M2x3(Q, number m of rows and the same number n of columns
same
the
1 We
columns. A
We M\342\200\236(K).
-1
it is
(1)
1 2
denning
A +
(B +
if the
x n
B,
number
matrix, as follows the
m
and
C) = (A
C are +
m x
B) +
n
of columns of A then
product
matrices,
C. is equal
be an n x AB where A is
B must
s
a
477
478
Appendix:Matrix 1 x
n matrix
=
matrix
the
dot
of
vectors or column A2 Example
We find
between a number and the (We shall not distinguish this product as as its sole entry.) You may recognize column vectors. Matrices having one row or one are row only only respectively.
that
7
an m x
A be
Let
entries
row
each
in
C =
product
is the product illustrated in A3 Example
number
vectors,
[3
The
matrix
n
AB is
of the
is m
an
row
i'th
= -15.
:(3)(1)+(-7)(4)+(2)(5)
2]
of A
B be
s matrix. Note that the number n of in each column of B. n of entries x s matrix. The entry of AB in the i'th row and j'th column of B as dennedby Eq. (2) and of A times the /th column and let
n x
an
number
as the
same
the
A2.
Example
Compute AB
Note that A is 2 x 3 and second row and third column is
all
Computing
eight
is 3
B
=
B)
entries of AB
in
-10
4
6]
17
1
x 4.
The entry
= 2 + 4+12
= 18.
will
be 2
in its
we obtain
fashion,
'2 ^
ABAB~
1
2
x 4. Thus AB
[1
this
1-1
14
4 6
1
(2nd row A)(3rd column
1
3 12
2-13
=
Solution
A4 Example
(2)
a-ib-i
+
a\\b\\
is a number. that
having
product
=
an]
a2
[a1
result
the
that
lxl
an n x 1 matrix:
B is
and
AB
Note
Algebra
9 6' 3
18
The product 2-13 1 is not number
2 1 4
6
5
4
of entries in a row of the denned, since the number entries in column of a of the second matrix.
For
square
matrices
denned. Exercise
of
10 asksus
the
same
to illustrate
first
both addition and the following fact.
size,
matrix
is not equal
to
the \342\226\262
multiplication
are always
Appendix: Matrix Algebra
Matrix
That
can
It
corner to the
BIn
where
be
is
is defined. by /\342\200\236 n matrix and
0
0\"
0
1
0
0
0
1
s matrix
and
If we
we can
conclude
e
the upper-left
B is any r x n matrix, then InA = as the number 1 does for multiplication
equation of
a matrix
consider
is unknown.
X
(A-1 A)X =
= A~lB,
A~l(AX)
\"1
acts much /\342\200\236
n x
an
known but AA~l = In, then
=
n x
any
the matrix
B are
and
A
A~lA
that
if A is,
multiplication
Let A
corner,
lower-right
to see that = B. That
is easy
when
are defined, as for A, B = AB + AC whenever
+ C)
A(B
with entries 1 along the diagonal from and entries 0 elsewhere.For example,
let
We
both products
when
= (AB)C and
expressions are defined. be the n x n matrix /\342\200\236
all these
and
commutative.
need not equal BA even be shown that A(BC)
AB
is,
M2(Z).
It
is not
multiplication
479
can find
the
n x n
an
A
AX = B, form matrix A~l such
that
A-1B,
=
InX
X =
A~1B,
A~lB,
X. Such a matrix A-1 actslike the reciprocal of a and we have found the desired matrix A-1. number: A-1 A = In and {\\/r)r = 1. This is the reason for the notation A is invertible and A-1 is the inverse of A. If If A-1 exists, the square matrix A-1 does not exist, then A is said to be singular. It can be shown that if there exists a = In also, and furthermore, A-1 such that A~lA = In, then AA~l there is only matrix
matrix
one
A5 Example
A-1 having
this
property.
Let A
=
2
9
1
4
We cancheckthat [2 [_1
9\"
\"2
=
4
-4
9]
1
1
4J
9
-2
Thus,
A~l
We
leave
a course in
the problems
linear
Associated
of
A
of
certain
and
denoted
of
=
9
-4
1 the
determining
-2 of A-1
existence
and
its
to
computation
algebra.
with each
square n x n This number
by det(A).
products of
the
numbers
that
matrix can
appear
A is
a number
be computed
in the
matrix
the
called
as sums A.
and
For
determinant differences
example,
the
Appendix: Matrix Algebra
480
2 matrix real number entries can be viewed 2 x
of the
determinant
,
that an
\342\200\224 be. Note
is ad
n
x
1 matrix
with
as giving coordinates of a point in \302\273-dimensional of such a single column matrix on the left by a real space W. Multiplication A produces another n x n matrix such single column matrix corresponding to another on the left by A thus a map of K\" into itself. It can gives point in W. This multiplication be shown that a piece of K\" of volume V is mapped by this multiplication by A into a \342\200\242 are important. that determinants piece of volume |det(A)| V. This is one of the reasons for n x n matrices A and B are of interest The following properties of determinants
Euclidean
text:
in this
1.
det(7\342\200\236)
2.
det(A\302\243)
3.
det(A)
=
1
=
det(A)
0 if
7^
det(\302\243))
and only if
4. If B is obtainedfrom then det(5) = - det(A) to
the
lower
diagonal.
In Exercises 1 through
\"-2
4\"
\"l +
i 4 3
\"1
3 \"4
3
of
A
is zero
2-f
-f
3
+
3-f
-i
\"3-
1
\342\200\224
-1
-1]
-1
2
1+i
-2+1
0
Ai
\342\200\224i
3
r
\"i
5
-4
2
2
1
i f
\"3f
0\"
7
3
-2 1
1
*-i
\"1
10. Give
11. Find
\"i
-1]
0
1 an
i example 0
-1
in Af2(Z)
1
0
showing
that
matrix
multiplication
-l , by
experimentation
if necessary.
4
3\"
6
1\"
-2f
i
&
8.
-
3
4\"
1 2
1
i-
1 + i
i
-2f
-1\" 1
if it is defined.
expression,
2
3-f\"
i
matrix
arithmetic
-3
1
-2
i
4
the given
9, compute
\"4
+
5
1
two rows (or two
interchanging
A t
\342\226\240 EXERCISES
1.
matrix
invertible
is an
columns)
of A,
above the main diagonal from the upper left corner this then det is the of the entries on (A) product right corner, are zero. The same is true if all entries below the main diagonal
If every entry
5.
A
A by
-r
i is not commutative.
Appendix:
12.
Find
\"2
0
0
4
0 0
-l 0\"
, by
0
experimentation
if necessary.
-1
\"
0
3
13. If
A =
10
4 14. Prove
that
if A,
0\"
0
-2
17 B e
,
find
det
(A).
8 are M\342\200\236(C)
invertible,
then AS
and
BA are
invertible also.
Matrix Algebra
481
Bibliography
Classic Works
1.
N.
Bourbaki,
58. 2. N. Jacobson, ..1953,and 3.
4. B. L. van
1949,and General
de Mathematique, BookII of Part
Lectures
in Abstract
Algebra.
Princeton,
I, Algebre.
N.J.:
Van Nostrand,
E. Sperner, Introduction 2nd Ed. New York: Chelsea,
and
G.
Birkhoff
1942-
vols. I, 1951,II,
der
Waerden,
Modern Algebra
to Modern
and Matrix
Algebra
Theory (English
1959. (English
New York:
translation).
Ungar, vols. I,
H, 1950.
Texts
Algebra
5. M.Artin, Algebra. Englewood Cliffs, N.J.: Prentice-Hall, 1991. 6. A. A. Albert, Fundamental Concepts of Higher Algebra. Chicago: Press,1956.
7.
Paris: Hermann,
1964.
HI,
O. Schreier translation),
Elements
and S.
MacLane, A
Survey
of Modern
Algebra,
3rd
Ed.
University of New York:
Chicago
Macmillan,
1965.
8. J.A.
Abstract Algebra, 2nd Ed. Lexington, Mass.: D. C. Heath, New York: Blaisdell, 1964. 10. T. W. Hungerford, Algebra. New York: Springer, 1974. 11. S. Lang, Algebra. Mass.: Addison-Wesley, 1965. Reading, 12. S.MacLane and G. Birkhoff, Algebra. New 1967. York: Macmillan, 13. N. H. McCoy, Introduction to Modern Boston: Allyn and Bacon, 1960. Algebra. 14. G.D.Mostow, J. H. Sampson, and J. Meyer, Fundamental Structures of Algebra. New 1963. McGraw-Hill, 15. W. W. Sawyer, A Concrete Approach to Abstract San Francisco: Freeman, Algebra. 9.
Gallian,
I. N. Herstein,
Contemporary
Topics in
1990.
Algebra.
York:
1959.
483
Bibliography
Group Theory
16.
W.
Theory
Burnside,
17. H. S.M.Coxeter
and
2nd Ed. New Order, of Groups of Finite W. O. Moser, Generators and Relations
York:
Dover, for Discrete
1955.
Groups, 2nd
Ed.
Springer, 1965. 18. M. Hall, Jr., The Theory of Groups. New York: Macmillan, 1959. New York: Chelsea, vols.I, 1955, The Theory of Groups (English 19. A. G. Kurosh, translation). Berlin:
II, 1956.
and
W. Ledermann, science, 1961.
20.
to the
Introduction
21. J. G.Thompson
and W.
775-1029. 22. M.A. Rabin, \"Recursive
172-194.
Feit,
Theory of Finite
Groups,
of Groups
of Odd
\"Solvability
Unsolvability of Group
4th rev. Order.\"
Pac.
Problems.\"
Theoretic
Ed. New
Inter-
York:
J. Math.,
13 (1963),
Ann. Math., 67 (1958),
Ring Theory
23. 24. 25.
26.
Studies in to Grobner Bases (Graduate W. Adams and P. Loustaunau, An Introduction R.I.: American Mathematical Society, 1994. Mathematics, vol. 3). Providence, E.Artin, C. J. Nesbitt, and R. M. Thrall, Rings with Minimum Condition. Ann Arbor: Press, 1944. University of Michigan No. 8). Buffalo: The Mathematical N. H. McCoy, Rings and Ideals (Cams Monograph JR.: Open Court, 1948. Association of America; LaSalle, 1964. N.H.McCoy, The Theory of Rings. New York: Macmillan, W.
' Field
27. E. Artin,
Theory Galois
28. O. Zariski
and P.
Number
29.
H. Hardy and E. M. Wright, Clarendon Press, 1960.
Dame,
N.J.:
Van Nostrand,
vol. I, 1958.
An Introduction
to the Theory of Numbers,
4th
Ed. Oxford:
1964. Reading, Mass.: Addison-Wesley, 1962. of Numbers, Reading, Mass.:Addison-Wesley, in Number Mass.: 2 1956. vols., Theory. Topics Reading, Addison-Wesley, to Number Theory. New York: Wiley, 1951. Introduction and H. S. Zuckerman, An Introduction to the Theory of Numbers. New York: Wiley,
32.
33.
T.
Nagell,
Ed. Notre
Numbers.
Algebraic
W J. LeVeque, W J. LeVeque,
34. I. Nivin
No. 2), 2nd
Theory
G.
30. S.Lang, 31.
Lecture (Notre Dame Mathematical Dame Press, 1944. Samuel, Commutative Algebra. Princeton,
Theory
of Notre
Ind.: University
Theory
Elementary
1960.
35. H.Pollard, Mathematical
36.
Theory of Algebraic Numbers (CamsMonograph of America; New York: Wiley, 1950.
The
No.
9). Buffalo:
The
Association
D. Shanks,
Solvedand
Unsolved
Problems
in Number
Theory.
Washington,
D.C.:
Spartan
Books, vol. I, 1962.
37. B.M. Stewart, 38. J. V Uspensky
Theory of Numbers, 2nd Ed. New York: and M. H. Heaslet, Elementary Number
Macmillan, 1964. Theory. New York: McGraw-Hill,
1939.
39.
E. Weiss, Algebraic
Number
Theory.
New York: McGraw-Hill,
1963.
485
Bibliography
Algebra
Homological
40. J.P.
Jans,
Rings and
Homology. New
41. S. MacLane,Homology. Other
42.
A.
Berlin:
York:
Springer,
Holt,
1964.
1963.
References
A. Albert
The Mathematical
(ed.), Studies in Association
Modern
Algebra of America;
vol. (MAA Studies inMathematics, Englewood Cliffs, N.J.: Prentice-Hall,
2). Buffalo:
43. E. Artin, Geometric 1957. Algebra. New York: Interscience, R. Courant and R. Robbins, What Is Mathematics!Oxford 44. Press, 1941. University 45. H. S.M.Coxeter, Introduction to Geometry, 2nd Ed. New York: Wiley, 1969. 46. R. H. Crowell and R. H. Fox, Introduction to Knot Theory. New York: Ginn, 1963. 47. H. B. Edgerton, 1977. Elements Press, of Set Theory. San Diego:Academic 1996. 48. C. Schumacher, Zero. Reading, Mass.: Addison-Wesley, Chapter
1963.
