Introduction to Modern Abstract Algebra

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This book is in the ADDISON-WESLEY SERIES IN MATHEMATICS

Ocnuultiftl/

Editor: LYNN H. LoOMI8

D A V I D M. BUR TON, University

of New

Hampshire

INTRO-DUCTION

TO

MOO'ERN ABSTRACT

ALGEBRA

ADDISON-WESLEY PUBLISHING COMPANY Menlo PIITk, California

.

London

.

Amatflfdem

•

ReedIng, M8ItIaChuIItItIt Don MIlls. 0nfarl0 S)IdIIey .

COPYRIGHT ® 1967 IIY AIlIII~{)N-Wt:~Lt:Y l'URLI~IIING COMrANY, INC. ALL RIGIITS Rt:St:RV}:/). TIII~ 1I00K, on l'AltT~ Tlnau:(w, MAY NOT III': Rt:I'IUJIlUCt:1> IN ANY FOUM WITHOUT WIUT'n:N I't:nMI~HION OF Til}: l'UIILISln:n. 1'liINT}:/) IN TIn: UNITt:!) STATES OF AMt:RICA. I'UIILlsllt:1> SIMULTANt:OUHLY IN CANAI>A. LIBRAUY OF CONGIlESS CATALOG CAIlII NO. 67-19426. , _ 0·201.00722·3 IJKLMNOPO-MA 19818

PREFACE

This book has been written with the intention of providing an introduction to model'll aht:!t.ract algebra for mathematics major8. The r('.ader is not presumed at thc Outllct to possess any previout:! knowledge of the concepts of modem algebra. Accordingly, our beginuing is somewhat elementary, with the exposition in the earlier sections proceeding at a leisurely pace; much of this early material may be covered rapidly on a first reading. An attempt. has been made to keep the book as self-contained as possible. To smooth the path for the unexperienced reader, the fir8t chapter is devoted to a review of the basic facts concerning sets, functions and number theory; it also serves as a suitable vehicle for introducing some of the notation and terminology used subsequently. A cursory examination of the table of contents will reveal few surprises; the topics chosen for discussion in COUr8es at this level are fairly standard. However, our aim has been to give a presentation which is logically developed, precise, and in keeping with the spirit of the times. Thus, set notation is employed throughout, and the distinction is maintained between algebraic systems as ordered pair8 or triples and their underlying sets of elements. Guided by the principII! thnt 0. st(~l"ly diet. of definitions and cxullIples SOOIl 1)(~c()llIel'l unpalatable, our eiTOl-ts are directed towards establishing the most important and fruitful results of the subject in a formal, rigorous fashion. The chapter on groups, fOl' example, culminates in a proof of the classic Sylow Theorems, while ring and ideal theory are developed to the point of obtaining the Stoile Representation Theorem for Boolean rings. Ell route, it is hoped that the, reader will gain an appreciation of precise mathematical thought and t.he current standards of rigor. At the eud of each section, there will be found a collection of problems of varying degrees of difficulty; these constitute an integral part of the hook. They introduce a variety of topics not treated in the main t.('xt, as well as impart much additional detail ahout material covered earlier. Home, especially in the latter seet.ions, pl'Ovide slIbl'ltantial extensionl'l of t.he'theory. We have, on the whol(', resist.('d t.h(' t.('mptation to lise the exercises to develop results that will be ncedl'd HllbsequC'nt.ly; aN a 1·(,~lIlt., the reader need not work all the problems in order t.o read the reHt of the hook. Problems whose solut.ions do not appear particularly straiJ!;htforward arc accompanied by hints, Besides the general index, a glossary of !!pecial Hymhol!! iN also included. v

vi

PREFACE

The text is not intended to be encyclopedic in nature; many importaut topics vie for inclusion and some choice is obviously imperative. To this end, we merely followed our own taste, condensing or omitting altogether certain of the concepts found in the usual first course in modem algebra. Despite these omissions, we believe the coverage will meet the needs of most students; those who are stimulated to pursue the matter further will have a finn foundation upon which to build. It is a pleasure to record our indebtedness to Lynn Loomis and Frederick Hoffman, both of whom read the original manuscript and offered valuable criticism for its correction and improvement. Of our colleagues at the University of New Hampshire, the advice of Edward Batho and Robb Jacoby proved particularly h~lpful; in this regard, special thanks are due to William Witthoft who contributed a number of incisive suggestions after reading portions of the galley proofs. We also take this occasion to express our sincere appreciation to Mary Ann MacIlvaine for her excellent typing of the manuscript. To my wife must go the largest debt of gratitude, not only for her generous assistance with the text at the various stages of its development, but for her constant encouragement and understanding. Finally, we would like to acknowledge the fine cooperation of the staff of Addison-Wesley and the usual high quality of their work.

Durham, New H amp8hirc March 1967

..D.M.B.

CONTENTS

Chapter 1 1-1

Preliminary Notion.

The Algebra of Sets

1-2 Functions and Elementary Number Theory Chapter 2

2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 Chapter 3

3-1 3-2 3-3 3-4 3-5 3-6 Chapter 4 4-1

1 13

Group Theory

Definition and Examples of Groups . Certain Elementary Theorems on Groupf! Two Important Groups Subgroups Normal Subgroups and Quotient Groups Homomorphisms The }..to-one, for 2 ~ ax:.! -- !! impliml x. = X2' Cont:I(Jqucmtly, the inVllftj(J of j exists and is the set of ordered pail~[-I = {(3x - 2, x) I x E R'}. It is preferable, however, to have[-t defined in terms of its domain and the image at each point of the domain. Observing that

ax. -

{(3x-2,x)lxER'} we choose to write

r

1

=

{(x,!(x+2»lxER'},

= {(x,!(x + 2) I x

In terms of functional vl\lueM, [-I(X)

E R'}.

= !(x + 2) for each x E R ' .

An important situation ariscs when we consider the behavior of a function on a subset of its domain. For example, it it! frequently advantageous to limit the domain 80 I,hnt t.he fUlldion beeomcM one-to-one. Suppose, in general, that j: X -+ Y it! an nrbitrary function and the subset A ~ X. The composition j i ... : A -+ Y is known as the restriction of [ to the set A and is, by established cllstom, denoted by f I A; dually, the funetion [ is referred to as an extension of f I A to all of X. For the reader who prllfers the ordered pair approach, flA= {(x,y)l(x,Y)E[andxEA}. 0

In any event, if the clement x E A, then (f I A)(x) = [(x) so that both [and coineide on the set A. It is well worth noting that while there is only one restriction of the given function [ to the subsct A, [ is not necessarily uniquely determined by [I A. The particular restrietion ix I A = i A , when viewed as a funetion from A into X, is termed the inclusion or injection map from A to X. Thl! lwxt ddinitinn 1'lIIilodil'fl lL fn~CJUlmtly employed uotational device. Observe that despite the use of the symbol j-1, the function [ is not required to be one-to-one.

[ IA

18

1-2

PRELIMINARY NOTIONS

Definition 1-14. Conllider a (unction f: X -+ Y. If A !;; X, then the direct image of A, denoted by f(A), is the subset of Y defined by

f(A)

=

{f(z} I z e A}.

On the ot.her hand, if B !;; Y, then the inverse image of B, denoted by f-l(B}, is the subset of X defined by f-l(B)

=

{z /f(z)

e B}.

It shall be our convention to omit unnecessary parentheses whenever possible. In regard t.o RinglctonR, for iIlRt,allce, we Rhl\lI write dirc(~t l\n

0, then there exist unique integers

0::; r
0 and

OS r < Ibl.

-b, we may take q = -q' to get

a

=

qIJ + r,

OS

r

< Ibl.

Let us now make the following definition. Deftnitlon 1-15. Let a, b E Z, with 0 ¢ O. Thc integer a is said to divide b, or 0 is a divi81w of h, in JoIymhols a I h, provid 1 having no divisol'H other than the improper ones is said to be a prime number; all integer a > 1 that is not prime is termed composite. Thus, according to our definition, 1 is neither prime nor composite. In particular, an integer a > 1 is composite if and only if there exist integers b, c with a = be, 1 < b < a,

l O}.

+

This set S is not empty. For example, if a ¢ 0, the integer lal = au bO will lie in S, where we ehoose u = lor -1 according as a is positive or negative. By the Well-Ordering Principle, S must contain a smallest element d > 0; that is to say, there exist x, y E Z for which d = ax + by. We assert that d = gcd (a, b).

22

1-2

PRELIMINARY NOTIONS

From the Division Algorithm, one can obtain integers q and r such that + r, ~ r < d. But then r will be of the form

°

a = qd

r

=

a - qd

=a= a(1

q(ax -

qx)

+ by) + b( -qy).

Were r> 0, this representation would imply rES, and contradict the fact that d is the least integer in S. Thus r = 0, so that d I a. A similar argument establishes that d I b, making d a common divisor of a and b. On the other hand, if e I a and e I d, then by Theorem 1-11(6), e I (ax + by), or rather, c I d. From these two statements, we conclude that d is the greateflt common divisor of a and b. It may be well to record the fact that the integers x a.nd y in the representation gcd (a, b) = ax + by are by no means unique. More concretely, if a = 90 and b = 252, then ged (90, 252)

=

18

(3)90 + (-1)252.

=

Among other possibilities, we also have 18

=

(3

+ 252)90 + (-1 -

90)252

=

(255)90

+ (-91)252.

There is a special case of Theorem 1-12 which will play an important role in the future; while it is, in effect, a corollary of the foregoing result, we shall single it out as a theorem. But first, a definition: two integers a and b, not both of which are zero, are said to be relatively prime (or prime to each other) if and only if gcd (a, b) = 1. For instance, the integers 8 and 15 are relatively prime, although neither is itself a prime. The.Nm 1-13. Let a, ,) E Z, lIot both zero. Th(m a Ilnd b are relatively prime if and only if there exiflt integers x and 11 such that 1

=

ax

+by.)

Proof. If a and b are relatively prime, 80 that gcd (a, b) = 1, Theorem 1-12 guarantees the existence of x and y satisfying 1 = ax + by. Conversely, suppose 1 = ax + by for suitable x, y E Z and that d = gcd (a, b). Since d I a, d I b, Theorem 1-11(6) implies d I (ax + by), or rather d 11. Because d is positive, this forces d = 1 [Theorem 1-11(2»), as desired.

In light of Theorem 1-13, one may easily prove Theorem 1-14. (Euclid's Lemma). If a I be, with a and b relatively prime, then a I e. . Proof. Since gcd (a, b) = I, there exist integers x and y for which 1 MUltiplying bye, wc obtain e

=

(ax

+ by)e =

a(cx)

+ (be)y.

=

ax

+ by.

1-2

FUNL'TIONS ANI) ELI!:MENTARY NUMB.m TIIEOItY

23

Now a I a trivinlly ILlld a I be by hypothcNi8, 80 that a IlIU8t divide thc sum acx + bey; hence a I e, as asserted. Corollary. If p i8 a prime and pi (ala2' .. a,,), then p I at for some k, 1 :::; k:::; n.

Proof. OUf proof is by induction on n. For n = I, the result obviously holds. Supposc, al:! the induction hypothesis, that n > 1 and that whenever p divides a product of less than n factors, then it divides at least one of the factors of this product. Now, let pi (ala2' .. an). If p divides at, there is nothing to prove. In thll contmry I!MC, p and al arc rdativcly prime; hence, by the theorem, p I (a2' .. an)· Since the product a2' •. an contains n - 1 factors, the induction hypothe8is implies p I at for some k with 2 :::; k :::; n.

Having developed the machinery, it might be of interest to give a proof of the Fundamental Theorem of Arithmetic. Theorem 1-15. (Fundamental Theorem of Arithmetic). Every positive integer a > I I~nll hll (lxpreHllCd u..'\ a P1"lltluct of prime!! i thi!! rcpretIClltu.tioll i8 unique, apart from the order in which the factors occur.

Proof. The first part of the proof-the existence of a prime factorization-is proved by induction on the values of a. The statement of the theorem is trivially true for the integer 2, since 2 is itself a prime. Assume the result holds for all positive integers 2 :::; b < a. If a is already a prime, we are through; otherwise, 4 = be for suitable integers b, e with 1 < b < a, 1 < e < 4. By the induction hypothesis, b = PIP2" • p"

with P.,

p, nil prime!!.

Hut theil, a = be = PI' ..

PTP~

.•.

p!

is a product of prinles. To establish uniqueness, let us supposc the integer a can be represented as a product of primes ill two ways, say

where the Pi and qi are primes. The argument proceeds by induction on the integer n. In the case n = 1, we have a = PI = ql(q2' •. q...). Since PI is prime, it possesses no proper factorization, so that m = 1 and PI = ql. Next, assume n > 1 and that whenever a can be expressed as a product of less than n factors, this representation is unique, except for the order of the factors. From the equality PIP2 ... Pn = qlq2' .• q... , it follows that PI I (qlq2' •• q...). Thus, by the preceding corollary, there is some prime qk, 1 :::; k :::; m, for which PI I qki relabeling, if Iwccssary, we may suppose, PI I ql. But then PI = qt,

24

1-2

PRELIMINARY NOTIONS

for ql has no divisors other than 1 and itself. Canceling this common factor, we conclude 1'2 .•. PA = q2 •.. qm' According to the induction hypothesis, a product of n - 1 primes can be factored in essentially onc way. Therefore, the primes Q2, .•• , qm are simply a rearrangement of the primes 1'2,'" , PA' The two prime factorization!'! of a are thus identical, (~ompleting the induction step. An immediate consequence of this theorem is the following: Theorem 1-16. (Euclid). There are an infinite number of primes.

Proof. Assume the Rtatement is false; that is, assume there are only a finite number of primes PIr 112, ... ,1'". Consider the positive integer a

=

(1'11'2' .. PA)

+ 1.

None of the primeR Pi divides a. If a were divisible by PIr for instance, we would then have p, 1(0 - P,1'a'" 1'A) by Theorem 1-11(6), or PI 11; this is impossible by part (2) of the same theorem. But, since a > 1, the Fundamental Theorem asserts it must have a prime factor. Accordingly, a is divisible by a prime which ifol not amonK our list of all primes. ThiR lugument shows there is no finite listing of till! prillle intOme familiarity with this idea, we pause to offer u selection of examples. Further examples appear in the exercises. Example 2-18. Let a be allY nonzero real number and consider the set G of integral multiples of a: G= {naJnEZ}.

The pair (G, -H, wher operation is commutative and associative. It is easy to H(,(, that fot· any Hl't A s;;; X (that iH, fot' any clement of P(X»,

A

~

0=

(A -

0) u (0 -

A) = A U 0 = A,

which proves that the empty 8('t 0 serves ns un identity element for~. Moreover, A ~ A

=

(A -

A) U (A _.- A)

= 0 u 0 = 0.

2-1

1>";)o'lNITION ANIl

l~XAMPL"~'i

OF GUOUPS

37

This implicH t.hllt, e:wh dl'lIWllt, of J>(X) ill it.!l own invc1'IIC. Conscquently, the mllthclllatil'nl HYHtl'l1l (f'(X), a) is n I:OJlllllut,nt.ive gnlUp.

Examp'e 2-21. As 1\ silllple exampl(! of a noneollllllutative group, let the set G consist of all ordered pail'R of real numbers with nonzero first component: G= {(a,b)la,bER',a~O}. Define the operation • on G by the formula

(a, b) • (e, d)

=

(ae, be + Il).

Thn nSH()(:intivit.y IIf t.he oJlI~mt,ioll flllluWH frolll tho f,ulli/inr properties of the real numbers, for wc have

[(a, b) • (e, tl») • (e,f)

+

=

(ae, be (l) • (e,J) = «ae)e, (be d)e f) = (a(ee), b(ee) (de f) = (a, b) • (ee, de + f) = (a, b) • [(e, d) • (e, f)].

+ + + +

It is readily verified that the pair (1,0) serves as the identity clement, while the inverse of (a, b) EGis (lla, -bla). To see that the group (G, *) is not commutative, merely consider the elements (1,2) and (3,4) of G: • (1, 2) • (3, 4) = (3, 10)

~

(3,6)

=

(3,4) • (1,2).

Example 2-22. For another example of a noneommutative group, take the set G as consist,iug of the six fUllctions f" /2, ... .J6, where for x E R' - {O, I}, we define l2(x)

x-I

x,

f3(X) = 1 -

= !, x

14(x) = - - ,

x

X

f6(x)

i5(X) = .r - 1 '

=

1

-1-'

-x

I.et the group operation \)(' that of functional composition. Thus, as an illustration, we have (/2 • is)(.')

= I2Us(x» = 1-

=

12

x = !a(x),

C~ x) =

1/(11_ x)

38

2-1

GROUP THEORY

whieh implies that. J2 hand, (f6 • !2)(x)

=

•

J6 = h. On the other

Table 2-2

it /2 fa !4

16(12(x»)

=16

G)

=

" ""

l/x

X

= - - = 1,,(x)

x -

I

fo

/2 fa !4 f6 fo f6 fs f4 fa /2 /2 fa fa f4 /l /2 f6 fs f4 f4 fa f5 f6 /2 it fll f" fo f4 fa /2 fo fo fs /2 it fa f4

1 1-

f6

"

'

so that 16 ·12 = is, which shows that the operation • is not commutative. The multiplication for (G, .) in this case is given by Table 2-2. Since functional composition is associative (Theorem 1-7), the system (G,.) is certainly a semigroup. The operation table shows that It is the identity element and the respective inverses are

It l

=

it.

1"41 = 16,

1"21 =!2, fil

=11'"

ri

l

= fa,

1;1 =14.

To encompass all the different groups above in a single concept obviously requires the formulation of the underlying group concept in the most general terms. This is preeisely the point we hope to convey to the reader; the value of contemporary mathematics lies in its power to abstract and thus to lay bare the structurally essential relations between superfieially distinct entities. Historically, thc notion of 11 group aro~ early in the ninet.L·(mth century out of attempts to solve polynomial equations. Galois waH the first to use the word "group" in any teehnical sense when he considered the group of permutations of the roots of such equations. A major achievement in the evolution of the theory was Klein's cln.."sification, in the IS70's, of the various branehes of geometry according to groups of transformations under whieh certain geometric. propertics remain invariant. It remained some time, however, before satisfactory group postulates, free of redundancy, were stated. Definition 2-11, first formulated in 1902, is attributed to the American mathematician E. V. Huntington. In the twentieth century, group theory has embraced all branches of mathematics and, incl(~ed, a wid" vari('ly flf ot.her fields. It is difJieult to give exampltJ8 without beeoming too tedlllical, but the theory of groups is now employed in the study of quantum mechanics, general relativity, and erystallography. In these areas, group theory is not only a tool with which calculations arc made but also a souree of coneepts and principles for the formulation of new theories. A recent example can be found in the physics of fundamental particles with the discovery of a new "elementary partide" whose cxistenee had been predicted from a e1a."sifil'ation scheme hased on groups. It is certainly appropriate to begin our investigation of mathematical systems with this concept.

2-1

DEFINITION AND EXAMPLES OF GROUPS

39

PROBLEMS

J. Determine which of the following binary operations on the set Q are associative and which are commutative. a) a * b = 0 b) a * b = l(a b) c) a * b = b d) a * b = a b- 1 2. Rllppose Lhe system (S, *) haR an identity element; show that if the equation

+ +

(a * h) * (c * d) = (a * e) * (h

* d)

holds for all pOH"ible choices of element!i a, h, c and d of S, then the operation i!! both aH:-Iociative and commutative. 3. Prove that the set of ordered pairs of real numbers together with the operation defined on 8 by (a, h) * (e, d) = (a e, b d 2bd)

+

4.

* *

+ +

constitute!! a commutative !lemigroup with identity. l,(~t 111'1 define a hinary operation * on the !!et S = {I, 2, 3, 4, 6} as follows: a

*h

= gcd (a, h).

For example, 6 * 4 = 2, 3 * 4 = J, etc. Show that (8, *) is a commutative semigroup. [Hint: Problem 17, Rection 1-2.) 5. ConRi!\er the three-element !let S = {a, h, e} and the operation mlllt.i"li(~ation table helow:

*

a h e

a h

abe h a e

c

e c e

* given

by the

Verify that the pair (8, *) is a commutative semigroup with identity, but not a group.

6. In the following illlltanceR, determine whether the systemR (G, *) described are (~ommul.ative groups. For tholie systems failing to be so, indicate which axioms are not satisfied. a) (} - Z+, (I * II - max {(I,ll:, h) a = Z, tl * b = mill {fl, bl (the Hlllall(!r of a and b), e) (} = n' , a * b = a h - ab, d) (J = Z+, a * b = Illax {a, b} - min {a, b}, e) (J = Z X Z, (a, b) * (e, d) = (a e, b d), f) G = R' X R', (a, b) * (e, d) = (ae bd, ad +. bd), g) (J = R' X R' - (0,0), (a, b) * (e, d) = (ae - bd, ad be). 7. RUJlPol'l(l that a E R' - {O, I} Ilnd conRider the set G of integral powers of a: G -= {a k IkE Z}. If· denotes ordinary multiplication, prove that (G,') is a group.

+

+

+

+

+

40

2-1

GROUP THEORY

+

8. u-t G = {I, (-1 iV3)/2, (-1 - iV3)/2}, where i 2 = -1. Show that the lIytltem (a, .) is a group. 9. Con!!idl'r the set a consisting of the fOllr functions/J,i2,Ja,/4: /I(x)

1

= x,

! 2(X) = -x'

hex)

=

-x,

x

with x E R' - {O}. Prove that (G,o) is a commutative group, where functional composition. 10. Prove that the symmetric difference opt'ration

:1

aB

= ( ..1

- B) U (B -

0

denotes

A)

discussed in Example 2-20 may also be defined by the formula

.\ a B

= (:I U B) -

(.\

n B).

11. Granting the associative law, show that the following two operation tables define groups:

...

-

.

I :i

5 7

I

3

5 i

-_._---

5 7 7 5 5 7 I 3 7 5 3 1 I

:~

a

I

...

a

b c

d

a b

a b c b a d c d b d c a

d c a b

c d

12. If G = {a E R'I -1 < a < I}, verify that (G, >II) form!! a commutative group, with the operation'" givl'n by a+b a>llb = - I fib

+

for a, bEG.

13. Let (S, >II) Ill' a sl'migroup wit.hout illl'nt.ity and e hl\ any ('It~ment not in 8. Define the olll'ration on the t«'t"" = S U {e) by ml'SnH of I.he ruleH 0

for all a, b E 8, aoe=a=eoa

for all

a E 8'.

Show that the pair (8',0) is a semigroup with the identity element e; (8',0) is said to be obtained by adjoining an identity. 14. Prove that the following weakened Het of axioms are actually equivalent to the classical axioms for a group as given in Definition 2-11: A group is a mathematical system (G, >II) for which 1) >II is an aN80ciative oper&tion, 2) there exists an element e in G such that a >II e = a for all a E a (existence of a right identity), 3) for ('.sch a E G, thero ('xists an ('lement a-I in G such that a'" a-I = e (existence of right inverse.'1).

2-2

CERTAIN ELEMENTAltY TUEOItEMS ON GROUPS

41

2-2 CERTAIN ELEMENTARY THEOREMS ON GROUPS

AI! wc rmllllrke Example 2-23. Lpt. r: dC'lI11l.ll tim HI·1. (If nil urdtln~1 pnil"!! of renl 1lI11nbers with nonzero first. ClolllpOIwnt. If t.he hinary operation • is defined on the sct G by the rule (a, b) * (e, cl) = (ae, be d),

+

t.hen (a, *) is n nOllc'omlllutntive ~roup [Example 2-21). The ici 1), Their

44

2-2

mWlll' TIIJ
This completes the induction stcp and therefore the proof of the theorem. Given four elements aJ, a2, aa, a4 E S, parentheses can be legally inserted and the elements mult.iplied (in the given order) five different ways, Theorem 2-4 Jl(~rmjt.H \If'! to cO/u'lude that all theHC produnts nrc cqual: «(al * a2) * a3) * a4)

=

(al * (a2 * aa» * a4 = a, * «a2 * aa) * a4) al * (a2 * (a3 * a4» = (al

* a2) * (aa * a4)'

As It matter of notation, it will bc our tendency to omit parentheses in writing products; the exception to this will be found 011 those occasions when we wish to emphasize the associative law or a certain grouping of elements. Theorem 2-5. (Cancellation Law). If a, b, and c arc elements of a group (G, *) such that either a * c = b * c or c * a = c * b, then a = b.

Proof. Sinee c E G, c- I exists in G. on the right Mid(· hy c-', we obtnin (a * c) * c:- I

l\lultiplying the equation a

=

*c = b *c

(b * c) * c-'.

Then, by the associative law, this becomes

or a * e = b * e. Therefore a implies a = b.

=

b. Similarly, we can show that c * a

Corollary. The only solution of the group equation x

*x =

x is x

=

=

c*b

e.

Pmof. Tlw eoneiusioll iM lUI illllllediate eonsequenee of the cancel1n.tion 1n.w and the fact that .c * x = x * e.

In n system (8, *), all element xES is said to be idempotent provided x * x = x. Whnt we have jU!5t shown is thut u group possesses exactly one idempot('nt clement, namely the group identity. For an illustration of a system in which eVl'ry element is idempotent the student is referred to Problem 4, Section 2-1.

46

2-2

GROUP THEORY

This last theorem allows us to cancel, from the same side, in equations involving group clements. Wc cannot conclude, however, that a * c = c * b implies a = b, unless the group is known to be commutative. An arbitrary binary operation need not satisfy the cancellation law. To see this, we consider the set G = {I, 2, 3} under the following multiplication table:

*

I

1 2 3

I 2 3 2 I 2 3 2 I

2 3

On examining the table, we observe t.hat 2 * 1 = 2 * 3; but obviously 1 ~ 3. The failure of the cancellation law in this instance results from the fact that when we mUltiply both sidcN of the equation 2 * 1 = 2 * 3 by 2- 1 = 2, the element 2 does not associate with the product 2 * 3; that is, 2

* (2 * 3)

~ (2

* 2) * 3.

Theorem 2-6. In a group (G, *), the equations a * x have unique solutions. Proof. First, x

=

a-I

=

band y

* b satisfies thc group equation a * x =

b, since_

a

* (a- 1 * b)

*a =

b

* a-I) * b = e *b = b. =(a

This shows that there is at least one solution in G; it remains for us to show that there is only one. Suppose there is some other element x' E G such that a * x' = b. Thcn a * x' = a * (a- 1 * b), so that by the cancellation law, x' = a-I *b,

or

x'

= x.

The second part. of til:

D{'finition 2-1;; ass!'!'ls I,hat. Hw lIlodular sum of t.wo {~ongruellcc (,lasses [a] and [b] is tl\(' Uni'll\(' mellliwr of Z" whil'h ('olltaills the ordinary sum a II. Ilow(w(~r, th!'r!! is It ImbUe prohlelll (·olllll'et.l'd wit.h t.his delillit.ioll. IlIlllollllud, as addit iOIl of ('ollll;rUl'n('e c:lItsses in Z,. is detin{'d in teruls of reprel:lCntatives from th(,sl' dusses, we must show that the operation +n doeM not depend Oil the two l'I'P!'l's('lItativ{'s dloS IInllmbig;uously defined, independent of the arbitrary choice of r('pr('sent at iv(·s.

+"

Example 2-26. Rllppose we {'onsider ('onJ!;I'UI'IH'!l modulo 7 and the typical mldit ion [a) -I- 7 [(j] = [3 n] = [9].

+

SinN! [:1] = [10] and [0] used

=

[101

[--I!)], the same allliWl'r should he obtained if one

+7 [-I;;] =

[10 - 15] = [-5].

While these results appear superficially different, both congruence classes [9] and [-!)] may be !'xpressed more simply as (2). Thus, although written in terms of different repreRellt.atives, either modular addition gives the same sum, [:2). Other pORsihl1' ('hoi(~es [-4] -h [-X] = [-121. [17) +7 [0] = (23), f3] +7 [13) = [Hi], also yil'ld the sum (2). We nt·(· 1I0W in a posit ion t.o prove one of UI(' prindpnl theorems of this section.

Theorem 2-13. For elwh posit.ive inl!'ger n, t.l1(' mathematical system (Z", t-,,) forms n ('omlllutative group, known as t.he (/1'OUp of inieyel's modulo n. J'l'()of. TllP ass()('iat ivily and ('ommut.ativity of t.\w opl'ralioll -1-,. nre 1\ direct consequenee of til(' sanl(' properties of the intl'geriS under ordinary addition.

2-:1

Indeed, if [a],

TWO IMI'OHTANT GROUPS

rbI,

57

[e] E Zn, then [all .. ([bj-I .. [c)) =Iaj I" [b I

r.J

+ (b -/ c)] = [(a + b) + c] = [a + b] +.. [r] = [a

([ll) I" [II)) I .. If').

Similarly, [a] 1-" [b] = [a

+ b] =

[b

+ a] =

[b) -1-.. [a).

By d('fiuition of -I,,, it is clear Um!. [0) iH the identij,y elmnellt. [a] E Z", then In - a] E Zn Ilnd [a]

+n [n

thlLt [a)-I = [n -- a]. mutative group.

110

- al

=

[a

+ (n

- a)]

=

Filll1l1y, if

[n] = [0],

Thill l'OIl1pl(lh~s the proof that. (Z",

+..) ill

n com-

InddmltnIly, Thron~m 2-Ja nlHO shows thnt for HV(' ..y positive integer n t1l1're ex iH\.l'I at. 11'11.,{1. 0111' eommut.at,ivl' gnmp with n "ll'Irll'lIl.s. If WI! adopt Uw c'ollvlmt.ioll of dl'lIiglmt,illg c~a"h I'ulIgru()IW(! dUIlI'; by it.H smaIlI'!;t. lIonn(·gnt.iv(~ I'(·pn,sellt.utiv(', 011'11 the ol)('ratioll t:lhle for, say (Z4' -h), looks like

+.

