Algebra

Algebra

This is a volume in PURE AND APPLIED MATHEMATICS

A Series of Monographs and Textbooks Editors: SAMUEL EILENBERG AND HYMAN BASS A list of recent titles in this series appears at the end of this volume.

Algebra Larry C. Grove Department of Mathematics The University of Arizona Tucson, Arizona

@

1983

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9 8 7 6 5 4 3 2 1

I

Contents

vii

ix xiii

Chapter I Groups 1. Groups, Subgroups, and Homomorphisms

2. 3. 4. 5. 6. 7. 8. 9. 10. I 1.

12.

Permutation Groups The Symmetric and Alternating Groups The Sylow Theorems Solvable Groups, Normal and Subnormal Series Products Nilpotent Groups Finite Abelian Groups Free Groups Generators and Relations Some Finite Groups Classified Further Exercises

1

12 16 19 22 27 29 32 33 37 40 41

Chapter I1 Rings I. 2. 3. 4. 5. 6.

7. 8.

Preliminaries: Ideals and Homomorphisms The Field of Fractions of an Integral Domain Polynomials Polynomials in Several Indeterminates Divisibility and Factorization The Chinese Remainder Theorem The Hilbert Basis Theorem Further Exercises

41 52 54 58 61 71 13

75

vi

Contents

Chapter I11 Fields and Galois Theory 1.

2. 3. 4. 5. 6. 7. 8.

Field Extensions The Fundamental Theorem of Galois Theory Normality and Separability The Galois Theory of Equations Symmetric Functions Geometrical Constructions Norm and Trace Further Exercises

19 86 90 96 102 109 115

119

Chapter IV Modules 1. Preliminaries 2. Direct Sums, Free Modules 3. Finitely Generated Modules over a PID 4. Applications to Linear Algebra 5. Computations 6. Tensor Products 7. Further Exercises

125 129 134 140 145 158 164

Chapter V Structure of Rings and Algebras I. Preliminaries 2. The Jacobson Radical 3. The Density Theorem 4. Artinian Rings 5. Further Exercises

172 176 180 183 189

Chapter VI Further Topics 1. Infinite Abelian Groups 2. P6lya- Redfield Enumeration 3. B U V 4. Integral Dependence and Dedekind Domains 5. Transcendental Field Extensions 6. Valuations and padic Numbers 7. Real Fields and Sturm’s Theorem 8. Representations and Characters of Finite Groups 9. Some Galois Groups

194 200 210 217 2 30 238 252 260 215

Appendix Zorn’s Lemma

289

References

29 1

Index

293

It is fairly standard at present for first-year graduate students in mathematics in the United States to take a course in abstract algebra. Most, but not all, of them have previously taken an undergraduate algebra course, but the content and substance of that course vary widely. Thus the first graduate course usually begins from first principles but proceeds at a faster pace. This book is intended as a textbook for that first graduate course. It is based on several years ofclassroom experience. Any claim to novelty must be on pedagogical grounds. I have attempted to find and use presentations and proofs that are accessible to students, and to provide a reasonable number of concrete examples, which seem to me necessary in order to breathe life into abstract concepts. My own practice in teaching has been to treat the material in Chapters I-V as the basic course, and to include material from Chapter VI as time permits. There are in Chapters I-V, however, several sections that can be omitted with little consequence for later chapters; examples include the sections on generators and relations, on norms and traces, and on tensor products. The selection of “further topics” in Chapter VI is naturally somewhat arbitrary. Everyone, myself included, will find unfortunate omissions, and further further topics will no doubt be inserted by many who use the book. The topics in Chapter VI are more or less independent of one another, but they tend to draw freely on the first five chapters. There are two types of exercises. Some are sprinkled throughout the text; these are usually straightforward and are intended to clarify the vii

...

Vlll

Preface

concepts as they appear. The results of those exercises are often assumed in the following textual material. The other exercises are at the ends of the chapters. They vary widely in difficulty, and are only rarely referred to later. Of course, not all of the exercises are new, and I am indebted to a wide variety of sources. My debts to earlier textbooks will be clear to those familiar with the sources, but particular mention should be made of the works of Artin [ 1-41, Van der Waerden [37], Jacobson [ 171, Zariski-Samuel [41], and Curtis- Reiner [8]. I have followed Kaplansky’selegant version of the Fundamental Theorem of Galois theory. I have learned more than I can reasonably acknowledge from my colleagues, past and present. I hope they know who they are and accept my gratitude. The same applies to a large number of students, who have suffered through several preliminary versions and who have prompted many improvements. I must single out Kwang Shang Wang and Javier Gomez Calderon, who ferreted out large numbers of mistakes, misprints, and obscurities by means of several careful rereadings. Finally, my best thanks go to Helen for all the typing and all the rest.

List of Symbols

I

a

ri X

0 V

Subgroup of Normal subgroup or Product Cartesian product. direct product Direct sum Join Tensor product over ring R Field of algebraic numbers a is a divisor of h: h is a multiple of a a and b are associates The group generated by commutators [x, y], I E A and 4' E B Ascending chain condition Alternating group on ti letters The autornorphwn group of group G The field of complex numbers Class functions on group G Centralizer of element Y in group G Column operation; add column i to column j Column operation; add r times column i to column j Conjugacy class containing group element x Centralizer in ring R of module M Descending chain condition Degree of polynomial Degree of representation Dihedral group of order 2m Matrix unit, ij-entry is 1 and others are 0 Endomorphism ring of abelian group A Simple field extension with primitive element ~1 Fundamental homomorphism theorem (for groups) Fundamental homomorphism theorem (for modules) ix

x

List of Symbols Fundamental homomorphism theorem (for rings) Fixed field of subgroup H Galois field with 9 elements Field of fractions of integral domain R Extension field of F generated by set S Derivative of polynomial f(x) Absolute value function on C p-adic valuation on Q Derived group (commutator subgroup) of group G Greatest common divisor Galois group of polynomial ,f(.x) Galois field with 9 elements Subgroup in the derived series of group G Galois group of field K over subfield F Subgroup of Galois group fixing elements of intermediate field L General linear group Hamilton’s ring of quaternions R-homomorphisms from module M to module N Radical of ideal 1 Inner automorphism group of group G Image of mapping f Integral closure of subring R in ring S Quotient of ideal I in ring R Set of irreducible characters of group G Jacobson radical of ring R Kernel of homomorphism j” Kernel of character Least common multiple Subgroup in descending central series of group G Minimal polynomial over field F of algebraic element u n x n matrices over ring R Submodule of M annihilated by ring element r Norm function from field K t o subfield F Normalizer in group G of subset A Orbit of elements under action of group G Group of permutations of set S Principal ideal domain Projective special linear group Field of rational numbers Quaternion group, order 8 Generalized quaternion group, order 4 m Field of real numbers Nonzero elements in ring R Group algebra of group G over commutative ring R Row operation; add row i to row j Row operation; add r times row i to r o w j Ring of algebraic integers in quadratic field Q(6)

List o f Symbols R-module generated by set S Ring of polynomials over ring R in indeterminate Y Field of rational functions over integral domain R in indeterminate Ring of polynomials over ring R in indeterminates x i . .. . , Y ,

Xi

Y

Field of rational functions over integral domain R in indeterminates Y 1 , . . . ,X " Ideal generated by subset S of a ring Subgroup generated by subset S of a group Principal ideal generated by ring element s Special linear group

,)

I*l [.Y,

L

L,,

4'1

Symmetric group on n letters Group with generating set S subject to set R of relations Stabilizer of element s under action of group G ith symmetric polynomial in indeterminates x i , .. . .Y, Transcendence degree of field K over subfield F Trace function from field K 10 subfield F Unique factorization domain Group of units in ring R with 1 Variation of a sequence of polynomials at element a of an ordered field Module over polynomial ring determined by linear transformation T of vector space V Order of group element Y Commutator of group elements .X and y Ring of integers Cycle index of permutation group G on set S Center of group G Subgroup in ascending central series of group G Ring of integers mod n Divisible abelian p-group; p-primary component of Q/Z

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Introduction

The conventions and notation of elementary set theory are assumed to be familiar to the reader. If { & : a E A } is any family of sets, indexed by a set A , we shall write n { S , : a E A } , or simply n,S,, for their Cartesian product. Thus n { S , : c r ~ A is } the set of all functions f:A + u { S , : c r ~A } for which f ( a ) E S,, all o! E A. If the family {S,} is finite, say {S,,. . . ,S,}, or countable, say ( S , , S , ,... 1, we may write S , x Sz x ... x S,, or S , x S, x ..., respectively, for the Cartesian product. In those cases the elements of the Cartesian product are conveniently represented as ordered n-tuples ( x l ,x , , . . . ,x,), or sequences (x, ,x 2 , . . .), respectively, where xi E Sifor each i. If S and T are sets we write S\T for the relative complement of T in S, i.e., S\T= { X E S:X 4 T } . The cardinality of any set S will be denoted by IS(. A binary operation on a set S is a function from the Cartesian product S x S to the set S. For our purposes a binary operation will often be called multiplication, with notation ( x , y ) X J ) ~, or addition, with notation ( x , Y ) H X y. A binary operation (say multiplication) on a set S is called associative if x( yz) = ( x y ) z for all x, y, z E S . We shall have occasion to use Zorn’s Lemma, an equivalent of the settheoretic Axiom of Choice. A brief discussion, with an example of an application, appears in an appendix. It is assumed that the reader is conversant with the material of a first course in linear algebra, including standard matrix operations and basic facts concerning vector spaces and linear transformations. The existence of a basis and dimension for a vector space are proved in the appendix. We shall denote the set of integers by Z, the rational numbers by Q, the real numbers by R,and the complex numbers by @. Frequent use will be made of the division algorithm in Z.Also, familiarity with Euler’s totient function 4 will be required on occasion. Details can be found in any book on elementary number theory or in almost any undergraduate abstract algebra book.

+

xiii

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Chapter I

Groups

1. GROUPS, SUBGROUPS, AND HOMOMORPHISMS A nonempty set with an associative binary operation is called a semigroup, and a semigroup S having an identity element 1 such that I x = XI= x for all x E S is called a monoid. Most of the algebraic systems discussed herein will be semigroups or monoids, but almost always with further requirements imposed, so the semigroup or monoid aspect will seldom be explicitly emphasized. One trivial consequence of the definition of a monoid deserves mention.

Proposition 1.1. The identity element of a monoid S is unique. Proof. Suppose 1 and e are identities in S. Then 1 = l e

= e.

A group is a set G with an associative binary operation (usually called multiplication) and an identity element I satisfying the further requirement that for each x E G there is an inverse element y E G such that x y = y x = 1.

Proposition 1.2. If G is a group and x element.

E

G, then x has a unique inverse

Proof. Let y and z be inverses for x. Then y

= y l = y(x2) = ( y x ) z =

The unique inverse for x

E

l z = z.

G is denoted by x - ' . Note that ( x - ' ) - '

Proposition 1.3. If G is a group and x , y 1

E

G , then ( x y f -

=y

-1

x

= x. -1

.

2

I Groups

Proof (xy)(y-'x-') and similarly ( y

= ((xy)y-')x-'

x ')(xy)

=

= (x(yy-'))x-' = (x1)x-I = xx-1 = 1,

1.

As Coxeter [7] has pointed out, the "reversal of order" in Proposition 1.3 becomes clear when we think of the operations of putting on our shoes and socks. If the binary operation of a group G is written as addition, then the identity element is commonly denoted by 0 rather than 1, and the inverse of x by - x rather than x - ' . It is customary to use additive notation only if x + y = y + x for all x, y E G. In general, a group G (multipkative again) is called abelian (or commutative) if xy = yx for all x, y E G. Wewritex' = 1,x' = x,x2 = xx,andingeneralx" = x"-'xfor 1 I n E Z . Define X - " = (x-')", again for 1 n E E.It is easy to verify by induction that the usual laws of exponents hold in any group, viz., X m X n = xm+"

and

(x")" = xmn

for all x E G, all m,n E Z.The additive analog of X" is nx, so the additive analogs of the laws of exponents are mx nx = ( m n)x and n(mx) (mn)x.

+

+

Exercise 1.1. Verify the laws of exponents for groups.

EXAMPLES 1. Let G = { 1, - l } E R, with multiplication as usual. Then G is a group. 2. Let G = E,Q, R, or @,withthe usual binary operation of addition. Then G is a group. 3. Let G = Q\(O), the set of nonzero rational numbers, under multiplication. Then G is a group. Similarly this'holds for R\{O} and @\{O}, but not for E\(O}. (Why?) 4. Let S be a nonempty set. A permutation of S (sometimes called a bijection of S ) is a 1- 1 function 4 from S onto S . Let G be the set of all permutations of S. If 4 , 8 E G, we define 48 to be their composition product, i.e., @(s) = 4(8(s)) for all s E S . Composition is a binary operation on G (verify), and it is associative, for if $,8, (T E G and s E S , then

I

Groups, Subgroups, and Homomorphisms

3

G has an identity element, the permutation 1 = 1, defined by I(s) = s, all s E S, and each 4 E G has an inverse 4 - l defined by cf~ ‘(sl) = s2 if and only if $(sJ = s1 (there are a few details to be verified). Thus G is a group; we write G = Perm(S). This example is of considerable importance and will be pursued much further. ~

5. As a special case of the preceding example take S = { I , 2,3,. . . ,n}. The group G of all permutations of S is called the symmetric group on n letters and is denoted by G = S,. If Cp E S,, it is convenient to display the function Cp explicitly in the form

For example, if n = 3, then 4 = (i i :) is the permutation that maps 1 to 2,2 to 3, and 3 to 1. The notation makes it quite simple to carry out explicit computations of the composition product. Suppose, for example, that n = 3 and 4 = (i i),8 = ($ :), Note from the definition of 48 in Example 4 that 8 acts first and 4 second. Thus 0 maps 1 to 3 and 4 then maps 3 to 1, and so the composite 48 maps 1 to 1. Similarly, $18 maps 2 to 3 and maps 3 to 2. Thus

@=(*

1 2 3 1 2 3 3 1)(3 2 I ) = ( ;

Observe that

1 2 3 1 2 3 0 4 = ( 3 2 1)(2 3 I ) = ( ;

; i).

: :)#@,

so S, is not an abelian group. It is easy to see that S, is, likewise, not abelian for any n > 3, although S, and S, are abelian.

6. Let T be an equilateral triangle in the plane with center 0.Let D, denote the set of symmetries of T , i.e., distance-preserving functions from the plane onto itself that carry Tonto T(as a set of points). Theelementsof D,arecalled congruences of the triangle T in plane geometry. With composition as the binary operation, 0,is a group. Let us list its elements explicitly. There is, of course, the identity function 1, with 1 (x) = x for all x in the plane. There are two counterclockwise rotations, 4, and 42,about 0 as center through angles of 120“ and 240”, respectively, and three mirror reflections 01, O , , 0, across the three lines passing through the vertices of T and through 0 (see Fig. I). It is edifying to cut a cardboard triangle, label the vertices, and determine composition products explicitly. The result is the “multiplication table” (Fig. 2) for 0,.

4

I Groups

I

3\

Figure 1

1

41

42

01 92

93 -

A routine inspection of the table shows that each element has an inverse, and also (if enough time is spent) that the operation is associative. Associativity is also clear from the fact that each element of D3 is a permutation of the points of the plane. Thus 0, is a group. If we let S = { 1,2,3} be the set of vertices of T, then each element of 0, gives rise to a permutation of S, i.e., to an element of the symmetric group S,. For example, 41H(: ;), O1 I-+(: l),etc. The result is a 1-1 correspondence between the group D3 of symmetries of T and the symmetric group S , . It is instructive to label the elements of S, accordingly [e.g., a1 = (i 3 ;), p1 = (i :),etc.], to write out the multiplication table for S , and to compare with the table above. 7. This time let T be a square in the plane, with center 0, and let 0,be its set (in fact group) of symmetries. There are four rotations (one of them the identity, through 0') and four reflections (see Fig. 3). The multiplication table should be computed. Again each element of D4 gives rise to a permutation of the set S = { 1,2,3,4} of vertices of T, i.e., to an element of S,. For example, the rotation d1through 90" counterclockwise about 0 gives the permutation a1 = (i f). Note in this case, however, that not all elements of S, occur. For example, (i f 2) is not the result of any symmetry of the square.

:

:

:

1 Groups, Subgroups, and Homomorphisms

5

Figure 3

'1,

'1,

8. The quaternion group Qz consists of 8 matrices f 1, fi, & j , multiplication, where 0

0-1 I. =

0

0

0 0 - 1

0

0 1

0

j=[y

0

0

0 0

0 0

k=[' 0 1

-'

0 0

1 0

0

0-1

0

0-1

1'

2 k under 0-1

0

0

0

and 1 denotes the 4 x 4 identity matrix. It is easy to verify that i 2 = j z = k Z = - 1 and that ij = k. All other products can be determined from those. For example, since ijk = k 2 = - 1 we have i 2 j k = - j k = - i, and hence j k = i. The chief advantage of presenting Q 2 as a set of matrices is that the associative law is automatically satisfied. 9. Klein's 4-group K consists of four 2 x 2 matrices:

I=[:

y],

a=[1

"1,

0 -1

]

b = [ -1 0 0 1 '

Its multiplication table is Fig. 4.

Figure 4

and

c=[-l

1'

0 -1

6

I Groups

10. Let T be a regular tetrahedron and let G be the set of all rotations of three-dimensional space that carry T to itself (as a set of points), i.e., all the rotational symmetries of T. Thus G consists of the identity 1, rotations through angles of 180" about each of three axes joining midpoints of opposite edges, and rotations through 120" and 240" about each of four axes joining vertices with centers of opposite faces. Thus IG( = 12. Exercise 2.2. Let G be the set of 12 rotational symmetries of a regular tetrahedron.

(1) Verify that G is a group and write out its multiplication table. (2) Each element of G gives rise to a permutation of the set of vertices of the tetrahedron, numbered 1,2,3, and 4. List the resulting permutations in S,. (3) Each element of G also gives rise to a permutation of the set of 6 edges of the tetrahedron. List the resulting permutations in S , . Exercise 1.3. Describe the groups of rotational symmetries of a cube (there are 24) and of a regular dodecahedron (there are 60). It will be helpful to have cardboard models. Many more examples will appear as we continue. It will be convenient at this point to introduce some concepts, some terminology, and some elementary consequences of the definitions. The cardinality IGI of a group G is called its order. If G is not finite we usually say simply that G has injiniinite order. An easy counting argument shows that the symmetric group S, has order n ! . A subset H of a group G is called a subgroup of G if the binary operation on G restricts to a binary operation on H under which H is itself a group. In that case the identity element of H must be the original identity 1 of G. (Why?) We write H 5 G or G 2 H to indicate that H is a subgroup of G. Referring to the additive groups Z, Q, and [w we have, for example, Z IQ, Q 5 R, and Z 5 R.

Proposition 1.4. If H is a nonempty subset of a group G, then H 5 G if and only if x y - ' E H for all x , y E H .

'

Proof. =5 : Obvious. e : Choose x E H and take y = x. Then x y - = x x - ' = 1 E H . Next take x = 1 and any y E H to see that l y - ' = y - ' E H .

Thus x ( y - ' ) - ' = x y E H whenever x , y E H , so the multiplication on G restricts to a binary operation on H , which is associative since the original operation on G is associative. Thus H is a group and so H I G. # H c G, show that H is a Exercise 1.4. If G is a finite group and subgroup of G if and only if x y E H whenever x E H , y E H .

Proposition 1.5. If { H a } is any collection of subgroups of a group G, then n.H, IG.

I

Groups, Subgroups, and Homomorphisms

7

Proof. Apply the criterion in Proposition 1.4. If G is a group and S is any subset of G, then by Proposition 1.5 we see that n { H : S G H I G} is a subgroup of G. It is the smallest subgroup of G that contains S; smallest in the sense that it is contained in every subgroup containing S. We write (S) for that subgroup and call it the subgroup generated by S. There is a useful alternative description of (S) if S # Let S - ' = { x ~ ' : x E S } . C h o o s e e l e m e n t s x , , x,..., , x,ESuS-'foranyk, 1 I k E Z , and form the product x l x z ... x k . The collection of all such elements is a subgroup of G (by Proposition 1.4) that contains S and is contained in every subgroup containing S, hence must be (S). A group G that is generated by a single element, G = (x), is called a cyclic group. For example, the additive group Z of integers is a cyclic group, since Z = ( 1 ) = ( - 1). A rotation of the plane about a point through angle 2n/n generates a cyclic group of order n for each n E Z,n 2 1 . If G is a group and x E G,then we define the order of x, written 1x1, to be I (x)l, the order of the cyclic subgroup generated by x. Thus either 1x1is infinite or 1x1 E Z,in which case 1x1 is the least positive integer n for which X" = 1.

a.

Proposition 1.6. Suppose x is an element of finite order n in a group G, and suppose xm= 1,0 < m E Z. Then n I rn.

+

Proof. Write m = nq r with q , r E Z, 0 I r < n. Then 1 = x"' = = xr, so r = 0. ~ " q =+ (xn)qxr ~

Corollary. If G = (x) is cyclic of finite order n and k I n, 0 < k (xnik)is the unique subgroup of order k in G. rn

E

Z,then

Proof. Clearly xnikhas order k. If x s has order k, then xSk= 1, so n 1 sk, say then xs = (x"")~E (xnik).

= sk. But

Exercise 1.5. If x and yare commutingelements (ie., xy = yx) in a group G, show that lxyl divides LCM(lx1, I y l ) ; equality holds if (x) n ( y ) = 1.

Proposition 1.7. A subgroup of a cyclic group is cyclic. Proof. Say H I G = (x). If H = 1 it is cyclic, so suppose H # 1. Choose xn E H with 0 < rn E Z and m minimal. If xk E H write k = mq + r, with q, r E Z and 0 I r < m. Then x' = x ~ - " = ' ~xk(xm)-qE H , so r = 0 by the minimality of m, and hence xk = (x"')~E H. Thus H = (xm) is cyclic. Exercise 1.6. (1) Suppose G = (x) is infinite. (a) If rn # k in H , show that x m# x k . (b) Show that G = (x) = (x-l), but that G # ( x k ) if k # 1, - 1. (2) Suppose G = (x) is finite of order n. (a) If m, k E Z,show that x"' = xk if and only if rn = k(mod n), i.e., n 1 rn - k. (b) Show that G = (x'")

8

I Groups

if and only if (m, n) = 1, i.e., m and n are relatively prime. Thus the number of different generators for G is 4(n), being Euler's totient function.

If H I G and x , y E G we say that x and y are congruent mod H, and write x = y(mod H ) , if y - ' x E H . It is easily checked that congruence mod H is an equivalence relation on G, so G is partitioned into equivalence classes. Note that x 3 y(mod H ) if and only if y - ' x = h E H , or x = y h for some h E H. Thus the equivalence class containing y is { y h : h E H } , which we write as y H and call the leji coset of H containing y . Note that x H = y H if and only if x = y(mod H ) . The number (possibly infinite) of distinct left cosets of H in G is called the index of H in G and is denoted by [G:H I .

Theorem 1.8 (Lagrange's Theorem). If G is a finite group and H IG, then [HI is a divisor of [GI. In fact IGI = [ G : H ] [HI. Proof. The mapping h H x h is a 1-1 correspondence between H and the left coset x H so (HI = lxHl for all x E G. Since G is the disjoint union of [ G : H ] left cosets, each with IHI elements, the theorem follows. A homomorphism f from a group G to a group H i s a function f : G -+ H such that f ( x y ) = f ( x ) f (y ) for all x , y E G. If f is 1-1 it is called a monomorphism; if it is onto it is called an epimorphism. If f is both 1-1 and onto it is called an isomorphism. In that case f - ' is also an isomorphism, from H to G, and we say that G and H are isomorphic. When G and H are isomorphic we write G z H . A homomorphism from G to G is called an endomorphism of G,and an isomorphism of G with itself is called an automorphism of G. We shall write Aut(G) for the set of all automorphisms of a group G. If f:G -,H is a homomorphism, then the kernel of f is defined as kerf = { x E G : f ( x )= 1 E H } .

Proposition 1.9. If f : G -+ H is a homomorphism, then kerf 5 G, and f is a monomorphism if and only if ker f = 1. Proof. Since f(1) = f ( 1 . 1) = f ( l ) f ( l), we may multiply by f(1) - ' to see that f(1) = 1. Thus

1 = f(1) = f ( x x - l ) = f ( x ) f ( x- l ) , so f ( x - l )

= f ( x )-

for all x E G. If x , y

f(xy-')

E ker

f , then

= f ( x ) f ( y - l )= f ( x ) f ( y ) - ' =

1 * 1 = 1,

so x y - ' E ker f, and thus ker f < G. For x , y E G we have f ( x ) = f ( y ) if and only if 1 = f ( x ) f ( y ) - ' = f ( x y - ' ) , i.e., if and only if x y - ' E ker f. If ker f = 1, then x y - ' = 1, or x = y , so f is 1-1. The converse is clear since f(1) = 1.

1 Groups, Subgroups, and Homomorphisms

9

EXAMPLES 1. Take G = R, the additive group of real numbers, and H = ( r E R:r > 0), with ordinary multiplication. Define f :G + H by setting f ( r ) = e'. Then f is an isomorphism and the inverse isomorphism is the natural logarithm function. 2. Let G be the group of symmetries of an equilateral triangle, with notation =1 as in Example 6, p. 3, and let H = { f 1) under multiplication. Define f(4) for each rotation 4 and f(8) = - 1 for each reflection 8. Then f is a homomorphism. (Verify.) 3. If G is an abelian group and n E L, then the function f : G -,G defined by f ( x ) = x" for all x E G is an endomorphism of G since (xy)" = x"y" for all x, Y E G. 4. Suppose (x) and ( y ) are cyclic groups with 1x1 = IyI. Then the function f:( x ) (y), given by f ( x k )= y k ,is well defined. That is clear if 1x1 is infinite, whereas if 1x1 = I yl = n and if x" = x k , then x " - ~= 1, so n 1 m - k, and thus also y m - k = 1, or y" = y k . It is easy to see that f is a homomorphism. If X" E ker f , then y" = 1, in which case X" = 1 since 1x1 = Iyl, so f is 1-1. It is clearly onto, so it is in fact an isomorphism. We have established that any two cyclic groups of the same order are isomorphic. 5. If G = S , and H is cyclic of order 6, then G and H are not isomorphic since His abelian and G is not. In general, in order to establish nonisomorphism it is necessary to exhibit some group-theoretical property that one of the groups has and the other does not have. 6 . The gist of the remarks in Examples 6 and 7, pp. 3-4, is that the group D, of symmetries of the triangle is isomorphic with the symmetric group S , , and the group D4of symmetries of the square is isomorphic with a subgroup of the symmetric group S,.

Exercise 1.7. Show that D4 is not isomorphic with the quaternion group QzProposition 1.10. If G is a group, then Aut G is a group with composition as multiplication.

Proof. Since Aut G is a subset of Perm(G) we may apply Proposition 1.4. If f,g E Aut G and x, y E G, then (fg-')(xy) =f(s-'(xy)) = f(g-'(x)g-'(y)) = f ( g - l ( x ) ) f ( g -' ( Y ) ) = ( f S - ' ) ( x ) ( f g - ' ) ( Y ) ,

so fg -' is a homomorphism. Also, fg - ' E Perm(G), so fg-' E Aut G and Aut G is a group.

10

If that

I Groups

1:G + H is a homomorphism, set K

= ker f . If x

E

G and y

E

K note

f ( x - ' Y X ) = f ( x - ' ) f ( Y ) f ( x )= f ( x ) - ' . 1 * f ( = 4 1, so x - ' y x E K . If we write x - ' K x for { x - ' y x : y E K } we have observed that x - ' K x c K for all x E G . Subgroups with the property just described are called normal subgroups. In general, then, if H I G we say that H is normal in G if x - ' H x G H for all x E G. Note that then H G x H x - ' E H , so in fact x - ' H x = H for all x E G. We write H d G, or G D H , if H is normal in G. Observe that if G is abelian, then every subgroup is normal. If G = S , , then the subgroup K = ((i 3 :)) is normal in G, but the subgroup H = ((i :)) is 123 1 2 3 1 2 3 - 1 2 3 not, since (1 3 2 ) ( 2 1 3 ) ( 1 3 2 ) - ( 3 2 1 ) # H . For any group G define the center of G to be Z ( C ) = { x E G : x y = y x , all y E G}.

It is easy to verify (Do so!) that Z ( C ) 4 G. If we had defined congruence of x and y in G modulo a subgroup H to mean that x y E H , then the equivalence classes would have been right cosets H x . If H 5 G, then X H H H x - ' is a 1-1 correspondence between the sets of left and right cosets of H in G.

'

Exercise 1.8. If H I G show that H 4G if and only if every left coset of H is also a right coset.

If H 4G write G / H for the set of all cosets of H in G. Note that IG/HI = [ G : H ] , and that IG/HI = lGl/lHl by Lagrange's Theorem if G is finite. If x H , y H E G / H , define a product ( x H ) ( y H )= x y H . The product is well defined since H 4 G, for if x H = uH and y H = OH, then x y H = x H y = u H y = uyH = uuH. It is a routine matter to verify that G / H , with the binary operation just defined, is itself a group with 1H = H as its identity element. It is called the quotient group, orfactor group, of G modulo H . Define a map 9: G + G / H by setting ~ ( x=) x H . Then XYH = XHYH = V ( X ) V ( Y ) , so 9 is a homomorphism, in fact it is clearly an epimorphism. We call 9 the canonical quotient map from G to G / H . Note that x E ker 9 if and only if x H = H , i.e. ker = H . When G is written additively and H 4 G we shall still write G / H for the quotient group of G modulo H , but the cosets of H in G are usually written in the form x + H rather than x H . For an important example take G = Z and let H = n7 = fnk:k E 7) for some n E Z, n > 0. Then, if x , y E (2, we have x = y(mod H ) if and only if x - y E H , i.e., if and only if n I x - y , which is the usual definition of V(XY) =

I Groups, Subgroups, and Homomorphisms

11

= y(mod n)from elementary number theory. Thus the cosets of H = nZ in Z are the classes of integers that are congruent mod n. By the division algorithm in Z every integer is congruent to one of 0, 1,2, . . ., n - 1, all of which _ _ _ are incongruent. if we write k for the coset k + nZ, then Z/nZ = {0,1,2,. . ., n - 1 }, a group of order n. This group is called the group of integers mod n and will be denoted by Z,. Note that Z,is cyclic, since Z,= ( 7 ) . x

Theorem 1.11 (The Fundamental Homomorphism Theorem). Suppose f :G -+ H is a homomorphism from G onto H , and that K = kerf. Then K 4G and G / K 2 H . Proof. We have already observed that K 4G. If x K = yK, then y - ' x E K , so 1 = f ( y - ' x ) = f(y)-'f(x), and hence f(x) = f(y). Thus if we set f ( x K ) = f ( x ) , then is a well-defined map from G / K onto H . Furthermore,

7

so f

f W Y K ) = f(XYK) = f ( X Y ) = f ( X ) f ( Y ) = f(XK)f(YK), is a homomorphism. Finally, k e r f = { x K : f ( x )= 1) = { x K : x E K } = K ,

the identity element of G / K , so ,f is an isomorphism by Proposition 1.8. Let us examine Therorem 1. I 1 and its proof a bit more closely. Note that x K = ~ ( x )where , q is the canonical quotient map from G to G / K , Thus we have defined T ( x K ) = .T(~(x)) = .f'(x), so that f is the composition product of 9 and f.That information is captured conveniently in the following diagram.

The shading indicates that the diagram commutes, which means simply that

f

=

fv.

Even if f : G + H is not an epimorphism, its image, which we denote by Im(f). is a subgroup of H , as is easily seen by Proposition 1.4. Of course, f may be viewed as a homomorphism from G onto im(f), and the Fundamental Homomorphism Theorem (FHT) gains the slightly more general form G/ker f~ im(f) I H. Diagramatically we have

G

where K

=

ker f ,

f

bIm(f) I H,

12

I Groups

Proposition 1.12. Suppose f : G + H is an epimorphism, with kerf = K . Then L o f -'(L) = { x E G : f ( x )E L } is a 1-1 correspondence between theset of all subgroups L of H and the set of all subgroups of G that contain K . Furthermore, L H if and only if f-'(L) U G. Proof. If L I H , then f ' ( L )I G by Proposition 1.4, and it is clear that K 5 f - ' ( L ) . Note that f ( f - ' ( L ) )= L, and hence L w f - ' ( L ) is 1-1. If K 5 M IG, set L = f(M). Then L _< H and it is clear that M 5 f-'(L). If x E f - ' ( L ) , then f ( x ) E L = f ( M ) , so f ( x ) = f ( y ) for some y E M . It follows that y - ' x E kerf = K I M , so x E y M = M , and hence f-'(L) = M . Thus L-f-'(L) is onto. The proof of the statement concerning normality is left as an (easy) exercise. ~

Corollary. If K U G, then all subgroups of G I K have the form M I K , where K IM 5 G , and M j K 4 G j K if and only if M U G . Theorem 1.13 (The Freshman Theorem). Suppose G is a group, H Q G, K Q H , and K Q G. Then H I K U G I K , and GIH z ( G / K ) / ( H / K ) . Proof: If x K = y K in G I K , then x K H = y K H = x H = yH since K _< H . Thus f : G j K + G / H , with f ( x K ) = x H , is well defined. Also, f ( x K y K ) = f ( x y K ) = xyH = x H y H = f ( x K ) f ( y K ) ,so f is a homomorphism from GIK onto G / H . Since kerf = { x K : x H = H } = { x K : x E H } = H / K , we see that H I K d G / K and by the FHT that ( G / K ) / ( H / K z ) G/H. 2.

PERMUTATION GROUPS

Suppose that S is a set, G is a group, and 4: G + Perm(S) is a homomorphism from G into the group of all permutations of S. Then we say that G is a permutation group on S , or simply that G acts on S. If i) is a monomorphism we say that G acts faithfully on S . It is customary to suppress mention of 4 when there is only one permutation action under consideration, and if x E G, s E S , we simply write x s rather than ~ ( x ) ( for s ) the image of s under the permutation 4 ( x ) .As a result we have a mapping ( x , s)H xs from G x S to S , which satisfies the following requirements: (a) ( x y b = x ( y s ) , and (b) IS = s for all x , y E G, and all s E S . Conversely, any mapping from G x S to S that satisfies (a) and (b) serves to define a permutation action of G on S if we set ~ ( x ) ( s=) x s for all x E G and all s E S . If G is a group acting on a set S and s E S we define the stabilizer of s in G to be Stab,(s) = ( x E G : x s = s}

13

2 Permutation Group

and we define the orbit of s under G to be

-

Orb,(s)

= {XS:X E

G}.

If we define s t in S to mean that s = xt for some x E G, then is an equivalence relation and the orbits are its equivalence classes. Thus S is a disjoint union of orbits. If OrbG(s)= S for some (and hence for every) s E S , we say that G is transitive on S . This means that given any s and t in S there is some x E G such that xs = t. h-

Proposition 2.1. If a group G acts on a set S and s E S , then StabJs) 5 G and [G:Stab~(s)]= (OrbG(s)l. Proof. Set H = Stab,(s). If x, y E H , then ys = s, so s = y - ' s , and hence ( x y - l j s = x ( y - ' s j = xs = s, so that x y - E H and H 5 G. We have xs = ys if and only if y - l x E H if and only if x H = y H . Thus X S H x H is a well-defined

'

1-1 correspondence between Orb,(s) and the set of left cosets of H StabG(S)in G.