Notations
G,a
S
membership,
0
empty
G
i,aiS {x
I
P(x)}
B C A
B CA AxB
Z
0 R
C
Z+,Q+,R+ Z*,Q*,R*,C*
1
set, 1
nonmembership, 1 set of all x such that P(x), 1 set inclusion, 2 subsets ^ A, 2 Cartesian of sets, 3 product integers,
rational
3
numbers, 3
real numbers, complex
3
numbers, 3
positive
elements
nonzero
elements
M
relation,
\\A\\
number
of elements
in
mapping
of A into
B by
0(\302\253)
0[A]
3
\302\2530
X = b(mo6.n) =\342\200\236,a
A, 4; as
of element
image
of set A
of \302\242, 5
cell containing
5 in a partition of n, 7
x g modulo
Rc
power set of A, 9 set of all z g C such that setof all x g R such that
+c
addition
U
4
correspondence,
the inverse function of Z+, 5 cardinality congruence
,nA)
order of gr
4 \302\242,
4 a under \302\242, under 0, 4
image
\342\200\242<->\342\226\240 one-to-one
0-'
Q, R, 3 Q, R, C, 3
of Z, of Z,
c, 16 roots of unity,
modulo
group of rcth
f/\342\200\236
|z| = 1, 15 0 < x < c,
18
S, 6
16
Notations
18 1,2,---,/1-1), \342\226\240 n \342\200\224 \342\226\240, 1} under addition modulo n, cyclic group (0, 1, \342\226\240 of residue classes modulo \302\253,137 group \342\226\240 \342\226\240 n \342\200\224 \342\226\240, ring (0, 1, 1} under addition and multiplication modulo n, 169
Z\342\200\236{0,
*, a
b
*
o, / o g, ax {S, *) ~, S ~ S' e
operation, 20
binary function binary
29 30
structures,
isomorphic
element,
identity
m x
22, 76
composition, structure,
32
matrices with entries from 5, 40,477 matrices with entries from 5, 40, 477 GL(\302\253, R) general linear group of degreen, 40 matrix determinant of square det(A) A, 46, 479 a~l, \342\200\224a inverse of a, 49 H < G;K < L 173 inclusion, subgroup inclusion, 50; substructure H
Mmxn(S)
M\342\200\236(S)
Sj n s2 n
n
n x n
\342\226\240 \342\226\240 \342\226\240 n
sn
SA
(
group of permutations
*
Sn
symmetric
group on n
n!
n
78
factorial,
D\342\200\236rcth
dihedral
A\342\200\236alternating
aH,a
+ H
Ha, H + a (G:H)
IT=i $, Si
left
coset
77
of A,
77
map,
identity
78
letters,
group, 79
group on n letters, of i? containing a, of H containing in G, 101
right
coset
index
of
Euler
phi-function,
#
Cartesian
product
93
97 a, 97
104, 187 of sets,
104
\342\200\242 \342\200\242 x S2 x \342\200\242 x S\342\200\236
n\"=1
Gi
Gi \302\251\"=1
lcm G 4>c 7T,
[B] \302\242-l
Ker(0) G/N;
R/N y ig
Z{G) C Xg
direct product of groups, 104, 105 105 direct sum of groups, least common multiple, 107 natural subgroup of fj\"=1 G,-, 107 126 evaluation homomorphism, onto ith component, 127 projection inverse image of the set B under 0,128 kernel
of homomorphism
factor group,
137;
129 \302\242,
factor ring,
class map, inner automorphism, 141 center of the group G, 150 canonical
commutator
residue
subgroup,
subset of elements
of X
242
139, 140
150
left fixed by g,
157
54
Notations
Gx
Gx
isotropy subgroup of orbit of x under G,
R[x]
polynomial
F(x)
field
<$>p(x)
FG
H R[[x]]
F((x)) F[x] V(S) {bi,
\342\200\242 \342\226\240 \342\200\242, br)
lt(/)
lp(/) irr(a, F) F)
deg(a,
F(a) [E : F]
G leaving
x fixed,
157
of F[x],
R, 200
in
201
of rational functions
in n indeterminates, 201 of degree p \342\200\224 1, 216, 217 polynomial of A, 221 223
cyclotomic endomorphisms
ring,
group
group algebra
the field
over
F, 223
225
224,
quaternions,
formal power seriesring in x over R, 231 formal series field in x over F, 231 Laurent \342\200\242 over F, 255 in xx, \342\226\240 of \342\200\242, x\342\200\236 ring polynomials variety
algebraic
of polynomials
ideal generated by
irreducible
over F,
260 for a
polynomial
of a
degree
5, 255
\342\226\240 \342\200\242255 \342\200\242, br,
/, 260
polynomial
of lt(/),
product
power
in
elements bx,
leading term of the
over F,
269
269
by adjoining a to field F, 270 over F, 283 \342\226\240 to F, 285 obtained by adjoining ax, \342\226\240 a\342\200\236 \342\226\240, obtained
field degree
\342\200\242 \342\200\242, a\342\200\236)field F(ax, \342\200\242
FE F
with coefficients
ring
of quotients
\342\200\242 \342\200\242, x\342\200\236)field F(*!, \342\200\242
End(A) RG
elements of
158
of E
algebraic closure of F in E, 286 an algebraic closure of F, 287, 288
Galois field of order p\", 300 product set, 308 HV N subgroup join, 308 normalizer of H, 323 N[H] free group on A, 341, 342 F[A] 348 group presentation, (xj : ri) 357 3\342\200\236 boundary homomorphism, GF(p\")
HN
C\342\200\236(X)
n-chainsofZ,
358
Z\342\200\236(X)
n-cyclesofZ,
359
B\342\200\236 (X)
n-boundaries
H\342\200\236 (X)
nth homology
5(n) C{n\\X) Z{n\\X) H{n\\X)
Hw (X) S\"
E\" X (X)
359 group of X, 361 coboundary homomorphism, 363 n-cochains of X, 363 of X, 363 n-cocycles of X, 363 n-coboundaries nth group of X, 363 cohomology n-sphere,
364
n-cell or n-baU, 364 Euler characteristic of X,
/*\342\200\236homology
(A, 3) dk
Hk(X,
chain
Y) a\\b
UFD
homomorphism 381
374
induced from
f :X
381
-*\342\226\240 Y, 375,
complex,
boundary operator, 382 relative homology group of chain
relative kth
Hk(A/A')
of X,
homology of simplicial a divides (is a factor of) b, 389 390 unique factorization domain, kth relative
complex complex
A modulo
A', 383 383
X modulo Y,
Notations
PID
ideal
principal
391
domain,
union of sets, 391 v
N(a)
fa.0
Euclidean
conjugation
408,410,455 isomorphism
subfield of E G(E/F)
{E:F}
automorphism index
401
norm,
norm of a,
of E
left
fixed
group
over F,
of F(ct)
with
F(fi),
416
by all at or all a e H, 419 of E over F, 420
428
Odd-Numbered
to
Answers
Not
Exercises
or Proofs
Definitions
0
SECTION
1.
{-V3,V3}
3. (1, -1,
5.
Not
2, -2,3, -3, 4,
-4,
5, -5,
(not well defined). A
a set
case
6, -6,
10,-10, 12,-12,15,-15,20,
can also
be madefor
the
7. Theset 0
9.
for
Asking
The
empty set
-20,
30, -30,
60, -60}
0.
set Q
c), (b, 1), (6, 2), (6,c),(c, 1),(c,2),(c,c) the line segment CD. through P and x, and let y be the point where it intersects = 2s. (Proofs are usually 17. Conjecture: omitted from answers.) n{0\\A)) = 12s\" = 2s\" = |K|. (The numbers x where 21. 102,105,10*\302\260 0 < x < 1 can be written to base 12 and to base 2 as as to base 10.)
11.
13.
(a, 1), (a, 2),
23. 1 29. 31.
33.
(a,
the line
Draw
25.
27.
5
Not an equivalence relation An equivalence relation; 0 An
well
52 = {0}, a
= {a,\342\200\224a} for each
nonzero a
eR
relation;
equivalence
1=(1,2,---,9),
10=(10,11,---,99),
\342\226\240 and 100= (100,101,\342\226\240 \342\226\240, 999), 10\"
35.
=
(10\",
10\"
+ 1,
-..,
10\"+1
in general \342\200\224
1}
i. (1,3,5,...),(2,4,6,.-.}
n. (1,4,7,...),(2,5,8,...),(3,6,9,.-.} \342\226\240 \342\226\240 iii. (1,6,11,\342\200\242 12, \342\226\240 \342\200\242}, \342\226\240}, (2,7, (3, 8,13, 37.
\342\226\240 \342\226\240 \342\226\240}, (4, 9,
14,
\342\200\242 \342\200\242 \342\200\242}, (5, 10,15,
\342\226\240 \342\226\240 \342\226\240}
two-to-twofunction that such a function into two distinct / should carry every pair of distinct suggests points Such a function is one to one in the conventional sense.(If the domain has only one element, a function cannot points. fail to be two to two, since the only way it can fail to be two to two is to carry two points into one point, and the set does not have two points.) Conversely, every function that is one to one in the conventional sensecarriesany pair of points into two distinct points. Thus the functions conventionally called one to one are precisely those that carry two points into two points, which is a much more intuitive unidirectional way of regarding them. Also, the standard way of trying The name
491
Answers to
492
to show a function a function is one to
is one
Exercises
Odd-Numbered
to one is precisely to more natural
becomes
one
does not
show
that it
in the
two-to-two terminology.
two points into
carry
just one point.
Thus,
proving
SECTION 1
3. -i
1. -i 7. 11.
5.
9. -4
17-15f 13.
2VT3
- + \342\200\224n
-Jli
21.
39.
V34
1
!
V2
V2h
) !
! \342\226\240
With
17.
=
-&iTL6l&m62) +
IZifelKcos^cos^
from Exercise38 and
at once
follows
the
(cos^
sin
equation
3. a, c. *
a one-to-one
correspondence.
52 +8^^008^)(]
IZ1HZ2I
=
|ziZ2l-
Not commutative,
associative.
is not
row: a;
row:
fourth
c,
b.
not associative
associative
Commutative,
Not commutative, not associative 8, 729, n[\"<\"+\302\273/2] No. Condition 2 is violated. No. Condition 1 is violated.
Yes
19.
23. a. Yes. b. Yes * and *' on S by a * b =? and a *' b = 25. Let S = {?,A}. Define 27. True 29. True 31. False. Let /(*) = x2, g(x) = x, and h(x) = 2x + 1.Then (/(*) g(x)) - h(x) = x2 - 3x - 1 but = x2+x + L h(x)) = x2-(-x-l) f(x) (g(x) 33. True 35. False. Let * be + and let *' be on Z.
SECTION3
1. i. 4> must 3.
is impossible for
2
1. e,b, a
21.
27. -Jl
we obtain
Multiplying,
5. Top row:
13.
2
-3
5 33. 1,7 0, f3 -w 7, f4 -w 4, f5 -w 1, f6 -e- 6, f7 -w 3 -<->\342\226\240 -<->\342\226\240 4 again, which -<->\342\226\240 -<->\342\226\240 0, and \302\2434 2, \302\2433 4, we must have \302\2432 \302\243
SECTION
9.
2
25.
4
23.
31.
-w f\302\260
and the desired result
11.
+^---
3-
19
Vl'
V2
Z1Z2
7.
\342\226\240
V3\302\261i\",\302\2612i', -V3\302\261i\"
29. 11 35. 37.
+ 4=^
)
V2
V2
V
15. 734(^1 VV34
23+7(
+ 4i
in.
4>(a *
No,
because
be one b)
to one. *' 4>(b)for
= 4>(a)
not map Z
does
5. 11.
Yes. 7. Yes = because No,
13.
No,
because
9.
ii.
must
be all
of 5\".
e S.
onto 11.
^ 1 for all
Yes
\302\242(/)
4>[S\\
all a, b
solution
/ e
F.
n
e Z.
A
for
all a,
be
S. (Other
answers
are possible.)
Answers to 15.
b. m
* n
=
*n
= mn
+ m+
a*
b =
-(ab +
b. a *
b =
3ab
19. a.
25. No. If {S,*) from
n;
a + b \342\200\224 2); identity 2 \342\200\224 \342\200\224 a
element 2
eL and a right
element
identity
2
-
element
b -\\\342\200\224; identity
left
has a
0
element
identity
2
element
identity
element
identity
eR, then eL = eR.(It is our
1. No. 9.
An
of solutions \342\200\242) (\302\243/10oo,
of the
equation
of
=
z1000
C under multiplication has 1000 elements. in {U, \342\226\240), four solutions one solution in (R,
1 in
form x*x*x*x=e
has
(R*, \342\200\242)\342\200\242 Yes 13. Yes
15. No.The matrix 17. Yes.
19.
(Proofs
21.
2, 3.
with
(It gets
c. for 4
harder
0 is upper
all entries
are omitted.)
25. a. F
answer
the
g. T
F
e.
no inverse.
\342\200\2241/3
elements, where
c. T
but has
triangular,
is not 4.) i. F
SECTION5 3.
Yes
11. No.
a. No.
19.
a.
21.
a. -50L
23.
All
25.
All matrices
27.
No. It
b.
Yes
17.
is not
a subset
even
b.
under addition. b. No. The zero constant
Yes
{it\"
\\ n
9.
e Z}
1
matrices
n
1
of the
29. c.
Yes
is not
in
1/2, 1/4
F. C. l,7T,7T2,l/7r,
1/7T2
n g :
for
0
of F.
function
b. 4, 2, 1,
25^0, 25L50
T
Q+ and
7.
closed
Not
4
Yes
5.
Yes
under multiplication.
closed
Not
13. Yes
39. a.
form
A\"
0
0
4\"
33.
4
31.
3
0 _22n
0
f or n
g Z
35, 3
2
i.
g- F
e. F
T
-22,,+1 +l
T
SECTION 6
1. g = 4,
r = 6
9. 4
21.
An
23.
infinite
3.
= -7,
r = 6
13. 2
11. 16 cyclic group
, z\342\200\236 :<3>. (4>:
proofs
5. No. .9, fails.
No. g? fails.
