First Course in Rings, Fields, and Vector Spaces [1 ed.] 0470990473, 9780852260647

Citation preview

iy

a 40

ete fe

f Ares

an Science Library JOINT UNIVERSITY LIBRARIES NASHVILLE

TENNESSEE

DATE DUE 24 OCT 1979 | OCT

2 2) 1979

‘APR 1 2

DEMCO

NO.

38-298

.

May 20

1987

First Course in Rings, Fields and Vector Spaces

First Course in Rings, Fields and

Vector Spaces

P. B. Bhattacharya

University of Delhi S. K. Jain Ohio University

A Halsted Press Book

JOHN

WILEY

New York

London

& SONS Sydney Toronto

Copyright © 1977, WiLey EASTERN New Delhi

LIMITED

Published in the U.S.A., Canada, Latin America and the Middle East by Halsted Press, a division of John Wiley & Sons, Inc., New York

Library

of Congress

Cataloging

in Publication Data

Bhattacharya, P B First course in rings, fields, and vector spaces. “A Halsted Press book.’’ Includes index. 1. Rings (Algebra) 2. Fields, Algebraic. 3. Vector spaces. I. Jain, S.K., joint author. Tie Title:

QA247.B47 5127.02 ISBN 0-470-99047-3

76-55303

Printed in India at Prabhat Press,-Meerut 250002 Nike

Preface

This book is based on the lectures given by the authors to the upper level undergraduates and beginning graduates at the University of Delhi, the University of California at Riverside, and Ohio University. This is a companion to our earlier book First Course in Group Theory. The prerequisite is mathematical maturity at the levels indicated above. We believe that (with a few additions and omissions depending on the interest of the instructor and the ability of the students) the material can be presented in a semester of fifteen weeks meeting every day. Chapters 1 to 7 deal with theory of rings, modules, and linear map-

pings.

Starting with the basic concepts, the reader is led up to and

including the theorems of Wedderburn-Artin, Goldie-Lesieur-Crosoit,

and Krull-Schmidt-Remak. The theory of fields is given in chapters 8 to 11 and can be read after chapter 3. The topics presented in the theory of fields includes the theory of algebraic extensions, splitting fields, finite fields, normal and separable extensions, the fundamental theorem of Galois theory and its applications to the solvability of algebraic equations and solutions of some of the ancient geometric problems. Certain essential topics in the study of linear algebra are also included. For example, in chapter 5 we have dealt with the theory of vector spaces, linear mappings and matrices, normal form, rank and nullity of a matrix. In addition the eigenvalue problems are dealt with in chapter 12, which can be read independently of other chapters (except possible for some sections on matrices in chapter 5). Most of the theorems in chapter 12 have been stated (without proofs) with many examples and solved problems. Steps for reducing matrices to

various

canonical

forms—diagonal,

triangular,

Jordan—have

been given to provide the reader with a working procedure which is not usually given so explicitly in texts on abstract algebra. Each chapter contains several examples to illustrate the concepts. Also, numerous solved problems in addition to the unsolved problems are included in each chapter to facilitate the reader’s understanding of the subject and to acquaint him with techniques for working the unsolved problems.

vi

PREFACE

Naturally we are indebted to many friends, colleagues and students for their suggestions and comments which were of great value to us. Our sincere thanks are also due to some

others, in particular V.K.

Goel and R.K. Jain (both of Ohio University) and S.K. Bhambri (University of Delhi), who have helped us in reading the proofs very carefully.

October 1976

P.B. BHATTACHARYA S.K. JAIN

Contents

Preface

. Some Fundamental Concepts and Definitions

Ideals and Homomorphisms . Unique Factorization Domains, Domains and Polynomial Rings

Euclidean

Rings of Fractions Modules and Vector Spaces Noetherian and Artinian Modules and Rings Composition Series and Krull-Schmidt Theorem Theory of Algebraic Extensions of Fields Normal and Separable Extensions Fundamental Theorem of Galois Theory . Some Applications of Galois Theory . Methods of Matrices Index

-

103 129 137 167 180 189 207 233

List of Symbols

is a Subset of

is essential (or large) in

Euler’s function

D

is not a subset of

is a proper subset of

(a, b)

divisor

a|b

does no contain

a+b

the empty

Rop

opposite

SA

(as a group) symmetric group of degree ; (as a ring) the ring of nxn

set

is a member of

*PUINAAN AMR

is not a member of composite, ffollowing g

a does not divide b ring of the ring

centre of R; left singular ideal

of

R

(section

2,

chapter 6 only)

if and only if

U (R)

for all there exists

a at aecartesian

adivides b

matrices Z(R)

implies

A

greatest common of a and b

contains

is isomorphic to

09

prime integer irreducible element

Zp or Z/(n) integers mod

product

C[0, 1]

product

®, (&)

sum

n

real-valued continuous functions on the closed

interval [0, 1]

direct product

cyclotomic polynomial of degree n

direct sum

orthogonal cular) to

group of units (invertible elements) of R

(perpendi-

cardinality of set § absolutevalueof numberS determinant of matrix S$

VP

radical of ideal P

0 (S)

left annihilator of §

r(S)

right annihilator of S

Hom(M,N) homomorphisms of M into N

image kernel

Homr (M, N) R-homomorphisms of M4 into N

dimension

determinant

Aut (G)

automorphisms of G

if and only if

End (M)

endomorphisms

of M

LIST OF SYMBOLS

Endr(M)

R-linear mappings

of M

F{S]

to M

ix subring

generated

by F

and §

dual vector space of V; conjugate transpose of

matrix V

F G(E|F)

rank of matrix A

algebraic closure of F

group of automorphisms of E leaving F fixed

dimension of E over F; degree of extension of E

fixed field of a group H of automorphisms

over

set of integers

F

localization of R at S

set of rationals

subfield and «

set of reals

generated

by F

AXON

set of complex numbers

ERRATA

Should read

Page

Line

13

17

different from following:

13

23

The set of invertible elements in

13

25

Let R be commutative.

2)

17

nonzero right ideals B of R} is an ideal in R

29

22-23

ois)

12

1 is right quasi-regular.

Show

the

If D is

Show that a ring R with unity is a division ring if and only if R has no nontrivial right ideals.

Let A,, Ay,..-, An be ideals in a ring R with unity, such that 4j-+ Aj; =R

43

31

bound in S, then (S, 1 and F, an arbitrary field. Show that the group algebra F(G) is not an integral domain. Solution.

Then and

WLetxEG be of order m>1.

f=1—x, g=14+x+...+x™-1, :

are

nonzero

elements

of F(G),

fg=1—x"=1-—1=0.

(b) Let G={x,=e, x2, ..., Xn} be a group of ordern, and field whose characteristic does not dividen. Then en (xy 4+X%.+...+%n) is an idempotent

in F(G),

where

F any

n-' is multiplicative inverse

of

n times

ar n=1+4+1+...+1EF. Solution. Since the n- exists.

characteristic

of F does

not divide n,n+0,

Consider (x,+%.+...+%n)? = (X,(Xy 1-2. + Xn) bX... Xn)... bXn (K+... +%n)) =a(X yeas ot Xn) tb(Xb da et on) + «:.. Oy ssh Xn)

=n(x,+...+2xn), since for any xEG, xx, XXq, ..., XXn is just a permutation of (x,,X,, ..., Xn).

Hence e is idempotent.

18

RINGS, FIELDS AND

VECTOR SPACES

(c) The algebra of quaternions over the field of rational numbers or real numbers is a division ring but over the complex field, it is not a division ring. Solution. Let H be the quaternion algebra over F where F is the rational

Then

or

the

real

field.

Let

OAq=ae+ityj+skE

q*=ae—Bi—yj—sk is such that qq*=(e2+B?+7?+8") eA0, since «, 8, y,5 are reals and not

all zero. If we set N(q)=e?+8?+7°+ 0°EF then

m=xG) q* (N(q) is called the norm of q).

Now, let F be the field of complex numbers.

Then

(4/—le+1i+0j+0k) (4/—1le—1i+0j+0k) =(4/—le+1i) (,/—le—1i)=—e+e=0 Hence in this case, His not an integral domain and so can not be a division ring. (d) The centre of the real quaternions is the real field. Solution,

Let «e+fi+yj+5kE€ Z(H), the centre of H.

Then (xe+Bi+ yj+d5k) i=i(«e+Bityj+dk). This gives ai—fBe—yk+8j=ai—Be+yk—sj and by definition of equality of quaternions we get y=0=s. Similarly, 8=0. So ae+Bityj+skE Z(H) implies B=O=y=8. Since ae is always in Z(H), the solution is complete. with «.

As remarked

earlier «e

is identified

Problems

1. 2.

Determine F(G) when F is Z/(2) and G is a cyclic group of order 3. If His the algebra of real quaternions, and N(q) is the norm of an element gE H then show that N(q)=0 if and only if g=0 and N(91-92)=N(Q1) -N(Q2).

CHAPTER

2

Ideals and Homomorphisms 1

IDEALS

Let S be a subring of a ring R. If a, bER, we define a=b (mod S)

if a—bES.

Then = defines an equivalence relation on R. The equi-

valence class containing an element a of R is easily seen to be the coset a+ S={a+s | sES} relative to the additive subgroup S. We shall write a for a+-S, and R/S for the set of equivalence classes. Our purpose is to make R/S into a ring. So we set a+b=a+b and ah=ab. We proceed to find necessary and sufficient conditions to be satisfied by the subring S so that these operations of addition and multiplication | of cosets are well-defined. Let a,=a, and b,= b,. Then a,—a,ES and b,—b,€S and thus This shows (4,—4,) +(b,— b,)ES, that is, (a,+b,)—(a,+h,)ES. that a,+5, =detby._ Hence addition of cosets is well-defined. We

assert that a,b,—=a,b,, that is, a,b,—a,b,ES, if and only if for all Then Assume a,b,—a,b,ES. rER, sES, rs as well as srES. Similarly choosing b,=b,=r, a=0 and a,=s we get srES. rsES.

To

prove

the

converse

of

this

assertion,

let us

write

Ob, —Ayb,=(G,—@2) b, +4, (b;—42) which is clearly in S, since a,—4,€S, b,—b,ES. Thus addition and multiplication of cosets a+b=a+b

ab=ab are well-defined if and only if the subring S satisfies the following condition: for all rER and sES, rsES and srES. It is then easily checked that R/S with the addition and multipliR cation defined above is a ring. The zero element is 0=S, and if R if Further R/S. ring the of unity the is has unity 1 then 1=1-S ab=ab=ab+S is commutative _ then R/S is also commutative; for

—pha+-S=ba=ba.

The ring R/S is called the quotient ring of R

In view of modulo S. Such a subring S of R iscalled an ideal of R. n of an definitio the below give the importance of this concept we ideal.

20

RINGS, FIELDS AND VECTOR SPACES

DEFINITION. ideal of Rif (i)

Ideal.

A nonempty

subset S of a ring Ris

called an

a, bES implies a—bES; and

(ii) a€S and rER imply arE€S and rac.

DEFINITION. Right (left) ideal. A nonempty subset S of a ring R is called a right (left) ideal if (i) a—beS for all a, bES; and (ii) arE S(raES) for all aE S and

re R.

Clearly a right or left ideal is a subring of R, and every ideal is both right and left and so an ideal is sometimes called a two-sided’ ideal. Trivially in a commutative ring every right or left ideal is twosided. In every ring R, (0) and R are ideals, called trivial ideals. 1.1 Examples of ideals (a) In the ring of integers Z every subring is an ideal. For let J be a subring of Z andac/,reEZ. Then r

times

ar=a+a-+...+a,

if r>0

ar=O1f r=0 —r times

—

and

ar=—a—a—...—a,

So in every case ar€J.

if r Ry

toa Then

X=(0; (0): syetetn

Ojs: Oy . -,0)=(a, Az)+

This gives a;=0 and hence x=0. we have shown that

+5 Ai-1, 0, Gj+1s+- +n).