=

As a first example take S = G and define 4 ( x ) y = x y to be the usual product in G. Then ker 4 = {x E G : x y = yfor all Y E G} = 1

so G acts faithfully. This faithful representation of G as a permutation group on S = G is called the lejt regular representation of G.

Theorem 2.2 (Cayley's Theorem). If G is any group, then G is isomorphic with a transitive group of permutations acting on a set S (viz., S = G). Proof. Only the transitivity remains to be verified. Given y, z take x = zy-' E G. Then x y = z y - ' y = z.

E

S = G,

If G is a group and x, y E G, then the conjugate of x by y is defined to be x y = y - ' x y . Note that x y z = (xy)'.

As a second example of a permutation action take S = G again, but this time if x E G and y E S, then the action of x on y is conjugation of y by x - i.e., $(x)y = y"-' = x y x - ' . If x, y E G and z E S , then (#)(xy)z= z(xY)r' = z Y - ' x - ' = ( z y - ' ) . - '

= 4 ( x ) ( z y - ' ) = &x)$(y)z,

so 4 is a homomorphism. Note that the kernel of the action is { x E G : y = x y x - I , all y E C} = Z(G), the center of G, so the action is faithful if and only if Z(G) = 1. If y E G, then Orb,( y) is called the conjugacy class of G containing y and is denoted by cl( y). The stabilizer of y E G is { x E G : x y = y x } , which is commonly called the centralizer of y in G and denoted C,(y). The next proposition is a direct consequence of Proposition 2.1.

14

I Groups

Proposition 2.3.

If G is a group and x

E

G, then Jcl(x)l = [G:C,(x)].

Note that Icl(x)l = 1 if and only if xy = x for all y E G, which is if and only if x E Z(G).If G is a finite group, let us choose representatives x i , x 2 , . . ., xk of the conjugacy classes having more than one element (if there are any), i.e., the classes not in the center of G. Thus G is the disjoint union of Z ( G )and the sets cl(xi), 1 Ii Ik, and we have the class equation for G: IGI = Iz(G)I +

k

k

i=l

i=l

1 Icl(xi)I = Iz(G)I + 1 [G:G(xi)I

By Lagrange's Theorem (1.8) and Propositions 2.1 and 2.3 each summand on the right-hand side of the class equation is a divisor of (GI. Proposition 2.4. If p E E is a prime, 0 < n E Z, and G is a group with (GI= p", then Z ( G ) # 1.

Proof. In the class equation IGl and all Icl(xi)lare divisible by p , so ( Z ( G ) ( must be divisible by p and hence must be larger than 1. Exercise 2.1. If G / Z ( G )is cyclic show that G = Z ( G ) is abelian. Conclude that if p is a prime and J G J= p 2 , then G is abelian. Next take S to be the set of all subsets of a group G and define an action of G on S by setting (P(x)A = xAx-' = A".' for x E G, A E S (we agree that 0"= 0). The elements of Orb,(A) are called the G-conjugates of A ; Stab,(A) is called the normalizer of A in G and is denoted by &(A). Proposition 2.5. If G is a group and A c G, then the number of distinct Gconjugates of A in G is [G:N,(A)]. Theorem 2.6 (The Isomorphism Theorem). Suppose H, K IG and K I N,(H). Then K H = H K IG, H 4K H , K n H 4 K , and K H I H z K / K nH . Proof. By KH we mean, of course, (xy:x E K , y E H}. Let us show first that K H IG. Take x, u E K and y, u E H . Then (xy)(uv)-' = X J T - ' U - '

= xu-'Uyv-'u-'

= ( X U - ' ) ( Y V ~ ' )E " -KH, '

since K I N,(H). Note that xy = yx-'x and yx = xy", so HK = K H . Since K 5 N,(H) it is immediate that H U K H . Define f : K + K H / H by setting f(x) = xH. Then f is a homomorphism; it is onto since xyH = xH, all y E K . Since kerf = {x E K :xH = H } = {x E K :x E H} = K n H, the result follows from the FHT. Exercise 2.2. (1) Suppose, in the context of Theorem 2.6, that HK is finite. We may conclude that I K H J * IK n HI = JKI [HI. Suppose more generally that A and B are arbitrary finite subgroups of G and define

2 Permutation Groups

15

-

AB = { a b : a E A and b E BJ (it may well not be a subgroup). Show that IABl . IA n €31 = IAl . (BI.[Hint: The relation (al, b , ) (a,, b,) if and only if a , b , = a,b, is an equivalence relation on the Cartesian product A x B. What are the equivalence classes?] (2) If C I B I G show that [G:C] = [G:B][B:C]. Conclude that if [G:A] and [G:B] are finite, then [ G : A n B] is finite. Suppose G is a group and H I G, and set S = { x H : x E G}, the set of all left cosets of H in G. Then G acts naturally on S, with &x)yH = xyH, all x,y E G. An element x E G is in the kernel of the action if and only if x y H = y H , or y-'xy E H , or x E y H y - ' for all y E G. Thus the kernel is K = n { H ' : z E G I , and the action is faithful if and only if all the conjugates of H intersect in 1. If [G:H] isfinite,say [G:H] = n,thenthepermutationactionof G o n S i s a homomorphism 4 from G into Perm(S), which is, of course, isomorphic with the symmetric group S, and has order n ! . By the FHT we see that G / K z Im(4), and so [G:K] I n ! by Lagrange's Theorem. Exercise 2.3. Suppose G is finite, H I G, [C:H] = n, and [GIt n ! . Show that there is a normal subgroup K of G, K # 1, such that K I H .

Theorem 2.7 (Cauchy). Suppose G is a finite group, p p 1 IGI. Then G has an element x of order p .

E

Zis a prime, and

Proof (McKay [26]). Let S be the subset of the p-fold Cartesian productG x G x " - x Gconsistingofall(x,,x2,..., x,)forwhichx,x ,... x , = 1, except that the element (1, 1, . . . , 1) is excluded. If ( x , , x 2 , . . .,xp)is to be in S we may choose x I , x,, . . ., x p - arbitrarily; then x p = xi!, . . . x;' is determined. Thus IS( = IGIP-' - 1, so p t ( S 1 . Let C = ( 2 ) be a cyclic group of order p (C has no a priori relationship to G). We may define an action of C on S by specifying that z ( x l , x 2 , ..., x p ) = (x2,x 3 , ..., x p , x , ) since x 2 x 3 . . . x p x l = (xl . . . x J X 1= 1. Since p is prime, each C-orbit in S must have either 1 or p elements by Proposition 2.1. If all orbits had p elements, then IS1 would be a multiple of p , so there must be an orbit with one element, which necessarily has the form (x,x,. . . ,x), x # I , hence xp = 1.

For an application of Cauchy's Theorem suppose G is a group of order 28. Then G has an element of order 7 and hence has a cyclic subgroup H of order 7. As in the discussion preceding Exercise 2.3, G acts on the set of left cosets of H , and there is consequently a homomorphism 4: G -+ S,. But IS,] = 24 and IGI = 28x224, so if K = ker 4, then K # 1. Since 1 # K = ( H " : x E G ) I H,of primeorder, it is clear that K = Hand the subgroup of order 7 must be normal.

0

Exercise 2.4. If IG( = p", p a prime, show that G has subgroups Go, GI, ... , G. with 1 = Go I G, I ... I Gn = G such that [Gi:Gi-,] = p , 1 2 i 2 n. [ H i n t : Try induction, choose G, 2 Z(G),and consider G/G, .]

16

I Groups

3. THE SYMMETRIC AND ALTERNATING GROUPS The symmetric group S,, acts on the set S = { 1,2, . . . , n), so if a is a fixed element in S,, then the cyclic group (a) also acts on S . Write T,, T,, . . . , Tkfor the (0)-orbits in S and define permutations ol, a , , . . ., C T ~as follows: a, acts as a does on ?; but acts as the identity on the remainder of S . Clearly a = a1c2... ak, and the permutations a l , a,, . ..,a k all commute with one another since the orbits TI,.. . , are pairwise disjoint. Since is a (a)-orbit, its elements are permuted cyclically by a, i.e., if s E and 1 TI = ni, then s is sent to as, which is sent to a2s, etc., until finally d " ' s is sent back to s by a. In general a permutation 4 that permutes a subset Tof S cyclically and fixes all elements of S\T is called a cycle (a k-cycle if I TI = k). A cycle 4 admits the convenient notation

4 = (s 4 s 42s '.. $k-'s), each element being mapped by 4 to the next element to the right except that

the last element is mapped back to the first, and elements not mentioned are understood to be fixed. For example, the permutation 4 = (i :) is a cycle in S,, and it can be written as 4 = (13524), or as 4 = (35241),etc. If cycles 41and 4, permute the elements of T, and T,, and if T, n T, = 0, we say that 4, and 4, are disjoint cycles. Disjoint cycles clearly commute with one another. A perusal of the above discussion serves to establish the next proposition. Proposition 3.1. If a E S,, then a can be expressed as a product of disjoint cycles. The expression is unique except for the order of occurrence of the factors, since disjoint cycles commute. It is customary to suppress 1-cycles (i.e., fixed points) when writing a permutation as a product of cycles. For an example take (T = (: $ 8 2 ) E S9. The (a)-orbits are { 1, 2, 7}, {3}, and 14, $ 6 , 8,9}, and a may be written as a = (127)(45869),the 1-cycle (3) being suppressed. Note that if 0 is a k-cycle, then 101 = k. If a = a l o 2... a,, disjoint, with ai a k,-cycle, then 101 is the least common multiple of k,, k,, . . ., k, (see Exercise 1.5). The inverse of a k-cycle is easily obtained by writing the entries in the cycle in reverse order. For example, if a = (12345),then a - l = (54321). A 2-cycle (ab) is called a transposition. Every cycle is easily expressed as a product of transpositions, e.g.,

: : : ::

( 1 2 3 . . . k ) = ( l k ) ( lk

-

1)...(13)(12).

Thus by Proposition 3.1 every permutation can be written as a product of transpositions. We say that a E S,,is euen if it is possible to write a as a product

3 The Symmetric and Alternating Groups

17

of an even number of transpositions; otherwise we say that o is odd. Thus a product c1o2 is even if o, and u2 are both even or if they are both odd, but 0102is odd if one of the factors is even and the other is odd. If we set H = { f l} under multiplication it follows that the function f : S, + H , defined by 1 -1

if o is even, if G is odd,

is a homomorphism. The kernel off is the normal subgroup of S,, consisting of all even permutations; it is called the alternating group on n letters and is denoted by A,,. Proposition 3.2. The alternating group A , has index 2 in the symmetric group S,, if n 2 2. Proof (Spitznagel [35]). If we show that the homomorphism f above maps S, onto H, then &/A, z H by the FHT, so IS,/A,,I = [$,:A,] = IHI = 2. Thus we need only prove the existence of an odd permutation. If the transposition (12) were even we would be able to write the identity permutation 1 as a product of an odd number of transpositions. Suppose we have done so, 1 = (ab) . . ., using the smallest possible number of as appearing. At least one more a must appear, since la # b, so suppose (ac) is the next one to the right. Note that (de)(ac)= (ac)(de)if (de) and (ac) are disjoint, and (dc)(ac) = (ad)(cd), so we may move the second a to the left and write 1 = ( a b ) ( a f ) " ; with the same minimality conditions met. But now if b = f we may reduce the number of transpositions by 2, and if b # f, then (ab)(af)= ( a f ) ( b f )and we may reduce the number of as, in both cases a contradiction. Thus (12) is odd and the proof is complete. Exercise 3.1. If o E S,, show that o can not be written once as the product of an even number of transpositions and another time as the product of an odd number.

Let G = (ala2 ... a k )be a cycle in S,,, let z be any other element of S,, and i Ik , and agree that consider the conjugate 7ozC1. Write b, = T(u,), 1 I ak+l = a,, b k + , = b , . Then zoz-'(b,) = 7oz-'(zai)

Furthermore,

or

TGT-'(S)

=

= s if s

to(ai)= z(a,+,) = b i + , ,

# b , , . . . , bk (Why?), and so

1I i Ik.

18

I Groups

and the conjugate is another k-cycle, obtained by replacing each a, by T(u,). For example, if CJ = (324) and T = (13524), then T C J T - ' = (541), since ~ ( 3 = ) 5, T(2) = 4, and T(4) = 1. When CJ E S, is expressed as a product of disjoint cycles suppose there are kj j-cycles, 1 Ij I n (so that k , + 2k, 3k, + ... + nk, = n). Then we say that 0 has cycle type ( k , , k , , .. .,k,,). For example, in S9 the permutation 123456789 CJ = (, 6 4 ) = (127)(45869) has k, = 1, k, = 1, k , = 1, and all other ki = 0, so its cycle type is (1,0, I , 0, 1,0,0,0,0).

+

,

Proposition 3.3. If

CJ E

S,, then the conjugacy class cl(a) consists of all

4 E S, that have the same cycle type as CJ.

Proof. Write CJ = C J ~ C J. .~ . o m as a product of disjoint cycles (include all 1-cycles throughout this proof). Then, for any T E S,, TOT-'

= TCJ,T-'

TO,,,-^,

5 ~ ~ ~ t - l

so T C J T - ~ has the same cycle type as CJ by the discussion above. Conversely, suppose that 4 E S, has the same cycle type as CJ.Say CJ

6

= ( u , o , ...)( blb, . . . ) . . * , = (c,c2 ...)(d 1d 2 . . . ) ...,

with the cycles appearing in order of increasing lengths in both cases. Define T E S, by setting T ( ~ J= c,, T ( h j ) = d ; , etc. As above we see that TCJT- = 4, so 4 E cl(CJ).

Corollary. If n 2 3, then Z(S,) = 1. Exercise 3.2. (1) Write out the conjugacy classes explicitly in S, and S,. (2) What are the conjugacy classes in A,? (3) Since 1 A,( = 12, any subgroup of order 6 would be normal. Use (2) to show that A , has no subgroup of order 6. Conclude that the converse to Lagrange's Theorem is false. Proposition 3.4. If n 2 5, then all 3-cycles are conjugate in A , . Proof. Let ( i j k ) be any 3-cycle. Then (ijk) = ~ ~ ( 1 2 3 ) 0for - ' some CJ A , we are finished. If not set T = 445). Then z E A , and

E

S,. If

0E

~(123)z-I= 0(45)(123)(45)6' = 0(123)0-' = (ijk).

Proposition 3.5. If n 2 3, then A , is generated by 3-cycles. Proof. If i, j, k, and m are distinct, then (ij)(ik)= (ikj) and (ij)(km) = (jmk)(ikj). The result follows.

A group G is called simple if its only normal subgroups are 1 and G. For example, any group of prime order is simple by Lagrange's Theorem. The

19

4 The Sylow Theorems

concept is of central importance in the structure theory of finite groups. The next theorem will be important later when we apply the Galois Theory to the solution of polynomial equations.

Theorem 3.6. If n # 4, then the alternating group A , is simple. Proof. A , , A , , and A , are simple since their orders are 1, 1, and 3. Suppose n 2 5 and take H d A,, H # 1. By Propositions 3.4 and 3.5 it will suffice to show that there is a 3-cycle in H in order to conclude that H = A,. Choose a prime p such that p 1 IHI, and choose an element Q E H of order p (Cauchy's Theorem, 2.7). Then o is a product of k disjoint p-cycles for some k. If p = 3 and k = 1 we are finished. Otherwise there are four cases to consider. Case 1: p > 3. Say CT = (alu2. . . u p ).... Then a(alu2a3)Q-'(a~a3a2)

= (a2a3a4)(ala3a2)

= (ala4a2)

H.

Case 2: p = 3 and k > 1. Say cr = (a,a2a,)(a,a5a,).... Then d a 1 a 2 u 4 ) o - '(aIa4a2)

= (a2a3a5)(a,a4a2) = (ala4a3a5a2)E

H,

and we are back in Case 1. Case 3: p = 2, k = 2m 2 2, and there is some letter a5 fixed by

o = ( ~ , a 2 ) ( 0 3 ~ 4.... ) Then

0.

Say

H.

dala2%)(r-1(a1%ff2) = ( a 2 a l a 5 ) ( a , a 5 a= , ) ( ~ 1 ~ 2 E~ 5 )

Case 4: p = 2 and k = 2m 2 2. If o = (a,a,)(a,a,), then we are back in .... Then Case 3, so suppose Q = (a1a2)(a3a4)(a5a6) a(a1a2a5)o-1(ala5a2)

= (a2a1ah)(a1a5a2)

= (ala5)(a2a6)

H.

Again we are back in Case 3 and the proof is complete. Exercise 3.3. Find a normal subgroup of order 4 in A , . Thus A , is not simple. 4. THE SYLOW THEOREMS We have observed (Exercise 3.2.3) that the converse to Lagrange's Theorem fails. The converse does hold, however, for prime power divisors of the order of a finite group by a remarkable theorem due to the Norwegian mathematician Sylow. If G is a finite group and p E Zis a prime, then a p-Sylow subgroup of G is a subgroup P such that IPI = pk is the highest power of p that divides IG/ (we write p' 11 ICl to indicate that pk I (GI but p k t ' 4 IGl).

Theorem 4.1 (The First Sylow Theorem). If C is a finite group and p E B is a prime, then G has a p-Sylow subgroup.

20

I Groups

Proof. Induction on IGI. The result is trivial if [GI = 1 or if p,+'lGl, so assume that p I [GI > 1 and that the theorem holds for all groups of smaller G, H # G, such that p , + " G : H ] , then H has a order. If there exists H I p-Sylow subgroup P which is also p-Sylow in G. Assume then that p I [ G : H ] for all proper subgroups H of G . By the class equation (p. 14) k

we see that p I IZ(G)l. By Cauchy's Theorem (2.7) there is an element x E Z(G) of order p . Set K = (x). Then K 4G since x E Z(G), and if p" (1 IGl, then pm-' 1) IG/KI. By the induction hypothesis G / K has a p-Sylow subgroup P, = P / K of order p " - l , where P I G. But then \PI = [PI1 . llyl = p", so P is p-Sylow in G.

As a consequence of Exercise 2.4 and the First Sylow Theorem we see that if p is a prime and pi divides [GI then G has a subgroup of order p i . Exercise 4.1. A group G is called a p-group ( p a prime) if every element of G has order a power of p . Show that a finite group G is a p-group if and only if IGI is a power of p . G is Proposition 4.2. Suppose P is a p-Sylow subgroup of G and that H I a p-group. Then H n N,( P ) = H n P . Proof. Set H , = H n N,(P). Clearly H n P I H,. By the Isomorphism Theorem (2.6) we have H , P / P z H , / ( H , n P), and so [ H , P : P ] = [ H , : H , n P ] = p" for some rn, since H , I H, a p-group. Thus IH,PI = [ H , P : P ] . IPJ= p" . /PI, and H , P is a p-group. Thus H , P = P since P is p-Sylow; hence H , I P, and so H , = H n P.

Theorem 4.3 (The Second Sylow Theorem). Suppose G is a finite group, p E Zis a prime, Pis a p-Sylow subgroup of G,and H I G is a p-group. Then H is contained in some conjugate P" of P. In particular, all p-Sylow subgroups of G are conjugate with one another. Proof. Let Y be the set of all G-conjugates of P; then H acts on Y by conjugation. Let Y,,Y,, . . . ,Y;,be the H-orbits in Y and choose P, E Yi, 1 I i 5 k . Then Stab,,(E)

=

{hE H : P b =

c}= H n NG(&)= H n fl

by Proposition 4.2, so IYil = [ H : H n 51 by Proposition 2.1 Note that IYI = [ G : N G ( P ) ]by Proposition 2.5, so p $ I Y I . Since 1 9 1= C:=,l~ql= [ H : H n p i ] and each [ H : H n e ] is a power of p we must have [ H : H n P , ] = p o = 1 for some i. Thus H n 4 = H a n d H IE = P" for some x E G.

cs=l

4 The Sylow Theorems

21

Corollary. If a finite group G has a unique p-Sylow subgroup P for some prime p , then P 4G. Theorem 4.4 (The Third Sylow Theorem). If G is a finite group and p E if is a prime, then the number of distinct p-Sylow subgroups of G is congruent to 1 modulo p . Proof. Let P be p-Sylow and again let Y be the set of all G-conjugates of P . Let Pact on 9 'by conjugation, with P-orbits Y l ,Y2,...,Yk. Note that { P ) is an orbit with just one element, say 9, = { P I . If 4 E q, 2 < i I k, then P, # P and then Stab,(e) = P n N , ( c ) = P n fl # P by Proposition 4.2. Thus 18.1= [P:Pn 41 is divisible by p if 2 < i I k, and so 1 9 1= 1 + 1.441 = l(mod p ) .

cf=,

Exercise 4.2. If P is a p-Sylow subgroup of G show that N,(N,(P)) NdP).

=

The Sylow Theorems can be used to show the existence of nontrivial normal subgroups in groups of certain orders, and hence to show that the groups are not simple. A few easy examples of applications of the theorems follow. For the examples let us write S1, S2, and S3 for the first, second, and third Sylow Theorems, respectively. EXAMPLES

1. Suppose (GI = 28. By S3 the number of 7-Sylow subgroups of G is in the list 1, 8, 15, 22, . . .. But by S2 and Proposition 2.5 the number must be a divisor of 28, and 1 is the only divisor of 28 in the list. Thus there is only one 7-Sylow subgroup, of order 7, and it is normal in G by the corollary to S2. 2. Suppose p and 4 are primes, with q < p , suppose [GI = p4, and suppose that p f 1 (mod 4). By S2, S3, and Proposition 2.5 the number of p-Sylow subgroups of G is 1 + k p for some k E Zsuch that 1 + k p I p4, hence 1 + k p 14. Clearly k = 0 since 4 < p , and so G has a normal subgroup P of order p . Likewise the number of q-Sylow subgroups has the form 1 + mq, with 1 + m4 1 p , so either 1 + m4 = 1 or 1 + m4 = p . But we have assumed that p f 1 (mod q), so G also has a normal subgroup Q of order 4 . Both P and Q are cyclic, say P = (x) and Q = ( y). Then x - 'y- ' x y E P n Q = 1 since P and Q are normal. Thus x y = y x and so lxyl = 1x1 IyI = p q and G = ( x y ) is a cyclic group. 3. Suppose the order of G is 56 = 23 . 7. If G has just one 7-Sylow subgroup it is normal. Otherwise G has 8 different 7-Sylow subgroups, and hence it has 8 . 6 = 48 elements of order 7. But that leaves only 8 more elements in G, which must constitute a unique (and hence normal) 2-Sylow subgroup. Thus G must have either a normal subgroup of order 7 or of order 8, and cannot be a simple group.

22

I Group

Exercise 4.3. Show that the only simple groups of order less than 36 are those of prime order.

Proposition 4.5.

If G is a simple group and J G J= 60, then G 2 A , .

Proof. By Exercise 2.3 G can not have subgroups of index 2, 3, or 4. By S3 G must have 6 5-Sylow subgroups, 10 3-Sylow subgroups, and either 5 or 15 2-Sylow subgroups. Suppose there is no subgroup of index 5 in G, so there are 15 2-Sylow subgroups. Let TI, T, be distinct 2-Sylow subgroups, and suppose Tl n T2 # 1; hence ITl n T21 = 2. Set T = TI n T2 and F = (Tl u T,). Clearly T I Z ( F ) , so T U F and F # G. But also IF1 > 4 and 4 I I FI since TI I F , so IF1 2 12. Consequently [ G : F ] I 5, a contradiction. Thus TI n T2 = 1. But then G has 45 nonidentity elements in 2-Sylow subgroups, 20 in 3-Sylow subgroups, and 24 in 5-Sylow subgroups, for a total of 1 + 45 + 20 + 24 = 90 elements, again a contradiction. Thus G must have a subgroup of index 5, and hence there is an isomorphism from G to a subgroup H of order 60 in S , . If H # A , , then H n A , would have index 2 and hence would be normal in A , , contradicting the simplicity of A , (see Exercise 12.22). Thus G z A,.

5. SOLVABLE GROUPS, NORMAL AND SUBNORMAL SERIES We know from the FHT that every homomorphic image of a group G is isomorphic with a quotient group of G. Among all possible abelian homomorphic images of G we will exhibit one that is in a sense maximal. We take this opportunity to present, by way of example, a notion of universality that will prove useful in later sections and chapters. A pair ( U , E ) is called universal for a group G (with respect to abelian epimorphic images) if U is an abelian group, 6: G + U is an epimorphism, and if given any abelian group A and homomorphism f:G + A , then there is a unique homomorphism g : U + A for which f = gs, i.e., the diagram

is commutative. We say that f can be “factored through” U . Proposition 5.1. If a universal pair ( U , E ) exists for a group G then U is unique (up to isomorphism).

5 Solvable Groups, Normal and Subnormal Series

Proof.

or E~

= glE

Let ( U , ,E

and

~ be )

23

another universal pair for G. We have

E = g2El. Thus E~

= g1g2s, and

E =

g2glE. But then we have

and by the uniqueness in the definition of a universal pair we see that g2gl = l,,, the identity map on U . Similarly glg2 is the identity map on U , , and so g1 and g 2 are inverse isomorphisms.

Exercise 5.1. If ( U , E ) is a universal pair for a group G and h E Aut(U) show that ( U , he) is also universal for G. Conversely if (U, E ~ is) universal for G show that = h~ for some h E Aut(U). If x , y E G, then their commutator is defined to be [ x , y ] = x - l y - ' x y . Note that [ x , y ] = 1 if and only if x and y commute. If f : G -+ A is any homomorphism from G to an abelian group A, then f ( [ x ,Y l ) = f h ' f W 1 f ( X ) j ' ( Y= ) 1

and every commutator is in the kernel of f . In particular, if (V, E ) is a universal pair for G,then every commutator [ x , y ] is in the kernel of E, since U is abelian. Exercise 5.2. If x , y , z E G show that [ x , y ] - = [ y , x ] and [ x , y]" = [x', y'], so inverses and conjugates of commutators are commutators.

For any group G define the derived group G' of G to be the subgroup of G generated by all the commutators, i.e., G' = ( [x, y ] : . ~y ,E G). It is clear from Exercise 5.2 that G' 4G. Note also that if x , y E G, then x - ' y - ' x y G ' = G', so xyG' = yxG', and GIG' is abelian.

Theorem 5.2. If G is a group, then G has a universal pair ( U , 8 ) . In fact we may take U to be C/C' and E : G + U to be the canonical quotient map. Proof. Suppose f : G + A is a homomorphism, with A abelian. Since every commutator is in the kernel of j' we have G' I k e r f . Thus if xG' = yG' or y - ' x E G', then f ( y - ' x ) = 1, or f ( x ) = f(y), and we may define

24

I

Group

g(xG’) = f ( x ) . Clearly g : U + A is a homomorphism (since f is a homomorphism) and gE = f.If also g1: U -+ A is a homomorphism such that glE = f, then glE = gE and so g1 = g since E is an epimorphism. It follows that (V, E ) is a universal pair for G.

Exercise 5.3. (1) Find G‘ if G = S, , S,, or A , . (2) If G’ I H I G show that H 4G. (3) Use Exercise 5.2 to show that if K U G, then K‘ U G. (4) Suppose f : G + H is an epimorphism, with ker f = K . Show that H is abelian if and only if G’ I K . In particular, a quotient group G / K is abelian if and only if G’ I K . Conclude that G/G’ is a “maximal” abelian epimorphic image of G in the sense that G’ is a minimal normal subgroup L for which G / L is abelian.

Since G’4G we see that G” = (G’)’ 4G by Exercise 5.2.3. Set G‘O) = G, G(’) = G’, G ( 2 )= G“, . . ., and in general G ( k +I ) = (G(k))’. Then G‘k)4G for all k, and the sequence G = G(0) 2 G(1) 2 G ( 2 )2 . . . 2 G(k)2 ... is called the derived series of G. A group G is called solvable if G‘k)= 1 for some G ( k in ) the derived series of G.Note that any subgroup of a solvable group is solvable. For example, if G is abelian, then G is obviously solvable since then G’ = 1. If G = S,, then G’= A , and G” = A; = 1, so S3is solvable. If n 2 5, then A , is nonabelian and simple, and A; 4A , , so A; = A , . Thus A , is not solvable if n 2 5. Exercise 5.4. (1) Find the derived series for S, and conclude that S, is solvable. (2) Show that S:, = A , if n # 2. Conclude that S, is not solvable if n 2 5. [Hint: (ijk) = ( j r k ) -‘ ( u s ) - ’( jrk)(ijs);see Proposition 3.5.1 A subnormal series for a group G is a sequence (finite or infinite)

G = Go 2 Gl 2 G2 2 ...,

where Gi+ 4Gi for all i. The subgroups Gj are called subnormal subgroups of G.A subnormal subgroup is not necessarily a normal subgroup; look at the group of symmetries of the square for an example. The successive quotient groups Gi/Gi+ are called the factors of the subnormal series. A subnormal series G = Go 2 G1 2 G2 2 - - - is called a normal series if Gi 4G for all i. Thus the derived series of a group G is a normal series, for example. The length (finite or infinite) of a subnormal series for G is the number of nontrivial factors Gi/Gi+1, or equivalently the number of strict inclusions in the series.

25

5 Solvable Groups, Normal and Subnormal Series

Theorem 5.3. A group G is solvable if and only if it has a subnormal series G = Go 2 G, 2 G2 2 ... 2 G, = 1 with abelian factors. Proof. +: A segment of the derived series has the desired properties. e: Suppose G = Go 2 G, 2 ... 2 G, = 1 is a subnormal series with abelian factors. Since Go/G, = G/G, is abelian we have G’ I G, by Exercise 5.2.4. Similarly G2 2 G‘, 2 (G’)’ = G”’ since G , / G , is abelian. InducGk for all k, so G(“) = 1 and G is solvable. tively Q k ) Theorem 5.4. Suppose K 4C. Then G is solvable if and only if both K and G / K are solvable. Proof. * : We observed earlier that any subgroup of a solvable group is solvable. Since the canonical quotient map q: G + G / K is an epimorphism every commutator [ x K , yK] in G / K is the image q ( [ x , y]) = [qx, qy] of a commutator in G. Thus (GjK)‘ = q(G’), and likewise ( G / K ) ‘ k= ) v(G(~)), all k, so G / K is solvable. -=: Choose subnormal series with abelian factors for K and for G / K , say K = K,,2 K , 2 ... 2 K , = 1 and G / K = G,/K 2 G , / K 2 ... 2 Gk/K = K . SinceGi/Gi+,z (Gi/K)/(Gi+,/K)bythe FreshmanTheorem(l.13)weseethat G

=

Go 2 ... 2 Gk = K

=

KO 2 K , 2 ‘ “ K , = 1

is a subnormal series for G with abelian factors, so G is solvable by Theorem 5.3. Exercise 5.5. Show that any finite p-group is solvable (use induction; see Proposition 2.4 and Theorem 5.4). A subnormal series G = Go 2 G, 2 G, 2 . * * 2 G, = 1 is called a composition series for G if each Gi+ is a maximal proper normal subgroup of Gi. Equivalently, by the corollary to Proposition 1.12, each factor Gi/Gi+ is a nontrivial simple group. For example, S, 2 A , 2 1 is a composition series, as is S, 2 A , 2 1 for all n 2 5. If G is a finite group it is clear (again using the corollary to Proposition 1.12) that any subnormal series with nontrivial factors can be “refined” (by inserting subgroups) to a composition series. For example, S, 2 A , 2 1 can be refined to S, 2 A , 2 K , 2 K , 2 1, where K , is the subgroup of order 4 in A , (Exercise 3.3), and I K,I = 2.

,

,

Theorem 5.5 (The Jordan-Holder Theorem). If G is a finite group and Go 2 G, 2 ‘ . ‘ 2 G,

G

=

G

= Ho 2

= 1

and H , 2 ... 2 Hk= 1

are composition series, then m = k and there is a 1- 1 correspondence between the sets of factors so that corresponding factors are isomorphic.

26

1 Groups

Proof. Induction on m. The theorem holds when m = 1, for then G is a simple group. Assume that the theorem holds for groups having a composition series of length m - 1. If G, = H I the theorem holds for G by the induction hypothesis. If not, set K , = G, n H , . Since GI and H , are maximal normal subgroups of G and G, # H , we have G , H , = G, and, by the Isomorphism Theorem (2.6) GIG, GIH,

G , H , / G , z H , / H , n G, = H , / K , , = G , H , / H , E G,/G, n H , = G,/K,.

=

In particular K , is a maximal proper normal subgroup in both GI and H , . Choose a composition series K , 2 K , 2 . .. 2 K , = 1 for K , . Then

G = Go 2 GI 2 G2 2 G3 2 ... 2 G,= 1, G = Go 2 GI 2 K2 2 K3 2 ... 2 K , = 1, G = H,, 2 H , 2 K , 2 K 3 2 ... 2 K , = 1, and

G

=

Ho 2 H I 2 H2 2 H3 2 ... 2 Hk = 1

are all composition series. By the induction hypothesis we may conclude that the first and second series have isomorphic factors (in some order), as do the third and fourth, and that m - 1 = s - 1 = k - 1. The second and third series have isomorphic factors by the remarks above, and the theorem follows.

For an example let G be the additive group H,,. It has composition series G G

2212 2 G2 = hi, 2 Hi = 2212 2 H2 = H I 2 2 GI =

= 42122 G3 = 0, = 6212

2 H3 = 0,

and

G = H i 2 2 K , = 3 2 1 , 2 K , = 6212 2 K3 = 0. The resulting factors are GIG, E U,,

GIH,

z H,,

G,/G, 2 U,, H , / H , z U,,

GIK,

z Z3,

K,/K,

G,/G3 r Z,, H 2 / H 32 Z , ,

and

z Z,,

K,IK3 z Z , .

By the Jordan-Holder Theorem each finite group G is associated with a finite collection of simple groups (viz., its composition factors), unique up to isomorphism. We say in general that a group A is an extension of a group B by another group C if B 4A and A / B E C . Thus each finite group is obtained by a sequence of extensions from its uniquely determined set of simple

6 Products

27

composition factors. It should be clear then that knowledge of finite simple groups is of extreme importance for the study of finite groups in general. There are several infinite families of finite simple groups and a handful of “sporadic” simple groups. The finite simple groups have only very recently been completely classified; the classification will likely stand as the major triumph of algebra in this century. 6. PRODUCTS

If G, and G , are groups it is a routine matter to verify that the Cartesian product G , x G , is also a group if we define a binary operation by setting ( x l , x 2 ) ( y l y,) , = (x,y,,x,y,). We find that G , x G, has identity ( 1 , l ) and that ( x l , x 2 ) - ’ = (x;’,~;~). Our discussion of a “product” for an arbitrary nonempty family of groups will be in terms of the notion of universality encountered in the previous section. If {G,:cYE A } is any nonempty family of groups, then a product of the G, is a group P together with a family p,: P + G,, all ci E A , of homomorphisms with the following universal property: given any group H and homomorphisms f,: H -+ G,, all (r E A, then there exists a unique homomorphism f : H + P such that p,f = f, for all ci E A , i.e., the diagrams

are all commutative. Exercise6.1. If A = {1,2) set P = G , x G, and define pl(x,,x2) = x , and p2(x1,x,) = x,. Show that ( P , { pl, p,}) is a product of G I and G 2 .