[0] [1]

[2] [a]

[0] [1]

[0] [1]

[1] [2]

[2] [3]

[21

[21 [a]

(a]

[0]

[3] [0] [1]

[0]

[1]

(2]

[3]

For simplir.it.y, it is convenient to remove the hrnckets in the designation of the eongruen(!e dallses of Zn. Thus we oft.1'1l write Zn = {O, 1, 2, ... , n - I}. With this notation, thc above operation table assumes the form

+4

0

0

0

1 2

3

1

2 3 2

:~

1

2

0

2 3

;~

3 0

0

1

2

I

For the seeollu of our I.wo exnmpleH, let us turn to the study of permutation groups. To this clld, supposc Umt N is n finite set having n clements which, for simplieit.y, we tnkl' to be t.he first n natural numbers; that is to SI\Y,

N= {1,2, ... ,n}.

58

2

GROUP THEORY

Definition 2-16. By a permutaliun of the set N is meant any one-to-oll mapping of N onto itself. In what follows, the totality of all permutations of the set N will be deno!.,·, by the Rymhol S". RiTl{'e the number of differ('nt, permut.ations of n object,!; i nl, the fift;t thing to note is that Sn ill itself a finite set with n! distinct element, Next, any permutation f E Sn may be described by

f= {(l'/(l», (2,/(2», ... , (n,/(n»)). While this is the Il(!eeptable funct.ional notation, it will prove to be more veni('nt to r('prcll('nl. f ill a two-line form

('Oil

where till' (~nrrl'Rpollding imag(,R uppear 1)('low eaeh integer. Clearly, the ord,·, of the ('ICIIII'lIts ill th(' t.op row of thiR lIymhol ill inulIllwriul, for th(l columll: may lw rl'arrnllged without affectillg t.he natufC of the function. Prccil:ll'l) speaking, if (J iR Illl arbitrary p(!rmutation of the integers, 1, 2, ... ,n, thcn .' could equally well b(' given by f -

(g(l)

g(2)

f(g(t»

f(g(2»

...

g(n»)

... f(g(n»)

From thill, WI' illf('r UIIlt. (,Iwh of the n! {>I'rmutntions in' 8" may be writ,tA'I' in n! ditTl'fCnt way!!. For illlltarU'e, the following two symbols both reprcsclIl the same clement of 8.:

(1 2 3 4) 2

4

:1

I

( 214:i).

'

4 2

1 3

Permutation!'!, h('ing fum'tions, may be multiplied under the operation of functional compol'litioll. Thus, for lK.·rmut!ltions I, U E S",

fog

=

(1 2 f(t)

( =

=

(1(\)

n) f(n)

f(2)

g(2)

l(g(1»

f(g(2) )

C(g;l)

2 l(g(2»

(1 2 ... 0

g(1)

g(2)

g(n) ) f(g(n»

f(g~»

)-

0

...

( 1 g(1)

g~) 2 g(2)

g~)

2-3

TWO IMPORTANT GROUPS

59

What we have done is to rearrange the columns of the first (left) permutation until its top row is thc same as the bottom row of the second (right) permutation; thp product /0 g is then the permutation whose top row is the top row of the se(!OII(I factor and whose bottom row il:l the bottom row of the first factor. With a little pmet.iee one can evaluate produet.l:I without luwillg to w.rite out. this intermediat!' preparation. }lany authors prefer to carry out the multiplication of P(·"llllll.ations in the oppo:-;it,(! onl"r (that ill, they apply the factors in It prouud frolll )(·ft. to right.), alld t.he read('r :-;hould h,· parl,il'tllllrly wlltehful for this, Before stating a theorelll whieh indieutell the algebmic nature of Sn under I his 1lI,,1 h()(1 of (!Ollll)()sit ion, we hop(~ to dnrify sOllie of till' foregoing "ointH Wit.ll IlII ('xILlllple,

Example 2-27. If 1.111' s(·I. N ('onsiHls of t.he illt.(·/.t(·rs I, 2, a! Ii Jll'rlllllf.nl.ions ill Sa, 1I1L1l1dy, 2

2 2

tlllln t.hcn! nrc

=C :;) , .ra=Ca :J) , fs =C :a)J , .f6 =C ;) .

=C :~) , f4 =C :) ,

II

a,

/2

:~

2

2

:l

2

2

2

2

:l

3 2

2

A t.ypical lIlultiplieation, say f4 0/6, proceeds as follows:

.f4 f6=C 0

=G On the other hand,

16 ·f4

WI!

2 :~

h:wl!

=Ca =G

flO

a)'·Ca D a 2

2 2

a

2

2

2

:)·G

2

:;) ·c ~)oC

2

:J)= C I

2

2 :l

:)=12,

2

:~)

a

2

2

~)=G :) =fa,

3

2

1

that. lIIult.ipli('ntion of pl'l'mutatiolls is not ('ol1lmutntivl',

Theorem 2-14. 1'1)(' pair (S", .) forms It /.troup, kllown as the symmell'ic f/l'OUP lin n Hymbo/s, wlli('" is IIOJJl:OIIUllutut.ivc fOJ' n ~ :t The proof of f his fad. is OIllil.t.('(1 in:l.'lIl1ll('h as !l mom gllllCral version of the thcorem will be l!;iv('11 shortly. III Jlus:sing, it is only 11I..'Cc8I!ary to note that the

(iO

2-:\

I:llOl'1' 'I'll EOHY

identity e1cmcnt for (,sn, 0) is the permutation

whil!. HCI'fl'sent (,ach of thCI!IC lIymmctricli by a corresponding permutation.

64

2-4

GROlTP 'fHEOIlY

15. Consider the set G consisting of the four permutations

(11 22 3344), (2I 21 a4 34),

~).

Show that (G, 0) constitutes a commutative group. 16. Form the sct G = {f,j2,f3,r,F',r) , where! is the permutation

! _

(1 2 3 4 5 6), 2

:J

4

a

6

1

and prove that thl' pair (0,0) ill a eommutative group. 2-4 SUBGROUPS

There nrc two Ktnnd:ml t.eehniqul's in at.ta(·king t.lw problem of t.he strueture of a parti('ular j!;roup. One method ('ulls for finding all the subgroups of the group, with the hopl' of gaining information about the parent group through its local structure. The other approach is to determine all homomorphisms from the given group into a more familiar group; the idea here is that the images will reflect some of the algebraic properties of the original group. On dOKl'r Kcrutiny. WI' KhaH KI'C that while HICKe lill('to! (If inveto!tigation aim in difT('rent dirpdions, t.hey nrc not entirely unrelnt(!d, hut rather'll.spects of th(! Kiln\(' probl(·m. For the moment, however, our attcntion ito! focused 011 analyzing It group by means of its subgroups; the question of structure-preserving mappings is a morl' sub tic matter and will be deferred to n lat.er section. From variolls examples nnd exercises, the reader may have noticed that "I'rlnill suhsl't.s of 1.1 ... 1'II'III1'II1s of n grouJI Il'llcl 1.0 III'W I!;rlluJls wlll'lI UlIt! n!Htri(·l.s tlw group olIPratioll to these subsets. It iH this situation in which we shall be primarily interested. Definition 2-19. Let (G, *) he n group and II ~ G be a lIonempty subsef of G. The pair (II,.) is snid to be 1\ subgroup of (0, *) if (II,.) is itself n group.

Eaeh group (0, *) hUH two oLvioUl~ subgroupH. For, if e EGis the idcntity clement of th(! group (0, *), t.hen both ({e}, *) and (0, *) arc subgroups of (G, *). These two sUhgfOUJlIl arc often referred to as the trivial subgroups of (G, *); all suhgroups hdw(,pn thelle two extrernefol are calIed nontrivial subgroups. Any subgrollp different from (0, *) is termcd lIroller.

Z. and Zo dcnote the setH of even and odd integers, rcspe(~ tively, then (Z., +) is a subgroup of the group (Z, +), while (Zo, +) is not.

Example 2-28. If

Example 2-29. Consid('r (Zo, +6), the group of integers modulo 6. If

II

~

{O, 2, 4},

2-4

SUBGROUPS

65

then (H, +6), whose operation table is given below, is a subgroup of (Zo, +0)'

+0

0 2 4

o

024

2 4

2 4 4 0

0 2

Example 2-30. Let (G, *) he the group of symmetl'ies of the square (Ree ExILlIlP'" 2 2·1), wlll'I'I' a :U uo , H IH ." H 2711 , H:lflfl , II, I', /)It 1)2] IUIII t.11I' OPI'I'ntion * l'onKiKt.!oI of following olle mot.ion by anot.her. Thi!ol group I~ollt,ltin!ol eight. nont.rivial subgroups. We leave it to the reader to V('rify timt the following sets comprise the ei(!ment.s of these subgroups: 0_

{If lilli, /(:11111, II, V}, {R IIlO, Raoo,

[)It

D 2 },

{Raoo, D 2},

{R 180, Ra6u},

{R 360 , H},

{R a60 , V}.

Suppose (H, *) is a subgroup of the group (G, *). Since the identity element of (H, *) satisfies the equation x * x = x, it mURt be the sn.me as the identit.y of UI(' pnnmt. group (a, *), for otherwi!olc we would hlWI~ two idl~mJlotcnt elenll'lIt.R in a, l~olltrlLry to Theorem 2-li. The identit.y demellt of a group thUl:l also serves 3.'> the identity element for any of its subgroups. Moreover, the uniqueness of the inverse elements in a group implies the inverse of nn element hE H in the subgroup (H, *) is the same as its inverse in the whole group (G, *). I'll l'fltnhliHII tlml. n l(ivl'lI !oIlIhH(·I. 1/ flf a, nlnlll( wil,h I,h,' illllul!l'Il IIJlI·rntiml of (a, *), eonHt.it.utl'!oI n subgroup, W(! IIlU!:!t verify tlmt nil the eonditions of Definition 2-11 arc sl\tiRfied. However, the asROdat.ivit.y of the operation * in II is an immediate consequence of its 3.'>sociativity in G, since H !; G. It is neeeRsnry then to show only the following: 1) a, bEll implies a * b E H (closure), 2) e E II, where e iR the identity element of (0, *), 3) a E II implies a-I E II. Needless to say, the saving in not having to eheck the associative law can prove t.o btl cOIl!olidl'l·able. A theorem whieh e!olt.ablishcs a single eonvf'nient criterion for determining subgroups i!ol given below. Theorem 2-17. Let (G, *) be a group and 0 "t; H !; G. Then (H, *) is a subgroup of (a, *) if and only if a, /J E II implies a * II-I E H. Pl'Oof. If (II, *) i!ol a !oIuhgl'oup and a, bEll, tlWII II-I Ell, 1UIIl!olO a * b- I Ell by the dosurl' "oJl(lit ion. COllvl·rMl·ly, suppose 1/ is tL 1101l(!Jllpty suhset of G

66

2-4

GROUP THEORY

which contains the element a * b- 1 whenever a, bE H. Since H contains at least one element h, we may take a = b to sec that b * b- I = e E H. Also, b- ' = e * b- I E H for cvery b in H, applying the hypothesis to the pair e, bE H. Finally, if a and b are any two members of the set H, then by what was just provl'd b- I also belongs to H, so that a * b = a * (b-I)-I E H; in other words, the sct 11 is dosed with resped to the operation *. Because * is 11.11 associative opl'ration in 0, II inherits the associat.ive law as a subset of G. All the group axioms arc satisfied and the sYi'lielll (11, *) is therefore a subgroup of (0, *). Definition 2-20. The center of a group ("'lit.

a=

{c E

a Ic * x

(a, *), dcnoted by cent G, is the = x

* c for all x

Ret.

E G}.

Thus cent. (J ('onsists of those clements which commute with every element of O. For example, ill the group of symmetries of the square, cent

a=

{R 1140, R360}'

The reader may already have deduccU that a group (G, *) is commutative if and only if c('nt. a = O. As ilIustratiolJs of the usc of Theorem 2-17 in determining when a subset of the clements of a group is the set of elements of a subgroup,..we present the following two theorems. Theorem 2-18. The pair (cent 0, *) is a subgroup of each group (G, *).

PmoJ. We finlt ohserve that cent Gis nonempty, for at the very least e E cent G. Now consider any two elements a, bE cent G. By the definition of center, we know that a * x = x * a and II * x = x * b for every clement x of G. Thus, if x E G, (a * b- ' ) * x = a * (b- I * x) = a * (x- 1 * b)-I = a * (b * X-I)-l

* (x * b- 1 ) (a * x) * b- I = (x * a) * b- I = = a

X

* (a * b- I ),

which implies a * b- I E cellt G. According to Theorem 2-17, this is a sufficient ('ondition for «(,pnt a, *) to he a subgroup of (a, *). Theorem 2-19. If (Hi, *) is IUJ arhit.rary indexed collection of subgroups of the group (a, *), then (nil;, *) is alHO It subgroup.

Proof. SilH'e tht' Sl't,; IIi all ('ontain t.Iw identity (·lement. of (a, *), the interscl'tion nil. ~~. 1\('xt" supposc a !\luI II Ilrc any t.wo clcmcnt8 of nH,; thcn

2-4

SUBGROUPS

67

a, bElli, where i rangeM over the index Met. The pair (Hi, *) being a subgroup, it follows that the product a * b- I 11.\:;0 belongs to Hi. As this is true for every index i, a * b- I E nIl.-, which implies (nHi , *) is a subgroup of (G, *). In regard to the group of symmetries of the square, we could take HI = {R 90 , R I!JO, R 270 , R 360 }, H2

= {R lIw • Rallo, Dt, D2}.

The system (HI n H 2 , *) = ({R UIO , R 360 }, *) is obviously a subgroup of this group, for its elements eomprise the eenter of thc grollp. III j!;PIII'r:d, wit.hollt fllrthpl" restric·tioll 011 the Imbp;roupll (1/;, *), it ill not true that the pair (Ulli, *) will 'IlJ.!:Uill he IL sllbg"oUII of (0, *). 0\1(' lIilllply eannot !!;uarallt('(l that. Ulli will c:olltain produets whose faetors come from different Hi. To give a eonerete illustmtion, both ({O, o}, +12) and ({O, 4, s}, +12) are subgroups of (Z 12+ 12), yet on takin!!; the union, ({O, 4, 0, 8}, + 12) fails to be so. The diffieulty in this ease is t.hat the modular sums 4 -h2 6 and () -/-t2 8 do not belong to the set {O, 4, 0, S}. By the way of an aualog to Theorem 2-1H, we have: Theorem 2-20. Let (IIi, *) be an indexed collection of subgroups of the group (G, *). Suppose the family of subsets {Hi} has the property that for any two of its members Hi and Hi there exists a set Hk (depending on i and j) in {H;} sueh that Hi S;;;; Hk and Hi S;;;; H k • Then (UH i, *) is also a subgroup of (G, *). Proof. By now the pattern of proof should be clear. We assume that a and b are arbitrary elemcnt.s of UIIi and Rhow that a * b- I E UHi. If a, bE UHi, then there exist subsets II i, II j cont.aining a and b. rel'lpectivcly. Aecording to our hypothesis, HiS;;;; H k and H j S;;;; Ih for some choice of lIt in {Hi}. Since (H k , *) is a subgroup and both a, bE H k , it follows that the product a * b- I belongs to H k • Accordingly, a * b- I E UHi as was daimed at the beginning. Asa particular case of the foregoing result, (~onsider just two subgroups (H I, *) and (H 2, *). Theorem 2-20 may be interpreted as asserting that (H I U H 2*) will again be a subgroup of (G, *) provided either HIS;;;; H 2 or H2 S;;;; HI' 'What is rather interesting is that thil:! condition is I1eeessary, as well as sufficient. The next theorem gives the details. Theorem 2-21. Let (HI, *) and (H 2 , *) be subgroups of t.hc group (G, *). The pair (H I U H 2,) is also a subgroup if and only if HIS;;;; H 2 or /J 2 S;;;; HI. Proof. I n view of the prececiing remarks, it il'l enough to I'lhow thut if (II, U 11 2 • *) is l~ tiubgl'Oup, then olle of the I'lets H I or 112 must be contained in the other. Suppose to the contrary that this assertion were false: that is, HI g; 112 and 112 g; II,. Then there would exist elementli a E III -- J/ 2 and bEll 2 - Ill'

!i8

2-,1

WUltI!' TIH:OItY

Now, if the procille! a

* IJ were n member of the set Il" wc (~ould infcr that b = a-I

* (a * b)

E Ill,

which it! c·lpurly not. t.rul'. On the other hand, the possibility a * bEll 2 yields the equally false conclut!ion

That it!, the den1t'nts a, b E HI U Il 2, but a * b Ii HI U Il 2. This conclusion is obviously untl'nublC', for it contradicts thc fact (II I U 1/2, *) is a group. Having arrived at n contradic~tion, the proof is compl(~te. The next t.opie of intC'rC'lit c'O!\ecrns eye·lil' subgrollps. To facilitate this disc'lIliliion, WI' fir:-;t int.rodll(·e some spceinlnotution. Definition 2-21. If (U, *) is an arbitrary group and 0 :;o
18 ~ II;

(Il, *) i:;

tL

~

G, then thc

subgroup of (0, *)}.

The set (8) dearly exists, for G itself is a member of thc family appcaring on the right; t.hat is, (G, *) is a (trivial) subgroup of (G, *) and 8 ~ G. In addition, sinee S is ('olltnillcd in eaeh of the sets being intersect&d, we always have thc inclusion S ~ (8). Theorem 2-22. The pair «S), *) is n subgroup of (G, *), known cithcr as tl\(' f'nveilJpiny .~ulJyl'Oul' for 8 or the RUIJI/roup yenerated by the set S. P/'(}of. The proof is !In imnwdiutl' consequl'lwe of Theorem 2-19.

Definition 2-21 implies thnt whenever (Il, *) is a subgroup of the group (G, *) with s ~ II, then (8) ~ II. For this reuson, one speaks informally of «8), *) us being the smallest subgroup which contains thc set 8. Of course, it ma:y well happen that (8) = G, and ill such a situation, the group (G, *) is said to bc (Jenel'ate(i by the subset S. For example, it is easy to sec that the group (Z, +) is generated by Zo, thc set of odd integers. We shall givl' an a\1ernative description of the sublict (8) which is frequently elllii('l' to work wit.h than Definition 2-21. III what follow:;, the symbol 8- 1 is u&-'gI'OUP whieh (~ontllins the set 8 Illust J1('(~!'sslll'ily ('olltain nil the l'I!'llI!'nts of [S). A ('asl' of spp(~i:t.I illlportlllH'!' aris!'s \\'11('11 S ('onsists of It singl(· plpllJ('nt a. In I.his sitllatioJl, it is US \I: II to \Hitp (It) im;tpad of (fn:) alld n,f!'1' t.o t.ht' Ilsso(~iatpd sUhg;I'Olljl (a), *) as the cydic SUIJUI'IJIlJl (/('naaled IllI a. The subs('1. (a) is rather !'asy to d!'s('I'il)('; as all ils produds involve the cI('llIent a or its inverse, (a) silllply I'('dw'('s to t he integral powel's of a:

It. is ('ntirdy possihl(' that. the gl'oup (a, *) is rty; that is, am = e, while a k ¢ e for 0 < k < m. The sct. 8 = :1', n, a 2, ' , , ,am-I} con8ists of distinct clements of (a). For a' = a', with 0 ~ r < 8 ~ m - 1, implies that a'-' = e, contrary to the minimalit.y of m. To I'olllplete the pl'Oof, it remains to show each member ak of the group «a),.) i;; {'qual to IlIl clem('nt of S. Now, by the division algorit hill, we Illay writ(' k ~ 1J1It ~I l' for SOIllI' integers q !lnd ,. with 0 ~ r < m. II CII(,(, , ak = (am)q. a' = e * a' = a' E S, This means 011' 1- QUOTn:NT GROUPS

75

17. Consider the group of symmetries of the square. Use Theorem 2-25 to obtain the subgroup J.!;cnprll.ted by II U K, where II = f RplO, R:wo) , K = {Raoo, Dd. IS. Ll't (0, *) hI'

is

Ii

sqIlIL),l'

It

(i.('.,

J.!;rollp of ord('r n, W}IPf(' n is odd. Prove thllt ('Iu:h "Ielnpnt. of G if rEO, thl'1l .x = y2 for ~Olll(, y ill G).

2-5 NORMAL SUBGROUPS AND QUOTIENT GROUPS

Althoul!:h WI' have dl'rivt'd sOllle interesting results eOllcel'lIing suhgroups, this e()(le(~Jlf" if unrcstrieted, is too gelwrul for muny purposes. To obtain (·Nt.ain highl.v dl'sirahlC' (:onehlsion>-l, :ulditiOlml assulllpt.ioJls t.hat p;o IlI'.vond I )I'Iinitioll:! I!I II I 11M I. I", iIlIPOM(·'1. Thus, in the Jll'esent S(,et.ion, we nurrow the field and focus attention on u rt'Ht.ri(,ted ('(ass of :mhgroupl'l which we I'IhnIl refer to ns normal subp;roups. FrollJ IL eon('l'pt.uul point. of view, sueh groups ure "noI'llHLI" ill the liense that t1H'Y make til(' )'t'sull,illg Hl('ory so Illueh ridH'r Ullin would oUlPrwi>-le he th(' ('!lS('. Whill' not. I'vl'ry suhgroup nl'l'd h(' of t.hiH t.YJlI', norlll:ll :.alhgroups O('plll' nOlwt.h"ll'lis with I'onsideruble frequenpy. It. will lioon he('ollle uPJlal'pnt. that, for the lIlajor part of OUI' work, t.he signitit':lnt. usped of this ('lILIiS of :lIIbgl'oups resides in the faet that they permit the ('onstruetion of algebraic structures known as quotient groups, Having already divulged some of the content of this section, let us now proceed to develop these idea.s in detail. As a starting point, we prove a sequence of theorems lmding to the conclusion that eneh subgroup induc:es a decomposition of the dementH of the parl'flt group into disjoint subsets known a.s cosets.

Deftnition 2-23. Let (H, *) be a subgroup of the group (G, *) and let a E G. The set a * II = {a * It I It E II} is called a left coset of ][ in G. The clement a is a representative of a

* H.

In u. similar fashion, we eun define the rip;ht eosets H * a of II. The right cosets of the same subgroup nrc in general different from the left cosets. If the grouJl operat.ion * of (G, *) is commutative, then dearly a * ][ = ][ * a for all a E 0, In HI(! suhsequent diseussiolls, we will generally consider only left cosets of a subgrouJl. It is obvious that, u parallel theory for right cosets may be developed. Before proeeedillg to all exampl£', we lihall make several simple observations. First, if e i:; thl' id!'lItity pll'nH'nt. of (0, *), thl'lI e*

II

=

[e

* It I Ii

so that 1I its('lf is a Ilies a *1/ * a-I s;;; 1/. To demonstrate the convenience of this result, we now prove the following assertion: (rent G, *) is a normal subgroup of each group (G, *). In terms of elements, it must be shown that if c E cent G and a is arbitrary in G, then a * c * a-I E cent G. But this is fairly obvious, since from the definition of the center of a group, a * c = c * a. It follows at once that a

* c * a-I =

c ... a'" a-I

=

c ... e =

C

E

cent G.

Examp'e 2-37, Let us return to the noneommutntive group (G, 0) of order 6 presentnd in Example 2-22. The reader may f! of the group (G. *) such that G = U/h Assume further that II, n IIi = {e} for i ;>C j, Prove that the parl-nt group (G, *) ill n"""I'C j, w.,e Prohlem 10. In (~a.~e i = j. choose any clement, c E G - II;. Then c aud c * a "OIlIllIUIl- wit,h IWI~ry ('lllment of IIi; in partkular, e = [e * a. b * al =[a, bl.]

22. Prove that if th!' qUoti"nt group (G/cent G, 0) is mutative group,

I~y('lil'.

then (G, *) ill a I:om-

2-6 HOMOMORPHISMS

Up t.o thiH point, in the t(~xt, w(~ hll.vI~ not eonHidcred ml\(lpinll;f1 from one group to Ilnothel'; in(l1('1, /I serV!'s IlS the identity clement for (O/H, ®), we must have ker (natll) = {a E G I natH(a) = H} {a E G I a

* /I

=

/I} = II.

Th(' InRt !'CllIlllit.y was ac'hievec! by t1w lise of TheorNIl 2-2fi,

It is pO>lsihl(', IUlcl sonwt.illll'H {!onvl'lIient, to phru.sc TIl(lorem 2-42 so that no rdl'rc'nc'C' is lIIaclc' to t.Il1' notion of CJllotil'nt p;roup: Theorem 2-43. Ll't (1/, *) be a nOl'mal Imbp;roup of the group (G, *). Then there exists a group (G', 0), and 11 homomorphism f from (G, *) onto (G', 0) su{~h that ker (f) = H. Of COUl'R(', we tak(' (G', .) to be the quotient group (G/II, ®) and

f = natH,

The usual custom is to refl'r to the function nntll as the natural or canonical Provided there is no danger of eOllfusion, we shall frequently omit the sub8cript II in writing this funl,tion. As a rl'bt.t'd rc'mark, it might be emphasized that the natural mapping is not generally one-to-Olw. For if a, b E (J arc clements such thaI, the product a-I * b is in H, tl1
* .1'2) = a * (XI * X2) * a-I = (a * XI * a-I) * (a * Xl * a-I) =

CTa(XI)

* CTa(X2)'

The next. t.hing to noti('c is that CT a maps the set G onto itself; specifically, for any e1 O'/f(lI) deRiglllLtes the naturnl mapping. diagram helps to visualize the situation:

The following

G'/I{H)

Observe that 1 nwr(·ly ILssigns to (·ach element a EO the coSet f(a) of(H) of 0' /f(II). Since both the functions f and nat/(II) are onto and operation-preserving, the composition of these gives us a homomorphic mapping from the group (G, *) onto the group (G'/f(ll), ®'). The main line of the argument is to show that ker 0) = H, for then the desired reRu!t. would be a simple eonl!l'quence of the Fundamental Theorem. Now till! ililmtity dlmll'nt of (a' /f(lI) , ®') iM jUHt the (:OKl.'t f(lI) = e' • f(II). This means the kernel of 1 ('onsists of those members of G which are mapped by ] onto f(lI); that is, ker

m=:a E G I](a) = f(lI)} fa E

=

G If(a) of(H)

= f(lI)}

:a EO I f(a) Ef(H)}

=

f-I (J(ll».

As we Ilrt· givI'n f.Imt, kl'f (f) ~ II, the lelllllll\ prm'eding Theorem 2-49 may b(' invoked t.o (,OIl1'lud(' II = f-I(J(ll». Hence, k('r (]) = H, which completes the proof. In applicat.ions of this theorem, we frmluently Htart with an arbitrary noml1l1 subgroup of (a', .) lind utilize inVl'rsc images rather than direct images.

2-7

1I:J

Corollary. If (II',.) is any normal suhgroup of the group (G', .), then (G//- 1 (1I'), ®) ~ (G'/H', ®'). I'mo/. By the ('llrollul'y to Tlwllrmll 2--:m, nUl pair (f-I(lI'), *) iK I~ Ilornml f:!lIhl(rIllip of (a, *). :'Itofl'llVl'r, kl'r (j) r;;/-I(II'), SIl 1.111' hYIlOUlC'siK of th(, UIC'()J'('III is ('()/IIJ1h,tdy sut isfil'd. This leads to till' iSOIllOl'phiKIIl

Since / is a mnpping onto

a', 1/' = /(j-I (1/'»,

and we arc done.

Our final theorem, a rather technical result, will be crucial to the proof of the ,Jor(lnll-Htildl'r Theorem. Theorem 2-51. If (//, *) lUlU (K, *) an! KuhgroliPH of 1.1)(' KrouP (a, *) with (K, *) llorllllLl,l.hen (II/II n K, ®) ~ (11* K/K, ®').

Proof. NeedlesH to Hay, it should be che(,ked Uu~t the quotient groupII appearing in the stat.emf'llt of the theorem nrc actually definc'd. We leave to t.he reuder the routine t.usk of verifying thltt (11 n K, *) ill U Ilormal Huugroup of (II, *), I.hnt.-(II * K, *) is 1\ group, lUui that (K, *) h~ llorlllal ill (1/ * K, *). Our proof is putt.erneu on t.hat of Theorem 2-flO. H(!I't~, the problem is to construct a homomorphism (I from the group (Il, *) ont.o t.he quotient group (/1* K/K, ®') for whieh ker «(I) = II n K, To n(,hil've this, cOllsili('r t.he function (I(h) = It * K, It E /I. Note, II = II * e ~ /I * K, so that a can be obtained by composing the inclusion map ill: /I .-. II * K with the natural mapping natK: H * K .-./1 * ~/K. In other words,

or, in diagrammatic language,

H oK/K

The foregoing factorization implies a is a hOlllomorphismund a(l/) = /I * K/ K. WI' next proceed to estahlish that. the kerlwl of a is precisely t.he sct II n K. FirHt., obHl'rw I hut. tI\(' cosl'!. K = e * K K(,\'V('S as till' id(,lltit.y ('1(,llwllt of I,h(! quotient gi'ouJI (1/ * K/K, ®'). This llIeans ker (a) = [It E III a(h) = Kl = Ut E /lIlt [It E /I lit E K} = II n K. The required

i~olllorJlhism

*K =

KJ

is /lOW cvili('nt from t.he Fundn/lwlltal Theorem.

114

2-7

GROUP THEORY

Example 2-54. As an illustration of this last result, let us return again to the group of integers (Z, +) anrl consider the cyclic subgroups «3), +) and «4), +). Both these Imbgroups are normal, since (Z, +) is a commutative group. Morc!Ovcr, it is fllirly obviolls tlmt (:J)

n

(4)

Thwrie .Jordan-Holder Theorem. For the sake of simplicity, we Hhall lilllit Olll'lielV!'H tll 1,111' I'II..'«~ of finite gJ'()\IJlH; 1.111' inl.l~J'(~Ht.ed fI~n.der ill referred to n lIIore gmwrtll lreutnwllt in [WJ. BI!('nuSt~ the theormn is ruthel' involved, it will be eOllvellicllt. to begin by introdueing t!Ollle special terminology. Definition 2-30. By n. dlain for a group (G, *) il> 1II('l\nt any finite sequence

of SUhsetH of

a,

(proper inclusions), descending from G to {e} with the property that all the pairs (Hi, *) are subgroups of (0, *). The integer n is called the length of the chain. What we are really interested in, and henceforth IIhall confine our attention to, are the so-called normal chains. These are chains in which each group (Hi, *) is a normal subgroup of its immediate predecessor (Hi_\, *). For grouP!! (a, *) with two or more clemellts there is always one normal chain, namely the trivial e1uLin a:J (e) ; however, this may v(~ry W('lI be the only such chain. Example 2-55. In UII' group (ZI2, ure normal dmillH:

+12)

of intl'gl'l'li modllio 12, the following

('hnilU~

Z12:::> (Ii) :::> (O},

z 12 :J (:J) :J (6) :::>

Z.2:::>(2):::>(4):::> (O},

(O},

ZI2:::> (2) :J (6) :J (O}.