Since itis clear that

R= >" Rf.

32

RINGS, FIELDS AND

VECTOR

SPACES

For the second part, we note that if xE R then x can be uniquely expressed aS a@,+d.+...+@n,a;€ Ai, 11, n,>1 then n, 7,=0 in Z/(n) and 0¢n,, 04n,. But this is a contradiction, since Z/(n), is a field. Hence 7 is prime. Conversely, for any prime p, we have shown earlier (example (c), section 1.1, chapter 1) that Z/(p) is a field and hence by the theorem 4.2 above (p) is maximal. (b) Let R=F{[x] be the polynomial ring over a field Fin a variable x. (We callan element p (x) € F[x] an irreducible polynomial over F if the degree of p(x)>1 and p (x)=p, (x) ps (x) where p,(x), Pa (x)EF [x]

An ideal M in F[x] is implies p,(x) or p,(x)EF). M=(p(x)) where p(x)€F[x] is an ideal if and only if polynomial over F. This follows by recalling the example 1.1, chapter 1, that each ideal in F[x]is also a principal proceeding as in (a) above. (c) If Ris the ring of 2x2 matrices over a field F

a maximal irreducible (d), section ideal, and

of the form

(a 4Iia, bE F, then the set

M={( 9 a | vert is a maximal ideal in R.

Clearly, Misanidealin R.

Let s={( 0

0

9) aert. Then Sisa

38

RINGS,

FIELDS AND

VECTOR

SPACES

P : 0 subring of R which is isomorphic to F under the mapping( i 0 )>a. P Further, the mapping / : R->S where f (‘6 j)=(5

an Die 0 )isan onto

homomorphism whose kernel is M. Thus by the fundamental theorem of homomorphisms, R/M=S. Since S isa field, it follows M is a maximal ideal in R. 4.4 We now introduce the concept of product of right (or left) idealsinaring. Let A, B be tworight idealsinaring R. Then the set S={ > ajbj|a;€A, b:€ B} isa rightideal. For, if x=a,b,+...+a,br finite sum

and y=c,d,+...+csd; belongto S then x—y is clearly in S. Also if tE€R then xt=a,(bt)+...+car(b-t)ES. Thus S is a right ideal inR. Thisright ideal S is called the product of A and B(in this order) and we writeitas AB. Similarly, we define the product of left ideals. We note the following facts about the product of right (or left) ideals. (1) For any right (or left) ideals A, B in R, ABC A (or ABCB). (2) IfA, Bare right (or left) ideals in a ring R then it is not necessary that AB=BA.

For let

Zi : ; R=( 0 = be the ring of 2x2 upper triangular matrices over Z. Then

G7zZ

0 0

4=(5 0} B=( 4 5) are right ideals in R. Such that 4B4(0) but BA=(0). Thus AB“ BA. (3) For any right (or left) ideals A, B,C in a ring R. (i) (AB)C=A (BC). (ii) A(B+C)=AB+AC and (A+B)C=AC+BC. (i) follows from the associativity of product in R, and (ii) from the distributive laws in R. (4) For ABCAQ)B.

any

ideals

A, B in R, AB

is alsoan

ideal

in Rand

If Ay, A,,...,A4n are right (or left) ideals ina ring R then their product A, A,...An can be defined inductively. Note that a typical element x€ 4,Ao...An is a finite sum: a,a,...@n+b, | Re 2) ne .. where ai, bE A;, i=1,2,...,n.

4.5 A prime integer p has the fundamental property that if p\ab where a and bare integers than p|a orp|b. In terms on the ideals in Z, this property may be restated as follows:

IDEALS

AND

HOMOMORPHISMS

39

If abE(p), then a€(p) or bE(p). Equivalently if (a)(b)C(p) then (2) C (p) or (5) C (p). This suggests the following: DEFINITION. Prime ideal. An ideal Pin a ring Ris called a prime ideal if it has the following property: If A and B are ideals in R such that ABC P then ACP or BCP. Clearly, (C) isa prime ideal in any integral domain. In fact, a commutative ring R is an integral domain if and only if (0) is a prime ideal. Also as discussed in the beginning of the section, for each prime integer p, the ideal ( p) in Z is a prime ideal. Asin Z,in the polynomial ring F[x] over a field F, a nonzero ideal P is prime if and only if itis of the form (p(x)) where p(x) is anirreducible polynomial over F. Thus a nonzero ideal in F[x] is prime if and only if it is maximal. However, we have

4.6.

TuHeoREM.

Jf R is a ring with unity then each maximal ideal is

prime. But the converse, in general, is not true. Proof. Let M be a maximal ideal and A, B be ideals in R

that acA,

ABC M. mEM.

Suppose Let

AEM.

DEB.

Then

Then

4A+M=R.

Write

b=ab-+-mbCAB+MCM.

such

1=a+m, Hence

BCM. This proves M is prime. The ring of integers has (0) as a prime ideal which is not maximal. For commutative rings we have 4.7. TuHeoremM. Jf Ris a commutative ring then an ideal P in R is prime if and only if abEP, aE R, bER implies aE P or bE P.

Proof. Let abEP and P bea prime ideal. We know (a) (d) consists of finite sums of products of elements of the type ma-+-ar and mb-t-bs where n, mE Z and r, sE R. Now (na-+-ar)(mb+bs)=nmab+

nabs+mabr-+abrs. Since abE P and P is anideal in R, we get (na--ar) (mb+bs) or finite sums of such like products are in P. Hence each element of (a)(b) is in P. Thus (a) (6)CP. But since P is prime, (a)CP or (b)CP. This implies a€P or bEP. Conversely if abe P, ac R, bER implies a€ P or bE B then we show P is prime. Let A, B be ideals in Rsuch that ABC P. Suppose A & P. Choose an element a€ A such thatag@P. Then ABC P implies aBC P and so abeP forall bE B. But then by our hypothesis bE P. Hence BCP. 4.8

Solved problems

(a) Let R be a commutative ring with unity in which each ideal is prime. Then R is a field.

40

RINGS, FIELDS AND VECTOR SPACES Solution.

If ab=O then (a)(6)=(0). But since (0) is a prime ideal,

(a)=(0) or (b)=0. Hence ab=0 implies a=0 or b=0, i.e. R is an integral domain. Next, if a@R then (a)(a)=(a"*) since R is commu-

tative.

But (a?) is prime.

contained

Thus

(a)C(a’).

in (a), we get (a)=(a’).

there exists an element

x@R

Since

(a?) is trivially

This implies a€(a?) and

hence

such that a=a’x, or a (1—ax)=0.

a0, we get 1=ax, ie. ais invertible. Hence R isa

If

field.

(b) Let R be a Boolean ring then each prime ideal PAR is maximal. Solution. Consider R/P. Since P is prime, abeP (equivalently ab=0) implies ae P or bEP (ie. a=0 or b=0). So R/P is an integral domain. Nowforall xER, (x+P)(x+P)=x*+P=x+P. So R/P is also a Boolean ring which is an integral domain. But we know that an integral domain has no idempotents except 0 and possibly

unity. Thus R/P=(0) or {0, 1}. However, which Hence (c) S={a,

R/P=(0) implies R=P

is not true. So R/P={0, 1}, which obviously forms a field. P is a maximal ideal. Let a be a non-nilpotent element in a ring R, and a’, a’,...}. Suppose P be a maximal in the family F of all

ideals in R which are disjoint from S. Then P is a prime ideal. (Note the statement does not say that P is a maximal in R. Precisely it means that there does not exist any ideal

YE F such that

YD P.) x

Solution. Let ABC P where A and B are ideals in R. If possible, let AEP and BP. Then ae and ee ee By maximality

ofP, (A+P)NS4#¢ and (BLP)AS# ¢. Thus "there exist positive integers i, j such that a@€A+P and aE B+P. Then aiai€(A+P) (B+P)=AB+AP+PB+PPCP,

since

ABCP

and also P isan ideal

in R. Thus PA S¢, a contradiction. Hence ABCP implies either ACP or BCP, proving that P is a prime ideal. (d) In the mn matrix ring Fn, over a field F, (0) is a prime ideal. Solution. By corollary 1.3, F, has no proper ideals except (0) or itself. Thus it follows immediately that (0) is a prime ideal. (e) Let R=C[0, 1] be the ring of all real-valued continuous functions on the closed unit interval. If Mf is a maximal ideal of R, then

there

exists

a real

number

7, O0 respectively where nR’ is any homomorphism of a ring R into a ring R’ such that every element f(s), sES, is a unit then

RINGS OF FRACTIONS

there exists a homomorphism f’: Rs->R’ such following diagram is commutative:

59

that the

i

Ree ee ee

be

ei

that is, f’h=f. We also say that any such homomorphism f factors through A. Let R be a UFD.

ie)

3. '

Show Rs is also a UFD.

Let R be a commutative ring with 1 having no infinite properly ascending chain of ideals in R. Show that the same holds for Rs.

2 RINGS WITH

ORE CONDITION

We discuss, in this section, a natural generalization of the ring of fractions of a commutative ring. Let R be an arbitrary ring. Call an element aE R to be regular if a is neither a left nor a right zero divisor.

DEFINITION. (Left) ring of quotients. The ring @Q is called a (left) ring of quotients if (i) Risa subring of Q and Q contains unity (ii) Every regular element aER is a unit in Q

(iii) Every element of Q has the form ab, a, bER, and a is regular. 2.1 Tweorem (Ore). Let R be a ring containing regular elements, Then R has a (left) ring of quotients if and only iffor each pair a,bER

with a regular there exists a pair a,,b,ER with a, regular such that a,b=b,a. (The condition in the theorem is called left Ore-condition. Clearly every commutative ring satisfies this condition.) Proof. In essence the method of constructing Q from R is very similar to that used to construct the rationals from the integers. We form a certain set of pairs of elements of R and define an equivalence relation on this set. So, first, assume

that the ring R satisfies left Ore-condition.

Let

a, b,c, dER with b, d regular. We say that the pair (a, b)~(c, ) if d,a=b,c where (1) b\d=d,b with b, regular. Then d, is also regular. Clearly xd,=0 implies x=0. To show d,x=0 implies x=0, we proceed

60

RINGS, FIELDS AND VECTOR

SPACES

as follows. Choose regular elements d, and wu such that (2) b,d=d,b and (3) ub,=vb,. From (1) and (2), we get ub,d=ud,b and vb,d=va,b. But then using (3), we get (ud,—vd,) b=0. This implies ud,—vd,=0. Thus d,x=0 gives vd,x=0 and so ud,x=0. Hence x=0, since both u and d, are regular. This proves our assertion that d, is regular. Next, we show that the relation ~ is independent of the particular

b,, d, which give the common left multiple of b and d. For if b,d=d,b, we choose e,, €, (both regular) such that e,b,=e,b,. Then e,d,b=e,b.d =e,b,d=e,d,b. This gives e.d,=e,d,, since b is regular. Now from d,a=b,c we get e,d,a=e,d,a=e,b,c=e,b2c. This gives d,a=b,c as és is regular. Thus we have shown that the relation ~ is independent of the particular choice of the common left multiple of a pair. It is straightforward to check that ~ is an equivalence relation. Let the equiva'ence class of (a, b) be denoted by a/b, and let Q denote the set of equivalence classes. In Q we define addition and multiplication by the following rules: a/b+-c/d=(d,a+b,c)/(b,d) where d,b=b,d with both b,, d, regular.

a/b - c/d=a,c/d,b show

where

that these operations

dja=a,d with d, regular. First one can are well-defined

and secondly that they

make Q into a ring which is a left ring of quotients of R. This is left as an exercise. Conversely, let Q be a left ring of quotients of R. We then show that

R

satisfies

the left Ore-condition.

Let a, bE R with b regular.