Proposition 6.1. Suppose {G,:a E A } is a nonempty family of groups. If a product ( P , { p m } )exists, then P is unique up to isomorphism and each p,: P + G, is an epimorphism. P r o o f . Let ( P I ,{ p : ) ) be another product. We have D

P;

D

P.

28

I Groups

for each o! E A, i.e., p i

= p,fand

p , = pkg = pafg for all o! E A. Thus we have

for all u E A, and so f g = 1 by the uniqueness in the definition of a product. Simifarly g f = I, so f:PI -+ P and g: P -+ P, are a pair of inverse isomorphisms. Fix G, and define f,:G, + G, by setting f , ( x ) = 1 E G, for all x E G , if B # a, but f, = i,, the identity map on G,. Then we have

for a unique homomorphism f and for all 8. In particular p t l f = f, = i,, so p a must be an epimorphism since i, is an epimorphism.

Theorem 6.2. If {G,: cc E A} is any nonempty family of groups, then a product of the G , exists. Proof. Take P to be the Cartesian product, P = n { G , : a E A}. Write the elements of,P as ( x , ) ~ or ~ ~simply , as (x,). Define a binary operation on P by setting (x,)(y,) = (x,y,). Then P is easily seen to be a group, with ( l ) a e Aas For each 01 f A define pa:P -+G, by setting identity and ( x u ) - ' = (x,'). p , ( ( ~ ~ ) , ~=, )x , E G,. Let H be any group and suppose that f,:H -+ G, is a homomorphism for all u E A . Definef: H + P by settingf(h) = ( f , ( / ~ ) ) , ~E ~P, so that p a f = f,,all a E A . If also g: H + Pis a homomorphism with p u g = f,, all o! E A, then pag = p , f , all a. Thus if x E H we have p,g(x) = g(x), = p,f(x) = f ( x ) , , all a E A, and hence f ( x ) = g ( x )for all x E H . Thus f is unique and P is a product.

IfA = {1,2,3,..., n},orifA = (1,2,3 ,... },weoftenwriteGI x G, x . . . x G,, or G , x G2 x G, x . * * for the product constructed above. Also, the product, as constructed, is often called the direct product. In general the homomorphism p , is called the projection of G, on the direct factor G,.

n

Theorem 6.3. Suppose G,,G2 Q G, G, n G, = 1, and GIG, = G. Then G r G , x G , . More generally, if G 1 , G 2,..., G , U G , if G i n ( U { G j : j # i ) > = 1 for 1 5 i I n, and if G , G 2 . . ' G ,= G, then G 2 G , x G 2 x ... x G,.

7 Nilpotent Groups

29

Proof. We sketch the proof when n = 2 and leave the proof of the more general result as an exercise. Each x E G can be expressed as x = x 1 x 2 ,with xi E Gi, uniquely since GI n G , = 1 . Note that [ x , , x , ] = x ; 1 x ; 1 x l x 2E G , n G , = 1, so x l x 2 = x z x I . If we define p i x = x i , i = 1,2, then p i is a homomorphism from G to G i . If H is any group and fi: H 4 Gi a homomorphism, i = 1,2, define f :H 4 G by setting f ( h ) = f,(h)f,(h). Then pif = A, i = 1,2. If also 9 : H + G is a homomorphism satisfying pig = &, i = 1,2, then pif = pig, i = 1,2. But then for any x E H we have

f ( 4= P , f ( x ) p , f ( x ) = Plg(x)P2g(x)= g(x), and so f = g is unique. It follows from Proposition 6.1 and Theorem 6.2 that G g G , x G, since G acts as a product for G I and G , . Exercise 6.2. Complete the proof of Theorem 6.3.

Under the circumstances of Theorem 6.3 we say that G is the internal direct product of its subgroups G1,G,,. . . ,G,. Exercise 6.3. Let G be the additive group Q of rational numbers. Show that G cannot be the internal direct product of two of its subgroups since any two subgroups have nontrivial intersection. Exercise 6.4. If G is the internal direct product of subgroups G, and G 2 show that G/G, E G, and G / G 2 z G , (use the FHT). Exercise 6.5. (1) Show that Z ( n u G d l= ) H,Z(G,). (2) Show that (G, x G2 x -.. x G,)’ = G; x G; x . . - x GL. (3) Under what circumstances is G, x G, x ... x G, solvable? 7. NILPOTENT GROUPS

The ascending central series of a group G is the sequence 1

=Z,

5 Z , = Z ( G ) 5 Z , 5 Z , 5 ...,

where Zi+,/Zi= Z(G/Zi) for all i. At times it may be necessary to stress the dependence on G and write Z i = Z i ( G ) .We see inductively, using the corollary to Proposition 1.12, that each Zi is normal in G , so the definition makes sense. If Z , = G for some n, then G is called a nilpotent group. Clearly every abelian group is nilpotent, since 2 , = Z ( G ) = G. The symmetric groups S, are not nilpotent if n 2 3 since Z ( S , ) = 1 and hence all zi = 1. Proposition 7.1. If G is a finite p-group, then G is nilpotent. Proof. We see by Proposition 2.4 that Z , # 1. If Z , # G, then G / Z , is also a p-group, so Z ( G / Z , ) # 1, again by Proposition 2.4, so Z 2 2 Z , and

30

I Groups

2, # 2,. Likewise if Z , # G, then Z , 2 2, and Z , # Z , . Since G is finite we conclude that Z , = G for some n.

Proposition 7.2. If G is nilpotent, then G is solvable. Proof. Since Z i + , / Z i= Z ( G / Z i ) it is abelian. Thus G = Z , 2 . .. 2 Z o = I is a subnormal series with abelian factors so G is solvable by Theorem 5.3.

2,- 2

Corollary. Finite p-groups are solvable. Exercise 7.1. Show by example that a solvable group need not be nilpo tent.

Proposition 7.3. If G is nilpotent and H I G but H # G,then N,(H) # H . Proof. Choose Z i in the ascending central series such that Zi5 H but 4 H . We show that Z i + ' 5 NG(H). If x E Zi+l, then x Z i E Z i + , / Z i= Z(G/Zi), so for any h E H we have x Z i h - 'Zi= h - ' Z i x Z i , and hence x 'hxh ' E Z i S H . Consequently x - hx E H , and so x E NG(H). ~

'

Corollary. If G is nilpotent and H I G is a maximal proper subgroup, then H 4 G . Proof. N G ( H )= G since H is maximal. Note that H = ((12)) is a maximal proper subgroup of S , that is not normal. Proposition 7.4. If G is a finite group and P G is a p-Sylow subgroup, then NG(NG(P))= NG(P). Proof. This was Exercise 4.2.2. If x E NG(NG(P))then both P and P" are p-Sylow subgroups of NG(P). By the second Sylow Theorem we have P = (P")' = PxY for some y E N G ( P ) .But then x y E N G ( P )and hence x E N,(P).

If H and K are subgroups of a group G denote by [H,K ] the subgroup generated by all commutators [x, y], x E H , y E K . Thus, for example, in the derived series for G we have G'k+l ) = [G(k),G'k'].Note that H 4G if and only if [ G , H ] IH . Exercise 7.2. Show that [H, K ] = [ K , HI.

Proposition 7.5. Suppose K 4G and K if and only if [ H , G ] s K .


5 G. Then H / K 5 Z ( G / K )

Proof. This is clear since [y, x ] = y - x - yx E K if and only if y x K = x y K for all y E H , x E G.

L,

Define the descending central series of a group G by setting Lo = G, = [ G , G ] , and in general Lk+ = [G, Lk] for k 2 1. We see inductively, by

7 Nilpotent Groups

31

,

Exercise 5.2, that Lk 4G for each k, and hence that Lk + I Lk for all k. It follows from Proposition 7.5 that Lk/Lk+ I Z(G/Lk+,),which is the reason that Lo 2 L , 2 L, 2 . * .is called a central series.

,

Theorem 7.6. A group G is nilpotent if and only if L,(G) = 1 for some n. Proof. *: Take n minimal so that 2, = G = Lo. Then, since Z,/Z,- = Z(G/Z,- 1), we see by Proposition 7.5 that Z,- 2 [G,Z,] = [G,Lo] = L,. The same reasoning shows inductively that Z,-k 2 Lk, 0IkI n, and in particular L, IZ, = 1. e: Take n minimal so that L, = 1 = Z,. Then [L,-,,G] = L, = 1, so L,-, I Z(G) = Z,byProposition 7.5.Next,[Ln-,,G] = L,-, I Z,,andwe may apply Proposition 7.5 with K = Z,, H = L,-,Z,. We conclude that L,_,Z,/Z, I Z(G/Z,) = ZJZ,. Thus L,-,Z, I Z,, and therefore zk for 0 Ik In, and Z, 2 Lo = G. L,-, IZ , . Inductively L,-k I

,

Exercise 7.3. Show that subgroups and homomorphic images of nilpotent groups are nilpotent.

Proposition 7.7. If H and K are nilpotent groups, then G nilpotent.

=

H x K is

Proof. Clearly L o ( G )= L,(H) x L,(K). If Lk(G)

=

Lk(H)

Lk(K)

for some k, then Lk+i(G)= [ H x K , Lk(H) x Lk(K)] = [H>Lk(H)I [K,Lk(K)] =

Lk+l(H)

“k+l(K).

The proposition follows by Theorem 7.6. Theorem 7.8. A finite group G is nilpotent if and only if it is the (internal) direct product of its Sylow subgroups. In particular if G is nilpotent, then each Sylow subgroup is normal, and hence is unique. Proof. +: Use Propositions 7.1 and 7.7. -: If P is p-Sylow in the nilpotent group G, then N G ( N G ( P ) = ) N G ( P )by Proposition 7.4, and so N G ( P )= G by Proposition 7.3. Thus P a G, and P is unique by the second Sylow Theorem. Let P,, P,, . . . ,P, be the distinct (nontrivial) Sylow subgroups of G . If i # j and x E y E 8, then [x, y ] E n 8 = 1, so x y = y x . It follows that n ( U P j :j # i ) = 1 (compare orders of elements), and also that IP,P,...P,l= IGI,soG = P,P,...P,.ThusP,,P,, ..., P,satisfy the hypotheses of Theorem 6.3, so G is their internal direct product.

e,

e

32

I

Groups

8. FINITE ABELIAN GROUPS

If G I , G 2 , .. . ,G , are additive abelian groups it is customary to write G,0G, 0 .. '0 G, for their direct product, and to call it the direct sum rather

than product. In like manner we speak of an abelian group as being an internal direct sum of subgroups as in Theorem 6.3, in which case we also write G = G1 0G 2 0 ... 0 G,.

Theorem 8.1 (Frobenius and Stickelberger). If G is a finite abelian group, then G is a direct sum of cyclic subgroups, each of prime power order. Proof. Since G is abelian, it is nilpotent, hence it is the direct sum of its Sylow subgroups, by Theorem 7.8. Thus we may as well assume that G is a p group for some prime p . Use induction on I GI.Choose a E G of maximal order, say la( = pk, and choose H 5 G maximal with respect to H n ( a ) = 0. Set G , = H 0( a ) I G.If GI # G we may choose an element x E G\G, such that p x E G, (this is Cauchy's Theorem, 2.7, applied to G/G,). Say p x = h + ma, with h E H and m E Z. Since pk is the maximal order for elements of G we have

o = p k x = p k - ' ( p x ) = pk-'h + p k - l m a ,

and so pk-'ma = 0 since H n ( a ) = 0. But then p ( rn, say m = pr, since la1 = pk. Consequently h = p x - ma = p x - pra = p ( x - ra) E H, but x - ra # H (since x 4 GI). By the maximality of H we must have ( H + (x - r a ) ) n ( a ) # 0, so there exist h , E H and s, t E Z such that sa = h , + t ( x - ra) # 0. Thus t x = -h, + (s + tr)a E G,.Note that pkt, for if t = up, u E Z, then t(x

- ra) = up(x - ra) = uh E H,

and then sa = h , + uh E H, contradicting the fact that H n ( a ) = 0. Thus p and t are relatively prime, so there are integers m and n such that mt np = 1. But then x = (mt + np)x = m(tx) + n ( p x ) E G , ,

+

a final contradiction. We conclude that G = G, = H 0( a ) , and an application of the induction hypothesis to H completes the proof. Theorem 8.2. Suppose G is an abelian group, JGI = p" for some prime p , and G = G, @ G 2 @ * * * G, 0 = H,0H2 H,, 0

.

a

.

0

witheachGiandHjcyclicandl < l G i l IIGi+,I,1
j . Then r = s and Gi

Proof. Let H = (x E G : p x = O}. Each G i , being cyclic, has a unique subgroup Ki of order p , so clearly H = K , 0 K , 0 * . 0 K,,and IH( = p'. But

9 FreeGroups

33

similarly IHI = p s , so p' = p s and r = s, i.e., any two decompositions of the sort indicated must have the same number of cyclic summands. Suppose now that lGil = [ H i [for 1 Ii Ik - 1, but that (say) lGkl < IHkI. If lGkl = q = p', then qGk = 0 but qH, # 0, and we have qk = qGk+ @ . . * @ qG,

= qHk

@ ... @ qH,,

which contradicts the first part of the proof. Theorem 8.3 (The Fundamental Theorem of Finite Abelian Groups). Suppose that G is a finite abelian group and that p l , p , , ...,p k are the primes that divide [GI. Then G = G, 0 G, 0 ... 0 G k , where G i is a pi-group, and G i = H i , @ Hi, 0 ... @ Hi,,,,i)

for each i, with Hij cyclic and 1 < lHijl I ( H i ( j +1 ) 1 for each j . The group G is determined to within isomorphism by the orders of the cyclic subgroups H,, 1 ~j Im(i), 1 Ii I k. Proof. The decomposition G = G, @ G, @ .. . @ G, is the representation of G as the direct sum of its unique pi-Sylow subgroups as in Theorem 7.8. The rest of the theorem is a consequence of Theorems 8.1 and 8.2.

Theorem 8.3 will be reestablished in considerably greater generality in Chapter IV.

EXAMPLES 1. Suppose G is abelian and IGI = 100 = 22 * 5,. Then G = G, @ G,, with lGll = 2, and IG21 = 5'. The possibilities for G , are Z, and Z2 @ Z,;the possibilities for G, are Z 2 5 and Z,@ Z 5 ,Thus there are (up to isomorphism) four different abelian groups of order 100, viz., Z,@ 7 2 5 , Z,0 Z,@ 2 2 5 , Z 4 @ Z,0 H,, and Z,0 Z,0 Z,@ Z5. 2. Suppose IGI = p 2 , p a prime. Then G is abelian by Exercise 2.1 and either G z Z p 2 or G z Z p @ Z,. Exercise 8.1. Describe all abelian groups of order n for n p 3 , and p4, where p is a prime.

= 24,200,1000,

9. FREE GROUPS

Recall that a semigroup is a nonempty set with an associative binary operation. It will be convenient to discuss free semigroups before proceeding to the discussion of free groups. If X and Y are semigroups, then a homomorphism from X to Y is a function f : X + Y such that f(xIxz) = f(xl)f(x2) for all xl, x2 E X . We say that X and Y are isomorphic if there is a 1-1 homomorphism from X onto Y.

34

I Groups

A semigroup X is free on a set S if there is a function j : S -+ X such that if we are given any semigroup Y and any function k : S -+ Y, then there is a unique homomorphism f :X + Y for which the diagram S

y/yx j

Y

is commutative, i.e., f j = k. Proposition 9.1. If a free semigroup exists on a set S , then it is unique up to isomorphism. Proof. See the proofs of Propositions 5.1 and 6.1. Exercise 9.1. If S = { 1 ) and N = { 1,2,3,. . .} is the semigroup of natural numbers under addition, verify that N is a free semigroup on S, with j ( 1 ) = 1.

Theorem 9.2. If S is any nonempty set then a free semigroup X on Sexists. Proof. Set X = S u ( S x S ) u ( S x S x S ) u ..., the union of all the Cartesian powers of.finitely many copies of S . If we define a binary operation on X by setting (a13 a23...7arn)(b19b2,.*.5

bk) =

* *

* , a m ,b1, *

*

-

9

bk),

then it is immediate that the operation is associative, so S is a semigroup. Take -+ X to be the inclusion map, j ( x ) = x. If Y is any semigroup and k: S Y a function define f : X -+ Y by setting f ( a , , .. .,a,) = k ( a , ) .. . k(a,). Then f is a homomorphism and f j = k. If also g : X -+ Y is a homomorphism and g j = k , then

j :S

-+

. .,a,) = d j a 1ja2...jam)= M a l )...(0,) = ( f j a J . . . ( f j a , )= f

so g = f.

=

m,. ., am),

(hja2..*j4

If A is a set and R is an equivalence relation on A let us write c~,(x)for the R equivalence class containing an element x of A. We write A I R to denote the set of all equivalence classes determined by R and call that set the quotient of A mod R. The function q: A AIR defined by setting q(x) = c~,(x)is called the quotient map. An example is provided by the canonical quotient map q from a group G to its quotient G I K by a normal subgroup K . -+

Proposition 9.3. Suppose Y is a semigroup and R is an equivalence relation on Y such that if x R y and zRw, then xzRyw. Then YIR is a semigroup if we define cl,(x) . cl,(y) = c~,(xY).If 2 is another semigroup and h: Y 2 is -+

9

FreeGroups

35

a homomorphism, and if h(x) = h ( y ) whenever xRy, then there is a unique homomorphism f : Y / R + Z for which the diagram

is commutative, Proof. The condition imposed on R is precisely what is needed to insure that Y / R is a semigroup and q is a homomorphism. If clR(x) = clR(y), then xRy, so h(x) = h ( y ) ,and we may define ,f(clR(x))= f(q(x)) = h(x). It is clear then that f is a homomorphism; it is unique since q is an epimorphism.

Definition: A group F is free on a nonempty set S if there is a function

4: S -+ F such that if G is any group and 8: S + G is any function then there is a

unique homomorphism f : F

+G

such that the diagram

is commutative. Exercise 9.2. If S = { u } let F be a set of distinct “powers” a“, n E Z,and define a”’. uR= am+“.Agree that uo = 1 and define $(a) = a = u’. Show that F is a free group on S. Thus a free group on a single “generator” a can be taken to be an infinite cyclic group ( a ) .

Proposition 9.4. If a free group F exists on a nonempty set S, then F is unique up to isomorphism, and 4: S + F is a 1-1 function.

Proof. Uniqueness is proved as usual. If 4 is not 1-1 suppose that

$(a) = 4 ( b )but a # b in S. Let ( a ) and ( b ) be free groups on { u ] and { b } , respectively, as in Exercise 9.2. Set G = (a) x ( b ) , the direct product, and define 8: S + G by setting O(a) = (a, I), O(b) = (I,b), and 8(c) = ( 1 , l ) for all c

other than a or b. Then there is a homomorphism f : F

-+

G, with 8 = f 4 . Thus

(a, 1) = Q(a) = f(4ca)) = .f(4(@)= W )= (1, b),

and so u

=

1, a contradiction since ( a ) is infinite cyclic.

Theorem 9.5. If S is a nonempty set, then there is a free group on S . Prooj’ (Meyer [27]). Choose a set S‘ with JS’I = IS( and S n S’ = @, and let s H s’ denote a 1-1 correspondence between S and S’. If we also write

36

I

Groups

T = S u S' onto itself. Let X be the free semigroup on T , as constructed in Theorem 9.2. If G is a group and g: X 4G is a homomorphism we say that g is proper if g(s') = g(s)-' for all s E S [it follows that g(t') = g(t)-' for all t E TI. Define a relation R on X by agreeing that x R y if and only if g ( x ) = g ( y ) whenever G is a group and g: X --+ G is a proper homomorphism. It is immediate that R is an equivalence relation, and that if x R y and z R w then xzRyw. Thus F = X / R is a semigroup and the quotient map q: X + X / R is a homomorphism by Proposition 9.3. Write X = q ( x ) for each x E X , so F = { X : x E X } . We show first that F is in fact a group. Choose a E S and any x E X . If g: X + G is any proper homomorphism, then g(aa') = 1, so g(aa'x) = g ( x ) = g(xaa'),and aa' . % = X = X * 2.Thus aa' is an identity for F (unique by Proposition l.l), and we write 1 = 1, = E'. Write x E X as x = a , a , ~ ~ ~ awith , , aiE T, and set y = a;...a;a;E X . For any proper homomorphism g: X + G we have (s')' = s" = s for each s E S , then tt-, t' is a 1-1 map of

g(xY) = g(al)'.'g(ak)g(ak)-' "'g(al)-'

=

= g(aa'),

so xyRaa', or Xy = X.7 = aa' = I,, and similarly 7 .X = 1. Thus 7 is an inverse for X and F is a group. Next, F is free on S . Let i: S 4 T and j : T X be the inclusion maps and define 4: S F by setting 4 = qji. If G is any group and 8: S -+ G any function we extend 8 to 8': T -+ G by setting 8'(s') = 8(s)-' for all s E S. Since X is free on T there is a unique homomorphism y : X + G such that y j = 8', and by Proposition 9.3 there is a unique homomorphism f :F -+ G such that f q = y . But then J4 = f q j i = yji = 8'i = 8. If there were another homomorphism h: F 4G such that h 4 = 8,or hqji = 8, then hqj = 8'. (Why?) Thus hq = y = fq since X is a free semigroup on T , and h = f by Proposition 9.3 (see the diagram below). The proof is complete. -+

-+

Since 4: S

-+

F is 1-1 (Proposition 9.4) we may (and shall) identify each

4(s) E F , and hence assume that S c F. Then each element of F is a product of elements of S and of elements 4(s'), s E S, which are the inverses in

s E S with

F of elements of S, so we have F = ( S ) . We usually write F = F' and call it the free group with S as its set of free generators. The definition of freeness thus says that F, has a set of generators (viz., S ) such that any function #: S -+ G, for any group G, can be extended uniquely to a homomorphism f:F, -,G.

10 Generators and Relations

37

Theorem 9.6. Suppose S and U are nonempty sets. Then F, E F, if and only if IS1 = IUI.

Proof: 3: Suppose first that one of the sets, say S, is finite. Since F, and F,, are isomorphic they have the same number of subgroups of index 2, and each such subgroup is the kernel of a homomorphism onto Z,. Any such homomorphism is obtained by choosing a nonempty subset of S (resp. U ) which is mapped to E Z,, and mapping the complementary set to 0 E Z,. Thus 21” - 1 = 2’”’ - 1, and hence IS1 = IUI. If S is an infinite set of free generators set T = S u S-I. Then IT1 = 21SI = IS[. Since any x E F, is a product of finitely many elements from T we have lFsl I 1 IT1 IT x TI IT x T x TI + ... =K,ITI = IS[.

+

+

+

Consequently lFsl = ISI. If S and U are infinite sets and F, z F,, then IS1 = lFsl = lFul = IUI. e : Let 0: S + U be a 1-1 correspondence. We may view 0 as a function from S into F,, so it extends to a homomorphism f : F, + F,,. Likewise 0 - I : U + S extends to a homomorphism g: F, + F,. But then the restriction of gf to S is just 8-’8 = 1, the identity map on S, and so g f = 1, the identity map on F,, since ( S ) = F,. Similarly f g = 1, the identity map on F,, so f and g are inverse isomorphisms. As a consequence of Theorem 9.6 we may define the rank of a free group F to be the cardinality of any set of free generators. Thus two free groups are isomorphic if and only if they have the same rank. A subgroup H of a free group F is also free (it is customary to agree that H = 1 is free on the empty set 0; this meets the requirements of the definition with 4 the “empty function”). However, even when the rank of F is finite a subgroup H may have infinite rank. For example, if F is free of rank 2, then its derived group F‘ has countably infinite rank. One source for proofs of these statements is Kurosh [22].

10. GENERATORS AND RELATIONS Theorem 10.1. If G is any group and S is a subset of G that generates G, then G is a homomorphic image of the free group F,.

Proof. Let 4:S + F, be the inclusion map and let 0: S + G be the inclusion map. By the definition of F, we may extend 4 to a homomorphism f :F, + G with f(S) = S . Since ( S ) = G it is clear that f is an epimorphism. Suppose G is a group, S E G , and ( S ) = G. As a consequence of Theorem 10.1 and the F H T we see that G z FJK for some K Q F,. If T C_ K , then each t E T is a “word” (i.e., product) in F, involving the generators s E S

38

I Groups

and their inverses. The effect of the canonical quotient map is to map each t E T to t K = K , the identity element in F J K , or, speaking loosely, to “set t = 1.” We say then that G has S as a set of generators which are subject to the relations { t = 1: t E T } . The remarks above suggest a means of describing groups abstractly. Let S be a set, let T c F,, and let K be the normal subgroup of Fs generated by T , i.e., the intersection of all normal subgroups of Fs that contain T. Then the presentation ( S I t = 1 , all t E T ) is defined to be the quotient group Fs/K. If G is a group and G has S as a set of generators subject to the relations { t = l : t E T } , so that G z F,/K, then we also say that G has the presentation ( S 1 t = 1, all t E T). For an easy example a cyclic group ( a ) of order n has the presentation ( u p = 1). Proposition 10.2. Suppose G, is a group with presentation ( S , I t = 1, all t E T ) , and that G, is a group. Suppose S , c G,, ( S , ) = G,, and s H s’ is a function from S , onto S , . Suppose further that the generators s’ E S , satisfy all the relations t = 1 , t E T ,in the sense that if each s E S , is replaced by the corresponding s‘ E S , in each word t E T, then the result is an element t’ E G, with t‘ = 1. Then there is a homomorphism from G , onto G,.

Proof. We may assume that G, = F s i / K , , where K , is the normal subgroup of F,, generated by T. The map s Hs‘ extends to a homomorphism f from F,, onto G, (since ( S , ) = G,), and K , = kerf contains K , since the elements of S , satisfy the relations for G,. Since K , < K , the mapping g : F , , / K , + G, defined by g ( x K , ) = f ( x ) is a well-defined homomorphism from G, = Fs,/K, onto G,.

For example, let G I = ( a l a ” = l ) , and let G 2 be cyclic of order rn, generated by b, where m I n. Then b“ = 1, so the generator of G, satisfies the relation defining G, and a H b extends to a homomorphism from G I onto G,. For another exampte consider the presentation G = (a,bla3 = 1, b2 = 1,abab = 1). The third relation can be written in the form a6 = ba- or ab = ba2.It follows that each element of G can be written in the form b i d , with 0 < i < 1 and 0 sj 2, and hence that IGI < 6. If we consider the elements o = (123) and z = (12), which generate the symmetric group S , , then c3 = z 2 = 1 and oz = (13) = TO’, so they satisfy the relations that define G. By Proposition 10.2 there is a homomorphism from G onto S , , and so [GI2 6 = IS31. Consequently IGI = IS,I = 6, and G z S , . Thus S3 has the presentation ( ~ , b \ ~ ~ = b1,ab=ba2). ’= Exercise 10.1. (1) Show that Klein’s 4-group has the presentation (a,bla2 = b2 = (ab)’ = 1).

’,

10 Generators and Relations

39

( 2 ) Show that the quaternion group Q , = { f 1, f i , kj, fk} has the presentation ( a , b l a 4 = 1 , a 2 = b*,ab = b a 3 ) ( t r y a ~ i , b w j ) . Given a presentation for a group G it can often be difficult to determine whether G is finite or infinite, and even to determine whether or not G = 1. In particular it can be difficult to find IGI. We present two more examples in some detail.

EXAMPLES 1. Let G = ( x , y 1 x y = y'x, yx = x 2 y ) . The relations do not suggest immediately that x and y even have finite order, so it is conceivable that G is an infinite group. However, y - l x y = y - ' y 2 x = yx = x2y = x x y ,

and by canceling x y we see that y - '

= x.

Thus

x y = 1 = y 2 x = y ( y x ) = y,

so also x

=

1, and hence G

= 1.

= ( a , b 1 am = b2 = 1, ub = ba '). In this case all elements of G can be written in the form b'aj, with 0 5 i I1 and 0 I j I m - 1 , so [GII 2m. If we can exhibit a group of order 2m with two generators that satisfy the relations defining G it will follow from Proposition 10.2 that IGl = 2m. We suggest two proofs that such a group exists. Consider first the group 0,of symmetries of a regular m-gon in the plane (assume here that m 2 3). It has a generating rotation o through angle 2n/m and it has m reflections. If 7 is one of the reflections then 0 and 7 generate 0, and satisfy the relations for G, and ID,l = 2m, so G 2 0,. For another approach let H , as a set (but not as a group), be the Cartesian product Z,x Z,,,. Define an operation on H by setting

2. Let G

~

(i,J).(6,n) = (F+ 6, (-

6).

l)kJ+

It can be verified (Do so!) that H is a group, and it is clear that I HI = 2m. If we set a = (O,]) and b = (1, then a and b satisfy the relations for G. In fact, the operation defined on H was inspired by the fact that ( b i a J ) ( b k a " =) bi+ka(-l)kj+n in G, as follows from the relations. Thus also G z H . The first approach is perhaps more natural and is geometrically satisfying. The second is more tedious but may have the advantage that it could be applied in some cases where a geometric interpretation is not available. In any event the group D,,, = (a,z = T~ = 1, o7 = TC ') is called the dihedral group of order 2m for any m 2 2. When m = 2, then D,,, = D, is isomorphic with Klein's 4-group.

o),

~

40

I Group

Exercise 10.2. (1) Verify that the rotation (r and the reflection T in Dm satisfy the relations defining G in the second example above. (2) Verify all the statements made about the group H in the second example.

11. SOME FINITE GROUPS CLASSIFIED

In this section we classify several finite groups (up to isomorphism) according to arithmetic properties of their orders.

1. If p is a prime and IG( = p 2 , then G is abelian, by Exercise 2.1. By Theorem 8.3 we conclude that G z H P 2 or G z B, @ Z,. 2. Suppose p and q are primes, with q < p , and suppose that (GI = p q . If p f 1 (modq) we saw in Example 2, p. 21, that G z Z P q . Suppose then that p = 1 (mod q). Choose elements x and y in G of orders p and 4, respectively, and set p = ( x ) , Q = ( y ) . Just as on p. 21 we see that P 4 G, and that [G:NG(Q)] = 1 or p. If [ G : N G ( Q ) ]= 1, then G = ( x y ) is cyclic, just as before. Suppose then that [G:NG(Q)] = p . Then x y = y - ' x y = x" for some k x n kif 0 < k E H , and in particular x = integer n, 2 I n < p . Thus y - k ~ y = y - ¶ x y q = x"', so n4 = 1 (modp). Consequently G has generators x and y that satisfy the relations xP = yq = 1, y - l x y = x", where 2 I n < p and nq = 1 (mod p ) . The group with that presentation, ( x , y I x p = yq = 1, y - ' x y = x"),with2 I n < pandnq E l(modp),doesinfacthaveorderpq,so G has that presentation. With the aid of some elementary number theory it can be shown that if m is any other integer such that mq = 1 (modp) and rn f 1 (modp), then the presentation (x, y 1 x p = yq = 1, y - xy = x"') defines a group isomorphic with G. 3. Suppose p is a prime and 1G1 = p 3 . (a) If G is abelian, then G z Z p s , Z,0Z P 2 , or Z,0 Z, 0 H, by the Fundamental Theorem of Finite Abelian Groups. (b) Suppose p is odd and G is not abelian.

(i) Assume first that G has no element of order p 2 . Set Z ( G ) = Z and note that Z has order p by Exercise 2.1, for otherwise G / Z would be cyclic. Thus I G / Z l = p 2 , so G / Z is abelian by Exercise 2.1, but is not cyclic (or else G would have an element of order p2). Thus C / Z E Z,0H,, so it has generators x Z , yZ with xPZ = y P Z = Z and x y Z = yxZ. Set z = [x, y] = x - l y - l x y E G. If z = 1, then G is abelian, since G = (x, y , Z ) , and so z # 1. But z E Z, SO Z = ( z ) and G has the presentation (x, y,z ( x p = yp = z

p

= 1, [x, y] = 2, [x,z] = [ y , z ] = 1).

12 Further Exercises

41

(ii) Assume next that G has an element x of order p 2 and set H = ( x ) . Then H is normal in G by the corollary to Proposition 7.3. Take y E G\H. Then y p E H since IC/HI = p . Say y - ' x y = x', where r f 1 (modp2).Then y-jxy' = x*' for all positive integers j , and in particular x = y - p x y p = x r p , so r p = 1 (mod p 2 ) . But also r p = r (mod p ) by Fermat's Little Theorem, so r = 1 (modp), say r - 1 = sp, s E Z.Choose j E Z,j > 0, such that j s = 1 (modp), and set z = y j . Then - lXz = -jxyj

= xrJ = x ( l + S P ) J =

1 +j s p

-

1 +(js- I ) p + p

We still have G / H = ( z H ) , so z p E H , say z p = x'. Then p JzI = p 3 . Say t = up, so z P = xuP. if i E H we have = (z-lxz)i

z-lxiz

=x

=

1 +pa

I t , for otherwise

( l + ~ ) ~ ,

or x i z = zx(l +p)i

Thus (zx-u)P

=z ~ - ~ z x - ~ . . . z= x -z~2 x - u ~ 1 + ~ l + P ~ ~ z x - u . . . z x - u

-z3x

- u ( l + ( I + P) + ( 1 + P ) 2 ) z x - u

. . .zx - u

- zpx-u(l+(I

+P)+...+(l

- zpx-u(l+(l -

+p)+(l+2p)+...+(l+(p-l)p))

- zpx

-u(p

+p)p-')

+ p ( 1 + 2 + 3 + ... + ( p - 1)))

- zPx-"(P+P'(P-I)12)

=

zPx-uP

= zPz-P

Set w = zx-". Then w-lxw = X ~ Z - ~ X Z= Xx tation ( x , w I xp2 = wp = 1, w -'xw = x1 + P ) .

=

1,

~~ ~ +and ~ ,G

has the presen-

(c) If p = 2 and G is not abelian, then G is isomorphic with the dihedral group D4 or the quaternion group Qz.

Exercise I I . 1. Prove 2(c) above. Exercise 11.2. Classify all groups of orders 12 and 20.

12. FURTHER EXERCISES 1. If G is a group and f : G -+ G is defined by f ( x ) = x-', all x E G, show that f is a homomorphism if and only if G is abelian. 2. If a group G has a unique element x of order 2 show that x E Z ( G ) . 3. Suppose G is finite, K 4G, H I G, and I KI is relatively prime to [ G : H ] . Show that K I H .