All subgroups are automatieally normal, since (ZI2' +12) is a commutative group. Among other t.hingR, this pnrtil:ular exnmp!e indinnt.efll t.hat II. p;ivcn chain be lellgthened 01' refined by the ill&!rtioll of admissable lIubl>Ct.!!. In technical terms, a second chain

lIIay

0= Ko:::> Kl :::> ... :J K m- 1 :J K". = {e} is said to

h(~ It

l'efinement of the (·hnin

a=

Hu:::>ll. J'" :::>Hn- I :::>lln

=

[e}

provided there exists a oJl(_~to-one function] from :0, I, ... ,n} into :0,1, ... , m)

118

2-8

mwul' THEORY

I'IlIl'h that If i ~~ K/(i) for all i. Whut w(~ :lrI~ rl'Cjllirillg, ill elT(!!'t., iH thltt every If i ('oilH:idl' with OIl(' of the Kj. The lellgths of the foregoillg ehains mll!;t ('\I'urly l J : I( I so. I(alll), fl, V} J : Hallllt Il} J .[ H:IIH1}.

It is worth lIutill/( that tIll' slIhjl;l'OllP (:/(:1110. II:, *) is no\. P;l'OllJl «(1, *), bllt only ill its illulH'diaLt' PI'('III'('(·I'ISiIl'.

11111'1111\1

ill till' pnrl'llt,

2-8

'rIlE JOnDAN-I10LD~;lt THEOHEM

lin

A normal ('IUlin for (G, *) which fails 1.0 Ill' a ('oIllJlosit.ioll ehain is

This ('huill admils allY olle of Ht'vt'ral I,t'finemellts; among others, tlwre is

Example 2-58. To see that not every group possesses u composition chain, merely ('onsider the additive group of integers (Z, +), We have previously observed that til(' normal subgroups of (Z, +) are the c:yclic :subgroups (n), +), II It nOlllll'gal iv(' intl'gPl', Rinf't' til(' i!lf'lUHiCIII (kn) o('iat.ed quotit'lIl. group, For this, r('('all that 11 group is said to h(' Hilllple if !.IH' ollly lIorlllal subgroups Ilr(' t he two trivial OI1(,S,

Theorem 2-52. A normal subgrouJl (H, *) of the group (G, *) is maximal if and only if the quotient (G/lI, ®) iH simpl('. Proof. This result follows immediately from the COI'I'(,Hpondence Theorem, We saw that to elU'h norlllal subgl'oup (K, *) of (G, *) with /l K 2 :>··· ::)K m_ 1 ::)Km

= {e}

hoth r('pre~ent (~()/IIJ1()silioll ehains fOI' the I:!uh/l:roup (Il" *). As the theorem is as!';lIllll'd true for (// t, *), whose order is l(~ss than that of (G, *), these two chains are necessarily equivalent. But (G/llt, 0) = (G/K l , 0)j hence, the given I'Omposit.ion ehains (1) and (2) lIlust, also be equivnlent. (' AHE 2. II. ~ K I' Eilhpr II,

n KI

=

k!,

0"

hy Th('Orp/ll 2-53, t.here exists

a ('ompositioll chain

Il, n K, ::) 1-, :> ... ::) 1-r _, :> l"r

= {e}

122

2-S

mlOt1P TIII';()(lY

for the tmbgroup (Ill nK 1 , . ) . Now, a(~(~rding to the lemma just established, (H InK 10 .) is a maximal normal subgroup of both (H 10 .) and (K 10 .). It then follows that the two chains .

nJlli (4)

are actually composition chains for (G, .). ApJX'aling to the lemma once more, we see that and Hence, the eomposition quotient groups ohtained from (:l) are isomorphie in pail"!! to those ohlllilwd frolll Uw e1min (4); let Ul! ngrcll to ubbrcviate tlii" situation by writing (;) ~ (4). Next, consider the following two composition ehains for the group (G, .): and

Since these ehains have their first two terms in common, we can concludp from case 1 that they must be equivalent. In a like manner,

and

are equivalent dmins for (G, .). Combining our results, we observe that

(1)

~ (:l) ~

(4)

~

(2),

from whieh it follows that the original chains (1) and (2) are equivalent. Let us quickly review what has just been learned. First, the preceding theorem implies that the composition chains of a given finite group (G,.) must all be of the same length. !\[oreover, by means of the quotient groups of any sueh chain, we are able to assoeiatl' with (G, .) a finite sequence of simple groups. Up to isomorphism, these simple groups depend solely on (G,.) and are independent of thl' partil'ular eomposition ehain from which they were originally obtained. The important. point is that the composition quotient groups will to some dl'grce mirror the properties of the given group and provide a hint to its algebraic strueturc.

2-8

123

Example 2-59. To illustrate the ,Tordall-lIolder Theorem, eOlll:!ider the group (Zoo, -ho) of int('gerl:! modulo 60 Ilnd t.he two composition chains: Zuo J (:l) J «j) J (12) J {O}, 7,'1111-)

(:!) -) (Ii)-)

(an)

~

:n}.

After suitable rearrangement, thc il:!omorphic pail'l:! of composition quotient groups are (Zao/(a), ®) ~ «2)/(f», ®), «a)/(f», ®) ~ (Z6o/(2), ®), «(j)/(12), ®) ~ «30)/{0}, ®), «12)/ {O}, ®) ~ «6)/(30), ®). All th('s(' flHOt.icllt, groUpH nrc simpl!', heing nYI'lie' groups of prime order: the first pair of iHolllor"hil' groupl:! iH of ol'llel' a, the HI~eOlI(I pair of Imler 5, while the remaining groupl:! are all of order 2. Note that the product of all these orden'! is HO. We 1I0W take a brief look at a wide c\al:!s of groups which contains, among others, all eommutative groups.

Definition 2-34. A group (q, *) il:! solvable if it has a normal chain (possibly of length 1) G=HOJHIJ"'JHn_lJHn= {e},

in which every quotient group (Hi_dH i , ®) [i = 1,2, ... , n] is commutative. AI:! a matter of language, we shall call such a chain a solvable chain for (G, *). It should be clear at onec that all commutative groups (G, *) with more than one element arc Holvltble, sinee in this ('/tRe, the trivial chain 0 J {e} is It solvable ehain. A:; IllI ('xum"le of It lloncomJl1utative solvable group, one need only eOIlHidl'r the symllletri(' group Oil three' :;ymbols, (Sa, 0); here, a solvable chain is

Sa

~ {C ~

:), G: D' G~

~)} ~ {(~

: :)} .

Reccntly, W. Fcit and ,J. Thompson succeedcd in proving the long-standing eonjecture that all fillite groups of odd order arc solvable. Thus, if we are interestcd in nonsolvable groups, t1wy will be found among the noncommutative groups of evell order. SOllie bal:!ic properties of solvable groupl:! arc given in our next theorem.

Theorem 2-55. If (0, *) is It solvablc group, then every subgroup of (G, *) is I:!olvahlc and evcry homomorphic image of (0, *) is solvable.

124

2-8

catoUl' TlU:OItY

Proof. Let

G = Ho:::> HI:::>' .. :::> H n_ 1 :::> Hn

=

{e}

be a fixed solvable chain for (G, .). Given a subgroup (K, .) of (G, .), define Ki = K n Hi Ii = 0, 1, ... ,nJ. We intend to prove that the chain

K = Ko:::> K 1 :::> ••• :::> K n_ 1 :::> Kn = {e}

is a Rolvahl{' chain for th~ subgroup (K, .). First, observe that the pair (K i , is Illlorlllul ~lIbJ(I'OIlP of (K i _ h .); in fact., for each a E K;-h (f.

Ki

•

a-

I

= (a. K i

•

a-I)

n

K

IIi. a-I) n K

S;;; (a.

.)

~

IIi n K

=

Kj.

Also, K; = K

n IIi

= K

n l1 i_ 1 n IIi = K'_ 1 n Hi,

so Umt

According to Theorem 2-51, we have the isomorphism

But the quotient group (K i - 1 • Hi/Hi, 0) is commutative, being a subgroup of the eomlllutlltive group (l1i_dIli' 0). This implim! all the quotient groups (Ki_dK j , 0) are commut.ative, as required. As for tlH' :-;"(""1(1 pnrt. of thl~ t1wormn, let f he a homomorphism from thl! group (0, .) Ollto t.he group (G', 0). Setting Il~ = f(lI i ) [i = 0,1, ...• nJ. we obtain a chain for (G', 0):

G'

=

H'o:::> Ht:::>···:::> H~-l:::> H~

=

{e}.

By the corollary to Theorem 2-39, the subgroup (H~, .) is certainly normal in (HLI, .). What little diffi(:ulty there is arises in showing the corresponding quotient group (HLdH~, 0') to be commutative. For this, we define mappings k Hi_dHi --+ lI';-dm hy taking fiCa • Hi)

=

f(a) • H~,

(i

a E Hi-l

=

It is easily seen that f; is well-defined; indeed, if

1,2, ... , n).

a. Hi = b· Hi, then

a-I. b E Hi, hence

f(a)-' • feb)

=

f(a-' • b) E f(H i)

=

H~,

which, in t.um, implies f;(a· IIi)

=

f(a) • Ht

=

feb)

0

H;

=

f;(b. Hi).

2-8

TilE JOIU)AN-HOI.ln;H TIlEOREM

125

Next, note that each/; is itself a homomorphism; for if a, b E Hi-I, f;«a*H i) 0 (b*H i» =j;(a*b*Hi) =f(a*b)oH~ = f(a) feb) m = (I(a) • m) 0' (f(b) • Hn. 0

0

Finally, because fell;) = 1I~, the function /; maps onto the set H~_dH~. From th('Hll faetH, W(l ILre ahle to eondu K~_I:::> K~

=

{e

* H}

be a solvable chain for the quotient group (G/H, 0). Using the Correspondence Theorem, we obtain a ,sequence of sets G = Ko:::> K 1 :::>···:::> K"_I:::> K,.

=

H

dL'Scllnding from (J to II with the prop(lrty that (K;, *) is a normal subgroup of (K i - 1 , *); in fact, Ki = nati?(Ki). Furthermore, by Theorem 2-50, (Ki-tlK;, 0)

~ (K~_dK~,

0'),

so that the quotient group (K i _tlKi ,0) is commutative. solva},)(!, it haH a solvabl(! ehain running from H to {e}, say

Since (H, *) is

11 = H o :::>H 1 :::>··· :::>H"'_1 :::>Hm = {e}.

By stringing t.heHll Hl'qlll!necll of setl! togcther, we can eonstruct a solvable ehain for (G, *): G:::>K 1 :::>·,· :::>K"_1 :::>H:::>Hl:::>'" :::>Hm_ 1 :::> {e}, Therefore, (G, *) is a solvable group. In conjunction, Theorems 2-51> and 2-.')6 tell us that if (11, *) is a nontrivial normal Hubgroup of (G, *), then (G, *) is a solvable group if and only if both (II, *) and (a/II, 0) art' HlIlvahlll gt'OllpH. The d"dHion r('garding Holvability is oft.en facilitated by this criterion.

126

2-8

WIOI'I' I'll EOH\"

TIl(' following I Iwort'llI giv('s fUl'tiwl' illsight. illlo t.he ('hains for 11 fillile solvahle group.

lIatuJ'(~

of t.hc c'omposit.ion

Theorem 2-57. Ld (0, *) he a finite solvable group. Then the quotiellt. groups of nny (·omposition c·hnin for (a, *) nrc ('yc·lie groups of p"ime or :e~ is the only c~omposit.ion ('hain. (This remark takes carC' of I he basis for the incluetion when order (] = 2,) In the contrary lI'::::>'''::::>lIn_I::::>lIn= {e} ancI

0/11 = K;)::::> K; ::::> ..• ::::> K:"_l ::::> K:" Taking Ki

=

=

{e

* ll}.

nat H' (Ki) [i = 0, 1, ... , m], it follows from Theorem 2-49 that

G = Ko::::> KI ::::>.,'::::> K"'_l::::> Km = H, where (K j , *) is a normal subgroup of (K i _ 1 , *), As before, Theorem 2-[>0 implies (Ki_dKi, 0) ~ (K~_dK~, 0'), when!'e (K; __ I//(;, 0) iH (c'yC'lic') of prillI(' order, Hooking theRe sequnllceH togethl'r, we ohl aill (aftc~r r(,IIIClvillg the H('c'cII\(1 oeeurrellce of II), a composil,ioll c,hain for (0, *), all of whosc~ quotient groUpH are c'yelie groUpH of prime order. By the ,Tordl\lI-l WIder Theor('m I he same is true of every eomposition chain. ilc·fore eOlH'luclilll!; Uw \lI'(,:';1'1I1. H('C'tiOll, \ct Ull :.;t.nt.1' n fcw I'('!mlts, omitt.illg t1wir proofs. WI' hav{' :.;eclI that ev{'ry pC'rmutation in Sn (n ~ 2) can be written III' a product of tmllspositiolls. While this expl' 1I(e'(n». Hilll'C' t.Ite t.1,,'or,'1lI it>! IlSSllIIlI't! t.o hold for t.hl! HuhJ,!;nlllp (('(It), *), if Jllo(e(a», then e'(a) has all element of order p, and wc are donc. On tltp ot.her hand, if 1d o(C(a» for evpry a fl eentdl, 11 must divide [a; C(a»); for JI is prime and divides 0(0) = [G; C(a)) o(C(a). But then, in the dass equation, 11 divides each term of t.he summation alld also divides o(G), so that p I o(cent (J). As (cent G, *) is n comlllutat.ive subgroup of (G, *), it follows (again, from the II!mom) t.hat. ,~ellt. (J ('ollt.aiIlH :lII 1'1('IIII'II!. of ordl'r fl.

Corollary. (0, *) is:t finit.1' p-J,!;roup if

:tll(i

only if o( a) = 11k, for some k

> o.

Proof. Rupposc (U, *) is a /I-group, but. q I o(G) for some prime q ¢ p. Then by Cauchy's Theorl'lll, 0 has lUi element of order q, I!ont.radil!tin)!; the fact that (0, *) is a {I-group. Thus, JI is the only prime divisor of o«(J), which implies o(G) = 11k (k > 0). Conversdy, if o(a) = pk, then from Lagrange's Theorem each clement of G has a power of 11 as its order.

As all appli'·at.ioll of C:ulI'hy's Tll

O. Theil

[G:C(a») = o(U)/o(C(a» llIu:;t 1)(' divi:.:ihl(~ hy p. ~ow, /I divide:; clwh term of the summation, as well a:; divicll':; o«(l), :;0 that. JlI n(c·cnt. G). Thi:; implies o(ecnt G) > 1. WI'

II0\\'

:;hifl our I'IlIJ1ha:;i:; frolll nrhit l'ary JI-lI;ronps t.o Sylow p-subgroups.

Definition 2-38. LC'I (G,.) be n fi II if,(' group IUld p It prime. A subgroup (1', .) of (0, .) is said to be a Sylow p-subymup if (P, .) is a p-group and is not properly ('()J\jaill!'d ill nny olh!'r p-subgroup of (G, .) for the same prime p. Example 2-63. Thp :;Ylllllwl.ric· group (.'01:1,.) hILS thrpc (eolljuglltc) Sylow ~-snhgl'llup:;,

Thu:;

It

slll·pili(·ally, t.he Mnhgrnup:.: who:.:c! sf't.s of dmmmts nrc!

Aylo", /I-:;uh!!:l'Oup of n giv('n group Iwed not be unique.

Theorem 2-66. For !'lll'h prim!' p,

th(~

finite group (0,.) has a Sylow

11-suhgroup. Proof. If o«(J) = I or ]I t 0(0), (re~,.) i:;, in a t.riviul I:!CIll~e, the required Sylow p-:;uhgroup. 011 til!' 01 her halld, wlumever 11 I n( G), Cmwhy's Thcorcl)l gllarlllli!'!':': til 0, fof' if Itl = 0, Ih('n 0(1') = 0(1'1 n I'), HO l' = PI n 1', or rather l' ~ 1'\. Bllt I hI' fae! that (1', .) iH a maximal p-subgroup would imply l' = PI, whieh iH impol'l'ihl('. l'\ote also that (1', .) will not be found among the conjlll!;at(' SlIhl!;l'OIlJlH of (1'1, .) by ell'n1l'nls of P; weI'(' P = a. PI • a-I for some a E P, the'n 1'1 = a-I * p. a ~ P, again h':u\inl!; to a eontradietion, l\OW, if t.hl' ('OIijlll!;ah' ~iUhl!;roups of (1', .) are no\' exhausted by (I', .) and the pk, eonjugates of (1'1,.) indul'et\ hy nl('mbers of 1', choose ano\'her conjugate (1'2, .) distini't frolll any yet ('numerat.cd. As above, we can obtain the pk2, > 0, di"linl'l eonjugat.(,H of (1'2,.) indll('ct\ by P. No ('onjugate of (1'2, .) hy an ,,11'1lI1'1I1 of I' will alHo b(, 11I1 (~onjllgal.(! of (1) ... ) by IlII clement of 1', for if a, iJ E P,

"'2

then (b- I • a) '" PI • (iJ- 1 • a)-I = 1'2, with b- I • a E 1', contrary to the choice of P 2 • If nce(,HHary, n'p('at thiH argument onee mor(', Since (G, .) is a finilc grollp, all the eonjllgates of (I',.) are found af\'er a finite number of HI('pH, Hay n sl('ps. Tlw ('onjllgate subgroups of (1', .) Ilrc these: (1', .) itself, the pk, eonjllgal('s of (/'), .) by elements of P, \'he 11 k• eonjuglltes of (1'2,.) by eleml'Tlts of P, ('Ie'. 'I'll(' t.otal numlwr of distinet (!onjugntcs is therefore

I I 1/" I 1/" I , .. I pk.

=

I ·1 mp == I (mod p),

k.

>

0,

At this st.age, an ohvious qucstion arises: are then! allY Sylow p-subgroups whieh nrc not. ('onjllgate suh!l;roups of (1', .)? The unswer will be sccn to be no. Let us llSSlIllW slwh It subgroup (R, .) does exist, anel arrive at a contradid.ion. As h('fOl"(, , we eount the eonjugates of (R, .). This is done by first lilldillJ,!; tI)(' l·olljllJ.!:at('~ of (H,.) indlJ('l'c\ hy ('11'1lH'nt.s of P; t.here nre (I(J»/o(N

n 1')-=

pi,

of them, whel'(> j. > 0 (if .il = 0, then R = 1', n eontradiction). Note that (R, .) i" its('lf jlll'huh-d in this eOllnt, sinee R = e • R • e-t, eEl', Following Ihc proc'('dllrl' ahov(', II\(' I.olnl III1111bN of (~onjugat(' subgroups of (R,.) is found to be j; > O.

2-9

SYLOW THEOREMS

137

Thc ilJlport.:Illt. point 1l('I'(' is t.hat no l('rll1 of this SIIIII /'('(ilw(,1-l to 1; if some j. = 0, t.Il(~n (ll, *) wOllld I)(~ a (~()njllgat.(! of (I', .), whidl is impoH!>iblc, But we havl' Jll'('viollsly dl'tl'rJIlilH'd Ihat. I hI' nllllll)('1' of ('I)njllgah's of any Sylow P-HIlItJ,!;I'OIiP i" (,OIlJ,!;I'tIl·/tt. to I IIlo!) mUtlt be a multiple of p. 7. For a finite p-grollJl (G, *), prove the following; a) Any homomorphic image of (G, *) is again a p-group. b) (G, *) has a normal subgroup of order p. [/lint: cent G contains an element of order p.) c) (G, *) is a solvable group. [Hint: Induct on o(G); show that (cent G, *) and (G/cent G, ®) are hoth !iolvable.) 8. Prove that if o(G) = p", then the group (0, *) has at least one normal subgroup of order pk for aU 0 :::; k :::; n. [llint: Proeeed hy induction on n and utilize part (b) of the last problem.] 9. Let (11, *) be a normal subgroup of a finite p-grollp (G, *) and suppose that 0(1/) = p. Establish the indus ion 11 ~ cent G. 10. Prove t.hat if 0(0) = p2, then (G,.) is a commutative group. (llint: Recall Problem 21, He(~tion 2-5.)

140

(mOlll' TIU;ORY

11. 1.et (G, *) Ill' a

2-9

/(1'11111' such that 0(0) - lIfO. Verify the following assertions: a) o(cent G) ,.. p .. -I. b) cent G contains at least p elements. c) Every subgroup of order p .. - l is normal in (G, *). 12:-netepnine all Sylow p-subgroups of the group (Z24, +24). 13. Let (P, *) be a Sylow p-subgroup of the finite group (G, *). Prove that if (II, *) is any subgroup of (G,*) with P
In a ring (R, +,.) with identity, we sayan element a E R is invertible if it possesses an inverse relative to multiplication. Multiplicative inverses are unique, when they exist, and shall be denoted as before by a-I. The symbol R* wiII be used subsequently to represent the set of all inyertible clements of the ring. Theorem 3-1. Let (R, +,.) be a ring with identity. Then the Jlllir (R*,·) forms a group, known aI:! the group of invertible elements.

Proof. The set R* is nonempty, for at thc very least the multiplicative identity 1 E R*; moreovcr, 1 serves as the identity clement for the system (R*, .). If a, bE R*, the equation!! (a· b) . (b- I . a-I) = a· 1· a-I = 1,

(b- I

.

a-I) . (a· II) = 1>-1·1· b = 1

show the product a· b is also a member of R*. Whenever a is invertible, then we have a· a-I = a-I. a = 1, indicating that a-I is in R*. The allllociative law for multiplication is obviously inherited since R* S;;; R, so (R*, .) is a group. Before proceeding to the proofs of the basie results of ring theory, we shall consider several exampleH. Example 3-2. Each of the following familiar sYHtems, where ordinary addition and multiplication, is Il. commutative ring:

(R',

+, .),

(Q,

+, .),

(Z,

+, .),

(Ze,

+ and·

+, .).

inuieate

3-1

144

The first three of t.he!!C rings have an identity element, the integer 1, for multiplication. Exampl. 3-3. Anotlll'r ('xlunple of a rillK is provided hy the sct R = {a

+ bval a, b E Z},

and the operations of onlinary addition and multiplication. The set R is obviously closed under these operations, for

+ bva) + (c + elva) = (a -I- bv'3) . (c + dva) =

(a

+ c) + (b + d)v3 E R, (ac + 3bd) + (ad + bc)va E (a

R,

whenever a, b, c, dE Z. We omit the details of showing that (R, +,.) commutative ring with identity element 1 = 1 + ova E R.

IS

a

Example 3-4. If P(X) is the power Het of HOnle nonempty !!Ct X, then both the triples (P(X), u, n) and (P(X), n, u) fail to be rings, Kince neither (P(X), U) nor (/,(X), n) forms a group. However, the reader may recall that

in Example 2-20 we showed the pair (P(X),~) to be a commutative group; here, the symbol ~ indi(~at.es the symmetri(~ diff('rence operation A

~

B

=

(A - B) U (B -

A).

Since (P(X), n) is dearly a nommutative semigroup, in order to establish that the triple (p(X) , ~, n) constitutes a ring it is only necessary to verify the left distributivity of n over~. For this, we require the set identity (Problem 6, Section 1-1) A

n (B

- C)

=

(A

n B)

-

(A

n C).

The argument now proeeedK as follows: for all subsets A, B, 0 A

n (B

~

C)

=

A

n [(B

- C)

u (0

~

X,

- B»)

IA n (B - C») u [A n (0 - B)] = ((A n B) - (A nO») u [(A n C) = (A n B) ~ (A nO). =

-

(A

n B»)

Example 3-5. Let R denote the tlCt of all functions I: R' -+ R'. The sum I g and product I . (f of two functions I, g E R are defined as usual, by the

+

equations (f

+ g)(a) = I(a) + !/(a),

(f. g)(a)

= I(a) . g(a),

a E

R'.

In other words, we specify the functional value of these combinations at each point in tlwir domain. That (R, +,.) is a (~ommutat.ive ring with identity

J)";~'INITION

AND ELJo:MENTAUY PROPEIlTlJo:S OF RINGS

145

follows from the fact that the real numbers with ordinary addition and multipliention comprisc 8u(!h a system. In particular, the multiplicative identity d('lIwnt ill the NIIIHtUllt fUllctioll whoHe value at each real number is 1. It iH int('rpsting to nof.(! that the triple (R, /-,0), where 0 indicates the operation of fUllctional eompositioll, fails to be a ring. The left distributive law

f

0

(g

+ /t) =

(f. g)

+ (f oIt)

does not hold in this case. Example 3-6. For a more interesting example, let (G, *) be an arbitrary commut.ative group and hom G be the set of all homomorphisms from (G, *) into itself. From Theorem 2-36, it is already known that (hom G, .) is a semigroup with identity. We propose to introduce a notion of addition in hom G such that the triple (hom G, +, .) forms a ring with identity. To achieve this, simply (lC'fillc the sum f + (J of two fUlletions f, (J E hom G hy the rule

(f

+ g)(a) = f(a) * g(a),

aE G.

With the possible exception of the closure condition, it is fairly evident that the pnir (hom G, +) is a commutative group; the trivial homomorphism acts as the zero element, while the negative of each function I E hom G is obtained hy taking (-- I)(a) = f(a)-l fOr nil a E G. To estnhlish (llmlllm, (!hooBrlnrbitmry elt!lIlcnts a, b E (J and fUllctiollti I, y ill hom G. Theil (f

+ g)(a * b) = I(a * b) * g(a * b) = I(a) * feb) * g(a) * g(b) = (j(a) * Il(a») * (f(b) * {f(b») = (f + g)(a) * (f + y)(b) ,

so that the sum J + g is also t\ homomorphitllll of (G, *) into itself. In J'(·gard to UI(l left distributive law, we lllwe, fol' /Lny a in 0, [f. (g

+ h)](a) = f«g + h)(a») = J(g(a) * /t(a») = J(g(a») * f(h(a») = (f. g)(a) * (f. h)(a) =

(f. g

+ I· h)(a).

Therefore, 1 «(J - f- It) = I· (J - f- I . It. The right distributive law follows in a similar manner, showing (R, +, .) indeed to be a ring. Necdless to say, eommutativity of the underlying group (G, *) is essential to our discussion; ill its absence, we could not even prove addition to be a. binary operation in hom G. 0

146

3-1

RING THEORY

Example 3-7. In Section 2-3, we considered the group (Z.. , +..) of integers modulo n. This group was obtained upon defining in the set Z.. the notion of addition of congruence classes:

fa]

+.. fb] =

[a

+ b].

A binary operation '" of multipliHation of e1a.~SC!! may f!qually well be introduced in Z .. by speeifying that for each pilir of clements [a], [b] E Z .. , [a]· .. fb]

=

fa· b].

This lat,t(1r definition pre!!l1ntR R. problem Rimilar to that of addition in that we InUloJt Hhow Umt !Jill rmmll.ing produd leI' 1/) iH illlil!)llm laws hold automatically ill S as a ('Ollsequl'll('(> of their validity in R. Sinee these laws arc inherited from H, there is 110 partirular n('('(>ssity of requiring them ill the definition of II. subring. In view of this observation, II. subring eould alternatively be defined as follows: the t.ripl!' (8, -1-, .) is It suhring of the ring (R, +, .) whenever 1) S iF; II nOIl(,llIpty HuhHet of Il, 2) (8, +) is a subgroup of (R, +), 3) S is closed undcr multiplication. Even this definition may be improved upon, for the reader may recall that if 0 ~ II ~ G, then the pair (II, *) is a subgroup of the group (G, *) provided that a, bEll implies a * b- 1 E H. Adjusting the notation to our present situation, we obtain a minimal set of conditions for detennining subrings.

+, .) be a ring and 0 ~ S ~ R. +, .) if and only if

Definition 3-8. Let (R, (S, is a 8ubring of (R,

+, .)

Then the triple

1) a - bE S whl'll('ver a, b E S (closed under differences), 2) a· b E .-; wiwlwVI'r a, 1) E'-; (closed under multiplication).

(R, +, .) has two trivial subrings; for, if 0 denotes the zero element of the ring (R, -1-, .), then both ({OJ, +, .) and (R, +, .) are sub rings of (R, +, .).

Example 3-9. Evpry ring

+, .), the triple (Z" 1-,') is U Hubrin!!:. while (Zo, +, .) i~ not. In particular, we infer that in It ring with identity, a suuring does not n('('11 to contain the identit,y element. 0 2 4 '0 +6 0 2 4 Example 3-11. COllsi(\l'r (Zo, I 0, '6), the - - - - - 0 0 0 0 0 2 4 0 ring of int('gl'rs modulo (i. IfS= ~O, 2, 4) , 2 0 4 2 2 2 4 0 then (8, +6, '6), whost, op 0, the notation na is merely an abbreviation for the finite sum a + a + ... + a, n summands. However, when there is an identity element present, it is possible to express na &8 the product of two ring elements: na = (nI)· a. DefInition 3-9. Let (R, +, .) be an arbitrary ring. If there exists a positive integer n such that na = 0 for all a E R, then the least positive integer with this property is called the characteristic of the ring. If no such positive integer exists (that is, na = 0 for all a E R implies n = 0), then we say (R, +,.) has characteri8tic zero.

The rings of integers, rational numbers and real numbers are standard examples of systems having characteristic zero. On the other hand, the ring (P(X), !:l, n) is of characteristic two, since 2A

=

A !:l A

=

(A - A) U (A - A)

=0

from every subset A of X. Let (R, +,.) be a ring with identity. Then (R, +,.) has characteristic n > 0 if and only if n is the least positive integer for which ni == O.

Theorem 3-6.

Proof. If the ring (R, +, .) is of characteristic n > 0, it follows trivially that = O. Were mi = 0, where 0 < m < n, then

nl

ma

=

m(I . a)

=

(mI) . a

= o· a =

0

+, .)