Then b-! and so ab-1€Q. Thus there exist c, d with d regular such that ab-'=d-1c, Hence da=cb as desired. This completes the proof of the theorem. As a special case of the above theorem we have:

2.2 TuHeoREM. Let R be an integral domain satisfying the left Orecondition. Then R can be embedded in a division ring Q which is a left ring of quotients of R. Proof. Since each nonzero element in R is regular, it follows from the construction of Q in the theorem 2.1 that if a/b is a nonzero element in Q then b/a is the inverse of a/b. This proves the theorem.

REMARK. If R is a commutative integral domain then the (left) ring of quotients of R is the usual field of fractions of R.

DeFINITION. Left Gre-domain. An integral domain satisfying the left Ore-condition is called left Ore-domain. Clearly a commutative integral domain is a left Ore-domain. Under what conditions a non-commutative integral domain is also a left

RINGS OF FRACTIONS

61

Ore-domain? We give below a sufficient condition for an integral domain to be a left Ore-domain in the following theorem which is a special case of a theorem of Goldie-Lesieur-Croisot. 2.3 THEOREM. Let R be an integral domain having no infinite strictly ascending chain of left ideals. Then R is a left Ore-domain and hence R can be embedded in a division ring which is a left quotient ring of R. Proof. We show that each pair of nonzero left ideals of R has nonzero intersection. If possible, let A, B be two nonzero left ideals in R such that ANB=(0). Let OAcEB. Then AM Rc=(0). Let m be a positive integer. We claim that the sum S° Ac? of left ideals Ac of i=i

R is a direct sum.

+amc”41=0, EA.

Let 0=a,c+a,¢?+...+a4,c™.

Then a,+a,c+...

—ayE AN Rce=(0). So a,=0. Continuing we get

all a;=0, 10

n=0 —n

times

=(—a)+(—a)+...+(—a) if n pr) fF" k

74

RINGS, FIELDS AND VECTOR SPACES =o pig fo; ', since

> px=identity k

=(ef)ii

This shows /—(f;;) is a homomorphism of rings. It is easily seen that this homomorphism is an isomorphism. As a special case of the above problems we have (d) Let Re,,..., Ren be a family of mutually isomorphic (as left R-modules) minimal left ideals generated by idempotents e,,..., €n respectively in a ring R with unity such that R= Re,®...@ Ren. Then R=D,, the nxn matrix ring over a division ring D. Problems

1.

If f:M—wN is an R-homomorphism R-module N, show that

of an R-module

M to an

(a) f(0)=0

(b) f(—x)=—f(x), xEM (c) f(x—y=f(x)—f(y), x, VYEM (d)

The set f(M)={f(m) |mEM)} isa submodule of N. (This submodule is also denoted as Im / and is called the homomorphic image of M under /.)

2.

Show that the set Ende (M) of all endomorphisms of an Rmodule M can be made into a ring by defining addition and multiplication suitably.

3.

If A and

4.

5.

B are

submodules

of an

R-module

M,

show

that

(A+ B)/A=B/(ANB) as R-modules. Let M be an R-module and xE€M be such that rx=0, rE R, implies r=0. Then show that Rx&R as R-modules. Let V bea vector space over.a field F generated by x,, X2,...,Xn and suppose that any relation a,x,+a,x,+...+4nxn=0, aEF

implies a,=O=a,=...=dn. Show that VF". 6.

7.

A linear transformation ¢ from R? to R? satisfies ¢(2, 1)=(—1, 1) and ¢G, 2)=(0, 5) Find ¢(1, 0), ¢(0, 1), ¢(@, b) for any a, bER. Two linear transformations ¢,, f2 from R* to R® satisfy (i) t,(2, 0, =, 3, 0), t, (3, 1, 0)=(4, 2, 1) and t,(0, 3, —1)=d, 1, —1) (ii). 4,(2, 1,0=(—1, 1,0). G, 4D @se2.,. 3) and #,(031, O=0, =F 2): If e,=(1, 0, 0), e2=(0, 1, 0), es=(0, 0, 1) find #, (e,), t, (es), ty (3), ty (a, b, c) and ft, (e;), ty (@2), t(€5), t2(@, b, c) for any

MODULES

8.

AND

VECTOR

SPACES

75

a, b, cER. Also find (t,+1,)(a, b, c) and (t, ty) (a, 8, c). Let M=K@K’=LOL’ be direct sums of submodules of M such Show that K’=L’. that K=L.

4 BASIS Throughout this section unless otherwise stated R is a nonzero commutative ring with unity 1, and M is a unital R-module. DEFINITION. Linearly independent set. Aset of elements x,, x9, ...5 Xn of M is called a linearly independent (/.i) set if for any a, a, ...,

Ore Ry Ses 0 mphies aa,

say

20.

i=1

The set {x,, X,, ..., Xn} is called linearly

dependent

if it is not

linearly independent. An infinite set (x;), i€ A, of elements in a module is called linearly independent if every finite subfamily is linearly independent. An example of a linearly independent set in an F-module F[x], the polynomial ring over a field F, is the set

6 De aisae

ee

The set {1, x, 1+, x?} is clearly a linearly dependent set. In the vector space V=R” over the real field R, consider the set {€,, €y, -~-, @n} where e; is the n-tuple in which all components except the i-th are zero and the i-th component is 1. This set {e,, e,, ..., en} is a linearly independent set in R” and is indeed a basis, according to

the following definition. DEFINITION. Basis. A subsetB of an R-module M is called a basis if (i) M is generated by B ; (ii) B is a linearly independent set. The basis {e,, €,, ..., en} of R” described above is called a standard basis of R". Of course not every module has a basis. For example, consider a cyclic group G and regard it as a Z-module. Then G hasa basis if and only if G is infinite. For, let G=(a) and ma, m € Z, be some basis element. Then A(ma)=0, A € Z must imply A=0. However, if G is a finite group of order n, then n(ma)=0, a contradiction. Hence G must be infinite if it has a basis. But if G=(a) is an infinite cyclic Gas Z-module. This proves our assertion. group then {a} is a basis of An R-module M is called a free module Free module. DEFINITION. words M is free if there exists a subset other In basis. if M admits a

76

RINGS,

FIELDS AND

S of M such that M is generated dent set.

VECTOR

SPACES

by S and S is a linearly indepen-

Let V be-a nonzero finitely generated vector space 4.1 TuHeorem. over a field F. Then V admits afinite basis. Proof. Let {x,, X2, .--» Xn} be a subset of V which generates V. If this set is not linearly independent, then

there exist «,, &, ...,@n in

F, not all zero, such that «,x,-++-o.x2-+...-+onxn=0. For simplicity we may, assume «, 40. Then x,=(.%2+ ...+8nXn where B.=—a!1e,, nes

C,—=—aj an. This shows that the set {x,,..., xn} also generates V. If {X,, ..., Xn} is a linearly independent set we are done, otherwise we proceed as before and omit one more element. Continuing like this we arrive at a linearly independent set which also generates V. Note that this process must end before we exhaust all elements; in the extreme case if we have come to a single element which generates V then that single element will form a basis since each nonzero element in V forms a linear independent set. Hence V does admit a finite basis. REMARK. By convention we regard the empty set as the basis of a zero vector space. This alongwith the above theorem shows that each finitely generated vector space has a basis with finite cardinality. 4.2 THEOREM. Let M be a finitely generated free module over a commutative ring R. Then all bases af M are finite and have the same number of elements. Proof. Let (e;),i€ A, be a basis of M, and let {x,, x5, ..., Xn} be a set of generators of M. Then each x; can be written as

=>

Ue,

oT Sn

i

and all but finite number of «;; are zero. Thus the set S of those e;’s which occur in expressions of all x;’s, j=1, 2, ..., n, is finite. Let us,

after reindexing, if necessary, write S={e,, @,, ..., er}. Since eachx € M can be expressed as a linear combination of X1, Xg, +++, Xn Over R and each of which in turn can be expressed as a linear combination of e,, e,,

..., er over R, it follows that M is also

generated by the set {e,, 3, ..., er}. However, {e,, e,, ..., er} being a subset of a linearly independent set, must be linearly independent and hence forms a basis of M; this being a subset of basis coincides with it. ' Next, let (€,, €2, ..., @m) and (ff, fa, .-.; fn) be two ordered bases of M. We show that m=n. Write m

n

K=>dX anjex and ej=> bj;fj. k=1

j=l

MODULES AND VECTOR SPACES

i)

Eliminating f’s between the above equations we get

C=

j=l

by Dany ek=S (SH i ani) ex k=1

k

7

==>) Cik Ce Where ei >. Djy'dz}.

k

j

By the linear independence of the set (e,, @., ..., @m), it follows cjj=1

and ci.=.0 if i4k. Now if weset A=the mn matrix (a,;) and B=the nxXm matrix (b;;) then by definition AB=(dx;) where dii=~ agy bji. ij But dki=cix.

Hence

AB=I,,,

the

m xm

identity matrix.

Henceforth

we shall omit the subscripts to denote the identity matrices. If possible, let m >.

Let us partition A, B as follows.

Sf AGy A-(47)

= B=(B,

Bs)

where A,, B, are both nxn matrices, A, is (m—n)Xn matrix, and B, is nX(m—n) matrix. From AB=/J, it follows

Pe

Ae

ae EB

4B)

Fe 4 A) (Ay B=(Fi ae} Thus A,B,=J, A,B,=0, etc. Then chapter 2, we have B,A,=J. Hence

Bit OMA,

by solved problem 3.5 (a),

OY.

(0° Z)(o* )=" Now from

we get

(0 (A) (a (0 2)lo a)=(o 1)(4:) & 9 (or i}

G H-(isjan mle mC, ta)

This gives

Hence

our

B,=0=A,

assumption m>n

and A,B,=I,

which

is false. Thus

is a contradiction.

m° qi; e, , =1, ..., 2, and hence the set { ia v3 €n} j=l

also generates F”. Since dimrF"=n, this set { ex, ee. en} must be a basis of F”, proving our assertion. Hence we conclude that every transformation of bases of F" corresponds to an invertible matrix and conversely. As an illustration of 5.1 consider a vector space V=F®? and bases B,=(e1, 22 €3), Bo=(S,, fo, fz) where the former is standard basis of Fe and fi=(1, 2, 3), £=(0, 1, 1), 4=(1, 0, —1). Then the matrix

of transformation from B, to B, is clearly the matrix P=} To

find

the

matrix

of transformation

1xed 0 1 10

from B, to B,, we

express each of e,, e, and e, as a linear combination Now if e,=(1, 0, 0) =anfitapfetaishs

3 ) -1 have

to

of the basis B,.

eo=(0, 1, 0) =o, fi +422 fot Ge fs €g3=(0, 0, 1)=31 fp +432 f2tOg3 Ss

then (1) (2)

1=a+443 0=24,,+4,.

(3).

0=34y,+ 42-43 esa)

a41;>2>

3

a.=—l,

aaa

ay3=4

Similarly a

te:

pad |

Q4=—F, Agg=2, Agg=% Q3;=%,

==

a32=

os

—1, 433—=

1

-—2

1

— 2

1

)It can be checked that PO =I=QP. —-1 5.2 Let U and Vbe two finite dimensional vector spaces over a field F with bases respectively B,=(e,, ..., en) and B,=(fy, .--,fm). Let A=(ajj) be a aes nxXm matrix. Then we can define a mapping Thus o-i(1

4 —2

f:U7>V bv (S aj ej)= ~.oj 2 aiifi, «jEF, which is easily shown to be well- defined and fides It is to be noted carefully that any linear mapping f is completely determined by the images, under f, of the

86

RINGS, FIELDS AND

VECTOR

SPACES

elements of a basis. Conversely, given a linear mapping f: UV, it determ

mines a unique n X m matrix A=(a;;) where f(ei)= > aij fj, i=1, 2,..., n. j=l The above discussion shows that there isa 1—1 correspondence between the set of nXm matrices over a field F and the set of linear mappings f: U->V where U, V are vector spaces of dimension n and m respectively over F. The 1—1 correspondence described above depends on the choice of bases of U(=F”) and V(=F”). Our purpose is to prove the following.