42

1 Group

4. If G is not abelian show that Z(G)is properly contained in an abelian subgroup of G. 5. If G is a group and 1x1 = 2 for all x # 1 in G show that G is abelian. Can you say more? 6. Suppose G is a group, H I G, and K I G. Show that H v K is not a group unless H I K or K I H. 7. (Haber and Rosenfeld [12]). Show that a group G is the union of three proper subgroups if and only if there is an epimorphism from G to Klein’s 4-group. 8. Suppose S is a subset of a finite group G, with IS[ > ICl/2. If S 2 is defined to be { x y : x ,y E S } show that S z = G. 9. If A , B I G and both [G:A] and [ G : B ] are finite show that [ G : A n B] I [G:A][G:B], with equality if and only if G = AB. 10. If [ G : A ] and [ G : B ] are finite and relatively prime show that G = AB. 11. Suppose G acts on S , x E G, and s E S . Show that Stab,(xs) = x Stab,(s)x 12. If A , B I G and y E G define the (A, B)-double coset A y B = { a y b : a E A, b E B } . Show that G is the disjoint union of its ( A , B)-double cosets. Show that 1 AyBl = [ A y :AY n B ] IB( if A and B are finite. 13. Suppose G is a permutation group on a set S , with IS1 > 1. Say that G is doubly transitive on S if given any (a,b), (c, d ) E S x S, with a = b if and only if c = d, then xu = c and x b = d for some x E G. (1) If G is transitive on S show that G is doubly transitive if and only if H = Stab&) is transitive on S \ { s } for each s E S. (2) If G is doubly transitive on S and IS1 = n show that n(n - 1) I (GI. 14. Permutation groups G, and G, acting on sets S, and S , are called permutation isomorphic if there exist an isomorphism 0: G, -+ G 2 and a bijection 4:Sl -+ S , such that ( O x ) ( d s )= ~ ( x sfor ) all x E G I , all s E S , . Sl

I

X

S,

I

Define two actions of a group G on itself as follows:

(i) the action of x E G is left multiplication by x ; (ii) the action of x E G is right multiplication by x I. Show that the two actions are permutation isomorphic. ~

43

11 Further Exercises

15. Suppose G is a transitive permutation group on a set S , and H = Stab&) for some s E S. Show that the usual permutation action of G on the set of left cosets of H is permutation isomorphic (see Exercise 14, above) with the original permutation action of G on S . 16. Suppose G is finite, p is the smallest prime dividing \GI, H I G, and [ G : H ] = p. Show that H 4G.

17. Suppose [G:H] is finite. Show that there is a normal subgroup K of G, with K I H , such that [ G : K ] is finite.

18. Suppose G is finite, H -< G, and G

=

U { H ” : x E G}. Show that H

=

G.

19. Let G be the group GL(2, C)of all 2 x 2 invertible complex matrices, and let H be the subgroup of all lower triangular matrices [;:I, ac # 0. Show that G = { H ” : x E G } (compare with Exercise 18 above).

u

20. Let T be the set of n - I successive transpositions (12), (23), (34),..., ( n - 1 n) in S,,. Show that ( T ) = S,. 21. If pis a prime, H I S,, and H contains a transposition o and a p-cycle 7, show that H = S,. (Hint: Show that there is no loss of generality in assuming o = (12) and T = (123. . . p ) . Then conjugate o by powers of 7.) Show by example that the result may fail if p is not prime.

22. Suppose H I S , , but H $ A,. Show that [ H : A , n H ] Observe that H A , = S,,.)

= 2.

(Hint:

23. If S = { 1,2,3,4,.. . ) let A , denote the (infinite) group of all o E Perm(S) such that there is a finite subset T c S for which o restricts to an even permutation of T and a(s) = s for all s E S\ T. Equivalently A, = { A , : n = 1,2,3,. . .}. Show that A , is simple.

u

24. Let o = (12) and 7 = (123... n ) in S,. Determine the centralizers of o and in S,,. (Hint: What must their orders be?)

7

0 E S, has cycle type ( k , , .. . ,k,) show that the conjugacy class of 0 has n ! / [ ( n i”)(n ki!)]elements. 26. Suppose K is a conjugacy class in S , of cycle type ( k , , . . . ,k,,), and that

25. If

K

A,,. If o E K write L for the conjugacy class of a in A,,. (1) If either k , , > 0 or k2,,,+ > 1 for some rn show that L = K . (2) If k , , = 0 and kz,+ I I 1 for all rn show that K = L u L’, where L’ is also a conjugacy class in A , and )L’I = JLI = IKl/2. G

27. Show that the group of rotational symmetries of the tetrahedron is isomorphic with A,. 28. Show that the group of rotational symmetries of the cube is isomorphic with S,. (Hint: Each rotation permutes the four diagonals of the cube.)

44

I Group

29. If 0 < n E Z,then a partition of n is a sequence ( k l ,k , , . . . ,k,) of positive integers such that k , I k , I ... I k , and k , + k , + ... + k , = n. For example, (2,3) and (1,1,1,2) are partitions of n = 5. Write P ( n ) for the number of distinct partitions of n [e.g., P(3) = 3, P(5) = 7, etc.]. Show that S , has P(n) conjugacy classes. 30. For any real number r write [r] for the greatest integer not exceeding r. If p is a prime show that the p-Sylow subgroups of S , have order p"', where m = [n/p] [n/p'] [n/p3] + ....

+

+

3 1 . Suppose G is a finite group, H 4G, and Pis a p-Sylow subgroup of H . Set N = N , ( P ) . Show that G = NH. (Hint: If x E G, then P" is p-Sylow in H.)

32. If G is a finite p-group and 1 # H 4G show that H n Z(G ) # 1. (Hint: H is a union of G-conjugacy classes.)

33. Suppose G is a finite p-group having a unique subgroup of index p . Show that G is cyclic. [Use induction and look at G/Z(G).] 34. Suppose G is a finite p-group. Show that Z ( G )is cyclic if and only if G has exactly one normal subgroup of order p . (See Exercise 32, above.)

35. Show that there are no simple groups of orders 104, 176, 182, or 312. 36. There is a simple group G of order 168. Show that G has 48 elements of order 7. 37. If p and q are primes show that any group of order p 2 q is solvable. 38. If G is a finite p-group show that all composition factors of G are isomorphic with H,. 39. If A and B are subnormal subgroups of G show that A n B is subnormal. 40. Show that an infinite abelian group G can not have a composition series. 41. Show that A , (see Exercise 23, above) has the composition series A , 2 1, but the subgroup H = ((12)(34), (56)(78),...) has no composition series. 42. If p is a prime, [GI = p 3 , and G is not abelian show that G' = Z(G). 43. A pair of homomorphisms K G A H is said to be exact at G if Im(f) = kerg. A sequence 1 4 K G -% H --+ 1 is called a short exact sequence if it is exact at each of K , G, and H. (1) Show that if K 4G, f:K -, G is the inclusion map and g: G --+ G / K is the canonical quotient map, then 1 -+ K G 4 G / K --+ 1 is a short exact sequence. (2) Show that 1 -+ K 3 G 5 H --+ 1 is short exact if and only iff is 1-1, g is onto, and Im(f) = ker g. Conclude that then K is isomorphic with a normal subgroup of G and that G/f(K) s H .

12 Further Exercises

45

(3) Suppose 1 -+ K -+ G -+ H 4 1 is a short exact sequence. Show that G is solvable if and only if both K and H are solvable (see Theorem 5.4). (4) Give an example of a short exact sequence 1 -+ K -+ G H -+ 1 for which K and H are nilpotent but G is not. 44. If G is a group and x E G define the inner automorphism f, by setting f,(y) = xyx- ',all y E G. Write I ( G ) for the set of all inner automorphisms of G. (1) Show that I ( G ) I Aut(G). (2) Show that Z(G) 2 G / Z ( G ) . (3) If I ( G ) is abelian show that G' I Z ( G ) .Conclude that G is nilpotent. 45. Show that Aut(S,) z S , . [ H i n t : Any automorphism permutes the elements of order 2 in S , , so IAut(S,)I I 6. Look at I ( S 3 )(Exercise 44 above).] -+

46. If G is cyclic of order n show that Aut(G) is abelian and has order $(n).If G is infinite cyclic show that IAut(G)I = 2. 47. If K is Klein's 4-group show that Aut(K) z S3. 48. If A 4G and B 4G show that G / ( An B) is isomorphic with a subgroup of C I A x GIB. 49. If G is a finite p-group that is not cyclic show that there is a homomorphism from G onto Z, x Z,. (Hint: Let A and B be distinct maximal subgroups of G and apply Exercise 48.) 50. If A , B I G show that [ A , B ] 4( A u B ) . 51. If G is a finite group in which every maximal subgroup is normal show that G is nilpotent. ( H i n t : Suppose to the contrary that P is a nonnormal Sylow subgroup and choose M I G maximal with N,(P) I M . If x E G\M consider P".) 52. Define the generalized quaternion group Q , by the presentation Q,

=

( a , b ) a 2 " = 1, b 2 = am,ab = b a - * )

for rn 2 1.

Show that lQml = 4m. Show that Q1is cyclic and Qzis the quaternion group introduced on p. 5. Show that Q 3 is not isomorphic with either the dihedral group Ds or the alternating group A , . 53. Find all positive integers n for which there is only one abelian group (up to isomorphism) of order n (e.g., n = 5,6, 10, and 105 have that property). 54. Suppose there is only one group (up to isomorphism) of order n. Show that (n, $(n)) = 1. (The converse is also true; the proof is more difficult.) 55. Use generators a = (12), b = (23), and c = (34) for S , and show that S4 has the presentation (a, b, c

I U'

= b2 = c 2 = (

~ b= )(uc)' ~ = (bc), = I ).

46

I Groups

56. If G = (a, b I a4 = b3 = 1, ab = ba3) show that G is cyclic of order 6. 57. A group G is called nilpotent of class n if L , - , ( G ) # 1 but L,(G) = 1. (1) Show that G is nilpotent of class n if and only if Z , - # G but Z , = G. (2) Show that the generalized quaternion group Q2,,(Exercise 52) is nilpotent of class n + 1. 58. Compute the centers and derived groups of the dihedral groups D,, and the generalized quaternion groups Q, (Exercise 52, above). For which values of m are the groups nilpotent? 59. Show that the dihedral group Dm has ( m + 3)/2 conjugacy classes if rn is odd and (m + 6)/2 conjugacy classes if m is even. Describe the classes explicitly. 60. Begin, at least, to classify all groups of order p4, where p is a prime. When p = 2 there are 14 isomorphism types, when p > 2 there are fewer. 61. If H and K are subgroups of G, with K 4 G, K n H = 1, and K H = G, then G is called a semidirect product (or split extension) of K by H .

,

(1) If

0

= (12) E S,,

n 2 2, show that S , is a semidirect product of A , by

(0).

(2) Showthat thedihedralgroupD, = (a,bla" = b2 = l , b - ' a b = a - ' ) is a semidirect product of A = (a) by B = (b). (3) Show that the quaternion group Q2 can not be expressed as a semidirect product (except with one of the subgroups trivial). 62. Suppose K and H are groups and 4 : H -+ Aut(K) is a homomorphism. Let G be the Cartesian product K x H as a set, but with binary operation (x, y)(u,u) = (x . $(y)u, yu). Show that G is a group; denote it by G = K 4, H , and call it the external semidirect product of K by H relative to 4. ShowthatK, = { ( x , l ) : x ~ K )G,H, d = ( ( 1 , y ) : y ~ H )IGandthatGisthe semidirect product of K , by H , as in Exercise 61.

1.

PRELIMINARIES: IDEALS AND HOMOMORPHISMS

A ring is an additive abelian group R with a second associative binary operation, multiplication, the two operations being related by the distributive laws x( y

+ z ) = xy +

XZ

and

(y

+ z)x = y x + zx

for all x, y, z E R . In particular R is a semigroup with respect to multiplication. If xy = yx for all x, y E R we call R a cornmufarive ring. If R has a multiplicative identity 1 = 1, # 0, then we say that R is a ring with 1. It is an easy consequence of the distributive laws that x . O = 0 . x = 0 for all x in a ring R . A subring of a ring R is a subset S of R such that the binary operations on R restrict to binary operations on S so that S is also a ring.

EXAMPLES 1. Let R = Z, the integers, with the usual addition and multiplication. Then R is a commutative ring with 1. 2. Likewise if R = Q, R, or @, with the usual operations, then R is a commutative ring with 1. 3. If R is the additive group Z, and if we define multiplication via 5.6 = ab, then R is a commutative ring with 1 = T. 4. Let R = Z,, Z,Q, R, @, or any ring with 1, and suppose 0 < n E Z. Let M , ( R ) be the set of all n x n matrices having entries from R, with the usual 47

48

I1 Rings

operations of addition and multiplication for matrices. Then M J R ) is a ring with 1 = I , the n x n identity matrix. It is not commutative if n 2 2 or if R is not commutative. 5. Let A be an additive abelian group and let R = End@), the set of all endomorphisms of A. We add endomorphisms in the obvious fashion, viz., if 4,O E R and a E A, then ( 4 O)(a)= 4 ( a ) O(a),and we take multiplication to be composition of functions. Then R is a ring with 1 (the identity function on A ) ; it does not tend to be commutative.

+

Exercise 1.1.

+

Verify that End(A)is a ring if A is an additive abelian group.

Proposition 1.1. If R is a ring and S is a nonempty subset of R, then S is a subring of R if and only if x - y E S and x y E S whenever x, y E S. Proof. 3: Obvious. c=: S is an additive subgroup of R by Proposition 1.1.4, and multiplication in R restricts to a binary operation on S. The associative law for multiplication and the two distributive laws hold for S since they hold for all elements of R, and so S is a ring.

For example, the set S = 2 2 of all even integers is a subring of R = Z; S is nor a ring with 1. The ring R = M,(R) has a subring S consisting of all matrices a E R. In this case both R and S are rings with 1, but 1, # 1,. For a more substantial example let H be the subset of M4(1w)consisting of all matrices

[no:],

1,

a, b , c , d E R. An application of Proposition 1.1 shows that W is a subring of M4(R).If we write 1 = I, the identity matrix in M,(R), and set

o],

0-1 0 0 i = l 0 0 0 0 0 0-1 0 0 1

0 j = [ l0

0-1 0 0 0 10

0 0 k=[:

0 0

1 0

0-1

+

‘1

0-1

0

0 ’

0

0

then each x E W can be written uniquely in the form x = a1 bi + cj + dk, where a,b,c,d E R. Note that ij = k = -ji, j k = i = -kj, ki = j = -ik, and i 2 = j 2 = k 2 = - 1. The ring W,called Hamilton’s ring of quaternions, is of considerable importance in the history of algebra. It will be discussed further as we proceed. Observe that Q2 = { f 1, k i, -kj, fk ) C W is the quaternion group introduced in Chapter I.

I

Preliminaries:Ideals and Homomorphisms

49

If R is a ring with 1, then x E R is called a unit if it has a multiplicative inverse, i.e., if xy = y x = 1 for some y E R . The set of all units in R is denoted by U ( R )and is called the group of units.

Exercise 1.2. (1) Show that U ( R )is a group. (2) Find U ( R )when R = Z and when R = L,. (3) If R = M 2 ( Z )show that U ( R ) is the group of all matrices integer entries such that ad - bc = 1. Generalize.

[::I

with

If R is any ring we write R* for R\{O). It should be noted that this notation is not universally used. Some authors, for example, write R* for the group of units in R , which we have chosen to denote by U ( R ) . If x , y E R* and x y = 0 we say that xis a leji zero divisor and y is a right zero divisor. The distinction between left and right disappears, of course, if R is commutative. Note that Zhas no zero divisors, and that Z,has zero divisors if and only if n is not a prime. If x = [ -2

-’] 4

and

y=[:

:]

in R = M 2 ( Z ) ,then x y = 0, so x is a left and y a right zero divisor. A commutative ring R is called an integral domain if it has no zerodivisors, or equivalently if R* is a (multiplicative) semigroup. If R is a ring with 1 in which every nonzero element has a multiplicative inverse [i.e., U ( R ) = R * ] , then R is called a division ring, or a skew ,field. A commutative division ring is called a Jield. Clearly a field is an integral domain. Note that Z is an integral domain but not a field; Z,is a field if and only if n = p , a prime; Q, R, and C are all fields. The next exercise will show that the ring W of quaternions is a division ring.

Exercise 1.3. (1) If x = a1 + bi + cj + dk E W, define X = a1 - bi cj - dk. Verify (by matrix multiplication) that X X = Xx = N ( x ) l , where N ( x ) = a 2 + b2 + c 2 + d 2 E R. Conclude that W is a division ring. (2) Show that X y = JX,and use that fact to show that N ( x y ) = N ( x ) N ( y ) if x, y E W. If R and S are rings then a function f :R + S is called a (ring) homomorphism if f ( x y ) = f ( x ) f(y) and f ’ ( x y ) = f ( x ) f ( y )for all x , y E R . As usual we call f a monomorphism if it is 1-1 and an epimorphism if it is onto. If f is 1-1 and onto we call it an ~ s o m o r p h ~In s ~that . case f - ’is also an isomorphism from S to R ; we say then that R and S are isomorphic and write R z S. The kernel of a homomorphism f:R + S is kerf = { x E R : f ( x )= 0). It is easy to verify, with the aid of Proposition 1.1, that kerf and Im(f) are subrings of R and S, respectively. The kernel satisfies a further condition.

+

+

I1 Rings

50

If x E kerf and y E R , then f ( y x ) = f ( y ) f( x ) = f ( y ) . O = 0, and similarly f ( x y ) = 0, so both y x and x y are in kerf. A subring I of a ring R is called a left ideal if given any x E I and any y E R we have y x E I . Analogously I is a right ideal if x y E I whenever x E I , y E R . If I is both a left and a right ideal in R we say that I is an ideal, or a two-sided ideal, in R . Thus the kernel of a homomorphism is an ideal. Proposition 1.2. A nonempty subset I of a ring R is an ideal in R if and only if x - y E I , zx E I , and xz E I for all x, y E I and all z E R .

Exercise 1.4. (1) If m E Z show that I

(4 If

R

=

M,(H)

and

I =

=

mZ

{[:

=

{ m k : k E Z)is an ideal.

:]:a,c~Z)

show that I is a left ideal but not a right ideal. ( 3 ) If R is a ring with 1 and I is an ideal (left, right, or two-sided) in R such that I n U ( R ) # 121 show that I = R . I

Proposition 1.3. If is any collection of ideals in a ring R, then n{lo:cc E A } is an ideal in R .

=

Proof. Apply Proposition 1.2.

Clearly statements corresponding to Proposition 1.3 hold as well for collections of left ideals or of right ideals. If X is a subset of a ring R, then, by Proposition 1.3, { I : I is an ideal of R and X E I } is an ideal, the smallest ideal of R that contains X as a subset. We call it the ideal generated by X and denote it by ( X ) . If an ideal J of R is generated by a single element a E R we write J = ( a ) and say that J is a principal ideal. If R is a commutative ring with 1, then the principal ideal ( a ) consists of all multiples ra, r E R , and we may write ( a ) = Ra. For example, in the ring Z of integers ( 5 ) is the principal ideal consisting of all multiples of 5. Since an ideal I in a ring R is an additive subgroup the quotient RII = { x + I : x E R } is also an additive group. Suppose x + I = r + I and y + I = s + I in RII, i.e., that x - r and y - s are in I . Then

0

x y - rs = x y - ry

+ r y - rs = ( x - r ) y + r ( y

-

s) E I

since I is an ideal. Thus x y + I = rs + I and we may define a multiplication on RII by setting ( x + l ) ( y + I ) = x y + I . It is immediate that RII is a ring; we call it the quotient ring or the residue class ring of R modulo I . The quotient map q: R + R / I is a ring homomorphism from R onto RII, with ker q = I .

Theorem 1.4 (The Fundamental Homomorphism Theorem for Rings, or the FHTR). If R and S are rings and f : R + S is a homomorphism, with

1 Preliminaries:Ideals and Homomorphisms

kerf = I , then there is an isomorphism g : R / I i.e., the diagram

+ Im(f)

such that gq

51 =f,

is commutative. Proof. As far.as the additive structure of R is concerned this is just the FHT for groups. All that remains is to verify that g is in fact a (ring) homomorphism. But S(("

+N

Y

+ 1 ) ) = S ( X Y + [) = g ( v ( x y ) )= m

y ) = f ( X ) f ( Y ) = 9(11(X))9(9(Y)) = Q(X + M Y + I ) ,

and the proof is complete. Similar remarks apply to the analogs for rings of Proposition 1.1.12, the Freshman Theorem (1.1.13),and the Isomorphism Theorem (1.2.6). We state the results and leave the completions of the proofs as an exercise. Proposition 1.5. Suppose R and S are rings and $: R + S is an epimorphism, with kerf = I. Then there is a 1-1 correspondence between the set of all ideals J in S and the set of all those ideals K in R such that I G K , given by J - f - ' ( J ) = K . In particular each ideal in a quotient ring R / I has the form K I I for some ideal K , I c K L R .

Theorem 1.6 (The Freshman Theorem for Rings). Suppose R is a ring, I and J are ideals, and J L I . Then I / J is an ideal in RIJ, and ( R / J ) / ( I / J )z R/I. Theorem 1.7 (The Isomorphism Theorem for Rings). Suppose R is a ring and I and J are ideals. Then I + J and I n J are ideals and (I + J ) / I 2 J / ( I n J ) . A ring R is called simple if its only (two-sided)ideals are 0 and R. An ideal I of a ring R is called maximal if i # R and the only ideals J of R for which I E J c R are J = I and J = R . Thus, by Proposition 1.5, I is maximal in R if and only if R/Z is a simple ring. Exercise 1.5. (1) If I = (n), a principal ideal in the ring Z of integers, show that I is maximal if and only if n = p , a prime. (2) Show that M2([W) is a simple ring. ( H i n t : Try working with the matrices [A g], [ g A], [y El, and [ y].) Generalize.

52

I1 Rings

Proposition 1.8. If R is a ring with 1 and I # R is an ideal in R, then R has a maximal ideal M with I G M . Proof. We apply Zorn’s Lemma (see the Appendix). Let Y be the collection of all proper ideals J in R with I G J , partially ordered by inclusion. Then 1 E 9, so Y # 0. If %? is a chain in 9, it is easy to check that L = U { J : J E %} is an ideal with 1 E L, and L # R since 1 4 L. Thus L E Y and Lis By Zorn’s Lemma there is a maximal element M in 9, an upper bound for %?. and M is by definition a maximal ideal in R . Proposition 1.9. Suppose R is a commutative ring with 1. Then R is simple if and only if R is a field. Proof. -: If 0 # a E R , then (a) = R . In particular 1 E (a),so 1 = ba for some b E R. Thus a is a unit and U ( R ) = R*, so R is a field. t: Let I E R be an ideal, I # 0. Choose a E I , a # 0; then a E R* = U ( R ) , so a - ’ a = 1 E I . But then if r E R we have r . 1 = r E I , so I = R and R is simple.

Corollary. Suppose R is a commutative ring with 1. Then an ideal I in R is maximal if and only if RIl is a field. Proof. I is maximal if and only if RII is simple if and only if R/1 is a field. Exercise 1.6. The corollary to Proposition 1.9 does not generalize to noncommutative rings in the way one might imagine. Give an example of a ring R with 1 and a maximal ideal I such that RIZ is not a division ring. Exercise 1.7. Show that a finite integral domain is a field. If R is a commutative ring and P # R is an ideal in R, then P is called a prime ideal if given any a, b E R with ab E P, then either a E P or b E P (or both). For example, if R = Z,then both 0 and ( 5 ) are prime ideals; ( 5 ) is also maximal but 0, of course, is not. Exercise 1.8. Let R be a commutative ring. (1) Show that an ideal P in R is prime if and only if RIP is an integral domain. (2) If R is a ring with 1 show that every maximal ideal is prime.

2. THE FIELD OF FRACTIONS OF AN INTEGRAL DOMAIN The ring Zof integers is an integral domain, the ring Q of rational numbers is a field, and Z is a subring of Q. Each r E Q can be written as a fraction r = a/b, with a, b E Z,and no proper subfield of Q has that property (the simplest reason being that Q has no proper subfields).The exact same remarks apply if we consider the ring 2 7 of even integers as a subring of Q.

2 The Field of Fractions of an Integral Domain

53

We discuss in this section the imbedding of an arbitrary integral domain R as a subring in a minimal field. It is clearly necessary that R be an integral domain if it is to be a subring of a field. A field of fractions for an integral domain R is a field F R with a monomorphism Cp: R -P F R such that if K is any field and 8: R -+ K a monomorphism then there is a unique homomorphism (necessarily a monomorphism) f:F R .+ K for which the diagram

commutes, i.e., 8 = f Cp. Proposition 2.1. If an integral domain R # 0 has a field of fractions then FR is unique up to isomorphism.

FR,

Exercise 2.1. (1) If R is a field show that R itself is a field of fractions for R. (2) Show that Q is a field of fractions for Z and for 22. Theorem 2.2. If R # 0 is an integral domain, then R has a field of fractions. Proof. Let X be the Cartesian product R x R * = { ( a ,b):a,b E R, b # 0).

--

-

- -

Define a relation on X by agreeing that (a, b) (c, d ) if ad = bc. It is easily checked that is a reflexive and symmetric relation. If (a, b) (c, d ) (e, f ), then ad = bc and cf = de. Thus adf = bc f = bde, or afd = bed, and hence (af - be)d = 0. But d # 0 and R is an integral domain, so af = be, or (a, b) (e, f ). Thus is transitive and hence is an equivalence relation. Denote the equivalence class containing (a, b) by a/b, and let FR be the quotient set XI-, i.e., F R = {a/b:a,b E R, b # 0 ) . If a/b, c/d E F R we define a / b + c/d = (ad + bc)/bd and (a/b)(c/d)= ac/bd. Routine but tedious calculations show that the resulting operations of addition and multiplication for F R are well defined and associative. FR is in fact a commutative ring with 1, with 0 = O/b for any b E R*, - ( a / b ) = -a/b, and 1 = aja for any a E R*. If a/b # 0 in FR then necessarily a # 0 (Why?) and so bja E FR . But then (a/b). (b/a) = 1 in F R , so FR is a field. Define Cp: R FR by setting Cp(r)= raja for any a E R*. Further routine but tedious calculations show that is a ring homomorphism and that Cp is 1-1. Suppose then that K is a field and that 8: R -P K is a monomorphism. Define f:F R + K by setting f ( a / b ) = 8(a)8(b)-' (it is well defined). Then f is a monomorphism since 8 is a monomorphism, and f Cp = 8.

-

-

-+

54

I1 Rings

Suppose g: FR + K is another monomorphism such that g$ Then if r/s E FR we have

= f$ =

8.

g(r/s) = g((ra/a).( s a / a ) - l ) = g(ra/a)g(sa/a)-’ = g($(r))s(4(s))- =

W ) w -’ = f(r/s),

so g = f and FR is a field of fractions for R . Since r~ ra/a is a monomorphism from R into FR we may identify r with ra/a and take the point of view that R is a subring of F’. Then each element of FR is a fraction r/s, where I, s E R and s # 0. It is clear from these remarks that FR is (to within isomorphism) a minimal field F in which R is a subring. The construction of FR can be generalized considerably. Let R be any commutative ring and let S be a subset of R* that is a multiplicative semigroup containing no zero divisors. Let X be the Cartesian product R x S and define a relation on X by agreeing that (a, b) (c, d ) if ad = bc.

-

-

-

Exercise 2.2. (1) Show that the relation just defined is an equivalence relation on X . (2) Denote the equivalence class of (a, b) by a/b and the set of all equivalence classes by R,. Show that R , is a commutative ring with 1. (3) If a E S show that { r a / a : rE R } is a subring of R , and that Z H ra/a is a monomorphism, so that R can be identified with a subring of R,. (4) Show that every s E S is a unit in R , . ( 5 ) Give a “universal” definition for the ring R , and show that R , is unique up to isomorphism.

The ring Rs is called the localization of R at S . The concept is important in algebraic number theory and algebraic geometry. Exercise 2.3. Suppose R is an integral domain and P G R is a prime ideal. (1) Show that both P and R\P are multiplicative semigroups. (2) If S = R\P show that U ( R , ) = Rs\R,P. Conclude that RsP is the unique maximal ideal in R , .

3. Polynomials If R is a ring, then intuitively a polynomial in one variable x with coefficients in R is an expression of the form f ( x ) = a,

+ a,x + a 2 x 2 + ’.. + anxn,

where a, E R and x is a “variable” that may be assigned values from R or from some ring that contains R as a subring. We present next a construction that makes these notions precise.

3 PoIynorniaIs

55

If R is a ring denote by P ( R ) the set of all sequences a = (a,) = (a,, a,, a 2 , .. .), where each a, E R and a, = 0 for all but finitely many values of i, 0 Ii E H . If a, b E P(R)we define a

+ b = (ai + bi),

Theorem 3.1. If R is a ring, then P( R ) is also a ring. It is commutative if and only if R is commutative and it is a ring with 1 if and only if R is a ring with 1, in which case l,,,, = (I,,O,O,. . .). Exercise 3.1. Prove Theorem 3.1. Suppose now that R is a ring with 1 and set x = (0, 1,0,0,. . .) E P(R).Note that x 2 = (O,O, 1,0,. ..), x 3 = (O,O, 0,1,0,. ..), . . ., and in general x" has 1 as its (n + 1)st entry and 0s elsewhere. We agree that x o = lp(R).The map a ~ ( a , 0 , 0...) , is a monomorphism from R into P ( R ) , and we may thus identify R with a subring of P ( R ) ,with 1, = I,,,,. From that point of view we may write a = (a,,a,,a,, ...) = a, + a,x a2x2 ...

+

+

for each a E P(R). We call x an indeterminate, and write R [ x ] for P ( R ) .We write f ( x ) [or g(x), etc.] for a = a, a , x + . . . E R [ X I ,and call f ( x )a polynomial with coeflcients a,, a , , . . . in R . If a, # 0 but a, = 0 for all m > n we say that f ( x ) has degree n and write degf(x) = n; in that case we call a, the leading coeficient of f ( x ) . If f ( x ) has leading coefficient 1 we say that f ( x )is a monic polynomial. The zero polynomial (O,O,O, ...) is denoted by 0, and we agree that degO = - co. Polynomials of degree 0 or - co are called constants-they are, of course, just the elements of R viewed as a subring of R [ x ] .

+

Proposition 3.2. Suppose R is a ring with 1 and f ( x ) , g ( x ) E R [ x ] . Then (a) deg(f(4 + d x ) ) 5 max{deg f(x),deg g ( x ) l ,and (b) deg(f(x)g(x)) deg f(4+ deg g ( x ) . Equality always holds in (b) if R has n o left or right zero divisors.

Proof. These statements are obvious consequences of the definitions of addition and multiplication in R [ x ] . Corollary 1. If R has no left or right zero divisors, then f ( x ) E R [ x ] is a unit if and only if f ( x ) = r, a constant, with r E U ( R ) . Corollary 2. REX] is an integral domain if and only if R is an integral domain.

56

I1 Rings

Exercise 3.2. Find U(R) if R = &[XI.

Theorem 3.3 (The Division Algorithm). Suppose R is a commutative ring with 1 and f(x), g(x) E R[x]. If g(x) has leading coefficient b, then there exist a nonnegative integer k and q(x), r(x) E R[x] such that bkf(x)= q(x)g(x) + r(x), with degr(x) < degg(x). If b is not a zero divisor in R, then q(x) and r(x) are unique. If b E U(R) we may take k = 0. Proof, If degf(x) < deg g(x) we may take k = 0, q(x) = 0, and r(x) = f(x). Assume then that degf(x) = m 2 deg g(x) = n, and say f(x) has leading coefficient a E R. Proceed by induction on m. Set bf(x) - ax"'-"g(x) = fl(x). Clearly degfl(x) < m, so we may write bk-'fl(x) = p(x)g(x) + r(x), where k - 1 is some nonnegative integer, p(x),r(x) E RCx], and deg r(x) < deg g(x). Thus bkf(x) = bk-'ax"-"g(x) + bk-'f1(x) = bk- 'aX"-"g(x) p(x)g(x) T(X) = (bk-'axm-" + p(x))g(x) + r(x).

+

+

The main result follows if we set q(x) = bk-laxm-" + p(x). Suppose b is not a zero divisor. If also bkf(x) = q,(x)g(x) + rl(x), with degr,(x) < degg(x), then ( 4 ( 4 - q,(x))g(x) = r1(x) -

w

If q(x) # ql(x), then the polynomial on the left-hand side has degree at least n since b is the leading coefficient of g(x), but the polynomial on the right has degree less than n. Thus q(x) = q,(x), and consequently t(x) = rl(x). Finally, if b E U ( R )we may simply multiply through by b-k and replace q(x) and r(x) by b-kq(x)and b-kr(x), respectively. Note that the proof actually suggests the "long division" algorithm learned in high school algebra. The polynomials 4(x) and r(x) are called the quotient and remainder upon division of f(x) by g(x). The division algorithm also holds when R is not commutative if we assume that b is a unit (Exercise 8.15). From this point on we will discuss polynomial rings R[x] only for R a commutative ring with 1. Noncommutative versions of the next results can be stated and proved but we shall only have occasion to use polynomials having coefficients from a commutative ring in later chapters.

Theorem 3.4 (Substitution). Suppose R and S are commutative rings with 1, that 4: R + S is a homomorphism with 4(lR)= l,, and that a E S . Then there is a unique homomorphism E,: RCx] -+ S such that E&) = +(r) for all r E R and E,(x) = a.

3 Polynomials

Proof. If f ( x ) = ro

+ r l x + ... + r,x" E R [ x ]

57

set

E a ( f ( x ) )= 4(ro) + + ( r l ) a + 4(r2)a2 + * * * + 4(rn)aflE S . It is an easy exercise to verify that Ea is a homomorphism with the right properties. If also F : R [ x ] + S is a homomorphism with F(r) = 4(r) for all r E R and F(x) = a, then F ( f ( x ) )= F(ro r l x ... r,x") = F(ro) F ( r , ) F ( x ) ... F(r,)F(x)"

+ + + + + +

4 ( r o ) + 4(r,)a + ... + 4(rn)afl = E a ( f ( x ) ) . Theorem 3.4 describes a universal mapping property of R [ x ] that could have served as a definition prior to the construction. As usual we may conclude uniqueness up to isomorphism. A particular case of Theorem 3.4 occurs when R is a subring of S, 1, = l,, and 4 is the inclusion map. Then E a ( f ( x ) )= r,, r l a ... r,a". This is simply the result of replacing, or "substituting," each occurrence of the indeterminate x by the element a of S . We write f ( a )for the result. The image of E, is a subring of S which we shall denote by R [ a ] . Thus R [ a ] = { f ( a ) : f ( x ) E R [ X I 1. =

+

+ +

Proposition 3.5 (The Remainder Theorem). Suppose R is a commutative ring with 1, f ( x ) E R [ x ] , and a E R . Then the remainder upon division of f ( x ) by g(x) = x - a is r = f ( a ) .

+

Proof. Writef(x) = q(x)(x - a) r by the Division Algorithm [note that r is a constant since degg(x) = 13. Substitute a for x to see that

+

f ( a ) = q(a)(a - a) r = q(a). 0 + r = r. Corollary (The Factor Theorem). Suppose R is a commutative ring with 1 , f ( x ) E R [ x ] , a E R , and f ( a ) = 0. Then x - a is a factor of f ( x ) ,in the sense that f ( x ) = q ( x ) .( x - a) for some q(x) E R [ x ] . If R and S are commutative rings with R G S and 1 , = l,, then an element a E S is called algebraic over R iff@) = 0 for some nonzero polynomial f ( x ) E R [ x ] . If a E S is not algebraic over R we call it transcendental over R (equivalently a E S is transcendental over R if E, is a monomorphism). For example, $ E R is algebraic over Z since f ( $ ) = 0 for f ( x ) = x 2 - 2 E Z [ x ] . However, 7c E R is transcendental over E-the proof will be given later in Chapter 111. Exercise 3.3. Show that

4 + f l E R is algebraic over H.