3.
fails. 3\302\276
The group
15.a.
to omit
4
7.
1.
practice
answers.)
SECTION
11.
+ 2;
mn
493
Exercises
/ g F.
no solution
=
1 has \302\242(/) \342\200\224 m \342\200\224 n
because
No,
17. a. m
Odd-Numbered
>>
>>;
(18) '(0)'
15.2
7.
5. 17.
6
60
19.
4
Yes
+),
and
two
solutions
in
Answers to
494
25.
33.
Klein
39. i(l
29. 1,17
27. 1,2,3,4,6,12
1,2,3,6 The
35.
4-group
Exercises
Odd-Numbered
37. Zs
Z2
+ iV3)andi(l-iV3)
41. -(73+ 0,2^-0, 51. (p -
\\){q
SECTION
7
-
2(-^
+
0,
2(-^-0
1)
3. 0,2,4,6,8,10,12,14,16
1. 0,1,2,3,4,5,6,7,8,9,10,11
5. ---,-24,-18,-12,-6,0,6, 12,18,24,
\342\226\240\342\200\242\342\200\242
b. a1
a. a3b
a
e e a
b
b
d
e
f
a
c
c
e
f
a
d
e
b
d
d
b
e
c
a
f
f
c
f d
a
b
e
e
11.
a
b
c
d
f
a
b
c
d
f
e
c
b
f
d
Choosea pair of generating directed arcs, call them sequences arcl, arc2and arc2, arcl lead to the same
vertex of the digraph, and see if the any (This corresponds to asking if the two corresponding group and only if these two sequences lead to the same vertex for every
The group is commutative if commute.) of generating directed arcs. is not obvious, since a digraph of a cyclic group might
generators
and arc2, start at
arcl
vertex.
pair
13.
It
one of which 15. 0
17. a.
of generators
any vertex or their
b. a4 = e,b2 = e, (abf SECTION
using a generating set of two
or more
elements, no
1
from
Starting
be formed
the group.
generates
a,
every
inverses
that
the graph that terminates at that same vertex to the identity and thus gives a relation.
through
path
is equal
= e
8
'12
3 4
,12
3 6 5 4)
5
1 2
6'
3'(
3
4
3
16
4
5
6^
2 5;
a represents
a product
/1
2
3
5
6\\
^2
6
1 5 4
3)
9.
7. 2 11.
4
(
13.
{1,2,3,4,5,6}
15. e, p, p2,
495
Exercises
Odd-Numbered
to
Answers
p^,
{1,5} their
p2
p
This gives our
is our/x^.
in the
elements
order p0,
Pi,
P2, Pi,
/ii,
Sx,
/x2, S2.
17.
24
19.
= (Po> Pi, /\302\276. = (Po> P2}, (Mi) = (Po> /*i}> P3}, (/\302\276) {po},(Pi) = (/\302\276) one of containing (Po, S2}. These are all the cyclic subgroups. A subgroup of the the \"turn the square over\" permutations /i\\, /x2, S\\, or S2 and also containing px or p3 will describe all positions the line of the we see that other elements so it table the must be the entire /xi, opposite only square group D4.Checking
(/x2) =
(po, /x2},
can
be in a
that
8.12, we
Table
to
Referring
=
(Si)
(p0) =
that
find
=
Si}, and (S2)
(po,
course, p0. We checkthat (p0, p2, /1.1,/1.2}is closedunder table opposite the /x2 gives the same subgroup. Checking possibility, using the same subgroup (p0, p2, Si, S2} as the only remaining
are p2, with/Xj proper subgroup subgroup. Checking the
multiplication
and is a
rows opposite
Sx
and
opposite
S2
gives
the
/x2, and, of
row
of the
reasoning.
21.
are \"elementary
a. These
When another matrix. matrices,\" resulting frompermuting the rows of the identity that in the same fashion P, the rows of A are permuted by one of these matrices of the three all 6 possiblepermutations were permuted to obtain P. Because matrix identity the entries 1,2, 3of the given column we see they will act just like the elements of S3 in permuting form a group because S3 is a group. permutation on the left
matrix A is multiplied the rows of the 3 x 3
23.
rows are present, vector. Thus they b. The symmetric group 25. D4 Z2
/0 1 2
,_
3\\ 1 /0 ^= : 2 2 3J, ^ A0=(^0
27. For Z4, The table fr0\"
rx
rx consists
\\r0 pa
left regular representation is m2 m3\\ r2 mx' /r0 rx I. Pi = I r2 r2 mx m2 m3y \\ri of the elements'of 53 in the order they the
for
is the
regular representation 31. Not a permutation 35. a. T c. T
37.
Not
X2 =
Q),
same
the
as the
1 2 3\\ 3 Q ^,
/0
^ table
for
Z4
X,
=
with n replaced
/0
1
2
3\\
^
Q
:
2j.
by
For S3, k\342\200\236.
p0
=
. , , , m2 m3\\ the bottom row in the permutation , etc-, where m2 m3 mxJ down the column under a in Table 8.8. The table for this right appear r2
m,
r0
for 53 with
a replacedby
pa.
IF
g. F
43.
No
3
a permutation
T
e.
table
3\\
2
Yes
9
SECTION
1.
same as the
33. 41.
monoid
A
53.
(1,2, 5},{3},(4, 6} 4, 5}, {6}, (7, 8}
3. (1,2,3,
5.
'12 12 4
5
11. (1,
3,
e Z} 7 2
3
3 4 5 6 7 3 7 8 6 2
0
3
13
4
13.
+ l\\n
{2\302\253|\302\253 eZ},{2n
4
5
6
5
8
6
6)(5, 8, 7) =
4)(2,
(1, 4)(1, 3)(2,6)(5,
7)(5,
8)
a. 4
b. A
cycle
c.
has
er
of length order 6; r
n
d. 6 in
Exercises 10 and
e. The
order
17.
30
4.
11, 8
of a permutation
the cycles.
15. 6
order n.
has
has order
in Exercise
12.
expressed as a product
of disjoint
cycles is the
least
common
multiple of the
lengths
of
Exercises
to Odd-Numbered
Answers
496
19.
(2, 3, 4)
I
*
(1, 4, 2)/
(1,4)(2,3)^ (1, 3, 4)
4)
-^(1,2,
(1,4, 3)^ 23.
3.
e. F
c. F
a. F
g. T
4)
T
i.
10
SECTION
1.
\\
N
1-(2,4,3)
(1,2)(3,
\\(1, 3)(2,4)
/
i
\\
4Z= {-.-,-8,-4, 0,4,8,---} 1+ 4Z= {---,-7,-3,1,5,9,---}
2+
4Z=
{---,-6,
-2,2,
3 +
4Z =
{---, -5,
-1,3, 7,11,---}
6,10,
\342\200\242 \342\200\242 \342\200\242}
1+.(2) = {1,3,5,7,9,11} \342\200\242 17 + (18) = {17,35} \342\200\242, 18}, (18) = {1,19},2 + (18)= {2, 20}, \342\226\240 Not the same. 7(Po, M2}, {Pu$ih lP2, Mi}, (P3, S2}. 9. {po, Pi], {Pi, P3}, (Mi, M2}, [Su S2} 11. Yes, we get a coset group isomorphic to the Klein 4-group V. (2)
=
8, 10},
(0,2,4,6,
5. (18)= {0,
1 +
Po
P2
Pi
P3
Mi
M2
Si
Po
Po
P2
Pi
P3
Mi
M2
s2 \302\2531
P2
P2
Po
P3
Pi
M2
Mi
s2
Si
Pi
Pi
P3
P2
Po
s2 \302\253i
M2
Mi
P3
P3
Pi
Po
Pi
\302\2532 \302\2531 Mi
Ml
Ml
M2
s2 h
M2
M2
Ml
M2
Po
P2
P3
Pi
\302\2531
s2
P2
Po
Pi
P3
Mi
P-2
Pi
P3
Po
Pi
\302\2532\302\2532 M2 \302\253i
Mi
P3 Pi
P2
Po
\302\2531\302\2531 S2
13. 3
19. a. T 21. G = 23.
s2
15. 24 c.
T
e. T
Z2, subgroup H = Z2. The number of cells Impossible.
g. T must divide
IF the order
of the
group, and
12 does
not divide 6.
Answers to Odd-Numbered Exercises
11
SECTION
1. Element
Order 1
(0,2)
2
(1,0)
2
(1,2)
2
(0,1)
4
(0,3)
4
(1,3)
4 4
The
is not
group
cyclic
7. 60 (1, 0)},{(0,0),
5. 9
2
9. {(0,0),
(0, 1)},
{(0, 0),
13.
(0, 1),(0, 2), (0, 3)} {(0, 0), (0,2), (1,0), (1,2)} 0), (1, 1), (0, 2), (1, 3)} {(0, Z20 x ^3, ^15 x ^4, ^12x Z5,
15.
12
11.
Order
Element
(0,0)
(1,1) 3.
497
(1,1)}
{(0, 0),
Zs
X Zj
X
Z4
17. 120
19. 180
21.
Z8,
23.
Z32,Z2
25.
Z2 x
Z4, Z2 x
x Z2
Z2
x Z8, Z2 x
Z2 x Z8, Z;xZ4x x Z2 x Z4, Z2 x Z2 x Z2 x Z2 x Z2 Z121, Z3 x Z3 x Z121, Z9 x Zn x Zn, Z3
x Z16, Z4
Z2
x Z2
Z9
x
b. i) 225
31. a.
It
of groups n)
is abelian
x Z3
x Zu x Zn
2
3
4
5
6
7
8
2
3
5
7
11
15
22
\302\253
number
Z4,
225 iii) 110 when the arrows on
both
rc-gons have
the same(clockwiseor counterclockwise)
direction.
b. Z2 x Z\342\200\236 c. When n is odd.
d. The dihedral 33.
35.
37. The numbers
SECTION
1.
group
D\342\200\236.
Z2 is an example. S3 is an example. are the
same.
41.
{\342\200\2241,1}
12
a. The only isometries of R leaving c fixed are the reflection a number c + x to c \342\200\224 x for all through c that carries x g R, and the identity map. b. The isometries of R2 that leave a point P fixed are the rotations about P through any angle 0 where 0 < 0 < 360\302\260 and the reflections across any axis that passes through P. c. The only isometries of R that carry a hne segment into itself are the reflection through the midpoint of the hne (see the answer to part (a)) and the identity segment map. d. The isometries of R2 that carry a line segment into itself are a rotation of 180\302\260 about the midpoint of the hne segment, the line segment, a reflection in the axis perpendicular to the line segment at its a reflection in the axis containing midpoint, and the identity map. e. The isometries of R3 that carry a hne segment into itself include rotations through any angle about an axis that contains across the line segment, and reflection the line segment, reflections across the plane any plane that contains to the hne segment at its midpoint. perpendicular
498
Answers to
X
P
X
X
P
W
w
P
p
px
/xy
w
M
V-Y
W
xp
xp
Y
W
w
xp
xp
Y
V-
7.
order
2
G/We reflection: order
Yes. The product 19. Yes. There is only toZ2. group of all a. No
31. a. 33. a.
one
b. b. b.
No
'
No
b.
No
b.
Yes
b.
Yes
39.
a. Yes.120\302\260 a. Yes.90\302\260, 180\302\260
41.
a. Yes. 120\302\260
b.
Yes
SECTION
1. 7. 11. 17.
do not
have
finite
isomorphic order
in the
180\302\260
c.
hence
all of
e. D^
d. No
Yes
c.
d.
e. D^
No
d.
No
e.
Yes
Z
c. No
Yes
c. No
No b.
37.
translations and glide reflections
because
identity)
c. No
No
Yes. 90\302\260, 180\302\260
35. a. Yes.
(and the
a group
isometries.
plane
25. 27. a. Yes 29. a. No
the identity and reflections. is a translation. translations is a translation and the inverse of a translation reflection /x across one particular Une L, and /x2 is the identity, so we have Only
of two
and rotations
reflections
Only
oo
13.
Rotations
17.
21.
2 or oo
n >
any
order
Reflection:
11.
\342\200\224I
order oo
Translation:
Rotation:
Exercises
Odd-Numbered
b.
c. No c. No c c. No
Yes
d.
No
1) and
(-1,
(0, 1) and
(1,1)
(1, 73)
13
Yes
3.
Yes
Yes
9.
Yes
13.
Yes Ker(0) =
19. Ker(0) =
0(25)
6Z;
0(20)
= (0,4,
21.
Ker(0)
23.
Ker(0) =
25.
2
33.
No
15.
Yes
7Z;
No
= 2 = (1,
2, 7)(4, 5, 6) 8,12,16,20}; 0(14)= (1,6)(4,
{(0,0)};0(4, 6) = 27. 2 29.
nontrivial
No
5.
homomorphism.
Z5 for a nontrivial 0. 5 does not divide 12.
But
7)
(2,18)
ForallgeG By Theorem 13.12,the image of cosets of a subgroup
the number
of 0 of a
would have
to be a
finite group is a divisor
subgroup of Z5, and of the order of the
group,
and
35.
Let
for (m, n) 4>{m, n) = (m, = p\342\200\236 for \302\253 g Z3, using
0)
37. Let0(h)
39.
Let
41.
Viewing
43. 51. 53.
Let The hk
4>{m, n)
499
Exercises
to Odd-Numbered
Answers
g Z2 x Z4.
our notation
text for elements of S3.
in the
= 2m.
let 4>(a) \342\200\224 (1, 2) for odd cr g Z>4 and 0(er) be the group of permutations, = (1,2) for odd a g S4 and 0(er) be the identity element for even a g S4. of
a g
for even
identity
D4.
