Essential Student Algebra: Volume Three: Abstract Algebra [Book 3, 1 ed.] 041227860X

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Essential Student

Algebra

——

VOLUME

cee) al

THREE

Abstract

Algebra

ea

ASS

Essential Student

VOLUME

THREE

Algebra

Abstract

Algebra T. S. BLYTH

& E. F. ROBERTSON

University of StAndrews

London

New York

CHAPMAN

AND

HALL

First published in 1986 by Chapman and Hall Ltd 11 New Fetter Lane, London EC4P 4EE

Published in the USA by Chapman and Hall 29 West 35th Street, New York NY 10001

© 1986 T. S. Blyth and E. F. Robertson Printed in Great Britain by J. W. Arrowsmith Ltd., Bristol

ISBN 0 412 27860 X This paperback edition is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold,

hired out, or otherwise circulated without the publisher’s prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser. All rights reserved. No part of this book may be reprinted or reproduced, or utilized in any form or by any electronic, mechanical or other means, now known or hereafter

invented, including photocopying and recording, or in any information storage and retrieval system, without permission in writing from the publisher. British Library Cataloguing in Publication Data Blyth, T.S. Essential student algebra. Vol 3: Abstract algebra 1. Algebra I. Title

II. Robertson, E. F.

512

QA155

ISBN 0-412-27860-X

Contents

Preface Chapter One : Semigroups and groups Chapter Two : Subgroups

19

Chapter Three : Quotient groups

38

Chapter Four : Group morphisms

47

Chapter Five : Isomorphism theorems

59

Chapter Six : Rings, integral domains, and fields

66

Chapter Seven : Quotient rings and ring morphisms

va

Chapter Eight : Polynomials

88

Chapter Nine : Fields of quotients

108

Index

119

Preface

If, as it is often said, mathematics is the queen of science then algebra is surely the jewel in her crown. In the course of its vast development over the last half-century, algebra has emerged as the subject in which one can observe pure mathematical reasoning at its best. Its elegance is matched only by the ever-increasing number of its applications to an extraordinarily wide range of topics in areas other than ‘pure’ mathematics. Here our objective is to present, in the form of a series of five concise volumes, the fundamentals of the subject. Broadly speaking, we have covered in all the now traditional syllabus that is found in first and second year university courses, as well as some third year material. Further study would be at the level of ‘honours options’. The reasoning that lies behind this modular presentation is simple, namely to allow the student (be

he a mathematician or not) to read the subject in a way that is more appropriate to the length, content, and extent, of the various courses he has to take. Although we have taken great pains to include a wide selection of illustrative examples, we have not included any exercises. For a suitable companion collection of worked examples, we would refer the reader to our series Algebra through practice

(Cambridge University Press), the first five books of which are appropriate to the material covered here.

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CHAPTER

ONE

Semigroups and groups

Abstract algebra consists of a study of many different kinds of algebraic structure. Roughly speaking, an algebraic structure

is a collection of objects (e.g. numbers, mappings, matrices) together with a set of rules on how to ‘combine’ objects (e.g. how to add and multiply two numbers, how to compose two mappings, how to multiply two matrices). Each algebraic structure can be regarded as a ‘game’ with its own players and its own rules. Quite different ‘games’ can have rules in common. For example :

(a) in the ‘game’ IR of real numbers we have the rules

at+(yt+z)=(z+y)+z

and

a(yz) = (zy)z;

(8) in the ‘game’ Map(£, E) of mappings f : E — E we have the rule

fo(goh)=(fog)oh; (7) in the ‘game’ Mat, (IR) of real n x n matrices we have the rule

A(BC) = (AB)C.

Clearly, these rules all have a similar description. To express this in a more formal way, we formulate the following abstract notion.

Definition By a law of composition (or binary operation) on a set E we mean a mapping * : E x E — E described by (z, y) r+ oxy.

VOLUME

2

3: ABSTRACT

ALGEBRA

Less formally, a law of composition on E is an operation x whereby elements z, y of E can be ‘combined’ to form another element zx y of E. A set on which there is defined a law of composition * will often be written as (EZ, x) and is called a groupod. Definition

We say that x is associative on E if

(Vz, y,z € E)

zx (yxz) =(zxy) *z.

When this associative law holds we say that (E,) is a semtgroup and denote each of zx (y*z) and (xx y)*z by zxyxz. In this case we also have, for example,

[(a* y)*2]xw=[2x (y*2)] w = zx ((y*z) «wu

= zx [yx (zx w)], so that, roughly speaking, we can calculate composite expressions either from the left or from the right. We can therefore agree to omit all brackets and write the above unambiguously as zxyxzxw. There is, in fact, a general result which says that, for an associative law, the way in which compound expressions are bracketed does not affect the result. The proof of this is somewhat involved, so we shall simply accept it. Of course, it is this result that allows us to interpret familiar expressions such as, for example, Zi tiet+t-++t+Iyn in an unambiguous way. Example

Subtraction is a law of composition on the set Z of

integers. Since, for example, 3 — (2 — 1) # (3 — 2) — 1, it is not associative, so (Z,—) is a groupoid that is not a semigroup. Example

Addition and multiplication of real numbers are as-

sociative laws of composition, so (IR,+) and (IR,-) are semigroups. Example

Composition of mappings is a law of composition on

Map(E, E) which is associative, so (Map(E, E),0) is a semigroup.

SEMIGROUPS

Example

AND

GROUPS

3

Multiplication is an associative law of composition

on the set Matnxn(IR), so (Mat, x,(IR), -) is a semigroup. Example

If n is a positive integer let Z, = {0,1,2,...,n—1}

be the set consisting of the first n natural numbers. operation +, on Z, by

Define an

Z+pn y = the remainder on dividing z+ y by n. It is clear that if z,y € Z, then

z+, y € Z,.

We show as

follows that (Z,,+,) is a semigroup, i.e. that (Vz,y,z € Z,)

(t+ny) tn 2z=2+n (y+n2)-

For this purpose, we observe that if p is the greatest integer such

that pn < z+y then z+y = pn+(z+,y), and so we have that Zt+tny=xr2+t+y-—pn.

Suppose also that q is the greatest integer such that gn < y+z, so that YtnzZ=—ytz2z-—

qn.

Now let r be the greatest integer such that rn < z+y—pn+z and let s be the greatest integer such that sn < a+y+2-— qn. These inequalities can be written

(r+p)n Definition Suppose that (E,) is a groupoid with an identity element e. Given z € E, we say that y € E is a left inverse of z if yxz=e;a right inverse of z ifrx y =e; and a two-sided

inverse (or simply an inverse) of z ifyxz=e=azxy. Example

Consider the groupoid (IR,x) where x is given by zrey=atytso’y.

Clearly, 0 is an identity element for this operation. To find a right inverse of z we solve zx y = O for y: it is clear that every z has a unique right inverse, namely i z

fa)

ape

Solving z x y = 0 for z, we obtain

Pe

-1+V/1-4y?

Sa 2y 0

Thus we see that :

:

if yF 0; ify =0.

6

VOLUME

3: ABSTRACT

ALGEBRA

y has two left inverses if 4y? < 1 and y #0; y has no left inverses if 4y? > 1; y has a unique left inverse if y = 0 or y = 2, the left inverse being respectively 0 and +1. The only element that has a two-sided inverse is the identity. 1.8 Theorem Let (E,x) be a semigroup with identity e. If z € E has a left inverse y and a right inverse z then necessarily Spa

Proof y=yxe=yx(z*«z)=(yxz)xz=exz=2z.9 1.4 Corollary If an element then that inverse 13 unique. >

of a semigroup

has an inverse

In dealing with an associative law x we shall denote the unique

inverse of z by x! (when it exists), except when x is written as addition in which case we shall write —z. that

Thus z+ is such

gxzt=e=2 xz.

An element that has an inverse is said to be invertible. It is clear that the identity element e is always invertible with e~! =e. Example There are semigroups (£, x) with identity e in which the only invertible element is e; for example, the semigroups

(IN, -) and (P(Z),/).

Example

In the semigroup (Map(£, E),o) the elements that

have left inverses are precisely the injective mappings; those that have right inverses are precisely the surjective mappings; and

those that have (two-sided) inverses are precisely the bijective mappings.

Example

In (Mat, ,(IR),-) the invertible matrices are pre-

cisely those of rank n; equivalently, those with non-zero determinant. If an element z of a semigroup (£, x) has an inverse then from

we see that z~/ also has an inverse, namely z. Since in a semigroup inverses (when they exist) are unique, it follows that

—1)-1 Seat (a9)

An important property of inverses is the following.

SEMIGROUPS

AND

1.5 Theorem

GROUPS

7

Let (E,x) be a semigroup with identity e.

If

21,22 € E are invertible then so 1s x1 * x2; moreover, ( If the law of composition of G is written additively, the above criterion becomes (Vz,y € H)

zx—yed.

The above result is particularly useful since it is often possible to show that a given structure is a group by showing that it is a subgroup of a known large group.

SUBGROUPS

Example

21

Let Map(IR, IR) be the set of all mapping f : IR > R.

It is clear that this forms a group under the operation of addition defined by

(Vee IR)

(f+ 9)(z) = f(z) +9(z),

the identity element being the constant mapping z +> 0 and the

inverse of f being —f, where (—f)(z) = —f(z) for every z € R. The subset Con(IR, IR) of continuous functions is a subgroup; for, as is shown in analysis, if f and g are continuous then so is f —g. Likewise, the subset Diff(IR, IR) of differentiable functions is a subgroup; for, again as is shown in analysis, if f and g are differentiable then so is f — g. Definition

The centre of a group G is the subset

Z(G)={g9EG; (¥rEG)

gr=29}

of elements of G that commute with every element of G.

Example Z(G) is a subgroup of G. In fact, it is clear that Z(G) # @ since it contains 1g. Also, if g € Z(G) then from gz = zg we have z = g~!2g and so zg"! = g!z,ie. gt €

Z(G). Consequently if g,h € Z(G) then (Vz €G) so that gh~+ € Z(G).

gh-42 = gth! = zgh* Thus, by 2.2, Z(G) is a subgroup of G.

Note that G is abelian if and only if Z(G) = G. Example Let G be the group of bijections f : € — C (the operation being composition of mappings). Consider the subset H consisting of those bijections f : € — € such that

(vz,we)

§ |f(z) — f(w)| = lz - ul.

Then H is a subgroup of G. In fact, if f € H then from

jz — w| = [f[F7*(2)] — FLF* (eI = [F7(2) - Fu) | we see that f—! € H. Consequently, if f,g € H then

[flg~*(2)] — fla? (w)II = 977(2)— 9*(¥)| = |e - BI shows that f og~! € H and hence that H is a subgroup.

VOLUME

.

22

Example

3: ABSTRACT

ALGEBRA

The set K of rationals of the form 1+ 2n 1+2m

where n,m € Z is a subgroup of the multiplicative non-zero rationals. In fact, given 1+ 2n

group of

1+ 2r

in K, we have

ae

1+2n

1+2s _

1+ 2(n+s+ 2ns) < K.

For the determination of finite subgroups, the following is a useful alternative to 2.2 : 2.3 Theorem A finite non-empty subset of a group G 1s a subgroup of G if and only tf it is closed under the operation of G. Proof The necessity of the condition is clear. As for sufficiency, suppose that S = {z),...,2m} is a finite subset of G that is

closed under the operation (multiplication) of G, ie. that S is a subsemigroup.

Given any z; € S, consider the products DjpZ1,2{TQ,...,LiLZm-

Since S is multiplicatively closed these products all belong to S; and by the cancellation law in G no two of them are equal. These products therefore describe the whole of S. Similarly, so do the products D1,

WQZji,.- +,

Lm Ty.

It is now immediate from 1.8 that S is a subgroup of G. > Example In the symmetric group S,, of all permutations on the set {1,...,n}, consider the subset A, of even permutations. Since the product of two even permutations is an even permutation, it follows by 2.3 that A, is a subgroup of S,, called the alternating group on n symbols. We shall show later that

|Ay| = 1\S,,| = inl.

SUBGROUPS

23

Example Consider the subset T, of S;, consisting of those permutations that fix a given element. Clearly, T,, is closed under multiplication and so is a subgroup of S,,. 2.4 Theorem The intersection of any collection of subgroups of a group G 1s a subgroup of G. Proof

Let C be a collection of subgroups of G, and let J be

the intersection of all the subgroups in C. Note first that

I ~@

since, by 2.1, it contains the identity element of G. If now z,y © I then z,y belong to every subgroup in the collection C,

whence so does zy! by 2.2. Consequently zy~! € I and so, again by 2.2, J is a subgroup of G. > Suppose now that S is a given subset of a group G. Consider the collection of all the subgroups of G that contain S. This collection is not empty since it clearly includes the group G itself. By 2.4, the intersection of all the subgroups in this collection is also a subgroup of G, and it clearly contains S. It is then the smallest subgroup of G to contain S. We denote it by (S) and call it the subgroup generated by S. Clearly, if S = 9 then

(S) = {1c}.

In the case where

S # § we determine (S) as

follows. 2.5 Theorem

If S is a non-empty subset of a group G then

(S) consists of all products of the form 2122-:-%pn where x; € S or zt € S for eacht. Proof Let H be the set of all products of the form 2122 --- zy where z; € S or o. € S for each z. If z,y € H, say x =

Z12%2°*-Zy, and y = yi¥2°-- Ym, then a

“

=

bees

ty} =2122°--Iny,, --- yp ¥, EH and so, by 2.2, H is a subgroup of G. Clearly, for every we have z € H and so S C H.

Consequently,

x€ S

(S) C H by the

definition of (S). Now every subgroup K of G that contains S contains also the inverses of all the elements of S, and hence all the elements of the form 2122°-:Z, where z; € S or a. es for each 1, i.e. contains H. Taking in particular K = (S), we

obtain H C (S). It now follows that H = (S). > A particularly important case of the above is obtained by taking the set S to be a singleton, say S = {g}. In this case we

24

.

VOLUME

3:

ABSTRACT

ALGEBRA

write (g) instead of ({g}). It follows immediately from 2.5 that for every g € G the subgroup generated by {g}, i.e. the smallest subgroup to contain the element g, is

CE Pan eh ede PS BAP Definition We say that a group G is generated by the subset S (or that S is a set of generators of G) if (S) = G. A group that is generated by a singleton is said to be cyclic.

Example For every n € Z the subgroup generated by {n} is the (cyclic) group (n) = {...,—2n,—n,0,n,2n, ...}. This is also written as nZ, the set of integer multiples of n. The abelian group Z has the following interesting property. 2.6 Theorem Every subgroup of the abelian group Z 1s of the form nZ for some neE Z.

Proof Suppose that H is a subgroup of Z. If H = {0} then clearly H = OZ. Suppose then that H # {0} and let n be a non-zero element of H. Since H is a subgroup we have —n € H and so, whether n is positive or not, we see that H contains

positive integers. Let Hi = {x € H ; z > 0}. Since the set of natural numbers is well-ordered (in the sense that every non-empty element, a of z by a. Now since

subset has a smallest element), H, has a smallest say. Given any z € H, consider the euclidean division We have z = aqg+r where g,r € ZandO

2.20 Corollary Proof

If G 1s of order n then (Vg € G) g” = 1.

Let g € G be of order k. Then k divides n by 2.19, so

n= kt'and hence g” = (g*)*= 1° =175 2.21 Corollary

All groups of prime order are cyclic.

Proof Let |G| = p where pis a prime. Given g € G with g #1, let g have order r. By 2.19, r divides p. Since p is prime and r # 1 we must have r = p whence the order of the subgroup

(g) is equal to that of G. It follows that (g) = G and so G is cyclic. > We end the present Chapter by examining certain properties of the equivalences Ly and Ry which, as we shall see, characterize these equivalences. Definition An equivalence relation E on a group G is said to be left compatible if

z= y(£) => (Vg € G) gx = gy(E), and right compatible if

z= y(E) =

(Vg € G) zg = y9(E).