If R is an integral domain with 1, then the field of fractions of R [ x ] is called the field of rational functions over R . It is commonly denoted by R(x). Thus R ( x ) = { f ( x ) / s ( x ) : f ( x g) ,( x ) E RCXl, g(x) z 01.

58

I1 Rings

Exercise 3.4. If R is a commutative ring with 1, f ( x ) E R [ x ] , and a E R , then we may substitute a for x and obtain f ( a ) E R . Thus f ( x ) determines a polynomial function f : R + R .

(1) If R is finite show that there must exist polynomials f ( x ) and g(x) in R [ x ] , with f ( x ) # g(x),such that the associated polynomial functions f and g are identical, i.e., f ( a ) = g(a), all a E R . (2) Find explicit examples of the phenomenon in (1) when R = Z,. (3) If R is an infinite integral domain show that the mapping f ( x ) ~ f assigning to each polynomial in R [ X I its corresponding polynomial function is 1-1. 4. POLYNOMIALS IN SEVERAL INDETERMINATES Let I = {0,1,2,3,. . .} and write I" for the Cartesian product of n copies of I , 1 I n E Z. If R is a ring denote by P,,(R)the set of all functions a: I" -+ R having finite support, i.e., having nonzero value for only finitely many elements of I " . If n = 1 such a function is just a sequence indexed by I and having only finitely many nonzero entries, so it is a polynomial as discussed in the previous section. Thus P,(R) = P ( R ) . Let us write 0 for the element ( O , O , . . . ,0) E I", and if i = ( i l , i,,. . .,i n ) E I" a n d j = ( j ,, j 2 , .. . ,j,) E I" define i

+j

= (it

+ j t , i, + j,, . . . ,in + j,) E I".

We define operations of addition and multiplication on P,,(R) by setting (a

+ b)(i)= a(i) + b(i),

(ab)(i)= C { a ( j ) b ( k ) : jk, E I", j

+k =i}

for all a, b E P,(R), i E I".

Theorem 4.1. If R is a ring, then PJR) is a ring; it is commutative if and only if R is commutative and it is a ring with 1 if and only if R is a ring with 1 . Proof. The proof is straightforward. We show that multiplication is associative as a sample and leave the rest as an exercise. If a, 6, c E P,(R) and i E I", then

((ab)c)(i)= z { ( a b ) ( j ) c ( k ) : kj , E I", j + k = i} = ~ { ( ~ { a ( u ) b ( uu) E : uI",, u + u = j } ) c ( k ) : jk, = C{(a(u)b(v))c(k):u, u, k E I", u + u + k = i } ,

E l",j

+ k = i>

4

Polynomials in Several Indeterminates

59

and j ) : u ,j E I", u + j = i} (a(bc))(i)= C{(a(u)(bc)( = C(a(u)~(b(u)c(k):u, k E I",u + k = j ) : u , . j E I", u j = i> = C{a(u)(b(u)c(k)):u,U, k E I", u u k = i } = ((ab)c)(i).

+ +

+

We remark that if R is a ring with 1, then the function 1: I" -+ R defined by l(0) = 1 E R, 1(i) = 0 E R if 0 # i E l",is the multiplicative identity of P,(R). Exercise 4.1. Complete the proof of Theorem 4.1.

For each r E R define a function a, E P,(R)by setting a,(O) = r a n d a,(i) = 0 if 0 # i E I". Then a, + a, = and a,a, = arsfor all r, s E R. Clearly the map I t - + a, is 1-1 from R into P,,(R),so it is a monomorphism. As usual we identify r E R with a, E P,,(R) and view R as subring of P,(R). Suppose now that R is a ring with 1. Let f?k be the n-tuple in I" having 1 as its kth entry and 0's elsewhere. Define xk E P,,(R) by setting xk(ek) = 1 and x k ( i )= 0 if ek # i E I". We see by multiplying that x i ( 2 e k )= 1 (where 2ek = e, + ek),and x i ( i ) = 0 otherwise, and in general that x r ( m e k )= 1, x r ( i ) = 0 otherwise, 1 I rn E z. We agree as usual that xf = 1 E P,(R). The functions xk all commute with one another. If we take i = ( i l , . . . , i n ) in I " and r E R and consider the element r x y *..x:: E P,,(R) we find that its value at i is r and its value is 0 elsewhere. Since any nonzero a E P,,(R)can be written uniquely as a sum of functions having distinct one-point supports it follows that each nonzero a E P,(R) can be written uniquely as a sum of elements each of the form .

.

rx;lxy.. .x;,

0 # r E R , i = (il, i,, . . . , i n ) E I".

An element of that form is called a monomial in P,,(R). The degree of a = rxy . . . xk is deg a = i , + i, . . . + in. More generally if a E P,(R), a # 0, and a is written as a = a , + . I . + a,, its unique expression as a sum of monomials, then we define the degree of a as deg a = max{deg ai:1 Ii Im}. As usual we agree that deg 0 = - KI. If all the monomial summands of a E P,,(R) have the same degree we say that a is homogeneous. The elements of P,(R) are called polynomials in the n commuting indeterminates x l r x,,. . ., x , . In accordance with tradition we write R [ x l , x 2 , ..., x,] for P,,(R), and denote its elements by f ( x , , x, ,..., xn), etc. It is often typographically convenient to write X for the n-tuple ( x l , x,, . . . , x,) of indeterminates, and write . f ( x , ,. . ., x,,) as f ( X ) when the number of indeterminates is clear from the context. When n = 2 or 3 it is customary to write x1 = x, x 2 = y, and x 3 = z. For example, f ( x , y) = x 2 y 3 - 3x"y E Z[x, y ] is homogeneous of degree 5.

+

60

11 Rings

Proposition 4.2. Suppose R is a ring with 1 and f(X), g(X) E R [ x l , - . . ,x , ] . Then

(a) deg(f(X) + S(X)) 5 max{deg f ( X ) ,deg S(X)l, and (b) deg(f(X)g(X)) I deg f(x)+ deg dx). Equality always holds in (b) if R has no zero divisors. Theorem 4.3 (Substitution). Suppose R and S are commutative rings with 1 and4:R+Sisahomomorphismwith4(lR) = l S . I f a , , a , , . . . , a , E S , then there is a unique homomorphism E = E,, ,,,,,, from REX,, x , ,..., x , ] to S such that E(r) = 4 ( r ) ,all r E R , and E ( x i ) = ai, 1 I i I n. Proof. We define E ( r x F x $ . . .x:) = $(r)a?a?. . .a:

for each monomial, extend to arbitrary polynomials in the obvious fashion, and proceed as for polynomials in one indeterminate. Thus R [ x , , . . ., x , ] could also have been defined abstractly in terms of the universal mapping property in Theorem 4.3, and is unique in that respect up to isomorphism. Other constructions are possible. Perhaps the most popular is first to construct R [ x ] by some means, then to construct ( R [ x ] ) [ y ] ,and then to say “et cetera” (i.e., apply induction). Because of the uniqueness one construction is as good as another, but the latter seems to have an aesthetic deficiency in that the indeterminates do not arise in a symmetric fashion. The finiteness of the cardinal number n played no essential role in the construction of the polynomial ring P,(R). Thus the discussion above would have served, with minor changes in notation, to define R[x,:cr E A ] for any nonempty index set A and commuting indeterminates x,. When R E S, with 1, = 1,. and 4: R + S is the inclusion map, then the homomorphism E of Theorem 4.3 allows us to substitute elements a,, . . ., a, E S in any polynomial f ( x , , . . . , x n ) E R [ x , , . . . ,x,] and obtain f(a,, . . . , a,) E S . The image of E is a subring R [ a , , . . ., a,,] of S . Elements a , , . . . , a, E S are called algebraically dependent over R if j ’ ( a l , .. ., a,) = 0 for some nonzero polynomial f(X) E R [ x , , . . . , x , ] ; otherwise they are algebraically independent. Exercise 4.2. Show that a, ent over h.

=

and a, = fi are algebraically depend-

When 0 exponents occur in monomials it is customary to suppress the corresponding indeterminate, in accordance with the usual convention that xo = 1. Thus, for example, we simply write S x ~ x ~ x as ~ x5x2x:. : With that notational convention we have R [ x , ] E R [ x , , x , ] c R [ x l r x 2 , x 3 ]G ....

5

Divisibility and Factorization

61

For an infinite sequence x l , x 2 , .. . of distinct commuting indeterminates we maywriteR[x,,x,,x, ,...] = U{R[x, ,..., x J : l I kEB},withtheobvious addition and multiplication, and obtain the polynomial ring in x,, x 2 , x3,. . . over R. It will be useful for examples. The field of fractions of R[x, ,. . ., x,] consists of all rational functions f ( x l , . . .,x,)/g(x,, ...,x,), g(x, ,. ..,xn) # 0. It is denoted by R(x,, ...,xn),and is defined, of course, only if R is an integral domain. 5. DIVISIBILITY AND FACTORIZATION

We assume throughout this section that R is an integral domain with 1. If a, b E R we say that a divides b, or b is a multiple of a, if b = ac for some c E R, and then we write a I b. If a I b and b 1 a we say that a and b are associates and write a b. Note that the only associate of 0 is 0, and the associates of 1 are just the units in R. N

-

Proposition5.1. Suppose R is an integral domain with 1 and a, b E R. Then a and b are associates (a b) if and only if a = bu for some u E U(R). Proof. We may assume a # 0. a : Since a b we have a = bc and b = ad for some c, d E R. Thus a = (ad)c = a(dc),so dc = 1 and c is a unit. -=: Ifa = bu, with u E U(R), then b a, but also b = au-', so a I b. It is clear that the relation of being associates is an equivalence relation. The equivalence class containing a E R is just aU(R). Note that if b E R and u E U ( R ) ,then u 1 b since b = u(u- 'b).If b E R is not a unit and the only divisors of b are units and associates of b, then b is called irreducible. Suppose that 0 # p E R, that p is not a unit, and that if p I ab for some a, b E R, then necessarily p I a or p I b. Then p is called a prime in R. Proposition 5.2. If p E R is prime, then p is irreducible. Proof. If not we may write p = ab, with a, b 4 U(R). Since p is prime p I a or p 1 b, say p 1 a, with a = pc, c E R. But then p = ab = (pc)b = p(cb),so cb = 1 since p # 0. But then b E U(R),a contradiction. It is important to note that the converse to Proposition 5.2 may fail to hold, as the following exercise shows. Exercise 5.1. Let R = { a + b P : a , b E Z } c C. N

I

(1) Show that R is an integral domain with 1 (it is a subring of C). (2) Show that U ( R )= { 1). ( 3 ) Show that 3 is irreducible in R. and b = 2 - p a r e both irreducible in R. (4) Show that a = 2 + ( 5 ) Conclude that 3 X 2 + and 3 $ 2 in R. ( 6 ) Conclude that 3 is irreducible but not prime in R.

62

I1 Rings

Recall that a principal ideal in a commutative ring R with 1 is an ideal ( a ) generated by a single element a E R, consisting of all multiples ra, r E R, so (a) = Ra. If R is an integral domain with 1 in which every ideal is principal, then R is called a principal ideal domain, abbreviated PID. The most prominent example of a PID is the ring Z of integers. If I is a nonzero ideal in Zlet a be the smallest positive element in I . If b E I we may write b = aq + r, with q, r E Z and 0 5 r < a. But then r = b - aq E I and r = 0 by the minimality of a, so b = qa E Za = (a),and I = (a). A common divisor of two elements a and b in an integral domain R is an element c E R such that c 1 a and c [ b. A common divisor d of a and b is a greatest common divisor (or GCD) if c 1 d for every common divisor c of a and b. For example a = 143 and b = 154 have d = - 11 as a GCD in the ring H of integers.

Proposition 5.3. Suppose R is a PID and a, b E R are not both zero. Then a and b have a GCD, denoted by d = (a, b). It is unique up to associates in R, and it can be expressed as d = (a, b) = xu + yb for some x, y E R. Proof. Let I = (a, b), the ideal generated by a and b. Then I = {ua + vb:u, v E R ) . Since R is a PID we may write I = ( d ) for some d E I , say d = x u + yb. Since a E I and b E I we have d I a and d I b. If c is any common divisor of a and b, then c I xu + yb, or c 1 d , so d is a GCD for a and b. If d , is another, then d I d , and d , 1 d, so d and d , are associates. Corollary. If R is a PID, then every irreducible element of R is prime.

Proof. Let p E R be irreducible and suppose p 1 ab for some a, b E R. If p ) a, then (p,a) = 1, so we may write 1 = x u yp for some x, y E R. Thus

+ b = ( X U + yp)b = x(ab) + (yb)p,

soplb. Suppose rn E Z, rn # 0 or 1. For an important class of examples consider the subring O [ f i ] of Q= obtained by substituting rn for x in O [ x ] . If m = k2n, with k, n E Z and k > 1, then clearly O[ m ] = a[$], so we will only consider the case where m is square-free, i.e., not divisible by any square k2 > 1 in Z. Exercise 5.2. Show that O [ f i ] = { r + s,,h:r, s E a), and that O [ f i ] is a field. It is thus its own field of fractions, and we will write Q ( f i ) rather than O [ f i ] . For any square-free integer m( #O, 1) we define the ring R, of algebraic integers in Q(Jm) as follows. If m = 2 or 3(mod 4), then R , = { a + b J - : a, b E h),and if rn = 1 (mod 4), then R , = { ( a + b@)/2:a, b E Z and a, b have the same parity) (having the same parity means that a and b are either both even or both odd).

i

5 Divisibility and Factorization

63

Exercise 5.3. (1) Show that R , is an integral domain with 1. (2) Show that O ( f i ) is the field of fractions for R,. ( 3 ) Show that R , is the set of all those r + s f i E O ( f i ) that are roots of a monic quadratic polynomial x2 c x + d E Z [ x J . [This is the reason for the variation in the definition of R , when m = l(mod 4).]

+

+

For any x = r s f i E O(fi) we define the norm of x to be N ( x ) = r2 - ms2. Note that N ( x ) = ( r + s , , h ) ( r - s f i ) . Thus if x , y E Q(fi) we have N ( x y ) = N ( x ) N ( y ) . Exercise 5.4. (1) Show that N ( x ) E Z if x E R,. (2) Show that u E U ( R , ) if and only if N ( u ) = & 1. (3) Use (2) to show that U ( R - , ) = { k l , L-i}, U ( R - , ) = { f l , +(1 f f l ) / 2 } , and U ( R , ) = { f 1) for all other negative square-free rn in Z.

We remark that if rn E Zis square-free and m 2 2, then U ( R , ) is infinite (see Exercises 8.32 and 8.33). Proposition 5.4 (Dedekind and Hasse, see [31]).

Let m be square-free in

Z,m # 0, 1, and let R , be the ring of algebraic integers in Q(Jk). Suppose that

for all nonzero x , y E R , such that y ',j x and (N(x)l 2 ( N ( y ) l there exist u, u E R , such that x u # yu and ( N ( x u - yu)l < IN(y)l. Then R , is a PID. Proof. Let I # 0 be an ideal in R , . Choose y E I , y # 0, so that IN(y)l is minimal. Given any x E I we wish to show that yl x . if not then y y x and IN(x)l 2 IN(y)l so we may choose u, u E R , such that xu - yu # 0 but ( N ( x u- yo)( < IN(y)l. But that is a contradiction since x u - yo E I and ( N (y)l was minimal. Thus y I x, so I = ( y ) .

Corollary. If m = - 19, then R , is a PID. Proof (Wilson [39]). Since - 19 = l(mod 4) we know that R - , 9 consists of all (a b f l ) / 2 for a, b E Z,a and b having the same parity. Note that N ( x ) 2 0 for all x E Suppose x , Y E R - , , , y I f x , and N ( x ) 2 N ( y ) . We need u, u E R - 1 9 such that x u # yu and N ( x u - yu) < N ( y ) , or equivalently N ( ( x / y ) u - u) < I . Since x / y E Q ( m ) \ R - 19 we may write x / y = (a b m 9 ) / c , where u, b, c E Z are relatively prime and c > 1. There are four cases to consider.

+

Q(m).

+

Suppose c = 2. Then a and b are of opposite parity since x l y 4 R - 9 . (i) But then u = ( ( a - 1) b F 9 ) / 2 is in R - 1 9 r so we may take u = 1 and obtain ( x / y ) u - u = 9, with 0 < N ( 9 ) = < 1. (ii) Suppose c = 3. Then either 3 $ a or 3 $ b , so a 2 + b2 f O(mod 3). Since a2 + 19b2 = a2 + b2(mod 3) we may divide by 3 and obtain

+

a

a2 + 19b2 = 3q + Y, r

=

1 or 2.

64 Set u

I1 Rings = a - b-

and u = q. Then

+ b m ) ( a - b,/-19)/3 = (a2 + 19b2)/3 - q = r/3,

( x / ~ )u u = (U

4

-q

and N ( r / 3 ) = or $. (iii) Suppose c = 4. Then a and b cannot both be even since (a, b, c) = 1. If they are of opposite parity, then a2 19b2 = a’ - b2 = l(mod 4), so we may divide by 4 and write a’ + 19b2 = 4q r, r = 1 or 3. In that case set u = a - b d q and u = q, and obtain (x/y)u - u = r/4, of norm & or A. If a and b are both odd, then u2 19b’ = a2 3b2 = 4(mod 8) and we may write u2 19b’ = 8q, 4. In that case set u = (a - b m ) / 2 and v = q l , so that 1 (x/~)u u = (a’ + 19b2)/8 - 41 = 7,

+

+

+

+

+ +

of norm b. (iv) Suppose c 2 5. Since (a, b, c) = 1 we may choose d, e, f E Zsuch that ad + be + cf = 1, then we may divide ae - 19bd by c to obtain q, r E Z such m and that ae - 19bd = cq r, with IrI I 4 2 . (Why?) Set u = e d u =q Then

+ fe.

(X/Y)U - v =

(a

+b

= (ae = (r

and N ( ( r + -9)/c)

+

+ d&G)/c - (4 - fJ-19) 19bd - qc)/c + (ad + be + c f ) p / c

+m

= (r2

(r2

m ) ( e ) / c ,

+ 19)/c2.If c > 5, then

+ 19)/c2 I (c2/4 + 19)/c2 = a + 19/c2 lb+%=$
and if c

=

5, then Irl I 2 , so (r2 + 19)/c2 5 (4

+ 19)/25 < 1.

Exercise 5.5. Use Proposition 5.4 to show that R - ,, the ring of Gaussian integers, is a PID (this is easier than the corollary above). An integral domain R with 1 is called Euclidean if there is a function d : R* -,H , with d(r) 2 0 for all r E R*, such that

(i) if a, b E R* and a I b, then d(a) Id(b), and (ii) if a, b E R , with b # 0, then there are q, r E R such that a either r = 0 or d(r) < d(b).

= bq

+ r, with

Clearly R = Z is Euclidean if we define d(r) = Irl for all r E Z*. For another example take R = F [ x ] , where F is a field, defining d ( f ( x ) ) to be degf(s). Then R is Euclidean by Proposition 3.2 and Theorem 3.3.

5 Divisibility and Factorization

65

--

Proposition 5.5. If R is Euclidean, then U ( R )= { x E R * : d ( x ) = d(1)). b in R * , then d(a) Id(b) and d(b) Id(a), so 1, so d ( u ) = d(1). If x E R* and d ( x ) = d(1) write1 = q x + r , w i t h q ~ R a n d r = O o r d ( r ) < d ( x ) = d ( l ) . B u t i f r # O , t h e n d(1) I d(r) since 1 I r. Thus r = 0, qx = 1, and x E U(R).

Proof. Observe that if a

d(a) = d(b). If u E U ( R ) ,then u

Proposition 5.6. If R is Euclidean, then R is a PID. Proof. Let 1 be any nonzero ideal in R. Choose b E I* with d(b) minimal. If a E I wemaywritea = bq + r,withr = Oord(r) < d(b).But r = a - bq E I, so d ( r ) < d(b) is impossible, and so r = 0. Thus a = qb E (b), and 1 = (b). Exercise 5.6. (1) Show that the ideal I = ( 3 , 2 is not principal in R - (see Exercise 5.1). Conclude that R - is not Euclidean. (2) If S is an integral domain with 1 set R = S[x, y] and let I = ( x , y). Show that I is not a principal ideal, so R is not Euclidean.

+ G)

Proposition 5.7. If m = - 2 , d(r) = IN(r)l for all r E R:.

-

1, 2, or 3, then R , is Euclidean with

Proof: Take a, b E R,, with b # 0. Then a/b E Q(fi), say alb = s + with s,f E Q. Choose c,d E Z nearest to s and t, respectively, so that 1s - c J I f and It - dl I$. Set q = c + d f i and r = a - bq, so that t f i

r / b = a/b - q

Thus N(r/b) = ( r - c)’

-

= (s

-

c)

+ ( t - d)&.

m(t - d)’. If m = - 2 or

0 I N(r/b) I $ - m/4 < 1,

-

1, then

-a

so either r = 0 or d ( r ) < d(b). If rn = 2 or 3, then IN ( r / b )I$, so IN(r/b)l I and again either r = 0 or d(r) < d(b). Exercise 5.7. Extend Proposition 5.7 by showing that R , is Euclidean [with d(r) = N ( r ) ] if m = - 3, - 7, or - 1 1. [ H i n t : Choose d E Z nearest to 2v and then choose c E Zso that c is as near to 2u as possible with c and d of the same parity. Then set q = (c + d f i ) / 2 . ]

a,

Proposition 5.8. Suppose m E Z is negative and square-free, but rn # - 1, or - 11. Then R , is not Euclidean.

- 2, - 3, - 7,

Proof. Suppose, by way of contradiction, that R , is Euclidean, with function d. We know by Exercise 5.6 that m # - 5, so m = - 6 or - 10 or else m I - 13. Choose h E RZ\U(R,) so that d(b) is minimal. Given any a E R , there are q, r E R , such that a = bq r and either r = 0 or d(r) < d(b). By the minimality of d(b) we may conclude that either r = 0 or r = f 1, since U(R,) = { f 1) by Exercise 5.4. Thus for any a E R , we may write either a = bq or a = bq & 1, and hence either b ( a or b l a f 1. Taking a = 2 we see in particular that b I 2 or b 1 3.

+

66 m

11 Rings

Let us see that 2 and 3 are both irreducible in R,. For example, if we could write

= l(mod 4 ) and if 3 were reducible in R,

+

+

3 = ((u u,/%)/2) . ( ( x y,/%)/2), and neither factor a unit in R,. Thus

with u, v, x, y E Z

N ( 3 ) = 3, = ((u2 - m u 2 ) / 4 ) . ( ( x 2- my2)/4),

and so (u2 - mu2)/4 = ( x 2 - m y 2 ) / 4 = 3 (since neither factor is a unit). But then u2 - mu2 = 12, which has no solutions in integers since m = -6 or - 10 or m I - 13 (try u = 0, 1,2), and we have a contradiction. The proof is similar that 2 is irreducible, and the proofs are similar and easier when m = 2 or 3(mod 4). Since 2 and 3 are irreducible we may conclude that b = f 2 or b = f3. But now if m = l(mod 4 ) take a = (1 + &)/2, so a + 1 = (3 + f i ) / 2 and a - 1 = (- 1 + fi)/2. Clearly none of a, a + 1, or a - 1 is divisible by 2 or 3 [since(l + ,/%)/4 $ R,,etc.], and we have a contradiction. If m = 2 or 3(mod 4 ) take a = 1 + ,,hAgain it is clear that none of a , a + 1, or a - 1 is divisible by 2 or 3, and the proof is complete. Corollary. R-

19

is a PID that is not Euclidean.

We remark that 2 is reducible in R - and 3 is reducible in R -

,.

An ascending chain I , c I , c I , E ... of ideals in a ring R is said to terminate if it is either finite or there is some index k such that lj = Ik for all j 2 k. A commutative ring R is called Noetherian if it satisfies the ascending chain condition (ACC), i.e., if every ascending chain of ideals of R terminates. Proposition 5.9. If R is a PID, then R is Noetherian. Proof. If I, E I , G I , C ... is any infinite ascending chain of ideals then I = {Ik:1 I k E Z) is easily checked to be an ideal in R. Thus I = (a)for some a E R. But then a E lk for some k, so I = lj = for all j 2 k, and the chain terminates.

u

We define a uniquefactorization domain (UFD)to be an integral domain R with 1 in which every nonzero nonunit element is a product of irreducible elements and every irreducible element is prime. Proposition 5.10. If R is a PID then R is a UFD. Proof. Let X be the set of all elements in R*\U(R) that can not be written as products of irreducible elements. We wish to show that X is empty. If not choose a , E X . If possible choose a, E X such that (a,) c ( a z )but ( a , ) # (a,). Then choose a3 E X such that (a,) c ( a 3 )but ( a , ) # ( a 3 ) etc. , The process must

5

67

Divisibility and Factorization

terminate, by Proposition 5.9, with some a, E X such that (a,) is not properly contained in any (x) for x E X. Since a, E X it is not irreducible, and we may writea, = a,+ ,h,witha,+, andbin R*\U(R).Oneof a,+, a n d b m u s t b e i n x , or else a, would not be in X, so assume a,, E X . But then (a,) E (an+1) and (a,) # (a,+ ,), a contradiction, so in fact X = @. An application of the corollary to Proposition 5.2 completes the proof. Theorem 5.1 1 (Unique Factorization). If R is a U F D and a E R is not 0 and not a unit, then a = plpz ".pk, where each pi is a prime. The representation of a as a product of primes is unique in the sense that if also a = q , q2...q,, with each q i prime, then m = k , and by relabeling if necessary we k. have pi qi,1 5 i I Proof. Only the uniqueness needs to be proved, and we use induction on k. Uniqueness is clear when k = 1 for then a is irreducible, so we assume the uniqueness of representation for a product of k - 1 primes. Since p l p 2 " ' p k =qlqz...qm we have p 1 1 q 1 q 2 . . . q m and hence p l l q i for some i, since p , is prime. Relabel the qs if necessary so that p I I q l .But then pI q 1 since both are irreducible, so q , = p , u for some U E U ( R ) . Thus pIp2"'pk = pluq,q,..-q,, and hence p 2 " - p k = q;q3...qm. where q; = uq, q,. By the induction hypothesis k - 1 = rn - 1, so k = rn, and we may relabel if necessary so that p , q; q 2 ,p , q 3 , .. . , Pk q k . If R is a U F D it follows easily from Theorem 5.11 that each nonzero nonunit a E R can be written as a = p;'p;'. . . P??

-

-

- -

-

-

-

where n, e l , e 2 , . . , e, E Z are unique and positive, and p , , p,,. . . , p , are distinct primes in R, unique up to associates. As a consequence of Proposition 5.10 and Theorem 5.1 1 we see that unique factorization into primes holds in Z (that fact is often called the Fundamental Theorem of Arithmetic), in F [ x ] for any field F , and in R - 1 9 , R -,, R- I , R,, and R, . Unique factorization does not hold in R 5 . If m < 0 it has been shown that R, is a PID if and only if rn = - 1, -2, - 3, - 7, - 11, - 19, -43, - 67, or - 163. Final details were supplied by Stark [36]. If m > 0, then R , is Euclidean [with d ( r ) = IN(r)I] if and only if rn = 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, or 73. I t is not known precisely which R,, rn > 0, are PIDs. Further historical information regarding these problems can be found in Hardy and Wright [14]. Proposition 5.12. If R is a U F D and if a, b E R are not both 0, then a and b have a GCD d that is unique up to associates. Proof. If a = 0, then b is a G C D for a and b. If a is a unit, then 1 is a G C D for a and b. If neither a nor b is 0 or a unit we may write and b = ~ p ( ~ p . .$ p ~p ,. a = p;lp". . .p i k

68

11 Rings

where p , . . . ., p k are distinct primes in R, 0 I ei E Z, 0I 1; E Z,and u E U ( R ) . Set g i = min{e,,fi}, 1 I i I k, and set

d = p ; ' p y . . .pz".

I

Clearly d I a and d I b. If c I a and c b, then only the primes p , , . . . ,p k (and their associates) can be prime divisors of c, so we may write c = u p t I p 2 . . .p t k , with 0 I h i e Z and V E U ( R ) .But then hi I yi, 1 I i I k, sincecja and cl b, and consequently c I d . Hence d is a GCD for a and b. If d , is another then d I d , and d , Id, s o d , d .

-

Recall that an ideal P in a commutative ring R is a prime ideal if P # R and if ab$ P for all a, b E R\P, or equivalently if R I P is an integral domain. If R is an integral domain with 1, then a principal ideal P = ( p ) is prime if and only if p is a prime element in R , as follows easily from the fact that p I ab if and only if plaorplb. If R is a commutative ring with 1 and I as an ideal in R write q: R -+ R / I for the canonical quotient map. It is a consequence of Theorem 3.4 that there is a homomorphism from R [ x ] onto ( R / l ) [ x ] given by

+

+ + rnxnw7(x) + + .. . + q(rn)x".

f ( x ) = ro r l x ... = q(r,) v ( r , ) x

That homomorphism is commonly called reduction of' coeficients modulo I . Exercise 5.8. Describe the kernel of the homomorphism f ( x ) ~ j ( x ) given by reduction of coefficients modulo an ideal I. Suppose R is a UFD. If 0 # f ( x )= a,

+ u l x + ... + a,x" E R [ x ]

we define the content of f ( x ) to be d = GCD(a,, a , , . . .,a,,). The content is of course only determined up to associates in R . If f ( x ) has content 1 we say that f ( x )is primitive. Observe that for any f ( x ) E R [ x ] we may write f ( x ) = d f , ( x ) , where d is the content of f(x) and f i ( x ) is primitive. In particular if f ( s )is irreducible in R [ x ] , then f ( x ) is primitive. Theorem 5.13 (Gauss's Lemma). Suppose R is a UFD and f ( x ) , g ( x ) E R [ x ] are both primitive. Then f ( x ) g ( x )is primitive. Proof. If not then there is a prime p E R dividing all coefficients of f ( x ) g ( x ) .Set S = R / ( p ) . Since p is prime ( p ) is a prime ideal, and hence S is an integral domain. We may reduce coefficients_mod(p ) in each polynomial in R [ x ] and obtain a homomorphism h ( x ) w h ( x )from R [ x ] onto S [ x ) . But S[xf is an integral domain (since S is), and ( f ( x ) g ( x ) ) - = f ( x ) i ( x ) = 0 in

5 Divisibility and Factorization

69

S [ x ] , so j ( x ) = 0 or g^(x)= 0. That means, however, that either p divides all coefficients of f ( x ) or p divides all coefficients of g(x), contradicting the fact that f ( x ) and g ( x ) are primitive. Proposition 5.14. Suppose R is a UFD and F = FR is its field of fractions. Suppose f ( x ) and g ( x ) are primitive in R [ x ] , and that f ( x ) and g ( x ) are associates in F I X ] .Then they are associates in R [ x ] . Proof. Since U ( F [ x ] )= F* we may write f ( x ) = ag(x) for some a E F*. Write a = b/c, with b, c E R , then c f ( x ) = bg(x) E R [ x ] .The content of cf ( x )is c and the content of bg(x)is b since f ( x )and g ( x )are primitive. Thus b c in R , and so a = b/c is a unit in R. But that means that f ( x ) g ( x ) in R [ x ] .

-

-

Proposition 5.15. Suppose R is a UFD, F is its field of fractions, and f ( x ) E R [ x ] is irreducible. Then f ( x ) remains irreducible when viewed as an element of F [ x ] . Proof. Since f ( x ) is irreducible in R [ x ] it is primitive. Suppose we could write f ( x ) = f i ( x ) f 2 ( x ) ,with f l ( x ) ,f 2 ( x )E F [ x ] , neither being constant. By “clearing denominators” and factoring common terms from coefficients we may write fi(x) = aigi(x),where a, E F , g,(x)E R [ x ] , and gi(x) is primitive, i = 1,2. Thus f ( x ) = a , a 2 g l ( x ) g 2 ( x ) ,so f ( x ) g , ( x ) g , ( x ) in F [ x ] , and g1(x)g2(x)is primitive by Gauss’s Lemma. Thus f ( x ) g l ( x ) g 2 ( x in ) R [ x ] by ) some u E U ( R ) , Proposition 5.14, which means that f ( x ) = u g 1 ( x ) g 2 ( xfor contradicting the irreducibility of f ( x ) in R [ X I .

- -

Theorem 5.16. If R is a UFD, then R [ x I , x 2 , . . , x , ] is a UFD. Proof. Since

R [ x , , x 2 , . . . ,x,l z (REX, x2 9

9 . .

.,x , -

11)CX,l

we may assume that n = 1; the general result follows by induction. We show first that each nonzero nonunit f ( x ) E R [ X I is a product of irreducibles, using induction on rn = degf(x). The case m = 0 is clear since R is a UFD. If rn > 0 we assume the factorization possible for polynomials of degree less than rn. Write f ( x ) = a f l ( x ) ,where a is the content of f ( x )and f , ( x ) is primitive. Then a E R is either a unit or a product of irreducibles. If f , ( x ) is irreducible the , factorization of f ( x ) is accomplished. If not write f l ( x ) = f 2 ( x ) f 3 ( x ) with degfi(x) < degf(x) = in, i = 2,3. By the induction hypothesis each of f i ( x ) and f 3 ( x ) is a product of irreducibles, and hence f ( x ) is a product of irreducibles. It remains to be shown that every irreducible in R [ x ] is prime. Suppose f ( x ) is irreducible (and hence primitive), and suppose f ( x ) l g ( x ) h ( x )in R [ x ] . By Proposition 5.15 f ( x )is irreducible as an element of F [ x ] , where F is the field of fractions of R . But F [ x ] is Euclidean, so it is a UFD, and hence f(x) is prime

70

I1 Rings

in F[x].We may assume then that f ( x )I g(x)in F [ x ] ,say g(x) = f ( x ) k ( x ) with , k ( x ) E F[x].By clearing denominators in k ( x ) and factoring out contents we may write g(x) = ag,(x) = (b/c)f(x)k,(x),

-

with g l ( x )and k l ( x )primitive in R [ x ] ,and so g,(x) f ( x ) k , ( x )in F [ x ] . But f ( x ) k , ( x )is primitive by Gauss's Lemma, and by Proposition 5.14 we have g,(x) f ( x ) k , ( x ) in R [ x ] , so g l ( x )= u f ( x ) k , ( x ) for some u E U ( R ) . Thus f ( x )1 g,(x) in R [ x J and hence f ( x )1 g(x)in R [ x ] since gl(x)1 g(x).

-

Corollary. If R is a UFD, then R [ x , y ] is a U F D that is not a PID. Proof. See Exercise 5.6.2. Proposition 5.17. If R is a PID, then every nonzero prime ideal P in R is maximal. Proof. Let I be an ideal, with P c I E R, and say P = (p), I = (a). Then p E (a), so p = ab for some b E R. Thus p l a or p J bsince p is a prime. If p l a , then a E (p) and I = P. Otherwise say that cp = b, c E R. Then p = ab = acp, so ac = 1. Thus a E U ( R )and I = R.