(or every element a E R. This would mean the characteristic of (R, is Jeaa than n, an obvious contradiction. The converse is established in much the same way. Corollary. In an integral domain, all the nonzero elements have the same additive order, which is the characteristic of the domain.

Proof. To verify this assertion, suppose the integral domain (R, +, .) has positive characteristic n. According to the definition o( characteristic, any a E R (a F- 0) will then poBse88 a finite additive order m, with m S n. But the equation

o=

implies ml

=

ma = (mI)· a

0, since (R, +,.) is free of zero divisors. We may therefore con-

3-1

DEFINITION AND ELEMENTARY PROPERTIES OF RINGS

153

clude from the theorem that n ::; m. Hence m = n and every nonzero element of R has additive order n. A somewhat similar argument can be employed when (R, +, .) is of characteristic zero. The equation 100 = 0 would lead, as before, to ml = 0 and consequently m = O. In this case, each nonzero element of R must be of infinite order. The last theorem serves to bring out another point. Corollary. The characteristic of an integral domain (R, or a prime number.

+, .)

is either zero

Proof. Let (R, +,.) be of positive characteristic n and assume that n is not a prime. Then n can be written as n = nln2 with 1 < ni < n (i = 1,2). We therefore have Since by hypothesis (R, +, .) is without zero divisors, either nil = 0 or n21 = O. But this is plainly absurd, for it contradicts the choice of n as the least positive integer such that nl = O. Hence, we are led to conclude that the characteristic must be prime. Now, suppose (R, +,.) is an arbitrary ring with identity and eonsider the set ZI of integral mUltiples of ihe identity ZI

=

{nIl n EZ}.

From the relations nl -

ml

=

(n - m) 1

and

(nI) . (ml)

=

(nm)I

one can easily sec that the triple (ZI, +,.) itself forms a commutative subring with identity. The order of the additive cyclic group (ZI, +) is simply the characteristic of the original ring (R, In case (R, +, .) is an integral domain of characteristic p > 0, p a prime number, we are able to show considerably more: each nonzero element of (ZI, +, .) is invertible. Before proving this, first observe that by Theorem 2-23, the set Zl consists of p distinct elements; namely, the p sums nI, where n == 0,' 1, ... ,p - 1. Now, let nl be any nonzero element of ZI, 0 < n < p. Since p and n are relatively prime, there exist integers r, 8 for whieh rp sn = 1. Therefore,

+, .).

+

1

=

(rp

+ sn)1 =

(rl)· (pi)

+ (d) . (nl).

As pi = 0, we obtain 1 = (81) . (nl), so that 81 constitutes the multiplicative inverse of nI in (Zl, +.. ). We shall return to a further discussion of the characteristic of a ring at the appropriate place in the sequel; in particular, the value of this last result will have to await future developments.

154

3-1

lUNG TIH:

0,

+

(a b)1' = a" b" for all a and b. 20. Suppose (R, ., 0) and (ll', .', .') urn t.wo rinll;H. J)('lim' hi nary operations lin the Cart.c>sian procluet Il X Il' as follow!!: (a, b)

+ (e, d)

=

(a. c, b.' d),

+, .)

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

0

+ and·

c, b.' d).

a) Prove t.hat th!' system (R X R', forms a rin!!;, "ailed t.he direct product of thc rinll;s (ll, *,0) and (R', .', .'). b) If the original rings are commutative with idl'ntity, show that the same must bc truc of (R X R',+, .).

156

3-2

RING THEORY

~2 IDEALS AND QUOnENT RINGS

In this section we introduce an important class of subrings, known as ideals, whose role in ring theory is similar to that of the normal subgroups in the study of groups. As shall be seen, ideals lead to the construction of quotient rings which are the appropriate analogs of quotient groups. Deftnltion 3-10. A subring (I,

+,.)

of the ring (R,

(R, +, .) if and only if r e R and a e I imply both r·

+,.) eI

0

is an ideal of and o· reI.

Thus we require that whenever one of the factors in a product belongs to I, the product itself must be a member of I. In a sense, the set I "captures" products. If (1, is a subring of (R, I is already closed under mUltiplication. For (1, +, .) to be an ideal, a stronger closure condition is imposed: 1 is closed under multiplication by an arbitrary element of R. In view of Definition 3-8, which gives a minimum set of conditiona on I for (I, +,.) to be a subring, our present definition of ideal may be rephrased as follows:

+, .)

+, .),

Definition 3-11. Let (R, +,.) be a ring and I a nonempty subset of R. Then (1, +,.) is an ideal of (R, +,.) if and only if 1) 0, bel imply 0 - bel, 2) r e Rand 0 e I imply both r· 0 e I and o· reI.

In the case of a commutative ring, of course, we need only require r·

0

e

I.

Before proceeding further, we shall examine this concept by means of several specific examples. Example 3-13. In any ring (R, +, .), the trivial subrings (R, +,.) and ({O}, are both ideals. A ring which contains no ideals except these two is said to be simple. Any ideal different from (R, is termed proper.

+, .)

+, .)

Example 3-14. The subring ({O, 3. 6, 9}, +12) is an ideal of (Z12. +12, '12), the ring of integers modulo 12. Example 3-15. For a fixed integer 0 e Z, let (0) denote the set of all integral multiples of o. that is. (0) = {no I n e Z}.

+.. ) to be an ideal of the ring of

The following relations show the triple «0). integers (Z. +, .): no - rna = (n - m)o. m(no)

=

(mn) 0,

n, m eZ.

In particular, since (2) = Z•• the ring of even integers (Z•• (Z, + .. ).

+..) is an ideal of

3-2

IDEALS AND QUOTIENT RINGS

157

Example 3-16. Suppose (R, +,.) is the commutative ring of functions of Example 3-5. Define 1= {fE R If(I) = O}.

For functions f, gEl and hER, we have (J - g)(l)

= J(I)

- gel)

=

0 - 0

=

0

and also (h· /)(1)

==

h(I) . f(I)

=

h(I) • 0

= o.

Since hothf - g and h· g belong to I, (1, +,.) is an ideal of (R,

+, .).

If condition (2) of the definition of ideal is weakened so as to only require that the product r· a belongs to I for every a E I and r E R, then we arrive at the notion of a left ideal (right ideals are defined in a similar way). For commutative rings, it is plain that every left (right) ideal is an ideal or; as it is sometimes called, a two-Bided ideal. We next derive several interesting and useful results concerning ideals of arbitrary rings.

3-7. If (I, +,.) is a proper ideal of a ring (R, +, .) with identity, then no element of I has a multiplicative inverse; that is, I n R* = 0.

_ Thearem

Proof. Suppose to the contrary that there is some member a ~ 0 of I such that a-I exists. Since I is closed under mUltiplication by arbitrary elements of R, a-I. a = 1 E I. It then follows by the same reasoning that I contains

r

r· 1 = r for every E R: "That is, R !; I. Inasmuch as the opposite inclusion always holds, I = R, contradicting the hypothesis that I is a proper subset of R. 3-8. If (It, +, .) is an arbitrary indexed collection of ideals of the ring (R, +, .), then IJ!> also is (nit, +, .).

Theorem

Proof. First, observe that the intersection nli is nonempty, for each of the sets Ii must contain the zero element of the ring. Suppose the elements a, bE nIi and r E R. Then a and b are members of It, where i ranges over the index set. As the triple (It, +,.) is an ideal of (R, +, .), it follows from Definition 3-11 that a - b, r· a and a· r all lie in the set h But this is true for every value of i, so the elements a - b, r· a and a· r belong to nI., which implies that (nI., +,.) is an ideal of (R, +, .).

Consider, for the moment, an arbitrary ring (R, 8 of R. By the symbol (8) we shall mean the set (8)

=

n{I I 8!; I; (1,

+, .) and a nonempty subset

+,.) is an ideal of (R, +, .)}.

The collection of all ideals which contain S is not empty, since the improper

158

3-2

RING THEORY

+, .)

idal (R, clearly belongs to it; thus, the set (8) exists and is such that 8 ~ (8). Theorem :l-S leads directly to the following result. Theorem 3-9. The triple «8), the ideal ymeraled by Ihe set S.

+,.) if! an ideal of the ring

(R,

+, .), known as

It is not.eworthy that whenever (I, +, .) is any ideal of (R, +, .) for which S ~ I, t.hell (8) ~ I. III view of this, olle frequently speaks of «S), +,.) ILS being the smallesl ideal to contain the set S. An ideal generated by a single ring clement, say a, is called a principal ideal and is designated by «a), +,.). A natural undt'rtaking is to determine the precise form of the members of (S). If we impot«) the requirement that (R, +, .) he a commutativo ring, it is a fnirly Hilllpl!' matter to e/wek Umt (8) is givell by (8) =

{1:ri"~i+ 1:njlljlriER;si,l!jE8;njEZ},

where the symbol 1: ill =

g • natker(Q)

that the funetiolls f and g arc themselves identical.

168

3-2

RING THEORY

As a final topic in this section, we look briefly at the problem of extending a function from a subsystem to..the entire system. In practice, one is usually eoncerned with extensions which retain the characteristic features of the given function. The ncxt theorem, for instance, presents a situation where it is possible to extend a homomorphism in such a way that its extension also has this property. But first, one more definition is needed (see Problem 17, Section 3-1). DefInition 3-14. The center of a ring (R, +, .), denoted by cent R, is the center of the multiplicative eemigroup (R, .) j that is,

cent R

=

{r E R I r· a

=

a· r for every a E R}.

We are now in a position to state and prove our theorem. Let (I, +, .) be an ideal of the ring (R, +, .) and I a homomorphism from (1, +,.) onto (R', +', .'), a ring with identity. If I s;;; cent R, then there is a unique homomorphic extension of I to all of R.

Theorem 3-20.

Prool. As a starting point, we choose u E I so that I(u) = 1', where l' is the identity element of the ring (R', +', .'). Since (1, +,.) is an ideal, the product a· " will be a member of the set I for each choice of a E R. It is therefore possible to define a new function g: R -+ R' by setting yea) = I(a· u), a E R. In particular, if the element a belongs to I, then g(a)

= I(a. u) = I(a) .' I(u) = I(a) . l' = I(a),

showing that the restriction y I I agrees with I. What we are observing, in effect, is that g extends the given function J to all of R. The next thing to establish is that both ring operations are preserved by the function g. The case of addition is fairly obvious: if a, b E R, then

g(a + b)

= I«a + b) . u)

= I(a· u + b . u) = I(a· u) +' I(b· u)

=

g(a)

+' g(b).

As a preliminary step to demonstrating that g also preserves multiplication,

note that

J«a' b) • u 2 )

= I«a· b· u) . u)

== I«a' b) • u) .' I(u) == I«a· b) . u).

In licht of this, we are able to conclude

g(a· b)

= I«a' b)· u) = I«a· b)· u2 ) = I«a· u)· (b· u» = I(a· u) .' I(b· u)

= g(a) .' g(b).

The third equality is justified by the fact that u E cent R, hence commutes with b.

3-2


Only the uniqueness of U remains unproved. Let us therefore suppose the function h is another homomorphic extension of f to the larger set H. Since f and h are required to coincide on I and, more specifically, at the element u, h(u) = f(u) = 1.

With this in mind, it follows that h(o) = h(o) .' h(u)

=

h(o· u) = f(o . u)

= g(o)

for all 0 in H, 80 hand 9 must be the same function. Hence there is one and only one way of extending f from the ideal (I, +, .) to the entire ring (H,

+, .).

PROBLEMS

1. Determine all ideals of (Z12. +12. '12). the ring of integers modulo 12.

2. Show by example that if (11,+.') and (12,+,·) are both ideals of the ring (R.+ • .). then the triple (11 U 12,+.') is not necessarily an ideal.

3. For any ideal (1.+,,) of the ring (R,+, .). define C(l) to be the set

C(I) - {r e R I r . a' - a . reI for all a e R}. Determine whether (C(l),+.·) forms a subring of (R,+ • .).

4. If (11,+,') and (12,+,') are ideala of the ring (R,+,') such that 11 n 12'" {O}, prove a· b = 0 for every a E 11, bE 12.

+, .). is a simple ring. ,+..,'..) of intepre modulo" It a

5. a) Verify thanlienng--orreal numbers (R', b) Prove that for each n e Z+, tho rin, (Z .. principal ideal ring.

6. Let (1,+,,) be an ideal of the ring (R,+.·) and define ann 1 :- {r E R ! r • a - 0 for all a E l}. Prove that the triple (ann 1,+.,) constitutes an ideal of (R,+ • •). called the

annihilator ideal of 1. 7. Let (1,+.') be an ideal of (R.+, .), a commutative ring with identity. For an arbitrary element a in R, the ideal generated by 1 U {a} is denoted by «(I, a), Assuming a Ii! I. show that

+, .).

(I,a) - {i+r·a!iEI.rER}.

8. Suppose (II, +,.) and (12,+,') are ideals of the ring (R.+, .). Define

/11+12 - {a+blaElt,bE I 2}.

+

Show that (11 12,+.·) is an ideal of the ring (R.+ • .); in Tact, (11 is the ideal generated by 11 U 12.

+ 12.+,·)

170

3-2

RING THEORY

9. In the ring of integers, consider the principal ideals «n),+,·) and «m),+,·) generated by nonnegative integers nand m. Using the notation of the previous two problems, verify that «n), rn) = «rn), n) = (n)

+ (m)

= ({m, n}) -

(d),

where d iR !.he grl'at.l'lit common rlivillor of nand m. 10. ConRider th,' ring (I'(X), ~,n) of ExamJllll3-4. For a fixed BuhllCt.~!;; X, definll the function I: I'(X) -+ P(X) by I(A) = Ii noS.

Show that I ill a homomorphism and determine its kernel. 11. THilizI! Prohl/'1Il 19, Hl'dinll 3-1, to Rhnw that in an integral domain (R, characteristic p > 0, the mapping

I(a) = aP ,

+, .) of

aE R,

+, .)

is a homomorphism of (R, into itself. 12. Given thatf is a homomorphism from the ring (R, onto the ring (R', prove that a) if (1, is an ideal of (R, then the triple (f(l), is an ideal of

+, .)

+', .'),

+, .) +, .), +', -') b) if (1', +', .') is an ideal of (R', +', -'), then the triple (f-l(1'), +,.) is an ideal of (R, +, .) with ker (f) !;;f- 1 (1'), c) if (R, +, .) is a principal ideal ring, then the same is true of (R', +', -'). [Hint: ForaER,f«a» = (f(a».] 13. Let I be a homomorphism from the ring (R, +, .) into itself and oS be the set of (R',+',·'),

elements that are left fixed by /:

oS = {a E R I f(a) = a}. Establish that (oS,+,·) is a subring of the ring (R,+, .). 14. For a fixed element a of (R, +, .), a ring with identity, define the le/t-muUiplicatio7J /unction/.: R -+ R by taking fG(x) = a· x,

xE R.

If F H denotes the set of all sllch functions, prove the ring analog of Cayley's theorem: a) The triple WH, forms a ring. where denotes the usual pointwise addition of funct.ions and. denotes functional composition. b) (R,+,·) ~ (FH,+, .). [Hint: Consider the mappingf(a) = fG.] 15. Let CR, be an arbitrary ring and (R X Z, be the ring constructed in Theorem 3-16. Establish that a) (R X 0, +',.') is an ideal of (R X Z, +', .'), b) (Z, ~ (0 X Z, e) if a is an idempotent element of R, thl'n the pair (-a, 1) is idempotent in the and (a, 0) .' (-a, I) "" (0,0). ring (R X Z,

+..)

+, .)

+

+', .')

+, .)

+', .'),

+', .')

3-2


171

16. Suppose (/,,+,.) and (/2,+,') are both ideals of the ring (R,+, .). Define the set 1,·12 by

+, .)

where:E denote!! a finitl~ Rum with olle IIr IlIOfe term",. Prove that (h, 12, is an ideal of (R, -+, '). 17. GiVl'II that. (/. -I, .) iH all idllal of the ring (R, -/-, .), Khow that a) whenevllr (ll, -I,,) ill Cllllllllutative with idellt.ity, then 1111 ill the quotiont ring

(Rl1, +, .),

h) the ring (Rl1, +,.) lIIay have ,livitlorll of Z(,fO, I,vell though (R, -t,.) does not hav" a.ny, c) if (R, +,.) is a principal ineal rinJ!;, then so is the quotient ring (RIT,+, .). IH. IAlt (ll, I , .) till " (!IIlIIlIIutu.tive rillK wit.h hlt'lltity u.nl! I"', N ,It'notl' the 81~t of nilpotent clements of R. Verify that a) the triple (N, is an ideal of (R, [lJint: If an = b'" = 0, consider

+,.)

+, .).

(a - b)n+"'.)

b) the quotient ring (RIN, +,.) has no nonzero nilpotent elements. 19. Assume (R, is a ring with the property that a 2 a E cent R for every element a in R. Show that (R, is a commutative ring. [Hint: Make use (a b) to prove, first, that a' b b . a lies in the of the expression (a b)2 center.) 20. Illustrate Theorem 3-18 by considering the rings (Zo, +6, '0), (Za, +a, 'a), and the homomorphismf: Zo -> Za defined by

+, .)

f OJ thus

for we plainly have a (a, b)· ( a 2 b2

+

'

a2

-b) + 00) = + = (a + + b -00 a + b2

2

2

a2

b"

l

b2

(1,0).

This shows that the nonzero elements of C have inverses under multiplication, proving the system (C, +, .) to be a field. The field (C, +, .) contains a subring which is isomorphic to the ring of real numbers. For if R' X 0 = {(a, 0) I a e R'}, it follows that (R',

+,.)

~

(R' X 0, +,.) via the mapping I defined by

I(a)

=

(a, 0),

aeR'.

(Verify this!) As the distinction between these systems is one only of notation, we customarily identify the real number a with the corresponding ordered pair (a, 0) j in this sense, (R', +, .) may be regarded as a subring of (C, +, .). The definition of the operations + and· enables us to express any element (a, b) E C a s ' (a, b)

=

(a,O)

+ (b,O) . (0, I),

where the pair (0,1) is such that (0,1)'

=

(0,1)· (0,1)

=

(-1,0).

lntto-

174

3-3

RING THEORY

ducing the symbol i as an abbreviat.ion for (0, 1), we thus have (a, b) = (a,O)

+ (b, 0) . i.

If it is agreed to replace pairs of the form (a, 0) by the first component a, this reprcsentation heeomcs (a, b) = a -1- bi,

with i 2 = - 1. In otlH'r WOrthl, the field (e, +, .) more than the familiar complex number system.

lUi

defined above is nothing

The following th('ormn shows that a field is wit.hout divisors of zero, and consequently it! a system in which the cancellation law for multiplication holds (see Theorem 3-5). Theorem 3-21. If (P,! , .) is a field ILnd a, b E F with a' b = 0, then cithcr a = 0 or b = O.

Prooj. If a = 0, the th('on'm is alreudy est.ublish(!d. So let us suppose that ;r6 0 and prove that b = O. By the definition of a field, the element a, being nonzero, must have a multiplicative inverse a-I E F. The hypothesis a· b = 0 then yields

a

o=

a-I. 0

=

a-I. (a· b) = (a-I. a) . b

=

1 . b =-b,

as desired. Since a field is a commutative ring with identity, and we have just proved that it contains no divisors of zero, we conclude that any field is an integral domain. There obviously are integral domains which are not fields; for instance, the ring of integers. However, an integral domain having a finite number of clements must necessarily be a field. Theorem 3-22. AllY finite integral domain (R,

+, .) is a field.

Pl'Ooj. Suppo!!C at, a2, ... , an are the members of the set R. For a fixed nonzero clement a E R, consider the n products a· alt a· a2, ... , a· an. These products are all distinct, for if a· ai = a· aj, then aj = aj by the cancellation law. It follows that each clement of R is of the form a· ai. In particular, there exililts some aj E R stwh t.hat a· aj = 1; since multiplication is commutative, we thus have ai = a-I. This shows that every nonzero element of R is invertible, so (R, +, .) is a field.

+.. , '..)

It was previously seen that for each positive integer n the system (Zn' is a commutative ring with identit.y. Our next result indicates for precisely what values of n this ring is a field. Theorem 3-23. The ring (Z", 1-", 'n) of integers modulo n is a ficld if and only if n is 1\ prime number.

3-3

FIELDS

175

Prooj. We first show that if n is not prime, then (Zn, +n, 'n) is not a field. Thus assume n = a· b, where 0 < a < nand 0 < b < n. It follows at once that [a)

'n

[h)

=

[a· b]

=

[n]

=

[0),

although hoth [a) .,& (0), [b) .,& [0]. This means that the system (Zn, +n, 'n) is 1I0t lUI intl'p;ral domain, and hcnce not a fidd. On Ow ol,hN hand, SIlPJlOSP that n is It prime number. To show that (Z", -I n, ',,) iH a fidd, it. sufliccs to pl'ove here that each 1l0llZt'ro e1emcnt of Z .. has a multiplicative inverse in Zn. To this end, let raj E Zn, where 0 < a < n. Aeeording to Theorem 1-1:~, Rinee a and n have no common factors, there exist intcgerH r and II lmdl tlmt

a·r+n·8=1. Thill implies tlmt. [a]· .. [rJ = [a· r] -1-.. [0] = [a· 1'] -1-.. [n· s]

= [a· r -I- n· 8] = [1], showing the congruence class r1"] to be the multiplicative inverse of [a]. Therefore (Zn, +n, 'n) is a field, aH required. There is an interesting relationship between fields and the la.ck of ideals; what we shall show is that fields have as trivia.l an ideal structure as possible. Theorem 3-24. Let (R, +,.) be a commutative ring with identity. Then (R, +,.) is a field if and only if (R, +,.) has no nontrivial ideals.

+, .) +, .)

Proof. ASHume first that (R, is a field. We wiHh to show that the trivial ideals ({O}, +, .) and (R, are its only ideals. Let us assume to the contrary that there existH some nontrivial ideal (I, +,.) of (R, +, .). By our assumption, the subRCt I is sueh that I .,& {01, and I .,& R. This means there is some nonzero clement a E I. Since (R, +, .) is a field, a has a multiplicative inverse a-I E R. By the definition of ideal, we thus obtain a-I. a = 1 E I, which in turn implies I = R, contradicting our choice of I. Conversely, suppose that the ring (R, +,.) has no nontrivial ideals. For an arbitrary nonzero clement a E R, consid!'r the principal ideal «a), +, .) generated by a: (a) = {r· a IrE R}. \

Now «a), +,.) cannot be the zero ideal, since a = a· 1 E (a), with a"& O. It follows from the hypothesis that the only other pORsibility is «a), = (R, that is, (a) = R. In particular, since 1 E (a), there exists an element ,. E R for whieh i'. a =, 1. Multiplication is eommulative, so that,. = a-I. Hence e:wh nonzero {'Iement of R hn" a lIlult.iplientive inverse in R.

+, .);

+, .)

176

3-3

RING THEORY

+, .)

In view of this last result, the ring of integers (Z, fails to be a field, since it possesses the nontriviaNdeal (Z., Theorem 3-24 is useful in revealing the nature of homomorphisms between fields.

+, .).

Theorem 3-25. Let f be a homomorphism from the field (F,

+, .)

onto the field (F', Then either f is the trivial homomorphism or else (F, and (F', +', .') are isomorphic.

+', ").

+, .)

Proof. The proof consists of noticing that since (ker (1), +, .) is an ideal of the field (F, +, .), either the set ker (J) == {OJ or else ker (f) = F. The condition ker (1) = {OJ impliesf is a one-to-one function, in which case (F, +, .) ~ (F', +', .') via f. On the other hand, if it happens that ker U) == F, then each element of the field (F, morphism.

+, .) must map onto zero i that is, f is the trivial homo-

Plainly, any ring with identity which is a subring of a field must in fact be an integral domain. We now tum our attention to the converse situation i specifically, one may ask whether each integral domain can be considered (apart from isomorphism) as a subring of some field. More formally, can a given integral domain be imbedded in a field? In the finite case, there is obviously no difficulty, since every finite integral domain already forms a field. Our concern with this problem arises from the desire to solve equations of the type a . x = b, a #- O. A major drawback to the notion of an in~l domain is that it does not always provide us with a solution. Of course, any such solution would' have to be unique for a· XI = b = a· X2 implies XI = X2 by the cancellation law. It hardly seems necessary to point out that when the integral domain happens to be a field, there is always a solution of the equation a . X = b (a #- 0), namely X = a-I. b. We begin our discussion of this question with a definition. Definition 3-16. By a aubfield of the field (F,

(F', +,.) of (F,

+,.) is meant any subrinl(.

+,.) which is itself a field.

+, .) of rational numbers is a subfield of the field Surely, the triple (F', +, .) will be a subfield of the field (F, +, .) provided For example, the ring (Q,

(R', +, .).

(1) (F',+) is a subgroup of the additive group (F,+) and (2) (F' - {O},·) is a subgroup of the multiplicative group (F - {O},·). Recalling our minimal set of conditions for determining subgroups (Theorem 2-17), we see that (F', +,.) will be a subfield of (F, +,.) if and only if the following hold: 1) F' is a nonempty subset of F with at lcast one nonzero clement. 2) a, b E F' implies a - b E F'. 3) a, b E F', where b #- 0, implies a· b- I E F'.

+, .)

It should come as no surprise that if (Fi' is an arbitrary collection of subfields of the field (F, +, -), then (n Fi, + •. ) is also a subfield.

3-3

I'IIILDS

177

The next theorem furnishes some clue to the nature of the field in which we wish to imbed a given integral domain. Theorem 3-26. Let the integral domain (R, field (F, If the set F' is defined by

, +, .).

F'

=

+, .)

{a· b- 1 I a, b E Rj b

~

be a subring of the

O},

then the triple (F', +,.) fonns a subfield of (F, +,.) such that R fact, (F', +,.) is the smallest subfield containing R.

~

F'. In

Proof. Note first that the definition of the set F' is meaningful; indeed, if a, b E H with b ~ 0, the product a· b- 1 must be in F by virtue of the fact (F, +,.) is a field. Since 1 = 1.1- 1 E F', F' ~ {O}. Now consider two arbitrary elements x, y of F'. We then have

for suitable a, b, e, d E R, where b x -

~

0, d

~

O. A simple calculation shows

y.= (a· d - b· c) . (b· d)-l E F'.

Also, if y is nonzero (that is, whenever c

~

0),

= (a· d) . (c· b)-l E F'.

X· y-l

In light of the remarks following Definition 3-16, this is sufficient to establish that the triple (F' , +, .) is .a subfield of (F, +, .). Furthennore, a =

a· 1 = a· 1-1 E F'

for each a in H, so that R S;;; F'. Any subfield of (F, +, .) which contains R necessarily includes all products a· b- 1 with a, b '" 0 in R, hence contains F'. Theorem 3-26 began with an integral domain already imbedded in a field. In the general case, it is actually necessary to construct the imbedding field. Since the expression a· b- 1 may not always exist, one must now work with ordered pairs (a, b), where b ~ O. Our thinking is that (a, b) will playa role analogous to a· b-1 • As a starting point, let (R, +, .) be an arbitrary integral domain and K the set of ordered pairs, K

=

{(a, b) I a, bE Hi b ~ O}.

A notion of equivalence may bc introduced in K as follows: (a, b)

= (e, d)

if and only if

a·d=b·e.

(We have in mind the foregoing theorem in which a· b- 1 if a· d = b· c.)

= c· d- 1

jf and only

178

3-3

RING THEOHY

It is not difficult to verify that the relation rl'lation in K; that is to sny,

=,

thus defined, is an equivalence

1) (a, h) = (a, h), 2) if (a, h) == (e, d), then (e,l1) == (a, h), 3) if (a, h) == (e, Ii) and (e, II) == (e, f), then (a, h)

==

(e, f).

The leaRt obvious statenlC'nt is (3). In this case, the hypothesis (a, h) and (e, rJ) = (e, f) implies that

e· f

a· d = h· e,

= d·

==

(e, d)

e.

l\Iultiplying the first of these equations by f and the second by h, we obtain

a . d . f = h . e . f = b· d· e, and, from the eomlllutativity of multiplication, a· f· d = b· e· d. Since d ~ 0, this flletor may be clUwelled to yield a· f = b· e. But then (a, h) == (e, f), as required. Next, we lahel those elements which are equivalent to the pair (a, b) by the symbol la, bl; in other words, la,

hI =

{(e, d) E K

=

{(e, d) E K

I (a, b) == (e, d)} I a . d = b· c}.

To empha.'1ize the similarity between what follows Ilnd the familiar construction of thc rational lIumi)('rs, many nuthors prefer to write alb in place of la, bl; the rcader will realize the difference is merely a matter of notation. The collection of all equivalence classes la, b] relative to == will be designated by F. From Theorem I-!i, we know that the elements of F constitute a partition of the set K. That is, the ordered pnirs of K fall into disjoint classes, with eneh dnss ('onsist illl( of (·quivnll'nt pairs, Ilnd noncquivalent pairs belong to different classes. Further, two Hueh classes la, hI and Ie, rll are identical if and only if a . d = h· e. },('t us proceed to introduee suitable operations of addition and multiplication in F. We do these by nlC'ans of the equations

la, hI

+' Ie, d]

=

la·

la,

bl·' Ir, til

=

[a· e, b ·11).

d

+ b· e, b . ri],

Note, incidentally, that sin('e b ~ 0 and tl ~ 0 imply b· d ~ 0, the right-hand sides of tiH's(' formulas are :l('tually eielllen!!; of F. WI! must, as mmnl, first justify that thpsp olwrntiolls an' well-defined. Otherwise expresS(·d, w(, 11('(,.1 to show that. t II!' sum nnd produet. are independent of til(! pnrt.i.·ular I'I('nl('lIts of H uspd ill til(' dp(init iOIl. To aehicve this, let

3-3

179

FIELDS

[a, b]

=

[a', b/] and [e, d]

=

[e ' , d/l. From the equations

a· b'

=

b· a',

=

e· d'

d· e' ,

it follows that (a· d

+ e· b) . (b' · el')

- (a ' · el'

= =

+ c' · b') . (b· el)

(a· b' - b· a'l . (d· d' ) + (c· d' - d· e' ) . (b· b') O· (d· d') + 0 . (b . b') = O.