Left and right compatibility can be expressed respectively as follows :

[z]z = [yle => (V9 € G) [grlz = [gy]zz; [z]z = [yle => (Vo € G) [zg]z = [yglz. 2.22 Theorem For every subgroup H of G the equivalence Ly is left compatible and the equivalence Ry 13 right compatible. Moreover, H=

[Ize = [t]Re:

36

VOLUME 3: ABSTRACT ALGEBRA

Proof If z = y(Ly) then z~'y € H and so, for every g € G,

(gx)-'gy=2 whence gz = gy(Ly).

'g *gy=27'yeH

Thus Ly is left compatible.

Similarly,

Ry is right compatible. Finally,

zeEH

«=> al t=ceH

—

2=1(Rqz)

shows that H = [1]r,,. Similarly, we have H = (1|z,,.0 In fact, as we shall now show, left and right compatibility characterize the equivalences Ly and Ry. 2.23 Theorem

If E is an equivalence relation on a group G

that ts left [right] compatible then there 1s a subgroup H (namely [1]z) such that E = Ly |E = Ry}. Proof We give the proof for right compatible equivalences, that for left compatible equivalences being similar. Suppose that E is right compatible. We observe first that

[1]z ts a subgroup of G. In fact, if y € [1|z then [y]z = [1] and so [I|z = [yt y|z =

[y-*1Je =[y“]z, ie. y-? € [1]z. If then z,y € [1]z we deduce from [y~*]z = [1|z that [zy]

= [zlJz = [z]z = [1]z, so

zy~* € [1]z and hence [1]g is a subgroup. We now observe that EF = Rj1),. In fact, we note that

[z]e = [ye > [zy "Je = [yy “le = (1a;

[zy "Je = [1Je => [2]e = [zy “ye = [lye = [yle. The first of these says that E = Rj,), and the second says that

Ri), => E. Thus we have FE= Rjy),, in the sense that z = y(E) if and only if z= y(Ri),). > The connection between subgroups and left [right] compatible equivalences is even stronger. 2.24 Theorem

For every group G the following are equipotent :

(1) the set S(G) of subgroups of G; (2) the set L of left compatible equivalences; (3) the set R of right compatible equivalences.

SUBGROUPS

37

Proof By 2.23 we can define a mapping 9 : R — S(G) by the prescription 3(Z) = [1]. Now # is injective since if [1]z = [1]r then

c=y(E) ie.

E =

F.

sy te [ile =([1]r —

z=y(F),

To see that & is also surjective, let H be any

subgroup of G. Then by 2.22 we have H = [1|r, = 0(Rx). Thus ¥ is a bijection and S(G),R are equipotent. Similarly, S(G), £ are equipotent. > Example Consider the group Z. Since this is abelian, the notions of left and right compatible for an equivalence relation coincide. Let H be any subgroup of Z. By 2.6 we have H = nZ

for some n € Z. The equivalence relation Ry [= Ly] is given by z=y

=>

cs-yEeH=nZ

12,y

differ by a multiple of n,

i.e. it is what is often referred to as the relation ‘mod n’. It is clear from this that z = 0 if and only if z € nZ, so that we have

(O]|r, = H = nZ, and that the ‘mod n’ class of z € Z is given

by

[zx] = {...,.c-—2n,z—n,z,2+n,2+2n,...}.

This can be written

[zs] ={z+h;

hEenZ},

which is precisely the description given in 2.14.

CHAPTER

THREE

Quotient groups

In the previous Chapter we saw how each subgroup H of G partitions G into left cosets and also partitions G into right cosets. In general, these partitions are not the same, for the left and the right cosets need not coincide.

Example Taking H = {e,c} in S3 (see the Cayley table on page 32) we have, for example, pH = {p,b} and Hp = {p,a}. Definition A subgroup H of a group G is said to be normal if it is such that cH = Hz for all rE G. Example

In an abelian group all subgroups are normal.

Example Every subgroup of index 2 1s a normal subgroup. In fact, if H is of index 2 in G then the Ly-classes are necessarily

H and G \ H; for one, namely [1]z,,, is H (by 2.14) and the Ly-classes form a partition of G. Similarly, the Ry-classes are necessarily H and G\ H. Hence Ly = Ry and so H is normal.

Example In S3 the subgroup H = {e, p,q} is normal. In fact, it has index 2 in S3 (see page 32). Definition If G is a group and z € G then a conjugate of zx is an element of the form grg~/ where g € G. 3.1 Theorem

If H is a subgroup ofG then for every

set

gHg"* = {ghg"*; of conjugates by g 13s a subgroup of G.

hE H}

g EG

the

QUOTIENT

GROUPS

39

Proof Clearly, gHg~! #9 h,k € H then

since it contains 1 = glg~!. If now

(ghg~*)(gkg~*)~* = ghg-49k—'g-) = ghk-'g-! € gHg™} shows that gHg™' is a subgroup of G. } A useful alternative criterion for normality is the following : 3.2 Theorem

A subgroup H of G is normal tf and only if

(Vg € G)

gHg- C H.

Proof Suppose that H is normal in G. Then gH = Hg for every 9 € G. Thus, if h € H, we have gh = kg for some k € H,

whence ghg-! = k € H.

Since this holds for all g € G, we

deduce that gHg™! C H. Conversely, if gHg~! C H for every g € G then clearly we have gH C Hg. Writing g~! for g, we have g-1Hg C H from which we clearly obtain Hg C gH. Thus gH = Hg for every g € G and so is normal. >

Example If G is a group, consider the centre Z(G) of G. We know that Z(G) is a subgroup of G. Now if z € Z(G) then zg = gz for every g € G and so gzg- 1= x € Z(G). Thus Z(G) is a normal subgroup of G.

Example The special linear group H = SL(n, IR) is a normal subgroup of the general linear group G = GL(n, IR). In fact, if A € H then detA = 1 and, for every X € G, det(X AX~*) = det A=1s0 that XAX-1€

H.

Example The alternating group A, is a normal subgroup of the symmetric group S,. In fact, if o € A, then o is an even

permutation (i.e. e(o) = 1 where ¢ is the parity map) and so, for every permutation p € Sy,

e(pap"*) = e(p)e(a)e(o"1) = e(0) =1 whence pop 1 is an even permutation, ie. pap ! € Ap. The concept of a normal subgroup is a very the sense that it allows us to construct new ones. In fact, given a normal subgroup H of that it is possible to define the structure of a of cosets of H.

powerful tool in groups from old G we shall show group on the set

40

VOLUME

3:

ABSTRACT

3.3 Theorem Let H be a normal subgroup of G. set of cosets ofH the prescription

ALGEBRA

Then on the

2H -yH =csyH defines a law of composition with respect to which the set of cosets of H forms a group. > Proof

First we show that the prescription stated ‘makes sense’

(or, more formally, is well defined). By this we mean that if cH = 2'H and yH = y’'H then zyH = z'y’ H; roughly speaking, whatever representatives we choose in the cosets we obtain the same answer. This follows from the fact that, since the subgroup H is normal,

syH = sy'H = Hzy' = H2'y' = 2'y'H. To see that the set of cosets, equipped with this multiplication, is a group we note that this multiplication is associative (since that of G is); moreover, 1H is clearly the identity element, and cH -2 tH=2cr

'H=1H=H

shows that the inverse of zH is x-1H.

Definition

When H is normal in G we shall denote the set of

cosets of H in G by G/H.

By 3.3, G/H is a group which we

call the quotient group (or factor group) of G by H. 3.4 Theorem [If G 1s finite and H is a normal subgroup ofG

then |G/H| = |G|/|H|. Proof This is immediate from Lagrange’s theorem (2.17) since by definition |G/H|= |G: H|. Definition An equivalence relation that is both left compatible and right compatible on a group G is called a congruence on G. By 2.23 we see that if FE is a congruence on G then there is a

subgroup H of G (namely, [1]¢) such that

EF= Ly = Ry. As

in 2.24, the assignment E ++ [1|¢ bijectively relates congruences with normal subgroups, so we have the analogue :

QUOTIENT

GROUPS

3.5 Theorem

41

In every group G the following are equipotent :

(1) the set of normal subgroups of G; (2) the set of congruences onG. We also have the following analogue of 3.3 : 3.6 Theorem Let E be a congruence on G. G/E of E-classes the prescription

Then on the set

[z]z -[ylz =([zy]e defines a law of composition group. ©

with respect to which G/E is a

Example If n is a positive integer consider the (normal) subgroup H = nZ of Z. The cosets are the ‘mod n’ classes. There are n of these, namely

0+nZ= {...,—2n,—n,0,n,2n, ...}, 1+nZ=

{...,-2n+1,-n+1,1,n+1,2n+1,...},

(n—1)+nZ={...,-n—1,-1,n—1,2n—1,3n—1,...}. The quotient group Z/nZ is thus of order n and the group operation is the addition defined by

(c+nZ)+(y+nZ) = (x+y) +nZ. The reader should pause to reflect that there are essentially three different interpretations of addition in the above formula! Example As seen above, the subset H = {e, p,q} is a normal subgroup of the symmetric group 53 and its cosets are H and

S3 \ H = {a,b,c}. We can write these cosets as H and aH.

42

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3: ABSTRACT

ALGEBRA

Now since aH -aH = a?H = eH = H, we see that the quotient

group S$3/H = {H,aH} has the Cayley table del

Bi A aH|iaH

(sl

ech H

which is essentially the same as that for S2. Example

In the group of symmetries of a square as described

on page 12, consider the subgroup H = {e,r?,z,y}.

By La-

grange’s theorem we have

IG:H| = |G|/|H| = 8/4 =2 so H is of index 2 and is therefore normal. The quotient group G/H has two elements, namely H and rH. The Cayley table

for G/H

is therefore H

We cH)

rH

Her rie

a

It is readily seen that K = {e, y} is a subgroup of H. The index of K in H is also 2, so K is a normal subgroup of H. The quotient group H/K also has two elements, namely K and zK.

The Cayley table for H/K is therfore

Kol tK

K

«cK

ek ick

2k K

Note, however, that the subgroup K is not normal in G. example,

For

hKh7* = {e,hyh—'} = {e,2} Z K. Thus we see that ‘a normal subgroup of a normal subgroup need not be normal’.

QUOTIENT

GROUPS

43

Example All subgroups of the abelian group Z24 are normal. Consider the subgroup H = {0,6, 12,18}. The quotient group Z24/H has 24/4 = 6 elements, namely

1+ 2+ 3+ 4+ 5+

H H H H H H

= = = = = =

{0,6, 12, 18}, {1,7, 13, 19}, {2,8, 14,20}, {3,9, 15,21}, {4, 10, 16, 22}, {5, 11,17, 23}.

The Cayley table for Z24/H is the following, in which z+ H is

denoted by [z] :

Thus we see that Z24/H has the same Cayley table as Ze.

Example

Consider the subset T of SL(2,€) consisting of the

matrices 0

2

zal; i

0

1

k=|4 5}

The subgroup of SL(2,€) generated by T has order 8 and is called the quaternion group. Indeed, since J~! = —J,K~1 —K and KJ = —JK it follows by 2.5 that

(T) = (Ie, -la, J,—J, K, —K, JK, — JK}.

=

44

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3: ABSTRACT

ALGEBRA

The Cayley table for (T) is

The subgroup W = {Ip, —I2} is normal, being the centre of (T). The cosets of WN in (T) are

N, JN ={J,-J}, KN ={K,-K}, JKN ={JK,-JK} and the Cayley table for (T)/WN is

This is the same as that of the Klein 4-group (see page 13). Example Let a,b € given by fa,(z) = az transformation. It is affine transformations we have

IR with a # 0. Let fa, : IR — +b. Such a mapping is called an readily seen that the composite is also an affine transformation;

las ° fed =

R be affine of two in fact

fadadtb:

The set A of affine transformations is then a group, for fio is the identity map and, as the reader can easily verify, fa, is invertible with aii! tae

=

fa-2, 0} then the exponential map f : (IR,+) — (IR™, -) given by f(z) = e? is a group morphism. Indeed, for all z,y € IR we have f(x + y) = e7t¥ = e7e¥ =

f(z)f(y)Example

[If IR* denotes the multiplicative group of all non-zero

real numbers then f : IR* — IR* defined by f(z) = |z| is a group morphism, for |zy| = |z| |y|. Example For every m € IN the mapping fm : (IN, +) — (IN, +) given by fm(n) = mn is a (semigroup) morphism, for

fm(n + p)= m(n+ p) = mn+mp= fm(n) + fm(P). Example Let G be the multiplicative group {1,—1}. Then the parity map e : S, — G described by o + e(c) is a group

morphism, for e(o 0 p) = e(c)e(p).

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Example If G is a group and g € G then the operation of conjugation by g can be described as a mapping ¢«, : G ~ G

given by ¢,(z) = gzg~*. Each ¢, is a morphism, for

¢g(zy) = gtyg7' = gzg*- gyg™* = Example

$g(2)sq(y)-

If G is a group and Z is a normal subgroup of G

then the natural mapping hy : G > G/H given by hx(z) = cH is a morphism.

In fact, by the way that multiplication is defined

on G/F we have, for all z, y € G,

ha (zy) = zyH = cHyH = by (z)ha(y). Example

Consider the mapping

f : Z —

Z, described by

f(m) = m* where m* € Z,, is such that m = m* (mod n), i.e. m* is ‘m reduced modulo n’. By the compatibility of the congruence ‘mod n’ we have

(m+ p)* = m+p=m* + p*(mod n), and hence (m+ p)* = m* +, p*, ie. f(m+p) = f(m) +p f(p). Thus f is a morphism.

Example

f : GL(n,IR) — IR™ defined by f(A) = detA is a

group morphism. Example The only group morphism f : Q — Z is the zero morphism, namely that described by z ++ 0. In fact, suppose

that 3: Q — Z is a group morphism and that 3(1)

4 0. Then

for every non-zero r € Z we have

9(1) = 8(1/r) + ---+ 8(1/r) =r 8(1/r) r

and so r divides §(1). But #(1) has only a finite number of divisors and so we must have #(1) = 0. For all non-zero p,qgE Z we then have

0 = p8(1) = p9(q/4) = pqd(1/q) = 99(P/a), whence 8(p/q) = 0 and so # is the zero map.

GROUP

MORPHISMS

49

If f : G — H isa group morphism then there are two subsets associated with f that are of paramount importance. The first of these is the image of f, namely the set

Im f = {f(z) ; x € G}, and the other is the kernel of f, namely the set

Kerf = {z€G;

f(z) =1y}.

We can depict these as follows :

4.1 Theorem [f f : G— H is a group morphism then Imf 1s a subgroup ofH and Kerf ts a normal subgroup of G. Proof Since f is 2 morphism, Im f is clearly closed under multiplication. Now for every z € G we have

f(z) f(1e) = f(z1e) = f(z) = f(1ez) = f(1e) f(z), so f(1g) is the identity element in Im f. Also,

f(z) f(z~*) = f(zz~") = f(1e) = f(27*2) = f(2~*) f(z), so the inverse of f(z) in Imf is f(z~+). Thus we see that Im f is a subgroup of H. As for Ker f, we note first from the above that since Im f is a

subgroup of H we must have, by 2.1, f(1lc) = 1x. Gotpeuently we see that Ker f4 9. If now z,y € Kerf then

f(zy~*) = f(=) f(y") = F(2)[F (y)]-* = tar Ly’ = La,

50

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ALGEBRA

so zy_! € Kerf and hence Kerf is a subgroup of G. To see that it is a normal subgroup, let z € Ker f and let g € G; then

f(gzg~*) = f(9) f(z) f(9~*) = F(9) “1a -[F(9)]-* = 1 and so gzg-! € Kerf. An interesting characterization following :

of normal subgroups is the

4.2 Theorem A subgroup K of a group G 1s normal tf and only uf it is the kernel of a group morphism f :G— H. Proof Let K be a normal subgroup of G. Then we can form the quotient group G/K. Consider the natural mapping 4x : G — G/K described by 4x (g) = 9K. We have seen above that this is a group morphism. Now, since cosets are equivalence classes, we have g€Kerhe

gK=K

=

ge k.