Corollary. If R is a PID and p E R is a prime, then R / ( p )is a field. Theorem 5.18 (The Eisenstein Criterion). Suppose R is a PID and f ( x ) = a,

+ a,x + a2x2 + . . . + a,x"

is primitive in R [ x ] . Suppose there is a prime p E R such that p ( a ifor 0I iI n - 1, but p X a, and p 2 $a,. Then f ( x )is irreducible. Proof. Set S = R / ( p ) ; it is a field by the corollary to Proposition 5.17. Then S [ x ] is Euclidean, and hence is a UFD. Reduce coefficients mod(p) ) R [ x ] onto S [ x ] . Suppose to obtain a homomorphism k ( x ) ~ k ^ ( xfrom f ( x ) = g(x)h(x),with g(x)= bo + b,x

+ ... + bkxk

and

h(x) = co + c l x + . ' -+ c,x"',

neither being a unit in R [ x ] . In particular, both g(x) and h(x) must have positive degree since f(x) is primitive. Then j ( x ) g^(x)&x)= q(a,)x", since p 1 a,, 0 Ii In - 1. Consequently x I G(x) and x I h(x) in S I X ] ,which means that q(bo)= q(co) = 0, or pI b, and p Ice. But then p2 I b,co = a,, a contradiction. We remark that Eisenstein's Criterion holds, more generally, when R is a UFD (see Exercise 8.30). Note that under the hypotheses of Eisenstein's Criterion f ( x )is irreducible not only in R [ x ] but also in F [ x ] , where F is the field of fractions of R, by

6 The Chinese Remainder Theorem

71

+ +

Proposition 5.15. For example j ( x ) = 6 2x 4x3 + 7x5 is irreducible in Q[x] since f ( x ) E Z[x] and f ( x ) satisfies the hypotheses of Eisenstein’s Criterion with p = 2.

6. THE CHINESE REMAINDER THEOREM If R and S are rings (not necessarily commutative), then they are in particular additive groups, so their direct sum R 0 S is defined. If we define (rl,sl)(r2,s2) = (r1r2,sIs2),then it is an easy matter to verify that R 0S is a ring. Clearly these remarks extend to any finite number of rings R , , R , , . . . ,R , . The direct sum R , 0R 2 0 . . . 0 R, is commutative if and only if each Ri is commutative, and has a 1 [ = ( 1,1,. . . , l ) ] if and only if each R i has a 1. Exercise 6.1. Show that

Z,and Z,0Z,are isomorphic rings.

Recall that if G is a group and H I G we write x = y(mod H ) for x, y E G if y - ’ x E H (or x - y E H if G is additive and abelian). In particular if R is a ring and I an ideal in R we shall write x = y(mod I ) if x - y E I.

If I and J are ideals in a ring R, then their product I J is defined to be the ideal generated by all ab, a E I, h E J . Clearly I J consists of all finite sums a161 a262 + ... + a&,, where ui E I, bi E J . Note that I J G I n J since I and J are ideals. For an example note that (6)(9)= (54) in the ring Zof integers, and that (6) n (9) = (18). These remarks extend in an obvious fashion to any finite collection I , , I,, . . . ,I, of ideals in R . We write nl=,l j for their product. This concept should not, of course, be confused with the notion of direct product.

+

Proposition 6.1. Suppose R is a ring with 1 and I , J are ideals in R such that I J = R . Then for any r l r r 2E R we may find an element r E R such that r - r , E I and r - r2 E J , i.e., r = r , ( m o d I ) and r = r,(modJ).

+

+ 6, with u E I and b E J , and set r = r2a + rlb. Then r - r I = r2a + r , ( h 1 ) = r2a - r , a = (r2 - rl)a E I

Proof. Write 1 = a

-

and r - r2 = r2(a - 1)

+ r,b = r,h

-

r2b = (rl - r 2 ) bE J .

Theorem 6.2 (The Chinese Remainder Theorem). Suppose R is a ring with 1 and I,, I,, . . . ,I,, are ideals in R such that Zj Ik = R i f j # k . Then given any r l , r 2 , ..., r,, E R we may find an element r E R such that x = r is a simultaneous solution for all the congruences x = rj(mod lj),1 I j I n. Any two solutions are congruent modulo I, n I, n ... n I,.

+

72

I1 Rings

Proof. For 1 < j 2 n write 1 = aj + b,; with aj E I , and bj E l j .Then

= R . By Proposition 6.1 we may choose s, E R such and hence Il + n3=21j and hence also s, = O(mod 1,) if that s, 5 l(mod I , ) and s t = O(mod n3=21j), 2 5 k In, since n3=21j G 1,. Similarly we may find s j E R such that s j = l(mod l j ) and s j = O(mod 1,) if k # j for all j , 1 I j I n. Set r = k = l r k S k * Then

c"

r - rj = x { r i . s i : i # j )

+ r j ( s j- 1) = O(mod l j ) ,

If s is another solution, then s s

= r j = r(modIj)

= r(mod

n;=,rj).

1 I j I n,

for all j , so

Note that the proof of the Chinese Remainder Theorem is constructive when applied to Z (or to any Euclidean ring). In fact if lj = (mi), then since (mj, n{m,:k z j } ) = 1 we can write 1 = cjmj + d j n ( m , : k # j ) by means of the Euclidean algorithm (see Exercise 8.26). We may then set sj = d j n { m k : k zj } and r = r l s l ... + Y,s,.

+

Exercise 6.2. Solve the congruences x

= 1 (mod8),

x

= 3(mod 7),

x

= 9(mod 11)

simultaneously in the ring Z of integers. Theorem 6.3. Suppose I , , I , , ...,I, are ideals in a ring R . Then there is a monomorphism f ' : R/(Il n I, n ... n 1,) 4 R / l l

If R is a ring with 1 and if Zj

0R J I , 0- - 0 . RJl,,.

+ I, = R when j # k, then f

is an isomorphism.

Proof. Define a map g: R + R / I , 0... 0 R/I, by setting g ( r ) = ( r + I,, ...,r + I,,). Then g is clearly a homomorphism, and ker g = I, n 1, n ... n I,. Thus the existence off is a consequence of the FHTR. If

R has a 1 and l j + I, = R for j # k, then g is onto by the Chinese Remainder Theorem, so f is an isomorphism.

If R

=

72 in Theorem 6.3, with I j = ( m j ) ,then lj

+ 1, = ( m i ) + (mk)=

( ( m j , mk)),

the principal ideal generated by the GCD of m j and m k .Thus lj + I k = Z if and only if (mi,m,) = 1, and in that case I j n 1, = IjIk = (mjmk). More generally I, n I, n . . . n I,, = (m,m2 . . . mn). As a consequence we see that if ( m j ,mk) = 1 forj#k,thenZ,,, *..., , r Z , , O Z , , O ~ ~ ~ O Z , n .

73

7 The Hilbert Basis Theorem

7. THE HILBERT BASIS THEOREM Analytic geometry involves the description of geometric objects by means of equations. For example, the set of common solutions to the equations x y + z = 1 and 2x - y - z = 3 is a line in the three-dimensional space Iw3. In general if F is a field, then an (affine)algebraic variety V in the space F" is the set of common solutions to a family of polynomial equationsf,(x,, x 2 , . . . , x,) = 0, where a ranges over some index set A and each f , ( X ) E F[x,, x,, . . . ,x,]. a E A, observe that every If I is the ideal in F [ x , , . . .,x,] generated by all &(X), g E I vanishes on the points in V , and that V could be described as the variety determined by the polynomials in I. This simple observation is the beginning of the rather deep subject called algebraic geometry, in which methods of abstract algebra are brought to bear on the study of geometry. Suppose I is an ideal in FCx,, . . . ,x , ] and V is the corresponding algebraic variety in F". The main result in this section is the Hilbert Basis Theorem, which states that the ideal I is finitely generated and has the consequence that the variety V can be described as the set of common solutions to a finite set of polynomial equations. A few preliminaries are necessary. Recall that a commutative ring is Noetherian if it satisfies the ascending chain condition (ACC) for ideals, i.e., every ascending chain I , c I, G I, L . . . of ideals of R must terminate. For example, every PID is Noetherian (Proposition 5.9). A ring R satisfies the maximal condition (for ideals) if every nonempty set Y of ideals in R contains an element I, that is maximal with respect to set inclusion, i.e., if I E Y and I, E I, then I , = I.

+

Proposition 7.1. A commutative ring R is Noetherian if and only if it satisfies the maximal condition. Exercise 7.1.

Prove Proposition 7.1.

Proposition 7.2. A commutative ring R is Noetherian if and only if every ideal in R is finitely generated.

Pro05 *: If I is any ideal in R let 9'be the set of all finitely generated ideals of R that are contained in I (e.g.,0 E Y ) .Let I , be a maximal element of Y: say with I, generated by r l , . . . ,r k . If I, # I choose r E I\I, and let J be the ideal generated by r l r . .. , r k and r. Then J E ,4p but I, 5 J and I, # J , contradicting maximality. Thus I = I, is finitely generated.

u

e: If I, E I, E I, E . . . is any ascending chain of ideals, then I = j I j is an ideal. Say I is generated by r , , . . . , r k rand say that ri E Ij(i),1 I i I k. We may assume the ri labeled so that j(i) Ij ( k ) for all i, and hence all ri E Ij,k).But then I = I j ( k ) and the chain terminates at j ( k ) .

Suppose R is a commutative ring with 1. If I is an ideal in R [ x ] and m is a nonnegative integer denote by I(m) the set of all leading coefficients of polynomials of degree m in I , together with 0. Exercise 7.2. (1) Show that I(m) is an ideal in R . (2) Show that I(m) E I(m + 1) for all m. (3) If J is an ideal with I c J show that I(m) E J(m) for all m.

Proposition 7.3. Suppose R is a commutative ring with 1, I and J are ideals in R [ x ] with I E J , and I(m) = J(m)for all nonnegative m E Z. Then I =J. Proof. If not, choose f ( x ) E J\I of minimal degree m 2 0. Since I(m) = J ( m ) there is some g ( x ) E I of degree m having the same leading coefficient as f ( x ) . But then 0 # f ( x ) - g(x) E J\I, and deg(f(x) - g ( x ) )< m, contradicting the minimality of deg f(x).

Theorem 7.4 (The Hilbert Basis Theorem). Suppose R is a commutative ring with 1. If R is Noetherian, then the polynomial ring R[x,, . . .,x,] is also Noetherian. Proof. By induction it will suffice to assume that n = 1 and write x1 = x. Let I. E I , E I , 5 * . . be an ascending chain of ideals in R [ x ] , and let 9’be the set of ideals {In(m):OIn E Z,0 I m E Z )

in R. By Proposition 7.1 there is a maximal element Ir(s)E 9’.Consider the infinite rectangular array IO(O) s I,( 1) G . . . G Io(s - 1) E . . .

nl 4 nl I,(O) G 1,(1) G ... E I,(s - 1) E ... nl nl nl Z2(0)s I,( 1) c . . . E I,(s - 1) E ’ * *

nl

nl

nl

The jth column is an ascending chain of ideals in R so it terminates, say at If(j)(j).Thus if i 2 f (j), then Ii(j ) = I,(j)( j). Set = max{f(O), f ( l ) ,. . . , f ( s -

11, r } ,

and suppose i 2 u. If 0 Ij I s - 1, then I i ( j ) = I,(j) = If(j)(j)since i 2 u 2 f ( j ) and the jth column terminates at f(j). I f j 2 s, then I i ( j ) 2 Zi(s) 2 Zr(s),so Ii( j ) = Ir(s) by maximality; and l , , ( j )2 I,( j ) =, Zr(s), so I J j ) = I,(s)

8

Further Exercises

75

by maximality, and li(j ) = I,,( j).We have thus shown that li(j ) = I,( j ) for all j 2 0 and hence, by Proposition 7.3, that Ii = I, for all i 2 u. Thus the chain I , c I , c I , c . . terminates at u and R [ X I is Noetherian. 1

Corollary. Suppose S is a ring with 1 , R is a commutative Noetherian subring with 1, E R , and s , , s2 ,..., s, are in the center of S. If R , = R [ s , , s,,.. .,s,], the (commutative) subring of S generated by R and { s , , . . . ,s,}, then R , is Noetherian. Proof. Clearly a homomorphic image of a Noetherian ring is Noetherian. We may map the Noetherian ring R [ x , , . . . , x,] homomorphically onto R , by means of f ( x l , ..., x,) Hf ( s , , . . . , s,) (see Theorem 4.3). Exercise 7.3. If R is a commutative ring with 1 and {x,:a E A } is an infinite set of distinct (commuting) indeterminates show that the polynomial ring R [ { x , : a E A } ] is not Noetherian.

8. FURTHER EXERCISES 1. If R is a commutative ring and a E R show that the principal ideal (a) has the form ( a ) = {ra na:r E R , n E Z).Describe the elements of (a) explicitly if R is not necessarily commutative. 2. Show that there is no ring R with 1 whose additive group is isomorphic with 62/77. 3. If R is any ring denote by R the additive group R 0 77, with multiplication defined by setting (r, n)(s,m )= (rs + mr + ns, nm).

+

,

Show that R is a ring with 1. If r E R is identified with ( r , 0) E R , show that R is a subring of R , . Conclude that every ring is a subring of a ring with 1. 4. (The Binomial Theorem). Suppose R is a commutative ring, a, b E R , and 0 < n E E. Show that

where (f) = n ! / k ! ( n- k ) ! is the usual binomial coefficient. ( H i n t : Use induction on n.) 5. Show that every ideal in Z,is principal. 6. If F is any field show that the ring M,(F) of n x n matrices over F is a simple ring. 7. Suppose R is a commutative ring, I , and 1, are ideals in R, P is a prime ideal in R , and I , n I , E P . Show that I , c P or I , G P .

76

I1 Rings

n

8. Suppose R is a commutative ring with 1 and x E { M :M is a maximal ideal in R } . Show that 1 + x E U ( R ) . 9. An element a of a ring R is called nilpotent if a" = 0 for some positive integer n. Show that the set of nilpotent elements in a commutative ring R is an ideal of R . 10. Find all nilpotent elements in H p k , then more generally in H,. (See 9 above.) 11. Suppose R is a ring with 1 , u E U ( R ) , a is a nilpotent element of R , and ua = au. Show that u U E U ( R ) . In particular 1 + U E U ( R ) for every nilpotent a. (Hint: Write (u + a)-' suggestively as l/(u + a) = u - ' / ( l + u - ' a ) and expand in a "power series." Then verify directly that the resulting element of R is an inverse for u + a.) 12. Determine the endomorphism rings of the additive groups H , Z,, and Q. 13. Give an example of a ring R with a prime ideal P # 0 that is not maximal. 14. Show that the ideal I = ( 2 , x )is not principal in Z [ x ] . 15. Suppose R is a ring with 1 and f ( x ) ,g(x) E R [ x ] , g(x) # 0. If the leading coefficient b of g(x) is a unit in R show that there are unique polynomials q(x), Y(X) E R [ x ] such that f(x) = g(x)q(x) + r(x), with degr(x) < degg(x). 16. Suppose R is a commutative ring with 1 and that f(x) E R [ x ] is a zero divisor. Show that there is an element a E R , a # 0, such that af(x) = 0. 17. Determine U ( Z , [ x ] ) .It may help to warm up by assuming that n = p k for a prime p . 18. Suppose R and S are commutative rings with R E S and 1 , = l,, and that R is an integral domain. If a E S is transcendental over R and g ( x ) is a nonconstant polynomial in R [ x ] show that g(a) is transcendental over R. , an are n 19. (Lagrange Interpolation). Suppose F is a field, a 1 , a 2 ..., distinct elements of F , and b l , b 2 ,..., b, are n arbitrary elements of F . Set pi(x) = n { x - a j : j # i ) and set

+

f ( x ) = C { b i p i ( x ) / p i ( a i1) :I i In ) . Show that f(x) is the unique polynomial of degree n - 1 or less over F for which f(ai) = bi, 1 I i 5 n. 20. Find f(x) E Q[x] of degree 3 or less such that f(0) = f(1) = 1, f(2) = 3, and f ( 3 ) = 4. 21. If R is a commutative ring with 1 and S = R [ x , y ] show that a , = x y and a2 = xy are algebraically independent over R . If a3 = x - y show that a , , a 2 , and a3 are algebraically dependent over R . 22. Suppose m and n are square-free integers, neither being 0 or 1, and that m # n. Show that R , and R, are not isomorphic rings.

+

8 Further Exercises

77

23. If n is a prime element of R , show that n is a divisor of exactly one prime p E Z. (Hint: Observe that n 1 N ( n )E Z.) 24. If R is a PID show that every nonzero prime ideal P in R is a maximal ideal. 25. Suppose R is a Euclidean domain, a, b E R * , a b, and d(a) = d(b).Show that a and b are associates. 26. (The Euclidean Algorithm). Suppose R is a Euclidean domain, a, b E R, and ab # 0. Write

I

a = bq1

6

+ rl,

= r1q2

= r2q3

rl

< d(b), 4 - 2 1 < @I)? 4 . 3 ) < 4r2), 4.1)

+ r2,

+ r3,

rk-2

= rk-14k

rk-l

=rkqkfl,

+ rk?

d(rk)

< d(rk-l)q

with all r i , q j E R . Show that rt = (a, b), and “solve” for t+k in terms of a and b, thereby expressing (a,b) in the form ua + ub, with u, u E R . 27. Use the Euclidean Algorithm (Exercise 26) to find d = (a,b) and to write d = ua + ub in the following cases: (1) a = 29041, h = 23843, R = Z; (2) a = X’ - 2x2 - 2~ - 3, b = x 4 3x3 (3) a = 7 - 3i, b = 5 3i, R = R - , .

+

+

+ 3x2 + 2x, R = Q[x];

+ + +

28. Suppose f ( x ) = 1 x x 2 ... + xp-l, where p is a prime in Z. (1) Show that f(x) is irreducible in Q[x]. (Hint: Write f(x) = (xp - 1)/ (x - 1); substitute x 1 for x.) (2) Show that (9) = c:2: ( p P ; ! l ) for all k < p.

+

29. If p E Zis prime and 1 < m E Z show that f(x) = xm- pis irreducible in Q[x]. Conclude that p”” is irrational.

30. Establish the Eisenstein Criterion for a polynomial f ( x ) over a UFD. [Hint: Assume f ( x ) = g(x)h(x), with p dividing the constant term of g(x). Investigate the first coefficient of g(x) not divisible by p.] 31. (Lipka [25]). Suppose p is a prime in Z, f(x) =

kp

+ a , x + azx2 + ... + a,x”

E

Z[x],

(ail < p. Show that f(x) is irreducible in Q [ x ] . Conclude, for and example, that x” + x + 3 is irreducible over Q if n > 2. [Hint: Show that every root of f ( x ) in 43 is larger than 1 in absolute value.]

I1 Rings

78

+ J2

U(R,). (2) If u E U ( R , ) show that either u < 1 or u > 1 + $(use the fact that if 1 < u < 1 +&then -1 < u - l < 1). (3) Use (2) to show that if u E U ( R , ) and u > 0, then u = (1 + 4)" for some n E Z. (4) Conclude that U ( R , ) E { k I } x (1 + $). 33. Determine U ( R , , ) . 34. Let R be the set of rational numbers alb with b odd. Then R is an integral domain. (1) Find U ( R ) . (2) Show that P = R\ U ( R )is a maximal ideal in R. (3) Find all primes in R. (4) Find all ideals in R and show that R is a PID. Is it Euclidean? 32. (1) Show that 1

E

(Remarks: R can be obtained from H by localizing at the prime ideal P.) 35. Suppose P # 0 is a prime ideal in R,. (1) Show that P n H is a prime ideal in H,so P n Z = ( p )for some prime p in h. (2) Set I = p R , c P and form the quotient ring RII. Show that RII, as an additive group, is generated by two elements of finite order; hence R I l is finite. (3) Show that there is an epimorphism from R / 1 to RIP and conclude that RIP is finite. (4) Conclude that every prime ideal in R , is maximal. 36. If R , , R , , . . . ,R , are rings with 1 show that

U ( R , @ R , @ ... @ R,) 37. Solve the congruences xri(modl+i),

=

U ( R , ) x U ( R 2 )x ..* x U(R,).

x-l(mod2-i),

xzl$i(mod3+4i)

simultaneously for x in the ring R - I of Gaussian integers. 38. Solve the congruences f(x)

= 1(mod x - l),

f(x)

= x(mod x 2 + I),

f(x)

= x3(modx + 1)

simultaneously for f(x) in F[x], where F is a field in which 1 39. If R is the subring of H[x] consisting of all f ( x ) = a,

+ 1 # 0.

+ a,x + ... + a,x"

such that ak is even for 1 Ik I n, show that R is not Noetherian. [Hint: Consider the ideal generated by {2xk:1 5 k E a}.]

Chapter I11

1

Fields and Galois Theory

1. FIELD EXTENSIONS

Suppose F and K are fields and that F is a subring of K . Then we say that F is a subjeld of K or that K is an exfension field of F. In that case K is a vector space over F. The dimension of K as an F-vector space is called the degree of K over F and is denoted by [ K : F ] . If [ K : F ] is finite we say that K is a finite extension of F . If K is an extension of F and S is any subset of K denote by F(S) the intersection of all extension fields L of F such that S E L E K . Clearly F(S) is the minimal field between F and K containing S as a subset. We call F(S)the extension of F generated by S . If a E K , then F(a) = F ( { a } ) is the simple extension of F generated by a, and a is called a primitioe element for F ( a ) over F. Note that F(a) is the field of fractions of the ring F[a] obtained by substitution of a for x in F [ x ] . Thus F(a) = ( f ( a ) / s ( a ) : f ( x ) , g ( Ex )FCxI,g(a)

z 0).

Proposition 1.1. Suppose Lis an extension field of F and K is an extension field of L. If A is a basis for Las an F-vector space and B is a basis for K as an Lvector space, then A B = {ab:u E A , b E B } is a basis for K as an F-vector space. Consequently [ K :F] = [ K :L ] [ L :F ] , and K is finite over F if and only if K is finite over L and L is finite over F. Proof. If c E K we may write c bj E B, and we may write

=

u,b,

+ u2b2 + ... + u,b,,

with

U ~ L, E

80

111 Fields and Galois Theory

cj

cj,i

with hi E F , ai E A . Thus c = ujbj = vjiaibj,and the F-span of A B is all of K . Suppose uijaibj= 0 for some finite set faibj} G A B and uij in F. For each j set uj = uijai E L, so that ujbj = 0. Then each uj = 0 since B is linearly independent. But then all uij = 0 since A is linearly independent, so A B is linearly independent. Thus A B is an F-basis for K . In particular the elements ab in AB are all distinct and ( A B (= JAJIBI.

Xi

cj

If F is a field let F, be the intersection of all the subfields of F. Thus F, is the unique minimal subfield of F ; it is called the prime Jield of F. There is a (ring) homomorphismf: Z + F, defined by settingf(n) = n * 1,1 E F , for all n E Z. If f is 1-1, then Im(f) is an isomorphic copy of Z in F,, so F,, being minimal, must be isomorphic with Q, the field of fractions of 8. We say then that F has characteristic 0. In many cases we shall identify F, with Q so F is an extension of Q. If f is not 1-1, then kerf is an ideal in Z,say kerf = (n).If a, b E Z and n(ab,thenO=(ab)l =(al)(bl),andsoeitheral = O o r b l =O.But thennla or n 1 b, so n is a prime. We take n > 0 and write n = p. Thus Im(f) z Z,, a field, and so Im(f) = F, E H,. We say then that F has characteristic p, and often identify Fo with Z,. If K is an extension field of F recall that a E K is called algebraic over F if f(a) = 0 for some nonzero f ( x ) E F [ x ] (otherwise a is transcendental over F ) . If U E K is algebraic over F choose a monic polynomial m , ( x ) E F [ x ] of minimal (positive) degree such that m,(a) = 0. Proposition 1.2. If a E K is algebraic over F, then the polynomial m,(x) E F [ x ] is irreducible and unique. If f ( x ) E F [ x ] and f ( a ) = 0, then m,(x) I f(x) in F [ X I .

Proof. If m,(x) is not irreducible write m,(x) = g(x)h(x), with neither factor a unit in F [ x ] . Then 0 = m,(a) = g(a)h(a), so g(a) = 0 or h(a) = 0, contradicting the minimality of deg m,(x). Divide f ( x ) by m,(x) and write f ( x ) = m,(x)q(x)+ r(x), where q(x), r(x) E F [ x ] and degr(x) < degrn,(x). Substitute a for x to see that r(a) = 0 and hence that r ( x ) = 0 since deg m,(x) is minimal. Thus m,(x) I f ( x ) . The uniqueness follows, for if k ( x ) were another polynomial in F[x] with the same properties, then m,(x) I k ( x ) and k(x) I m,(x), so k ( x ) m,(x) in F [ x ] . But two rnonic polynomials are associates only if they are equal.

-

The polynomial m,(x) is called the minimal polynomial of a over F. We shall write m,Jx) for m,(x) when it is important to emphasize the underlying field F . Proposition 1.3. Suppose K is an extension field of F and a E K is algebraic over F , with minimal polynomial m ( x ) = mo,F(x).If degm(x) = n, then [ F ( a ) : F ] = nand {l,u,a', ..., a " - ' ) isan F-basis for F(a).

81

1 Field Extensions

Proof. If b,1+bla+b2a2 then f ( a ) = 0, where

+ . . . +b , - , ~ " - ~ = 0 ,

with

biEF,

+ +

+

f ( x ) = b, b l x ... b , - l ~ " - lE F [ x ] . That is in conflict with the definition of the minimal polynomial m(x)unless all bi = 0. Thus { 1, a,. . .,a"- 1 is linearly independent. We know that F(a) = { f l a ) / g ( a ) : f ( x ) , s ( xE)FCXl3S(Q)z 0). If f ( a ) / g ( a )E F(a), then (m(x),g(x))= 1 since m ( x )is irreducible and g(a) # 0, so we may write 1 = b(x)m(x) c ( x ) g ( x )for some b(x),c ( x ) E F [ x ] . Substitute a for x to see that 1 = c(a)g(a)and hence f ( a ) / g ( a )= f(a)c(a).Thus F ( a ) = {h(a):h(x)E F [ x ] } . But if h(x) E F [ x ] we may write h ( x ) = rn(x)q(x) r(x), with q(x)and r(x)in F[x]and deg r f x ) < deg m ( x ) = n. Substitute a for x t o see that h(a) = r(a). Thus { 1, a,. . . ,a n -' } spans F(a) and the proof is complete.

+

+

Note as a consequence that if a is algebraic over F , then F ( a ) = F [a]. An extension K of F is called algebraic over F if every a E K is algebraic over F. Proposition 1.4.

If K is a finite extension of F , then K is algebraic over F .

Proof. Let ~ = [ K : F ] E ZI f. a E K , then the set { l , a , a 2,..., a"'} is linearly dependent, so there are b,, b,, . . . ,b, E F , not all 0, such that bol

+ b,a + b,a2 + . . . + &am = 0.

But then if we set

f ( x ) = b, we have f(a)

=

+ b l x + ... + b,xm

E

F[x]

0, so a is algebraic over F .

Proposition 1.5. If L is an algebraic extension of F and K is an algebraic extension of L, then K is an algebraic extension of F.

+

Proof. If u E K there is a nonzero polynomial f ( x ) = b, . . * + b k x k in L [ x ] such that f(a) = 0. By Propositions 1.1 and 1.3 each of the fields F(bo),F(b,,bl), . . . , F (b,,bl, ..., bk) = L' is finite over F. But a is algebraic over L', so L'(a)is finite over L', and hence over F . Thus L'(a)is algebraic over F by Proposition 1.4, and in particular a is algebraic over F . Proposition 1.6.

If K is an extension field of F set E

Then E is a field.

= {aE

K : a is algebraic over F } .

82

I11 Fields and Galois Theory

Proof. If a, b E E , then a k b, ab, and a / b (if b # 0) are all in Ffa, b),which is finite over F . Thus a k b, ab, and a / b (if b # 0) are all in E and E is a field.

For example, if F = Q and K = C, then the elements a E C that are algebraic over Q are called algebraic numbers. By Proposition 1.6 the algebraic numbers constitute a field, which will be denoted by A. Exercise 1.I. Show that [A :Q] is infinite.

If f ( x ) E F I X ] ,K is an extension field of F , a E K , and f(a) = 0, then we say that a is a root of f ( x ) in K . Recall from the Factor Theorem (Proposition 11.3.5, Corollary) that if a is a root of f ( x )in K , then x - a 1 f ( x ) in K [ x ] . If ( x - a)kI f ( x ) but (x - a)"" $ f ( x ) we say that a is a root of f ( x ) with multiplicity k.

Proposition 1.7. If K is an extension of F and f ( x ) E F [ x ] , with degf(x) = n, then f ( x ) has at most n roots in K . Proof. Since K [ x ] is a UFD we may factor f ( x ) in K [ x ] as a product of irreducible factors, which are unique up to associates, and a E K is a root of f ( x ) if and only if x - a (or an associate) is one of the irreducible factors of f ( x ) .Since the sum of the degrees of the irreducible factors is n the proposition follows. If F is a field, f ( x ) E F [ x ] has degree n > 0, and K is an extension of F such that f ( x ) has n roots (counting multiplicities) in K , we say that f ( x ) splits over K . Note that then f(x) is a product of n linear factors (i.e., factors of degree 1) in K [ x ] . It will be convenient to agree as a matter of convention that any constant polynomial in F [ x ] splits over every extension K of F. Proposition 1.8. If F is a field and f ( x ) E F [ x ] has degree n 2 1, then there is an extension K of F such that f ( x ) has a root a in K and [ K : F ] I n. Proof. Let g(x) be a nonconstant irreducible factor of fix) in F [ x ] , say f ( x ) = g(x)h(x).Then ( g ( x ) )is a prime ideal, and hence a maximal ideal, in F [ x ] by Proposition 11.5.17. Consequently K = F [ x ] / ( g ( x ) )is a field. The map b H b ( g ( x ) )is a monomorphism from F into K so we may (and do) identify F with its image, a subfield of K . Set a = x ( g ( x ) )E K . Then

+

+

f ( 4= f ( x ) + ( 9 W )= g ( x ) h ( x )+ ( g ( x ) )= 0 in K . Note that K is in fact the simple extension F(a), so

[ K : F J = degg(x) I degf(x)

=

n

by Proposition 1.3. Corollary. If f ( x ) E F [ x ] has degree n, then there is an extension K of F, with [ K : F ] I n ! , such that f ( x ) splits over K .

I

Field Extensions

83

Suppose F is a field and 9is a set of polynomials in F [ x J An extension K of F is called a splitting field for 9over F if every f ( x ) E 9splits over K , and K is minimal in that respect. Equivalently K is a splitting field for . 9over F if every f ( x ) E 9splits over K , and if S is the set of all roots in K of all f ( x ) E 9 then K = F(S). Note that if F = { f ( x ) }contains a single polynomial f ( x )of degree n, then by the corollary to Proposition 1.8 there is a splitting field K for f ( x ) over F with [ K : F ] 2 n ! .

Exercise 1.2. Find a splitting field K

G

C for f ( x ) E Q [ x ] if f ( x ) =

x 3 - 1.

A field F is called algebraically closed if every nonconstant f ( x ) E F [ x ] has a root in F, and consequently splits over F. The so-called Fundamental Theorem of Algebra, which will be proved later, asserts that the field C of complex numbers is algebraically closed. An algebraic closure of a field F is an algebraic extension K of F that is algebraically closed. Thus for example A is an algebraic closure of Q and C is an algebraic closure of R. Exercise 1.3. Show that K is an algebraic closure of F if and only if K is a splitting field over F for F = F [ x ] . It is an easy consequence of Theorem 11.3.4 that if Fl and F, are fields and

4: F, + F2 is an isomorphism, then $ extends to an isomorphism (also called

$) between F1[ x ] and F , [ x ] , with $ ( x ) = x.

The next proposition, which is rather technical, will be needed in discussing isomorphism of splitting fields. Proposition 1.9. Suppose Fl and F2 are fields and 6: F, -, F, is an isomorphism, which we assume extended to an isomorphism between F, [ x ] and F2[x] via $ ( x ) = x . Suppose K , is an extension of Fl and a , E K , is algebraic over F , , with minimal polynomial m l ( x )E Fl [ X I . Set m 2 ( x )= $ ( m , ( x ) )E F 2 [ x ] and suppose K , is an extension of F2 in which m 2 ( x )has a root a,. Then F , ( a , ) 2 F2(a2);in fact $ can be extended to an isomorphism by means of $(al) = a,. Proof. Note that m , ( x ) is the minimal polynomial of a, over F,. Let us indicate two proofs, the first somewhat abstract, the second quite explicit.

I. Note that ( m i ( x ) )is the kernel of the substitution epimorphism Ei: + & ( a i ) obtained by substitution of a, for x, so that

&[XI

&(a,)

z F,[.x]/(rn,(x)),

i = 1,2.

Let ~ i : F i [ x I+

&CxI/(mi(x))

84

111 Fields and Galois Theory

-

be the canonical quotient map, i

FlCXl

=

1,2. In the diagram 4

F2CxI

rlt

note that ker(v],$) = (m,(x)), so by the FHTR there is an isomorphism 6 as indicated. It follows that F,(a,) z F,(a,). 11. To be more explicit note that by Proposition 1.3 we have

F,(u,)

=

(bo

+

F,(u,)

=

(cO

+ c l a 2 + ... + c,-,u;-~:c~E F,},

and

+ ... + b n - , d - ' : b i E F,)

b 1 ~ 1

where n = degm,(x) = degm,(x). We may extend 4 t o an isomorphism between F,(a,) and F,(a2) by setting $(al) = a,, and hence

4(bo + ... + b,-

') = $(b,)

+ + +(b,,- ,)a;-', ..*

In the special case where F, = F, = F , K , = K , = K , and 4 is the identity map on F we say that a , and a, are conjugates over F. Equivalently a,, a, E K are conjugates over F if and only if they have the same minimal polynomial over F .

Corollary to Proposition 1.9. If K is an extension field of F and a,, a, E K are conjugates over F , then there is an isomorphism $: F ( a , ) + F(a,), with 4(a,) = a, and $(b) = b for all b E F. Theorem 1.10. Suppose Fl and F, are fields and $ : F , + F2 is an isomorphism. Suppose fl(x) E F, [XI,f2(x) = 4(fl(x)) E F2 [XI,and K , , K , are splitting fields for f,(x) and f2(x), respectively. Then $ can be extended to an isomorphism 6: K , + K , . Proof. Induction on degf,(x). The result is clear if degf,(x) I 1, so suppose degf,(x) > 1 and assume the result for polynomials of lower degree. Let a, E K , be a root of a monic irreducible divisor of fl(x) in F, [x] and let a, be a root of the corresponding irreducible factor of f 2 ( x ) , a, E K , . By Proposition 1.9 we may extend to an isomorphism 41:Fl(a,) + F,(a,). Write fi(x) = (x - al)yl(x)and f2(x) = (x - a,)g,(x), with gi(x)E F,(a,)[x]. Then #l(g,(x)) = g,(x), Ki is a splitting field for g,(x), and deg g,(x) < deg f,(x). By the induction hypothesis (and hence 4) can be extended to an isomorphism O:K, + K , .