Thus, by the definition of equality of classes, [a· d

+ e· b, b· d] =

+ e' · b', b' · d/],

[a" d'

proving addition to be well-defined. In mueh t.he [a· e, b·

dl =

~mnle

way, one can show that

[a' • e' , b' . d').

The next lemma estabHs}les the algebraic nature of the triple (F, +',

./).

Lemma. The system (F, +',.') is a field, generally known as the field of quotients of the integral domain (R, +, .).

Proof. It is an entirely straightforward matter to establish that the triple (F, is a commutative ring. We leave the reader to make the necessary verifications at his leisure, and merely point out that [0, bl serves as the zero element while [-a, b] is the negative of [a, b]. That the equivalence class [a, a], where a is any nonzero element, constitutes the multiplicative identity is evidenced by the following:

+', .')

[a, a]·' [e, d]

=

[a· e, a· d]

=

[e, d],

with [e, dl arbitrary in F. To show that every nonzero clement of F has all inverse under multiplication, suppose that [a, b] is not the zero of (F, +', .'). Then a ~ 0, whence the class [b, a] is a member of F. Accordingly, [a,

bl ., [b, a] =

[a· b, b . a]

Since the product a· b is not zero, [a· b, a . [a, b]-I

=

=

bl

[a· h, a • b].

is the identity clemcnt, so that

[b, a].

We wish to show next that the field (F, +', ./) contains a subsystem isomorphic to (R, this will estnblish the requil1'd imbedding t,heorcm.

+, .);

Theorem 3-27. Thc integral domain (R,

field of quotients (F,

+/, ./).

+, .)

CIlII

be imbedded in its

180

RING THEORY

Proof. Consider the subset F' of F consisting of all elements of the form [a, 1],

where 1 is the multiplicative identity of (R, +, .):

F'- {[a,I]laeR}.

It is readily checked that the triple (F', +', .') is a subring of (F, +', .') and, in actual fact, is an integral domain. Now, let /: R -+ F' be the onto mapping defined by lea) -

[a, 1]

(or each a E R. Since the condition [a, 1] - [b, 1] implies a· 1 - 1· b or a - b, we see that f is a one-to-one function. Moreover, this function preserves addition and multiplication: f(a

+ b) = [a + b, 1] =

f(a· b)

=

Accordingly, (R, +,.)

[a· b, 1]

~

(F',

=

[a, 1]

+' [b, 1] = f(a) +' f(b),

[a, 1]·' [b, 1] == f(a) .' feb).

+', .') under f, and the proof is complete.

Several remarks are in order. First, note that any member [a, b) of F can be written in the form [a, b) = [a, 1] .' [1, b) = [a, 1] .' [b, 1]-1.

Since the systems (R, +, .) and (F', +',.') are isomorphic, one customarily identifies the element [a, 1] E F' with the element a of R. The above equation then becomes [a, b] = a·' b- 1 • The point is this: we may now regard the set F as consisting of all quotients a.' b- 1 , with a and b '" 0 in R. It should also be observed that for any a '" 0, [a, 1] .' [b, a]

=

[a· b, a]

=

[b, 1].

Again writing [a, 1] simply as a, we infer that the equation a·' s == b always has a solution in F, namely s = [b, a) = b·' a-I. A final fact of interest is that the field of quotients (F, +" .') is the smallest field in which thc integral domain (R, +, .) can be imbedded, in the sense that any field in which (R, +, .) is imbeddable contains a subfield isomorphic to (F, +', .') (Problem 14 of this section). The field of quoticnts constructed from the integral domain (Z, +, .) is, of course, the rational number field (Q, +, .). Definition 3-17. A field which does not have any proper subfields is called a prime field.

+, .),

Example 3-27. The field of rational numbers, (Q, is a prime field. To see this, suppose (F, +, .) is a subfield of (Q, +, .) and let a e F be any nonzero element. Since (F, +, .) is a subfield, it must contain the product

3-3

FIICLDS

181

= 1. In tum, n = n· 1-1 E F for any n in Z; in other words, F contains all the integers. It follows then that every rationaillumber n/m = n. m- 1, m ;o! 0, also belongs to F, so that F = Q.

a· a-I

Exampl. 3-28. For every prime p, the field (Zp, +p, .p) of integers modulo p is a prime field. The reasoning here depends on the fact that the additive group (Zp, +p) of (Zp, +p, .p) is a finite group of prime order, and therefore has no nontrivial subgroups. We r.ondudc this section by showing tha.t the rational number field and the fields (Zp, +1" .p) are, in a certain sense, the only prime fields. The proof relies heavily on earlier results. Theorem 3-28. Any prime field (F,

+, .)

+, .),

is isomorphic either to (Q, the field of rational numbers, or to one of the fields (Zp, +p, .p), where p is a prime number.

Proof. Let 1 be the identity clement of (F, f: Z-Fby

+, .)

and define the mapping

fen) = n1 "

for any integer n E Z. Then I is a homomorphism from (Z, +, .) onto the subring (f(Z), +,.) consisting of integral multiples of 1 (Example 3-22). By Theorem 3-18, we see that (Z/ker (f),

+, .) ~ (f(Z), +, .).

But the triple {ker (/h +,'.) is an ideal of (Z, +, .), a principal ideal ring. Whence, ker (f) = (n) for some nonnegative integer n. The possibility that n = 1 may be ruled out, for otherwise I would be the trivial homomorphism j that can only happen if F = {O}. Note further that if n ;o! 0, then n must in fact be a prime number. Suppose to the contrary that n =, nln2 where 1 < ni < n (i = 1,2). Since n E ker (I), (nIl) . (n21)

=

(nln2)1 = n1

=

0,

yielding the contradiction that the field (F, +, .) has divisors of zero. (This result is not entirely unexpected, because n is the characteristic of (F, and as such must be prime.) The preceding discussion indicates that two possibilities arise: either \ 1) (f(Z), +, .) ~ (Z/(p), +, .) == (Zp, +p, .p) for some prime p, or 2) (f(Z), +, .) ~ (Z/(O), = (Z, +, .).

+, .)

+, .)

+, .)

. Turning to a closer analysis of these C8.ses, suppose first that (f(Z), ~ (Zp, +p, .p), with p prime. Inasmuch as the ring of integers modulo a prime forms a field, the subring (f(Z), +, .) must itself be a field. But (F, +, .) con~ (Zp, tains no proper subfields. Accordingly, feZ) = F and (F,

+, .)

+", .,,).

182

3-3

JUNll TIfI
+, .)

+, .)

It is now 1\ JI\m'ly routinc mattcr to Khow that the fields (F, and (Q, are isomorphi(' UlI(I(~r the mapping fI{nfm) = (nl) . (ml}-l; we leave this as an exercise.

Since each field has a prime subfield, we get the following subsidiary result. Corollary. Every field contains a subfield which is isomorphic either field (Q, +,.) or to one of the fields (Zp, +p, 'p), p a prime.

to the

PROBLEMS

I. If, + and· denote ordinary addition and multiplication, for which of thp. following 1.1. field? sets It' is (F, 1.1.) F = {a - bv'21 a, b e Z} b) F = {a+ b01 a, beQ} c) F = {a b0 c~ I a, b, ceQ} 2. In a fi!'ld (F, Hhow t.hat 1IH1 !''Iuation (,2 = a impli('!! eithl'r a - 0 or a 3. Define two binary operations * and on the set Z of integers by letting

+, .)

+ + +, .),

0

a* b = a

+b -

a,bei.

1,

Prove that the triple (Z, *, 0) forms 1.1. field. 4. In the field (e, of eomplex numbers, define the mapping rule I(a, b) = (a, -b); in other words,

+, .)

/(a+

bi) =

I: e -. e by

the

a - bi.

Determine wheth!'r the function 1 is 1.1. homomorphillm. 5. A division ring is 1.1. ring with identity in which every nonzero element has a multiis a division ring, prove that (cent H, plicative inv!'rRe. As.'Juming (H, forms a field. 6. Let (H, be an integral domain and consider the set Zl of all integral multiples of the identity: Zl = {nl In e Z}.

+, .)

+, .)

+, .)

Vl'rify that (ZI, +,.) is a fi!'ld if and only if (H, +,.) has positive characteristics. 7. a) Prove that ev!'ry field is a principal ideal rinl/:. b) Consid!'r the set of numbers H = {a by'21 a, b e Z}. Show that the ring (H, ill not a field by exhibiting a nontrivial ideal of (H,

+, .)

+

+, .).

CERTAIN SPECIAL IDEALS

183

S. Derive the following rellults: a) The identity element of a subfield is the same as that of the field. h) If Wi. it! an indexed collection of 8uhfiC'ld8 of the field (F. then (n Pi. -\ •. ) is ulso a subfil'ld of W. 9. Lei. I he a homomorphism from the field W. + .. ) into itllelf and K be the set of ell'ments left fixed by I: K = {a E P I Ita) = a}.

+.. )

+..).

+.. ).

Given K # to}, verify that the triple tK. + .. ) ill a Muhfield of (l 1. Then the principal ideal «n). +,.) is maximal if and only if n is a prime number.

Theorem 3-29. Let (Z,

Prool. First, suppose «n), +,.) is a maximal ideal of (Z,

n is not prime, then n = nln2, where 1 < nl ideals «n1), and «n2), are such that

+, .)

+, .)

~

n2

1 i this also is untenable, since n is prime, not composite. We therefore conclude that «n), +,.) is a maximal ideal.

+.. ),

An additional illustration may be of some interest: Let R denote the collection of all functions I: R' -+ R'. For two such functions I and g, addition and multiplication are defined by the formulas ' U

+ g)(x) = I(x) + g(x),

(f. g)(x)

= I(x)g(x),

xER'.

CERTAIN SPECIAL IDEALS

185

Then (R, +, .) is a commutative ring with identity (see Example 3-5). Consider the set M of functions in R which vanish at 0:

M = {fE R 1/(0) =

OJ.

Evidently, the triple (M, +,.) forms an ideal of (R, +, .); we observe that it is actually a maximal ideal. For, if I e M and i is the identity map on R', one may easily check that (i 2 + 12)(x) ."" 0 for each x E R'. Hence, the sum i 2 + 12 is an invertible element of R. This implies that (M, f)

;2

(i, f)

=

R,

and therefore (M, f) = R for every Ie M [here (i, f) designates the elements of the ideal generated by i and I; that is, (i, f) = {r. i + 8 • I 1 r, 8 E R} J. Our immediate goal is to obtain a general result assuring the existence of suitably many maximal ideals. As will be seen presently, the crucial step in the proof depends upon the maximal element principle, or what is commonly termed Zorn's Lemma. It would take us somewhat far afield to do much more than merely formulate thislcmma as an axiom; the reader who wishes to pursue the topic further is directed to the comprehensive discussion in [38]. For ease of reference, let us recall that by a chain is meant a collection e of sets such that A, BEe implies either As;;; B or B S;;; A. We now state Zorn's Lemma in a form best suited to our present needs. Zorn '. Lemma. Let (1 be a nonempty family of subsets of some fixed set with the property that for each chain e in (1, the union ue also belongs to (1. Then (1 contains a set which is maximal in the sense that it is not properly contained in any member of (1.

The significant point, needless to say, is that this lemma asserts the existence of a certain maximal element without actually giving a constructive process for findjng it. Theorem ~30. (Krull-Zorn).

In a commutative ring with identity, each proper ideal is contained in a maximal ideal.

+, .), a commutative ring with

Proof. Let (I, +,.) be any proper ideal of (R, identity. Ddino n family of subsets of R by taking (1

=

{J 1 IS;;; J; (J,

+,.)

is a proper ideal of (R,

+, .)}.

This family is obviously nonempty for I itself belongs to (1. Now, consider an arbitrary chain {l;} in (1. Our aim, of course, is to establish that ul; is again a member of (1. Notice first that UI; ."" R, since 1 e I, for any i. Next, let the elements a, bE UIi and r E R. Then there exist indices i ~nd j for which a Eli, bE Ii' As the collection {Ii} forms a chain, either Ii S;;; Ii or else Ii S;;; Ii; say, for definiteness, Ii S;;; Ii' But (Ij, +,.) is an ideal,

ISH

:J-4

It! Nil 'I'll EOilY

the difference a - bE I j S;;; uh For the same reason, r· a E I j • This shows the triple (uh +, .) to be a proper ideal of the ring (R, +, .). Finally, IS;;; uI i , hellce the union UI. E Cl. Thus, on the basis of Zorn's Lemma, the family Cl contains a maximal element lIf. It follows directly from the definition of Cl that the triple (M, +, .) iR a proper ideal of the ring (R, +,.) with IS;;; M. We assert (lIf, +,.) is in faet It maximal ideal. To see this, suppose (.I, is any ideal of (R, for whi('h 111 C.l S;;; R. Sinl'c AI is a mnximal I'lenH'nt of the fnmily a, the set .I cnnllot belong to a. Accordingly, the ideal (.I, +, .) must be improper, whieh implies .I = R. We therefore cOllclude (ill, +,.) is a maximal ideal of (R, +, .), ('ompleting tIl n + m (i j = k > n + m implies either i > n or j > m), whence /. g is once again a polynomial. A routine computation, which we omit, establishes that this multiplication is associative as well as distributive with respect to addition; in fact, these properties follow almost at once from the corresponding properties in (R, +, .). All this may be summarized in a theorem.

k

+

Theorem 3-40.

0/ polynomial8

The triple (poly R, +, .) forms a ring, known as the ring

+, .)

over R. Furthermore, the ring (poly R, is commutative with identity if and only if (R, +, .) is a commutative ring with identity.

At the risk of belaboring the obvious, let us point out that while the operations in (poly R, +, .) are defined in terms of those of (R, +,.), these are entirely new operationR; we have used the same symbols only to avoid unnecessary notational complications.

198

3-5

If 8 rl!prcHcmtH Ule HPj, of all constant TJolyrtomials,' that is, the set

=

8

{(a, 0, 0, ... ) I a Ell},

+, .)

then it is not particularly difficult to show that (S, constitutes a subring of (poly R, +, .) whieh is isomorphic to (R, +, .); one need simply consider the mapping that scnds the eonstant. polynomial (a, 0, 0, ... ) to a. In this sense, (poly R, +,.) eont.ains (R, +,.) as a suhring. As 1\ result of the uforementioned imbedding, we shall no longer distinguish between an clement a E R and the constant polynomial (a, 0, 0, ... ) of poly R. By developing IlOme additional notation, it is possible to express polynomials in Hw fl1ll1ilillr form usually PIII'Oulltl'rpci ill f'lf'mnntary COllrHHS in algebra. As a firHl. HI.I·JI ill this din·d.iulI, wC' Id a.c del:liI(IIILf.c~ til(! polynomial (0, a, 0, 0, ... ).

That is, ax is the spcdfie nu'mber of poly R which has the element a for. its 8('cond term nnel 0 for nil other terms. :\[ore genemlly, the symbol ax", n ~ 1, will dC'lIot.e tlw polynomial (0, ... , 0, a, 0, ... ), wll(~re

t.he ('Iem('nt a iH t.Iw (n

have

+ I )Ht. tc'rm in this polynomial; for

ax 2

=

example, we

(0, 0, a, 0, ... )

and

ax:!

=

(0,0,0, a, 0, ... ).

Utilizing these conventions, ench polynomial

mny be uniquely represcnted in the form

f = (an, 0, 0, ... ) -I- (0, alt 0, 0, ... ) -I- ... = an

-I- alx -1 a2x2

+ (0, ... ,0, a", 0, ... )

+ ... + a"x",

with, of course, ao repl:wing (an, 0, 0, ... ). Thus, there is no harm in regarding t.he polynomial rillj!; (poly n,l, .) IlS consisting of all formal expressions

where the ('1('nI('IIts all, aI, ... , an (t.he cl)('.jJidcnts of f) lic in R. WC' Hhould pllll'liaHizp Ihal It('c'ordilll( to our dl'finil.ion, x is Himply 1\ new symbol, or indctcrmina7l', totally 1I11Tf'lated to the rillg (R, +, .), and in no Hellse r('pn'spllts all ('\('nwlIl of H. To indimte the indeterminnnt x, it is com mOil prac~tic'e to write Rf.rJ for \.Iw 1'11'1. poly R, ulld f(.c) for any member of the Rl1mC'. From 1I0W 011, W(' Hlmlllllalw exc·lllsivl' use of this IIotation.

3-5

POLYNOMIAL RINGS

199

Remark. If t.IlC rillg (U, +, .) has a llIult,ip\i('utive identity 1, many authors will identify the polynmnilll Ix with the ill(letOtIC I(x), g(x) E Rlx] with deg/(x) = n and 0, then f(x) can be expressed in C[x) as a product of n (not necessarily distinct) linear factors. Throughout the remainder of the section, we shall focus our attention on polynomials with coefficients from a field (F, +, .). In this important and interesting case, the associated polynomial ring (F[x], +, .) is an integral domain (but not a field I); in fact, (F[x), +, .) is a principal ideal domain. Theore~ 3-46. If (F,

+, .) is a field, then the ring (F[x), +, .) is a principal .

ideal domain. Proof. By Theorem 3-41, it is already known that (F[x1, +,.) is an integral domain. To see that any ideal (I, of (F[x], is principal, we need .

+, .)

+, .)

oiuy mimic the argument of Theorem 2-24: If 1= {O}, the result is trivially true, since I = (0). Otherwise, there is some nonzero polynomial p(x) of lowest degree in I. For each polynomial f(x) E 1, we may use the Division Algorithm to writef(x) = q(x) . p(x) rex), where either rex) = 0 or deg rex) < deg p(x). Now, rex) = f(x) - q(x) . p(x) lies in 1; if the degree of rex) were less than that of p(x), a contradiction to the choice of p(x) would result. With this possibility ruled out, rex) = 0 and f(x) = q(x). p(x) E (p(x»; hence, I ~ (p(x». But the opposite inclusion clearly holds, so that I = (p(x».

+

Corollary. A nontrivial ideal of (F[x), +, .) is maximal if and only if it is a prime ideal.

3-5

POLYNOMIAL RINGS

205

+, .)

By custom, the prime elements of the principal ideal domain (F[:c], are referred to as irreducible polynomial8. Translating Definition 3-20 into the language of the present section, we see that f(x) is an irreducible polynomial if and only if f(x) is of positive degree and in any factorizationf(:c) = g(x) • hex), with g(x), hex) E F[x], either g(x) or hex) must be a constant polynomial (recall that the invertible elements of F[x] are precisely the nonzero constant polynomials); the constant polynomials are neither reducible nor irreducible. Let us record this observation as a formal definition. Definition 3-23. A nonconstant polynomial f(x) E F[x] is said to be irreducible in F[xl if and only if f(x) cannot be expressed as the product of two polynomials of positive degree. Otherwise, f(x) is reducible in F[x].

+, .)

The dependence of this definition upon the domain (F[x], is essential. It may well happen that a given polynomial is irreducible when viewed as an element of one domain, yet reducible in another. One such example is x 2 1; it is irreducible in (R'[x], but reducible in both (C[x], where x2 1 = (x + i)· (x - i), and (Z2[X], wherex 2 1 = (:c 1)· (x+ 1). Thus, to ask merely whether a polynomial f(x) is reducible is incomplete and meaningless. For the ques~ion to make sense, one must indicate what c0efficients are to be allowed in the factorization. It is often quite difficult to decide whether a particular polynomial is irreducis a finite field, say one of the fields ible relative to a specific field. If (F, of integers modulo a prime, we may actually examine all of the possible roots. To cite an illustration, x 3 x 1 is irreducible in Z2[X); in this case, the only possible roots for a polynomial are 0 and 1, but 0 +2 0 +2 1 pi! 0, 1 +2 1 +2 1 = 1 ;oil O.

+, .),

+

+, .),

+

+, .), +

+

+, .)

+ +

+

Example 3-35. Any linear polynomialf(x) = ax b, a;oll 0, is irreducible in F[x]. Indeed, since the degree of a product of two nonzero polynomials is the sum of the degrees of the factors, it follows that a representation

ax

+b=

with 0 < deg g(x) < 1,0 < deg hex) polynomial has degree at least 2.

g(x) . hex),

1 and that the thoorem holds for all polynomials of degree less than n. If f(x) is irreducible in F[x], WI' are through Ilfh'r factoring out. its IC'uding (·oeffi(·ient. Otherwise, .r(x) ill rt'.hll'ihl(~ and it. ill pORllihl(l t.o writ(~ f(x) = !I(x), hex) with y(x), hex) E F[x], 0 < d('g y(x) < 11, 0 < d('g hex) < n. Therefore, by the induction' hypothesis, (/(.r:)

=

flJ •

OI(X)' 02(X)' .. O,(;!:),

hex) = a2 . h, (.1') • h2(.r) ... h.(x),

whC're a" a2 are nonzero elemen!.s of F (in fact, the leading (:oefficients of o(x) and h(x») al\ll (I.(:c), ".(x) art' irr('(hwihle monic polynomial!:!. Thus, /(./,)

(11'1'

(I:.d . f/.(./,)· .. Yr(./') . h,(.c) .. · 1I.{.r)

i!:! a faelorizllt ion satisfyinl!; the condit.ions of the t.heorC'lll. Tht' IIniqll('n('~s of th(' flldors still rt'lll11inl< t.o he shown. W(, again proceed hy indudion on 1/ = th·gf(x}. Th(' r('slllt is lrivial for n = 1: if

POLYNOMIAL RINGS

207

then a, = a2 and a, . b, = a2' b 2, whellC~e b, = b2 by the cancellation law for multiplieation. Now, let n > 1 and f(;l') = c· p,(.r) . P2(X) ... Pr(x) = d· q,(x) . Q2(X) •.• q,(x)

be any two faetorizations of f(x) into irreducible monic polynomials. Surely = d, for euch is the leading coefficient of f(x). Furthermore, Pl(X) divides the produc!t q,(x) . q2(X) ..• q.(x) and must thercfof (,' givell hy

+, ,)

4>(a

-1- bA)

= a

+ bi,

H(·forl· pro('('pding, two ('0111111('111.,.; 1t"P in OI'd("', l'it"Ht, EXlllllpl(~ :J-:17 HhoW8 that there ('xist finite fields other than the fields (ZI" -h, ',,) of integers modulo It prime 71, TIll' flu,t that. til(' ficlt! of this example' IUlS 2 3 = 8 e'1e'ments is typical of til(: 1!;(,II('ral ,.;iflmtion: givl'll II prime Ilullllwr 7' Ilnd It positive int.eger n, HIP"p is PXI""t Iy Oil(' (lip t.o isolllorphi";Jll) fipld with [I" pic:nwnt.s. Jlldcml, if j(.r) i,.; /lily irrPlhwiIJl1' polYllolllial of dl'gl"l'l' It ill /'plJt t.llt· tjuot.ilmt lidll (/'p[x]!(J(:r», -1-, ,) eonHiHts of all polynomials bo bRA b,,_lA,,-I, where bk E Zp; Hillee there arc only 7' choice'S for eaeh coefficient bk , we thus obtain Il finit(: fi(~ld with p" members, HI'I'0I1d, til(' I·OIlHt.l"lldioll of TllI'orclII a-4!l yiddi'l lUI extelllSioll of a field (P, 1 , .) ill whil,h a giv('n polynomial f(x) E PIx] splitH off olle lillt'llr fll(,tor. By repeated application cU this procedure, we can build up an extension (F', of (F, in whichf(x), thought of as a member of F'[x], factors into a product of linear factors; that is, the field (F', is large enough to contain all the roots of f(x) (tedmieally srll'aking, the polynomial splits completely in F'[x]). We phrase this result ill the form of an existence theorem.

+

+ .. , +

+, .)

+, .)

+, .)

Theorem 3-50. If f(x) E "'[x] iH a polynomial of po::;itive degree, then there exi,.;t.s UII cxt,ellHion lield W', +,.) of (p, +,.) ill which f(x) factors completl'ly int.o lil1l'llr polynomial::;,

Proof. The proof iH by iuduction on n = deg/(x). If n = 1, f(x) is already linC'ur and (F, +, .) is itself th(l required field. Therefore, assume that n > 1 and t.hat t.iw tlworem is t.ruC' for all polynomials of degree less than n. Now, the polynomial f(x) IlIlIst have some irreducible factor (I(x). By Theorem 3-49, there is lUI ('xtension liPId (/"1, +,.) of (F, +,.) in which y(x), and hence f(x), has a root al; specifieally, 1'\ = P[x]/(g(x»). Thus, f(x) can be written ill F l[X] as f(x) = (x - al) , fl (x). Since degfl (x) = n - 1, there exists, by our induetioll hypothesiH, an (·xt.ension field (F', -+ , .) of (F 1, in which fl (x) = ao(x - a~)(.r. - aa) ... (x - a,,), with ak E F', ao ¢ O. From this, we see thatf(x) call he completely factored into linear factors in F'[x].

+, ,)

Corollary. Lt't, f(x) E F[x] with deg f(x) = n tension of (F, in which f(x) has n roots.

+, .)

>

O. Then there exists an ex-

Example 3-39. To illustratl: this situation, let us consider the polynomial f(x.) = X4 - !ix 2 t- (j = (x:.! - 2) . (x 2 - 3) over the field (Q, of rational lIumbers. F"om EXatllpll':3- ali, x 2 - 2 (mid similarly x 2 - 3) is already known

+, .)

212

3-5

BING THEORY

to be irreducible in Q[~]. So we first extend (Q,

Fl

+, .) to the field (F 1, +, .), where

2)'== {a + bX I a, b eQ; X' - 2 == O},

= Q[z]/(x' -

and obtain the factorization I(z)

= =

(z (x -

+ X) • (x' - 3) V2) . (x + V2) . (x 2 X) • (x

(Aa X' .... 2, one customarily identifies X with

3).

V2.)

However,l(x) does not factor completely, since the polynomial Xl - 3 is irreducible in F I[X], For, suppose to the contrary that X2 - 3 has a root in F 11 say e + dV2, with e, d E Q. Substituting, we find that (el

+U 2 -

3)

+ 2cdV2 == 0,

hence el

+U'

- 3

== 0,

cd== O.

This latter equation implies that either e = 0 or d = 0; but neither c nor d can be zero, since otherwise we would have d ' = 3/2 or cl == 3, which is impossible. Thus X2 - 3 remains irreducible in Fl[X]. In order to factor I(x) into linear factors, it is necessary to extend the c0efficient field further. We therefore construct the extension (F" where

+, .),

The elements of F I may alternatively be expressed in the form (a

+ bV2) + (c + dV2)V3 == a + bV2 + cv'3 + dV6.

It follows at once that the original polynomial factors in FI[x] as I(x)

=

+

+

(x - X) • (x X) • (x - 1') . (x 1') == (x - 0) . (x + 0) . (x - v'!) . (x + v'!).

Observe that the four roots all lie in F ••

+, .)

+, .)

Let I(x) E F[x). An extension (F', of (F, is said to be a 'Plitting field for I(x) over F provided I (x) can be factored completely into linear factors in F'[x], but not 80 factored over any proper subfield of (F', containing (F, Loosely speaking, a splitting field is the smallest extension field in which a given polynomial decomposes as a product of linear factors. To obtain a splitting field for any polynomiall(x) E F[x] of positive degree, we need only return to Theorem 3-50 and consider the family (Fi, of all subfields of

+, .)

+, .).

+, .)

3-5

POLYNOMIAL RINGS

+, .)

213

(F', in which f(x) factors completely (the theorem guarantees the existence of such extensions); then (nF" serves as a splitting field for /(x) over F. Having thus indicated the existence of splitting fields, it is natural to inquire about their uniqueness. As a final topic for this section, we shall prove that any two splitting ficldll of the I!&mc (nonconstant) polynomial are isomorphic; this being 110, one ill justified ill using the definite article and speaking of the splitting field of a given polynomial. Before presenting the main theorem, two preparatory results of a somewhat technical nature are needed. As previously noted, if r is a fixed element of a field (K, +,.) and F!',;;; K, we write F(r) for the set of finite sums:

+, .)

F(r)

=

{ao

+ al . r + ... + ~. r"l aAo E F, n

~

I}.

Lemma. Let f(x) be an irreducible polynomial in F[x] and r be a root of /(x) in some extension field (K, of (F, Then (F(r), ~

+,.)

+, .).

+,.)

(F[x]/(/(x», +,.) under an isomorphism whereby the element r corresponds (j(x». to the coset x

+

Proo/. The mapping II; F[x] - K defined by setting IIf(x) = /(r) is easily checked to be a homomorphism of (F[x], +, .) onto the ring (F(r), +, .). It follows at once that f(x) E ker (II) = {g(x) E F[x] I g(r) = O},

whence (/(x» !; ker (II). One observation is quite pertinent: the possibility that ker (II) = F(x].does not arise, since the identity element of (F[x], +,.) is not mapped onto zero. Asf(x) is assumed to be irreducible in F[x], «(f(x» , +, .) is a maximal ideal of (F[x], +, .), so that (f(x» = ker (II); in other words, (f(x» simply consists of all polynomials in F[x] having the element r as a root. Thus, by the fundamental homomorphism theorem for rings (Theorem 3-18), there exists an isomorphism 8 of (F[x]/(I(x», +,.) onto (F(r), +,.) such that II = 8 nat(/(:(ao) + t/>(al)Y + ... + t/>(a..)y" be the corresponding polynomial in F'[y]. Then1'(Y) is irreducible in F'[y]. Furthermore, if r is a root of f(x) in some extension field of (F, +.. ) and r'is a root of f'(y) in some extension field of (F', +', .'), then t/> can be extended to an isomorphism 4> of (F(r), onto (F'(r'), with 4>(r) = r'.

+, .)

+'.. ')

214

UlNH '1'ln;OTty

Proof. Let us first extend 4> t.o a mapping f, between the polynomial rings (F[x), and (F'[y], by taking

+, .)

f,g(x) =

+', .') f,(b o -I- blx + ... + II"x") =

4>(b o)

+ I/>(bdy + ... + 4>(b,,)y"

for every polynomial g(x) = bo + blx + ... + b"x" E F[x). Using the fact that I/> ill an isomorphism, it is an easy matter to verify t.hat f, is an isomorphism of (F[x), +,.) onto (F'[y], .')j we leave the render to supply t.he necessary dctni\s. Not i('(! that. for !lny polynomilll Vex) ill FIx}, an clement 0 E F is a root. of g(x) if and only if 4>(0) iN a root of 4iu(x). Indeed, if (I(x) = bo + blx + ... + b..x", we hnvc

+',

(f,(J(;rJ)(r/>(fl»

(ll) 1 .,. 14>(I'")'4>(tt)" "-c

1/>(11 0 -I b l • a -I ••• 1- b,.· a")

~: 4>(u(a»,

from which the nr-;s('rtion follows. In pnrti('ular, the givl'n polynomial f(x) is irrcdueiblc in FIx} if and only if f'(y) = f,f(x) is irredueible in F'[y]. Now, by !.Ill' fort'going lemma, we know that there exist isomorphisms

a: F(I')

-+

F[.rl/(J(x»

and

(3: F'(r') -+ F'[y]/ (J'(y».