0(er)
SECTION 14
1.
3
3.
9.
4
11.
21. a. When
4 3
working does
probably
b. We must 23. a. T
show
c. T
5. 2 7. 2 13. 4 15. 1 with a factor group G/H, you would let a and & be elements of G, not elements not understand what elements of G/H look like and can write nothing sensible Let aH and bH be two elements of G/H. that G/H is abelian.
29. (p0,/xj,{p0,/x2}, 35.
Example: Let
1.
3.
Z2
Z2 x Z
15.
Z(53 x
19.
a. J
5.
Z4
D4) =
13. Z(D4)=
{(p0,
p0),
c. F
21. {/gF*1/(0)=1} 23.
Yes.
Let f(x)
Thus /iT
25.
has
= 1 for order
31.
a. {e}
9.
13. 17.
/^}.
Then N
7.
Z
is normal
in G,
Z,xZ,
C = (po,
but H n
p2)}>
using
'e. F
the notations
= -1
for these
i. J
g. F
0 and /(x)
9.
Z3 x Z
x
N
=
H is
not normal
in G.
Z4
p2}
for
x < 0.
in Section
groups
Then /(x)
\342\226\240 =
f(x)
8, C =
1 for all
A3
x, so /2
x (p0, p2}.
g K*
but
/ is not
in K*.
F*/i<:*.
b. The
U
of complex
H =
whole
((1,0))
numbers of absolute to K
is isomorphic
value
1
= ((0,2)), but
is isomorphic
G/H
group
= {mi,m2,di,d2,C},Xn = {C}, ={C},Xn = {s2, s4,ml.m2, mi,m2, C, ^,,7\302\276}. X\342\200\236, C, P2, P4), = (2, 4, d,, d2, C], Xs, = (1, 3, d\\,d2, C}. Xh j2, i3, s4], [mum2}, [du d2), {C},[Pu P2, P3, P4} {1, 2, 3, 4}, {\302\242,, A transitive G-set has just one orbit. a. (J,, j2, S3, s4} and (P,, P2, P3, P4} b. The set of points on the circle with center at the origin and passing through c. The cyclic subgroup {2n) of G = R a. K = g0Hg0l. H and K should be conjugate of G. b. Conjecture: subgroups XP0 Xw
7.
(p0,
16
SECTION
3.
=
U
The multiplicative group Let G = Z2 x Z4. Then isomorphic to Z2 x Z2.
1.
(/)o,
x >
2 in
29.
27.
and let #
The student them.
15
SECTION
11.
G=
concerning
i. T
g. T
T
e. and{p0,/x3} N = S3,
of G/H.
=
=
X,XP1 {ji,j3,
P
to
Z4
while
G/K
is
Answers to Odd-Numbered Exercises
500
19.
X
a
a
a
There
c
0 a a 1 a b 2 a a
b
a
b
c
a
b
c
a
b
c
a
b
3
a
b
a
a
b
c
4
a
a
b
b
c
a
5
a
b
a
c
a
b
are four of
them: X,
Y,
Z, and ;
17
SECTION
5.
3. 2
1. 5
7. a. 45 9.
b
Z b
Y
11,712
b. 231 b. 6,246
a. 90
SECTION 18 5. (1,6) not a field
1
3.
1.0
7.
Commutative
ring,
9.
Commutative
ring
with
unity,
not a field
ring
with
unity,\"
not
11. Commutative 13. No. [ri \\r 15.
17.
21. 23. 25.
e K}
no unity,
is not closedunder
(1,1),(1,-1),(-1.1),(-1,-1) All nonzero q e Q Let R = Z with unity
0(1)= (1,O)^1'.
a field multiplication.
19. 1,3 1 and
R' = Z x Z
with
1' =
unity
(1, 1). Let,
R
= : Z -\302\273 Z where 0t(\") = \302\253 \302\2421(/1) 0, \302\2422 where ^(/i,m)=0,^:ZxZ-*Z where 02(\", m) = n = , x Z \342\200\224\342\226\272 Z where 03(\302\253, m m) - /3)(X + /3) of two matrices 27. The reasoning is not correct since a product (X either matrix be 0. Counterexample:
-\302\273 R'
(n, 0).
Then
xZ-^Z
0 1] 0 1 0 0
1
31.
a =
33.
a. T
2, b
=
may be the
/3.
0_
= 3 in
Z6
e. T
c. F
g.
T
i. r
SECTION 19 1.
by
, -\302\273 Z where
0
11.
be denned
0,3,5,8,9, 11 a4
+ 2a2b2
17. a. F
+ b4
c.
F
3.
a6
e. J
7. 0
5. 0
No solutions
13.
+ 2a3b3+b6
2. F
i. F
9. 12
zero matrix
0 without
having
19. 1. Det(A)
2. The column
= 0.
3.
The row
5.
A
vectors of A
Exercises
501
are dependent.
of A
vectors
4. Zero is an
are dependent.
Odd-Numbered
to
Answers
of A.
eigenvalue
invertible.
is not
SECTION 20
1. 7.
3
3.
or 5 =
\302\242,(1) 1 = \302\2425(2) 1
(p(j)
=
6
\302\2425(15)
= 2
\302\2425(10)
=
4
\302\2425(16)
=
\302\2425(11)=10
= 2
= 4
\302\2425(6)
16 +
19. 1
21. 9
65Z, 29 +
c.
23. a. F
=
10
\302\2425(28)
= 22
\302\2425(29)
8
\302\2425(22)
\302\2425(17)=16
\302\2425(23)
= 6
\302\2425(24)
=
No
12
=
18
=
12
= 28
=
8
13.
=
\302\2425(30)
8
solutions
e.
65Z
55 +
g. F
T
\\.
F
to the
ring D of all
that can be expressedas a quotient
numbers
rational
of integers
with denominator
of 2.
when \\#e try to prove the transitive property in the proof of Lemma5.4.2,for multipUcative cancellation trouble ~ (2,4) since (1)(4)= (2)(2) = 4 and (2, 4) ~ (2, 1) since may not hold. For R = Z5 and T = (1,2,4}we have (1, 2) to (2, 1) because (1)(1) ^ (2)(2)in Z5. (2)(1) = (4)(2)in Z5. However, (1, 2) is not equivalent
22
7.
16 0,
1, 2,
21. 0, x
+ 5,
f(x)g(x) = 6x2+Ax+6 + 1, f(x)g(x) = x3 + 5x 2 11. 0 9.
+ 5x
7
\342\200\224 2x \342\200\224 x2 \342\200\224 x2 \342\200\224 x4 \342\200\224 5x3.
5,
25.
a. They
27.
b. F
10,
c.
are the
25,
units
(Other
5x,
g. T
e. F
J of D.
c. F[x]
b. 1,-1
31. a. 4,
27
SECTION 23
1. q(x) = x4+x3+x2+x-2, r(x)
3. ^(x) =
5. 2,3
15. 0,2,4
13. 2,3
3
23. a. T
11.
\302\2425(27)
=
65Z, 42 + 65Z,
T
1. f(x) + g(x) = 2x2 3. f{x)+ g(x)= 5x2
9.
12
\302\2425(21)
= 20
It runs into
SECTION
5.
\302\2425(26)
=
\302\2425(20)
ki + qii I \302\2421,426 Q} It is isomorphic
somepower
17.
8
21
SECTION
17.
\302\2425(25)
=
No solutions
17. 3 + 65Z,
i.
18
8
11. 1+4Z, 3+4Z
2
=
=
\302\2425(18)
\302\2425(12)
(p-1)(4-1)
5.
14.
\302\2425(19)
= \302\2425(14) 6
\302\2425(9)
\302\2425(5) 4
15.
= \302\2425(13) 12
= 2
\302\2425(4)
15.
6,7, 10, 11, 12,or
= 6
= \302\2425(8) 4
\302\2425(3)
9.
of 3, 5,
Any
= 4x
+3
6x4+7x3+ 2x2-x+2,r(x)=4 7.
3,10,5,11,14,7,12,6
+ l)(x-2)(x+2) (x-l)(x (x 3)(x + 3)(2x+ 3)
answers are possible.)
i. T c. 1,2,
b. Z2 x
3, 4, 5, 6
Z2, Z3
x Z3
x Z3
13.
It is of
Yes.
degree 3 with
Exercises
to Odd-Numbered
Answers
502
no zeros
in Z5.
2x3+x2+2x+2
15. Partial 19. Yes. p 25. a. T
irreducible over R, but 21. Yes. p = 5 e. T g. T
g{x) is
answer:
= 3 T
c.
27. x2+x+ l
29.
x2
2, 2x2 +
+ l,x2+x+2,x2+2x+
31. p{p-If/2
9. 11.
not irreducible over C.
IT
2, 2x2
+ x
+ 1,2x2
+ 2x + 1
24
SECTION
1.
it is
3. 2e + 2a + 2b 5. y le + 0a + 3b e R, ^ ^ 0} R*, that is, {ax + Oi + Oy + 0\302\243 ]^ e. F \\. a. F c. F g. T = {0}. e. 0 e End(A) c. If |A| = 1, thenEnd(A)
19. a. K b.
=
0
-i_ by S
I, we must
the matrix
(3/50)\302\243
T is not
in Iso(A).
with
b and
coefficient
C the
by
matrix
with
c and the
coefficient
that
check
B2 =
CK = c. We
-
(l/50)y
0'
\\i
Denoting
7.
should checkthat
6 is
-I, C2 =
-I,
K2
B,KB = C, CB= -K, KC
=
= -B,
-I,
and BK
= -C.
one to one.
SECTION25 1.
x <
a <
3. m 5. i.
x2
< x3
is positive + \302\253V2
acedb
if m
> 0 and
li. ecbad
< 0, \302\253
e R.
or
if
m >
0 and m2 >
ii. ecbad
i. caedb
13. debac
11. dbaec
15. a.
any a
\342\200\242 \342\200\242 \342\226\240 \342\226\240 \342\226\240 < x\" \342\226\240 for
li.dceab
7. i. dabce 9.
<
IF
g. T
e. T
F
c.
T
SECTION26 1.
There are 0(1,0) \302\242(1, 0) \302\242(1, 0)
0) \302\242(1,
3.
(0) =
nine
possibilities: \302\242(0,1)
= (0, 0)
or (0,1),
= (0, 0) or (1, 0), \302\242(0, 1) = (0, 0), and \302\242(0, 1) = (0, 0), (1, 0), (0, 1),or (1, 1). \302\242(0, 1)
{0},Z12/{0}^Z12
(1)= (0, (2)
just
= (1, 0) while = (0,1) while = (1, 1) while = (0, 0) while 1, 2,
= (0, 2,4,
~
3, 4, 5, 6, 7, 8, 9, 10,11}, Z12/(l)
6, 8,10},Z12/(2)~ Z2
{0}
(3)={0,3,6,9},Z12/(3)~Z3
<4) =
9.
<6) =
{0,4,8},Zl2/<4)~Z4
{0,6},Z12/(6)~Z6
het\302\242: Z-\302\273Zx
Z be given
by 0(h)
=
e Z. (\302\253,0)for\302\253
2n2,
< 0 or if \302\253
and 2n2
2 x 2 identity
matrix
by
to Odd-Numbered
Answers
11.
are not of real interest
and R/[Q}
R/R
because
R/R
is the ring containing
only
the zero
Exercises
503
element, and R/[G} is isomorphic
toi?. to Z4, which has a divisor 2 of 0. an integral domain. Z/4Z is isomorphic {(n,n)\\n e Z}. (Other answers are possible.) of Z32 is of Z12 is {0, 6}. The nilradical of Z is {0} and the nilradical 31. The nilradical (0,2,4,6,8,---,30}. 35. a. Let i? = Z and let N = 4Z. Then -Jn = 2Z ^ 4Z b. Let i? = Z and let N = 2Z. Then ^N = N.
13.
Z is
15.
SECTION 27
1. (0, 2,4} and (0, 3} are both prime and maximal. 3. {(0,0), (1,0)}and {(0, 0), (0, 1)} are both prime and maximal. 15. 2ZxZ 17. 4Z 5. 1 9. 1,4 7. 2 with p = 2. 19. Yes. x2 \342\200\224 6x + 6 is irreducible over Q by Eisenstein 21. Yes.Z2 23.
x {0}
x Z3
No. Enlarging Zp and
domain
the
to a
you would
of quotients,
field
have
to have a
field containing two different
is impossible.
which 7Lq,
SECTION 28
1. -3x3+ 7x2y2z -
-5x2yz3+2xy3z5
+ 3y + 10z3 - 3x3 7. 10z3 - 2z2y2x + 2z2yx2 + 3y-lx 9. 1 < z < y < x < z2 < yz -< y2 < xz 9 9 9 3 \342\226\240 \342\226\240 < x z < x y < x < \342\226\240 xy 3.
2x2yz2
5.
2z5;y3x
11.
-
-
3yV
15.
()>5+
19.
{1}
-7x
2xy2z2
5z3yx2 +
lzy2x2
+ 5/z3
8z7
algebraic
(x + y,
-
y3
-
3yz3 (y2z3
< z3 <
8xy
< y2z
yz2
<
- 4xz+ 2yz + 38 - 2z,y2z2+ 3)
+ 3,-3y
+ 18}
variety is {(1, 3), (\342\200\224-, 6)}.
The algebraic 27.
17.
< x2
{x-1}
23. {2x+y-5,y2-9y 25.
13.
;y3,;y3+z,x-;y4) 21.
The
- 4x
< xy
y + 1} variety
of one
consists
e. T
c. J
a. T
where point (a, \342\200\224a) g.