5.3. THEOREM. Let f: UV be a linear mapping and A be its matrix with respect to the pair of the bases B,, Bz of U, V respectively. Then the matrix A’ of f with respect to the new pair of bases BL. B, of UF is given by A'=PAQwhere P,Q are matrices of transformations of B, to B, and By to B, respectively. 5a)" Ba=Ohys © on) Proof. Let Bye), By =(e

SFereke'y

en)s Bs =(/,-

Then as in 5.2 A=(q;;) is an nm

: AD!

matrix, where

m

(1) LCDS

dijfj; t=),

Also A’=(qj;) is

Seen

nm matrix where

Q) fed=> ay t=1, ohn

j=l Let P=(pij), Q=(qij) be matrices of transformation of B; to B, and B, to B, respectively. Then as in 5.1, we have n

Gy)

G=S- pifenti=l, omen j=l U

.

m

(4) f= Cafe Jal, ...,m. k=1

where (qj; ) is the inverse of Q. From (3) we get

f(Gi)=&X Pi f(Cs) j=1

=>) > Pi ay fi, by (1) J=1 j=1

MODULES m

AND

m

=>

VECTOR

SPACES

87

n

> Dd pian gifs

kel ue

fst

j=l ,

>) ce fy» Say. k=1

Thus the matrix of f w.r. t. bases Bj, B3 is the matrix (cj,)=PAQ-. We now prove the converse of the above theorem.

5.4 THEOREM. Let B,=(e,,...,€n), Bo=(f,, ---,fm) be bases of vector spaces U(=F") and V(=F") respectively. Let f: UV be a linear mapping whose matrix with respect to this pair of bases (B,, Bz) is A. If P, O are any invertible nxn, mxm matrices respectively then PAQ- is the matrix of the same linear mapping f: U->V with respect to the pair of bases By =(e1,. eg e,) and B} =(fi, bis See) where ej =>" Diz ei and f ;= dX dikfe jal kml

Proof. fle=f(X pisei)=>dX puf es) j=l

m

=>

j=l

n

m

Vpyanfe=dX

k=1 j=1 ae

m

n

XK Vepyandiahr

J=1 k=1 j=1

,

=o cif, where c= >) pij Qik Gk 1=1

Thus (ci) =PAQBi and B35.5

i, k is the matrix of f with respect to the pair of bases

Solved problems

(a)

Let ~:R°—R? be the projection 7(q, b, c)=(@, b).

“Let By=

(€,, 2, €3) and B,=(f,, fz) be standard bases of R? and R?® respectively. Then the matrix corresponding to ™ w.r. t. bases B, and B,

boy (P20

is

0

Solution.

1

x(e,)=7(1, 0, )=(1, )=1f,+0f2

x (e)=7(0, 1, 0)=(0, 1)=0/,+Ife +2 (0, 0, 1)=(0, 0,)=0f, 7 (e3)=7

ji tage Osa Thus matrix ofx is ( Oar

88

RINGS, FIELDS AND

VECTOR

SPACES

(b) In the above example (a) the matrix of za w.r.t. B, ={(1, —1, 0), (1, 0, —1), (1, 1, D} and B,={(4, 3), (3, 2)}

—5 —2 1

is A'=PAQ=1=|

=

bases

#! 3 —1

eter) 1

where

—1 0

P=

0 —1

e(S jpom(3 2

Solution.

To find P=(p;;), we write (1, —1, 0)=1e,—e,+0e,, (1, 0, —1)=1e,+0e,—e,

(1, 1, I)=e,+e.+e, 1

Thus P=|

—l1

1

QO

1

1

0

—1 1

To find Q-1 we write

(1, 0)=4n (4, 3)+-9i2 (3, 2) (0, 1)=qai (4, 3)+-@22 (3, 2) Then

1=49

+3qi2, O=39i +2¢h

This gives gj, =—2, giz =3

Similarly g4 =3, q3, =—4 Thus

Supe o7=| 3

Then

Pee gd’ (1, lids:

3 oH

Ob

—5 (= 1

Taye foe lg

gO ecru tlah0 a | ‘ ee NG. 0

7 ‘} by actual multiplication of matrices. —1

(c) Let (e,, e,) be the standard basis of R2. Let (ey €) be a basis of R° obtained by rotating the vectors e,, e, through an angle @ in the counterclockwise direction. Then the matrix of transformation of

basis (c,, ¢) to a basis (e’, c4) is Ghee ia a Solution.

Write

.

€1=e; Cos 0—e, sin 0 €,=e} sin 0+e; cos 0

MODULES

AND

VECTOR

SPACES

89

Thus the desired matrix is cos 0 —sin 6 ks 0:

0s )

(d) Let W be the vector space of real valued differentiable functions generated by {e', ce}. Let D:W-sW be a linear mapping where D(f)=f',

the derivative of f£ Then the matrix

associated

with D

relative to the basis B, where B=(c‘, e*) is

Solution.

(0 2)

(a2 D (et)=et=1e'+0e D (e7*)=2e?t =0e! + 26%

the matrix is fe 3)

6

NORMAL

6. In this section we theorem 5.4.

FORM give

an

OF A MATRIX important

application

of the

6.1 THEOREM. (Normal form of a matrix). Let A be annXm matrix. Then there exist invertible matrices P, Q such that

rao-(%

where I, is the rXr identity matrix. Prod. Let (e;,.-..€n) And,( fr. 2.97) be vases. of F” and “F* respectively. As in 5.1 let f: F"-+F™ be a linear mapping corresponding to the given matrix A. Now ker fis a subspace of F”. We choose a basis of ker f and extend it by theorem 4.8 to a basis of F”. Let (U1). 2.5 Up, Urti;.++, Un) be that basis where (uU,41,..., Un) is a basis of kerf. Then { f(),..., f(ur)} is a linearly independent set in f” and we extend this set to a basis of #”. Corresponding to these new

bases of F” and F”, the matrix of f is clearly Gr on Further, know by theorem

that PAg=(5

we

5.4 that there exist invertible matrices P, Q such

:),and this proves the theorem.

To get a practical method of reduction we define elementary operations on matrices.

Let R,,..., Rm be the Elementary operations on rows. DEFINITION. on rows are operations following the Then rows of a given matrix A. called elementary operations:

90

RINGS, FIELDS, AND VECTOR SPACES (i) Interchange R; and R;. We denote this operation by Rie Rj. (ii) Replace R; by AR; where AAO. We denote this operation by

R;—>)AR;j.

(iii) Replace R; by R; +R;

(j4i).

We

denote this operation by

Ri Ri +pR;.

The elementary denoted.

operations on columns

are similarly defined

and

We observe that the effect of any of these on the rows (columns) of a matrix A is the same as multiplying A on the left (right) by a certain elementary matrix, namely the matrix obtained by applying the given elementary operation to the rows of the identity matrix. Thus if

v2 3 A={| 10 1 bs ST Sy then R,«>R, corresponds to left multiplication by O40 oid

Or EO

Tie:

DOr

OF O48 O. ef eo LOO

Ne Aaa 1 es slat el —1 0 1

—1 0 ji aloes eB a reas

We want to illustrate that any given matrix can be brought to the form (0 a by elementary

row

and

column

operations.

We first

transform it to echelon form, namely the matrix having in each row the first nonzero element (if any) 1 and for any such occurrence of 1, all the elements which lie below it in the same column or an earlier column zero. After bringing into echelon form, we perform, if necessary, column operations to complete the solution. The column operation shall not be needed if A is invertible (—nonsingular). In case A is invertible we shall arrive at PA=J and then Q can be chosen to

be identity. This will also then provide a method for computing the inverse of a matrix if it exists. 6.2

(a)

Solved problem

Reduce the matrix

Al

0 fo 3 nee elu yeseer Ged Pyle ae lO RTO RORY

MODULES.

VECTOR

al

SPACES

70 (0 0 )

to the normal form Solution.

AND

Consider the two matrices

Oy ies ha 9 2-1 0 2-6

ae were aang

ypHg

26071

1

Te O50 0 1 0

Ph EHP g 64

iy?

Ora

@

re OCS

and first determine an invertible 4x4 matrix P such that PA is in the echelon form. For this purpose we reduce to the echelon form by elementary operations on the rows of A and at the same time we also perform the same set of elementary operations in the same order on rows of the 4x4 identity matrix to get the matrix P. Perform R,R, in both and get Obit (O50 2-6 0 2—1 0.

alo

Pout

Dont)

.0,

6
R* bea linear mapping defined by f(v)=av where | v € R®* and a is some fixed element in R. Find the matrix of f where the bases are taken bases.

3.

as standard

Consider a vector space V=F® over a field F. In each of the

MODULES

AND

VECTOR

SPACES

following find the matrix of transformation basis B,. (a) Bailey O)a (=F 3 0), (0, +1,..2)} Be

12... 15 1,0; 1,0),

95

ofa basis B, to a

1.1}

cb) B, =4(3, 2, 1), 0; —2, Seah

ey

B,={(1, 1, 1), (—1, 2, 4), @, —1, 0)}.

4.

Sea

In each of the following let D=d/dx be the derivative. Let V be a vector space of real-valued differentiable functions on the interval [0, 1] generated by a basisB. Find the matrix associated with D relative to the basis B.

CE

Cage a 8

(b) {1, 7},

(c)

ah t, ek tae

(d)

aet A=( ; ey

{sin t, COS

th.

) Find matrices P, Q such that

p4o=( Peet, A oF 6.

For each of the following matrices A find matrices P, Q such that

Pee) PAg=( 0 0 )

—1 1 a

a2 ~

5 2 0

o EZ Van Se

Ee EE

RF OWA ON’ DRUNK RFP YR ONQWN Qe

7

ROW

RANK

AND

Let f: U->V be a linear transformation

space U to an m-dimensional vcctor earlier section thatker f andf(U) tively subspaces of U and V.

DEFINITION.

of an n-dimensional vector

space V. We had seen in the

(also written as Im f) are respec-

Rank of a linear transformation.

subspace f(U) of V is called the rank f: U-Vv.

NULLITY

The dimension of the

of a linear transformation

96

RINGS, FIELDS AND VECTOR SPACES e(/) will denote the rank of f.

DEFINITION. Nullity of a linear transformation. The dimension of the subspace ker fof U is called the nullity of linear transformation f: UV. r(f) will denote the nullity of f The relation between the rank and nullity of a linear transformation is given in the following theorem. 7.1 ‘ RANK-NULLITY THEOREM. Let f: UV be alinear mapping where U and Vare finite dimensional vector spaces respectively over a field F. Then

e(f)+r(f)=dime U. Proof.

By the fundamental theorem of linear maps U

ker ¥ as f{G) Hence by the problem 6 (iii) in the section 4, dimp U—dimp (ker f) =dim,y (f(U)). This proves the theorem. We now describe the relation between the rank of a linear mapping f:F"-F"™ and the number of linearly independent rows of the matrix of f w.r.t any pair of bases B,=(e,,...,éen) and B=

(fi. -» fm) Of F" and F* respectively. 7.2

THEOREM.