Furtlwrmof(', it. is not. diffi('ult to 8how t,hat, thl'1' 0, where the prime number 11 is the character· istic of the field. Granting this, Theorem 3-53 may be interpreted as assertilll! that any two finite fields having the same number of elements are isomorphi(', PROBLEMS

+, .)

1. For an arbitrary ring (R, with identity, prove that a) the polynomial 12:" E cent R[x], b) if (1, is an ideal of (R, then (l(x], is an ideal of the polynomial

+, .)

+, .), +', .').

+, .),

+, .)

ring (R(x], c) if (R, +,.) and (R/, +/,.') are isomorphic rings, then (Rlx], +,.) is isomorph ii' to (R/(x],

+, .)

2. Given (R, is an integral domain, show that a} the only invertible elements of R(x] are the constant polynomials determined by the invertible elements of R, b) the characteristic of (R(x],+,·) is equal to the characteristic of (R,+, .). 3. Prove that no monic polynomial can be a zero divisor in (R[x],

+, .).

3-5

POLYNOMIAL RINGS

217

4. Show that the relation - defined by taking f(x) - g(x) if and only if deg fex) = deg g(x) ill an equivalence relation in the set of nonzero polynomials of R(x]. 5. a) Let P be the set of all polynomials in Z(x] with constant term 0: P = {a\x

+ a2x2 + ... + a"x" I a. E Zj ft ~

+, .)

I}.

+, .).

Establillh that the triple (P, is a prime ideal of (Z[x1, b) Show the principal ideal «12:),+,') is a prime ideal of the polynomial ring (Z[x], but not a maximal ideal. 6. If (Il, +,.) iM a commutative ring with identity, prove that the polynomial I ax is invertible in R[x] if and only if the element a is nilpotent. 7. Let (R, +,.) be a commutative ring with identity and the element r E R be fixed. a) If R(r) denotes the set

+, .),

+

R(r) = U! follows: if f(x)

= ao + alx +

... + anx" E

HlrJ.

Rlx),

then !Jf(x) = al

+ 2a2X + ... + na"x,,-I.

For any f(x) , g(x) E Rlx) and any a E R, verify that a) /J(f(x) h) fI(af(x»

I g(x»

+ !Jg(x), of(x) . g(x) + f(x) . og(x).

= flf(r)

= a!Jf(x) ,

c) fI(f(x)' g(x» = nonz('fO t('rms of f(x).)

[/lint: Induct on the number of

15. ),e·t W,I, -) he a commutativ(' ring with identity, and let r E R be a root of the nonz('ro polynomial f(x) E R[x). We call r 0. multiple root of f(x) if f(x)

=

(x -

r)" . g(x),

n>

I,

wh('re g(x) E Rlx) is a polynomial such that g(r) ~ O. Prove that an element r E R i.. 0. multiple root of f(x) if and only if r iii 0. root of both f(x) and of(x). 16. Determine a) 0.1\ irreducibl{' Jlolynomials of degree 2 in Za[x], and b) all irreducible polynomials of degree 3 in Z2[X).

+, .)

17. Let (f', b(' a fie>l.l anll/(x) E F[x) be a Jlolynomial of degree 2 or 3. Establish that f(x) is irre>ducihle in FIx) if and only if f(x) ha.'i no root in F. Give an l'xample which >!howH that this fI'sult IU'C't\ not he true if t\e·g fIx) ~ 4. IS. Prov!' tluLt if f(x) is all irn'dllcible polynomial in Z[xl. then f(x) i>! also irredllcibl(' whf'n rf'gardf'd 1\." all 1'1C'lIll'lIt of Qlx). 19. Df'riw Iht· following analog of Euclid's LI'llIma: L.. t (R, +,.) be a commutative ring with iell'nlhy and f(x) be all irrC'ducihlt' polynomial ill Rlx). If f(x) divides the prulhH't g}(x) . U2(:r) .•• g,,(x), then !(x) divi.l(·s gk(X) for some k, 1 ::; k ::; n.

+, .)

20. In regan 1 to ProblC'm 7, show that the ring (R(r), is a field if and only if ker (I/» = (f(x» where I(x) is an irreducible polynomial in R[x).

219 21. a) Prove thl' Eisl'nstein Criterion for irreducibilil,y: If f(x) = ao -I- alx

+ ... -I- anxn E Z[X)

and p is a prime number such that pi ak (k = 0, \, ... , n - I). pt an. p2 tao. then fix) is irrl'ducible in Z[x). [!lint: Proof is by contradiction.) h) 1l1 splittin~ fi('ld (i = I, 2), establish that

(Zp(rll,+,') ~ (X,,(r2).-I-•. ). 30. Derive Fermat's Little Theorem: If p is a prime number and a ¢ 0 (mod pl. then a,,-l == 1 (mod pl.

3-6 BOOLEAN RINGS AND BOOLEAN ALGEBRAS

The thcory of Boolean algebras is IlII algehmie eOlillt.erpart to thc logical theory of the I'l\lelllllll of propm;itiolU;. It.s origius lit' in the work of the English mat.lwnmti('ian Gcorge Boole (ISI!i-1Sfl4), who first attcmpted 1.0 give a systematic t.reatnH'nt of lop;i(~ by abstract methods. Siu('(' su(~h n strueture ma.y be viewed as t,hc postulational ahHt.met ion of the rules gov!'rning the algebra of sets. it providcs a suitable topie with whi('h to eonelude our discussion of twoopcrat.ional systemll. Indecd, as will be evidenced shortly, Boolean algebras may be subsumcd undcr thc gcncral theory of rings.

220

3-6

RING THEORY

We begin by studying the properties of a special class of rings, which we shall designate as Boolean rings. '. Definition 3-24. A Boolean ring (R, +, .) is a ring with identity every element of which is idempotent; that is, a 2 = a for every a E R.

It should be pointed out that the existence of an identity is frequently omitted in the definition of a Boolean ring. (One can show that if the number of elements in a Boolean ring is finite, then an identity element always exists.) The definition given here, however, will be more convenient for the applications we have in mind. Let U8 pause to give several examples of the concept just introduced. Example 3-40. The ring of .integers modulo 2, (Z",

ring, since 0·" 0

= 0 and 1 ." 1 =

+:h .,,),

forms a Boolean

1.

Example 3-41. The ring (P(X),.1, n) of subsets of a nonempty set X is easily verified to be a Boolean ring. In this case, we have A n A = A for every subset A ~X. Example 3-42. For It. 1~1!H obvious example of a Boolean ring, let R consist of all (unctioll8 from lUI arbitrary noncmpty set X to Z" with the operations defined pointwise; specifically, if I and g are in R, then

(J

+ g)(x) =

(f· g)(x)

=

I(x)

+" g(x),

f(x) ., g(x),

(x EX).

Under these operations, the triple (R, +,.) is a commutative ring with identity; the proof is straightforward and will not be given in detail. To establish the idempotency condition, we proceed as follows: If the function fER is such that f(x) = 0, then (J2)(X) = I(x) '"/(x) = 0. 2 0 = O. While if lex)

=

1, then (J")(x) =

In any event, (J')(x)

= f(x)

lex) ·,/(x)

= 1·,1 = 1.

for every x in X, whence'"

= f.

Our ultimate purpose is to prove the celebrated theorem of Stone which asserts that each Boolean ring may be represented by a ring of sets. Looking forward to this result, we first develop a number of the fundamental properties of Boolean rings necessary to the proof. While the conclusions obtained are somewhat restrictive (Boolean rings have an almost embarrassingly rich structure), they bring together much of the material developed earlier. 3-54. Every Boolean ring (R, +, .) is a commutative ring of characteristic 2.

Theorem

3-6

221

BOOLEAN RINGS AND BOOLEAN ALGEBRAS

Prooi If a and h are arbitrary elements of H, then

+ h)2 = a2 + a· h + h· a + h2 = a + a· h + b· a + b, and hence a . b + b . a = O. In particular, setting a = h, we obtain 2a = a + a = a 2 + a 2 = 0 a

+h =

(a

+, .)

for every a E H. This shows that (H, is of characteristic 2. But then by adding a . b to both sides of the equation a . b + b • a = 0, we obtain

We proved earlier that, in any commutative ring with identity, maximal ideals are automatically prime ideals. For Boolean rings, the converse is also true.

+, .) be a Boolean ring. A proper ideal (1, +, .) +, .) is prime if and only if it is a maximal ideal. Proof. It is sufficient t.6 show that if the ideal (I, +, .) is primo, then (1, +, .) is also maximal. To see this, suppose (J, +, .) is an ideal of (H, +, .) with the property 1 e J s; H; what we must prove is that J == H. If G is any element of J not in 1, then a· (1 - a) = 0 E 1. Using the fact that (1, +, .) is a prime Th.....m 3-55. Let (H,

of (H,

ideal with a

e 1, we conclude

1- aE IeJ.

As both the elements a and 1 -

(J

1= a

lie in J, it follows that

+ (1 -

a) E J.

The ideal (J, +,.) thus contains the identity, and consequently J == H. A natural undertaking is to determine which Boolean rings are also fields. We may dispose of this question rather easily: up to isomorphism, the only Boolean field is the ring of integers modulo 2. Theorem

3-56. A Boolean ring (H, +, .) is a field if and only if (H, +, .)

~

(Z2' +2, '2)'

Proof. Let (H, then have

+, .) be a

Boolean field. For any nonzero element a

E

H, we

This argument shows that the only nonzero element of H is the identity; in other WOrdR, R = {O, I}. But any two-element field is isomorphic to (ZI' +1, '1)' The opposite direction of the theorem is fairly obvious.

222

3-6

lUNG TIIF..olty

+, .)

Corollary. A proper ideal (I, of the Boolean ring (R, ideal if nnd only if (RI I, +, .) ~ (Z2' +2, '2).

+, .) +I

Proof. Fil'8t, note t.hut the quotient ring (HI I, since (a + 1)2 = a 2 + I = a

+, .) is a maximal

is itself a Boolean ring,

for each clement a in R. By Theorem 3-32, (I, +,.) is a maximal ideal if and only if (HI I, is a (Boolean) field. An appeal to the above theorem now completes the proof.

+, .)

The next theorem is a major olle and re1luires a preliminary lemma of some difficulty. Lemma. Let (H, I,') he a Boolean ring.

there exiHtH a homomorphiHm f from (H, such thatf(a) = 1.

Proof. Let (I, that is, The set I

+, .)

For cnch nonzero clement a E H, +, .) onto the field (Z2, +2, . 2)

be the principal ideal generated by the clement 1 + a, 1= {r·(l+a)lrER}.

R, sinee the identity is not a member of I. Indeed, if 1 E I, then

~

1

=

r· (1

+ a)

for some choiee of r in H; this means

+ a)2 = (r· (1 + a») . (1 + a) = 1· it folloWH that a = 0, contrary to hypothesis.

1

=

r· (I

(1

+ a),

from which BCCIUlHe (/, ~I •. ) is a proppr iil(,lLl, Theorem a-ao aHHurefl the ()xistence of 1\ mnximal ideal (111, +,.) of (fl. + .. ) sud, that I ~ M. In light of the result just proved, the associated quotient ring (RI 1II, will be isomorphic to (Z2, +2. '2) vin KOIlW homomorphism g. W(~ may thl'refom define I~ funet.ion f: U --+ Z2 hy t.aking f = !I natM, where nlLtM ill Himply the lIatural mapping of U Ollto Illlll. The situation ill conveniently depicted by the following diagram of maps:

+, .)

0

'~/Z',~

._

RIM

The rcmllinil('r of till' proof amounts t.o showing that t.he function I, flO i1efillC'i1, hUH t hI' 11I'/)pI'rlil's ILH.'!I'rted in till! statement of the tlwormn. Plainly, f is hoth 1\1\ unto mup and a homomorphism I)('illg the composition of two such

223

:~-6

functions. Rillce 1 + a E 1 ~ Af, the ('oset 1 1

+a+

Af

=

Af, so that

+2 f(a) = f(1) +2 f(a) = f(l + a) = g(1

+ a + 111) =

But, 1 -h f(a) = 0 if and only if f(a)

=

(!(III) = O.

1, which finisheR the proof.

An immediate consequence of this lemma is the following corollary. Corollary. Every Boolean ring (R, +,.) iR a semisimple ring; that is to say, rad R = {O}.

I'roof. In order to n.rrivl! at a cOIlt.rll.didion, w(~ aliHulnc t.hltt a E rad Il with a "., O. Thrn t.here exi8t.8 a homomorphism f from (R, -1--, .) onto (Z2, +2, '2) for whi +, .)

Proof. To begin the attack, let H denote the collection of all homomorphisms from (R, +, .) ont.o the field (Z2, +2, '2)' Next define It flllWtioll h: R -. P(H) hy assigning to caeb element a E R t.hose members of II which assume the value 1 at a; in other wordll, h(a) = {fEHlf(a) = I}. While the notation is perfectly clear, let us emphasize thll.t h is a set-valued function in the sense that its functional values are certain subsets of H. By means of this function, we shall establish the isomorphism mentioned in the theorem. Let us now give some details: FOl' any fEll, the productf(a) '2 feb) = 1 if and only if both f(a) = 1 and feb) = 1. This being so, we eon('lude h(a· b) =

UE

H I f(a . b) = I}

U E II I f(a)

UE

'2

II I f(a) =

feb)

I:

11 n U E

II I feb) =

Il

= h(a)

n h(lI) ,

224

RINO THEORY

showing that the function h preserves multiplication. The verification that h(a

+ b) =

h(a) 11 h(b)

is equally straightforward, depending chiefly upon the observation that the sum/(a) +2/(b) = 1 if and only if one of /(a) or /(b) is 1, while the other is 0; the reader may easily fill in the steps for himself. These remarks demonstrate the fact that h is a homomorphism from (R, +,.) into the ring of sets (P(H), 11, n). All that is needed to complete the proof is to show that h is a one-to-one function or, what amounts to the same thing, that ker (h) = {O}. But this follows immediately from the preceding lemma which asserts that the set h(a) is empty if and only jf a = 0, whence,

=

ker (h)

{a E R I h(a)

= 0} =

{O}.

The pieces all fall into place, and we see that the ring (R, +, .) is isomorphic to a subring of (P(H), 11, n). The definition of a Boolean algebra which we are about to present is based on a structure introduced by E. V. Huntington in 1904. A variety of other sets of postulates could be chosen that would define the algebra equally well; indeed, few areas of mathematics have received more diverse postulational treatment. Aesthetically speaking, it seems desirable to build our theory on is few assum~ tions as possible. The axiom system quoted below was therefore selected with the intention that no axiom could be derived from the others. Definition 3-25. A Boolean algebra is a mathematical system (B, V, 1\) consisting of a Donempty set B and two binary operations V and 1\ defined on B such that

(PI) Each of the operations V and 1\ is commutative; that is,

aVb=bVa,

al\b=bl\a

for all a, b E B.

(P 2 ) Each operation is distributive over the other; that is,

a V (b 1\ c)

=

(a V b) 1\ (a V c),

a 1\ (b V c)

=

(a " b) V (a 1\ c)

for all a, b, c

e B.

(P.) There exist distinct identity elements 0 and 1 relative to the operations V and 1\, respectively; that is,

aVO=a,

al\l=a

for all a E B.

(P,) For each element a E B, there exists an element a' E B, called the complement of a, such that

a V a' = 1,

a 1\ a' = O.


225

Example 3-43. An obvious example of a Boolean algebra, but nonethe1eas an

important one, is the system (P(X), u, n), where X is a nonempty set. It is apparent that we should take 0 = 0, 1 = X, and whenever A!;;;; X, A' = X-A. More generally, if B is any family of subsets of X, including 0, which is closed under unions and complements, then (B, U, n) will be a Boolean algebra, in fact, a Boolean subalgebra of (P(X), n, U). Example 3-44. For an illustration quite removed from the algebra of sets, consider the set B of positive integral divisors of 10, that is, B == {I, 2, 5, IO}. Given elements a, bE B, we define a V b to be the least common multiple (lcm) of a and b, a A b to be the greatest common divisor (gcd) of a and b:

a V b

=

lcm (a, b),

a A b

= gcd (a, b).

The tables for these operations are given below.

1

V

1 2 5 10

5 10

A

1

1 2 5 10 2. 2 10 10 5 10 5 10

1 2 5

10 10 10 10

10

1 1 1 1

2

2

5 10

1

1 1 1 2 5 5 5 10

2

1 2

The formal verification that the triple (B, V, A) constitutes a Boolean algebra is left as an exercise. (We should caution the reader that in this example the integer 1 plays the role of the identity element for the operation V, while the integer 10 serves as the identity element for the operation A.) A quick inspection of the foregoing tables will reveal the various complements to be l'

=

10,

2'

=

5,

5'

=

2,

10'

=

1.

We ca.ll attention to the fact that a' is simply the quotient when 10 is divided bya. The first thing one notices on inspection of the axiom system for a Boolean algebra is the perfect symmetry or duality between the properties of the two operations V and A. That is to say, if V and A are interchanged in the axioms and if, at the same time, 0 and 1 are also interchanged, then the properties are merely permuted amongst themselves. This principle of duality permits us to state all theoreIns in dual pairs (unless, of course, a statement happens to be its own dual) and guarantees that the proof of one of the pair of statements will be sufficient for the establishment of both; the proof of the dual theorem is obtained by maki~g the appropriate interchange of symbols in the proof of the original theorem. We now proceed to deduce from the postulated properties of the operations in a Boolean algebra a series of further properties, including, for instance, the

226

3-fi

lUXG THEORY

aSHOciative laws for V and 1\. Dual statements are placed side by side; ill view of the principle of duality, only one statement from each dual pair need be proved. In order to condense the demonstrations, we shall arrange the steps 80 far as possible one under another, citing to the right those I>ropositions used in passing a('ross succcssive equality signs. Theorem 3-58. III !LilY Boolc!L1I ILlgchra (8, V, 1\), the following prolK!rtil'H

hold: 1) The c1cmellts 0 and 1 are unique. 2) For each clement a E B, a V a

= a,

al\a=a.

=

a 1\ 0

:J) For l'll.{'h dement a E B, a V 1 4) For eaeh pair of elements

1,

= o.

a, b E B,

a V (a 1\ b)

=

a,

a 1\ (a V b) = a.

Proof. To establish (1), we need only appeal to Theorem 2-1. The proof of (2) is indicated below:

a=aVO = a V (a 1\ a / )

= =

(a V a) 1\ (a Va')

(a V a) 1\ 1

=aVa

(by P a) (by 1>4) (by P2) (by P 4 ) (by Pal.

We obtain (3) as foIlows: l=aVa'

= a V (a' 1\ 1) = (a Va') 1\ (a V 1) = 1 1\ (a V 1) =aVI

(by (by (by (by (by

P4) )la) P2) P4)

Pal.

The proof of (4) requires the use of (3): a=al\l = a 1\ (b V 1) = (a 1\ b) V (a 1\ 1)

= (a 1\ b) V a = a V (a 1\ b)

(by Pa) (by 3) (by P 2 ) (by P a) (by PI)'

3-6


227

We did not include the 8II8Ocint.ive laws for V and 1\ among our axioms for a Boolean algebra, as is frequently done, since they arc logical consequences of the properties listed. This is demonstrated in our next theorem. Theorem 3-59. In any Boolean algebra (B, V, 1\), each of the operations V IUlIl 1\ is I\IIHOdat.iv(·; thl~t. is, for pVl'ry t.ripll· of I'll'tnc'nts a, b, c e B, a V (b V c)

=

(a V b) V c,

a 1\ (b 1\ c)

=

(a 1\ b) 1\ c.

Proof. First, set x = a V (b V c) and y = (a V b) V c. We wish, of course,

to prove that x = y. Note that

la

a 1\ x = (a 1\ a) V

1\ (b V c»)

= a V fa 1\ (b V c») =a

(by 1'2) [by Theorem 3-58(2») [by Th(.'Orem 3-58(4)1

and also a 1\ Y = [a 1\ (a V b)] V (a 1\ c)

a V (a 1\ c)

=

=a

(by P 2 ) [by Theorem 3-58(4») [by Theorem 3-58(4)].

Therefore a 1\ x = a 1\ y. Now, a' 1\ x

=

(a' 1\ a) V [a' 1\ (b V c)]

=

0 V [a' 1\ (b V c)]

= a'

1\ (b V c)

(by P 2) (by Ph P,,) (by P a)

and also

a' 1\ y

=

[a' 1\ (a V b») V (a' 1\ c) = (a' 1\ a) V (a' 1\ b») V (a' 1\ c) = [0 V (a' 1\ b)] V (a' 1\ c)

= Therefore a' 1\ x

(a' 1\ b) V (a' 1\ c)

= a'

(by (by (by (by

P 2) P2) Ph p.) Pa).

1\ y. From these observations, we conelude that

(a 1\ x) V (a' 1\ x) (a 1\ a') V x

= =

(a 1\ y) V (a' 1\ y) (a 1\ a') V y

IVx=IVy

x=y

(by Ph P 2) (by P,,) (by Pa),

proving the associative law for the operation V; that 1\ is also associative follows by a dual argument.

228

3-6

BING TBJlORY

As yet nothing has been said about the properties of complementation. In the next group of results, we sIuill prove, among other things, that each elem6nt has & unique complement; thus, ' may be viewed as & function from B into itself (as & matter of fact, onto the set B).

Theorem 3-60. In any a E B Boolean algebra (B, V, 1\), the following hold: 1) Each element a E B has a unique complement. 2) For each element a E B, a" = a. 3) 0' = 1 and l' = o. 4) For all a, b E B, (a V b)'

=

(a 1\ b)'

a' 1\ b',

=

a' Vb'.

Proof. For (1), asaume there are two elements x and y of B such that

= 0,

a V x = 1,

a 1\ x

a V 11 = 1,

al\y=O.

We then have (by Pa) (by hypothesis) (by P,) (by p.) (by hypothesis) (by Pa).

x=xl\1

= x 1\ (a V y) = (x 1\ a) V (x = (a 1\ x) V (x = 0 V (x 1\ y)

1\ y) 1\ y)

=xl\y

In the 8&11le manner, 11 = 11 1\ x = x 1\ y, 80 that x = 11. Accordingly, any two elements asaociated with a as specified by axiom p. must be equal; in other words, the complement a' is uniquely determined by a. From the definition of the compleinent of a, a V a' = 1 and a 1\ a' = O. Hence, by Pit and a' 1\ a = O. a' V a= 1 From this, we conclude that the element a is the complement of a': (I," = (a')'

=

a.

Using the uniqueneaa of the complement and the relatioDB

o VI = it follows that 0'

=

1. But then 0

1,

01\1=0,

= 0" =

1'.

3-6


229

Finally, we prove the first statement of (4). Note that, since complements are unique, it is enough to establish (a V b) V (a' A b') = 1,

=0

(a V b) A (a' A b')

Now, (a V b) V (a' A b')

=

[(a V b) Va'] A [(a V b) Vb']

= [(a Va') V b) A [a V (b Vb')] = (1 V b) A (a V 1) =lAI

=

1

(by P 2 ) [by Ph Theorem 3-59] (by p.) [by Theorem 3-58(3)] (by Pa).

Furthermore, (a V b) A (a' A b')

=

(a' A b') A (a V b)

= [(a' A- b') A a] A [(a' A b') A b)

=

b'l

A [a' A (b' A b)] = (0 A b') A (a' A 0) =OAO [(a' A a) A

=0

(by PI) (by P 2) [by Ph Theorem 3-59] (by Pl. p.) [by Theorem 3-58(3») (by Pa).

These considerations imply that (a V b)' = a' A b'. The last three theorems do not, in any sense, exhaust the theory of Boolean algebras; we could continue to deduce a large number of results. This brief outline should serve, however, to give some idea of the structure of the system, as well 8.8 to I>repare the way for our final objective-that of showing that every Boolean algebra can be transformed into a Boolean ring and vice versa. As a first step in this direction, we indicate how, by the introduction of suitable definitions of addition and multiplication, a Boolean algebra may be converted into a Boolean ring. The argument relies heavily on the results of the previous three theorems. Theorem 3-61. Every Boolean algebra (B, V, A) becomes a Boolean ring

(B,

+, .) on defining addition and mUltiplication by the formulas a

+b=

(a A b') V (a' A b),

a· b = a A b,

(a, b E B).

Proof. It is obvious that addition as defined above is commutative, for

a

+b=

(a A b') V (a' A b) = (b' A a) V (b A a')

=

(b A a') V (b' A /I) = b

+ a.

230

3-6

RING THJo:OllY

Furthl!rmore, a

+0 =

(a A 0') V (a' A 0)

= (a A 1) V (a' A 0) =aVO=a and a

+a =

«(,I

A a') V (a' A a)

=

0 V 0

=

O.

This shows that the element 0 acts as the identity for the system (B, +), while each element is its own (additive) inverse. To establish the associativity of addition, let us first perform a preparatory calculation: (a

+

b)'

=

[(a A b') V (a' A b)]'

=

(a A b')' A (a' A

bY

= (a' V b) A (a Vb') = [(a' V b) A a) V [(a' = (a A b) V (a' A h').

V h) A b']

Utilizing this relation, we then have (a

+ b) + e =

[(a

+ b)

A e') V [(a

= [«a A b') = (a A b' A

(a'

V

+ b)'

A b») A

A e)

e]

V

_

[«a

A b) V

(a'

A

b'»

A

e)

e) V (a' A b A e) V (a A b A e) V (a' A b' A e).

Note, however, that the foregoing expression is symmetric in a, band e; that is to say, it is unaltered by permuting these clements. Thus, after interchanging a and c, we obtain (a

+ b) + c =

(c

= a

+ b) + a + (b + e).

From all this, one may infer that the pair (B, +) is a commutative group. Turning next to a discussion of multiplication, it is evident that both the commutative and associative laws hold, while 1 serves as the mUltiplicative identity. Because a2

=

a· a

=

a A a

=

a,

each element a in B is also idempotent. Finally, to establish that the triple (B, +,.) is a Boolean ring, it remains only to verify that multiplication is distributive over addition. We may dispose of this rather easily, since a· (b

+ e) = =

a A [(b A c') V (b' A c)]

(a A hAc') V (a A b' A c),

a-Ii

JlOOLEAN IUNUR ANI> 1I01ll••:AN ALUEDUAS

231

wher('IUI

a· b

+ a· e =

(a /\ b) + (a /\ e) /\ b) /\ (a /\ e)'] V [(a /\ b)' /\ (a /\ e)] = [(a /\ b) /\ (a' V c')l V [(a' Vb') /\ (a /\ e)] = (a /\ b /\ a') V (a /\ b /\ e') V (a' /\ a /\ e) V (b' /\ a /\ e) = (a /\ b /\ e') V (a /\ b' /\ e).

= [(a

The proof of the theorem is therefore complete. We now reverse this process; in other words, we start with a Boolean ring and transform it into a Boolean algebra by suitably defining the opera.tions V and /\. Theorem 3-62. Every Boolean ring (B, (B, V, /\) on defining

a V b

= a + b + a· b,

+,.) becomes a Boolean algebra

a /\ b

= a' b,

(a, b E B).

Proof. That V and /\ are both commutative followli immediately from the commutativity of the operations in (B, A simple calculation will show that /\ is distributive over V :

+, .).

a /\ (b V e)

= = =

a . (b + e + b . e) a· b + a . e + (a· b) . (a· e) (a· b) V (a· e)

=

(a /\ b) V (a /\ e).

The verification of the other distributive law relics on the fact tha.t, since 2x = 0 for every clement of the ring (R, +, .), it is ullnecessary to distinguish betwccn addition and subtraction:

(a V b) /\ (a V e)

(a + b + a ' b) . (a + e + a . e) = a+a·b+a·b+a·e+b·e+a·b·c+a·c +a·b·c+a·b·c = a+b·c+a·b·c

=

=a

V (b· c)

=

a V (b /\ e).

If 0 and 1 are the additive and mUltiplicative identities of (B,

a V 0= a+O+a'O= a,

+, .), then

a/\l=a·l=a

for every a E B. Filllllly, we hllve

+ a) = a + (1 + a) + a . (1 + a) = 1 + 4a = (1 + a) = a' (1 + a) = a + a 2 = 2a = 0,

a V (1 a /\

1,

232

3-6

RING THEORY

which implies 1

+ a is the complement of a in (B, V, 1\), that is a'=l+a.

Theee computations show that the postulates of Definition 3-25 are all 88.tisfied and, consequently, the triple (B, V, 1\) is a Boolean algebra. Taken together, Theorems 3-61 and 3-62 indicate that the theory of Boolean algebras is equivalent to the theory of Boolean rings. What is to be considered remarkable is the identification of a. notion a.rising out of questions of logic and set theory with a. system amenable to the powerful techniques of modem algebra.

PROBLEMS

+, .), every triple of elements a, h, c E R satisfies (a + h) • (h + c) • (c + a) = O.

1. Prove that in a Boolean ring (R, the identity

2. If a Boolean ring (R,+,·) has at least three elements, show that every nonzero element except the identity is a divisor of zero. [Hint: For a, hER, consider the product (a h) • a • h.] 3. Prove that any ring (R,+,·) in which each element is idempotent can be imbedded in a Boolean ring. [Hint: Let R' - R X Z2 and mimic the argument of Theorem 3-16.] 4. a) Let (R, +,.) be a commutative ring with identity and S the set of idempotents of R. For a, h E S, define the operation * by taking

+

a * h .. a

+h-

2(a . h).

Prove that the triple (8, *, .) forms a Boolean ring, known &8 the idempoUnt. Boolean rifl(J of (R, b) In particular, obtain the idempotent Boolean ring of (Z12,+12, '12)' 5. Suppose (1,+,·) is an ideal of the Boolean ring (R,+, .). Show that (1,+,.) is a proper prime (maximal) ideal if and only if for each element a in R, either a E I or 1 - a E I, but not both. 6. Given (R,+,·) is a Boolean ring. For an element a E R, define the set I(a) by

+, .).