Thus Kerhx = K, i.e. K is the kernel of a group morphism. The converse was established in 4.1. >

Definition A group morphism that is injective will be called a monomorphism; a group morphism that is surjective will be called an eptmorphism; and a group morphism that is bijective will be called an 1somorphism. If f : G — H is a group morphism then it is clear that f is an epimorphism if and only if Im f = H; in other words, if Im f is as large as it can be. It is natural to investigate the dual of this, namely the situation where Ker f is as small as it can be. 4.3 Theorem

A group morphism f : G — H is a monomor-

phism tf and only tf Ker f= {1c}. Proof

Suppose that f is a monomorphism

and let z € Ker f.

Then f(z) = 1y = f(1q) and so, since f is injective, z = 1g. Conversely, suppose that Ker f = {1c}. Then if f(z) = f(y) we have

f(zy7*) = f(z) F(y7*) = F(z) [F(y)-* = 1k and so zy! € Kerf = {1c} whence zy~! = 1g and hence z=y.

Thus f is injective and so is a monomorphism. ©

GROUP

MORPHISMS

Example

The exponential map f : (IR,+) —

monomorphism. Example

hy :

61

(IR™,-) is a

In fact, e* = 1 if and only if z = 0.

If H is a normal subgroup of G then the natural map

G— G/H

is an epimorphism.

Example The mapping f : Z — Z,, described by ‘reduction modulo n’ is an epimorphism. Example

We have seen that for every g € G the operation of

conjugation by g, namely z ++ ¢,(z) = gzg~1, is a morphism. Now ¢g is surjective since for every z € G we have

¢g(g~*zg) = gg *agg7' = =. Also, ¢g is injective; for if ¢,(z) = ¢g(y) then gzg~*= gyg™? and so z = y by cancellation. isomorphism.

Thus, for every g € G, ¢, is an

Definition When there is an isomorphism f : G — H we say that the groups G and H are tsomorphic and write G ~ H. We can think of isomorphic groups as being ‘essentially the same’ in the sense that one can be derived from the other by ‘relabelling’ the elements. In particular, finite groups are isomorphic if and only if they have the same Cayley table except for the labelling of the elements.

Example

Sz ~ Z2 ~ {1,-—1}.

In fact the respective Cayley

tables are

so G ~ Z, under the isomorphism 1-0, Example

t+1, —1++2,

—21+>3.

Z ~ 2Z under the assignment n ++ 2n.

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Example (IR,+) ~ (IR*, -). In fact, an isomorphism is given by z + e®. In contrast, note that (Q, +) and (QT, -) are not isomorphic.

In fact, if there existed an isomorphism f : (Q,+) —

(Qt, -) then there would exist r € Q with

2= f(r) = f(r + 31) = (GNP which is impossible since z? = 2 has no solution in Q. Example If T is the circle group and R is the group of rotation matrices then T ~ R. In fact, the mapping f : T — R given by f(z) = Rargz

is an isomorphism. To see this, we write the elements of T in the form z = e*® for 0 < 9 < 2x where by definition 8 = arg z. Then from 2122

=

etP1 et02

=

e(91 +02)

we have arg z;22 = arg z, + arg 22 and consequently f (z122) = ares, sd —— Jfees zy+arg 29 = Race Zz Rarg z3

= f (21) f(z2) and so f is a group morphism. is an isomorphism.

Clearly, f is a bijection and so

Further examples of group isomorphisms can be given using

the notion of a cartesian product group. If (G,*q) and (H,«z) are groups then the cartesian product set G x H can be made into a group by defining

(91, h1) * (g2, he) = (91 *G ge, hi *H he). Clearly, the identity element is (1g, 1) and the inverse of (g, h) igtg7 jh). Example

€ ~ IR x IR under the assignment z+ ty ++ (sz, y).

Example

If G,H are groups then G x H ~ H x G under the

assignment (g,h) ++ (h, g).

GROUP

MORPHISMS

Example

53

R* x Z2 ~ IR*. In fact, let f : Rt x Z, — IR* be

given by f(z,y) = (-1)¥z.

Since Zz = {0,1} we have f(z,0) = z and f(z,1) = —z, from which it follows that f is a bijection. Moreover,

f(zz',yt+y')= (—1)9+' ca!

= (=1)¥2 (-1)""2 = f(z,y)f(z',y’),

so f is also a morphism.

Hence f is an isomorphism.

Concerning isomorphisms of groups, we note the following properties that are like those of an equivalence relation :

(1) For every group G we have G ~ G; indeed, the identity map on G is clearly an isomorphism.

(2) If

G~ H then H ~ G;; indeed, if f : G — H is an isomor-

phism then for all h1,h2

€ H we have

f[f~* (a1) f-*(ho)] = FF" (ra IF LFA *(A2)] = hihe = f[f~*(Arh2)] whence, f being a bijection, f—+(h1)f~!(h2) = f— (Ai h2) and so f~? is also an isomorphism. (3) If G~ H and H ~ K then G ~ K;; indeed, if f: G— H and g: H — K are isomorphisms then so is gof:G— K. We can thus separate groups into isomorphism classes. As we shall now show, the class of the cyclic group Z, consists of all cyclic groups of order n, and that of the cyclic group Z consists of all infinite cyclic groups. 4.4 Theorem Every finite cyclic group of order n 1s 1ts0morphic to Z,, and every infinite cyclic group 13s tsomorphic to Z.

Proof Let G = (g) be of order n. Then by 2.7 we have G=

EO

a

i

tee

a

54

‘

VOLUME

3: ABSTRACT

ALGEBRA

with the convention that g°? = 1. Consider now the mapping f :Z, —G

described by

f(t) =9°. It is clear that f is a bijection. a+7n then

flitag) =fitij—n) =" = 99 (97*)" =9'9" = F(t)F(9)Thus we see that f is an isomorphism. Suppose now that G = (g) is of infinite order. we have G=

{Ac

Py sapgneeh

afg%

where all the powers of g are distinct. f :2Z-—G described by

Then by 2.9

ca}

Consider the mapping

fli)=9. Clearly, f is a bijection; and since

fits) = 9? = gig? = F(%) f(s) we see that f is an isomorphism. > 4.5 Corollary If G,H are cyclic groups of the same order then G and H are tsomorphic. > 4.6 Corollary phic to Z,.

The group G, of n-th roots of unity ts ts0omor-

4.7 Corollary If p 1s a prime then to within tsomorphism the only group of order p 1s Zp. Proof

By 2.21, a group of prime order is cyclic. >

Some cartesian products of cyclic groups are cyclic : 4.8 Theorem

Z,,xZ, ~ Znn if and only tf

m,n are coprime.

GROUP

MORPHISMS

55

Proof

It is clear that |Z, x Z,| =

that Z,, x Z, ~ Zmn

|Zmn|.

By 4.4, we have

if and only if Z,, x Z, is cyclic, ie.

and only if (1,1) is a generator (of order mn).

if

Suppose that

(1,1) has order k. Then we have

(Ae ees: Fy] Ne Ne k

el) ee k

H(t) + © 2 + (1)1)=(0;0) a es k

and so m|k and n|k. The smallest integer k with these properties is lm(m,n). But Icm(m, n) hcef(m, n) = mn, and so we

deduce

that

(1,1) is of order mn

if and only if

hef(m,n) = 1.9 Example

Z; x Z45, Lad Zs x Zs, x Zo of Z25 x Zp. In fact, the

first two are isomorphic by 4.8, and the second and third are not isomorphic since no element of Zs x Zs x Zo has order greater

than lcm(5,9) = 45 whereas by 4.8 the third is isomorphic to Z225 so is cyclic of order 225.

It is possible to obtain a structure theorem for any finite abelian group G in terms of a cartesian product of cyclic groups each of prime power order (i.e. of order p* where p is a prime dividing |G|). However, this is too complicated to give at this stage.

In Chapter One we described, up to isomorphism, all groups of order at most 4. By 4.7 we see that, up to isomorphism, the only group of order 5 is Zs. Using 4.4 we can now describe all groups of order 6. We have seen in Chapter One that the symmetric group $3 is of order 6, as is Z,. Up to isomorphism, this completes the list : 4.9 Theorem

Up to isomorphism,

the only groups of order 6

are S3 and Ze.

Proof Let G be of order 6. If G is cyclic then by 4.4 we have G ~ Z,. Suppose then that G is not cyclic. By 2.19, every element of G other than 1 has order 2 or 3. Suppose that every element (other than 1) has order 2. If

a,b € G are two such elements then the (distinct) elements

56

1,

.

VOLUME

3: ABSTRACT

ALGEBRA

a,b,ab form a subgroup of G; for a? = b? = (ab)? = 1 give

a7 =.apbs =

and tab= Definition By a permutation group we mean a group of bijections on some set. The importance of permutation groups is shown by the following result.

4.10 Theorem

[Cayley] Every group is isomorphic to a sub-

group of a permutation group. Proof

Given a group G, consider the set F={d,;

9€G}

GROUP

MORPHISMS

57

of all the left translations A, : G + G described by A(z) = gz. By 1.8 and the cancellation laws we see that A, is a bijection on G for every g € G. Now for all g,h,z € G we have Ag{An(z)] = Ag(hz) = ghar = Xgh(z)

and so Ag oAn = Agn. Thus F is a closed subset, and hence a subsemigroup, of the group (Bij(G, G), 0). Clearly, 41 = idg is the identity element of F, and since Ag oAg-1 = Aga-1 = Ai the inverse of Ag is Aq-1. Thus we see that (F,0) is a subgroup of

the group (Bij(G, G), o). Consider now the mapping f : (G, -) — (F,o) given by f(g) = Aq.

For all g,h € G we have f (gh) = Agh = Ag 0 An = F (9) © f(A) and so f is a morphism. By the definition of F, this mapping f is surjective. Moreover,

geKerf

rA,=ide

(Vz EG) gr=2t —

g=l1e

and so Ker f = {1c} whence, by 4.3, f is also a monomorphism. Thus f is an isomorphism and so G ~ F. > 4.11 Corollary Every group group of realn Xn matrices. Proof Suppose that G is of left translation 4,, which is G, induces a permutation on space over IR and so defines a

of order n 13 tsomorphic

to a

order n. Then for every g € G the a permutation on the elements of a basis of an n-dimensional vector linear transformation T, and hence

a matrix. The result follows from the fact that Mat(T, o T,) = Mat T, - Mat T,. > Example

Consider the group Z3 whose Cayley table is

58

.

VOLUME

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ALGEBRA

The permutations associated with the left translations Ag, A1, A2 are respectively : O

1

a 27’

(| 1 °) TP Cros?

( 1 2g

aH ei)"

These give rise to linear transformations on the vector space IR°, in fact those that are defined respectively by the following

effects on a basis {€0, €1, €2} :

fi =id fe(eo) = e1 fa(e1) —="€2

fo(e2) = €0 fs(eo) = e2 fs(e1) = e0 fs(e2) = e1 The respective matrices are then

Iz,

001 A=]1 00], O 1

010 B=l/0 071 daDecd

and, as can readily be seen, they multiply according to the table

which is the same as that for Zs.

CHAPTER

FIVE

Isomorphism theorems

We now describe an important isomorphism that is induced by a group morphism.

5.1 Theorem

[First isomorphism theorem] If f : G — H is a

group morphism then

Im f ~ G/ Ker f. Proof Let K = Kerf. We observe first that if sK = yK then from z = zinc € tK = yK we have z = yk for some k € K = Kerf whence

f(z) = (yk) = f(y) f(*) = f(y). This means

that we can define a mapping

3 : G/K

—

Imf

by the prescription 0(zK) = f(z). This mapping is a group morphism, for we have

0(xK -yK) = 0(zyK) = f(zy) = f(z) f(y) = 0(2K) 9(yK). It is clear from the definition that 0 is surjective.

It is also

injective; for if 3#(zK) = ¥(yK) then f(z) = f(y) whence f(zy~*) = f(z)[f(y)]-* = 1 and so zy~! € Ker f = K which gives zK = yK. Thus we see that # is an isomorphism. Example

Consider the mapping

f : Z —

Z,, described by

f(m) = m* where m* € Z,, is such that m* = m (mod n). We have seen above that f is a group morphism.

Clearly, f is sur-

jective. Now Ker f = {me Z; m* = 0} = nZ. Consequently, by 5.1, we have Z/nZ ~ Z,,.

60

.

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Example Let T = {z € € ; |z| = 1} be the circle group. Consider the mapping 8 : IR > T given by 8(z) = e?”*. This mapping is surjective, and is a group morphism since o(z pe y) = e2t(zty)t

=

e2hzi 2ryt _

9(z)9(y).

Now Ker? = {z € IR; e?"** = 1} = Z and so, by 5.1, we have IR/Z~ T. Example If G is an abelian group consider the set H of elements of G that are of finite order. If z,y € H, sayz” =1=y”", then (ey. 2).

and so zy_

=

sca Goh Mi —= Cava at Coad ce

=]

€ H. Thus H is a subgroup of G. Taking G to be

the circle group T = {z € C; |z| = 1}, we deduce from (e°)" =1

cosnd +isinn’ = 1

nv0=2mr

8=2na

(mMEZ)

(a€ Q)

that H = {e?™* ; a € Q}. Consider the mapping f : Q — H described by f(a) = e?"%*. Clearly, f is surjective and is a morphism. Now Ker f = {a € Q; e?7% = 1} = Z and so, by 5.1, we have Q/Z ~ H. Example

Permutations multiply according to the scheme even

odd

even

| even

odd

odd

odd

even

and so the mapping f : S, — Zp» described by 0

if o is even;

fo) ={9

if o is odd,

is a morphism with Im f = Zz and Ker f = An, the alternating group. By 5.1 therefore, we have S,,/A, ~ Z2 and consequently

|S, /An| = 2. But |S,/A,| is the index |S, : A,| of An in Sp. Thus, by Lagrange’s theorem, we have |S,| = 2|An|. This is a powerful, yet elegant, way to show that |A,| = inl.

ISOMORPHISM

Example

THEOREMS

61

If G is a group and g € G then we have seen above

that the mapping ¢, : G — G described by ¢,(z) = gzg™! is an isomorphism. These mappings are called the inner automorphisms on G. Now for all g,h € G we have

Salsn(z)] = ¢g(hth-*) = ghah-*g™* = ¢g,(z) and so ¢g 0 ¢, = ¢g,. Thus the set I(G) of inner automorphisms on G is a semigroup under composition. Clearly, ¢; = idg is the

identity element of J(G), and the inverse of ¢g is ¢,-1 € I(G). Thus (J(G), 0) is a group. Consider the mapping ¢ : G > I(G) described by ¢(g) = ¢,. That ¢ is a morphism follows from

$(gh) = son = Sg Osh = $(g) oS (A), and clearly ¢ is surjective by definition. Ker¢. We have

ge Ker¢

Let us now identify

¢.= ¢(9) = ide

(VzEG)

grgt=z

(VzEG)

grt=24,

and so it follows that Ker¢ = Z(G), the centre of G. It now follows by 5.1 that I(G) ~ G/Z(G). Example We have noted that f : GL(n,IR) — IR* defined by f(A) = det A is a group morphism. It is in fact an epimorphism since for every non-zero real number z the n x n diagonal matrix D, whose diagonal entries are all equal to 1, except the leading entry which is z, is such that det D, = z. Now

Ker f= {Ae GL(n, IR) ; det A= 1} = SL(n, IR)

and so, by 5.1, we have GL(n, IR) /SL(n, IR) ~ IR*. Example

If G,H

are groups then the projection map pg

G x H — G given by pc(g,h) = g is clearly an epimorphism. Now

Ker pg = {(9,h) ; 9= 1c} = {(1e,h); hEH} = {1g} x H and so, by 5.1, we have (G x H)/({1c} x H) ~G. We shall now use the first isomorphism theorem to describe the subgroups of a quotient group. For this purpose, we establish the following result.