+

1

Field Extensions

85

If K , and K , are extensions of a field F , and if there is an isomorphism 8: K , + K , such that 8(b)= b for all b E F we say that 8 is an F-isomorphism from K , to K,. Corollary to Theorem 1.10. Suppose K , and K , are splitting fields over F for a polynomial f ( x ) E F [ x ] . Then there is an F-isomorphism 8: K , -,K,. Theorem 1.11. Suppose F, and F, are fields and 4:Fl + F, is an E F , [ x ] , F2= 4(9,) E F , [ x ] , and K , , K , are isomorphism. Suppose 9, F,,respectively. Then I$ can be extended to an splitting fields for F,, isomorphism 8: K , + K , .

Pro05 Let Y be the set of all ordered pairs (Fa,4a),where Fa is a field, F, E Fa E K , , Fa + K , is a monomorphism, and 4aI F, = 4. Then Y is partially ordered by means of (Fa,4a)I(F,, 4fl)if and only if Fa 5 F, and 4, I Fa = 4a;Y # since (F,, 4 ) E Y . By Zorn’s Lemma there is a maximal element (F,,O) in Y . If F, # K , , then there is a polynomial f , ( x ) E Fl that does not split over F,. Since f,(x) does split over K , we may view f,(x) as an element of F,[x] to see that there is a splitting field L , for f , ( x ) over F,, with F,sL, EK,, and correspondingly there is a splitting field L, for O ( f , ( x ) )= f 2 ( x ) over 8(Fo), with O(F,) E L, E K,. By Theorem 1.10 we may extend 8 to 8’:L, + L,. But then (F,,8) s (L,,O’) and (F,,8) # (L,,O’),contradicting the maximality of (F,,8). Thus F, = K , . But then 8(K,) is a splitting field for 8(51)= F,within K , , so 8(K,) = K , by the minimality of splitting fields, and the theorem is proved. Corollary. If F is a field and K , , K , are both algebraic closures of F , then there is an F-isomorphism 8: K , -, K , . Proof. An algebraic closure is just a splitting field for d Exercise 1.3).

= F[x]

(see

Exercise 1.4. Suppose K is an algebraic extension of F.

(1) If F is a finite field (e.g., F = BP)show that K is finite or countable. (2) If F is infinite show that (KI = I F [ . (This exercise may require some review of set theory. For a start see Theorem 2, p. 3 of Kamke [ 191.) Theorem 1.12. If F is a field, then F has an algebraic closure. Proof. Consider first a deceptively easy but invalid proof. Let Y be the set of all algebraic extension fields K of F , partially ordered by set inclusion. Then Y # 0since F E 9. By Zorn’s Lemma there is a maximal element F in 9. If f ( x ) E F [ x ] is not constant, then , f ( x )has a root a in some extension of F. If a 4 P, then F(a)is an algebraic extension of F by Proposition 1.5, but F(a) # F, and the maximality of F is violated. Thus F is an algebraic closure for F.

86

I11 Fields and Galois Theory

Surprisingly (perhaps) the difficulty with the above ”proof” is that Y is not necessarily a set (see Exercise 8.9).In order to repair the proof it is necessary to obtain a class Y of algebraic extensions of F that is large enough to apply Zorn’s Lemma, but so that 9 is a set. To that end we choose an uncountable set X such that F c X and IF1 < (e.g., we could take X = F u 2F u R). We then impose field structures on subsets K of X in all possible ways such that K becomes an algebraic extension of F. Note that if L is any algebraic extension of F , then by Exercise 1.4there is an algebraic extension K of F , with F c K c X, and L z K . We may now take 9 = { K G X : K is an algebraic extension field of F}, which really is a set, and the proof can proceed as above.

1x1

2. THE FUNDAMENTAL THEOREM OF GALOIS THEORY

If K is an extension of a field F, then an F-autornorphisrn of K is an automorphism 4 of K such that $(b) = b for all b E F. The set G = G ( K :F) of all F-automorphisms of K is a group, called the Galois group of K over F. For example, if F = 54 and K = C, then JGI= 2; the nonidentity element 4 in G is the complex conjugation automorphism, i.e., 4 ( a + bi) = a - bi. For another example let F = Q and K = Q( fl). Any 4 E G must carry to a conjugate, i.e., to another root of m(x) = x3 - 2. But also $( 3 )must be real since K G 54, and the other two roots of m(x) are not real. Thus 4( 3)= so 4 = 1, and G = 1. If L is an intermediate field, F c L G K , we set

9

3,

9 L = G ( K : L )= {$ E G : 4 ( a ) = a for all a E L } Note that 9 L I G = G ( K :F). For any H I G we set

9 H = { a E K : 4 ( a )= a for all 4 E H 1, thejixedjield of H . Clearly 9H is a field intermediate between F and K . Note that 9F = G,9 K = 1, and 9 1 = K . Observe, however, that F G can be larger than F [see the example above, with K = Q( r 2 ) ] . If K is an extension of F with Galois group G we say that K is a Galois extension of F if 9 G = F . Equivalently K is Galois over F if and only if for each a E K \F there is an element 4 E G ( K :F) such that $(a) # a. Exercise 2.1. If rn E Zis square-free and rn # 0,l show that K = Q ( f i ) is Galois over F = Q.

If F E L E K , then the field 9 9 Lis called the closure of L. We say that L is closed if L = 9 9 L . Dually, if 1 IH I G = G ( K : F ) ,then the closure of H is 9 9H , and H is closed if H = 99H .

2 The Fundamental Theorem of Galois Theory

Proposition 2.1. Suppose F c E Then

G

L E K and 1 I J IH

87

s G(K : F ) .

(1) YL I YE and 9 H c B J , (2) H I Y B H and L G B Y L , and (3) Y L = Y B Y L and F H = Y Y F H . Proof: We leave the proof of ( 1 ) and (2) as an (easy) exercise. Note that Y L I 9B(YL)by(2).Butalso L G 3 9 L b y ( 2 ) , s o Y ( B Y L ) I YLby(l),and hence YL = YFYL. The proof that 9 H = B Y F H is entirely analogous.

Proposition 2.2. If K is an extension of F and G = G ( K :F), then all subgroups YL are closed and all intermediate fields B H are closed. Moreover B is a 1- 1 inclusion-reversing correspondence between the set of all closed subgroups of G and the set of all closed intermediate fields between F and K . Proof. See Proposition 2.1 above. Note that 9 is a 1-1 correspondence as indicated since it has Y as an inverse.

Proposition 2.3. Suppose F c E Then [YE:YL] I [L:E].

c

L c K , and suppose L is finite over E.

Proof. Induction on n = [L: El. The result is obvious when n = 1. If there is a field M properly between E and L, then by Lagrange's Theorem, the induction hypothesis, and Proposition 1.1 we have

[Yb:gL] = [YE:VM][Y'M:SL]

s [M:E][L:M]

= CL:EI.

Assume then that there are no fields properly between E and L. Thus if a E L\E we have L = E(a). Let m(x) = mo,E(x),so deg m(x) = n. If & E YE form the left coset $9L. If 0 E YL, then e(a) = a, so @(a) = &(a),i.e., each element of the coset 4YL sends the root a of m(x) to the same root &(a) of m(x). Different cosets &,YL and &,YL give different roots bl(a) and &(a), for otherwise 4;' &,(a) = a, which would mean that &; E 9 L since L = E(a), contradegm(x) = dicting the fact that q51gL # cj29L. Consequently [YE:9L] I n = [L:E]. Proposition 2.4. Suppose F E K, G = G(K:F),1 I J I H I G , and [ H : J ] is finite. Then [ B J : 9 H ] I [H:J]. Prooj Note that if I$ E H , 0 E J , and a E F J , then &?(a) = &(a), so the image of a is constant as 40 ranges over the left coset & J . Choose left coset representatives & 1 = 1, $J,,...,&~ for J in H . If [ F J : F H ] > n = [ H : J ] choose a,, a,, . . . ,a, + E BJ , linearly independent over 9H . Consider the homogeneous system of n linear equations in n + 1 unknowns x l , x,, .. . ,x,+ given in matrix form by A X = 0, where X = (xl,. . .,x, + I and the coefficient matrix A has ijth entry c#&zj). The matrix A represents a linear transformation T :K"" + K", so T has nontrivial kernel, i.e., there is a nontrivial solution to

88

111 Fields and Calois Theory

the system of equations. Choose a nontrivial solution with the least possible number of nonzero entries. By relabeling we may assume that the solution has the form ( b l ,b,, . . . ,b,, 0,. . .,O), 0 # bi E K , and we may assume (by dividing) that b , = 1. If all bi were in 9 H , then the first equation in the system would be a dependence relation for ( a l , .. .,a k ) ,so we may assume further that b, $9 H. Choose Q, E H such that 4 ( b 2 )# b,, and apply 4 to each equation in the system A X = 0. By the remark at the beginning of the proof the equations are simply rearranged, since { Q,Q,,, . . . ,Q,Q,"} is a new set of coset representatives for J in H . Consequently

( 4 ( b ,Q,(bZ),...~Q,(b~),O,...~O) ~) is also a solution vector for the system, with Q,(bl)= 4(1) = 1 and Q,(b2)# b,. If the new solution is subtracted from the old we obtain a third nontrivial solution having k - 1 or fewer nonzero entries, contradicting the minimality of k. Theorem 2.5. (1) Suppose F E E s L s K , E is closed, and [ L : E ] is finite. Then L is closed and [ S E : 9 L ]= [ L E I . (2) Suppose 1 I J I H I G = G ( K , F ) , J is closed, and [ H : J ] is finite. Then H i s closed and [ 9 J : 9 H ] = [ H : J ] . Proof. (1) [ L E I = [ L : 9 9 E ]I [ B 9 L : 9 9 E ] I [9E:teL] I [ L : E ] by Propositions 2.1,2.3, and 2.4. Thus equality holds throughout. The proof of (2) is similar to (actually it is "dual" to) the proof of (1).

Corollary. (1) All finite subgroups of G = G ( K : F )are closed. (2) If K is Galois over F, F G E 2 K , and [E: F ] is finite, then E is closed, and consequently K is Galois over E . ProoJ (1) The subgroup 1 is closed. (2) Since K is Galois F is closed, and hence E is closed by the proposition. Thus E is the fixed field of the Galois group YE = G ( K :E ) , which says that K is Galois over E.

If F E E E K we say that E is stable if Q,(E) c E for all 4 E G = G ( K : F ) . Note that if E is stable, then also Q,-'(E) _c E, all Q, E G, so E = & K 1 ( E ) c b(E)E E , and in fact @ ( E ) = E for all Q, E G. Proposition 2.6. ( 1 ) If F s E G K , and E is stable, then 4eE 4G. (2) If H U G, then 9 H is stable. Exercise 2.2. Prove Proposition 2.6.

Corollary. (1) If F c E c K and E is stable, then the closure 9 S E is stable. (2) If H -4 G, then 3 9 H 4 G.

2 The Fundamental Theorem of Galois Theory

89

Proof: (1) Since E is stable YE Q.G, and hence 9 9 E is stable. (2) Since H Q G we have F H stable, and hence 3 9 H Q G. Proposition 2.7. If K is Galois over F , F is Galois over F .

cL

E K , and L is stable, then L

Proof. If a E L\F, then $(a) # a for some $ E G = G ( K : F ) . But $ restricts to an F-automorphism 8 of L since L is stable, so 0 E G ( L : F ) and O(a) # a. Thus L is Galois over F .

Theorem 2.8. Suppose K is Galois over F , f(x) E F[x] is irreducible, and f(x) has a root a E K . Then f(x) splits over K and all its roots are distinct. Proof. We may replace f(x) by an associate if necessary and assume that = f(x) is monic, hence f ( x ) = mo.F(x).For each $ E G observe that $(I(.)) f(x) and hence $(a) is also a root of f(x) in K . Let a, = a, a 2 , .. . ,ak be all the distinct roots of f(x) of the form 4(a), # E G. Thus k I n = degf(x). Set g(x) =

n k

(x - Ui).

i= 1

Since each 4 E G merely permutes the roots a, we see that $(g(x)) = g(x)for all $ E G. It follows that g(x) E F[x] since F is the fixed field for G. But g(x) is monic, g(a) = 0, 1 I deg g(x) I degf(x), and f(x) = m,(x). Thus g(x) = f(x) and the theorem follows. Corollary 1. Suppose F F. Then L is stable.

c L c K and Lis both Galois and algebraic over

Proof. If $ E G ( K :F ) and a E L, then $(a)is a root of m(x) = m,,,(x) since $(rn(x)) = m(x). By the theorem m(x) splits over L, so 4(a), being one of its roots, must lie in L.

Corollary 2. If K is Galois and algebraic over F and F E L stable if and only if L is Galois over F .

_c

K , then L is

Proof. t: Corollary 1. *: Proposition 2.7. Proposition 2.9. Suppose F c L c K and L is stable. Then the quotient group G / % L is isomorphic with the subgroup of G ( L : F )consisting of those 8 E G ( L : F )that are restrictions to L of elements of G, i.e., those 0 E G ( L : F ) that may be extended to automorphisms of K . Proof. If 4 E G,then 8 = 4 I L is in G ( L : F )since L is stable. The map f: 4~ 4 I L is clearly a homomorphism from G into G ( L : F ) ;its kernel comprises those # E G such that 4 I L = 1, i.e., kerf = YL. The proposition follows from the FHT.

111 Fields and Galois Theory

90

Theorem 2.10 (The Fundamental Theorem of Galois Theory). Suppose K is a finite Galois extension of F , and G = G ( K :F ) is its Galois group. Then the operation 9 of taking fixed fields is a 1-1 inclusion-reversing correspondence between the set of all subgroups of G and the set of all intermediate fields L between F and K . If J I H I G, then [ H : J ] = [ 2 F J : 9 : H ] ,and in particular I GI = [ K :F ] . Furthermore, H d G if and only if 9 H = L is Galois over F, in which case G ( L :F ) z G / H . Proof. Since K is a finite Galois extension of F all intermediate fields are closed and all subgroups of G are closed, by Theorem 2.5, and [ H : J ] = [ 9 J : @ B H ] for J I H 5 G. The statement about normality is a consequence of Proposition 2.6 and Theorem 2.8, Corollary 2. If H G and L = . 9 H , then by Proposition 2.9 we have G / H = G / g S H = G/YL isomorphic with a subgroup of G ( L :F). Note, however, that ( G / H (= lGl/lHl = (GI/IYL(= IGl/[YL:gK] = [ K : F ] / [ K : L ]= [ L : F ] = (G(L:F)I,

so G / H 2 G ( L : F ) .

Corollary. If K is a finite extension of F and 1G(K :F)l Galois over F.

=

[ K : F ] , then K is

Proqf. It is clear from the definitions that G = G ( K : F )= G ( K : 9 G ) , so 9 G = 9 G ( K : S G ) ,which means that K is Galois over 9 G , and hence JGI = [ K : 9 G ] by the theorem. Thus [ K : F ] = [ K : F G ] ,so F G = F and K is Galois over F.

3. NORMALITY AND SEPARABILITY Over fields of characteristic p > 0 it is possible to have irreducible polynomials having multiple roots in an extension field. For example, let F = Z,(t), t being an indeterminate. Let f ( x ) = x 2 + f ; it is irreducible in F [ x ] . (Why?) By Proposition 1.8 f ( x ) has a root, call it $,in some extension field K of F. But then in K [ x ] we have ( x = x2 - 2 J x t= x 2 t = f ( x ) , so $is a root of f ( x ) with multiplicity 2.

+

fi)’

+

Exercise 3.1. Suppose F is a field of characteristic p > 0. (1) Show that ( a + b ) p = a p + bP and ( a - b)P = a p - bP for all a, b E F . (2) Let K = F ( t ) for an indeterminate t and let f ( x ) = x p - t E K [ x ] . Show that f ( x )is irreducible and that f ( x )has just one root with multiplicity p in any splitting field. If F is a field, then an irreducible polynomial f ( x ) E F [ x ] is called separable if f ( x ) has distinct roots in a splitting field (the definition is of course

3 Normality and Separability

91

independent of the splitting field used since any two are F-isomorphic by the corollary to Theorem 1.10). In general a polynomial f(x) E F [ x ] is called separable if each of its irreducible factors is separable. If F is a field and f ( x ) = a,

+ a , x + ... + a k X k E F [ x ]

the deriuative of f(x) can be defined formally as f‘(X)

= a,

+

2U2X

+ 3 a 3 X 2 -k ’ . ’ + k U k X k - ’ .

Of course j ’ ( x ) = 0 if f ( x ) is a constant. Exercise 3.2. If f ( x ) , g(x) E F [ x ] show that the derivative of f ( x ) is f ’ ( x ) + g’(x) and the derivative of f(x)y(x) is f ( x ) g ’ ( x ) + f’(x)g(x).

+ g(x)

Proposition 3.1. Suppose F is a field, f ( x ) E F [ x ] , and degf(x) > 0. Then

f ’ ( x ) = 0 if and only if char F = p > 0 and there is a polynomial g(x) E F [ x ]

such that f ( x )

= y(xp).

Proof. This becomes clear upon inspection of the coefficients of f ’ ( x ) . Proposition 3.2. Suppose F is a field and f ( x ) is a nonconstant polynomial in F [ x ] .

( I ) If f’(.x) = 0, then every root of f ( x ) in any extension field has multiplicity 2 or greater. (2) If f ’ ( . x ) # 0 and ( f ( x ) ,f’(x)) = I , then f ( x ) has no repeated roots in any extension field. Proof. ( 1 )

Let a be a root of , f ( x ) in an extension field K of F , and write

f(x) = (x - a ) g ( x )in K [x]. Then

0 = f’(x)

=

g(x) + ( x

and so g(x) = -(x - a)g’(x). Setting - a)’ I f ( x ) in K [ x ] .

Y

=

-

a)g’(x)

a we see that g(a) = 0 and hence

(x

( 2 ) Suppose to the contrary that f ( x ) has a root a with multiplicity 2 (or x ) K [ x ] and more) in an extension K of F. Write f(x) = (x - ~ ) ~ g ( in differentiate to see that a is also a root of ,f’(x). By Proposition 11.5.3there are h ( x ) , k ( x ) in F [ x ] such that h ( x ) f ( x )+ k ( x ) f ’ ( x )= 1. Substitute x = a to obtain 0 = 1, a contradiction. Corollary 1. If f(x) E F [ x ] is irreducible, then f ( x ) is separable if and only if f ’ ( x ) # 0. Corollary 2. If char F = 0, then every polynomial is separable Corollary 3. If char F = p > 0, then an irreducible polynomial f ( x ) is inseparable if and only if it has the form f ( x ) = g ( x p )for some g(x) E F [ x ] .

92

111 Fields and Galois Theory

If F is a field, K is an extension of F, and a E K is algebraic over F , then a is called separabZe over F if its minimal polynomial m,.,(x) is separable. If K is algebraic over F , then K is called a separable extension of F if every a E K is separable over F. Thus for example every algebraic extension of a field F of characteristic 0 is a separable extension by Corollary 2 above. Proposition 3.3. Suppose K is a splitting field over F for some set 5 F [ x ] , that f ( x ) E F [ x ] is separable and irreducible of degree k > 0, and that f ( x ) splits over K . If G = G(K:F) and if a E K is one root of f ( x ) , then {$(u):$ E G ) is the set of all roots of f ( x ) in K . In fact, if L = F(a) and H = 9 L IG, then [C : H ] = k, and if c j , , . . . ,(Pk are coset representatives for H in G, then $la,. . .,&a are all the roots of f(x).

d

Proof. Note that G acts as a permutation group on the roots of f ( x ) in K, since f($c) = $f(c) for all c E K , 4 E G. In particular every &z, 4 E G, is a root of f(x). Let b E K be any root of f(x). Then there is an F-isomorphism 8: F(a)-+ F(b), with 6(a) = b, by the corollary to Proposition 1.9. By Theorem 1.1 1 0 extends to an F-automorphism 4 E G, and $(a) = b. Thus the set of roots in K of f ( x ) is precisely { $ ( a ) : $ E G}. Clearly StabJa) = S L = H, so there are [G:H] distinct roots of f ( x ) , and [ G : H ] = k since .f(x) is separable and irreducible of degree k. If i # j , then cjiu # cPja, or else c$;'$~E Stab, (a) = H, and then = djH.

Theorem 3.4. If F is a field and K is an algebraic extension of F , then the following statements are equivalent. (a) K is Galois over F. E F[x]. (b) K is a separable splitting field for some set (c) K is a splitting field for some set F2of separable polynomials in F [ x ] .

Proof. (a) = (b): Set Fl = frn,(x): a E K ) c F[x]. Each m,(x) E PI splits over K and has distinct roots by Theorem 2.8. Thus K is a separable splitting field for 9,. (b) * (c): Take F2 = F1. (c) (a): Assume first that [ K : F] is finite and let G = G ( K :F ) . Choose an irreducible factor f ( x ) of some g(x) E F2,with degf(x) = k > 1. Let a E K be a root of f(x), set L = F(a), and set H = 9 L . Then [ G : H ] = k = [ L : F ] by Propositions 3.3 and 1.3. Now use induction on [ K : F ] , and consider the setting L c_ K . Since K is a splitting field over L for g2, and [ K : L ] < [K:F], we have by induction that K is Galois over L, and hence [K:L] = IG(K:L)I = I9LI = IHJ.Multiplyby[L:F] = [G:H]toseethatIGI = [ K : F ] . Thus K is Galois over F by the corollary to the Fundamental Theorem of Galois Theory. Suppose then that [ K : F ] is infinite, and take any a E K \ F . Then [F(a):F ] is finite, so a is contained in a splitting field M G K over F for some finite

3 Normality and Separability

93

subset of P2.But then [ M : F ] is finite, and M is Galois over F by the first part of the proof, above. Thus $(a) # a for some $ E G ( M : F ) .By Theorem 1.1 1 $ extends to an element 0 E G ( K :F). Since 8(a) # a we have K Galois over F.

Corollary. Suppose K is an algebraic extension of F and char F = 0. Then K is Galois over F if and only if K is a splitting field over F for some set of polynomials in F [ x ] . An extension field K of F is called normal over F if every irreducible polynomial in F [ x ] that has a root in K splits over K . The next theorem shows that the notion of normality characterizes splitting fields, and also characterizes stability within an algebraic closure.

Theorem 3.5. Suppose K is an algebraic extension of F and F is an algebraic closure of F with K E F. Then the following statements are equivalent. (a) K is normal over F. (b) K is a splitting field for some set 8 E F [ x ] . (c) If 4 E G ( F : F ) ,then $ ( K ) E K . Proof. (a) *(b): Let 8 = { m , ( x ) : aE K } . It is clear from the definition of normality that K is a splitting field for 8over F. (b)+(c): Say K is a splitting field for 9 = { f a ( x ) : aE A ) E F [ x ] . Take 4 E G ( F : F ) and f , ( x ) E 9,and suppose a E K is a root of f,(x). Then 0 = +(f,(a)) = f,(4(a)) since .f,(x) E F [ x ] , so 4(a) is also a root of f,(x), and consequently 4(a) E K . Since K is generated over F by elements such as a we have & K ) c K . (c) *(a): Take f ( x ) irreducible in F [ x ] , with f(a) = 0 for some a E K . Let b be any other root of f ( x ) in F. Then there is an F-isomorphism 4: F(a) -,F(b) with +(a) = b by the corollary to Proposition 1.9, and 4 extends to an automorphism 8 E G ( F : F ) by Theorem 1.11. But then 8 ( K ) E K , so O(a) = b E K and f ( x ) splits over K .

Corollary. If K is an algebraic extension of F , then K is Galois over F if and only if it is both normal and separable over F. Proof. Theorems 3.4 and 3.5.

Let K be an extension of F . A normal closure of K over F is a field L 2 K that is normal over F and minimal in that respect. A Galois closure of K over F is a field L 2 K that is Galois over F and minimal in that respect.

Theorem 3.6. If K is an algebraic extension of F , then K has a normal closure Lover F which is unique up to K-isomorphism. If [ K :F ] is finite then [ L :F ] is finite. If K is separable over F , then L is a Galois closure for K over F.

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111 Fields and Galois Theory

Proof. Let {as:BE B ) be a set of generators for K over F. Set 9 = (m,,,,(af:fiE B ) , viewed as a subset of K[x], and let L be a splitting field for B over K . Since some of the roots of elements of 9generate K over F and the remaining roots generate Lover K it is clear that Lis also a splitting field for B over F. Thus L is normal over F by Theorem 3.5. If K is also separable over F , then L is Galois over F by Theorem 3.4. If K c M E L and M is normal over F, then one root (viz., as) of each m,,(x) E 9is in M (in fact in K ) , and hence each ma,(x) E B splits completely in M since M is normal over F. Thus M = L and L is a normal closure for K over F . If L' is another normal closure for K over F , then L' is also a splitting field for 9 over K , so L' and L are K isomorphic by Theorem 1 . 1 1 . If [ K :F] is finite we may assume that B is finite and hence [L:F] is finite. Proposition 3.7. If G is a finite subgroup of the multiplicative group F* of a field F , then G is cyclic. Proof. Since G is the direct product of its Sylow subgroups it will suffice

to show that each Sylow subgroup P of G is cyclic. Set rn = max{ 1al:a E P I , and choose b E P with Ibl = m. Then 1, b, b 2 , ...,b"- are m distinct roots of f(x) = X" - 1 E F[x], so they are all the roots by Proposition 1.7. If c E P , then cm = 1, so c is a root of f(x), and therefore c = bk for some k. Thus

P = (b).

Recall that an extension K of F is called simple, with primitive element a, if K = F(u). Theorem 3.8. If K is a finite extension of F, then K is simple over F if and only if there are only finitely many intermediate fields L between F and K . Proof. *: Say K = F(a) and write m(x) for the minimal polynomial m,,,(X). If F G L E K set f(x) = ma.Jx), so f(x) I m(x) in K[x] by Proposition 1.2. If f(x) = bo + b , x .,. x'set M = F(b0,b,,...,bk-,)G t.Then W Z , . ~ ( X ) = f ( x ) and K = M ( a ) , so [ K :M] = degf(x) = [ K :L] by Proposition 1.3 and M = L. Thus L is completely determined by f ( x ) ,and the number of intermediate fields can be no larger than the number of monic factors of m(x) in K [XI, which is finite since K[x] is a UFD. e: We may assume that F is infinite since K * is a cyclic group if K is finite, by Proposition 3.7. If a, b E K , with b # 0, set L = F(a, b). Consider all elements c E L obtained by forming c = a + bd, d E F . Since F is infinite but there are only finitely many intermediate fields there are elements c1 = a + bd, and c2 = a + bd, # c 1 such that F(c,) = F ( c 2 )= E c L. But then c, - c2 = b(d, - d , ) E E, and 0 # d , - d2 E F , sob E E. Also, a = c, - bd, E E , so E = L, is., F(a, b) = F(c,).Now choose a E K so that [ F ( a ) : F ]is maximal. If F(a) # K we could choose b E K\F(a), and then write F(a, b) = F(c) as above, contradicting the maximality of [F(a):F ] . Thus K = F(a) is simple.

+ +

3 Normality and Separability

95

Theorem 3.9. If K is a finite separable extension of F , then K is simple over F . Proof. Let L be a Galois closure for K over F (Theorem 3.6).Then G(L:F ) is finite and has only finitely many subgroups. Thus there are only finitely many fields between F and K by the Fundamental Theorem of Galois Theory, and K is simple over F by Theorem 3.8. Exercise 3.3. Find a primitive element over Q for a splitting field K G C for the polynomial f ( x ) = x4 - 5 x 2 + 6. It is fashionable to assert that the Fundamental Theorem of Algebra is actually a theorem of analysis, presumably because all known proofs depend on “analytic” properties of C or R. The proof that appears below, attributed by S . Lang [23] to E. Artin, provides interesting applications of some of the general field theory results obtained thus far. The only result needed from analysis is the fact that a polynomial of odd degree with real coefficients has a real root, a consequence of the Intermediate Value Theorem of calculus since such a polynomial clearly takes on both positive and negative values.

Theorem 3.10 (The Fundamental Theorem of Algebra). The field C of complex numbers is algebraically closed. Proof. If f ( x ) E @ [ X I let u be a root of f ( x ) in some extension of @. Let K be a Galois closure of C(u)over R (Theorem 3.6) and set G = G ( K : R). Let H be a 2-Sylow subgroup of G and let L = F H . By the Fundamental Theorem of Galois Theory we have [ L :R] = [G: H I , an odd number. By Theorem 3.9 we may write L = R(b) for some b E L, so the minimal polynomial mb&) is irreducible over R and of odd degree. By the remarks preceding the proof that degree must be 1, and hence L = R,which means that G = H , a 2-group. Thus G , = 9@ = G ( K : C )is also a 2-group. If G , # 1 choose G, I G , such that [ G , : C 2 ]= 2 (see Exercise I. 2.5), and set M = 9 C 2 , so that [M:@] = [ G , : C,] = 2. But any polynomial of degree 2 over C has roots in C by the quadratic formula (verify that every a E @ has a square root in C), so such a field M cannot exist! The contradiction shows that G , = 1, hence K = @, and a E @, completing the proof. As a final application in this section of the ideas above we describe all finite fields. Note that a finite field must have characteristic p for some prime p E Z. For any field F of characteristic p the mapping 4 p : uH u pis a monomorphism; it is called the Frobenius map on F.

Theorem 3.11. Suppose F is a finite field with q elements, having prime field F, E Z,.Then q = p”, where n = [ F : F , ] , and F is a splitting field over Fp for the polynomial f ( x ) = xq - x . Conversely if 0 < iz E 7 and p is a prime,

96

Ill Fields and Galois Theory

then there is a field F with q = p" elements, and F is uniquely determined up to isomorphism. The Galois group G(F:F,) is cyclic of order n, with the Frobenius map 4, as a generator. Proof. Note that q = p" is a special case of the fact that any vector space of dimension n over a field with p elements has p" elements; it consists of all linear combinations of n basis elements so there are p choices for each of n coefficients. The multiplicative group F* has order q - 1, so each a E F* is a root of x 4 - - 1, and hence each a E F is a root of f ( x ) = xq - x . Thus F is a splitting field for f ( x )over F,. For the converse take F, = B, and let F 2 Fp be a splitting field for f ( x ) = x 4 - x E F,[x]. Note that f ( x )has q distinct roots by Proposition 3.2, since f ' ( x ) = - 1. But if a, b E F are roots of f ( x ) ,then a f b, ab, and a/b (if b # 0) are also roots of f ( x ) , so the set of q roots of f ( x ) constitute a field over which f ( x ) splits, and IF1 = q. The uniqueness follows from Theorem 1.10.The Frobenius map 4 pis in G(F:F,) since F is finite, and 4:(a) = aPk for all a E F if 0 I k E Z, so clearly 4: = 1. If 4, had order k < n, then all elements of F would be roots of g ( x ) = xPk- x, contradicting Proposition 1.7. Since IG(F:F,)( < [F:F,] = nit follows that G(F:F,) = (4,).

Corollary. If F and K are both finite fields, with F extension of F .

G

K , then K is a Galois

Proof. If IKl = q view f ( x ) = x4 - x as a polynomial in F [ x ] . Since f ( x ) has distinct roots it is separable, and K is a splitting field for f ( x )over F. Apply Theorem 3.4.

A field with q = p" elements is called a GaloisJield and will be denoted by 4 [another common notation is G F ( q ) ] .

4. THE GALOIS THEORY OF EQUATIONS If F is a field and f ( x ) E F [ x ] let K be a splitting field for f ( x ) over F . Then the Galois group of f ( x ) over F is defined to be G = G ( K :F ) . The genesis of Galois theory was E. Galois's study in 1832 relating properties of f ( x ) to the interplay between subgroups of G and fields Lintermediate between F and K . In his work F and K were taken to be subfields of @. Note that the Galois group of f ( x ) over F depends (up to isomorphism) only on f ( x ) and F , since all splitting fields for f ( x ) over F are F-isomorphic. If S c K is the set of (distinct) roots of f ( x ) ,then G acts as a permutation group on S . The action is faithful since F ( S ) = K , so if 1st = n we may (and often shall) view G as a subgroup of the symmetric group S,,. Exercise 4.1. If f ( x ) E F [ x ] and K is a splitting field for f ( x ) over F, denote by S the set of distinct roots of f ( x )in K and let G = G ( K : F ) .If f ( x )is

4 The Galois Theory of Equations

97

irreducible over F show that G is transitive on S . If f ( x )has no repeated roots and G is transitive on S show that f ( x ) is irreducible over F. A simple radical extension of a field F is a field K = F(a), where u" E F for some positive integer n. Thus a is a root of a polynomial of the form x" - b, where b E F. For example, if char F # 2 and [K:F] = 2, then K is a simple radical extension of F , as is easily seen by the familiar process of completing the square. If F is a field and 0 < n E Z consider f ( x ) = x" - 1 E F[x]. If char F = p > 0 we assume that p X n. Let K be a splitting field for f ( x ) over F . Since f ' ( x ) = nx"- ', and p X n if char F = p , we see by Proposition 3.2 that f(x) has n distinct roots in K [ f ( x )and f ' ( x )are relatively prime since they have no roots in common]. The roots of f ( x ) are called nth roots of unity. The nth roots of unity constitute a multiplicative subgroup of K *. The subgroup is cyclic by Proposition 3.7, and its generators, which have multiplicative order n, are called the primitive nth roots of unity. By Exercise 1.1.6 there are $(n) primitive nth roots of unity, where 4 is Euler's totient function. If [ is a primitive nth root of unity in K , then 1, [, [', . . . , i n -' are all of the nth roots of unity, and K = F ( [ ) ,a simple radical extension. Exercise 4.2. If r] . . . + $ - 1 = 0.

E

K is an nth root of unity,

r]

# 1, show that 1

+ r] +

r]2 +

Proposition 4.1. Suppose F is a field, 0 < n E Z, and if char F = p , then p J' n. Then the Galois group G of f ( x ) = x" - 1 over F is abelian. Proof. By the remarks preceding the proposition we may take G = G ( K : F ) ,where K = F ( i ) and [ is a primitive nth root of unity. If 4, 8 E G, then $([) and O ( [ ) are roots of f ( x ) ,so 4(() = ('and O ( c ) = i j for some integers i and j . Thus 4O(i) = iiJ= O$(i). Since K is simple over F any element of G is completely determined by its effect on (, so 46 = 6 4 and G is abelian.

If f ( x ) = x" - 1 E F [ x ] as above let roots of unity in K and set

C2, ..., [,(")

be the primitive nth

@,(x) = n { ( x - i i ) :1 Ii I+(n)},

the nth cyclotornic polynomial. If r] is any root of f ( x ) = x" - 1, then r] is a primitive dth root of unity, where d is the multiplicative order of q in K *, so d is a divisor of n by Lagrange's Theorem. Consequently X" -

Thus

1 = n { @ d ( X ) : O < d E ?i and d I n}.

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111 Fields and Galois Theory

and cD,(x) can be determined recursively. Note that cDl(x)= x - 1, so cD2(x)= @3(x)=

x2 - 1 -x+l, x-1

~

x3- 1 x-1

~

= x2

+ x + 1,

x6 - 1 = (x - l)(x l)(X*

+

+ x + 1) = x 2 - x + l ,

Since each successive cyclotomic polynomial is obtained by division we see inductively, by the division algorithm for F [ x ] ,that cDn(x)E F [ x ] for all n. In fact, if F = Q it follows that cD,,(x)has coefficients in Z.