Let A=(a;;) be the

nm matrix of a linear mapping

f: F* >F" w.r.t the pair of bases B,=(e,,. £., en) and B,=-(f,,..., fm) of F" and F™ respectively. Then the number of linearly independent elements in the set {f(e,), ..., f(en)}, that is the rank of f, is equal to the number of linearly independent rows of the matrix A (rows are regarded as members of the vector space F). Proof. ©Write’ -c=(an; Gp; 2... Gnet, (isl, Soe pale C1,.-+, Cn are the rows of the matrix A. Let W, be the subspace of F” generated by f(e;), i=1,..., n and W, be the-subspace of F*

generated by cj, defined by

i=1,...,

nm. Consider

o (S oj fe)

Now >) a f(e)=0@>) i=l

isl

Dp

eee

mM,

os

i=1

mapping

“iC;

SY aaj fr=0> LY aia;j=0, j=l

n

jas;

a

“ic; =0.

i=1

o : W,-W,

MODULES

AND VECTOR

SPACES

97

Hence o is a well-defined and 1—1 mapping; and it is clearly a linear mapping. Thus by the fundamental theorem of F-homomorphisms W,=0(W,). Therefore dimrW, D5 ui f(ei) i=1

i=1

isa 1—I1 linear mapping of W, into W, and thus dimp W,- Gisfi, f=i j=1 i=1,..., 7m then as discussed in 5.1, B= (e,, ark e,) and Bolu Ce fe) respectively are also bases of F” and F”. Then by theorem 5.3, the matrix of f with respect to bases Bj, Bj, is PAQ. Invoking theorem

7.2 again we obtain that the row

rank

of PAQ=rank of f. Hence

the row rank of A is same as the row rank of PAQ.

7.4

THEOREM.

pag=(5

The row rank of amatrix

A is the integer r where

a P, Q being suitable invertible matrices and I, the rxr

identity matrix. Proof. Follows by theorem 6.1 and theorem 7.3.

8

COLUMN

RANK

OF A MATRIX

Let F be a division ring. Then F” can be considered both as a left and a right vector space over F. (If F is a field then the two concepts are the same.) Thus if A is an nm matrix over F, then we shall regard the columns of A as elements of the right vector space F”.

DEFINITION.

Column

rank of a matrix.

Let A be an nXm

matrix

RINGS, FIELDS AND VECTOR

98

SPACES

over a division ring F. Then the maximum number of linearly independent columns considered as vectors of the right vector space F* is called the column rank of A.

If Fis a field then the qualifying word right before vector space in the above definition may be omitted. Our purpose in this section is to prove that if A is any xm matrix over F then the maximum number of linearly independent rows of A, the rows being considered as elements of the left vector space F”, is equal to the maximum number of linearly independent columns of A, the columns being considered as elements of the right vector space F”. Let

U,V

be

right vector

spaces

of dimensions

n, m

respec-

tively over F, and f: UV bea linear mapping. Let (e,,...,én) be a basis of U over Fand (f,,...,fm) be a basis of V over F. If m

f(en=>d fj ay, =1, ...,0 j=l then the matrix A=(q;;) is called the matrix of f with respect to this pair of bases. Thus f determines a unique mXn matrix. Conversely, any given m Xn matrix A=(a;;) determines a linear mapping f: USV, defined by the above system of equations. Thus as in the case of a left vector space there is a 1—1l correspondence between linear mappings f : UV and the set of mxXn matrices over F. It can be shown exactly as in theorem 5.3 that the matrix of f:U-V with respect to new pair of bases Bi=(e1, eee and B, = ‘eeae al of U and V respectively is G-!AP where the matrices n

P=(pij) and Q=(qxi) are determined by the equations e;=S° e; pj,

71

m

i=1,...,n, and f, =>) fignw, k=1, ...,m. I=1 linear mapping between

Further,

the rank ofa

right vector spaces canbe defined as before

and it can be proved as in theorem 7.2 that the rank of fis equal to

the column rank ofany mXn matrix which represents f. Asa consequence it follows as in theorem 7.3 that if.A is any mxn matrix

and P, Q are any invertible mxm,nxXn matrices respectively then the column rank ofA is equal the column rank of PAQ. We are now ready to prove 8.1

THEOREM.

Let A be an

mXn

matrix

over a division

ring F.

Then the row rank of A (that is the maximum number of linearly independent rows of A, the rows being considered as elements of the left vector space F") is equal to the column-rank of A.

MODULES

Proof. equal to invertible a matrix

AND

VECTOR

SPACES

99

We have shown that the row (or column) rank of 4 is the row (orcolumn) rank of PAQ where P, QO are any matrices of orders mXm,nXn respectively. Further, for A over a division ring F we can, as in theorem 6.1, select

invertible matrices P, O such that PAQ=(5 identity matrix.

5)where J, is therxr

Since it is clear that the row rank as well as the

column rank of ( e # is r, the proof of the theorem is completed.

In view of the theorem 8.1, the row rank, or the column rank, is also called the rank.

8.2

THEOREM.

Let

yy XTAygX—

Am1X4+

eee

+ aynX,=0

Gm2X_+ 2+ e+ AmnXn=0

be a system of the linear homogeneous equations inn unknowns with the coefficients from a field F. Then the set of solutions forms a vector space over F of dimension n—r, where r is the rank of the coefficient matrix

A=(ai)). Proof.

The given system of equations can be written as A (X45, Xos0%=>.Xn)* =O.

Consider the matrix A as acting on the vector space F”. (We write the elements of F” as columns.) Then A is a linear mapping of vector space F” to vector space F™. That the set of solutions is a subspace of F” is obvious. The solution space is then the null-space of the transformation A. Hence by the Rank-Nullity theorem rank of the linear transformation A- nullity of linear transformation A=n,

that is,

the rank of the matrix A+dimension of the solution space=n Thus the dimension

of the solution

space==n—r,

proof. 8.3

(a)

Solved problems

The rank of a3x4 matrix 2 8 fol ye FeO Osea S22 SABRE Gib $0 21

is 2 and the nullity is 1.

é

completing

the

RINGS, FIELDS AND

100

SPACES

VECTOR

Solution. We perform elementary row and column operations on Ato reduce it to the normal form. First we perform row operations simultaneously on the matrix A and 3x3 identity matrix / till A is reduced to the echelon form

A=

2t—% 1 —1 5 —20 —4 —22 mo NL Se ee

=|

1 206 0.1 0 1 ie oa |

Performing R,—2R, we obtain B=

2 —8 Lre+l 1—4 —6 —20 mes ete Daca

J=|

1 00 —2 1 0 OO od

Performing R,R,, we obtain 1 —4 —6 —20 —2 1 0 C= 2 —8 1 —1 K= AS ale a= Sok Was 720 OS Overt Performing R,—2R, and R,+3R,, we obtain 1 —4 —6 —20 —2 10 D=}] 02 Oy. 13, 239 L= > == 258 0 O—13 —39 =G2 90s Performing R,-+R, and ;4;R, in succession we get,

1 —4 —6 Ds

Oe

—20

O20

—2

ti

ih

oe

The above matrix £ is in echelon form butas in the previous section we shall reduce it further to the reduced echelon form in which all elements above the leading element of any nontrivial row are equal to 0. So performing R,+6R,, we get 0

1—4 Jas

0

0

One

oe

—2

050730

ae

3

Bs

—

A ee

ly

0

+3;

0

Pte

F is now in the reduced echelon form and NA=F. We now consider the matrices 1 —4

0 -—2

Fat 0i 5 0 A eS 0 00 O

and

J= oo oF (eee eK)oO © or less) eo heres a

MODULES

AND

VECTOR

SPACES

101

and perform elementary column operations on both till Fis reduced to the normal form.

Performing C,+4C,, C,+2C,, Cy—3C,, we get ‘ein eta

FH-[-0 of 01 0 eres

Qi

2

pen 00 fe 1-3 O0e:

Oe

uh

Performing C,F* be a linear mapping defined by f(a, b,c)= (2a, 3b, c, 0). Find the rank and nullity of f Show that LetA, B be two matrices which can be multiplied. the rank of AB1

and if p(x)=

DP, (X) Po (x) with p, (x), P2(x)€ F[x] then either p, (x) € For p,(x)EF. It is worth pointing out that irreducibility of a polynomial f(x) € R[x] over aring R depends on the nature of R. For example, x2—5 € Z[x] is irreducible over Z but is not irreducible over Q(«/5). Similarly, x?--1 € R[x] is irreducible over R but is not irreducible over C. We call a polynomial f(x) € R[x] reducible over a ring R if f(x) is not irreducible over R. 1.2 Proposition. Let f(x) € F[x] be a polynomial of degree>1. If f(«)=0 for some « € F, then f(x) is reducible over F. Proof. By 1.1 (i) we can write f(x)=(x—«' q(x)+r where r € F. Then 0=f(«)=r. Thus f(x)=(x—«) g(x) since f(x) is of degree>1, q(x) & Fand hence f(x) is reducible. An element « € E where E is a field containing F as a subfield is called a zero or aroot of a polynomial f(x) € F[x] if f(@)=0.

138

RINGS, FIELDS AND VECTOR SPACES

The converse of 1.2 holds for a certain class of polynomials as given in the following proposition. 1.3 Proposition. Let f(x) € F[x] be a polynomial of degree 2 or 3. Then f(x) is reducible if and only if f (x) has root in F. Proof. If f(x) is reducible then f(x)=f,(x) f2(x) where f,(x), fox) are nonconstant polynomials each of which has a degree less than 3. But this implies that either f(x) or f,(x) must be of degree 1. Let

S(x)=ax+b

with a~40. Then f,(—ba-)=0 and hence f(—ba)=0

proving that position.

—ba-!

is a

root

of f(x).

This

proves

the

pro-

Recall that a polynomial f(x) € Z[x] is primitive if the greatest common divisor of the coefficients is 1. DEFINITION. Monic polynomial. A polynomialaj+a,x-+...—+dnx" over aring R, with a,40 is called monic if a,=1. Clearly every monic polynomial f(x) € Z[x] is primitive. The following lemma was proved earlier.

1.4 Lemma. If f(x), g(x)€ Z[x] are primitive polynomials their product f(x) g(x) is also primitive.

then

1.5 Lemma. (Gauss). Let f(x) € Z[x] be primitive. Then f(x) is reducible over Q if and only if f(x) is reducible over Z. Proof. If f(x) is reducible over Z, then f (x) is reducible over Q. Conversely, suppose f(x) is reducible over Q. Let f(x)=u(x) v(x) with u(x), v(x)E Q[x] and u(x) EQ, v(x) FQ. Then f(x) =(a/b) u'(x)v'(x) where u’(x) and v’(x) are primitive polynomials in Z[x]. Then bf(x)= a(u'(x) v'(x)). The gcd of coefficients of bf(x) is b and the gcd of the coefficients of a u’(x)v'(x) is a, by lemma 1.4. Hence b=a and so f(x)=u'(x) v(x). Therefore, f(x) is reducible over Z [x].

Note.

deg u’(x)=deg u(x) and deg v’(x)=deg v(x).

1.6 Corotiary. Let f(x) € Z[x]. Then f(x) is reducible over Q if and only if f(x) is reducible over Z. Proof. Write f(x)=bf'(x) where f’(x) € Z[x] and is primitive. Now apply the above lemma to f’(x).

1.7 CoROLLary. Let f(x)=a)+a,x+...+x" € Z[x] be monic with a,40. If f(x) has a zero in Q then f(x) has a zero m€ Zandm ing

Proof. If n=1, the result is trivial; so assume n>1.: Since f(x) is monic it is primitive; so Gauss lemma applies. If f(x) has a zeroin Q, then f(x) has a linear factor over Q and hence is reducible over Q and hence by Gauss lemma it is reducible over Z. Then by the

ALGEBRAIC

EXTENSIONS

OF FIELDS

139

notein the proof of Gauss lemma f(x) hasa linearfactor over Z, say Bax with «, 8 € Z. Then ay+a,x+.. .+dn—y xP 2+x"=(8+ax)(by+ -ee$bnax") with bo, ..., bnr€Z. Then ay=by, so B|a,. Also 1=abn-1 So a=+1

and b,-;=+1.

Therefore,

8-+ax=8-+x.