I(a)

>=

{I 1(1,+,,) is a maximal ideal of (R,+, .); a E I}.

Verify that the sets I(a) have the following properties: a) I (a) ,. ., whenever a ,. O. b) l(a h) - l(a) ~ l(h). c) l(a' h) - I(a) n I(h). d) l(a) - I(b) if and only if a-h. [Hint: a E 1 if and only if 1 - a E 1.]

+


233

7. In reference to Problem 6. prove that if M

=

{I 1(1.+.') is a maximal ideal of (H.+.

·n.

+..)

then the ring (H. is isomorphic to a subring of (P(M), A, n). [Him: Consider the mapping I: H -+ P(M) given by I(a) - I(a).] 8. Establish that there is no Boolean ring having exactly three elements. 9. In any Boolean ring (R. an order relation ~ may be defined by taking a ~ b if and only if a . b = a. If the elements a. b. c. d E H, establish the following order-properties : a) a ~ a, 0 ~ a:S 1 for every a E H. b) a ~ band b ~ c imply a ~ c. c) a:S b and b ~ a imply a-b. d) a :S c and b ~ d imply a . b ~ c· d. e) b· c = 0 implies a • c = 0 if and only if a :S b. 10. a) Let (H. be a Boolean ring and I be a nonempty subset of H. Show that (1,+,') is an ideal of (H,+,') if and only if i) a, bel imply a bel, ii) a E I and r E H with r :S a imply r . a E I. b) If the set la is defin.ed by I • ... {r E HI r ~ a}. verify that the triple (1., forms an ideal of (H,

+, .).

+..)

+

+, .)

+, .).

11. Suppose that (8,+.,) is a subring of a Boolean ring (R,+, .). Prove that any homomorphism I from (8, + •. ) onto the field (Z2, +2, '2) can be extended to all of (H. + .. ). [Him: The ideal (ker (f), +,.) is contained in maximalideal (M,+, .), where (H/M,+.·) ~ (Z2, +2, '2).] 12. For elements a. _~. and__c_~t a Boolean algebra (B. V, A), prove that a) (a A b) V (b A c) V (c A a) - (a V b) A (b V c) A (c Va), b) a A c = b A c and a A c' == b A c' imply a - b, c) a = b if and only if (a A b') V (a' A b) - 0, d) a ... 0 if and only if (a A b') V (a' A b) - b, e) a A b = a if and only if a V b = b. 13. Let the set B consist of the positive integral divisors of 30, that is, B

=

{I, 2, 3, 5, 6.10,15, 3O}.

If V and A are defined by a Vb,. lcm (a, b),

a A b - gcd (a, b),

show that the triple (B, V, A) is a Boolean algebra. 14. Given that X is an infinite set. Let B be the family of all subsets A !;;; X such that either A or X - A is finite, plus ~ and X. Determine whether the triple (B, U, n) forms a Boolean algebra. 15. Prove that if (B, V, A) is a Boolean algebra having identity element 1 for the operation A, then every Boolean subalgebra must contain 1. Contrast this with the case of rings.

234

RING THEORY

16. By means of Theorem 3-61, convert the Boolean algebra (B, V, A), as defined below, into a Boolean ring. d

V

abc

a b c d

abc d b b b b c b c b d b b d

A

abc

d

a a a a b abc cae c dad a

a d a d

B == {a, b, c, d}

17. Suppose a Boolean algebra (B, V, A) is made into a Boolean ring (B, +,.) via Th('orem 3-61, and then (B, +,.) is convertt"d baek to a Boolean algebra (R, V \,1\ \) via Tlworem 3-62. Verify that (R, V, 1\) ... (B, VI, 1\ .). 18. Suppose (B, V, 1\) is a Boolean ring and 0 ;o!i I\; B. The triple (I, V, 1\) is said to be a (Boolean) ideal of (B, V, 1\) if and only if i) a, bEl implies a V bEl, ii) aE I and bE B imply a 1\ bE I. a) Prove that every Boolean algebra (B, V, 1\) has two trivial ideals, namely, ({O}, V, I\) and (B, V, 1\). b) If (Ii, V, 1\) is a collection of ideals of (B, V, 1\), show that (nIi, V, 1\) is also an ideal. e) Prove that if (I, V, 1\) ill an ideal of (B, V, 1\) and 1 e 1, then 1 - B. 19. Let (P(X), U, n) be the Boolean algebra of subsets of a nonemptlt set X and Xo be any element of X. Prove that a) if I is the family of all subsets A \; X such that Xo ~ A, then the triple (1, U, n) is an ideal of (P(X), U, n), b) if X is infinite and I is the family of all finite subsets A \; X, then the triple (1, U, n) is an ideal of (P(X), U, n). 20. Let (B, V, 1\) and (B\, V \,1\ \) he two Boolean algebras and I a mapping from B into B.. Then I is Haid to be a Boolean homomorphiam from (B, V, 1\) into (Bl' VI, 1\.) provided I(a V b) = I(a) V. I(b), I(a 1\ b) = I(a) 1\. I(b) , I(a') "" I(a)'

for all elements a, bE B. (The formation of complements may be regarded as a unitary operation.) Show that such a function has the following properties: a) 1(0) ... 0.,/(1) = 1\. b) (f(B), V., 1\.) is a suhalgebra of (B., VI, 1\ .). c) If a :S b is taken to mClI.n a 1\ b .. a, then I(a) :S I(b) whenever a d) The triple (ker(!), V, 1\) is an ideal of (B, V, 1\), where

k('r (f)

=

{a E B I/(a)

:S

b.

= Od.

e) If (It, VI, 1\ 1) is an ideal of (Bl, VI, 1\ .), then (f-l(It), V J 1\) is an ideal of (B, V, 1\).

CHAPTER 4

VECTOR SPACES

4-1 THE ALGEBRA OF MATRICES

The theory of matrices has long occupied a strategic position in various branches of mathematics, physics, and enginccring. Only in reccnt years has its importance in the social and biological sciences as well hecome apparent. Thc subject t.oday has become an indispensable tool ill :mcll new field!! as game theory, linear programming, and statistical decision theory. Part of the reason for the widening applicability of matrix theory is no douht the role it plays in the analysis of di!!eretc'obHcrvations and the case with whieh matric operations may bl~ progmmmed for modern highapeed computen;. We do not intend to give a complete nnd systcmatie aeeount of the problems of matrix theory and its diverse applications. Rather, the operations and the basie properties of veetors and matrices arc approached from an algebraic point of view with the aim of illustrating some of the conccpts of the previous chapters. One result of sueh a study will be t.hc formulat.ion of a mathematical sy!!tcm, somewhat more complicated than tho!IC st.udied earlier, known as a v~tor spa(!e. The basic definition whieh atl1rts U!! ofT ill that of II. vector, the fundamental object in our study. Deftnitlon 4-1. An n-componcllt or n-dimensional vector over a field (F, is un ordered n-tuple (aI, a2, ..• ,an) of clements ak E F.

+,.)

The clements ak E F ure called the comTllInents of the vector and we say n is its dimension. Clearly, the set of all one-dimensional veetors over (F, can be identified with F itself. To have to write out the whole vector is somewhat awkward; hereafter, we will condense our notation and designate the vector with (~OmpollentR ak by (a,,). It is hardly necessary to point out t.hat two n-component V(,(,t.()rtI (ak) and (Ilk) Iln~ (~qlllll, ill whidl ellS(. w/' writ./' (nk) ~ (lJd, if Illld only if t.hcir eorrcspolldillg eompOltelltll arc equal: (ak) = (I)k) if and only if ak = bk for k = 1, 2, ... , n.

+, .)

Definition 4-2. n) The sum of two n-component vectors (ak) and (b k) , denoted by (at) + (b,,), is the veetor obtained by adding their corresponding components. Thus, (at) + (b,,) = (a" + bk). 235

236

.4-1

VECTOR SPACES

b) The product of a vector (a,,) and an element r of F, denoted by rea,,), is the vector obtained by multiplying each component of (a,,) by r. Thus, real)

=

(r· a,,).

Here, we conform with the standard practice of using the plus sign in two different contexts, for vectors and for their components. It should be perfectly clear in any given situation whether we are adding vectors or elements of F. Note, incidentally, that the difference of two vectors may be expressed in terms of the operations already given:

For a simple illustration of these ideas, consider vectors over (R', +, .); in this case, we have (1,2,3) - 2(1,0, -1)

=

(1,2,3)

+ (-2,0,2) =

(-1,2,5).

Definition 4-3. Any vector whose components are all zero is called a

zero

vector and is represented by the symbol O. Let V,,(F) denote the set of all n-component vectors over an arbitrary field +, .). Inasmuch as vector addition enjoys the basic additive properties of its components, the following theorem concerning the algebraic nature of the pair (V,,(F), +) is obvious. (F,

Theorem 4-1. The system (V,,(F), +) is a commutative group, having the zero vector of dimension n as its identity element and (-a,,) as the inverse of a vector (a,,) E V,,(F).

The operation of multiplication of vectors by elements of F, as defined above, has the following properties: if r, • E F and (a,,), (b,,) are vectors in V,,(F), then

+

+

1) (r .)(a,,) - r(a.) .(aAl), 2) r[(a,,) (b,,)] = rea,,) r(b.), 3) r[.(a,,)] = (r· 8) (a.,) ; l(a.) = (a,,); O(a.) =

+

+

o.

Verification of these facts is not particularly difficult and is left to the reader. Vectors may also be combined under a rule of composition known as inner product multiplication. Definition 4-4. The inner product of two vectors (a,,), (b,,) E V,,(F), denoted

by (a,,)

0

(b.), is defined to be (aAl)

0

(b,,)

=

.

L al . bl;. 1-1

According to this definition, inner product multiplication • may be regarded as a function from V,,(F) X V,,(F) onto F; that is, the inner product of two vectors is an element of F. Note also that the product of two nonzero vectors

4-1

THE ALGEBRA OF MATRICES

237

may be zero, as in Va(R'), where (1, 2, -3)

0

(3, 6, 5)

=

1 . 3 + 2· 6 + (-3) . 5

=

O.

However, one should not jump to hasty conclusions concerning divisors of zero, for on the right-hand side we have the real number zero and not a 3-eomponent zero vector. While failing to be even a binary operation, inner product multiplication nonetheless enjoys some interesting properties, several of which are listed in the next theorem. Theorem 4-2. If rEF and (at), (bt ), (Ct) are vectors in V .. (F), then 1) (a.,) 0 (bAo) = (bAo) 0 (at), 2) 0 (a,,) = 0 = (a,,) • 0, 3) r[(a.,) (b t )] = (r· at) • (b t ) = (a,,) • (r· bAo), 4) (a.,). [(b,,) (Ct)] = (aAo) • (bAo) (al;) • (CI;). 0

0

+

+

Proof. Let us illustrate the type of argument involved by establishing the last statement; the remainjng parts of the theorem are left as an exercise. We proceed as follows: (a.,)

0

[(b.,)

+ (CI;)] =

(al;) • (bl;

+ CI;) =

..

L aAo' [bl; + Ct]

"-1 .. .. = E a., . b., + E a., . c.,

"-1

"-1

= (a.,) • (b.,) + (a.,) • (c.,). Definition 4-5. By an m X n matriz (plural: matrice.) over the field (F, we mean a function from {I, 2, ... ,m} X {I, 2, ... I n} into F.

+,.)

In the case of matrices, one customarily arranges the functional values block fashion in a table made up of m rows and n columns. S~cifically, if the value of the matrix at the ordered pair (i,j) is denoted by aij, where 1 ~ i ~ m, 1 ~ j ~ n, then the matrix is indicated by the following rectangular array: (

all

a12

~21

~22

a... 1

a...2

... a l .. ) ...

~2. . .

... a.....

Abusing terminology, we shall hereafter refer to the above display of mn elements of F as the matrix itself (in the strict sense, this display is simply a representation of the matrix). Further, we will call 4;, the ijth entry or element of the matrix, and we shall speak of the integers m and n, the number of rows and columns, as its (limenBiona.

238

4-1

VECTOR SPACES

Note that clements are located ill the matrix by the use of double subscripts. the first subscript indicating ttHl row. and the second subscript the column in which the element is found. For instance. the element a23 is in the second row and third column. To avoid cumbersome notations. it is convenient to abbreviate a matrix aH (ai;)",X", to be read "the matrix of dimension m X n whose elements are the ai/s." When the numbers of rows and columns are clearly understood. we may instead simply write (aij). If m = n. the mntrix is snid to be square of order n. Definition 4-6. Two m X n matrices (ai;) and (b i ;) are equal, for which we write (ai;) = (b i ;) , if and only if their corresponding elements are equal; Lhllt iH, flij • l)ij fill' ILII i ILIIII j.

Since the rowl'! of all m X n mlttrix may be regarded as clements of the vector space V,,(F) , it iH 1I0t HlIl'priHing that the operations defined in V,,(F) have natural generalizations to matrix operations. Definition 4-1. a) The Hum of two m X n matriceH (ai;) and (bij). denoted

by (aii) + (b i ,), is the matrix obtained by addillg their corresponding elements. Thus, (aij) + (b i;) = (ai; + bi ;). b) The product of the matrix (ai;) and the element rEF. denoted by r(ai;), is the matrix obtaitwd by mult.iplying each element of (aij) by r. Thus, r(ai;)

=

(r· ail)'

Observe that by its definition, addition is a binary operation on the set of all matrices of a given size; that is, the sum of two n X m matrices is again an n X m matrix. Example 4-1. Taking (ll',

+, .)

as the base field, let

~) ,

-6

A=e

3

0

4

B= (:

-1

:).

Then

2A

+ = (:

-12

B

0

-8 = (:

-1

:)+(: :).

4 -1

~)

A matrix each of whose elements is zero is called a zero matrix and is denoted by O. Accordingly, a zero matrix need not be square. For the zero matrix whose dimensions are those of (ai;), we have (ai;) (ai;)

+0

=

-I- (-I)(aij)

= 0 + (a;j). = 0 = (-I)(a;j) + (aij). (ai;)

4-1


239

+, .)

Let. us denote the set of all m X n matrices over the field (F, by the symbol M ",n(F). The following theorem establisheH the algebraic properties of (M",,,(F), +) under matrix addition.

Theorem 4-3. The system (JIf ",,,(F), +) is a commutative group, with the m X n zero matrix as the identity element and (-aij) as the inverse of a matrix (aij) E M ",,,(F). Proof. Definition 4-7 indicates that each property of matrix addition is derived from the correHpondillg ndditive property in the field (F, +, .). For instanoe, to establish the commutative law, let (aij), (bij) EM",,,(F). Then

The rest of t.he proof proceeds along similar lines and is left to the reader. Although multiplieation of matrices by a field element is not a binary operation on M ".,,(F) lunless, of couI"s(~, m = n = 1), this operation has several int.eresting features. Specifieally, if (aij), (b ii ) E JIf ".n(£] -1·2+0-3 -1·0+0·2 -1 . 1 + 0· (-1)

1

-[-: :-~l· 6

4-2

3X3

0·1 +2· (-1)

)1

4-1


241

We next dispose of one natural question that arises here, namely, the question of commutativity of matrix mUltiplication. First of all, given an m X n matrix (ao/), the matrix products (a'/)' (bu) and (b'/)' (O(/) are both defined if and only if (b ii ) is an n X m matrix. When the latter condition holds and it is a.t least possible to form these two products, (aii) • (bi/) and (bii) • (4;i) will be of different dimensions unless m = n. Even if this is the case, where it is meaningful to ask whether (ail) . (bi i ) a.nd (b ii ) . (aii) are equal, matrix mUltiplication will not as a rule be commutative. One need only consider the computation

Due to the asymmetric way in which two matrices combine in a product, such an outeome is not totally unsuspected. It is quite possible, of course, that a particular pair of matrices may commute. I·'or the zero .matrices of appropriate dimensions, (aii)' 0 = 0 and O· (ail) = O. In particular, if both (aii) and 0 are square matrices of the same order, then (aii) . 0 = O· (aii) = O. If (a;j) is any square matrix, then that part of the matrix consisting of the the elements ali is called the (main) diagonal of the matrix. DefInition 4-9. The identity matrix oj order n, designated by I .. , or simply I when there is no chance of confusion, is the square n X n matrix having ones down its diagonal and zeros elsewhere.

It is helpful to have some notation for the elements of the identity matrix. Consequently, we will denote the element in the ith row and jth column of I .. by the symbol o;/t where 'il

and thus write Ifl

={ =

I

for i

= j,

o

for i

~

(the Kronecker delta)

i,

(O;i)' To illustrate,

For each positive integer n, the set of all square matrices of order n over the field (P, will be represented by M fI(F), rather than M .... (F). The identity matrix I .. serves as an identity element for the operation of matrix multiplica-

+, .)

242

4-1

VI j and atricll" upper triangulm- if ail = 0 for i ~ j. Let T,,(F) and n(F) denote the sets of all upper triangular and strictly upper triangular matrices of order n, respectively. Prove the following: a) A matrix (ail) e T .. (F) is nonsingular if and only if 4i1 '" 0 for i - I , 2, ... , n. b) The triple (T,,(F),+,.) isasubringofthering (M ..(F),+,·). c) Each matrix (aii) e T:(F) is nilpotent; in particular, (41/)" - o. d) (T:(F),+,') is an ideal of the ring (T ..(F),+, .).

14. The trampoae of a matrix (41/), designated by (ail)', is the matrix whose ijth entry is the jith entry of (a,;), that is, (aii)' .. (alf)' Given matrices (ail), (h#) e M ..(F), verify that a) (41/)" - (41/), b) [r(ai/) .(hi/)]' ., r(ai/)' .(hi/)', r,. e F, c) [(4i/)' (bi /»'

-

+

+

(bi/)'· (4ii)',

d) whenever (a#) is nonsingular,

80

also ill (a#)', with [(44/)'1- 1

-

(4#)-11'.

15. Show that the field (S, +,.) of Example 4-3 is isomorphic to the complex numbers (C, under the mapping

+, .)

j I\_b 4

h) =

(a, b).

a

16. In the ring (M2(C),+, .), let D be the set of all matrices having the form

+, .)

is a division where 4 is the complex conjugate of a. Prove that the triple (D, ring, but not a field. 17. Show that for any element a e F, the following matrices are both idempotents in (Al2(F),+,'): and

4-2

ELEMENTARY PROPERTIES OF VECTOR SPACES

249

18. A matrix (a.;) E M,,(F') is said to be aymmetric if (a.;)' - (atj) and ,kew-.ymmetric if (ai;)' ... -(ail)' Establish the following assertions: a) If (a.;) and (bij) are symmetric matrices, so also is r(ai;) .(bo;). b) The products (ai;) . (ail)' and (ai;)' . (aij) are both symmetrio. c) If (aij) and (b ij) are symmetric, then (aij) . (bij) is a symmetrio matrix if and only if (ail) . (bii) - (bi;) . (ail)' d) The diagonal elements of a skew-llymmetric matrix are all sero. e) Every (square) matrix can be written as the sum of a symmetrio and skewsymmetrio matrix.

+

19. Let the set 0 be comprised of the following matrices:

I -

U

=

(~

:),

8 - (_: :),

e-:), e_:), V

=

y

=

T - (-: _:),

w - (: :),

(0 -1). -1

0

a) Prove that the pair (0, .) forms a group. b) If II = {I, 8, T, U}, show that (H, .) is a normal subgroup of (0,·) and find the cosets of H in O. 4-2 ELEMENTARY PROPERnES OF VECTOR SPACES

-

We saw in the last section that the collection Mn(F) of square matrices of order n over a field (F, +, .), together with the operations of matrix addition and multiplication, constitutes a ring. At the time, our third matrix operation, mUltiplication of a matrix by an element of F, seemed relatively unimportantparticularly, since it failed to be even a binary operation on M n(F). However,' by abstracting the essential features of this operation, we now define a mathematical structure having the set M .. (F) [more generally, the set M ..... (F)] under matrix addition and multiplication by a field element as a model. Basically, this is a matter of combining two different algebraic systems into a single entity known as a vector space. Due to the availability of a number of excellent texts on the subject, there is no need for us to develop the theory of vector spaces in any great detail. Instead, our goal shall be to give a survey of some, but by no means all, of the main ideas. The pace will frequently be brisk and much is left to the reader. Definition 4-10. A vector space (or linear space) over a field is an ordered triple «V, +), (F, -t-,.), .) consisting of 1) a commutative group (V, +) whose elements are called vector., 2) a field (F, whose elements are called acalar.,

+, .)

250

4-2

VECTOH SPACES

3) an operation of scalar multiplication conneeting the group and field which satisfies the properties: 0

a) for each c E F and x E V, there is ctefinpd an element cox E V; that

is, V is closed under left multiplication by scalars;

+

+ C2)

h) (CI c,) (el'

C2)

+

0 x = (CI 0 x) (C2 0 x); x = CI 0 (C2 0 x); y) (c x) (c y);

0

+

d) co (x = e) 1 o:r = x, wh('re I is the fi('ld id('utity element. 0

0

While the addition symhol has been used in two contexts in the above definition, to d!'signate the operation of the group and one of the operations of the fi!'ld, no eOllfllsioll should arise from thiN pmdil'e. It will alwaYII he dear in any giv('n :-;it.uatioll wlwtlwr v('(·t.ors or sealars are being added. When both veetor additioll and sealar multipliention are involved in an expression, we follow our usual understanding in omitting parentheses: multiplication takes precedence over addition. In the Sl'qu!'l, a vector RP:WC over the fi!'ld (F, +,.) will be denoted merely by l"(F), rather than the corred but cumbersome notation (V, +), (F, .),0). The ('onveniellC'e rmmlt.ing from this convention more than outweighs its lack of pl'('('i8ion. For fu.-Owr simplicity, we shall hel'('aftcr drop the and write cx for the procluet r. x. It should be apparent that a veetOl' space is markedly diffel"Cnt from the previous systems we have discussed in that the products of scalar multiplication employ clements from both F and V. Part (3) of the definition relates the possible wayR th('se prorlu('tR ('ombine t.he operation + of (V, +) with + and· of (P, 1-, .). l'\ot.1' all'lo tlmt t.!IC' hYIHIt.llPsil'l Ix = X il-l quite el!RCntinl; without it, every field and c'ommutative jl;roUJl would yield a vector space under the trivial scalar multiplicat.ion ex = 0 for all C E P, x E V. Before discussing the implications of t.he axioms of a veetor space, let us give It sel('ction of ('xamples. The formal verification that each example de-. sl'ribed aetually I'ollstitutes a vee\.or SPIlC'!! iH left as all exercise.

+,

0

0

Example 4-5. Let t.he commutative grollp be (Vn(P), +), where Vn(F) is the IlCt of all n-('ompOII
if and only if ak

= bk for all k e Z+.

252

4-2

VECTOR SPACES

Vector addition and multiplication by a scalar c e F are perfonned componentwise: (ah a2, aa, ... )

+ (b., b1h b., .•.) =

(at

+ bt, a2 + b2, a. + ba, .•• ),

c(a., a2, Ga, ••• ) = (c· a.. c . a2, c· aa, ... ). Using these operations, V.(F) becomes a vector space over (F,

+, .).

Example 4-10. As a final, and not quite 80 simple, example of a vector space, consider a commutative group (V, +) in which every nonzero element baa order p (p a prime); that is to say, px = 0 for all x e V. If In] e Z. and x e V, we take the product [n)x to mean [n]x = x

+ x + .. : + x

(n summands).

With scalar multiplication defined in this way, V(Z.) may be regarded as a vector space over the field (Z., +., .•). A further comment on notation: To avoid a proliferation of symbols, 0 will be UIIed to designate the zero element both of (V, +) and of (F, +, .). The additive inverae of a scalar c e P is denoted by -c, while the inverse of a vector x e V is also represented by its negative, -x. These conventions should lead to no ambiguity if the reader attends closely to the contexHn which the notation is employed. For the sake of brevity, we shall frequently speak of a vector space over a field F when, in aCtual fact, we mean over a field (F, Some immediate consequences of Definition 4-10 are embodied in our first theorem.

+, .).

Theorem 4-7. If V(F) is a vector space and x 1) Oz = 0, 2) cO = 0, 3) -(CI:) = (-c):t = c( -:t).

e

V, c e P, then

Proof. To establish (1), we use the field result 0 + 1 = 1. Then Oz + x = Ox + Ix = (0 + l)x = l:t = :t

= 0 + x. Since (V, +) is a group, the cancellation law yields Ox = o. The proof of the second part of the theorem follows from the group result x. We have

o+ x =

cO

+ CI: =

c(O

+ x) =

CI:

= 0

+ CI:.

Again the cancellation law gives the desired conclusion. Finally, to obtain (3), observe that

o=

Oz = [c + (-c)]x =

CI:

+ (-c)x.

4-2

ELEMENTARY PROPERTIES 01' VECTOR SPACES

This means that (-c)z

=

253

-(ex). Similarly,

0= cO = c[z

+ (-z)] =

which proves c(-z)

=

ex

+ c(-z),

-(ex).

As the reader should expect by now, a formal investigation of vector spaces involves consideration of such notions as subsystems, operation-preserving functions, quotient structures, etc. Following our standard pattern of presentation, we begin the study with the question of subsystems. In the case of vector spacc!!, the subvcctor spaces are customarily referred to as subspacee. Definition 4-11. Let V(F) be a vector space over the field F and W!; V '¢ 0. Then W(F) is a BUbBpacB of V(F) if, under the operations of V(F), W(F) is itself a vector space.

with W

Since W!; V, much of the algebraic structure of W(F) is inherited from V(F). The minimuJD conditions that W(F) must satisfy to be subspace are: 1) (W, +) is a subgroup of (V, +); 2) W is closed under scalar multiplication.

The usual criterion for deciding whether (W, +) is a subgroup of (V, +) is to see if W is closed under differences. The second of the above conditions implies that -z = (-1) z will belong to W whenever z is an element of W. Because z - y = z + (-y), condition (2), together with the closure of W under addition, is sufficient to guarantee that W be closed under differences. This observation allows U8 to recast Definition 4-11 as follows: Definition 4-12. W(F) is a subspace of the vector space V(F) if W is a nonempty subset of V such that 1) z, yEW implies z YEW, 2) z E Wand c E F imply ex E W.

+

Example 4-11. Every vector space V(F) has two trivial subspaces, namely V(F) itself and the zero subspace {O} (F). Subspaces distinct from V(F) are

said to be proper. Example 4-12. Consider the set

W of vectors in Va(F) whose components add

up to zero:

If (alt a2, aa) and (b l , b2, ba) are arbitrary elements of W, then their sum (al + bit a2 + b2, aa + ba) is such that (al

+ bl) + (a2 + b2) + (aa + ba) =

(al

+ a2 + aa) + (bl + b2 + hi)

= 0+0 = O.

254

.

4-2

VECTOR SPACES

This estnhlishcH the eJosure of W under addition. It is equally clear that W is closed under scalar multiplication, hence W(F) is a subspace of V(F). Example 4-13. Let W denote the collection of all elements from the space

M 2(F) of the form

It follows immediately from the definition of the matrix operations in M 2(F) that W(/") is a subspaee, for

k( a b)_( k'a k.b)EW -b a -(k·b) k'a . The two conditions of Definition 4-12 may be combined into a single easily applied criterion. Theorem 4-8. W(F) is a subspace of the vector space V(F) if and only if W £;;; V and ex dy E W whenever x, YEW, e, d E F.

o¢

+

Proof. If W(F) is a suhspace, t.hen hy definition W is nonempty and contains ex + fly for all x, YEW, e, dE F. Conversely, if thi!! condition holds, W must contain Ix +- ty = x + y and ex + Oy = ex for every x, YEW, C E F. Accordingly, W is c1osl'd wit.h rl"slled In t.lm vl'dor HIl'U~!l ClI)(!rUUoIIH.

We next consider operations on the subspaces of a vector space that produce other subspaccs, the mOlolt important of which are sum and intersection. If U and W are nonempt.y suhsets of thc veetor space V(F), their (linear) sum is . defined to be the set IT

+W =

{u

+ W I U Ell, W E

W}.

The following simpl II".

4-3 BASES AND DIMENSION

Perhaps the most far reaching notion in the study of vector spa(!es and the prindpal t1wme of this 1I(,dion ill thnt of a hllllis. As will 1:10011 be evident, this concept is preeisely the tool needed to formalize the more or less intuitive idea of what is meant by the dimension of a space. A convenient starting point is the following definition. Deftnition 4-16. Let V(F) be II. vector space over the field F. A finite set {X., X2, ••• , Xn} of vectors from V is said to be linearly dependent (over F) X2, ••• , X,,; otherif the zero vector is a nontrivial linear combination of wise, the set {XI, X2, ••• , xn} is termed linearly independent.

x.,

Aecording to our dnfinition, {xt. and only if there exist scalars c., C2, that

,xn } ill a linearly uepenuent set if r.n E / 1) is a linear combination of the preceding ones, x" X2, ••• ,X"_l'

x"

Proof. Suppose the vectors x" X2, nontrivial linear relation

••• , Xn

are linearly dependent,

80

there is a

where not all the c's are zero. Let k be the largest integer for which CAl ¢ O. If k = 1, we would have C1Xl = 0, hence Xl = 0, contrary to assumption. Thus k> 1 and

Since C;l exists in F, it follows that x"

= =

Ct l (-I)(C1Xl (-Ct l . CI)XI

+ CaXa + ... + C"-lX"_l) + (-Ct l . Ca).r2 + ... + (-Ct l . C"_l).l'k_l.

The vector x" i8 therefore expressible 8.8 a linear combination of its predC(~el!8Ors, as claimed. The converse is almost obvioml: If the vector Xli: depends linearly on XIt X2, ••• , Xk-It 80 that

26()

4-3

VECTOR SPACES

for suitable scalars b/c E P, then btxt

+ ... + bk-1Xk-1 + (-I)xk + OXk-t + ... + Ox" =

BecaullC tim (~oeffieicrtt of Xk if! nOllzero, {Xt, IICt of vcetors.