T
a
**>
1.3247.
IF
SECTION 29
1. 5.
x2 x12
- 1
3. x2 - 2x + + 3x4 + 12x2+ 5 2 , 62 , = x4 --x2 ;deg(a,
- 2x
7.
Irr(a,
9.
Algebraic,
Q)
F) =
deg(a,
11.
Transcendental
13.
Algebraic,
deg(a, F)
Algebraic,
deg(a,
15.
17. x2+x
23.
a.
2
+ 3x8-4x5
T
25. b. x3 + 27. It is the
Q) = 4
2
=2
F) = 1
+ 1 = (x-a)(x + 1+a) c. J
x2 + 1
= (x
e. F
- a)(x -
monic polynomial
in
g. F -
a2)[x
F[x]
of
IF
a + a2)] minimal degree having (1 +
a as a
zero.
y3
< xz2
< xyz
<
prime
fields
Answers to Odd-Numbered Exercises
504
30
SECTION
1.
3.
5.
((0, 1), (1, 0)},((1,1),(-1, 1)}, ((2, 1), (1, 2)}. (Other answers = (0,0,0) No. 2(-1, 1,2)-4(2,-3,1) + (10,-14,0) 7.
(1}
9.
(1,72,72,(7\302\2763} IT g. F of V generated by S is the intersection of all A basis for F\" is
(1,(}
c. T
T
15. a.
17. a. The
subspace
19.
answer:
Partial
are possible.)
F
e.
subspaces
of V containing
((1,0,---,0),(0,1,---,0),---,(0,0,---,1)}
where 1 is
25. a.
A
the
identity
multiplicative
of F.
homomorphism
b. Partial
c. 0 is an
answer: The kernel (or nullspace)of 0 is {a e V | 0(a) = 0}. of V with V if Ker(0) = (0}and 0 maps V onto V.
isomorphism
SECTION31
3. 4,(1,73,72,76}
1. 2,(1,72}
5.
6,
(1,72,
72,
9.
9,
(1, 72,
74, 73,
a.
Partial
2\"
IF
g. F for n
e Z+ are
obtained.
32
SECTION
All odd-numbered answers
proofs
require
and are not
here.
listed
33
SECTION
1.
of degree
Extensions
answer:
79, 7l8,736}
e. F
c. F
F
19.
23.
76, 7l2,
13.2,(1,72}
11. 2,(1,72}
7. 2,(1,76}
(72)2, 72(72)2}
72(72),
Yes
3.
Yes
5.
6
7. 0
SECTION34 1.
a. K=
(0,3,6,9}.
b. 0 + ^
3.
5.
= {0,3,6,9),1 + =
^ 10}, 2 + K = (2,5,8,11}. (1,4,7, = + + 2, K) K)=1. /x(0 /x(2 a. HN = (0,2, 4, 6, 8, 10, 12, 14, 16, 18,20, 22}, H n TV = (0, 12}. b. 0 + N = (0, 6, 12, 18},2 + TV = (2, 8,14,20}, 4 + N = (4, 10,16, 22}. c. 0 + (tf n/\\0 = {0,12},4 + (tf nA0= (4,16},8+ (tf nA0 = (8,20}. d. 0(0 + A0 = 0 + (H n N), 0(2 + A0 = 8 + (H n A0, 0(4 + A0 = 4 + (tf a. 0 + tf = (0, 4, 8,12,16, 20},1 + tf = {1, 5, 9,13,17, 21}, 2 + H = {2,6,10,14, 18, 22}, 3 + H = (3,7, 11,15,19,23}. b. 0 + A\" = {0, 8, 16}, 1 + K = (1,9, 17}, 2 + K ={2, 10,18},
c.
+ K)
3
+ i<:=
A
+ K
7 +
=
^ =
= 0, /x(l
(3, 11,19}, {A,
12, 20}, 5
(7,15,23}.
+ K
= (5,13,21},6 +
A\"
=
(6,14,22},
n TV).
S.
Answers to c. 0 +
K={0, 8,
16}, 4
20}.
+ K={4,12,
= (0
505
Exercises
Odd-Numbered
4 + K] = ((0,8, 16}, 4, 12, 20}} (1 + ^) + (H/K)={l + K,5+ K} = ((1, 9, 17}, (5,13, 21}} (2 + ^) + (H/K)= {2 + K,6 + K] = ((2,10,18},(6,14,22}} + K}= ((3,11, 19},(7, 15, 23}}. (3 + ^) + (H/K) = {3 + K,l + H) = (0 + K) + (H/K),\302\242(1 + H) = (1 + K) + (H/K), \302\242(0 = + H) = (3 + K) + (H/K). + + +H) (2 K) \302\242(2 (H/K), \302\242(3
d.
e.
= H/K
+ (H/K)
+ K)
(0
+ K,
35
SECTION
250Z < 10Z < Z of
(0} <
refinements
The
(0}
< 10Z
< Z and
(0}
< 250Z
< 25Z < Z of
0 <
25Z < Z are
isomorphic.
given series refinements
The The
are isomorphic.
(4800Z) x Z < (240Z) x Z < (60Z) x Z < (10Z) x Z < x (4800Z) < Z x (480Z) < Z x (80Z) < Z x (20Z) <
((0, 0)} <
((0,0)} <
Z
Z x Z Z
of the
x Z of
first series and the second seriesare isomorphic
refinements.
(16) < (8) < (4) < (2) < Z48 < (8) < (4) < (2) < Z48 (0} < (24) < (12) < (4) < (2) < Z48 (0} < (24) < (12)< (6) < (2) < Z48 (0} < (24) < (12) < (6) < (3) < Z48 (0} <
(0} < (24)
9.
< A3
x (0} <
x (0)
((po,
((p0, 0)} <
i. The
19.
21.
x Z2
(po}
isomorphic to Chain
Z2
53
A-j
x Z2
theorem
Jordan-Haider
(po} < (po, P2}
Yes.
<
applied to the
Pi, P2, Pi] and are thus abelian. < [Po,
Chain
(3)
(12) <
(0} <
(12) <
5
(12)< (12)
(0}
(12) < (4)
< (2) < Z24
groups
< D4
is a
\342\200\242 < \342\226\240 \342\226\240.
Z\342\200\236 implies
composition
the Fundamental (actually
Theorem of Arithmetic.
a principal)
series and
all
factor
groups
are
(4) (12) <
<
< (6)
(12) < (6) < (3)
< (6)
< (3) < : J24
< : '24
Z4
2z
^24
Isomorphisms
~
(12)/(0}
~
(12)/(12)
(12)/(0}~ Z2, ~ (0}. (3)/(3)
(12)/(12)~ (6)/(6)~ (0}, ~Z2,
(2)/(4)-(6)/(12)
Z24/Z24
SECTION
1.
3
5.
TheSylow
etc.
~
(6)/(6) ~ (0}. ~ ~ (12)/(12) (0}, (12)/(12)
~ Z24/Z24
(4)/(12) Z24/(2)
~
~
~
(12)/(12)
Z24/(3)~ (3)/(6)
~
Z3 Z2
(0}
36
3. 1,3 3-subgroups
are ((1,
2, 3)), ((1,2,
4)).
((1, 3,
4)),
and
((2,3,4)).
Also
(3,4)((1,2,3))(3,4)
= ((1,2,4)),
Answersto Odd-Numbered Exercises
506
37
SECTION
conjugate classesare (po}> (P2}> (Pi, P.i}, {Mi,/^2},{^i, 52}. + 2 + 2+ 2 3. a. T c. F e. T \\. F g. T e. This is somewhat a matter of opinion. 9. 24 =1+6 + 3 + 8 + 6 1.
a. The
b. 8
= 2
SECTION38
1. ((1,1,1),(1,2,1),(1,1,2)}
3.
No.
7.
2Z <
1) +
m(4, 1) can never
rank
r = 1
n(2, Z,
yield
number for
an odd
first
coordinate.
SECTION 39
1. a. a26Vc36~2, 3. a. 16 5.
c.
Partial
answer:
13. c.
A
group on
SECTION
blop
36
c. 18
b. 36
a. 16
11. a.
ac~6a~4^3a
b. a-'iWa\"1,
62c~3a~36-2a-2
b. 36
{1} is a basis for
S is isomorphic
c. Yes
Z4. to the
free group
F[S] on
5.
40
1. (a : a4 = \\);(a,b: a4 3. Octicgroup:
=
\\,b =
: a \342\200\224 1, b4 = 1, c = 1).(Other
a2); (a,b,c
1
a
a2
a3
b
ab
a2b
a3b
1
1
a
a2
a3
b
ab
a2b
a3b
a
a
a2
a3
1
ab
a2b
a3b
b
a2
a2
a3
1
a
a2b
a3b
b
ab
a3
a3
1
a
a2
a3b
b
ab
a2b
b
b
a3b
a2b
ab
1
a3
a-1
ab
ab
b
a3b
a2b
a
1
a 1
a2b
a2b
ab
b
a'b
a'1
a3b
a3b
a2b
ab
b
a3
a 1 a~
Quaternion
group: The
sameas the
table
a3
1
a3
a
1
for the
octic group
except
that the
answers
16 entries
are possible.)
in the
lower right corner are
Answers to Odd-Numbered Exercises
5.
a2
a
1
a3
a3
a2
a
1
1
a3
a2
a
a
1
a3
a2
la1 =
Z2l.(a,b
507
= 1,\302\2763
= a2b)
l,ba
SECTION 41
1. a. 2P{P3 No
b.
+ AP6 c. Yes
- 3P,P,
3. Ci(P)=
Z,(P)= B,(P)
5.
Z,(X)
is
generated
7.
> 0.B0(P)
= Ofon
Hi(P)
and
= 0.
Since
P2.
oriented n-simplexis an ordered sequance \342\226\240 \342\226\240 is given by of Pl P2 \342\226\240 P\342\200\236+1
a. An
4P3P6-
> 0. B0(X)~
= Ofon = B,(X) = H,(X) generated by the two 0-cycles P{ by the coset P, + B0(X).
=
C,(X) and
=
~ 5P3P4 +
+ 3P2P4
3P2P3
Z0(P)
Z and
5P4P6
~
Zand
is generated
Z0(X)/B0(X)
is generated
by the
\"indentifies
by the
-
0-chain P2 P{ with
~ Z.
0-cycle P. H0(P) Pi-Z0(X)
P2,\"
H0(X)
~ Zx Z is
~ Z and
\342\226\240 \342\226\240 P\342\200\236+i.
PtP2-
The boundary
b.
n+l \342\226\240 \342\226\240 \342\226\240
UPlPl
ft+l)
\342\226\240 \342\226\240 \342\226\240
(-V'+lPlP2
Y,
Pi-lPl+l
\342\226\240 \342\226\240 \342\226\240 P\342\200\236+l-
/=1
c. Each
a. S<'0 J J2m>a-
13.
H{\"\\X) = Z{n\\X)/B{n\\X) ~ Z and is Hi0\\S) generated
by (Pi
H^XS) = 0 ~
H{2\\S)
Z and
oriented
rc-simplex
is a face of the
simplex.
= ^mi&l\"Xoi)
11.
)
of an
of the boundary
summand
individual
is generated
+P2 +
+ P4)
P3
+ {0}
+ B(2\\S)
by PiP2P3
SECTION42
1. 3. 5. 7. 9.
11.
1.
~
H0(X)
Both
~
= 0 x Z. H\342\200\236(X)
~ Z. tf2(X)
Z. tf2(X)
e. r
-
Z x Z
=
H\342\200\236(X)0
~ Z. tf\342\200\236(X) = 0
~ Z x Z x Z. H2(X) ~
> 1.
forrc
Z.
g.~r
x Z x Z. H2(X)
Z x ~
show that X(X)
counts
i.
> 2. \302\253
2.
F
Z.
= tf\342\200\236(X)0for\302\253 >
Z.
tf\342\200\236(X)0
=
for
2.
\302\253 > 2.
= 1.
hold for a square region, for such a region onto 2-cells,for each can be mapped continuously It
for
for\302\253 >
43
will
5. H0(X) ~
7.
Z. H,(X) ~ Z Z x Z. ^(X)
tf0(X) ~ Z. tf,(X) c. F a. J ~ Z. H0(X) tf,(X) ~ tf0(X) Z. tf,(X)
SECTION
3.
~
H0(X)
2-2\302\253
Z
x Z.
H^X)
~ Zx Zx
Z2
is homeomorphic
It obviously does not a map has no fixed points.
to E2.
the other, and such = Oforrc > 1. x Z2. H\342\200\236(X)
hold
for two
disjoint
9.
~
H0(X)
Exercises
to Odd-Numbered
Answers
508
\342\226\240\342\226\240\342\226\240 = Oforrc xZxZ2. tf\342\200\236(X)
Z. fli(I)~ZxZx
(g-l)
> 1.
factors
11.
be a vertex
Let g
c
2-chain
2-simplexes of X,
of all
consisting
the same way,
all oriented
so that
\342\202\254 Z2(X).
a. /,0 /,i
is given
by /,0(g
is given
by
/,2 is given
b. /,0is /,
Let g /,i
= g
+ B0(6).
+ B^X))
=
+ B^X))
= 2nb
nb
+ Bi{b).
= 0.
+ Bi(X))
as in (a).
+
by ftl((ma
nb)
+ B^b).
as in (a).
be a
vertex
is given is given
/,o
+ S0(X))
/^((ma+nb)
by /,2(c
is given
i
/,2 is 13.
be the
let c
of &, and
/,2 is trivial,
on
&.