Hence

6 or —f is a root of f(x). 1.8 THEOREM. (Eisenstein criterion). Let f(x)=a,ta,x+...+ anx" € Z[x]. If there is a primep such that p* + a, p|ao,p|a,..., P|@n-1, Pt Gn, then f(x) is irreducible over Q. Proof. We will show f(x) is irreducible over Zand then the result

follows from the above

corollary 1.6. Suppose f(x)=(bp+b,x+...

+brx") (Coteyx+... esx’) with bi, ce; € Z, b+AOKes, rayta,x4+...+anx"+(f(x))—> ag ta,b+.. -a,B"

Let us fenpts thisisomorphism of F(«) to F(f) by o. Clearly o(«)=8. Consider the diagram

Ree FG

F(a) ———>F@) We

know

o can

be extended

to an

isomorphism o* from F to

F(®@)=F. Then o* induces a ring homomorphism.

4: F[x]>F [x] given by 7 (a)ta,x+...+arxt)=0* (ay)-+0* (a,) x4... +0% (ar)x". We note y (f(x))=f(x). Since n(x—a«)k=(x—B)*, we set (x—B)* is a factor of f(x) and so k’>k. Theroles of « and 8 can be interchanged

to show k>k’. Hence k=k’.

160 6.7

RINGS, FIELDS AND VECTOR SPACES CoroLtary.

f(x)=a

> (x—«)*

Jf f(x) € F[x] where

a; are

is

irreducible

roots of f (x) in its

over

F,

then

splitting field

i=1

over F. Proof.

Obvious from proposition 6.6.

6.8 Solved problem (a) Let K=F(x) be the field of rational functions in one variable

x over a field F of characteristic 3. (Indeed F(x) is the field of fractions of the polynomial ring F[x].) Then the polynomial y?—x in the polynomial ring K[y] overK is irreducible over K and has multiple roots. Solution. If y?—x has a root in K then thereexists g(x)/h(x) in K with h(x)~0 such that (g(x)/h(x))®=x, that is, g3(x)—xh? (x). But this implies that 3 (degree of h(x))+1=3 (degree of g (x)), which

is impossible. Thus y3—x€K[y] is irreducible over K. Now if 8,, 8, are two roots of y*—x in its splitting field then B}=x=63.

But then

(8,—B2)® =P? +(—1)?B3=0 and hence B,—f$,=0. This shows that y®—x has only one distinct root whose multiplicity is 3. This completes the solution. In the next section we shall show that an irreducible polynomial over a finite field has all roots simple. Hence it will follow then that the only fields over which an irreducible polynomial may have multiple roots are infinite fields of characteristic p40. Problems

1.

Verify that (f(x)+g (x))’=f" (x) +8" (x).

2.

Let Fbe a field and a,hcF.

F@)EF{x],

Show that for any polynomial

: flath=> a S®™ (a)h* kao *

where f

3.

(a) =(f >) (a))L ands (Ota)i (6) kh2, se

(f “ (a) is called kth derivative of f(x) at x=a). f(x)EF[x]hasaroot « of multiplicity n>1 ifand only iff «)—0, k=1,...,n—1

and f™) («)0, where f() («) is the ith derivative

of f(x) at x=a as defined in problem 2. 7

FINITE

FIELDS

Let F be a finite field with q elements. Consider the homomorphism

ALGEBRAIC

EXTENSIONS

OF FIELDS

161

f:Z>F sending | onto 1. If kerf=(0), then Z is embeddable in F which we

have assumed to be finite. But this is impossible. Thus kerf(0). Since Z is a principal ideal domain, there exists a positive integer n such that kerf=(). Then Z/(n)=Im fC F. Thus Z/(n) is an integral domain. But then 7 must be prime. Otherwise n=n,n, with n, a;(x?)!. Since i=0 Set bi=al Pr. These (x) has multiple roots if and

only if f(x)= rsbixr=( bix')P, a contradiction

since f(x) is

=0

irreducible. Thus F(x) saul have distinct roots. The next problem is a converse of the Theorem 7.6.

(c) If the multiplicative group F* of nonzero elements of a field F is cyclic then F is finite. Solution. Let F*=(«) where « generates F*. If F* is finite then F is finite and we are done. So assume F™ is an infinite cyclic group. Case 1. The characteristic of F is p>0.

In this case F=F,(«) where Fy, is the subfield {0,1,2,..., p—l} of F. Consider 1+-«. If 1+a=0 then «?=1, a Sono ya since F* is infinite. If 14+-«A0then1+«€F* and so 1+a=«" where ris some positive or negative integer. In either case 1+«—a" yields a polynomial over F, with «as its root. Thus « is algebraic over F, and so [Fp («) : Fp]=degree of the minimal polynomial of « over F,=r, say.

Then F=F,(«)

has p* elements,

a contradiction.

So either the

characteristic of F is 0 or F* must be finite. Case 2. The characteristic of F is 0. Here O4—1EF. So —1=ar where r is some positive or negative integer. In any case this gives that « is algebraic over F, and as before we shall get that F=F, («) is finite, a contradiction. Hence F* must be finite and so F must be a finite field. (d) The group of automorphisms of a field with p” elements is. cyclic of degree 7 generated by 9 where 9 (x)=x?. Solution. Let F be a field with p" elements. Let Aut (F) denote the group of automorphisms of F. Clearly the mapping 9: F->F defined by 9 (x)=x? is a homomorphism. Since x?=y? implies x=y, gis 1—1 and hence onto. Thus p€ Aut (F). We note that »"=identity because

on (x)=xP"=x

for all xeF. Let d be the order of the element » of

the group Aut(F). We have 94(x)=xP4 for all xe F. Hence each x€F is a root of the equation

ALGEBRAIC

EXTENSIONS

OF FIELDS

165

xP4—_x=0 This equation has p4 roots. It follows that d>>n, whence d=n (Why?). Let « be a generator of the multiplicative cyclic group F*. Then F=F, («) where Fy, is the subfield of F with p elements. Let F(x) be the minimal polynomial of « over F,. Clearly the degree of f(x)=n. We are interested in counting the number of extensions of the identity mapping

A: Fy>F to the mapping A*: F>F

This will then give any automorphism of Let 8 be a root of define an extension of

us all the automorphisms of F since clearly F keeps each element of F, fixed. f(x). Given an element g(«)€F(«)=Fla], we A by

g («)—g (8). This is in fact well defined, i.e. independent of the choice

of poly-

nomial g(x) used to express our element g(«)E F(a). Indeed if h(x)EF[x] such that g(«)=h («), then (g—A)(«)=0, whence f(x)| (g(x)—h(x) ). Hence g (%)=hA(8) since f(8)=0.

Thus for each root B

of f(x), we have constructed an automorphism of F which sends « to 6. Since for any automorphism o of F, o(«) is againa root of f(x),

it follows that the number of automorphisms of F is equal to the distinct roots of f(x). However, by solved problem (b), f(x) has all its roots distinct. Thus the order of the group Aut(F) is n. We

had

shown

in the beginning

that

there

exists an

element

o€ Aut (F) such that order of ¢ is n. Hence Aut (F) is a cyclic group generated by 9. Problems

1.

If F isa

finite field of characteristic p, show that each element

a of F has a unique pth root ?/ain F. 2. 3. 4.

5.

Construct fields with 4, 8, 9 and 16 elements. Find generators for the multiplicative groups of fields with 8, 13, 17 elements. Find generators for the group of automorphisms of fields with 4, 8, 9, 16 elements. Let F bea field with 4 elements. Find irreducible polynomials over F of degrees 2, 3, 4.

166

RINGS,

FIELDS

If F is a field and f: FF

AND

VECTOR

SPACES

is a mapping defined by

Tisx ou seu f@)=0 if x=0, show that fis an automorphism of F if and only if F has at most 4 elements. If F isa finite field then HU{0} is a subfield of F for each subgroup H of the multiplicative group F* if and only if F* is either 1 or a prime of the form 2"”—1 where n is a positive integer. Let F be a finite field with g=p” elements and p42. Show that a nonzero element of F has a square root in F if and only if al¢-22—1],

CHAPTER

9

Normal and Separable Extensions 1

NORMAL

EXTENSIONS

Let (fi(x))je4 be a family of polynomials of degreeD1 over a field F. In the previous chapter we had defined splitting field of a polynomial over F. Now by a splitting field of a family (fi(x) Jie, of polynomials we shall mean an extension E of F such that every fi(x) splits into linear factors in E[x], and E is generated by all the roots of the polynomials f;(x), iG A. If A is finite and our polynomials are f,(x),..-. fn(x) then a splitting field for them is a splitting field for the single polynomial f(x)=f,(x)..-fn(x) obtained by taking the product. The proof of uniqueness (up to isomorphism) of a splitting field of a single polynomial can be extended to prove the uniqueness (up to isomorphism) of a splitting field of a family of polynomials over a given field and is left as an exercise. We showed in solved problem (i), section 5, chapter 8, that if E is a splitting field of a polynomial f(x)€ F[x] then each irreducible polynomial p(x)€ F[x] over F having one root in E has all roots in E. The next theorem proves a set of equivalent statements which includes a converse of this fact.

Let E be an algebraic extension of a field F, conTueorem. s tained in an algebraic closure F of F. Then the following condition are equivalent: (i) Every irreducible polynomial in F[x] which has a root in E splits into linear factors in E. (ii) E is the splitting field of a family of polynomials in F[x]. (iii) Every embedding « of E in F which keeps each element of F fixed maps E onto E. (In other words o may be regarded as an automorphism of E.) Proof. (i)=(ii). Let «EE and let pa(x) be its irreducible polyit nomial over F. By (i) Pa(x) splits into linear factors in E. Thus 1.1

follows immediately that E is a splitting field of the family (px(*)), ak.

168

RINGS, FIELDS

AND

VECTOR

SPACES

(ii)> (iii). Let (f(x)), iGA, be a family of polynomials of which E is the splitting field. If « is a root of some f;(x) in E, then for any embedding o of E into F which keeps each element of F fixed we know o(«) is a root of f;(x). Since E is generated by the roots of all the polynomials f; (x) it follows that «maps E£ into itself. Thus by theorem 3.8, chapter 8, o is an automorphism of E. (iii)>(i). Let p(x)E F[x] be an irreducible polynomial over F

which has aroot «€E. Let BEF be another root of p(x). We show BEE. Since «, B are roots of the same irreducible polynomial p(x), we have isomorphisms

F («) =Flx\l(p (x) =F @).

Let o: F(«)—>F (8) be the isomorphism given above. Then o(«)=8 and o(a)=a for alla€ F. By theorem 4.5, chapter 8, o can be extended to an embedding o*: E->F.

But then by (iii) o* is an automorphism and hence o*(«)=o(«)=8 € E. This completes proof of the theorem.

DEFINITION. Normal extension. An extension E of a field F is called normal extension if EF satisfies any one of the equivalent statements of the theorem 1.1. 1.2

Examples of normal extensions

(a) Cis a normal extension of R. (b) R isnota normalextension of Q. For x*—2 € Q[x] is irreducible over Q, has a root ne in R but does not split into linear factors in R since it has complex roots. (c) If a=cos x/4+i sin 7/4 then Q(«) is a normal extension of Q. This follows from the fact that Q(«)is splitting field of x4+-1 € Q[x]. (d) In general, any extension E of a field F such that [E : F]=2 is anormal extension. Let «€ E,a@F. Let p(x) be the minimal polynomial of « over F. Then [F(«): F]=degree of p(x). Since [E: F («)] [F («):F]=[E: F]=2, we must have [E: F(a)]=1 and [F(«):F]=2. Thus E=F(«) and the degree of p(x) is equal to 2. Since p(x) has one root « € £, it must have its other root also in E. Hence E£ is splitting field of p(x) € F[x] and is thus a normal extension of F.

1.3 Solved problem Let E be a finite extension of F. Then Eis a normal extension of F if and only if E is a splitting field of a polynomial f(x) € F[x].