:

62

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ABSTRACT

5.2 Theorem Let G,H be groups and let epimorphism. If K is a subgroup of H let

f: G —

ALGEBRA

H be an

fT (K) ={ceEG; f(z) € K}. Then f~ (K) ts a subgroup of G, Ker f ts a normal subgroup of 7 (one K = f*(K)/ Ker f.

Proof Note first that f~ (K) # @ since f(ic) = 1y € K. If now z,y © f~ (K) then f(z) € K and f{y) € K. Since K isa subgroup we also have [f(y)]~* € K and hence

f(zy~*) = f(z) f(y*) = F(a)[f(y)-*

EK

and consequently zy! € f~ (K). Thus f* (K) is a subgroup of G.

For every z € Kerf we have f(z) = ly € K andsoze f (K), whence Ker f C f*(K). Since Kerf is a normal subgroup of G, it is then also a normal subgroup of f*(K).

Consider the mapping ¥x : f~ (K) > K given by 0x(z) = f(z), ie. Ux is the restriction to the subgroup f~ (K) of the group morphism f. Clearly, }x is a morphism. Now by hy-

pothesis, f is surjective. Thus every k € K is of the form f(z) for some z € G, whence it follows that 3% is surjective. Also,

2éKerdy

f(z) =0x(z) >in, =>

2.6 Kerf

and so Ker¥, = Ker f. Applying the first isomorphism theorem, we obtain

K =Im0x ~ f7(K)/Ker8x = f7(K)/Kerf.

©

5.3 Corollary (Correspondence theorem] Jf N is a normal subgroup of G then every subgroup of G/N is of the form A/N where A ws a subgroup ofG such that N C A. Moreover, every normal subgroup of G/N 1s of this form where A ts a normal subgroup of G such that

NCA.

ISOMORPHISM

THEOREMS

63

Proof In 5.2 take H = G/N and f = by. If K is asubgroup of G/N then A = hy (K) is a subgroup of G containing the normal subgroup Kerhy = N, and K ~ A/N. Indeed K = A/N since for every a € A we have aN = hy(a) € K, so that A/N C K; and for every k € K we have k = gN for some g € G where, in

fact, g € A since hw (g) = gN EK. For the last statement it suffices to note that if K is a normal

subgroup of G/N then A = 4(K) is a normal subgroup of G. In fact, if a € A then for all g € G we have

hw(gag-') =gN-aN-g 'NEK since aN = 4n(a) € K and K is normal in G/N. Thus gag"! € hy (K) = A and so A is normal. }

Example

The subgroups of Z/nZ are of the form mZ/nZ

where m divides n. In fact, the subgroups of Z are necessarily of the form mZ for some m € Z, and since n = nl € mZ we have that n is a multiple of m. We can also consider quotient groups of quotient groups, i.e. given a normal subgroup H of G we can form the quotient group

G/H and then, given a normal subgroup N/H of G/H, form the quotient group (G/H)/(N/#). 5.4 Theorem

[Second isomorphism theorem] If N and H are

normal subgroups of G with

H C N then N/H

is a normal

subgroup of G/H and

(G/H)/(N/H) ~ G/N. Proof By 5.3, every normal subgroup of G/H is of the form N/H where N is a normal subgroup of G containing H, so we

can certainly form the quotient group (G/H)/(N/#H). We now observe that if zH = yH then y-'x € H C N and so zN = yN. This means that we can define a mapping 0 :

G/H — G/N by the prescription 0(zH) =

zN=0(cH)=N

=

ceNn,

64

VOLUME

3: ABSTRACT

and so Ker9 = {zH ; t € N} = N/H.

ALGEBRA

Applying the first

isomorphism theorem, we conclude that

G/N = Im8 ~ (G/H)/ Ker 8 = (G/H)/(G/N). Example

©

If m|n then (Z/nZ)/(mZ/nZ) ~ Z/mZ.

We end our present discussion of groups by establishing another important isomorphism theorem. For this purpose, we introduce the following notation. If H,K are subgroups of a group G then we define the complez HK to be the set product

HK = {hk; hE H, ke K}. Since 1g belongs to every and K C HK. Similarly 2.5, both HK and KH generated by HUK (i.e.

H and K). In general,

subgroup of G it is clear that H C HK we have H C KH and K C KH. By are contained in the subgroup H V K the smallest subgroup containing both

HK and KAZ

are distinct from H v K.

For example, in the Cayley table for S3 given on page 10 we

have that both H = {e,a} and K = {e,b} are subgroups but HK = {e,a,6, ab = p} is not since it is not closed. However, we have the following situation : 5.5 Theorem I[f H,K are subgroups of G and if one of them 1s normal then

HK=KH=HvkK. Proof Suppose that H is normal. Then given h,k, hok2 € HK we have

€ HK and

hy ki (hoke)~! = hykykythz?. But since H is normal we have kikz*hz* € kik,> +H = Hkykz? so there exists hg € H such that kik, 'hz' = hgk,k,*. Consequently we see that hyky(hok2)~*

—

hyhgkykz*

= HK

and so HK is a subgroup of G, whence it follows that HK = HV K. Similarly we have KH is a subgroup, so KH = HVK.>

ISOMORPHISM

THEOREMS

5.6 Theorem

65

[Third isomorphism theorem] Let H and K be

subgroups of G with H normal. and

K/(H

Then

HONK is normal in K

K) ~ (HV K)/H = KH/H.

Proof We observe first that (H V K)/H = KH/H is the subgroup of G/H consisting of all cosets khH where kh € KH.

Since khH = kH, it follows that (H V K)/H consists precisely of all cosets of H of the form kH where k € K. We can there-

fore define a mapping f : K — (H Vv K)/H by the prescription f(k) = KH. Clearly, f is a group morphism and is surjective. Now Ke Kerf kH=H > a; X* and > 6; X* to be equalif n = mand a; = b; for every 1=0

#=0

2. We add and multiply polynomials in the usual way. The set

R[X] of all such polynomials is a commutative ring with a 1.

RINGS,

INTEGRAL

DOMAINS,

AND

FIELDS

69

Example Just as we can consider the set € of complex numbers as the set of objects of the form a+ bi where a,b € IR with 1? = —1, so we can consider the set IH of quaternions as the set of Bee of the form a + bt + cj + dk where a,b, c,d € IR with

1? = 9? = k? = -1 and ij = —ji= k (so that we then also have 7k= Ws =tand ki = -i1k = 7). We define a+ bi + cj + dk and a’+b't+c'j+d'k to be equal ifa=a',b=b',c=c',d=d'. We can define an addition on IH by

(a+ bi +cj + dk) + (a’ + bt + c'7 + d’k)

=(a+a)+(b4+0')t+(c+e')j7+(d+d')k, and a multiplication on IH by pretending that we are dealing with real numbers, i.e. we multiply out the product of two such sums as though they were real numbers and use the given relations to express the product in the form a+ 621 + 47 + 6k. The reader can verify that in fact

(a+ bi+cj+dk)(a’ + b'%+c'j + dk) = (aa’ — bb! — cc’ — dd’) + (ab’ + ba’ + cd’ — de')i + (ac’ + ca’ + db’ — bd’)j + (ad’ + da! + be’ — cb')k. Then (IH, +, -) is a non-commutative ring with 0+ 0: + 07 + 0k as zero element and 1+ 02 + 07 + Ok as identity element.

6.1 Theorem

[/f (R,+, -) ts a ring then

(1) (¥VzeE R) 20=0=0z; (2) (Vz,yER) z(-y) = —zy = (—-z)y;

(3) (Vz,yER)

(—2)(-y) = zy.

Proof (1) We have 20+0 = 20 = z(0+0) = 20+20 and so, by the cancellation law, 0 = 20. Similarly,

0 = Oz for every x € R.

(2) We have zy + z(—y) = z[y + (—y)] = 20 = 0 and so z(—y) = —zy. Similarly, (—z)y = —zy for all z,ye€ R.

(3) By (2), (-2)(—y) = —[(-2)y] = —[-zy] = zy. ©

If R is a ring and z € R then, for n € Z, we define nz as follows : if n > 0 then nZ=Z+z+:::'+2, ———— n

if n= 0 then nz = 0, and if n < 0 then nz = —|n|z. Then the following result is also basic and will be used without

further reference.

70

VOLUME

6.2 Theorem

3: ABSTRACT

ALGEBRA

[If R is a ring then

(1) (Vm,nEZ)(VrER) (2) (Vm,n€Z)(VzE R)

(mM4+n)z=mz+nz; (mn)z = m(nz) = n(mz);

(3) (Vm e Z)(Vz,y ER)

m(zy) = (mz)y = z(my).

Proof It is clear that (1) and (2) hold for all m,n € IN, and by

definition we have, for m > 0, (—m)z = —(mz). It follows that (1) and (2) hold for all m,n € Z. As for (3), we note first that the result is trivial if m = 0. Suppose, by way of induction, that

it holds for m € IN. Then we have

(m+ 1)zy = m(zy) + zy ” {z(my) + zy = 2(my + y) = 2[(m+ 1)y];

(mz)y + zy = (mat z)y = [(m+

1)zly.

Thus (3) holds for all m € IN by induction. Taking n = —m in

(1), we obtain (Vm € IN)(Vz € R) (—m)z = —(mz). Thus, if m > 0, we have

(—m)(zy) = —[m(zy)] = —[(mz)y] = [-(mz)y] = [(-m)z]y, i.e. the equality m(zy) = (mz)y holds for all m € Z. Similarly, so does the equality m(zy) = z(my). > Example

Consider the abelian group (Q/Z,+).

We can use

6.2 to prove that only the trivial multiplication yields a ring

(Q/Z, +, -). In fact, let

[2]=2+z. q

q

Suppose that - is a multiplication on Q/Z which makes this abelian group into a ring. Given n € Z, consider the expression

*(lr] Lal): On the one hand this is n | . +] = B . Br and on the n n n n AO. nit 1 1 other it is || n | = || -0 = 0. Thus we see that, for n n n every ne Z,

EE

RINGS,

INTEGRAL

DOMAINS,

AND

FIELDS

Ok

It follows that for all & - € Q we have q

and so the only ring multiplication possible on (Q/Z, +) is the trivial multiplication. Another useful elementary result is the following. 6.3 Theorem [Binomial theorem] Jf R is a ring and z,yER are such that zy = yz then, for all integers n > 2, n-1

n

m=1

\™

Proof The proof is by induction. chor point n = 2 since

The result holds for the an-

(z+y)?=(z+y)(z+y) =2(z+y) + (z+) =2?+ayt+yrt+y?

= 274+ 2ryt+y’. For the inductive step, suppose that the result holds for n. Then

(x) (zty)"** = (z+ y)(c+y)" =2(zt+y)" + y(z+y)”. Now a(x + y)” =

n—1

htt +4 >

(2)atm

m=1

=f pagel

ig os|

at),m=1

n \™

“yn—m+1,m y".

ay

72

VOLUME

\

ABSTRACT

3:

ALGEBRA

Also, since z commutes with y we have that y commutes with z+ y; for y(z+y)=yrty?

=syty?

=(rc+y)y.

Consequently y(z + y)” =e zy

ll

n—-1 sY

3 Mes

3 ———

n ( jeer.

| 3 8} Seed

As

3 te

ty

A}

of

nd

me

Substituting in (x) and using the fact that

(2) savast=(aa we obtain the required result for n+ 1. Note that if R has a 1 then defining z° = 1 = y° we can write the above formula as

crane8("erm

We now note that in a ring R it 1s possible to have ab = 0

soutgpuitalcseteal with a

#0 and b # 0. For example, in the ring Mat22(IR) we

When this happens, we say that a,b are zero divisors. Rings without zero divisors are characterized as follows : 6.4 Theorem equivalent :

[f R 1s a ring then the following statements are

(1) R has no zero divisors;

(2) R\ {0} ts closed under multiplication; (3) every element of R \ {0} ts right cancellable in the sense that 1f ax = br with

z #0 thena=b;

(4) every element of R\ {0} ts left cancellable in the sense that if za = zh withz #0 thena=b.

RINGS,

INTEGRAL

DOMAINS,

AND

FIELDS

73

Proof The equivalence of (1) and (2) is clear. (1) = (3) : If az = bz with z # 0 then (a — b)z = 0 and z0, by (1),we have a—b=Osoa=b. (3) => (2) : Suppose that a # 0 and b # O. If ab = O then ab = 0b and, by (3), we have the contradiction a = 0. Hence ab #0.

The proofs of (1) = (4) and of (4) > (2) are similar. } Definition By an integral domain we shall mean a ring with a 1 in which there are no zero divisors. Example

(Z,+, -) is a commutative integral domain.

Example

If R is a commutative

the ring R[X].

integral domain

In fact, if f = ag

then so is

+ayX + ---+a,X"

and

g = bo +61X+ ---+b,X™ are elements of R[X] with a, #0 and b,, # 0 (i.e. the degrees of f and g are n and m respectively) then by 6.4 we have a,b,, # 0. Consequently, fg 4 0 whenever

f #0 and g £0

so, again by 6.4, R[X] is an integral domain.

If R is an integral domain then it is clear from 6.4 that the

set R \ {0} of non-zero elements in R forms a (multiplicative) subsemigroup in which the cancellation laws hold. It is natural,

therefore, to consider those integral domains R for which R\ {0} is a group.

field.

Such a structure is called a division ring (or skew

Example The ring IH of quaternions is a division ring. In fact the reader can easily verify that the product of two non-zero elements of IH is non-zero, so IH is an integral domain. Also, by analogy with complex numbers, every quaternion has a conjugate, that of g= a+bi+c7+dk being g= a—ht—cj—ck. The reader can also verify that ifg=a+bhi+cj+dk#0 then

qq = 9q = a? — (bi +07 + dk)? =a? +b? +c? +d? £0. It follows that q~1 exists and is given by

ay iP

1

1

2

@4+R4e24+a7

If we strengthen the axioms still further, we can consider the notion of a commutative division ring. This is a very important algebraic structure and is called a field. Thus a field is an

algebraic structure (F,+, -) such that (1) (#,+) is an abelian group;

74

VOLUME

3: ABSTRACT

ALGEBRA

(2) (F \ {0}, -) is an abelian group; (3) - is distributive over +. Example

(Q,+, -),(IR,+, -),(€,+, -) are fields.

Example

Consider the subset F of IR defined by

F = {a+ bV3; a,b € Q}. It is easy to see that F is a commutative ring with a 1. The

zero of F is 0 = 0+ 0V3 and the identity is 1 = 1+ 0/3.

If

now a+ bV3 € F \ {0} then a,6 are not both zero in Q, and

a? — 3b # 0 (otherwise \/3 would be rational). Consequently (a + bV3)7}

1

a—

bV3

a+ bV/3 a — b/3 a b

a2 —3b2

a2 — 3b2

V3 € F \ {0},

and hence we see that (F \ {0}, -) is a group. Thus (F,+, -) is a field. In contrast, note that the set

I= {a+bvV3; abe Z} is a commutative integral domain that is not a field. For example, the element /3 € J has no inverse in J , for there is no p+qvV3 €I such that V3(p + qv3) = 1. Indeed,

1 = V3(p + qV3) = pv3 + 3q requires p = 0,3q = 1 which contradicts the fact that g € Z. Example

(Zs3,+3, -3) is a field. In fact, the Cayley tables are

and it suffices to check that (Z, \ {0}, -3) is an abelian group, which is clear.