+ B0(X)) = Q + B0(b). + rc&) + Bi(X)) = rc& + Bi(b), by /,i((ma since both H2(X) and H2(b) are 0. /,0(g
by
where m = 0,
1.
SECTION44 5.
For Theorem
44.4, the
condition
fk_i
=
dk
that
implies
d'kfk
/W(BW(A))CBW(A').
Then Exercise14.39shows that fk^l induces a natural This is the correct way to view Theorem 44.4. For Theorem 44.7,ifwe use Exercise 14.39,the fact that
3*: (Ak/A'k) 7.
-+
sequence
is
[H2(a) =
0] -^
~ Z ~
-\302\276.[H0(X)
c + S2(X)of H2(X)
a generator
Z]
of #2(X,
a) and is an
9,2 maps i',i maps
everything
(image
9,i maps
11.
3,2)=
the generator
a + Bi(a) onto
~
x Z] -\302\276[tfj(X,
dk
(ma +
a natural
induces
-\302\276. [H^a)
Z]
a)
a) =
-\302\276[ff0(X,
Bk_x(A').
homomorphism
~
Z]
onto
+ Bt(X) rc\302\243)
+ B{(X, a)
a, i',0
g + S0(X)onto
maps
(image
(a + 0b)
S0(X,
a) in
B0(a)
[H2(Y) =
0] -\302\276[H2{X) -\302\276[tfi(X) -\302\276[ff0(X)
Z]
(image
f,2)
= 0.
Z. y',2) \342\200\224
+ Bi(X),
onto
g +
ff0(T. a), so (kernal
The answer is formally with that in identical Partial answer: The exact homology sequence
~
0].
so ;',! is
an
Exercise
S0(X), 7,0)
so
is
if0
= (image
into,
isomorphism
Ci(a)) + Bi{X, a), so (kernal jtl) = (image Z. 0, so (kernal 9,0 = (image 7,.1) \342\200\224
g +
Z]
-\302\276[tf0(a)
+ (\302\2536 onto
~
C2(a))+ B2(X,a)
(kernel 7,2) =
Thus isomorphism. onto 0, so (kernal
+ Ci(a)) (\302\2536 a vertex g of d^) = 0.
7,0 maps 9.
that
shows
9,2) = 0.
/\302\273i maps
(image
A'k_^
Zk_i(A')/
the generator
onto
(c +
For
c
[H2(X, a)
~Z]^>
[H2{X)
-\302\276[Hx(X)
maps
dk(A'k)
into
Zk\342\200\236i(A)/Bk~i(A)
(A^/A'm).
The exact homology
y',2
of
homomorphism
an
i',0 ~ Z.
isomorphism,
Z. i',0) \342\200\224
44.7.
is = 0] ~ ~
~
Z]
~
Z] -^-
-\302\276[H2{X,
Y)
Z]
-\302\276[#i(X,
7)
Z]
-\302\276[ff0(X,
F) =
0].
-\302\276[#i(7)
and (kernal i4l) =
~ Z
x Z]
[flo(y) ~ Z x Z]
and (kernal i',0) =
Answers to Odd-Numbered Exercises
The
1.
Yes
9.
In Z[x]
: only
InQ[x]
:4x-
P\\Q\\ of Fig. 42.11gives
the edge
that
Q, a
17.
2ax2
a. T
of H \\(X,
Y).
13. 198,-198
^ 0.
\342\200\224 3ax
+ 6a
is primitive
for all
c. T
of Q
0 in
Z7 because
g. F
IF
a ^
T
e.
element
nonzero
is a unit.
Indeed
18ax2
every such element a
i.
31.
to a generator
5. No. 7. Yes -2x + 7 7 14,x- -, 6* - 21,-8*+28 - 7, lOx - 2, 6x + 1, 3x - 5, 5x - 1
\"primitive\" because every
is akeady
21. 23.
rise
interesting.
- 7,
2x
11. 26,-26 a g
Note
to you.
very
No
3.
In Z\342\200\236 [x] : 2x
It
are
these maps 3\342\200\2362>
with
45
SECTION
15.
is left
of exactness
verification
Starting
509
must appear in every factorization into Either p or one of its associates in Q[x] but not in Z[x]. 2x + 4 is irreducible Partial answer: x3 - y3 = (x - y)(x2 + xy + y2)
\342\200\224 12ax
is a unit
+ 48a
is primitive
for all
in Z7.
irreducibles.
SECTION 46
1.
Yes
7.
61
No. (1)
3.
is violated.
9. x3+2x-l c. J e.
13.
a. T
23.
Partial
answer: The equation
J ax
Yes
66
i.
g. J
= b has a
solution in
T
for nonzero Z\342\200\236
a, b
if and \342\202\254 Z\342\200\236
only
*
b.
divides
n'mZ
5.
11.
47
SECTION
3. 4 + 3(=(1+20(2-0 5 = (1+ 20(1-20 7. 7 5. 6 = (2)(3)= (-1 + V=5)(-l-V=5) 15. 3 order c. i) order 9, characteristic characteristic 2, fi) 5 iii) order 5, characteristic 1.
3.
1. V2,-V2
7. -JI + 72, -VT -V2
15.
a. Q
17. Q(V2, 25. a. 3 27.
+ 3V5 V3,
5.
+ V2,3-V2
-
\\/l
V2,
19.
V5)
b. They
V2
--v/l
1,
= 2, ff3(a)
ct3(2)
ff3(l o3(2
+
c. T
e.
F
=
-a,
2a, o3(2 g. T
;, V2
9.
V3
21. Q
Q(V3,7l0) are the same maps.
= 1 - a, 0-3(1 + 2a) = 1 \342\200\224 2a; Z3(a)|tj31 = Z3 2a) = 2
V2 +
- V2
13. -V2 + V45 c. Q b. Q(V6)
\342\200\224
+ a)
Yes
3
+ y/1,
0-3(0)= 0, 03Q)=
29. a. F
37.
2
48
SECTION
11.
i
ff3(2a)
+ a)
=
-2a,
= 2
IT
- a,
-
f,
-V2
+;,
-72
- ;
if
the positive
gcd of a
and
Exercises
to Odd-Numbered
Answers
510
SECTION49 1.
3.
5.
The identity map of E onto E; = VI r(V3) x given by r(V2)
t,
given
by
t2
given
by
t3
given
by
t4
given
by
The
identity given
t2
given
by
t3
given
by
t4
given
by
t5
given
by
7. a.
Q(7T2)
15.
3. 4
2 Let
-V5
VI
F = Q and
7. 1
5. 2 Q(V2).Then
E =
has a zero
but \302\243,
in
does not
split in
13.
9. 2
- 5x2 + 6 =
= x4
fix)
23.
=
r(V5)
50
SECTION
1.
by
X\\
= -VI
= VI r, (V5) = xx (V2) = VI tj (V3) = -VI r2(V2) = VI r2(V3) r2(V5) = VI = VI r3(V2) = -VI r3(V3) r3(V5) = VI = -VI r4(V2) = -VI r4(V3) r4(V5) = -V5 of into Q(VI VI itself; map = ori, Ti(V3) = \342\200\224V3 where ai = VI ^(\302\242^) = V5 where a2 = V2(-l r2(ai) = a2, r2(V3) +i'V3)/2; r3(ori) = ct2, t3(V3) = -VI - iV3)/2; = or3, r4(V3) = V5 where or3 = V2(-l r4(or1) = = or3, -VI r3(ori) r5(V3) b. Ti given by T\\(y/7t) = i^Jn, t2 given by x2(s/7t) = \342\200\224iy/n
(x2
1<[\302\243:F]<\302\253!
-
2)(x2
- 3)
E.
a. 6
SECTION 51
1. a = V2 3. a = V2 7. 15.
= (V2>\\ V2
=
(-)a3 - (-)a,V3
=
= 2/(V2~V2).V2 + VI
/(x) = x4 zerosof multiplicty
V2 =
\342\200\224 Ax2
b. The field
+ 4
= (x2
\342\200\224Here
(\342\200\224)a
/(x)
2)2.
answers
(Other
(V2)2.
is
not
are possible.)
(Other answers
(-)a3.
an irreducible
are possible.)
polynomial.
Every
irreducible
1 only.
c. F[xp]
F
SECTION 52
1.
Z3(y\\z9)
5.
a. F
SECTION
3.
e. F
g. T
IT
53
1. 8 9. The group 11.
Z3(y\\z2)
c. F
3. has
a. Letai=V2.
8
5.
4
two elements, the
7. identity
+ a2 = ^2~1 i^,
2 automorphism and
= \302\2533
The maps
are
Po, where
po is the
pu where
( of
Q(0 and a
that
~'^.
^-1
identity
pi(a,) = a2
such
and
map; piO'VI
= ;'VI
cr(i) =
\342\200\224i.
factor
of /(x)
has
Answers to
p2(ai)
= a3
[i\\, where/xi(a!)
= ai
where /\302\276.
/x2, where /x2(ai) /x3, where /x3(ai) b.
5.3. The
in (a) was
notation
{Po>
to coincide
chosen
with
Pb Pi)
and
p2(( -/3)
=
* V3;
=
and/xi(i'V3)
= a3 and
/^ (\302\243-/3)= and /x3(\302\243a/3)= 0\302\276
=
the notation
^2)
{Po<
for S3 in
iPo<
511
Exercises
Odd-Numbered
\342\200\224
i'V3;
\342\200\224\302\243-/3; \342\200\224 \302\243
-/3-
Example 8.7.
^3i
~{p0]
diagram
Group
K = K,
=
=
0(^)
'('\342\226\240^3) ^,p\342\200\236p2)
=
^,}
v^
\302\2532)
=
Q(\302\2533)
^p0.fl3!
Field diagram 13.
The splitting
identity map of Q(\302\243-/3), and 15.
a. F
25.
Partifl/
V L))
G(K/(E
Q(\302\243-/3), and the \342\200\224 i
er(\302\243-/3)=
er, where
e. J
c. J
a/wwer:
is
of (x3 \342\200\224 1) g Q[x]
field
g.
F
group is cyclic of
-/3. i. F
= G(K/E) n G(K/L)
SECTION54 3.
-\302\2768 + 4x5 \302\243,
+
-^2 \302\243):
Q(^2,
Q(^2):-/2,
-
x4
+ 2x4 +
28x2+
1;
2;
Q(\302\243^2)):\302\243(^2),;c4-2;
Q(V2,
0: V2
+
- 2x2 +
x4 \302\243,
9;
i+i()2)):)2 + ;(^),i4 + 8; I-1(^2)):
Q(V2): -/2, x2-
^2-1(^2),
x4 +
S;
2;
+ 1; Q(\302\243):\302\243,x2
Q((V2):(V2,x2+2;
5.
Q: l,xThe
group
1 is cyclic
of order
5, and
its elements are
oi ^2-
^2
where -/2 is the
?(^) real
5th root
K\\y%
of 2.
o-i
K\\y%
04
f4(^2)
order
2 with
elements: (, where
(
is the
512
7.
Answers to
The
9.
a.
\342\200\224 1 over Q is the same as the splitting 54.7. (This is the easiest way to answer the
of x8
field
splitting
in Example
contained
Exercises
Odd-Numbered
of x4 + 1
over (
so a completedescription
problem.)
MS*
b.
^-2\302\276
field
S3
55
SECTION
b. 400
16
3.
2160
5.
7.
over Z2 is x2
$3(x)
$8(x) over 9. a. T
11.
Z3 is x4
c. F
= x-
$1(x)
$2(x) = =
$3(x)
X X2
+ 1. + 1 = (x2 + + x
t.
x + 2)(x2
+ 2x
+ 2).
i. F
g- r
T
1 1
+
+X
+ 1
= X2 + 1 $4\302\253 = X4 + X3
$5(x) =
x2
+
+ X + 1
X2
- X+ 1
SECTION56 1.
No. Yes,
3.
a. r
K is an
of la
extension
by
e. r
c r
radicals.
i.
g. r
APPENDIX
'2
-3+2(-
1'
2 7 5
16
0
-18
\342\200\224 1 \342\200\224 4i\"
2
-f
0
-;\342\226\240
24
8
\342\200\224 8i-
8f
8
-i
1
-3 4
11.