NORMAL

AND

SEPARABLE

EXTENSIONS

169

Solution. By hypothesis E=F (a, ..., %n) where w€E are algebraic over F. Let p;(x) be the minimal polynomial of «; over F. Assume first E isa normal extension of F. Then p;,(x) splits in E since it has one root a; € E. Thus f(x)=p,(x). . .pn(x) © F[x] has all TOOtS Ine. Since B= F'(G5. 15%), ANd. 445. 4,0, are some.of:. the roots of f(x), E must be the splitting field of f(x). The converse follows from the theorem 1.1. Problems

1.

Which of the following extensions are normal over Q?

(a) Q(v—3) (b) Q(5v7) () Q(v—1) (d)

Q(x) where x is not algebraic over Q.

Is R(/ —3) normal over R ?

4.

Let E be a normal taining F. Then E ple to show that K Let F=Q(vV 2) and

5.

tension of F, F is a normal extension of Q but £ is not normal extension of Q. Show that every finite extension of a finite field is nornal.

WN

2

extension of F and K be a subfield of E conis normal extension over K. Give an examneed not be normal extension of F. E=Q (4/2). Show that £ is a normal ex-

SEPARABLE

EXTENSIONS

Separable polynomials. An irreducible polynomial f(x) DEFINITION. € F[x] over a field F is called a separable polynomial if all its roots aresimple. Any polynomial f(x)€ F[x] overa field Fis called separable if all its irreducible factors are separable. A polynomial which is not separable is called inseparable. Let E be an extension of a field F. Separable element. DEFINITION. An element «GE which is algebraic over F is called separable if its minimal polynomial over F is separable. An algebraic extension E of a field F is called separable if each element of E is separable over F.

2.1

REMARKS.

(a)

It follows by corollary 6.5, chapter 8 that any

polynomial over a field of characteristic zero is separable.

is a field of characteristic 0 then separable.

Thus if F

any algebraic extension of F is

170

RINGS,

FIELDS

AND

VECTOR

SPACES

(b) By solved problem 7.9 (b), chapter 8, irreducible polynomials over finite fields have distinct roots. Hence any algebraic extension of a finite field is separable. (c) It was shown in solved problem 6.8 (a) chapter 8, that if K= F(x), is the field of rational functions over F in a variable x then

the polynomial y3—x€K [y] is irreducible over K. Also y?—x has all its roots equal each being «, say. Hence K(«) isnot separable extension of K.

DEFINITION. Perfect field. A field F is called perfect if each of its algebraic extensions are separable. Examples of perfect fields are (i) fields of characteristic zero (ii) finite fields. We remarked that infinite fields of characteristic p>0O have inseparable extensions. Thus such fields are not, in general, perfect. Anextension E of a field F is called simple if E=F («) for some “EE. 2.2 THeoREM. Jf E is a finite separable extension of a field F then E is a simple extension of F. Proof. Wf Fis a finite field then by corollary 7.7, chapter 8, each finite extension E£ of F is simple. So suppose now that Fis infinite. Since E is a finite extension of F, E=F(a,, ..., an) where ai;CE, 1(i).

NORMAL AND SEPARABLE EXTENSIONS

177

(ii)> (iii). Follows from theorem 3.5. (iii)> (i) By relation (2) in the beginning of the proof of the theorem

[E: E)|=| G(E/F) | Thus by (ili) we get [E: Ey]=[E: F]. Hence E,=F, theorem.

proving

(iii)>(ii). This completes the proof of the

Solved problems (a) The group G(Q(«)/Q) to Klein’s four-group.

where

«'=1

and «1

is isomorphic

Solution. Clearly #®—1=(«—1)(1+a+o2+o3+a*) and so « is a root of a polynomial p(x)=14-x+x?+x3+x1'EQ|[x]. Since p(x) is irreducible over Q, [Q(«): Q]=4. Also the roots of x5—1 are 1, «, a2, «3, «4. So O(a) is the splitting field of x°—1GQ[x] and hence a normal extension of Q. Thus

G(Q («)/Q)=[Q («): Q]=4. Since {1, «, «2, a3} is a basis

of Q(«) over Q, a typical element of

Q (2) is

Ag ta,%+aya?+ a,0°. The four Q-automorphisms of Q(«) are indeed the following: 641 Ay taya+ dy? aga?—>dyt ay%+ aya?+a,03

By 1 Ap Aya +A?

63 Ap

,09—>ay +4, 07+ a,04 + aza® =d)+a,0?+a,a4-+a,a

tae a_u?+ dgx3—>dy+a,0° +a,0°+a,0° =a)

+a,03+a,0%+a,04

+ aga >aq+ aya*+ doae+asa™ 641 Ug tae +a,07 = Ay ta 04+ a,a3+a,02

Since each element different from identity is of order 2,G(Q («)/Q) is isomorphic to the Klein’s four-group.

(b) Let E=Q(4/2, ©) where o°=1, wA1. Let o, be the identity automorphism

of £ and

o,

be

automorphism

of E such

o,(w)=w? and o,(¥/2)=of/2. If G={o}, 02}

then

Eg =Q(4/2e").

that

178

RINGS, FIELDS AND

VECTOR SPACES

Solution. Now «€ Eg if and only if o, («)=a. following diagram of towers of fields

Consider the

Q6Y2, )

basis of {1, a}

QW/2) basis {1, ~/2, $/4}

Q Thus {1, 9/2, ~/4, w, w%/2, w*/4} is a basis of Q(¥/2, ) over

Q.

So let

a=a+bi/2+ c/4+da+ea/2+ fow/4 be an element in Q (*/2, w). Then o,(a)=o implies b(8/2—w¥/2)+ ¢(4/4—0?/4) +... =0 The above after rearranging gives

d+ (b—e)¥/2+-2c¥/44 (2d)o+(e—b)w¥/2+cw¥/4=0. From this we get that a is arbitrary, b=e, c=0, d=0, and fis also arbitrary. Therefore « is of the form a=at+b~/2+bo/2 +fors/4 =a+(—

b)arV/2+ fos

Hence Eg=Q(#/2w2). Problems

Lee,

ECE E=Q(#/2, ©) be an extension of a field Q where wr=1, o#1. For each of the following subgroups S; of the group

G(E/Q) find Es,.

(a)

S,={1,o,} where o,:i+—i

(b)

S,={1, o,} where o, SvV2>-v2

(c)

S,={1, 63} where a

(d)

S,={1,o,} where SOAS

(ce)

S,={1, ¢,, o,} where

3

a +2

12)

Oo>w? >a?

a ae a>®

® and o,:4V2> 02 o a>

NORMAL

AND

SEPARABLE

EXTENSIONS

179

Let E be a splitting field of x*—2€Q[x] over Q. Show that G(E/Q) is isomorphic to the group of symmetries of a square. Find subgroups of order 4 and their fixed fields. [HINT: E=Q(¥/2, i). If o is the automorphism of E given by (4/2) =it/2 and o(i)=i then o generates a subgroup of order 4,

Also if t is the automorphism given by 1(#/2)=~/2 and t(i)=—i, then we can describe all elements of G(E/Q) in terms of o and t by means of the relation o4=1=7?, there are three subgroups of order 4, namely,

Cen, Op. o 9 or} Caitl, o8,%,0°T} Ca tl, o-, ara th Q (i), QV2), QV 2) are

Cay Caz-]

ta=o%r, Also

respectively the fixed fields of Cy

CHAPTER

10

Fundamental Theorem of

Galois Theory Galois theory is concerned with relationships between subfields of a field and subgroups of its group of automorphisms. The fundamental theorem on Galois theory establishes a one-to-one correspondence between certain subfields of the splitting field of a polynomial and the subgroups of the group of automorphisms of its splitting field. This correspondence may transform certain problems about subfields of fields into more amenable problems about subgroups of groups. For example, this will be used in the next chapter to derive conditions for the solvability by radicals of the roots of a polynomial. DEFINITION. Galois group of a polynomial. Let f(x)€ F[x] bea polynomial over a field F andK be its splitting field over F. Then the group G(K/F) of F-automorphisms of K is called the Galois group of f (x). DEFINITION. Galois extension. A finite normal and separable extension E of a field F is called Galois extension. For example, if f(x)€F[x] is a polynomial over a field F of characteristic zero then its splitting field E over F is a Galois extension of F. 1.1 THEOREM. (Fundamental theorem of Galois theory). LetE be a Galois extension of F. Let K be any subfield of Econtaining F. Then the mapping

|

K-+G(E/K) sets up a one-to-one correspondence from the set of subfields of E containing F to the subgroups of G(E/F) such that

(i) K=Egzyk) (ii) (iii) (iv)

For any subgroup H of G(E/F), H=G(E/Ez) [E:K]=|G(E/K)|, [K: F]=index of G(E/K) in G(E/F) K is anormal extension of F if and only if G(E/K) is anormal subgroup of G(E/F) (v) JfK is a normal extension of F then G(K/F)=G(E/F)/G(E/K).

FUNDAMENTAL

Proof.

THEOREM

OF GALOIS

THEORY

181

By definition of normality, it follows that E is a normal

extension of K. Thus Kis the fixed field of G(E/K). This proves (i). The proof of (ii) follows from theorem 3.5, chapter 9. Note that care needs only the fact that Eis a finite separable extension of F. Since

FEis a normal

extension

of F and also of K, we have

by

theorem 3.6, chapter 9

[E: F]=|G(E/F)| and [E: K]=|G(E/K)|.

Thus [E :FJ=[E: K] [K: F]

gives

|G(E/F) |=| G(B)K) |[K : F].

This proves [K : F]=index of G(E/K) in G(E/F), as desired. Now we proceed to prove (iv). Recall thatK is a normal extension of F if only if each embedding o:K>F

which keeps each element of F fixed, maps K onto K (theorem 1.1, chapter 9). We assert that K is a normal extension of F if and only if for each c€ G(E/F), o(K)=K. If K is a normal extension of F and o€G(E/F) then o restricted to Kis an embedding of K into F and

hence into F. Thus by the result mentioned above o(K)—K. Conversely, let o: KF be an embedding which keeps each element of F fixed. By theorem 4.5, chapter 8, o can be extended to o* :E>F. But then o*(£)=E£, since Eis a normal extension of F (theorem 1.1,

chapter 9). Thus o* €G(E/F). So by hypothesis, o*(K)=K. Soo(K)=K, since o* is an extension of o. Thus we have shown that if o: K->F is an embedding which keeps each element of F fixed then o(K)=K. Then by the result cited in the beginning, we get that K is a normal extension of F. This proves our assertion stated above. Therefore, K is a normal extension of F if and only if for all o€ G(E/F), and kEK, o(k)EK. Then for all t€ G(E/K), t(o(k))=o(K).

This implies (o-!t0)(k)=k for all KEK. Hence oc

G(E/K). This

proves G(E/K) is a normal subgroup of G(E/F). Retracing the steps back it is clear that if G(E/K) is a normal subgroup of G(#/F) then a(o(k))=o(k) for all tEG(E/K), for all cE G(E/F) and for all KEK. is the Since we do know Lis a normal extension of K and hence K K is Hence o(k)EK. implies fixed field of G(E/K), thus t(o(k))=o(k) a normal extension of F as desired. Finally, we prove (v). Let K be anormal extension of F. By above discussion for all o€ G(E/F), o(K)=K. Thus cinduces an automorphism

182

RINGS,

FIELDS

AND

VECTOR

o* of K defined by o* (k)=s (k), KEK. Clearly, the mapping

SPACES

co €G(K/F).

Consider

f: G(E/F)+G(K/F) defined by f(c)=o*. Let c,, o,€ G(E/F). Then (7 of )(k)=of (3 (k)) =of (6,(k))=(c,0,) (k). Therefore, (c,5,)*=of o3 . Thus f is a homomorphism of G(E/F) into G(K/F). Now ker f={c€ G(E/F) | o*=identity}. But o*=identity if and only if o* (k)=k for all KEK. Thatis, o(k)=k

for all KEK and so cEG(E/K).