X2, ••• ,

O.

x .. } cOrtHtitutell a dependent

Theorem 4-18. If VW) iH a finitely generated vcctor space, say

then V is spanned by a linearly independent subset of these vectors. X2, •.. , In} is already indepencient, nothing n(J(~ds to be pmv('d. OtlwrwillC, Th('orem 4-17 implies that ROmf! ve"f,or Ik is 1\ linear (:omhina(.ion of .rt, .1'2 •••• , Xk_I' By hypothesis, any vector x in V can be written as a linear ('ombination of the n vectors Xt. X2, ••• , Xn; in this combination, Xk may hf! rt'pln('f!d hy n litmllr ('ombillation of Xt. X2, ••• , Xk-t. thereby showing x E [Xt, ... , Xk-t, Ik+t. . . . , In). The net relmlt is thn,t the n - 1 vectol'l! Xt, ... , Xk_l, Xk+t. . . . , In generate the space V(P). Next, examine t.he set {xt. ... , X/o-t, Xk+t. ... , xn} and repeat the process of removing a vector if it can be written as a linear combination of its predecessors. Continuing in this way, we eventually reach a subset

Proof. If th(' S4't [x I.

wiwrI' 1 ~ it < i2 < ... < im ~ n, of the original set of n vectors, still having lilwnr spall V aJl(1 su(,h thnt no Xi. is a linear combination of the preceding vect.ors. That the set· {Xi" Ii 2 , , •• ,Xi.. } is linearly independent follows immediately from Theorem 4-17, Example 4-18. As a simple illustration of some of t1wsc ideas, let us observe

that

l' 3 = [(1, 1,0), (1,0, 1), (0, 1,0), (1, 1, 1)].

The rml.soninJl; h('re is jmltifif!d by t.hc fa('t t.hat. e:wh of t1w unit, vectors et. e2, ea iM It lillcllr ('(lIllhillul.ioll of I.JII·MI~ Vl'dOrM: el

=

(1, 1,0) -

(0,1,0),

e2

=

(0, 1,0),

Ca =

(1, 1, 1) -

(1, 1,0).

Since V 3(P) is spanned by the ullit vedors, we infer that every clement of V 3 I\lUSt. 1)(':\ lilwar ('omhinatioll of (I, 1,0), (1,0,1), (0,1,0), (1,1,1), The lilJ('ar dl'l)('JI(\I'III'(~ or iIlC\I'JlC'tHII'II('c of t.his SC't. of ve(~t.ors is equivalent III Ilw exist.ellc'(! or 1I00I1'xisll'lI('(' of S('lIl:trl; (~I, 1'2, ra, (:4 (1I0t. nil zero) sudt thnt I't(1, 1,0) 11'2(1,0,1) 11:3(0,1,0) 11'.(1,1,1) Chl'ekillg ('OIllPOlll'lItl'l, we note that. the CI

+

C2

-I 1'4 = 0,

Ct

Ck'S

-I- C3 -I-

=

O.

must. sat.isfy the three equations

1'4

=

0,

4-3

267

BASES ANI> DIMENSION

In terms of an arbitrary choice of for instance, C4 = 1 leads to 0(1,1,0) -I- (-1)(1,0, I)

C4,

+

a solution is Cl = 0,

(-1)(0, 1,0)

C2

=

+- 1(1,1,1) =

-C4, C3

=

-C4;

(0,0,0),

whi('h, of eounw., impli!'s that t.he given set of gem'rut,or!! is linearly dependent. To obtain a linearly independent subset of the aforementioned vectors, we need only removc the element (1,1,1). In fact, (1, 1,1) is the only vector that can be written as a linear combination of those preceding it: (1, 1, 1)

Any

memh(~r

= (1,0,1)

+ (0,1,0).

(ai, a2, aa) of V 3 ean obviously be expressed as

wh('nec Va

=

[(1,1,0), (1,0,1), (0, 1,0)1.

The next, somewhat technical, result is the key to all that follows. theorem 4-19. (Steinitz Replacement Theorem). Let W(F) be 8. finitely gellerau~d subsplwe of t.Il(, vc(:tor I'lpnce V(F), W = Ix., :1:2, ••• ,xnl, and let {Y1o Y2, •.. , Ym} be any linearly independent subset of W. Then m of the Xt'S, say X10 X2, ••• , X m, may be replaced by Y .. Y2, .•• ,Ym, so that W = [y., . .. ,Ym, Xm+1o'" ,xnl, in particular, m ~ n.

Proof. Since {X., X2, ••• , x,.} spans W(F), the vector Yl E W can be expressed as 8. linear combination of the Xk'S:

Not all the cocffieients ak = 0, for otherwise Yl = 0, contradicting the linear independenee of t.he I!f1t [lit, 112, .•. ,11m}. R('illdexing, if llN·CR.'!firy, we may ILSSllIII(l thllt al ;t6 O. Nuw solve fUl' the V('I~t(IJ' of I in \.(~rllltl of 11 .. X2, ••• , .1:,,:

This relation permits us to replace a linear combination of X10 X2, ••• , Xn by a linear combination of the vedol'H YI, X2, .•• ,X"' and leads to the conclusion that W = rUI, 3'2, •..• x,,1. Rep(ml.l.he repilwenwn\, PI'OCNl:i with the vllC'l.or!1'l nlHl the Het [U .. .1:2 • ••• , x,,}. B('('IUI!'!e 112 belong!! to the Kubspac(' spanlled by y., X2, ••• , X n , we must have

for suitable scalars b., b2 • ••• , bn E F. The coeffieiellts b2 • ••• , b.. cannot all

268

4-3

VECTOR SPACES

be zero, for this would imply that b l r& 0 and, in tum, that bl 1l1

+ (-1)111 + Oys + ... + 011", = 0,

contrary to the independence hypothesis. Hence, one of the coefficients bl, ... , b" is nonzero; let us, for simplicity, take this to be bl. As before, we can solve for X2 in terms of the vectors YIo Y2, Xa, •.• , x" to obtain

W

=

[1Il> YI, Xa, ... , x,.1.

Continue in this manner: At each stage a y-vector can be introduced and an x-vector deleted so that the new set still spans W(F). If m were larger than n, after n steps all the Xk'S could be removed and the set {1Il> 112, •.• , 11..} would span W(F). Accordingly, the vector 11"+1 E W could be written as a linear combination 11"+1 = CIYI + C2112 + ... + c,.1I" with not all the c" being zero, since 11,,+1 po! O. Once again a contradiction to the independence of 111, Y2, ••. ,11.. would arise. It follows that the 1I'S must be exhausted before the x's (that is, n ~ m) and W = [YIo ••• , Y"., x..+1, ••• , x ..1. Of course, the possibility that n = m is not excluded; in this situation, the set {x..+1, ••• , x,.} is empty and the vectors 111, YI, .•. ,1/" themselves span W(F).

Corollary. If a vector space V(F) is spanned by n vectors, thea any set of n 1 vectors from V is linearly dependent; in particular, any n 1 vectors of the n-tuple space V,,(F) are dependent.

+

+

Proof. A set of n + 1 linearly independent vectors from V would be impossible, since the theorem would imply n ~ n + 1, an obvious contradiction.

The Replacement Theorem has several notable consequences, but these will have to await further developments. DefInition 4-17. A bam for a vector space V(F) is a linearly independent· subset of V that spans the entire space V(F). Example 4-19. The familiar unit vectors e" e2, ... , ell of V,,(F), the space of n-tuples of scalars, form a basis (see Examples 4-14 and 4-17). Hereafter, we shall refer to this particular ba.sis as the natural or standard ba8i8 for V,,(F). Example 4-20. For a more general example, consider the vector space V.(F) of infinite sequences of elements from F. If

k

=

1,2, ... ,

then the set 8 = {e" e2, ...} is linearly independent. These vectors do not constitute a basis for V.. (F), however, since it was shown earlier that the linear span [81 is a proper subset of V•. Example 4-21. Let M,,(F) be the vector space of n X n matrices over a field

F. This space has a basis consisting of the n 2 matrices EiJr where EiJ is the

4-3

BASES AND DIMENSION

269

square matrix of order n having 1 as its ijth entry and zeros elsewhere. Any matrix (aij) E M .. (F) can obviously be written as (aij)

=

(ll1EU

+ auEu + ... + a,...E .....

Moreover, (ai;) = 0 if and only if all = au Ell, Eu, ... , E .... are linearly independent over F.

= ... = a,... =

0, hence,

Example 4-22. One final example: Consider F[x], the vector space of polynomia.ls in x with coefficients from F. A basis for F[x] is formed by the set

S

=

{I, x, x 2 ,

••• ,

x .. , . ..}.

+

+ ... +

By definition, each polynomial p(x) = aol alX a,.x" of F[x] is a linear combination of elements from S. The independence of S follows from the fact that, for any finite subset the relation holds if and only if Cl

=

C2

= ... =

CIc

= o.

These are but a few examples of the more frequently encountered bases and should amply illustrate the concept; as we continue our discussion, additional examples will appear. We ought to point out several things. First, Theorem 4-18 may be rephrased 80 as to assert that any vectOr space which is spanned by a finite subset, linearly independent or not, possesses a finite basis. Since a basis S for a vector space V(F) is by necessity a linearly independent subset of V, it is possible to express each vectorj Vasa linear combination of elements from S in exactly one way. The u ique scalar coefficients which occur in this repreeentation are called the c dinateB of x with respect to the given basis. Thus the notion of a basis enables us to coordinaliie the space. • Finally, let us observe that a. given vector space may have more than one basis. Example 4-18, for instance, shows that the vectors Xl

=

(1, 1,0),

X2

=

(1, 0, 1),

xa

=

(0,1,0)

constitute a second basis for Va(F). In this case, an arbitrary vector (alJ all, 4a) of Va can be written as (aJ, all, aa)

=

(al - aa)Xl

+ 4aX2 + (all -

al

+ aa)xa.

While the coefficients in the above linear combination are uniquely determined, they obviously differ from those which represent the same vector relative to the standard basis ell e2, ea; roughly speaking, a vector has different coordinate. with respect to different bases.

270

4-3

We next dispose of a natural question that arises here: is it possible to obtain a basis for a given vector space? A closely related question is this: presuming one can seleet two differ('nt bases for a space, must each contain the same number of c1!'n1('nt.s? When our vector space is the zero space {O} (F), no subset is linearly independent and certainly no basis exists. On the other hand, t.he coming th!'or!'m guarantees that a nonzero space will always have a basis. The proof ill II. stl'llightforward application of Zorn's Lemma. Theorem 4-20. a basis.

(Ba.~i.~

Theorem). Every nonzero veetor space V(F) possesses

I'roof. Let a he the family of all linearly independent Hubsets of V. If x "" 0, then {:r} E «, so t.hat. (t is plainly nonempty. Our immediate aim iH to show that for any chain of Sl't.s f11 j} in a. their union UA j also helongs to a. To do so, we aSSlIllH' th .. vl'etors .rl, X2, ••• , .rll E UA j and that. UH' lilwar combination ('I.rl 1 r2'(2 1 ... 1 r":e,, = O. Now, each vllct.or Xk lieH in some member Aik of {/1,}. As : A,: forms a chain, one of the sets Ai" A ;2' .•.• A i. contains all the others. call it A j'. This means that the given vectors Xlr X2, ••• , Xn arc nil in A j'. Bill. the linear ilHlependence of A j' then impli(~s that the Hcalar coefficients rk = 0 for all k = 1, 2, ... ,n. Thus the union uA i is itself a linearly independent subset. of V, whence a ml'mher of thc family a. The hypoth!'s('fol of Zorn's Lemma being Ratifolfied, there exists a maximal clement S in a. As a m('mber of a, S is a linearly independent ""subset of V, folO to complet!' the proof it remains simply to establish [S) = V. To see this, let x be any v!'dor of V not in S. Because th!' set S' = S U {x} properly contains S, it mUfolt be linearly dependent (the maximality of S enters here). Therefore, for folome finite suhs!'t :!lI, 112,' .. ,lIm1 of S. It dependence relation

exists in which not all the cocffieicnts are zero. Were a = 0, we would contradict· th~ lill
Proof. Let {XI, :r2, ... , xn) he one basis for thl' spaee V(F) and {YI, Y2,' .. ,Ym} be unother. Beeause the vedors YI, !/2, ... , 11m are linearly independent and V(F) is spannell hy :rl, X2, ..• , Xn , it follows from t.h(~ Rephwcmcnt Theorem t.hat t.he integer m is no larger than n, tlmt. is, m ~ n. Simply revertling the roles of {XI, X2, , .. ,xn } and {Yb Y2, .•. , Ym) in the argument, we also condude n ~ In, whClII'e n = m. Theorem 4-22 ean be used to define dimenllion. For if II vedor space V(F) has a finite basis, then every ballill for the IlJlace wiII have the sallle finite number of clemen til. This unique integer is called t.he dimension of V(F) anti dellignatcd by dim V. Although UII' notion of hmlis iH IlOt ltpplimhlc to the z(~\'O space, it is (~IIKt.omary t.o t.n'llt. :0) (P) lUI finit.e-dinwllliiollal, wil h Zf'ro dimellsion. A (nollzcro) v('ctor sJllwe ill ~tid to be injinite-dill/.fIt.~il}Ttal if it. ill not spanned by any fillite sllhiil~t., that. is, if it iH not finite-dimellsiollal. Not every spnce is of finil.t, dilll('IISioll, II..'! ('videlll·(·d hy F[.rj, t.I\(' vpdo\' KpIU'(' of' polynomials ill x. III lIIost of what. follows, we lihall for 1.1", sakI' of I·(llll'ppl.llal HilllJlli"ily eOllfille ollr alll'lItioll t.o fillitl!-dilll('IIHiolllll VI'I'I.OI· liJlIU~I'S. '1'111' IIl'xl I hl'fl 1'('11 I illdi":lII'K t.Iml al II':llit r'JI' lillill'-,liIlII'IIHioll:l1 HJlIU'I'H, Oil£' "lUI f01'1II haiil'" without. 100 I11I1I'h dilli,·ulty.

Theorem 4-23. Let \'(F) 1)(' It fillite-dillll'lI"ional V('dOl'''JlIU'I', liay dim V = n. Thl'll I) I'vl'ry iiI't. of n Vf'f'lors whil'h KJlall V iH a haHiH, /I lilll'ady illlll'Jllmcll'lIt V\'dorH frolll \' is

:!) ('V\'ry Ii!'t of

Il

hllSiK.

272

4-3

VECTOR SPACES

Proof. Suppose V = [Zl, Z2, ..• ,Zt&J. According to Theorem 4-18, some subeet of {Zh Z2, ... ,Z,.} is a basis for V(F). But the previous result implies that this subeet must contain n elements and accordingly is the entire eet {Zh Z2, ... , z,.}. For a proof of the second assertion, let Zh Z2, ... ,Z.. be n linearly independent vectors of V. If Z E V is arbitrary, the set {z, Zh ... , z,.} is dependent, since the mwcimalllumoor of linearly independent vectors in V is n. Thus some nontrivial dependence relation exists among theee vectors: ex

+ C.Zl + ... + C,.z,. = o.

Were the coefficient c = 0, a contradiction to the linear independence of Zh Z2, ... ,Z.. would arise. Hence c ~ 0, and we may solve for the vector Z in terms of Zit Z2, ... , z .. to obtain

This argument shows that

making the eet {Zit Z~h ... ,z,.} a ; ; . for V(F).

...

Theorem 4-18 told us that it is ·ble to chooee a basis for a vector space V(F) from any set of generators of V. In the opposite direction, the coming theorem asserts that any linearly independent subset of V is either a basis or else can be extended to a basis for V(F), in the sense that vectors may be added to it to fonn a basis. Theorem 4-24. If V(F) is a finite-dimensional vector space and {ZltZ" ... ,Z,.}

is a linearly independent subset of V, then there exist vectors 1I,.+h ... ,II... such that {Zl, ... ,.z,., 1I.. +h ... ,II..} forms a basis for V(F). Proof. The proof is short: Since V(F) is finite-dimensional, it has a finite basis IIh IIllt •.. ,II..· As V = [lIh 112, ... , II..], we may apply the Steinits Theorem to replace n of the II's by z's and obtain a eet {ZI, •.• , z,., 1I"+l, ••. , II..} whose linear span is atill V. But any generatin, set with m elements in an m-dimensional space is a basis.

Corollary. Every basis for a subspace of a finite-dimensional vector space can be extended to a basis for the entire space. . Example 4-23. As a particular case of this last point, consider the vector space M,(F) and the three matrices

4-3

BABES AND DDlJ!lNBION

273

To check the linear independence of these matrices, let

On equating corresponding entries, we see that Cs - 0,

which implies CI = C2 = Ca = O. Theorem 4-24 now tells us that the given matrices will form at least part of a basis for M 2(F). We shall leave the details to the reader and content ourselves with this one comment: the addition of any fourth matrix having a nonzero element in the (2, 2)-position will yield a linearly independent set. For instance, among the many possibilities, the matrices

are linearly independent and, being four in number, must therefore be a basis for the 4-dimensional space M 2(F). This illustrates the wide latitude of choice for bases of M 2(F). We terminate the present section by giving some useful results concerning the dimensions of subspaces of a given vector space. The reader should first prove the following theorem. Theorem 4-25. If W(n is a subspace of a finite-dimensional vector space V(F), then 1) W(F) is also finite-dimensional with dim W S dim V. 2) dim W = dim V if and only if W = V. The next theorem relate~ the dimensions of the sum and intersection of two subspaces. Theorem 4-26. If U(F) and W(n are 8ubspaces of a finite-dimensional vector I!plwe V(F), then dim (U

+ W) = dim l! + dim W -

dim (U

n W).

In particular, dim (U E9 W)

= dim U + dim W.

Proof. Before entering into the details of the proof, we observe that by Theorem 4-25 the four subspaces involved in the statement of our theorem all have finite dimension. Now, let {xa. X2, ••• ,x,,} be a basis for (If n W)(F). Accord-

274

4-3

VECTOU SPACES

ing to Theorem 4-24, there will exist vectors u., U2, ••• , u'" such that {X., •.. , Xn, u., ... , u"'} is a basis for the subspace U(F) and vectors 1/'., 11'2, ••• , //', ~\wh that {Xb' •• , X"' WI, ••• , w,} i~ a bu.sis for the subspace W(F). Combine these two bases into a single set

Becausc the finlt m + n vectors of the foregoing set are a basis for U(F) alld th(' last TI -I r vect.ors are a basis for W(F), any c1f1ment of the sum (! + W may I)(~ cxprN;sc!\ ItS It lincar eomhirmtion of the vectors of thi" sct, that is

F / W = [UI, ... , U"',

XI, ••• ,

xn ,

w., ... ,w,l.

We wish to show that the vectors on the right are also linearly independent and conRequent.iy a 1>I\sis for the subspace (U + W)(F); once this is established, it would follow that dim (fT

+ W)

=

m+n

= dim

II

+r =

(m

+ dim W

+ n) + (n + r) - dim (11

- n

n W).

Thereforc, let us suppose

Settinp; 2 =

C111'I

+ ... + C,II'"

we would t.hen have

I fIlII / ... t- bnxn) or 2 E [X., . .. , In, ttl, ••. , uml = [T. Since the vector 2 ulao belongs to W, it lIIust hI' 1\ lilll'ar (·llIl1hiIlHj.ion of Hw hnsiIl1'1!'lIl1'lIt.S X" X2, ••• , Xn of I,he Imbllpal'1' z~·

- (alul -/ .... / a,nu",

(i! n W)(/"), lIay,

so that

But the set {XI, ••• , X"' till, ... , til,} is linearly independent, being a basis for the subspace lV(F), hell(~e d l = ... = dn = CI = ... = Cr = O. In conjunction wit-h the IinC'ar indl'pcIHI!'n('e of fx., ... , Xn, U" ••• , u m }, this fon~es al = ... = am = II. = ... = b" = O. We havl' thus sueeccded in provinj!; that t.I\I~ vcclol"ll I l l , " " Itn" .r" ... , .r", !t., arc linearly independent, !l.S required.

w" ... ,

Theorem 4-27. If W(F) is a subspace of It finite-dimensional vcetor spa(:c V(F), then the quotient space (V /W)(F) is allID finito-dirnelll~ional and

dim V = dim W

+ dim V/W.

4-3

BASES AND DIMENSION

275

Proof. Let {Xl. X2, ••• ,XII} be a basis for W(F). By adding vectors Yt. Y2, .•. , y"" we may extend this set to a baliis {x!, ... , X,,, Yh .•• ,y",} for the whole !!pacc V(/t'). Any vector x E V can then be written in the form x = atXl a"x" b1Yl b",y", for appropriate choice of coefficients. Since x - (b'Yl b",y",) E W, the coset x + W is expressible

+ ... +

+

+ ... + + ... +

as

In othn!" words, i1w demcnl.li Yl + W, 712 + W, ... ,!1m + W span the quotient Rpuee (V /W)(P). The remainder of t.he proof amounts to showing th('HC coset!! to be linearly independent and heIl(~e a basi!! for (V /W)(F). To see thill, we suppose

where, of course,

t')

+W =

W is the zero clement of V /W. Thus

elY I

and must he

:~

+ C2Y2 + .. , + CnY", E W

linpur ('ombilllLtioll of the basis v('ctors

Xl, X2, •• , ,

x" of W, say

But the lincar indepcndence of the set {x!", . , x"' YI, .•. , y",} then implies 1:, ~ •.• = em = (I, = ... = till = O. The foregoing urgulllcnt. indicates that the quotient space (V /W)(F) has a basis eonsisting of the m cosets Yl + W, Y2 + W, ... , y", + W. The conclusion of Uw U\('orem followH imml'diutcly from this, sirwil dim V /W

=

m

=

(m

+ n)

-

n

=

dim V - dim W.

Example 4-24. We iIIustmte the above by looking again at Va(F) and the one-dimensional subspace W(F), where W

=

{(a, 0, 0) I a E F}.

In this case, the quotient space (Va/W)(F) has dimension 2, for the equation of Theorem 4-27 reads dim V 3/W

=

dim V 3

-

dim W

=

3 - 1 = 2.

To actually obtain a basis for the quotient space, one need only employ the procedure of the theorem. First, extend the basis of W(F)-that is, the vector (1,0, O)--to a basis for the entire space V(P), say by adjoining vectors (0, 1,0) and (0,0,1). The corresponding cosets (0,1,0)

+W

and

(0,0,1)

+W

276

4-3

VECTOR SPACES

then serve as a basis for (Va/W)(F). Indeed, it is easy to show that any element (CllI Cl2, CIa) W of Va/W m&y be written in the form

+

(CllI CI" CIa)

+ W = CI,[(O, 1,0) + W) + Cla[(O, 0, 1) + W).

PROIUMS

In the problems below, F will denote an arbitrary field.

1. For each of the following vector spaces, determine whether the lets listed are linearly dependent or independent. a) V.(F): {(I, 0, 0, 0), (1,1,0,0), (1, 1, 1,0), (1,1,1, I)} b) F[%]: {%' % - 1, %2 - % - 2, %2 + %+ I} c) Va(Za): {(4, 1,3), (2,3, 1), (4, 1, On d) Va(C): {(I, 2 i, 3), (2 - i, i, 1), (i, 2 3i,2)}

+

+

+

2. Prove that if each vector form % - 41%1 42%, are linearly independent.

+

% e [%1, %" ••• ,%..] + 4,.2: .. (4t F),

+ ...

e

is uniquely representable in the then the vectors %1, %2, ••• , %..

3. Given vectors %1, %2, ••• ,%.. e V, establish the &8IIertioDB below: a) If %i ... %j for some i ~ j, the set {%1, %2, ••• ,%..} is linearly dependent. b) If {%1' %2, ••• ,%..} forms a linearly independent set and aI, a2,'" •• , a ..-1 then

e F,

is also an independent set of vectors. c) If {%l' %2, ••• ,z ..} is linearly independent while {%1, ••• , %., %.. +1} is linearly dependent, then the vector %.+1 e [%1, %" ••• , %J. 4. Show that the vectors (3 - i, 2 2i,4), (2,2+ 4i, 3), and (1 - i, -2i, 1) form a buis for the apace Va(C) and determine the coordinatee of each of the .tandard basis vectors (1,0,0), (0,1,0), and (0,0,1) with respect to this basis. .

+

o.

a) Find a basis for the vector space C(R') and all bases for the space Va(Z,), b) For what values of 4 do the vectors (1 a, 1, 1), (1, 1 a, 1), and (1,1,1 a) form a basis of V,eR'>?

+

+

+

is a basis for the vector space Va(R'>. Verify that the sets and {%1, %1 + %2, %1 %, %a} &lao serve as bases of Va(R'). Is this situation true in the space Va(Za)?

6. Assume

{%1, %2, %a}

{%1 + %2, %, + %a, %a + %I}

+ +

7. Prove that the subspace of F[%] consisting of all polynomials of degree at most n is finite-dimenaional. 8. If diag Jf.. denotes the set of all diagonal matrices of order n (over the field F), show that (diag M ..)(F) is a subspace of the vector space M ..(F) and determine its dimension. I

9. Prove that if W(F) is a proper subspace of the finite-dimensional vector space V(F), then dim W < dim V.

LINEAR MAPPINGS

10. Assume the space V(n is finite-dimensional with basis {~l, ~2, ... ,z,,}. If W.(n is the subspace generated by the vector Zk (Ie - I, 2, ... , n), verify that V - WIE9 W2E9 .•• E9 W". 11. Let U(n and W(n be subspaces of V,,(F) such that dim U > n/2, dim lV > ,,/2. Show Un w pi {OJ. 12. Suppose {~l, ~2, ... ,~,,} is a basis for the subspace U(n of V(F) and {Ill, 112, ••• ,II.. } is a basis of the subspace W(F). Given that the set {~l, ... , X", Ill, .•. ,II..} forms' a basis for the entire apace V(F), prove that V - U E9 W.

+, .)

13. Let (F[x), be the ring of polynomials in X over F and p(~) E F[~) be a polynomial of degree n. If «p(~», +, .) is the principal ideal generated by p(x), establish that (F[~1I(p(~»)(F) is a veotor space of dimension n. [Hint: Consider the cosets 1 + (p(~», Z + (p(x», ••• ,X,,-1 + (p(x) ).] 14. Determine the dimension of the quotient space (Va/W)(F), where the set W is defined by W - {(a, h, a+ h) I a, h E F}. 15. In the vector spaCe M,,(n, let T ..(F) be the subspace of upper triangular matrices [matrices (a.,) such that a., - 0 for i > j) and T~(/") be the subspace of lower triangular matrices [matrices (ai,) such that a.i == 0 for i < j), Find dim T", dim T~, dim (T" n n), dim (T.. + T~), and verify the truth of Theorem 4-26 in this particular case. 4-4 LINEAR MAPPINGS

In this section, which is our last, we examine the vector space analog of the familiar homomorphism concept. Since a vector space V(F) is comprised of two algebraic systems, a group (V, +) and a field (F, +, .), there may be BOrne initial confusion as to what operations are to be preserved by such functions; the answer is only those operations which explicitly involve vectors: vector addition and scalar multiplication. Traditionally, vector space homomorphisms are called linear mappings or linear transformations, and we adhere to this terminology. DefInition 4-18. Let V(F) and W(F) be vector spaces over a field F. A function f: V - W is said to be a linear mapping from V(F) into W(F) jf f(x

+ y) = f(x) + f(y)

and

f(cx)

= cf(x)

for all vectors x, y E V and all scalars C E F. The set of linear mappings from V(F) into W(F) wiIl subsequently be designated by L(V, W). Simply put, a linear mapping from the space V(F) into W(F) is a homomorphism from the additive group (V, +) into the additive group (W, +) which, at the same time, preserves scalar multiplication. This is plainly equivalent to the single requirement that f(az

+ by) = af(x) + bf(y)

278

VECTOR SPACES

for all x, Y E V and a, b E F. Suffiee it to say, the above definition makes sense only when both vector spaces are taken over the same field. Before proceeding to the theory of linear mappings, let us illustrate to some extent the great variety of possible examples. Example 4-25. Let M m,,(F) be the vector space of all m X n matrices over a field F, and let (ai;) he!\ fixed m X m matrix (again, over F). Define a function I: 1If rn" -+ 111 mn by

Then I is seen to be a linear mapping, because fHII,j) I

II«(:,}»

= (aij)' [r(lli,)

-

r{a,;)· (/Iii)

I II(C,,»)

I-

lI(aj/) . (r.ii)

.,.,- r!«aij» /- 1I/«r.iJ»' Example 4-26. In the Rpll('e V",(F) of infinite Heqlwnees of elements from a field F, we define thc shifl function f as follows: for each sequence x = (at. a2, aa, ... ), take I(x)

=

(a2' a3, a4, ... ).

The reader may easily verify that I is a linear mapping from VlI (F) to V..,(F). In fact, any power of I is again linear, Hinee

Example 4-27. Next consider F[x], the space of polynomials in the indeterminant x with coefficient,s from F. A lillear mapping 011 F[x] is given by means of Uw Ho-{:nlInd dijJerentiat1:on functilln. That. iH, for lUI arhitrary polynomial p(x) = ao --j- alx + a2x2 + ... -f- a"x" ill ~'[x], let I(p) = a1

+ 2a2x + ... + nanx,,-l.

Example 4-28. One more example: Suppose W(F) is a subspace of the vector space V(F). Then the familiar natural mapping natw: V -+ V IW defined by taking natw (x) = x + W

is a linear tranHformation. By virtue of the definition of the operations in (V IW)(F), we easily dlCck that natw (a.x

+ by) =

+ by + W = a(x + W) -I- bey + W) = ax

a

natw (x)

+ b natw (y).

One fact which follows almost immediately from Definition 4-18 is that if dim V is finite, then any linear transformation f from V(F) into W(F) is com-

4-4

LINEAR MAPPINGS

279

pletely described hy specifying its values on a basis for V(F). For suppose {Xli X2 • ••• ,;r n } is It basis of the space V(F) j then Mch vector X in V has the form for suitable t;("alars ak E F. By the linearity condition, extended to n vectors,

Thus,! is completdy determined onee its cITed on the bu.'lis vectors Xl, X2, ••• , Xn is known. The next. theorem indicates that this effeeL may be prcserihed nrhil.rllrily.

Theorem 4-28. 1,,'1. V("'IIII'HIII\I'"

:.r I, .r~, ...• ;r,,:

1"(/

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