0
1
-
-1 0
6f
-2
-
2f
13. -48
F
(x3
\342\200\224 2x over
Q gives a counterexample.)
is
Index
Abel,
Niels
Abelian
39, 174, 324,471
Henrik,
455
extension,
39
Abeliangroup(s),
free, 334
for
of,
finitely-generated, 108 torsion free, 113, 142
Absolute value,
faithfull, 155 on a group, 154 transitive, 86, 155
Aschbacher, Michael,
of a field,
modulo
Al-Banna, Abu-1-'Abbas Al-Tuse Sharaf al-Din,
288
254,
380
closure, 287, 288 closure of F in E, 286
Algebraic element over F,
267
Algebraic
integer,
Algebraic
property,
Algebraic
variety,
Alphabet,
Alternating
255
for
Annulus, 367
n-, 364
letters,
91
abelian
an ideal,
334
group,
Bijection,
255 Frenicle 109
4
58,
Chain(s),
288,
150, 318
358
Chain complex, 380
for a vector space,278 Betti number,
Center of a group,
Centroid, 115
345
Bessy,Bernard
341
82
Cell,6
275
for a free abelian GrObner, 261
closed field, 287, 292
77
70
digraph,
Cayley's theorem,
group,
16
group on n
Cayley
Stefan,
104
Augustin-Louis,
Cauchy's theorem, 322 Arthur, 70, 81, 347 Cayley,
288, 289 114
for a finitely-generated
Algebraic number, 268
Algebraically
Cauchy
178
206, 471
4, 5
Cardinality,
Ball, 364
463
laws, 41,
Girolamo,
Cardano,
Basis
283
extension,
Algebraic homotopy, 388
161
formula,
Cartesian product, 3,
of a ring, 232 Axiom of choice, of reflection, Axis
Banach,
fixed-point theorem, 376 149, 330 William,
Burnside,
Cancellation
141
inner,
fundamental theorem of, group, 223
Algebraic
66, 141
Frobenius, 421
206
289, 345
403
Burnside's
of, 418,419
field
fixed
Algebra
Algebraic
23, 37
357
416
of a group, 77
ibn
358
homomorphism,
Boundary
Brahmagupta,
Automorphism 2jt, 16 n 18, 64
modulo
of, 31 11, 20 M., 91
property
operation,
Brouwer
Associative operation,
Addition
homological,
149
389
Associates,
structural Binary
Boundary of a simplex, Bourbaki, Nicholas, 4,
condition,
392, 401
Action
Algebraic
318
series,
Ascending chain
29
structure(s),
algebraic
isomorphic, 30
Bloom, David Boolean ring, 177
207, 419
central
Ascending
;
theorem
395
Artin, Emil,
13
Binary
fundamental
Arithmetic,
theorem
structure
288
law,
Antisymmetric
70 Arc of a diagraph, Archimedian ordering, 230
de, 185
subcomplex
of, 381
Chain condition, ascending,
392,
401
descending, 401 Characteristic
of a
ring, 181
513
Index
514
Chief
315
series,
Coset,97
8
equivalence,
residue modulo H, residue modulo n, Class equation, 328
Closedinterval,
137 6
set under an operation, Closed surface, 371
15,21
F
in
Cochain,
363
Cocycle,
363
446
15, 21,
E, 447
228
Commutative
numbers, 205
143,
150
criterion,
215
comparable,
288
conjugate
over F,
Composition, function, 22, 23
28, 48,
Composition series, 315
1
class, 328 complex
Conjugate
subgroups,
416
Conjugate elements over F, 416 141, 143
Conjugation, 141
416
Connected component, 365
Connectedspace,365
Constant polynomial, number,
293
Constructible polygon, 466 Content of a polynomial, 396 function,
70
377
Contractible space, 365
228
positive,
prime, 394 primitive,
441
separable
over F,
438
inseparable over
function,
457
F,
105
1
Empty
set,
Empty
word, 341 220
Endomorphism,
relation, 13
Equality
328
108
Equation,
of rings,
169
Equivalence class, 8
105
sum,
spaces, 281
of vector
Peter
Dirichlet,
Lejeune,
Discrete frieze group,
174
union
of G-sets, 167
160
law,
Division algorithm for
Divisor, greatest
algorithm, 404
Euclidean
domain,
Euler
374
characteristic,
Euler formula, 13 phi-function, theorem,
Evaluation
104, 187 187
homomorphism,
171,201
173
256, 389 common,
401
Euclidean norm, 401 13, 39, 186, 468 Euler, Leonard,
Euler's
210, 220
F[x],
Division ring,
Euclidean
Euler
for Z, 60
118
185,403
Euclid,
463
7
relation,
Escher M. C,
116
Disjoint sets, 6
class,
Equivalence
of a polynomial, 89 cycles,
Disjoint
over F, 444 F, 267
symmetric
Elementary
group, 79, 86 of a vector space over
Distributive
199
176, 245 158
internal,
Disjoint
348
288
nilpotent,
Elementary contraction, 341
70 of,
Discriminant
isomorphism,
maximal,
transcendental
product,
Direct
numbers,
of, 38
totally
external, 108
H, 137
Conjugate
46
280 Direct
Congruence
Conjugate
443
Digraph, 70
Dimension
independent transcendental, 473
of, 84, 87, order of, 59
198
matrix,
176, 183
32, 38
identity,
orbit
chain condition, 401 of a square matrix, 46,
Diagonal
of, 23
associativity
199
479, 480
vertex Dihedral
358
simplicial,
283
of a polynomial,
Derivative
416
418
fixed,
irreducible, 389
269
of a polynomial,
arc of,
absolute value of, 13 conjugate of, 416
Continuous
Eisenstein
inverse
over F,
Determinant
3, 12
Complex number,
Constructible
174,241,419
1
Descending
Commutator subgroup, 143, 150\302\273 elements, 288 Comparable
Consequence,
109
group,
Descartes,Rene,
23
operation,
Commutative ring, 172
Conjugation
464
extension,
an extension,
of
478
Commensurable
modulo n,
297
algebraic over F, 267
Cyclic subgroup, 54, 59
Degree of a
group, 363
vector,
modulo
the cube,
Element(s), 1
54, 59
Cyclicgroup,
Dedekind, Richard,
113
Complex,
Doubling
idempotent,
polynomial, 199
Commutator,
Doublecoset,103
of, 89
Definitions,
Column
Dot product, 478
89
Decomposable
Cohomology
390
factorization,
unique
Cyclotomic polynomial, 217, 465
Coefficients
torsion,
ideal, 391
principal
Cycle(s), 89, 359,380
Cyclotomic
Codomain, 4 of a
39
length
of F in
Closurecondition, Coboundary, 363
137
Crelle,August,
Cyclic extension, 456
E, 443,
inseparable
totally
Coset group,
homologous, 367
287 228
286,
algebraic,
179
integral,
disjoint,
Closure
4
of a function,
Cross cap, 378
Closed
217
401
Euclidean,
97
right,
9
in an ordering, separable of
103
double,
left, 97
367
homology,
one-to-one, 4
Correspondence,
conjugate, 328
of a polynomial, of zero, 178 Domain
341
elementary,
Contraction,
Class
Even
62, 395
permutation,
92
Exact sequence, 385
126,
515
Index
Exact
of a pair,
sequence
homology
386 455
283
algebraic,
cyclic, 456
283 43
group,
group, 69
Fixed
418
elements,
Fixed field, 418,419
of, 456
separable, 438, 443
Formal Laurent series, 231 Formal power series,230 Free abelian group, 334
270
simple,
inseparable, 444
totally Extension
Face of a simplex, 256, 389 Factor, of a polynomial,
Factor
theorem,
Faithfull action,
rank
256
Fermat
Frobenius
155
Frobenius homomorphism, 244
Fermat's last theorem, Fermat's
\342\200\224 a2 + b2
p
\302\273
theorem,
411
closure of, 287,288 in E, 286 closure closed, 287, 292
algebraic algebraically
automorphism of, extension
418
265
of,
fixed, 418,419
formal Laurent series, 231 formal series, 230 power
Galois, 300 perfect,
250 quotient in a,
subfield
range of,
128
446 438, 443
E, 443,
of, of, 270
432
theorem, Fundamental
173
37
39
abelian,
letters, 93
on n
series of, 318 of, 66, 141 automorphism of automorphisms, 420 central
ascending
150,318
of, 58,
in a,
commutator
209
of cosets, 137
109
decomposable,
dihedral, 79, 86
308
457
direct
homomorphism
140, 242
of, 105
product
direct sum of, 105 discrete frieze, 116 of, 220
endomorphism
137, 139
theorem
of algebra,
theorem
of arithmetic,
finitely-generated,
of
freeon A,
factor, finite,
theorem
finitely-generated
143,
cyclic,54,59
395 Fundamental
skew, 173
149
L. Jr.,
cohomology, 363
254,288 Fundamental
Hermann, 275 divisor, 62, 395
common
center
4, 128
Fundamental
194,
of,
on F,
128
two-to-two, 10
functions, 201
separable extension simple extension skew, 173 strictly
104, 187
symmetric,
179
separableclosurein
splitting,
phi-,
restricted,
prime,
of rational
4
114
116
alternating
rational, 201
440
of quotients,
image under,
4 one-to-one, polynomial
379
Glide reflection,
Group(s),
104, 187 image of A under, 82, inverse of, 5
onto,
250, 339
Grobner basis, 261
of, 4
inverse
457
for a vector space,276
Genus,
Griess,Robert
23
Eulerphi-,
algebraic
group,
General polynomial of degreen, set, 68, 69 Generating Generator(s), 54, 59, 68, 69 for a presentation, 348
Greatest
elementary symmetric, 457
173
196, 407 40
integer,
General linear
Grassmann,
of, 22,
composition
domain
Ferro, Scipione del, 471 Field,
4
22, 23
composite,
302,
Gauss's lemma, 396 Gaussian
nontrivial,
continuous, 377
471
Lodovico,
Ferrari,
421
substitution,
codomain of,
390
184
Fermat's theorem,
automorphism,
Function(s), 4
468
prime,
421
211
Fermat, Pierre de, 185
300
group, 451
54, 59, of a principal ideal, relation on, 73, 348
Georg, 324
Frobenius
Galoisfield,
of a group,
390 116
Frobenius,
174, 302,
317,464
free, 342
342
of,
Friezegroup,
139
342
342
Frey, Gerhard,
149, 330
Walter,
Feit,
Free group,
118
132,
408,464
334
rank of, 336 Free generators,
108
357
Factor group, 137, Factor ring, 242
119,376
subfield, 418
basis for,
265
field,
External direct product,
Joseph A.,
Gauss, CarlF., 38,108,298,
Fixed point, Fixed
155 204
of,
Galois,Evariste, Galois
of a map, 425 470 radicals,
by
sub-, 159 Gallian,
space, 277
vector
Finitely-generated
index of, 428
orbits of, 158
union
Finite presentation, 348 448
159
isomorphic,
transitive,
Finite-dimensional
normal,
join
extension,
Finite
degree of, 283 finite, 283 finite
basis
Finite
degree of, 283 464
cyclotomic,
Finite
160
union of,
disjoint
condition, 401 for an ideal, 256
Finite-basis
154
G-set(s),
270
simple,
Extension(s), 265 abelian,
Field extension, 265
groups, 108,338
abelian
43
free, 342
342
free abelian,
frieze, 116
334
69
Index
516
projection, 127,237
(cont.)
Group(s)
of a ring,
451
Galois,
general linear,
40
generator(s)
68, 69
54, 59,
of,
380
361,
homology,
of, 125
homomorphism
109
indecomposable,
of, 141
inner automorphism
51
Klein 4-,
of rc-boundaries,
ofrc-chains, 358, 380 359, ofrc-cycles,
380
of, 50
over F, 452 348
of, 347,
139
relative
homology,
of, 83
383
series of, 311 149
317
solvable,
n
78
letters,
torsion
free, 142
wallpaper,
117
.
Group
action, 154 algebra,
Group
table,
Group ring, 223 43
Sir William
Hamilton,
224, 275 Hilbert,
left, 35,43
Hilbert basis theorem, 256 Otto, 317, 347 spaces,
Homeomorphic
128
under
a map,
Index of E
group,
relative, 383
of
of, 364
Homomorphism, 30, 125,171 358
boundary,
coboundary,
kernel
of,
theorem for, 129, 171,238
140,242
348
presentations,
172 312
series,
16 29
structure,
45, 132
a group,
of a ring,
172
of a vector Isomorphism
space, 282 extension
theorem,
theorems,
307-309
425, 428 Isomorphism
Isotonicity, 229
101
157
subgroup,
Isotropy
21
Infinite set, 5
Inner
244
fundamental
of
109
59
Join
of extension
fields,
133,194
automorphism,
Jordan,
Camille,
Jordan-Holder
456
308
of subgroups,
4, 133
Injection map, 4,
evaluation, 126, 171,201 Frobenius,
159
ofaG-set, 159
57
ordering, 228, 231 order,
Injection,
363
G-sets,
198
a subgroup,
Infinite
binary structures, 30
Isomorphic
of a binary
over F, 428
Induced
property
114
Isomorphic
conjugation, 416
transcendental
Induced operation,
361, 380
iaF[x], 214
Isomorphism,
12
subgroup,
Improper
389
element,
Isomorphic
82, 128
number,
Indeterminate,
355 380 Homological algebra, class, 367 Homology
5
Isomorphic rings,
elements, 473
Homeomorphism,
map,
Isomorphic
82, 128
Independent
355
function, 5
Inverse
Isomorphic groups, 45
35
inverse,
Holder,
invariance
32, 38
element,
Identity
Indecomposablegroup,
168
Inverse
Isometry,
Improper subset, 1
David,
Homology
28,48,
Improper ideal, 246 Rowan,
173 479
Irreducible polynomial, 214 for a over F, 269
element,
Imaginary
Half-open interval, 15
43
Irreducible
of, 254
Image of A,
223
element, 38
of an left,
right, 254
right,
Group
141
Inverse image under a map, 128 Invertible matrix, 479
176, 183
142
torsion,
311
series,
of a matrix,
250
Idempotent
79, 114
half open, 15
multiplicative,
trivial, 246
50
on symmetric of symmetries,
9
closed,
product of, 254 quotient of, 254 radical of, 245 sum
subgroup of,
Interval
Inverse
of, 245
principal,
representation
401
Invariant
prime, 248
regular
simple,
for, 401
247
nilradical
quaternion, 352
quotient,
for,
maximum condition for, 401 minimum condition for, 401
117
crystallographic,
of a polynomial
108
Invariant subgroup,
maximal,
presentation
Internal direct product, 59, 69
254
left,
P-, 322 plane
for,
improper, 246
352
octic, 79, order
394
of, 194
field of quotients unit in, 389 Intersection,
condition
finite-basis
on, 401
norm
prime element of,
chain condition descending finite basis for, 256
380
359,
condition
associates in, 389 Euclidean
241
ascending chain 392, 401 basis for, 255
45,
isomorphic,
30, 125
29,
374
179
domain,
Integral
Homomorphism property, 388 Homotopy, algebraic, Ideal(s),
prime, 62,
relatively
126
trivial,
408
rational,
171,237
39, 132, 317 316
theorem,
141
algebraic,
463
Kernel, 129, 171,238 of a linear transformation,
Gaussian,
196, 407
Khayyam,
Integer(s), 3
Omar,
206
282
517
Index
Klein
371
bottle,
Klein
Kronecker's
108, 174,266 266
theorem,
Kummer, Ernst,
241, 390
108,
Lagrange, Joseph-Louis,38,77,
96,
100,146
theorem of,
390
series, formal, 231
Laurent
477
sum of,
477
trace
100,471
Lame,Gabriel,
square,
transpose of, 55 upper-triangular, Maximal
element, 288 ideal,
McKay, J.