Hence ker f=G(E/K).

Then by the

fundamental theorem on homomorphisms

G(E/F)_

(1)

GER)

mst G(K/F)

Further by (iii) we have GEE)

(2)

ae =[K: F]

Also since K is normal over F

(3)

|G(K/F)|=[K : F]

Hence by (1), (2) and (3), we get G(E/F) |

G(EIK) =o (KIF).

This completes the proof of the theorem. 1.2

Solved problems

(a) Iff(x)€ F[x] has r distinct roots in its splitting field E over F then the Galois group G(E/F) of f (x) is a subgroup of the

symmetric group S,.

Solution.

Let f(x)=ay+a,x+...++anx"€ F[x] has r distinct roots

Oi, Kiss ss rm Cys

Clearly for each c€G(E/F), o(a;) is again a root of J (x). Also if

aj Aaj, then o(aj;)4a(«,). Thus o(«,), o (a), ...,0()

isa permutation of «,, ...,«,. Let us set ag,( =o i (ai), i=l, ...,7r. Then doS, and we have a mapping

f:G(E/F)>S, given by f (6)=¢s. If o, t€ G(E/F) then

X( $5q) = %$q($_(i1)) =F (Xp, (iy) =6 (1 (ai) =07 (ai) =ag,, cy

Hence (¢o¢-) (i)=¢0, (i) since «;’s are all distinct. Thus fis a homomorphism. To show f is 1—1, let ¢o=identity. Then X54) =%;. SO

FUNDAMENTAL

THEOREM

OF GALOIS THEORY

183

o («;)=«; for all i=1,...,r. Thus o=identity, since BSF (Gy. 45 Gr) Hence f isan embedding of G (E/F) into S,. (b) Let F bea field of characteristic 42. Let x*—a€ F[x] be an irreducible polynomial over F. Then its Galois group is of order 2. Solution. Clearly if « is one root of x2—a then —« is the other root. So «A—« since characteristic of FA2. Thus x2—a is separable over F. The splitting field F(«) of x?—a over F is finite, separable and

normal extension of degree 2 over F. Thus |G (F («)/F) |=2. (c) Let F bea field of characteristic 42 or 3. Let f(x)=x®+bx-+e be a separable polynomial over F. If f(x) is irreducible over F then the Galois group of f(x) is of order 3 or 6. Also the Galois group of J (x) is S, if and only if A= —4b?—27c? is not a square in F, that is there does not exist any element «€F such that «2=A. Solution. If f(x) has a root «EF then f(x)=(x—«)g(x) where g(x)E€ F[x]. In case g(x) has aroot in F then f(x) splits into linear factors in F itself and so the Galois group of f(x) is of order 1. In case g(x) is irreducible over F then the splitting field E of f(x) over F is extension of degree 2 and so|G(E/F)|=[E : F]=2. Thus if f(x) is irreducible over F, then |G(E/F)|1 or 2. Since f(x) is also separable all its roots are distinct and hence by solved problem (a), G(E/F) Yn)/E)S=Sn.

Finally, the fact that K=E shows that every symmetric function can be expressed as a rational function of the elementary symmetric functions s,, ..., Sn. This completes the proof.

4.2 Solved problems We express the following symmetric polynomials as rational functions of the elementary symmetric functions

204

RINGS,

FIELDS

AND

VECTOR

SPACES

(a) xd-+xe +5 (b)

(x1 —%q)? (X2—%3)* (%3—%)?

Solution.

(a)

(x 4%. 4% 3)?—2 (%4%_2+%y%3+%y%y)

(x¢+xS+x2)=

=Si— 25,

where s, and s, are elementary symmetric functions of x,, x, and x3. (b) By simple computation it can be checked that J

Fae

Sy

5 4 Va

Xa

Sy

> Va

yt

ha

Sy

3,

are the roots of x*+3ax+8=0 where

Ase’

Ces

2

oy og 251 Ss 7

3

PSa5 b=

S182

si 3

Then the cubic equation whose roots are (y,—Y2)", (¥2—)’s)”s (¥3—1) IS (1)

(30+ y)?+ 9a (30-+y)?+ 278?=0.

Now

(X,—X)" (%2—%3)? (%3— 4)? =(¥1—Y2)® (Vo—Vs)? (Vs—- 1)” = product of all the roots of (1)

= —27 (8? +423), Problems

1.

Express the following symmetric functions as rational functions of elementary symmetric functions

(a) xi+23+35 (b) X1x3+-75%3-+23%1

CC) C79) (%8+-%8) (3427) (d)

(%1+%)8 (xg+%5)8 (%34+%)%.

5

RULER

AND

COMPASS

CONSTRUCTIONS

The theory of fields provides solution to many ancient geometric problems. Among such problems are the following: 1. To construct, by ruler and compass, a square having the same area as a circle. 2. To construct, by ruler and compass, a cube having twice the volume of a given cube. 3. To trisect a given angle by ruler and compass. 4. To construct, by ruler and compass, a regular polygon having n sides.

SOME

APPLICATIONS

OF

GALOIS

THEORY

205

To do this we must translate the geometric problem into an algebraic problem. We shall regard the plane as the coordinate plane R? of analytic geometry. If P, is a subset of R?, we shall construct new points from P, by using ruler and compass. We shall say that a line is constructible from P, if it is the line through two distinct points of Py. A circle is constructible from P, if it is the circle passing through one point of P, and whose centre is another point of Py». A point is constructible from P, if it is common either (i) to two distinct lines constructible from Py or (ii) to a line and a circle each constructible from Py). Now suppose that the coordinates of points in P, belong to a subfield K of the field of real numbers. Then by writing the equations of lines and circle and solving them simultaneously we get in the case (i) the coordinates of the new point belong to K, while belong to K constructible which is the

in the case (ii) the or to K(4/a) where number u lies in result of a series of

coordinates of the new point either a@ K and a>0. It follows that any a subfield

K,, of the real numbers

adjunctions

O=hoC KIC

hoC..- Ga,

with each K; equal to Ki-;(+/ di), ai, a positive number in Kj-}. So’ [K}? Ki4|=1 or 2/'Thus

(Kat heise. for some nonnegative integer m. We shall call a real number u constructible from Q by using ruler and compass only. Thus we have proved.

if it is constructible

5.1 THeorem. Any constructible real number u is algebraic over the rational numbers, and its degree over the rational numbers is a power of two. We now apply theorem 5.1 to prove the impossibility of certain constructions by ruler and compass.

5.2 Solved problems (a) It is impossible to construct a square equal to the area of a given circle by using ruler and compass only. Solution. If ais the side of the square to be constructed, and r is the radius of the given circle, then we must have a?=xr?. This means we have to construct x. But it is known that x is not algebraic over rational numbers. So by theorem 5.1 we cannot even consider x for being constructible number from Q by ruler and compass.

206

RINGS, FIELDS AND

VECTOR

SPACES

(b) Itis impossible to construct a cube equal to of the given cube by using ruler and compass only. Solution. We can assume that the side of the Let the side of the cube to be constructed is x. Then have to construct the number 2/3 (the real cube

twice the volume

given cube is 1. x?—2=0.

So we

root of 2). Since

x3—2 is irreducible over Q, [Q(243):

Q]=3a

power of 2.

Thus by theorem 5.1, 2" is not constructible from Q by ruler and compass. (c) There exists an angle which cannot be trisected by using ruler and compass only. Solution. We show that the angle 60° cannot be trisected by ruler and compass. Now if this angle can be trisected by ruler and

compass then the numbsr cos 20° is constructible from Q. This is equivalent to the. constructibility of 2 cos 20° from Q. Set a=2 cos 20°. Then from cos 36=4 cos?0—3 cos 8, we deduce a?—3a—1=0.

Since the polynomial x*—3x —1€ Q [x] is irreducible over Q, and has a root a, it follows that

[Q (a): Q]=3F power of 2. Thus by theorem 5.1 a=2 cos 20° or equivalently angle of 20° can-

not be constructed the solution.

by ruler and

compass from Q. This completes

Problems

1.

2.

3.

Show that the angle 27/5 can be trisected using ruler and compass. Show that it is impossible to construct a regular 9-gon using ruler and compass.

Show that it is possible to trisect 72° using ruler and compass.

4.

Prove that the regular

5.

compass. Show that it is impossible to construct a regular polygon of

17-gon is constructible with ruler

7 sides with ruler and compass.

and

CHAPTER

Methods

12

of Matrices

This chapter is concerned with the methods of obtaining certain canonical forms of matrices over the complex field. Most of the theorems in this chapter have been stated (without proofs) with many examples and solved problems.

1

CHARACTERISTIC VALUES AND CHARACTERISTIC VECTORS

We call a matrix A=(a;;) real or complex according as all aj; are real or complex. Given a real or complex matrix A, in many applications we are interested in vectors ¥~0 and numbers A such that AX=2XX. For n=2, 3 this means that if we consider the linear transformation of R"-+R” which changes the vector X to the vector AX, X and AX are colinear. For example

ACL EG)

A nonzero vector X (i.e. »x/ matrix) which satisfies the equation AX=)X is called a characteristic vector or eigenvector and the associated

number A is called a characteristic root or eigenvalue

of A.

In the above example (: )is an eigenvector and 3 is an eigenvalue 1 pea

of (> Now

ae )

AX=AX

holds

if and

only if (A4—AZ) X=0

whereJ is the

identity matrix. Since (A —AJ) X¥=0 represents a homogeneous system of n equations in ” unknowns, the system shall have a nontrivial solution if and only if the rank of (A—AJ) Vm}. We cons-

truct an orthonormal basis {u,, U2, .. .Um} as follows: y

Set u,=— yy

,

} %

SA

Vo,

Ve =Ve—| V2 Uy|Uy, aT

| ’

,

v3 =V3—| V3 Uy|U,;—|V3 Us |Ue, 3

; yi, Vm—|Vin Uy |---|

V3

Tye]

Vin Um—y|Um—1, Un= i]

of W. Then it can be shown that {1, Ua, . .-» Um} isan orthonormal basis 2.2

(a)

Solved problems

Let W be a subspace of R‘ generated by 2 1 0 y=

0

1

Ve>=

0 1 2

214

RINGS, FIELDS AND

VECTOR

SPACES

0 and

y=

: . Then u,=1/4/2

and

u,=

0 ee Ke COO Oo oro form an orthonormal basis of W. Solution. It can be easily seen that 2v,+(—1)2v,+17,=0. Thus the set {v,, v2, v3} is a linearly dependent set. But {v,, v,} is clearly a linearly independent set and so {v,, v,} is a basis of W. We now apply Gram-Schmidt method to find an orthonormal basis. 1 Set

“==

v3

Vy =3

4

1

V2 =V.— (v2 uy )u,=

9.

1

0

0

1

—2

2

rae

eae |ve

0

1

0

-

0

1

0

0

1 0

This completes the solution (b) Let W bea subspace of R¢ generated by

Ve

1

1

1eee Vg =

0:

1

0

0

and

v=

1

,

Then u,=4

0

; 1

i

pine

Bl a~2 St

0

|and =

—2/¥/6 |form an orthonormal

1/6

1/+/6

|

basis of W. Solution. Observing that v,, v2, v3 are linearly independent, we can apply Gram-Schmidt method to find an orthonormal basis of W with a basis {1,, v2, v3}. The solution is straightforward.

METHODS

OF MATRICES

215

Problems

1.

Apply Gram-Schmidt method to find orthonormal bases of the subspaces generated by each of the following sets.

S Giae,

(b)

1 0 iFe

1 1 0

0 (c)

2.

1

1 ret 1k 2

0 0 1

|

0

1 LL 0 } 1

Lete,,..., em be nonzero

|

0 a 3 5

vectors

in R®

such that e7 e;=0,

i~j, 1