In fact, as we shall see in due course, the ring

(Z,, +p, ‘p) is a field if and only if p is prime.

RINGS,

INTEGRAL

DOMAINS,

AND

FIELDS

75

Definition By a subring of a ring R we shall mean a nonempty subset S of R that is closed under both addition and multiplication and is also a ring with respect to the ‘inherited’ addition and multiplication. It is clear that the distributive laws are inherited by subsets that are closed under both addition and multiplication. Consequently, we have : 6.5 Theorem A non-empty subset S of a ring R 1s a subring of R tf and only if

(Vz,y ES)

z-yE€S

and

szyE€S.

Example

J = {a+ 6/3; a,b € Z} is a subring of IR.

Example

If FR is aring then its centre is

Z(R) ={ceER;

(Vr € R) ar =1rz}.

Clearly, 0 € Z(R) and so Z(R) #9. Also, if z,y € Z(R) then (2 —y)r = a2r—yr=rz—ry=r(z—y), so z—y € Z(R); and zyr = zry = rzy, so zy € Z(R). Thus Z(R) is a subring of R. Clearly, R is commutative if and only if R = Z(R). Example

The subset R of Matzx2(IR) consisting of the matri-

ces of the form

a

b/2

—bV/2

a

is a subring of the ring Matz x2(IR). In fact, if M and WN are of this form then so are M — N and MN.

Example

The subset H of Mat2,2(€) consisting of the matri-

ces of the form ath —c+td

c+td a-—lht

is a subring of the ring Mat2,2(C). Example The subset of the polynomial ring IR[X] consisting of all polynomials f = ap + a1X + ---+4,X” whose constant

term do is 0 is a subring of IR[X].

76

:

Example

VOLUME

3: ABSTRACT

ALGEBRA

If R is a ring then e € R is said to be 1dempotent if

e? = e. For example, in the ring Mat2x2(IR) the matrix

| is idempotent.

If e € R is idempotent then the subset eRe =

{ere ; r € R} is a subring of R with an identity; for ere — ese = e(r — s)e € eRe, ere -ese € eRe, e =

eee € eRe, and

e.ere = ere'= ere.c. Example If R is a ring then z € R is said to be nilpotent if xz” = 0 for some positive integer n. For example, in the ring

Mats x3(IR) the matrix

A=

1 0 © oo 0

ore

is nilpotent, for A? = 0. If R is commutative and has a 1 then

the set N(R) of nilpotent elements of R is a subring. 0 € N(R), and if 2" = 0 = y™ then by 6.3 we have

In fact

(2—yyrim = "SP (PE) artery) =o, r=0

since each summand with r < m has a factor z” = 0 and each summand with r > m has a factor y” = 0. Also, (zy)"” =

(z")™(y™)" = 0. 6.6 Theorem A subset S ofZ 1s a subring of Z tf and only if at 1s of the form nZ for some ne€ Z. Proof By 2.6 every subgroup of Z is of the form nZ. If p,q € nZ then p = na,q = nb for some a,b € Z and hence pq = n7ab € nZ. Consequently, by 6.5, nZZ is a subring of the ring Z. Since every subring is by definition a subgroup, it follows that every subring of Z is of this form. >

CHAPTER

SEVEN

Quotient rings and ring morphisms

If R is aring and S is a subring of R then, since the addition in R is commutative, S is a normal subgroup and so we can form the quotient group R/S. This consists of the cosets of S in R, namely the classes modulo the equivalence relation E's given by

z=y(Es) We shall now

consider when

z-yesS.

this equivalence relation is left

[right] compatible with the multiplication in R. For this purpose, we note that

z=0(Es) and that, for every

=>

zES

re R,

rz=r0=O0(Es)

=>

rzeS.

Thus, if Es is to be left compatible with multiplication we require S to satisfy the property

zES=>(VreR)

rzeS.

Similarly, for S to be right compatible with multiplication we require S to satisfy the property

zES=>(VvVreR) zreS. These observations lead us to the following notions.

78

VOLUME

Definition

3: ABSTRACT

ALGEBRA

If X,Y are non-empty subsets of a ring R let

XY ={zy;

ce X, ye Y}.

Then a non-empty subset J of R is called a left tdeal of R if it is an additive subgroup of R such that RI C IJ, a right tdeal of R if it is an additive subgroup of R such that JR C I, and a two-sided ideal of R if it is both a left ideal and a right ideal. We shall often use the term ideal to mean a two-sided ideal. Example

The set M of matrices of the form

Hg

where a,6 € IR is clearly a subgroup of Mat2x2(IR). Since w

«|10

|: Be

a

0

wa+a2b

slate wat

lem

we see that M is a left ideal of the ring Mat2y(IR). It is readily seen that M is not a right ideal. Example

The set N of matrices of the form a 0

b 0

is a right ideal, but not a left ideal, of Mat2x2(IR).

Example The only two-sided ideals of Matnxn(IR) are {0} and Matnxn(IR) itself. In fact, suppose that J is a two-sided ideal with J # {0}. Let A = [a,;] € J be such that a,, # 0. Let B = diag{a;,',...,a7,'} and let E;; denote the matrix which has a 1 in the (t,7)-th position and 0 elsewhere. Then since J is a two-sided ideal we have E,,-BAE,, € J, where BA has a 1 in the (r, s)-th position. Observing that E,,X retains the r-th row of X and reduces all other rows to zero, and that XE,, retains the s-th column of X and reduces all other columns to zero, we deduce that E,,BAE,, = E,, and hence that E,, € J for all r,s. Consequently, In = Eqy + +--+ Enn

and hence, for every X, Math xn(IR).

€ J

X = XJ, € J. Thus J is the whole of

QUOTIENT

RINGS

AND

RING

MORPHISMS

79

Example A real nxn matrix [a,;]| is upper triangularifa;; = 0 whenever 2 > 7. The set T;,(IR) of such matrices is a ring, and the set T* (IR) of upper triangular matrices with a;; = 0 for every

i is an ideal of T,(IR). In fact, if A € T*(IR) and B € T,(IR) then n

[AB]is = » Qinbes = > aindas = 0, k>t

so AB € T*(IR). Similarly, BA € T*(IR). Example

The set J of continuous functions f : IR — IR such

that f(0) = 0 is an ideal of the ring (Map(IR,IR),+, -). In fact, if f € IJ then for every g : IR — IR we have (fg)(0) = f(0)g9(0) = 0g(0) = 0, so fge I. 7.1 Theorem nZ 1s an ideal of the ringZ, and every ideal of Z is of this form. Proof It is clear that nZ is an ideal and, by 6.6, every ideal of Z is of this form. > As the following result shows, the ideal structure of a division ring (and hence that of a field) is particularly simple. 7.2 Theorem

A ring R with a1 is a division ring if and only

if it has no left or right ideals other than {0} and R. Proof Suppose that FR is a division ring and that J of R with I # {0}. Given x # 0 in J we note that and so, for every y € R, we have y= yl = yx 42 € and hence R = J. Similarly, if J is a non-zero right

is an ideal 2! exists yx HI CI ideal then

R=dJ. Conversely, suppose that {0} and R are the only left or right

ideals of R. Given a € R \ {0}, the set Ra = {ra; r € R} is a left ideal of R which is distinct from {0} since a = la € Ra. By the hypothesis we therefore have Ra = R and so there exists y € R with ya = 1. Thus a has a left inverse. Arguing similarly

with the right ideal aR = {ar ; r € R}, we see that a also has a right inverse. It follows by 1.3 that a is invertible. Consequently

R \ {0} is a group and so R is a division ring. > In Chapter Two we showed how quotient groups were related to subgroups of the group. We can, of course, extend these results to rings.

80

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3: ABSTRACT

ALGEBRA

For an ideal J of R we shall denote the set of cosets of J as

usual by R/I. The elements of R/J can of course be described

by

zt+I={x+i1;7€

J}.

The following result can be established along the same lines as 3.3. 7.8 Theorem Let I be an ideal of the ring R. Then on the set of cosets ofI the prescriptions

(c+ D+(ytD=(et+y)+l

(c+ DI(y+D=sytl

define laws of composttion with respect to which the set of cosets of I 1s a ring. >

By 7.3, R/I is a ring which we call the quotient ring of R by the (two-sided) ideal J. We can also define a ring congruence to be an equivalence relation EF on a ring R that is compatible with addition and compatible on both the left and the right with multiplication. The following result is now immediate. 7.4 Theorem Let E be a congruence on the ring R. the set R/E of E-classes the prescriptions

[z]e + [yle = [2 + ylz, define a law of composition ring. > Example

Then on

[zlel[yle = [zyle

with respect to which R/E

is a

nZ is an ideal of Z so we can form the quotient ring

Z/nZ. This quotient ring has an identity, namely 1+ nZ. Example 4Z is an ideal of 2Z so we can form the quotient ring 2Z/4Z. This quotient ring has only two elements, namely

0+4Z and 2+ 4Z. If we write these as [0] and [2] respectively then the Cayley tables are

[0] [2] [0] [2] [2] }[2] [0]

[o} [2] [0] [0] [0] [2] [0] [0]

Thus 2Z/4Z is a trivial ring. We now tie in these notions with that of a structure-preserving mapping from one ring to another.

QUOTIENT

RINGS

AND

RING

MORPHISMS

81

Definition If R and S are rings then a mapping f : R > S is a ring morphism if

(Vz,yeR)

f(z+y)= f(z) + f(y) and f(zy) = f(2)f(y).

Note that since a ring morphism f : R — SQ is in particular a group morphism we have that f(0r) = Os. If, however, R and S have identity elements then it is not necessarily the case

that f(1r) = 1s; for example, the assignment z ++ Og is a ring morphism, called the zero morphism. A ring morphism f :R—S such that f(1pz) = 1s will be said to be 1-preserving. Example Let FR be a ring with a 1 and let f : Z — R be given by f(n) = nlp. Since nlp + mlz = (n+ m)1p and, by the distributive laws, nlp -mlpz = (nm)1p we see that f is a 1-preserving ring morphism.

Example

Given z € IR define ¢, : IR[X] — IR by

6z(@o + ayX + ---+a0,X")

=anp +ayz+

---+ay2”.

We shall prove later that ¢, is a 1-preserving ring morphism, called evaluation at x or the substitution morphism. Example The only ring morphisms f : Z — Z are the zero morphism and the identity morphism. In fact, if f 4 0 then a simple inductive argument shows that

(Vn € IN)

f(n) =nf(1).

If now n > 0 then we have f(—n) = —f(n) = —nf(1) and hence we see that

(vn€Z) —f(n) = nf(1).

But we also have f(n) = f(n1) = f(n)f(1) and so, by cancella-

tion, we deduce that f(1) = 1. Thus we have (Vn € Z) f(n)= n, so that f is the identity map. Example

br:

[If J is an ideal of the ring R then the natural mapping

R— R/I given by h;(r)

=r + J is a ring morphism.

As with groups, we shall use the terms monomorphism, eptmorphism and isomorphism when talking of a ring morphism that is respectively injective, surjective, and bijective.

3: ABSTRACT

VOLUME

82

Example

ALGEBRA

Consider the set S of matrices of the form a

b

Ma» = Mi 4 where a,b € IR. It is readily seen that Ma» — Mc,a = Ma-—c,b-a and that Ma,Mc,a = Mac-—bd,ad+be and so S is a subring of

Mat2x2(IR). The mapping 3: S — € defined by

8(Ma») =a+bi is then a ring morphism. It is in fact an isomorphism since it is both surjective and injective. Thus S ~ € as rings, and so S is a field. Note that the matrix that corresponds toz € € is Mo,

and M3, = M_10 = —Iz. Also, the subset {Mao ; a€ IR} isa subring of S that is isomorphic to the field IR. Concerning ring morphisms, we have the following important results. 7.5 Theorem If f: R—S ts a ring morphism then Imf 1s a subring of S and Kerf 1s an tdeal of R. Proof

Im f is a subring of S since it is a subgroup and

f(z) f(y) = f(zy) € Im. As for Kerf, this is a subgroup of R. If

ze Kerf andyER

then f(zy) = f(z) f(y) = Of(y) = 0 and so zy € Kerf. Thus Ker f is a right ideal. Similarly it is also a left ideal of R. > 7.6 Theorem A non-empty subset I of a ring R 1s an ideal of R tf and only tf it 1s the kernel of a ring morphism f: R—-S. Proof

If J is an ideal of R then we can form the quotient ring

R/I. The natural morphism 4; : R — R/I then has the property that Ker; = J. The converse follows from 7.5. > We can now extend the first isomorphism theorem for groups to a corresponding result for rings.

7.7 Theorem and f: phism

[First isomorphism theorem] If R,S are rings

R-— S 1s a ring morphism then there 1s a ring isomor-

Im f ~ R/ Kerf.

QUOTIENT

Proof

RINGS

AND

RING

MORPHISMS

83

Let Ker f= J. Then, as in 5.1, we can define a mapping

0: R/I — Im f by the prescription 8(r+ I) = f(r), and # is an isomorphism from the group (R/JI,+) to the group (Im f, +). It therefore suffices to show that 0 preserves multiplication, and this follows from 8[(r+I)(s+D)] = O(rs + I)

= f(rs)

= f(r)f(s) = (r+ I1)0(s+lI). Example We know that the morphism f : Z — Z, given by f(m) = m* where m* = m (mod n) is a group epimorphism with Ker f = nZ. It is in fact a ring morphism since, by the compatibility of ‘mod n’ with respect to multiplication, we have

(mp)* = mp = m*p*

(mod n)

and so (mp)* = m* -, p*, ie. f(mp) = f(m) -n f(p). Thus, by 7.7, we have the ring isomorphism

Z, =Imf ~ Z/ Ker f = Z/nZ. Example

Consider the ‘evaluation at 0’ map ¢ : IR[|X] — IR

given by

$o(ao + a,X+ -+-+a_,X") = ag. Clearly, this ring morphism is surjective. Now Ker ¢ consists of all polynomials of the form a,X + ---+a,X", ie. all polynomials whose constant term is 0. Equivalently, this is the ideal

(X) of all (polynomial) multiples of the monomial X. By 7.7 we then have the ring isomorphism

IR = Im ¢p~ IR[X]/ Ker¢o= IR[X]/(X). Example set E.

We have seen that (P(E), A,N) is a ring for every

Let F be a subset of E and let F’ be the complement

of F in E. Consider the mapping 9 : P(E) — P(F"’) defined

VOLUME

84

3 : ABSTRACT

ALGEBRA

by 0(X) = XN F’. Clearly, we have #(X NY) = #(X)N o(Y). Also,

O(X)AWY) =(KXN F'N(YNF)'jul((Xn FY aYaF' =(XNF'n(Y'UF)JU[((X’UF)NYNF' =(Xni ny ju ony ar) =((XNY')u(X'NY)| nF

= 9(XAY). Thus 9 is a ring morphism.

Now # is clearly surjective; and

Ker9 = {X €P(E); XN

F’=$9}=P(F).

It then follows by 7.7 that we have the ring isomorphism

P(F’) = Im8 ~ P(E£)/Kerd = P(£)/P(F). We have seen above that Z/nZ ~ Z,. Some important results of a number-theoretic nature can be derived from a consideration of the invertible elements in the ring Z,. 7.8 Theorem

r+nZ 1s invertible in the ring Z/nZ if and only

af hef{r,n} = 1. Proof If hcf{r,n} = 1 then there exist z, y € Zwith cr+yn = 1. It follows that (zr + yn) +nZ=1+nZ, ie. that

(c+ nZ)(r + nZ) + (y+ nZ)(n+nZ) =1+4 nZ. Since

n+ nZ=0+

nZ, this reduces to

(c+ nZ)(r+nZ) =1+nZ, whence r + nZ is invertible in Z/nZ. Conversely, suppose that r+nZ is invertible in the ring Z/nZ.