288
antisymmetric,
178
Mersenne
167
distributive,
Minimal
Leastcommon
Left
83
Length of a cycle,89
order, 260
Lexicographical
127, 282
transformation, kernel of, 282 277
independent
Linearly
364
vectors over F,
277
358
H-cycle,
359, 380
425
46,479,
480
orthogonal,
55
288
of a
Nullstellensatz,
betti,
480
of power
products,
of a ring,
228 114,
259
356 55
matrix,
p-group, 322 322
ordering, 288
Partial
6
Partition,
periodic,
Pattern,
Perfectfield,
440
pattern, 117
Periodic
76
Permutation,
subgroup, 323
multiplication,
254
odd,
complex, 3,12
76
92
orbits of, 84, 87
sign of, 135
109
commensurable,
117
Peano, Giuseppe, 275
even, 92
algebraic, 268 of, 46,
diagonal
132, 141
subgroup,
Number(s)
of, 479
invertible, 479 main
finite, 448
extension,
maximal, 149
diagonal, 46 inverse
410
455
Normalizer of,
260
228
natural,
cells of, 6 of \302\253,333
Euclidean, 401
Normal
477
determinant
61
subgroup,
Norm
Normal
restricted, 308 Matrix,
Nontrivial
Normal series,311
projection, 127 of, 4, 128 range
231
lexicographical,
p-subgroup,
Emmy, 168,419
over F,
of, 4
inverse
228,
induced,
Orthogonal
245
multiplicative,
41, 133, 194
228
Oriented rc-simplex,356
element, 176, 245
Nilradical,
Main diagonal of a matrix, 46,480 Main theorem of Galois theory, 451 Map, 4 injection,
Ordered ring,
364
n-sphere,
Noefher,
extension of,
59
Orientation,
n-chain,
Nilpotent
390
Joseph,
Liouville,
359, 380
rc-boundary,
rc-cell,364
dependent vectors over F,
Linearly
Multiplicative norm, 410 of a zero, 436 Multiplicity H-ball,
50
element, 59
of an
partial,
Linear combination, 276 Linear
Order of a group,
158
Archimedian, 230 173
298
Ferdinand,
Lindemann,
84, 87,
Ordering
inverse,
Multiplicative
304
Norman,
Levinson,
104,105
by components, n, 169
permutation, 76
Gerson, 77
25
well-defined,
Orbit,
term, 260
modulo
341
Letter,
Levi ben
23
induced, 21
infinite,
Multiplication
representation,
regular
269
42
37
23, associative, 11,20
binary,
commutative,
107,407
Left identity, 35, 43 Left inverse, 43 Left
over F,
Multiple, least common, 67,
254
ideal,
for a 53
subset,
Monoid,
371
4
Operation
prime, 185
Monic polynomial,
law, 41
cancellation
Left coset, 97
One-sidedsurface, Onto function,
Minimum condition, 401 MObius strip, 372, 373
67,
multiple,
149
401
condition, 322 H\342\200\236
4
function, 4
One-to-one
polynomial
Minimal
107, 407 Left
One-to-one correspondence,
247
273
288 transitive, 288
reflexive,
79, 352
Odd permutation, 92
36
Maximal
322
R. J.,
Octicgroup,
46
Matrix representation,
Maximum
268
transcendental,
Nunke,
Maximal normal subgroup,
Law
cancellation, 41,
rational, 3 real, 3
133
of,
12
imaginary,
479
singular,
Kronecker, Leopold,
293
constructible,
product of, 478
51
4-group,
87
permutation,
pinched, 388
Permutation
205
matrix,
Phi-function, Plane,
translation
87
104, 187 of, 114
Index
518
Plane
114
Polygon, constructible,
466
199
Polynomial(s),
coefficients of, content of,
199
degree of,
199
derivative discriminant
of,
of, 463 256
51
modulo
Restricted
205
theorem,
irreducible, 214
in
division algorithm, 60 179
a field,
of ideals, 254 Quotient
over F,
for a
Quotient,
in the
273
269
83
of, 232
automorphism
Boolean, 177 181
of,
172
173
division,
220
of endomorphisms,
396
primitive,
35
identity,
Right regular representation, Ring(s), 167 additive group of, 168
282
space,
law, 41
97
commutative,
Quotient ring, 242 Quotient
coset,
characteristic
139
group,
cancellation
Right
Right
Quaternions, 224
269
Right
Right ideal, 254 352
group,
308
map,
Ribet, Ken, 390
16
Quaternion
137
H,
modulo n, 6
Qin Jiushao, 403
217,
83
regular,
right
Residue class
subgroup,
Pythagorean
83
regular,
matrix, 36
127
structural, 11, 31
443
general of degreen, 457 group over F of, 452 irreducible for a over F, irreducible over F, 214 minimal
map,
algebraic,
of, 256
monk,
Projection
127, 237
Property
Eisenstein, 215 factor
homomorphism,
Proper subset, 1
217,465
divisor of,
left
Projection
Proper
396
cyclotomic,
Representation
259
Projective plane,
199
constant,
478
of matrices, power,
119, 376
fixed,
Point,
117
group,
crystallographic
Plane isometry,
241
factor,
reducible, 214
Rabin, Michael, 348
formal
power series, 230
separableover
Radical(s)
group,
223
F, 438 over radicals
solvable
by
splitting
field
Rank,
209
228
element,
Power product,
259
348
for,
348
Regular
ideal, 248
right,
Primitive element theorem, nth root of unity,
Primitive
67,301
Primitive
polynomial, 396
Principal
ideal,
250
ideal
Principal
domain,
Product
Cartesian,
3, 104
direct, 105, 169 of
ideals,
254
288
348
Row
348
Relator,
algorithm,
rcth, 67,
301
vector,
Ruffini,
478
Paolo, 471
3
Scalar, 275
Schreiertheorem,
7
7
transitive, 7 Relative homology group, Relatively prime, 62, 374 Remainder
18
Rotation, 114
equivalence, 7 symmetric,
391
3, 73,
173
of,
primitive
consequence of, equality,
simple, 253
Hth, 18
83
Relation(s),
241
of, 245
Roots of unity, 83
representation,
reflexive,
generator of, 250 Principal series, 315
441
quotient,
unit in a, 173,389 with unity, 172 zero, 172
left, 83 441
element,
law,
201
200,
radical subring
Reflexive relation, 7
250
Primitive
114
Reflexive
185
Mersenne, Prime
214
glide, 114
468
Prime field,
of polynomials,
prime ideal of, 248
axis of, 114
Prime, 394 Fermat,
ordered, 228
polynomial,
Reflection,
of, 245
nilradical
Reduction modulo n, 127 Refinement of a series, 311
348
finite,
generators isomorphic,
of, 247
ideal
maximal
408
3
Reducible
Presentation, 347, 348
172
isomorphism of, 172
word, 341
Reduced
230
242
isomorphic,
Rational number, 3
Real number,
Power series, formal, Power set, 8
128
Rational function, 201 integer,
171, 237
homomorphism,
ideal of,
342
336,
Rational
of, 259
ordering
ideal,
Range of a map, 4,
on F,
function
Positive
by, 470 245
extension
of an
term ordering zero of, 204, 255 Polynomial
F, 470
of, 432 of, 260
348
in the division 60, 210, 220
Sefer Yetsirah, Semigroup,
383
314
77
42
Separable closure 443,446
of F
in
E,
Separable element over F, 438 438, 443 Separable extension, over F, 438 Separable polynomial
519
Index
Sequenceof
the circle, 297 field, 173
Squaring
groups
exact, 385
exact homology,
Structure(s)
386
Series
isomorphic, 30
315
chief,
315 formal Laurent, 231
formal
Subcomplex,
230
power,
simplicial,
312
isomorphic,
of, 4
104
closedunder
an operation,
21,35,
disjoint, 6 element of,
1
of, 59,
69
partial ordering of,
288
well-defined, 1
135
modulo
F, 470
space)
Span, 276
field,
determinant main
trace
diagonal of, 133
genus
422 of, 46,
Sylow
479, 480
of, 46,
480
477
law,
Transitive
relation, 7
288 of SA, 86
114
Translation,
of a matrix,
Transpose Transposition,
90
Triangulation,
364
Trivial
55
of an
angle, 297 126
homomorphism,
Trivial ideal, 246 Trivial
51
subgroup,
Two-to-two function, 10
of, 379 Union
341
of sets,
Mejdell, 324 325
p-subgroup,
group
on n letters,
factorization
Unique
457
function,
391
of G-sets, 160 domain,
390
Unit, 173, 389 Unity,
elementary, 457 Symmetric
155
Transitive
Trisection
2jr, 16 n, 18, 64
Sylow theorems, 324, 325 Symmetric
267
Trichotomy, 228,229
Sylow, Peter Ludvig
Square matrix, 477
288
Surjection, 4 Syllable,
Sphere, 364 Splitting
for,
one sided,371
over
over F, 268
155
G-set,
Transitive subgroup
Surface, closed,371
53
group, 317
number,
Transitivity, 229
53
of matrices,
Skew field, 173
Solvable polynomial Space (seetopological
1 1
modulo
element
Transcendental
Transitive
105 of ideals, 254
479
Transcendental
Transitive action,
direct,
complex, 358 matrix,
177
by a,
1
Sum
357
Smallest subset,
series, 311 173
upper bound
356 boundary of, 357
444
Trace of a matrix, 133 Trace over F, 455
51
proper,
extension,
inseparable
Totally
Subspace of a vector space,281
253
Simplex,
Solvable
444
p-, 221
smallest,
group, 149
Simple ring,
Totally inseparable element over F,
minimal, 53
270
F
E, 447
in
51
improper,
permutation,
Simple extension,
Singular
of, 323
generated Subset,
112
368
pinched, 387 Totally inseparable closure of
51
Subring,
142
group,
Torus,
132, 141
trivial,
Shimura, Goro,390
Simplicial
normal, 149
Subnormal
of, 391
faceof,
Torsion subgroup,
torsion, 112
subset of, 1
Simple
Torsion
308
Sylow
partition of, 6 of, 76 permutation power, 8
of a
isotropy, 157
join of,
proper,
intersection
Torsion coefficient, 113 free, 113, 142
Torsion
P-, 322
68, 69
infinite, 5
Sign
141
normalizer
generating,
union
101
normal,
of, 374
355
homeomorphic,
mapping of, 375 of, 364 triangulation
index of,
nontrivial,
G-, 154
143
59
invariant,
maximal
1
empty,
365
Euler characterictic
51
improper,
Cartesian product of, 3,
component of, 365
contractible,
150
143,
cyclic,54,
on, 20
operation
365
connected,
conjugate, 141,
Set(s), 1
G., 330
John
Topological space(s),355 connected
commutator,
subnormal, 311
cardinality
Thompson,
Subgroup(s), 50
of, 311
binary
382
fixed, 418
principal, 315 refinement
Term ordering, 260
173
Subfield,
311
normal,
390
Richard,
Taylor,
381
Sub-G-set, 154
invariant, 311
390
Yutaka,
Taniyama,
Structural property, 11,31
composition,
114
Tartaglia, Niccolo, 471
of, 29
isomorphism
79,
Table, group, 43
29
algebraic,
binary
ascending central, 318
7
relation,
Symmetric
Symmetries, group of,
skew
Strictly
172
18,301
Hthrootof,
78
primitive
nth
root
of,
67, 301
Index
520
for a subset, 288 48 matrix,
bound
Upper
Upper-triangular
Van derWaerden,
Vector(s), 275, 478 linear combination linearly
dependent
linearly
independent
of, 276 over F, 277 over F,
277 row,
478
dimension direct
group,
sum
274, 275
278
280
finite-dimensional, 277 isomorphism
of, 282
operation,
Well-defined set, 1 Weyl,
Hermann,
275
348
problem,
lemma,
25, 137
313 314
Zermelo, Ernst, 289 Zero
of, 436
multiplicity
226
Weierstrass, Karl, 266 Well-defined
Word
Zassenhaus
38, 174, 419 Joseph Henry
theorem,
341
reduced, 341 Zassenhaus, Hans,
117
298
Maclagan, 224 Wedderburn
over F, of, 281
38, 81
190
341
empty,
Heinrich,
Wedderburn,
Vector space(s),
390
Andrew,
Wiles,
Word(s),
198
Wantzel, Pierre, Weber,
basisfor,
70
Von Dyck, Walter, Wallpaper
Weyl algebra, 222 Wilson's theorem,
Viete, Francois,
478
column,
Vertex of a digraph,
282
of a simplex
255
algebraic,
Variety,
B. L.,419
linear transformation of, 281 subspaceof,
of a polynomial, 204, Zero divisors, 178 Zero
ring,
172
Zorn, Max, 289 Zorn's
lemma,
289
255