Then there exist

s+ nZ such that (r+ nZ)(r+nZ) =1+nZ,

so that cr +nZ=1-+nZ and consequently there exists y € Z such that zr —1=ny. We then deduce from zr — yn = 1 that any integer that divides both r and n must divide 1, ie. that

hcf{r,n} = 1.0

QUOTIENT

RINGS

AND

RING

MORPHISMS

:

85

7.9 Corollary The generators of the additive group Z, are the invertible elements in the ring Z,. Proof

This is immediate from 7.8 and 2.12. ©

7.10 Corollary Proof

Z, is a field if and only tf p ts a prime.

By 7.8 and the isomorphism Z/pZ ~ Z, we see that Z,

is a field (i.e. every non-zero element is invertible) if and only if hef{r,p} = 1 for r=1,...,p —1. Clearly, this is so if and only if p is a prime. > 7.11 Theorem p then

[Fermat] If p 1s a prime and n does not divide

n?—1 = 1(mod p). Proof By 7.10, Z, \ {0} is a multiplicative group of order p—1. By 2.20, we have a?~+ = 1 in Z, for alla # 0. Thus, ifn = n* (mod p) then the compatibility of ‘mod p’ gives n?-1 = (n*)?—-1 = 1 (mod p). > 7.12 Corollary Proof

Ifp is a prime then (Vn € Z) n?

=n (mod p).

This is immediate from 7.11 since if n is a multiple of p

then n = 0 (mod p), and in this case the result is trivial. > Just as the first isomorphism theorem for groups (5.1) can be used to establish the correspondence theorem for groups (5.3), so can the first isomorphism theorem for rings be used to establish a correspondence theorem for rings. For this purpose, we use the following ring-theoretic analogue of 5.2. 7.13 Theorem

morphism.

Let R,S be rings and f : R — S a ring epr-

If J is an ideal of S then f~(J) 13s an ideal of R

and

J ~ f~(J)/ Ker f. Proof By 5.2, f~ (J) is a subgroup of R. Now if z € f* (J) then f(z) € J so, since J is an ideal, for every

r € R we have

f(rz) = f(r)f(z) € J whence rz € f~ (J). Thus f~(J) is a left ideal. Similarly, it is also a right ideal. In view of 5.2 and 7.7, it suffices to prove that the mapping 8; : f~ (J) — J given

by 9;(z) = f(z) is a ring morphism; and this is clear from its definition. >

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86

3: ABSTRACT

ALGEBRA

7.14 Corollary [Correspondence theorem] If J 1s an ideal of R then every subring of R/J is of the form I/J where I 1s a subring of R with

Proof

J CI.

Adapt the proof of 5.3. >

7.15 Theorem [Second isomorphism theorem] If I and J are ideals of R with J CI then I/J is an ideal of R/J and (R/J)/(I/J) = R/I. Proof

Adapt the proof of 5.4. >

Naturally, we now seek an analogue of 5.6. For this purpose, we note that in order to establish 5.6 we introduced the complex

(or set product) HK = {hk ; h € H,k € K}. dealing here with additive analogous notion.

groups,

Since we are

we consider the following

Definition If J, J are ideals of a ring R then by the sum of I and J we mean the set

T+J={ft+j7;1E1,j7€ J}. By the additive version of 5.5, we have that J+J is asubgroup of R and is the smallest subgroup to contain both J and J. It is in fact also an ideal of R since r(t +7) =rit+trj E1+J and

(¢+9)r=ir+jrel+J.

It follows that

I+ J is the smallest

ideal to contain both J and J, i.e. that it is the ideal generated by IUJ. 7.16 Theorem ideals of R then

[Third isomorphism theorem]

Jf I and J are

I/(In J) ~ (I+ J)/J. Proof

Adapt the proof of 5.6. >

We now consider two types of ideal that are particularly important.

Definition By a mazimal ideal of a ring R we mean an ideal J with J # R such that there is no ideal J of Rwith I CJC R. By a prime ideal of R we mean an ideal J with I 4 R such that if zy€ I thenz el orye l.

QUOTIENT

RINGS

AND

RING

MORPHISMS

7.17 Theorem Let R be a commutative I be an tdeal of R. Then

87

ring with a1 and let

(1) I ts mazimal sf and only if R/I is a field; (2) I ts prime tf and only if R/I ts an integral domain. Proof (1) To say that J is maximal is equivalent to saying that the set of ideals that contain J is {J, R}. By 7.14, this is equivalent to saying that the quotient ring R/J has only two

ideals, namely {0+ J} and R/I. By 7.2, this is equivalent to R/I

being a division ring, i.e., since R is commutative, a field.

(2) R/J is an integral domain if and only if it has no zero divisors, i.e. if and only if

(c+ I)(yt+ J) =04+2=>24+71=0+l ory+J=0+1. This is equivalent to sye€Il=>zeloryel, i.e. to J being prime. > 7.18 Corollary In a commutative ring with a1 every mazimal ideal 1s prime. >

Example Let A be the set of continuous functions f : [0,1] > IR. Then (A,+, -) is a commutative ring with a 1. The mapping ¢: A— IR described by ¢(f) = f(0) is a ring epimorphism and so, by the first isomorphism theorem, we have

IR ~ Im¢ = A/Ker¢. Since IR is a field, we have by 7.17 that

Kers ={f€A; f(0) = 0} is a maximal ideal of A.

Example We have observed that IR[X]/(X) ~ IR. By 7.17, (X) is then a maximal ideal of IR[X]. Similarly, we have Z[X]/(X) ~ Z. By 7.17, (X) is then a prime ideal of Z[X] that is not a maximal ideal.

CHAPTER

EIGHT

Polynomials

In this Chapter we shall formalise the notion of a polynomial and apply some of the previous results to polynomial rings. The reader will be familiar with a polynomial as ‘something of the form .

f =a9+0,X + agX7 4+ ---+a,X" where do,..-.,@,, belong to the ring R’, and X is ‘something that is called an indeterminate’. In order to present the notion of a polynomial in a more explicit way, we require the notion of an infinite sequence of elements of a ring R, and by this we mean a mapping f : IN — R. Such a mapping is determined by

fo=f(0),

fi=f(1),

.--,

=f),

and so, for convenience, we shall agree to denote such a sequence

by Re =

fos fiscee sds; eee

Let us temporarily denote by S(R) the set of all such infinite sequences which have the property that only finitely many

of

the a; are non-zero. If f,g € S(R), say

f= Goliad

9 = (905 915-++5 Gi ++);

then we say that f and g are equal, and write f = g, if f; = g; for every 2. We define the sum and product of two such sequences

by (fo, fi, ---) + (90,91, ---) = (fo + 90, f1+ 91, ---); (fo, fi, se) é (90,91;

whereh; =

)> fig. j+k=1i

sat) =

(ho, hi, Sap

POLYNOMIALS

8.1 Theorem

89

(S(R),+, -) is a ring.

Proof It is clear that (S(R),+) is an abelian group with identity element 0 = (0,0, ...). To see that (S(R), -) is asemigroup, write f = (fo, fi, ...) as simply f = (f;) and observe that, for every 2,

((fa)hk=Jt+k=t XO (falsre=

Jt+k=t

(DO fog)hs ‘pt+q=j

oy

fp9qhk,

pt+qt+k=t

and similarly

[F(gh)k=ptr=i X folgh)e=

XL fo( D gahe) = ‘q+k=r

ptr=t

=

DL

fp9qhk-

pt+qt+k=1

Consequently, (fg)h = f (gh). Finally, the distributive laws hold since, for every 2, [f(9 + A)]: =

: a Ai(g + h)k = j+k=1

, ys. (f5 9% + fjhe) jt+k=1t

=

»: fio +

j+k=i

dU

fyh

j+k=1

= (f9)i + (fh) and so f(g+h) = fg+ fh, and similarly (g+h)f =gf+hf. Suppose now that R has a 1. If X = (0,1,0,0, ...) then it is readily seen that

ee

ne 1 (eee),

eee

— (0,0)... 0,10, us), —$=[{>—

ecw,

n

and that S(R) has an identity, namely X° = (1,0,0, ...). Con-

sequently we see that every element (ao, a1, a2, ...) of S(R) can be written as agX° +a,X+

a2X? +

-+++a,X”

VOLUME

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90

3 : ABSTRACT

ALGEBRA

where a, # 0 and a, = 0 for allk > n. We call X an andetermi-

nate, each element of S(R) a polynomial in the indeterminate X with coefficients in the ring R, and S(R) the ring of polynomials over R. With this notation involving an indeterminate, we shall

write R[X] instead of S(R). If nis the largest integer such that dn # Oin f = (ao, a1, ...) then we call a, the leading coefficient of f, and n the degree of f which we write deg f. The polynomials of degree 0 are then the non-zero constant polynomials. We say that f is monic if its leading coefficient is 1. It is common practice to write simply ag for ag X°. 8.2 Theorem Let R be a commutative ring with a1. Then for every z CR the substitution map ¢, : R[X] + R given by

¢z (ao + ayX + +--+ 4,X") = a9 +aiz+ ---+an2" 18 @ ring morphism. Proof If f=aj9+a,X+---+a,X” and g=bo +0;X+---+ by,X™ then the product fg is given by ai

fg9= bd

k=0'

in which at =

a; for

k

aj,__;b) X*

“j= 10)

0 < 7 < n and a; = O otherwise,

bi, = b;, for 0 < 7 < mand bi, = 0 otherwise.

and

That the group

morphism ¢, is a ring morphism follows from the identity

3 ¥ abjaitd ="S (55 afb) a*. nom

t=17=0

aig:

n+m,k k=0'

k

~j=0

To establish this, consider the rectangular array (with m+

1

rows and n+ 1 columns) whose (:,7)-th position is occupied by a;b;a' Tt? = a;2° -b;2’. The left-hand side of the above equality is the sum of all the elements in this array. Now extend the array to m+n-+ 1 rows and m+n+1 columns by defining a; = 6; = 0 fort >n+1 and j = m+1. Then the right-hand side of the above equality is the sum of all the elements in the second array, this being achieved

POLYNOMIALS

91

by summing the SW-NE diagonals. Because of the zero entries in the second array, these sums are clearly the same and the identity holds. > Our objective is to show that, with appropriate restrictions where necessary on the ring R, the polynomial ring R[X] satisfies properties similar to familiar properties of the ring Z. For this purpose, we recall that Z has the following properties :

(a) [Euclidean division]

If a,b € Z with b > O then there

exist g,r € Z with 0 < r < b such that a = bg +1; moreover, the quotient g and the remainder r are unique.

(6)

If a,b € Zand d = hcf{a,b} then there exist z,y E Z

such that az + by = d.

(y) [Unique factorization]

Every integer n > 2 can be writ-

ten as a product of powers of primes, this product being unique up to the order in which the factors are written. We begin, therefore, by showing that if R is a ring with a 1

then R[X] satisfies a property that is analogous to (a). This is simply a formalisation of the ‘long division’ process of elementary algebra.

8.3 Theorem

Let R be a ring with a1

and let f,g € R[X| with

g monic. Then there exist q,r € R(X] such that f = gq+r and either r = 0 or degr < degg. Proof If deg f < degg then taking g = 0 and r = f we obtain the result. Suppose that deg f > deg g. We establish the result in this case by induction on deg f — deg g. Suppose then that

f =antayX+---tamnX™,

g = botbiXt---+bp-1X"-* +X"

(where deg f = m and degg = n). If deg f — degg = 0 then m =n f =

and f can be written in the form gan

+ [(4@n—1 iJ REET

nNPda

We

ag

(ao ay boan)|

‘in which case the property holds, with q the constant polynomial an and r= (an-1 — bn—14n)X"* + «+++ (ao — boan).

3: ABSTRACT

VOLUME

92

For the inductive step, suppose

ALGEBRA

that the result holds for all

f*,9* © R[X] with O < degf* — degg* < deg f — degg. We have f = ga,,X™~" +h where h is (er =1 Stee)

x

sights Sa

aes

sa bGGm)A

“+++ ao,

and degh < m—1< m=degf. If degh < degg then no more proof is required; if, on the other hand, degh > degg then 0 < degh — degg.< degf —degg and so, by the induction hypothesis, r = 0 or degr < degg. Consequently

f= 9amnX">” + gk

h = gk + r where either

+r = o(amX” " +k) +r

as required. > 8.4 Corollary

If R is an integral domain then q,r are unique.

Proof Suppose also that f = gg* + r* where degr* < degg. Then we have (q—q*)g = r*—r. Suppose, by way of obtaining a contradiction, that g— q* # 0. Then, since g 0 by hypothesis

and since R[X] is in this case an integral domain, we have that r* —r #0. Thus

deg[(q — 9*)g] = deg(q — q*) + degg > degg > deg(r* — r), which contradicts the equality (q — q*)g = r* —r.

Hence we

must have gq — g* = 0, from which it follows that r*

-r=0.¢

8.5 Corollary If F is a field and f,g € F[X] with g #0 then there exist unique q,r € F[X] such that f = ggt+r withr =0 or degr < deg g. Proof Let the leading coefficient of g be a and let g* = a~1g. By 8.3 and 8.4 there exist unique g*,r such that f = g*q* +r with r = 0 or degr < degg. The result follows on taking g = at

g*. ros

We shall often refer to 8.5 as the division algorithm for polynomials over a field. It plays an important role in the study of

divisibility in F[X]. Before giving a few illustrative examples, we establish the following useful result.

POLYNOMIALS

8.6 Theorem

93

Let 3: R — S be a ring morphism.

Then the

induced mapping 0* : R[X] — S[X] given by 8* (ao +arX+ ---+a,X") = B(ap) + 9(a1)X+ - +++ 9(an)X” 18 @ ring morphism. Proof

If f=a9+a,X+

---+a,X" and g = bp +b)X4---+

bmX™ then it is clear that 3*(f + g) = 9*(f) + 9*(Q), so that 0* preserves sums. Now m2 we

3

(fg) = o| Gd Il >

I

3 we 3 com Il

I

(2 a,b; a4-sbs))X* x")

Me 8(ax—36;)) X*

It ° ° a S&S

3 af 3 Co

II

°

k

tie9(ax—3)9(bj)) X* 9(a;)X*) (> 9(b,) x?)

a w& Il

uf

x

o

ay Pog

= 9*(f)9*(9), and so 1 also preserves products. >

Example

Consider f,g € Z|X] given by

f =X*—3X°4 2X74 4X 1, g = X?— 2X43. By the usual long division process, the reader can verify that

f =9q+91r where q = X?-X—3 andr=X+8. Suppose now that we consider f,g as elements of Z;[X], ie. the coefficients are elements in the field Z;. The euclidean division in this case can be deduced from the above by using 8.6. If §5 : Z— Zz is

the morphism described by ¥5(n) = n* where n* = n(mod 5) then we simply apply the induced morphism ¥§ to the equality f =9q+1 to obtain f = gq* +1r* where g* = X7+4X+42 and r* = X+3.

Example

In Z[X] we have

X12 —14

= (X*+ 1)(X® — X® + X*° - 1).

Applying ¥% to this we obtain, in Z[X],

X12 +1 = (X° + 1)(X° + X°+ X° +1).

|

Thus we see that, in Z[X], the quotient on dividing X1? by

X°+1is X° + X® + X° +1 and the remainder is —1 = 1.

VOLUME

94

Example

3: ABSTRACT

ALGEBRA

Suppose that we wish to discover for which values of

n the polynomial X? +3 divides X°+9X?— 18X in Z,[X]. We

first perform euclidean division in Z[X]| and obtain

X® + 9X? —18X = (X? +3)(X° — 3X? +9) —9(X+ 3). By applying the morphism ¥%* to this, we see that X?-+3 divides

X® +9X? —18X in Z,[X] if and only if 9X + 27 = 0 in Z,[X]. This is the case if and only if 9 = 0 = 27(mod n), ie. if and only ifn =3 orn=9. We now derive some important consequences of 8.3.

8.7 Theorem [Remainder theorem] Let R be a ring with a 1. If f € R[X] and a € R then the remainder on dividing f by X —a is f(a). Proof By 8.3 we have f = (X —a)q+r where r = 0 or degr < deg(X —a) = 1, ie. degr =0. Thus r is a constant polynomial. Applying the substitution morphism

¢, to the equation f =

(X — a)q +r we obtain

f(a) = sa(F) = ¢a(X — a)¢a(q) + sa(r) = 0.q(a) + r(a) = r(a),

and so, since r is constant, the remainder is f(a). > 8.8 Corollary

X —a divides f if and only if f(a) =0.>

8.9 Corollary

If R is an integral domain and f € R[X] ts of

degree n then f has no more than n roots. Proof The proof is by induction. If deg f = 1, say f = agpt+a,X where a; # 0, let z, ybe roots of f. Then ag +a,z = 0 = ag+ayy gives a,;Zz = a,y and hence z = y. For the inductive step, suppose that the result holds for all polynomials of degree n. Let f have degree n+ 1. If f has no roots there is nothing to prove. Otherwise, let a be a root of f. Then f = (X —a)g where deg g = n. Since g has at most n roots, it follows that f has at most n+1.

POLYNOMIALS

95

Note the restriction that R be an integral domain in 8.9. If this is removed then the result is no longer true: For example,

in the ring Z,[X] of polynomials with coefficients in Z, (which is not an integral domain) the polynomial 2X is of degree 1 but has two roots, namely 0 and 2.

Example Consider X*+4 € Z;[X]. Substituting 1 for X gives 1+4=5=0(mod 5) and so, by 8.7, X —1 is a factor of X*+4 in Z5|X]. We have

X*4+4=(X—-1)(X°+ X?4+

X+1).

By the same reasoning, we see that X° + X? + X + 1 has the factor X + 1, and

X° 4X? 4.X41=(X+1)(X? +1) = (X + 1)(X— 2)(X+ 2). Thus, in Zs[X], we have

X*+4=(X—1)(X +1)(X — 2)(X +2) = (X+ 4)(X+ 1)(X+ 3)(X4 2). Example

Consider

f = X* + 3X? +2 € Z.[X]. This has no

roots in Zs. In fact, if f factorizes into a product of polynomials of lower degree then clearly it must have a factor of the the form

X — a for some a € Z,. By 8.8 we would then have f(a) = 0. But, testing each element of Z;, we have

f(0)=2, f(1)=1, F(2) =2, Fy

2)

ay

est)

Thus no element a of Z, is such that f(a) = 0. Definition

Let R be a ring with a 1. If a € R then the subsets

Ra = {ra; r € R},

aR={ar;reéR}

are, respectively, left and right ideals of R, called the princtpal

left [right] ideal generated by {a}. When these ideals coincide (e.g. when R is commutative) they are denoted by (a), which we call the principal ideal generated by {a}. An ideal J of R is said to be principal if I = (a) for some a € R. A principal ideal domain is a commutative integral domain in which every ideal is principal.

96

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Example Z is a principal ideal domain. In fact, we know that

every ideal of Z is of the form (n) = nZ for some n € Z. 8.10 Theorem domain.

[f F is a field then F[X] ts a principal tdeal

Proof We know that F[X] is a commutative integral domain. Suppose that J is an ideal of F[X] with I # {0} = (0). Let d be a polynomial in J of least degree, and let d* be the monic polynomial obtained from d by dividing the coefficients of d by

its leading coefficient. We claim that I = (d*). To establish this, we first observe that since d* € I we have (d*) C J. To obtain the reverse inclusion, let f € J and observe that, by 8.3, f =d*q+r where r = 0 or degr < deg d* = degd. Now, since I is an ideal, we have r = f — d*q € J and so, by the choice of d, we cannot have degr < degg. Hence we must have r = 0,

whence f = d*q € (d*) and so I C (d*). > Note that 8.10 is no longer true if we replace ‘field’ by ‘ring’.

For example, in Z[X] the set of polynomials that have an even constant term, i.e. those of the form 2a9 +a,;X + ---+a,X”, is

an ideal of Z[X] but is not a principal ideal. In fact, this ideal can be written

{2p+ Xq; p,qge€ Z[X]} and is the ideal generated by {2, X}. Let us now consider the notion of a highest common factor

for polynomials and an analogue of property (f) on page 91. Definition Let f,g be non-zero polynomials. By a common factor of f,g we mean a polynomial d that divides both f and g. We say that a common factor d of f,g is a highest common factor if it is such that every common factor h of f,g divides d. If F is a field and f,g are non-zero elements of F[X] then all highest common factors of f,g have the same degree. For if a,f are each highest common factors of f,g then a|f gives

dega < degf, and fla gives degP < dega.

It follows that

a = k6 for some constant polynomial k. This observation allows us to focus on a particular highest common factor, namely one that is monic. There is at most one such; for if a, 8 are monic highest common factors then a = kf gives k = 1 and soa= f. We call this the highest common factor. Its existence is assured by the following result.

POLYNOMIALS

97

8.11 Theorem Let F be a field and let f,g be non-zero polynomials over F. Then the highest common factor d of f,g exists

and can be expressed in the form d = af +bg where a,b € F[X]. Proof

Consider the set

I= {af + bg; a,b € F[X}}. Clearly, J is an ideal of F[X] and so, by 8.10, we have I = (h) for some h € F[X]. Suppose that

hh Sag + ay X + - +++ an) X"-* +4,X"-

(a, #0),

and define

Pe

eK ee Xe n

an

Xe

an

Then d is monic and J = (d). Now since f = 1-f+0-g EI we have d|f, and similarly g = 0-f +1-g € I shows that d|g. Thus d is acommon factor of f,g. If now t is a common factor of f,g

then t divides af + bg for all a,b € F[X] and so in particular t divides d which, being in J, is of this form. Thus d is the highest common factor of f,g. > It should be noted that the highest common factor of f,g €

F(X] is the monic polynomial of least degree that is expressible in the form af + bg, and is the monic polynomial of greatest degree that divides both f and g.

In order to compute the highest common factor of f, g € F[X]| we make repeated use of the division algorithm (8.5). First, we divide f by g to get

f=9q9+r. If r ~ 0 then degr < degg so we divide g by r to get g=rq

+11.

If r; 4 0 then degr; < degr so we divide r by r; to get aoe li! da Re1 If ro £ O then degr2 < degr, so we divide r; by rz to get r1 = 7293 + 13,

98

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and so on. Since deg g is finite, this process must terminate in some remainder being 0, i.e. the process terminates with, say,

Tk-3 = Tk-29k-1 + Tk-1; Tk-2 = Tk-19k + Tk, Tk-1 = TRQk+1Using the notation z|y to mean ‘z divides y’, we see that the last equation gives r,|r~-1, then the preceding equation gives rT |Tk—2, then the one before that gives r,|r~—3. Working back through the equations in this way, we see that r; divides r and g and hence also f. Thus rx, the last non-zero remainder in the above procedure, is a common factor of f and g. The above equations also show that if h divides both f and g then h divides r, whence it divides r}, whence it divides r2, ..., whence it divides r,. Thus we see that r, is a highest common factor of f,g. We therefore have the following result. 8.12 Theorem Let r, be the last non-zero remainder obtained by applying repeatedly the division algorithm to non-zero polynomials f,g € F[X]. If a ts the leading coefficient of rz then the highest common factor of f,g 1s a trz.

Example

Consider, in Z3[X], the polynomials

fi XA

OKO

WAS

1)

gat2Ne aX? qlioy tis!

By long division (remembering that the coefficients are in Zs) we have f =9q +r where

g

g= 2X +2,r=2X742X;

=1rqi +11 where gq; = X,r1 = 2X +2;

r=171q92 + ro where gz = X,r2 = 0. Thus, by 8.12, the highest common factor of f,g is 2~1r,. Now

in Zs we have 2~' = 2, so 2-!r, = 2(2X + 2) = X4+1.

By

8.11 we can express this in the form af + bg. To obtain this, we simply ‘roll up’ the above equations : ba Ee

Boe

=9-(f-—ga)u

=(1+qn)g—-af = (2X? + 2X + 1)g— Xf,

POLYNOMIALS

99

so the highest common factor can be written

X+1=271(2X?74+2X+4+1)g-2°-'Xf = (X?+X+2)g—2Xf. Example

Consider, in Z;[X], the polynomials

f= Xo + 2X2

OX? 44X44,

g@=X° + 3X +3.

By long division (remembering that the coefficients are in Zs) we have f =9q+9r where

g=

X74+4X4+2,r=X+3;

g=1qi +11 where gq; = X,r; = 3; r = 11qQ2 where gg = 2X + 1.

The highest common factor of f,g in Z5[X] is then 3-!r, = 1. To illustrate 8.11 again, we have Tie — Gi TG)

= 9—(f —gq)X =(1+9X)g9—Xf = (X° +4X? 4+2X + 1)g- Xf, and so, since 3~1 = 2 in Z, the highest common factor can be written 1=3>'r, =2r

= (2X° 4+ 3X? 4+4X4+2)g—2Xf. We now seek to find a property analogous to (7) on page 91, i.e. to the unique factorization of the integers. For this purpose, we require a polynomial analogue of a prime number. Definition A polynomial is said to be reducible if it can be written as a product of two polynomials each of positive degree. A polynomial that is not reducible is said to be trreductble.

100

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Example X?—2€ Q[X] is irreducible. In fact, if X? — 2 were reducible over Q then we would have

X? —2= (X —a)(-XB) for some a, € Q. Applying the substitution morphism ¢q to this equation gives a? —2 = 0, whence we have the contradiction J/2 =aEQ.

Example

X?-— 2 € IR[X] is reducible. In fact, we have

X? —2=(X —V2)(X+ V2). Example

f = X°+3X+2€ Z,[X| is irreducible. In fact, if f

were reducible then clearly it would have a linear factor X — a for some a € Zs. But, testing each element of Z;, we have

f(0.)=2, #1) =1, FQ) =1, #(3) = f(-2)=3, f(4) = f(-1)=. Consequently f has no roots in Zs and hence no factor of the form X — a.

8.18 Theorem Let F be a field, Ifp€ F|X] ts irreducible and divides the product fg € F[|X] then p divides f or p divides g. Proof Suppose that p does not divide f. Then p, f have no non-constant divisors and so the highest common factor of p and

f is the constant polynomial 1. By 8.11, we can find a,b € F[X| such that 1 = ap+bf, and then we have g = gap+gbf. But, by hypothesis, p divides fg. Consequently p divides the right-hand side of this equation, whence it divides g. >

8.14 Theorem

[Unique factorization] Every non-zero polyno-

mial f over a field F can be expressed in the form

f =apip2-:-pr, where a € F \ {0} and each p; € F(X] ts monic and irreducible. Moreover, this factorization 1s unique up to the order of the factors.

POLYNOMIALS

101

Proof If f is irreducible then there is nothing to prove. Suppose then that f is reducible, so that f = apg where a € F\ {0} and p,q are each monic and of degree at least 1. If either p or q is reducible then we can apply the same argument and extend the factorization. Since deg f is finite, this process terminates in a factorization of the form stated. As for the uniqueness, suppose that

apip2-::pe = f = bq192 °°: dmare two such factorizations of f. Consider p; and the product

qi(92 °°: qm). By 8.13 we have either pi|qi or pilqgo---Qmargument with p; and the product

Repeating this

q2(q3 a chee

we see that either pi|g2 or pilgs---qm.- It follows in this way that, for some 7, we have p:|q;. But each of these polynomials is irreducible, and so we must have p; = q;. Cancelling these from the equation apip2:-:pr

= bqig2:*- dm,

we can then repeat the argument for p2, and so on. The outcome is that k = m, each p; is one of the q;, and a=b. > An important property of irreducible polynomials is the following, which is the analogue of the fact that if p € Z is prime

then Z/pZ is a field.

8.15 Theorem

Let F be a field and let f € F|X] be irreducible.

Then f[X|/(f) is a field.

Proof Let (f) = I and suppose that g ¢ I. Then g,f are coprime, in the sense that their highest common factor is 1.

Thus there exist a,b € F[X] with 1 = af + bg. We now apply the natural morphism 4;, namely the assignment z+ this to obtain

14+J=(a+]I(f+1+(b+D(94+2)

z+ I, to

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VOLUME 3 : ABSTRACT ALGEBRA

which, since f € J and therefore f + J = 0+ I, reduces to

14+I=(b+1)(9+J). Thus we see that every

g+ J

40+]

has an inverse and so the

quotient ring f[X]/I is a field. > When f € F[X| is irreducible, it is natural to ask what the field F[X]/(f) looks like. We now give a convenient way of describing this field. Suppose that

f=sadyta, X04: 40,21” XxX" is monic, irreducible, and of degree n. Given any g € F[X], we can write g = fqg+r where r = 0 or degr < deg f. Applying the natural morphism h,,) to this, we obtain g + (f) =r + (f) where degr < n— 1. Consequently we see that

F[X]/(f) = {bo + 1X + ++++bn-1X"7* + (f) 5 0: © FH. To express this in a more convenient way, write

z=X+(f). Then by induction we have z* = X*+(f) and, for every a € F, az* = aX* + (f). Thus we can write

bo + by X + +--+ bp X14 (f) =bo +biet

---+bp-y2"?.

Now since f +(f) = 0+(f) is the zero of F[X]/(f) we also have Gg +ayz+

-+++a,-12" 1+2" =0.

We conclude, therefore, that we can regard the quotient field

F[X]/(f) as the set {bo + biz+ +++

+bp_-1z"1;

each b; € F}

subject to the ‘rule’ z” = —ap — ayt—

-+:—an_y2"™?.

POLYNOMIALS

Example

103

Let f = X?+X+1€

Z[X]. Then f is irreducible,

for if it were not so then f would have a linear factor X — a,

which is not possible since f(0) = 1 and f(1) = 1. By 8.15, the quotient ring

Z,/(X? + X

+1)

is then a field. By the above discussion, we can regard this field as the set

F=

{a+ bz; a,b€ Zo}

subject to the ‘rule’ sz? = —r—1(=2z+1

in Z). Now F has

four elements, namely 0=0+

02,-4=14-02,

s=0+

12;) 1+'c = 1+ 12.

Using the rule z? = 1 +1 we can compute the Cayley tables for F. For example, since we are computing modulo 2,

a(i+2z)=24+22?=2+24+1=22+1=1. The Cayley tables in question are

Note that the above field can also be represented by matrices. To see this, we must locate a matrix K, of size 2 x 2 say, such that K? = K + Ib. Such a matrix is

x-[t of

104

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Clearly, F can be represented as the subring of Mat2x2(Z2) given by

{alz +bK ; a,b € Zp}.

Example

X°+X+1€ Z2[X] is irreducible. In fact, if it were

not so then it would have a linear factor X — a, and this is not

possible since f(0) = 1 and f(1) =1. Thus the quotient ring Z2[X]/(X* + X

+1)

is a field. We can regard this as the field

F = {a+bz+ cz? ; a,b,c € Zo, } subject to the rule z*> = z+ 1. This field has eight elements. Example

X? + 1 € IR[X] is irreducible, for otherwise it would

have a linear factor X — a and we would have a? + 1 = 0 where a € IR, which is not possible. Thus

IR[X]/(X? + 1) is a field. We can regard this as the field

{a+ bz; a,b€ IR} subject to the rule z? = —1. In other words, we have

IR[X]/(X? +1) +. Example

Consider the problem of expressing a

1+ ¥3+ V27 with a rational denominator.

We begin by observing that X°—3

is irreducible in Q[X], so the quotient ring Q[X]/(X® — 3) isa field that is isomorphic to

F = {a+ bz+cz? + dz* + ex‘ ; a,b, c,d € Q}

POLYNOMIALS

105

subject to the rule z° = 3. Now 14+ */34 9/27 = 142429 this corresponds to the class

and

1+ X + X° + (xX*— 3). What we want, therefore, is the element that corresponds to the

inverse of this class in the field Q[X]/(X® — 3). Let f = X°—3 and g = X°+ X+1. We first determine the highest common factor of f,g. By long division, we have f =9q+4r where q = X* —1,r

= —X?74 X-2;

g9=1qi +; where gq, = —(X + 1),r) = —1. So the highest common factor of f,g is (—1)r; = 1. Now l=rq-g

=(f-gq)n-g9 =nf—(1+4a)9 = —(X+ 1)f —[1-(X

+1)(X? — Ig,

and so

1 = —(X+ 1)(X° — 3) + (X° + X? — X —2)(X° +

X41).

It follows from this (by passing to quotients modulo X® — 3) that the inverse of the class of X° + X +1 is the class of X° + X?* — X — 2, which corresponds to the element 2° + x? — x — 2. Thus we have

a

18

1 a

77

hp

EA

ED?

ieee)

In general, the problem of determining whether or not a given polynomial f is irreducible can be difficult. However, when the

coefficients are all real, ie. when f € IR[X], it is possible to say precisely when f is irreducible, and we step into the field €

to do so! Let us note that (as we have seen) it is possible for a polynomial to be irreducible over a field F but reducible in some larger field G; for example, X?—2 is irreducible over Q but reducible over IR. Also, the field € has the remarkable property

that every polynomial f € €[X] with deg f > 2 is reducible; in fact, if f is of degree n then f factorizes completely into a product of n linear factors. This is often called the fundamental theorem of algebra.

106

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8.16 Theorem A monic polynomial f over IR 1s trreducible tf and only tf it 1s of the form X —a for some a € RR, or of the form X? — 2aX +a? +b? for some a,b EIR with b #0. Proof We prove first that if f = X? — 2aX + a? + 6? where a,b € IR with 6 # O then f is irreducible. Suppose, by way of obtaining a contradiction, that f is reducible. Then there exist p,q © IR with

f =(X —p)(X—q) =X? — (p+ q)X +p, from which we obtain equations give

p+ q = 2a and pq = a* + b?.

These

p* —2ap+a*+b?=0 and so we have that

p = 4(2a + \/4a? — 4(a? + 6]. Since 6 # 0 by hypothesis, we have the contradiction p ¢ IR.

Suppose now that the monic polynomial f € IR[X] is irreducible and that f is not of the form X — a where a ER.

z=a+h€C

bea

Let

root of f (this exists by the fundamental

theorem of algebra). If f = > a;X* then 0 = f(z) = oF:a;z* gives, on taking complex conjugates,

Ov o aj2* = $0 a;z* = f(z). i=0 i=0 Thus Zz € C is also a root of f. Consequently

(X — z)(X —z) = X? —-(24+2)X +22 = X? —2aX +0? +0? is a divisor of f. Since f is irreducible over IR we must have

z€C\R, ie. 8.17 Corollary

z=a+bi with b £0, and the result follows. The irreducibles in IR[X] are the polynomials

of degree 1 and the polynomials pX*+qX+r where q?—4pr 0. Thus if g? — 4pr < 0 the polynomial X? + q¢X +r must be irreducible. Conversely, if pX* + ¢X + r is irreducible then so is

Kar ox 4 p

P

so that, by 8.16, there exist a,b € IR with b 4 0 such that

gq =

—2a,

= a? + b?. ||] as}

It follows that q? — 4pr = 4a?p? — 4p?(a? + bi

—4p*b? < 0.

co

Example Consider X??—1 € IR[X]. Over € the equation z?? = 1 has solutions z, = e2"*#/2P = e"™*/P for n = 0,1,...,2p — 1. Thus, over C,

x?

= T] (X—errle) 2p-1

;

n=0

= (X —1)(X —e™/P)...(X — er Uet/P), Grouping together conjugate pairs, we obtain by 8.16 the following factorization over IR :

KP? — 1S (X= (X

p-1

ahs

keri

+1) T(x — F/?) (x =e”) k=1

= (X

p-1 kr -1)(X +1) [J (x? — 2X cos a + 1). k=1

CHAPTER

NINE

Fields of

quotients

In this Chapter we shall obtain a construction theorem that not only will show how the field Q of rationals can be obtained from the integral domain Z of integers, but also can be usefully applied to the integral domain IR[X]. Basically, we shall be interested in the embedding of a commutative integral domain D in a field F, i.e. in finding, for a given D, a field F and aring

monomorphism f : D > F (so that

D~Imf

C F).

In what follows, given a commutative integral domain D, we shall denote by D* the set of non-zero elements of D. 9.1 Theorem If D is a commutative integral domain then the relation ~ defined on D x D* by

(a,b) ~ (c,d)

ad=be

1s an equivalence relation.

Proof Since D is commutative we deduce that (a,b) ~ (a,6),

and that if (a,b) ~ (c,d) then (c,d) ~ (a,b). Thus ~ is reflexive and symmetric. Suppose now that (a,b) ~ (c,d) and that (c,d) ~ (e, f). Then we have ad = bc and cf = de, so

adf = bef = bde. By cancellation of d # 0, we obtain af = be, i.e. (a,b) ~ (e, f). Thus ~ is also transitive. } We shall denote the ~-class of (a,b) € D x D* by . Thus we have that a

Cc

ry

qt

ad =

be.

FIELDS

OF QUOTIENTS

9.2 Theorem on the set

109

Let D be a commutative integral domain.

Then

Qp = (D x D*)/~ the prescriptions ue a =|

ad+

d

bc

bd

a@c_

’

b

ac

d_ bd

define laws of composition with respect to which (Qp,+, ‘) ts a field. Moreover, the mapping 8:

D— Qp given by

O(a) = 1s @ ring monomorphism.

a’

a

c

Proof If aes U and ae

we

then (a,b) ~ (a’,b') and (c,d) ~

(c’,d’), so ab’ = ba’ and cd’ = dc'. Consequently

(ad+ bc)b’d' = ab'dd! + bb'cd’ = (a'd’ + b'c')bd ad + bc

and hence (ad + bc,bd) ~ (a’d’ + b'c’,b'd’), ie 1d!

b!

Se b'd'

-

bit

U

Thus the given addition is well defined. Similarly,

ac - b'd’ = ab! -cd' = ba’ - de' = a'c' - bd (atc! bd) ; and so (ac, bd) ~~ (a’c’,b'd'), ive.

bea id

a’c' ae

: Thus the given

multiplication is also well defined.

That (Qa, +, -) is a field follows from a routine verification of the field axioms:

(1) (Qp,+) ts an abelian group. In fact, each of =Re ( f(a)[f()|"* = 9(5) = IF

f(a) = f(b)

—>

cif

—

eae

aren co,

We can, of course, apply the above construction theorem to the commutative integral domain Z. What we obtain is no surprise in view of the notation that we have chosen for the elements of Qp in the above: the field of quotients of Z 13 the field Q of rational numbers.

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‘\

In view of the fact that is a monomorphism,

8 : D — Qp given by ¥(a) =

it is common

ela

practice to identify D with

Im 9, i.e. identify a with —. In this way we can regard D as a subdomain of the field Qp. Note that then Qp is the smallest subfield of Qp that contains D. For, if K is a subfield of Qp that contains D then for all a,b € D we have that K contains I and rr If

b

b~ 0 then K must contain the inverse of = namely

1 1 5 Hence K contains — -= = — and consequently

K = Qp. In

particular, we regard Z as asubdomain of the field Q in precisely this way, and Q is the smallest subfield of Q that contains Z. Another commutative integral domain with which we are familiar is the polynomial ring D[X] where D is any commutative integral domain. We can therefore also consider its field of quotients Qp;x)- A typical element of Qp;x] is an equivalence class which we can write as

f° agtayX

g

+---+a,X4"

bo Hb X + +++ by X™

We shall call Q pj xj the field of rational forms over D and denote

it henceforth by D(X). In particular, if F is a field then we can

consider its field of rational forms F(X). 9.4 Theorem [If D 1s a commutative integral domain and Qp

ts its field of quotients then D(X) = Qp(X). Proof Recall that D(X) is the field of quotients of D[X], the ring of polynomials over D, and Qp(X) is the field of quotients of Qp[X], the ring of polynomials over Qp. A typical element of the latter is 5

Beye

Oy £ shoo Days

g

bono,

Ley

Ewa

—+—X+:---+—xX™ Bo

If A= |] a; and B= +=0

Rg

S Sadetts 2

fon

Bm

[[ f; then ABS € D[X] and ABg € g=—0

D{X], and the above quotient can be written as pal € D(X). 9

FIELDS

OF QUOTIENTS

113

Thus we see that

But we also have D[X] C Qp[X] € Qp(X), and D(X) is the field of quotients of D|X]. Since the only subfield of D(X) that contains D[X] is D(X) itself, we deduce that D(X) = Qp(X). > Our objective now is to show how rational forms can be decomposed into a sum of so-called ‘partial fractions’.

Given any element f € F(X) with deg f > degg, we can divide f by g to obtain f = gg+r where r = 0 or degr We then have

< degg.

r ‘heal

g

g

In what follows we shall therefore consider only those rational forms ;where deg f < degg.

We shall also consider only the

case where g is monic, for otherwise we have only to divide both f and g by the leading coefficient of g. We begin with a particular case.

9.5 Theorem Let F be a field and let p,q € F[X] be such that hef{p,q} =1. Then there exist a,b € F[X] such that

Proof By 8.11 there exist a,b € F[X] such that bp + ag = 1. Since clearly neither p nor g is 0 we have pq # 0 and so we can 1 multiply this equation by —. > Pq It is immediate from this result that a rational form, the denominator of which can be expressed as the product of relatively prime polynomials p and q, can be expressed as a sum of two quotients, the denominators of which are p and q. In fact, we have immediately from the above that

f eiafeoof Pq

P

q

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\

We can of course repeat the argument if p or q factorizes into

relatively prime factors : if p= pip2 where hcf{pi,p2} = 1 and = q1q2 where hcf{gq1,q2} = 1 then we obtain a b b bade Sil, Pods sat qnedad Pq

P1

P2

41

q2

Clearly, this process can be repeated until the denominators in the summands are powers of distinct irreducible polynomials

(8.14).

Consider, therefore, an element of F(X) of the form h pm where p is monic and irreducible with deg p = k, and degh < deg p™ = mk. By the division algorithm we have h=p™~1q, +1, where degr; < (m-— 1)k; r, = p™ 2q2 +12 where deg rz < (m-— 2)k;

Tm—2 = PGm-1 Tm-1

=

+ Tm-1 Where degrm_1 < k;

Im-

It follows from these equations that

hu,

&

Qm-1 , Im

pe oi tg Pe eat

These observations give the following result.

9.6 Theorem g 1s monic

[Partial fractions] Let f,g € F(X] be such that

and degf

< degg.

Let g =

Ilp;'' be the de-

composition of g as a product of trreductble Sarma

Then

; € F(X) can be expressed as a sum of rational forms whose

denominators are the powers po (A; = 1,...,m;) and whose numerators are either 0 or have degree less than that of the corresponding trreductble p; in the denominator. > The remarks preceding 9.6 provide a practical method of computing partial fractions. However, in practice there are short

FIELDS

OF QUOTIENTS

115

cuts that can reduce the labour. Before giving some examples, let us remark that the above results are of most practical value in the calculus. Here we deal with rational functions, i.e. functions f : IR — IR of the form ay

it p> ao +ayr+

---+a,z"

bo tbya+---+b_2™ These are often confused with rational forms, and it is worth while considering the difference. If F is a field and n

f=

-

4a,X* € F[X]

+=0

then we can define a function 0; : F — F by the prescription

97(z) = ¢2(f) = F agai +=0

Such a function is called a polynomial function.

It is not hard

to prove that the mapping 0 : F[X] — Map(F,F) given by 0(f) = 8; is a ring morphism. What is important about this morphism is the following. 9.7 Theorem

9% 1s a monomorphism if and only if F is infintte.

Proof If F is infinite and f € Ker then 0s = 0 and so 9 f(z) = O for all z € F, whence every z € F is a root of the polynomial f. But every non-zero polynomial has only finitely many roots. Hence we must have f = 0. Thus ® is injective. Conversely, suppose that F is finite. Then the polynomial

f=

[J (X — a) is a non-zero element of F[X] and belongs to ac€F

Ker# since every element of F is a root of

f. Thus Ker f 4 {0}

and so # is not injective. > In the case where F = IR, the error of confusing a polynomial with the corresponding polynomial function (and also a rational form with the corresponding rational function) is therefore

pardonable! The importance of 9.6 as applied to IR(X) lies in the fact that, via 8.16, it implies the following result in analysis.

116

VOLUME

9.8 Theorem

3 : ABSTRACT

ALGEBRA

Every rational function Go +:042+ th Onzl bo +byz+ ---+bmz™

a

can be integrated in terms of elementary functions. > Example

Consider the rational function f : IR — IR given by z® — 227 + 32-5

sd oad OS LET

:

By the analogue of 9.6 we have

ite) a

A

B

Cc

eee ee

D

which can be written in the form

z° — 227 + 32 —5 = A(z — 1)?(z — 2) + B(z — 1)(z — 2) + C(z — 2) + D(z —1)°. Substituting z = 1 in this we obtain C = 3, and substituting xz = 2 we obtain D = 1. This equation therefore reduces to the identity

2° — 227 + 32 —5 = A(x — 1)?(z — 2) + B(x — 1)(x — 2) + 3(z — 2) + (x-1)°. Putting z = 0 we obtain —5 = —2A+2B-—7 so that A-B = —-1; and comparing the coefficients of x? we obtain 1 = A+1. Thus A=0 and B= 1 and hence

a°—227+32—-5 1 ee aelasa feet (c-—1)8(r-2) ~~ (2-1)? (x-1)8 © 2-2" Example

Consider the rational function f given by e

ze +1

2) =Grae)’

FIELDS

OF

QUOTIENTS

a7

by the analogue of 9.6 we have

eel es ee

A

B Pes.

Cz+D z2+1’

which we can write in the form

x +1= A(z—1)(2z? +1) + B(x? + 1) + (Cx+ D)(z—1)?. Substituting z = 1 we obtain 2 = 2B, so € and substituting z = 7 we obtain

B = 1. Stepping into

—i+1=(Ci+ D)(i—1)? = (Ci + D)(—21) = 2C — 2Di and soC = D= 2 Finally, comparing coefficients of z* we obtain 1= A+C,so A=1-C= on Thus

a

oe

fie eek) Example

1

+1

D(a=Tene 1222 1):

Consider the rational function f given by

fe) = Beet The roots in € of z2? = —1 are 2, = e(2*+1)"#/2P where k takes the values 0,1,...,2p—1. have

By the analogue of 9.6, we therefore

fi

zP+1

2p-1

Ak

fm tz

To determine the constant A,, multiply throughout by z — z,

and take limz.z,.

Observe that the right-hand side becomes

A, and the left-hand side becomes ZL— 2

Tze g2P — ae

1

Zk

2p22P io

2p

=

1 P21;

x

2p rao

\t—-Ze

+)

te

AS

u(z_ + 2) — 2242

2p k=0

(2 — 2%)(2 — %)

We can now exit from € by summing in conjugate pairs :

1 z2P +1

are

Pe

Xe

=t—zR

(2k + 1)x Z COs aT - 1

5) 4 eee

P k=0 52 _ 2208

Get lz As soe

118

%

VOLUME

3:

ABSTRACT

ALGEBRA

9.9 Theorem

Every positive rational number less than 1 can : , a be expressed as a sum of‘partial fractions’ of the form — where

p 1s a prime

P

andO