Exercise 4.3. ( 1 ) Compute the first twelve cyclotomic polynomials. (2) Find all n for which deg cD,,(x) = 2. Theorem 4.2. If p is a prime and n cD,(x) is irreducible in Q [ x ] .

= p",

then the cyclotomic polynomial

Proof. Let us write f ( x )for cD,(x). Note that f ( x ) = (XP" - l)/(XP". - 1 ) =

p m - 1(

P-1) + XP--l(P-2)

+

... + X P - - I

+ 1,

since the roots of x P m - - 1 are roots of unity of orders lower than n = p". If f ( x ) is reducible in Q [ x ] it is reducible in H[x] by Proposition 11.5.15, so suppose f ( x ) factors as f ( x ) = g(x)h(x)in Z [ x ] . Since f ( 1 ) = g ( l ) h ( l )= p , a prime, we may assume that g ( 1 ) = k 1. If n, = 1, n2,..., n, are the integers between 1 and n = p" that are relatively prime to p set

k ( x ) = n { g ( x " ' ) 1: I i I s}. If [ E C is any primitive nth root of unity then { l n 1l :I i 2 s} is the set of all primitive nth roots of unity, i.e., the set of all roots of f ( x ) .Some ["J is a root of g(x),so k([) = 0. Since [ was arbitrary every root of f ( x )is a root of k ( x )and hence f ( x )1 k ( x ) in Z [ x ] , say k(x) = f ( x ) . a ( x )with a(x)E Z [ x ] . But then k(1) = g(1)" =

k1

= f ( l ) a ( l )= p . a ( l ) ,

a contradiction.

It should be remarked that in fact @,,(x)is irreducible in Q [ x ] for all n. For a proof see Van der Waerden [37]. Proposition 4.3. Suppose F is a field, 0 < n E Z,and if char F = p , then p ,'j n. Suppose F contains a primitive nth root l of unity and 0 # b E F . Then a

4

The Galois Theory of Equations

99

splitting field K for f ( x ) = x ” - h over F is a simple radical extension of F and the Galois group G of f ( x ) is abelian.

Proof. Let a E K be a root for ,f(x). Then a, i a , [,a,. . . ,( “ - ‘ a are n distinct roots of f ( x ) in K , so K = F(a), and a” = b E F so K is a simple radical extension. The proof that G is abelian is very similar to the proof of Proposition 4.1 (in fact G is cyclic in this case). A field K is called an extension by radicals of a field F if there is a sequence Lo G L , c L , c ... c L, of fields, with Lo = F and L , = K , such that Liis a simple radical extension of Li - 1 I i I n. A polynomial f ( x ) E F [ x ] is said to be soltiable by radicals over F if there i s an extension K of F by radicals such that f ( x ) splits over K . Thus for example if char F # 2 and deg f ( x ) = 2, then f ( x ) is solvable by radicals. For F = Q that fact goes back at least to Eudid, and we have the quadratic formula for expressing the roots of f(x) in terms of its coefficients by means of “algebraic operations,” i.e., the usual field operations together with the extraction of square roots (radicals). In the first half of the sixteenth century the Italian mathematicians del Ferro, Tartaglia, and Ferrari obtained formulas for the roots of polynomials of degrees 3 and 4, again in terms of the coefficients via field operations and extraction of roots (see Van der Waerden [ 3 7 ] ; also see Section VI. 9). As a consequence polynomials of degree 3 or 4 over Q (and over many other fields) are solvable by radicals. The search continued for formulas for the roots of polynomials of degrees higher than 4 until early in the nineteenth century, when P. Ruffini in Italy and N. H. Abel in Norway showed that not only could there be no general formulas for degrees 5 or greater, but that not all polynomials were even solvable by radicals. Galois subsequently set their explanations and examples in the general framework of Galois theory and gave necessary and sufficient conditions for solvability by radicals. If F E E L K and F G L C K define thejoin of E and L to be E v L = F(E u L), the smallest subfield of K that contains both E and L. If G is any group, J I G, and H I G,define thejoin of J and H to be J v H = ( J u H ) , the smallest subgroup of G that contains both J and H .

Exercise 4.4. Suppose F G E H I G.

K,F EL

C_

K, G

=

G ( K :F ) , J I G, and

(1) Show that 3 ( E v L ) = 9 E n 9 L and 9 ( J v H ) = 3 J n F - H . ( 2 ) Show that [ E v L : F ] I [ E : F ] [ L : F ] . Proposition 4.4. Suppose F and L are fields, F c K , c L, F c K , G L, and K , , K , are both extensions of F by radicals. Then the join K , v K , is also an extension of F by radicals.

100

Ill

Fields and G a b s Theory

Proof. Let a , , a,,. . . ,a, be the "radicals" that are successively adjoinec to reach K , , and let b,, b,, . . . ,b, be the corresponding elements for K , . Ther K , v K , = F(u,,. . . , a mb,,. r . ., b").

Proposition 4.5. If K is a separable extension of F by radicals and L is a Galois closure for K over F , then L is a separable extension of F by radicals. Proof. Recall (Theorem 3.6) the construction of L as a splitting field = {m,,(x)>,where { a , ,..., a,> is an F-basis for K . Set G = over K of 9G(L:F).By Proposition 3.3 {4(ai):4E G, 1 I i I n} spans L over F . If G = { 4 , , 4 ,,..., 4k} set K i = 4 i ( K ) , 1 I i I k. Then L = K , v K , v ... v K k , and each K iis F-isomorphic with K , so K i is an extension of F by radicals. Thus L is an extension of F by radicals by Proposition 4.4.

Theorem 4.6 (Galois). Suppose char F = 0 and f ( x ) E F [ x ] is solvable by radicals. Then the Galois group of f ( x ) is a solvable group. Proof. Let F = Lo E L , c ... c Lk = K be a sequence of simple radical extensions, with L, = L i - , ( a i ) ,a;' E L,- such that there is a splitting field L for f ( x ) over F , with L c K. By Proposition 4.5 we may assume that K is Galois over F . Let G = G ( L : F ) ,the Galois group of f ( x ) . Since L is Galois over F we have by the Fundamental Theorem of Galois Theory that G = G ( L : F )z G ( K : F ) / G ( K : L )so , by Theorem 1.5.4 it will suffice to show that G ( K : F )is solvable. Set n = nln2 "'nk, let M be the splitting field for x" - 1 over K , and choose a primitive nth root [ of unity in M . Note that all n,th roots of unity are in F ( [ ) , 1 I i Ik. Since K is Galois over F it is a splitting field for some g ( x ) E FCx]. But then M is clearly a splitting field over F for ( x " - l ) g ( x ) , so M is Galois over F as well. As above G ( K : F ) is a homomorphic image of G ( M : F ) ,so it will suffice to show that G ( M : F )is solvable. To that end set

,,

F

=

Mo, F([) = M , ,

,

M,

=

M,(u,),..., M k + , = M,(ak) = M .

Now each Mi+ is a splitting field over Mi, 0 I i I k, because the appropriate roots of unity are present after i = 0, and hence Mi + is Galois over M i .Define subgroups Ho,H1,...,Hk+, of G ( M : F ) by setting Hi = G(M:Mi), so that Ho = G ( M :F ) and Hk+ = 1 . Since M is Galois over Mi and Mi+ is also Galois over Miwe apply the Fundamental Theorem of Galois Theory to see that

,

,

G(M:Mi+,) = Hi+,
and that Hi/Hi+, 2 G(Mi+,:Mi), which is abelian for each i by Propositions 4.1 and 4.3. Thus G ( M :F) is solvable by Theorem 1.5.3, and the theorem is proved.

4

The Galois Theory of Equations

101

For a concrete example set f ( x ) = x 5 + 5 x 3 - 20x + 5 E Q [ x ] . Then f ( x ) is irreducible in Z [ x ] by the Eisenstein Criterion (Theorem 11.5.18), and hence also in O [ x ] by Proposition 11.5.15. Since f ' ( x ) = 5 ( x 2 + 4 ) ( x 2 - 1) we see by inspection of the graph off (x) that f ( x )has exactly three real roots a l , a * , and a 3 .Thus it has two nonreal roots a4 and a, in @,whichare necessarily complex conjugates of one another. Let K c C be a splitting field for f(x) over Q, and let G = G ( K :Q) be the Galois group of f(x), viewed as a subgroup of S, . Since 5 I [ K :Q ] ,and hence 5 I JGI,there must be a 5-cycle in G. By Theorem 3.5(c)the complex conjugation automorphism of A restricts to an automorphism of K . Its effect on the roots of f ( x ) is to fix a,, a 2 ,and a,, and to interchange a4 and a 5 , so G also contains a 2-cycle. It follows (see Exercise 1.12.21)that G = S,. But S, is not solvable since S ; = A , , which is simple but not abelian. Thus f ( x ) is not solvable by radicals. Exercise 4.5. Show that f ( x ) = x 5 - 2 x 3 - 8x radicals over Q.

+ 2 is not

solvable by

Suppose F is a field containing a primitive nth root of unity, K = F(a) is a simple Galois extension with [ K :F ] = n, and the Galois group G = G ( K :F ) is cyclic, say G = (4) of order n. If iis any nth root of unity in F define the Lagrange resolvent of [ and a to be L(i,a)= a + [+(a)

+ [ 2 4 2 ( a ) + ... + i n - 1 @ - 1 ( a ) .

Exercise 4.6. If F, K , and G are as above and L(i,a) is a Lagrange resolvent show that

4(W, 4) = L ( i ?@(a))= i-l L ( i ,4, and conclude that the nth power of L ( i , a) is in F. Proposition 4.7. Suppose F is a field containing a primitive nth root [ of unity, K = F ( a ) is a Galois extension of F with [ K : F ] = n, and the Galois a) lies in K\F group G = G ( K : F )is cyclic. Then the Lagrange resolvent for some i. Proof. Say G = (4). Then

1 L@,a) = 1 c p ( b J ( a )= c $'(a) 1

n- 1

n-ln-1

n- 1

n- 1

i=O

i = o j=o

j=O

i=O

([j)i

= nu

(see Exercise 4.2). Since F has a primitive nth root of unity the characteristic of F cannot divide n, so nl # 0 in F. Thus nu E K \F since a $ F, and consequently at least one of the summands L ( t i , a )must lie in K\F. Proposition 4.8. Suppose p E Z is a prime, F is a field containing a primitive pth root of unity, and K is a Galois extension of F , with [ K :F ] = p. Then K is a simple radical extension of F .

102

111 Fields and Galois Theory

Proof. Choose a E K\F. Then K = F(a) since [ K : F ] is prime. The Galois group G = G(K :F) is cyclic since IGI = [K :F ] = p. By Proposition 4.7 there is a Lagrange resolvent b E K \ F , so K = F(b),and by Exercise 4.6 we have b P = c E F.

Proposition 4.9. Suppose f(x) E F[x] has Galois group G over F, and E is an extension field of F. Then the Galois group of f(x) over E is isomorphic with a subgroup of G. Proof. Say L is a splitting field for f(x) over E, and that f(x) has roots u1,u2,. ..,a, in L. Then K = F(a,, . ..,an)is a splitting field for f(x) over F. If 4 E G ( L : E ) then , 4 permutes the roots a,, . ..,a,, so 4 ( K ) = K, and 4 acts as the identity on K only if 4 fixes each root ai,which is only if 4 = 1 E G(L:E). Thus Cp H 4 1 K is a monomorphism into G(K :F).

Theorem 4.10 (Galois). Suppose F is a field of characteristic 0, f(x) E FCx], and the Galois group of f(x) is solvable. Then f(x) is solvable by radicals over F. Proof. Let K be a splitting field for f(x) over F, set G = G(K : F ) , and say [ K : F ] = n. Let L be a splitting field over K for xn - 1, and let [ E L be a primitive nth root of unity, so L = K ( [ ) . Set E = F ( [ ) , then clearly L is a splitting field for f(x) over E. If we set H = G(L:E), then H is isomorphic with a subgroup of G by Proposition 4.9, and hence H is also solvable. By Theorem 1.5.3 H has a subnormal series H = Ho 2 H, L . * .2 Hk= 1 with abelian ,/Hi is factors, and by refining to a composition series we may assume that Hicyclic of prime order pi, 1 I i I k. In the setting of E E L we set Li= 9 H i , 0 I i Ik, so E = Lo c L , c ... 5 Lk = L and [ L j : L i - , ]= p i . Since G(L:Li) = Hid Hi- = G(L:Li - ) we have LiGalois over Li- and Li- contains a primitive pith root of unity (it is a power of [). By Proposition 4.8 Liis a simpie radical extension of Li-I , 1 I i I k. Thus L is an extension of E by radicals, and hence also of F since E = F([).

,

,

5. SYMMETRIC FUNCTIONS If F is any field let xl, x2,. . .,x, be distinct indeterminates over F and set K = F(xlrx2,. . .,x,), the field of rational functions. Set f(x) = (x - xl)(x - x2)...(x - x,) E K[x]; f(x) is called the general polynomial of degree n. Multiply the linear factors of f(x) together to obtain

f(x) = xn - c 7 + - 1

+ g2xn-2

-

'.. + (-1Ygnr

5 Symmetric Functions

103

so that

a, = X,X,X3"'Xn

In general ak = ~ { x i , x i z ~ ~ ~I x ii,k < : ~i, < ... < i, I n},

a homogeneous polynomial of degree k in xl, x, ,. . . ,x,. The coefficients ok are called the elementary symmetric polynomials in xl, x,, . . . ,x,.

n

Exercise 5.1. (1) Write out the elementary symmetric polynomials for 3, 4, and 5. (2) How many monomial summands are there in a,?

=

Each element 4 of the symmetric group S, defines an F-automorphism of K by permuting the indeterminates xl,. . . ,x, as follows: if g(xI,xz,. ..,x,) E K , then 4g(x,

For example if

., xn) = Y(x+,~,,x+(2),. . . ,x+(n)).

>~ 2 1 . .

4 = (132) and g(x,,x,,x,)

+

= x: - ~ 2 x 3 x:, then

d)g(x1,x2,x3) = g(x3,x1,x2) =

-

+

Set F, = F(a,,a, ,..., a,) G K and let F, be the fixed field in K of S,. Clearly F, c F1since each (T, is invariant under permutations of x x, ,. . .,x,. The elements of F, are called the symmetric rational functions in x,, x2,. . .,x, over F . Note that f(x) E F , [ x ] , and that K = F,(x,,x,,. . .,x,), so K is a splitting field for f(x) over F,. Thus [ K : F , ] I n! by the corollary to Proposition 1.8. Since f ( ~ )has distinct roots, K is Galois over F, by Theorem 3.4. Since G ( K : F , ) 2 S, we have IG(K:F,)I = [ K : F , ] 2 n ! . and so n! 2 [K:F,] 2 [K:Fl] 2 n! and equality holds throughout. Consequently F, = FI and every symmetric rational function can be represented as a rational function in the elementary symmetric polynomials al, a,, . . . ,on (this statement will be refined later, in Theorem 5.3). Also the Galois group G = G ( K :F,) of the general polynomial f(x) is the symmetric group S,. The next theorem is an immediate consequence of this fact and Theorem 4.6. Theorem 5.1 (Ruffini, Abel). If F is any field of characteristic 0 and n 2 5, then the general polynomial of degree n is not solvable by radicals over the field of symmetric rational functions in x,, . . . ,x, over F .

Exercise 5.2. Show that the general polynomial f(x) is irreducible in FOIIXl.

104

III Fields and Galois Theory

Theorem 5.2. If H is any finite group then there exist a field Land a finite Galois extension K of L such that H is (isomorphic with) the Galois group G ( K :L). Proof. By Cayley's Theorem (1.2.2)H is isomorphic with a subgroup HI of a symmetric group S , (we may take n = [HI).Take F , F,, and K as above so that G ( K :F0) = S , . Set L = F H l . Then by the Fundamental Theorem of Galois Theory H , = 3 L = G ( K :L).

It is a famous unsolved problem to determine whether every finite group H occurs as the Galois group of a finite Galois extension of the rational field Q. Shafarevich [34] gave an affirmative solution in 1954 for solvable groups H , but the general question remains open. We describe next a constructive procedure for expressing arbitrary polynomials in xl, x 2 , .. .,x, in a particular form involving the elementary symmetric polynomials ol, 0 2 , .. . , o n .It will be convenient to replace the field F by an arbitrary commutative ring R with 1. Note that o,,.. .,on can be viewed as elements of R [ x , , . . .,x,]. As usual each 4 E S , acts on elements of R[xl, . .. ,x,] by permuting the indeterminates, and we say that g(x,, . . .,x,) is a symmetric polynomial over R if $g(x,,. . .,x,) = g(x,,. . . ,x,) for all 4 E S , . Set f,(x) = f(x), the general polynomial of degree n, with coefficients in R [ x , , . . . ,x,], and note that f,(x) is monic of degree n, has x, as a root, and has 1, - ol, 0 2 , .. . as its coefficients. Next set f n-

1(x) = fn(x)/(x - x n ) .

On the one hand, of course, f , - l ( x ) = (x - xl)..-(x - x,-,).

On theotherhandifweviewf,(x)asx" - o1xB-' + ... wemayobtainf,-,(x) by the division algorithm for polynomials with coefficients in the ring R [ o , , . . .,on,x,]. Observe that 1,-l(x)is monic of degree n - 1, has x,- as a root, and has coefficients that are various polynomials over R in ol,0 2 , ... ,on, and x,.

Next set ffl-2W

= ,L

-

-L11,

and in general set fk(')

= f k + l(x)/(x

- x k + I),

15k 5n

-

1.

Just as for f , ( x ) and for f,- l(x) we see in general that .fk(x)is monic of degree k,

5 Symmetric Functions

105

has xk as a root, and has coefficients that are polynomials over R in ... . .,xk + 1 . For example, if n = 3 we have

ul? 9 On?xn, Xn- 1 , .

f2(x) = (x - xJ(x - x2) = x2

+

(-@I

+ x3)x + ((72 -

X3U]

+ x:,

and fl(X) = x

-

+

x1 = x

+ x2 + x3).

(-Q1

If we now set x = Xk in fk(x) we may solve for xf: as a polynomial of degree 1 or less evaluated at x,., the coefficients being polynomials over R in o l , . . . ,rr,, x,, . . .,xk+ 1. Taking n = 3 again for an example we have k

-

X ] = (71

- x2 - x3,

x: = ( 0 1

-

x; = alx:

x3)x2 - ((72 - x3G1

- C2X3

+ a3.

+ xi),

Given any polynomial g(Xl7.. ., xn) E RCx1, ~

2 , .. . >xnI

we may replace each x1 by u1 - x2 - ... - x, so that x 1 no longer appears. Then we may replace each x: by its expression in terms of x2(to the first power only), o,,...,on,x, . . . ,x3so that x2finally appears only to the first power, if at all. Continuing, we may make replacements so that Xk appears only to the power k - 1 or less for each k. Ultimately we will have expressed the original polynomial g(xl,. . . ,x,) as a sum of terms, each having the form h(a,,. . ,,(7,)xpxy . . .x:.,

where 0 I k2 < 2,O I k, < 3,..., 0 I k, < n, and h ( a l , . . .,a,) is a poly. . ,on. nomial in n variables over R evaluated at ol,. For example, with n = 3 again, if g(xI,x2tx3) = x: - x 2x 3

+ x:?

then the process above gives g(x,,x,,x,)

= olx:

- ((71

+ (72)x3 - (71x2 +

((7:

-02

+ (73).

Exercise 5.3. Apply the above process to

+ x: + x;,

(1) g(x,,x,,x,)

= x:

(2) g(x,,x2,x,)

=XI -

2 X2

+ x3-

and

3

Theorem 5.3 (The Fundamental Theorem on Symmetric Polynomials). Suppose R is a commutative ring with 1 and dx,,

. . ., x,)

E

R [xi,. . ., xn1

106

111 Fields and Galois Theory

is a symmetric polynomial. Then there is a polynomial f(xl,... , x , ) in R[x, ,..., x,] such that g(x, ,..., x,) = f(ol ,..., on), i.e., g(xl,..., x,) is a polynomial over R in the elementary symmetric polynomials. Proof. By the procedure above we may write g(xl,. . . ,x,) as a sum of monomials h(a,,.. .,o,)xP...x?. Since g(xl,. . . ,x,) is symmetric it is invariant under the various transpositions (1 2), ( 1 3), . . . ,(1 n), from which it follows that k, = k, = ... = k, = 0 since x 1 does not appear in any of the monomials and each coefficient h(a,,. . . ,a,) is symmetric. The theorem follows.

Theorem 5.4. Suppose f(x) = bo + b l x + ... + b , ~ "E Z[X] has roots a,,a, ,..., a, in @, and suppose g(xl,x, ,..., x,) is a symmetric polynomial over Q of degree d. Then g(al,a,, . . .,a,) E Q, and if g(xl,. . . ,x,) has coefficients in Z then bfg(a,,. . .,a,) E Z. Proof. Note that = b,

f(x)

. n { ( x - ai):1

I iI

n},

so bi = k b,oi(al,.. . ,a,) for each i, and hence oi(al,... ,a,) = fbi/b, E Q. Thus g(a,, . . .,a,) E Q by Theorem 5.3. If we set h(x) = f ( x / b , ) ,then the roots of h(x) are b,al, b,a2, ..., b,a,. Note that b;-'h(x) is monic in Z [ x ] , say b;-'h(x) = co + C , X = n{(x

and so ai(b,al,.. .,b,a,)

=

-

+ ....+ C , _ ~ X ~ - '+ X" b,ai):1 I i I n} E Z[x],

fciE Zfor each i. If

g(x1,. . . y . x n )

E z[x,,

. . .,xnI,

then it is a sum of homogeneous symmetric polynomials over Z of various degrees r I d. If k(xl,.. . ,x,) is one of those homogeneous summands, then

b ; k ( a , , .. . ,a,) = b f ' k ( b , a , , . . . ,b,a,)

E

Z,

again by Theorem 5.3, and so b f g ( a , , . .,a,) E Z. Recall that an algebraic number is a complex number that is algebraic over

Q. All other complex numbers are called transcendental.

Theorem 5.5 (Lindemann, 1882). The real number n is transcendental. Proof (Niven [28]). If not, then in. is also algebraic (where i2 = - 1) and there is an irreducible polynomial f,(x) E Z[x] such that f , ( i n ) = 0. Let al = in,a,,. . . ,a, be all the roots of f l(x) in @. Since ein= - 1 we have

+

n{(l e"J):l ~j I n} = 0.

5 Symmetric Functions

107

Multiply to see that

o=

1 +eai+...+eatB+

C , a ,+o,+...+eai+a2+-..+-a.

i<j

Set g2(x) =

n{(~ (ai + a j ) ) :1 -

_< i

(*)

< j I n}.

If 4 E S, apply 4 to g2(x) in its factored form by permuting the subscripts of the ais, and note that each 4 E S, leaves g2(x) invariant. Thus the coefficients of g2(x) are symmetric polynomials (over Z)in a,, . ..,a,, and hence g,(x) E Q[x] by Theorem 5.4. Multiply by a suitable integer to obtain f2(x) E ZCx] having all ai + aj as its roots, 1 Ii <.j I a. Similarly we may obtain f3(x) E Z[x] having all ai a j ak as its roots, 1 I i < j < k I n, etc. Thus

+ +

n{j,(x): 1I j I n} E Z[x]

has as its roots all of the exponents q , a i+ a j , etc., in (*) above. Suppose Ij 5 n} has 0 as a root with multiplicity k - 1, and set

n{fi(x): 1

f(x) = (H{j,(x): 1 Ij I n})/x"' = c,,

+

C1.Y

+ ... + c r x r E Z[x],

c,, # 0, c, # 0. Thus the roots b , , b,, . . . ,b, E C of f(x) are just the nonzero exponents in (*), and (*) becomes ebl

+

eb2

+ . . . . + ebr + k

= 0,

with 0 < k E Z. Let p E Z be some prime and set g(x) = ( c y l / ( p

-

l)!)x"-I(f(x))P.

Then set h(x) =

where g'''(x)

=

C{g"'(x):O

I j I p ( r + I)},

g(x) and g"'(x) is the jth derivative of g(x).

Exercise 5.4. (1)

Show that d dx

-(e-"h(s))

(2) Show that

= -e-"g(x).

108

111 Fields and Galois Theory

and g(j'(0)E pZ

if p I jI p

+ r p - 1.

(3) If p > max{k,co,cr}conclude that k . h ( O ) E Z and ( k . h ( O ) , p ) = 1. Continuing the proof we apply Exercise 5.4.1 and the Fundamental Theorem of Calculus to see that -{:e-'g(t)di

Change variables,

t = sx,

-x{I

= e-"h(x) - h(0).

and get e-"g((sx)ds = e-"h(x)- h(0)

or

Substitute x

= b , , b,,

. . . ,b, in succession and add to obtain

1 h(hj)+ k * h ( O ) r

I

C (h(bj)

-

ebJh(0)) =

j= 1

j= 1

Note next that g ( b j )= g'(bj)= ... = g(P-l)(bj)= 0

for all j ,

since bj is a root of multiplicity p , and that ( p - l)!g(x) E Z[x]. The pth and higher derivatives of ( p - l)!g(x) are all in Z[x], and each has all its coefficients divisible by p or more consecutive integers. It follows (Why?) that each has all its coefficientsdivisible by p ! , and so all g ( j ) ( x )j, 2 p , are in pZ[x]. g(j'(b,) is a symmetric polynomial over Zin b , , . . . ,b,. For each j 2 p , By definition of g(x) each coefficient of g(j)(x) is divisible by ciP-', and degg(j'(x) Irp - 1. By Theorem 5.4 and the observations above

zm=

m= 1

Thus

g(J'(b,) = p k j E pZ

for all j 2 p .

6 Geometrical Constructions

109

Since k . h(0)E Z and p $ k . h(0)we conclude that

0#

2 h(bj)+ k . h(0)

E 7.

j= 1

But

Note that and that

If M is an upper bound for Ic:bjf(sbj)l, 0 I s 5 1, then limp-m M p / ( p - l)! = 0 (e.g., since the series M " + ' / n !converges by the ratio test). It follows that each integrand in (**) converges uniformly to 0 as p -, co, and hence the integral can be made as small as we like by a suitable choice of p . That, of course, is incompatible with the sum of the integrals being a nonzero integer, and the theorem is proved.

1

6. GEOMETRICAL CONSTRUCTIONS The ancient Greeks left unsolved several problems involving geometrical constructions. Their point of view was that the only tools allowed for the construction of geometrical figures are compasses and (unmarked) straightedges. It will be convenient to take the point of view of analytic geometry in discussing what can and what can not be constructed. If we choose two points in the plane we may label one as the origin (0,O)and the other as (l,O), thus choosing a unit distance and determining an x-axis. With a straightedge we may draw line segments joining previously constructed points. With a compass we may draw circles having previously constructed centers and radii. We say that a point is constructible if it can be obtained as a point of intersection of such lines and/or circles. Recall from high school geometry that a perpendicular to a line can be drawn at a given point on that line. Clearly then any point (a,b), with a, b E h,is constructible. A real number is called constructible if it appears as a coordinate for a constructible point in the plane. Denote by K the set of all constructible real numbers. Thus for example Z E K .

1 10

111 Fields and Gnlois Theory

Proposition 6.1.

The set K of constructible real numbers is a field.

Proof. Take a, b E K, with b # 0. The following pictures (Fig. 1) indicate how to construct a b, - b, ah, and alb, and hence to conclude that K is a field.

+

I

b

a

I

I

I

a f b

0

-b

b

b

Figure 1

In order to describe the field K somewhat more explicitly let us suppose we have a field L, with Q s L E K, and ask what sort of numbers in K can be obtained by constructions using points with coordinates in L (think of the elements in L as having been constructed previously). Intersecting two “L-lines” amounts to solving two linear equations with coefficients in L, so the solutions will be in L and no new numbers are obtained. Suppose we intersect two “L-circles”, i.e., solve the following equations simultaneously:

( x - a)2 + ( y - b)2 = u, (X -c

+

) ~ ( y - d ) 2 = V,

where a, b, c, d, u, v E L. If the squares are expanded and the second equation is subtracted from the first we obtain 2(c - a ) x

+ 2(d

-

b ) y = u - u - a’

-

b2 + c 2 + d 2 ,

the equation of an L-line whose intersection with either circle is the same as the intersection of the two circles. Thus it suffices to consider the result of

6 Geometrical Constructions

11 1

intersecting a line and a circle, say (x - a)’

+ (y

-

b)’

and

=u

cx

+ d y = u,

with a, b, c, d, u, u E L and either c # 0 or d # 0. We may solve for x or y in the linear equation, substitute into the quadratic equation, and then solve the quadratic equation. Ultimately then the new numbers obtained are roots of a quadratic equation and hence are either in L or in an extension L’ having degree 2 over L. We have established half of the next theorem. Theorem 6.2. A real number a is constructible if and only if there is a sequence Lo _c L , 5 L, c . . . c Lk of subfields of R such that Lo = Q, [ L i : L i - , ]= 2 for 1 _< i 5 k, and a E L,. Proof. It will suffice to show that a square root can be constructed for any constructible number, and we appeal again to a picture (Fig. 2) to indicate the construction.

Figure 2

Exercise 6.1. (1) Verify the constructions above for ab, alb, and @. ( 2 ) Construct the number 1 + (2 + $)”’ with straightedge and compass. Perhaps the best known construction problem left by the Greeks was that of trisecting an arbitrary angle. Note that an angle 4 can be constructed if and only if cos 4 is a constructible number (see Fig. 3).

0

Figure 3

cos+

1

112

I11 Fields and Galois Theory

For the trisection problem we suppose that we are given an angle 8, and we wish to construct the angle 4 = 0/3,so we assume cos 0 = cos 34 as known (and constructible), and ask whether cos 4 is constructible. By elementary trigonometric identities we have cos 6' = 4cos3 4 - 3 cos Cp, so if we set a = cos 8, then x = 2 cos 4 is a root of f ( x ) = x3 - 3x - 2a. If f ( x ) is reducible [over Q(a)],then its roots lie either in Q ( a ) E K or in an extension of degree 2 over Q(a), and hence are constructible by Theorem 6.2. However, if f ( x ) is irreducible over Q(a),then each of its roots lies in an extension of degree 3 over Q(a), so the roots cannot be constructible, by Theorem 6.2 and Proposition 1.1. For example, if 8 = n/2, then a = cos(n/2) = 0, so f ( x ) = x(x2 - 3)and 0 can be trisected. However, if 8 = n/3, then a = cos(n/3) = 5, and f ( x ) = x 3 - 3x - 1, which is irreducible over Q, so 4 3 can not be trisected. It follows in particular that there can be no general procedure for trisecting angles with compass and straightedge. It is of some interest to observe that with a very slight relaxation of the construction rules any given angle can be trisected. All that is needed is a straightedge that has a point marked at unit distance from one end. Suppose the angle 0 has its vertex at the origin and one side along the x axis. Draw a unit circle centered at the origin and slide the marked straightedge so that its end is on the x axis, the unit mark is on the circle, and the edge passes through the point where the other side of the angle meets the circle, as pictured in Fig. 4.If the resulting angles are labeled as in Fig. 4, then CI + a ( n - p) = n, so = ~ Cand I , also /I+ + (n - a - 0) = n, and hence 0 = 28 - CI = 3a. The problem of duplication (i.e., doubling the volume) of the cube is very ancient and has an extensive history. An oracle on the island of Delos apparently suggested in about 430 B.C. that a plague might be lifted if a new altar for Apollo were constructed having the same cubical shape as the old altar but of twice the size (i.e.,volume). As a result the problem is often referred to as the Delian problem. If the old altar is assumed to have had an edge of length 1, then the assignment was to construct a new edge of length a such that a3 = 2, or a = 3.Since x 3 - 2 is irreducible in Q[x] we have [Q($):Q] = 3, and hence J2 is not constructible by Theorem 6.2 and Proposition 1.1.

+

Figure 4

6 Geometrical Constructions

113

Exercise 6.2. A straightedge with a point marked at unit distance from one end can also be used to solve the Delian problem. Let A B be a segment of unit length. Drop a perpendicular ray from A and another ray from A inclined at angle 120" to AB. Slide the straightedge with its end on the latter ray and passing through B until the unit mark lies on the perpendicular to AB, say at point P. Show that then PB has length 3 (see Fig. 5). Each side of a regular polygon with n sides subtends a central angle of 2n/n, so a regular n-gon can be constructed with straightedge and compass if and only if cos(2n/n) is a constructible number. Note for example that a regular 18-gon can not be constructed since its central angle would give a trisection of n/3. If n = 2k,k 2 3, the regular n-gon is easily constructed by bisecting angles after a square has been constructed. If n = ab, with (a, b) = 1, and if an n-gon has been constructed we need only join every bth vertex to construct an a-gon. On the other hand if both the a-gon and the b-gon are constructible write 1 = ra sb, with r,s E Z, so that 2njn = 2nr/b + 2ns/a, to see that the n-gon is also constructible. As a result we need only consider the constructibility of a regular n-gon in case n = pk, p an odd prime. Most of the following discussion is valid for arbitrary n, however. a primitive n t h root of unity in @. Then [ = cos(2x/n) + Let [ = eZniin, isin(2njn) and [-'= cos(2n/n) - isin(2n/n), so cos(2njn) = ([ + n/2. We conclude that a regular n-gon is constructible if and only if [a([+ [):Q] is a power of 2. First consider a([).It is a Galois extension of Q, being a splitting field for x" - 1, and we know that [Q([):Q] = 4 ( n ) by Theorem 4.2. Set L = Q([ + c Q(c) and H = 3 L . If 0 E H , say with O ( [ ) = [', 1 I r < n , then 0([ [) = [' [-' = 5 + r a n d so cos(2nrln) = cos(2n/n), which holds only for r = 1 and r = n - 1 (look at the graph of the cosine function). Thus either

+

c=

+

c)

+

Figure 5

114

I11 Fields and Galois Theory

c,

0([) = ( (so 6' = 1) or O(i) = and ]HI = 2. Consequently [ Q ( i + c):Q] = 4(n)/2, and the regular n-gon is constructible if and only if $(n) is a power of 2. Assume that n = pk, p an odd prime. Then $(n) = $ ( p k ) = p k - l ( p - l), which is a power of 2 if and only if k = 1 and p = 2' + 1 for some positive c E Z. If c has an odd prime factor, then 2' + 1 is easily factored algebraically, so c = 2", and hence p = 2'" + 1, a so-called Fermat prime. We have proved the following theorem, which was first proved by Gauss in 1796 when he was 19 years old. Theorem 6.3. A regular n-gon, n 2 3, can be constructed by straightedge and compass if and only if n = 2 k p , p , * - * p ,where , 0 I k E Z and the p i are distinct Fermat primes of the form 2'"' + 1,0 I mi E Z. In general the numbers 2'"' + 1 = F,, 0 I m E Z,are called Fermat numbers, since Fermat conjectured in 1650 that every F, is prime. The first five are indeed primes; F, = 3, Fl = 5, F2 = 17, F3 = 257, and F4 = 65537. However, Euler observed in 1732that F, = 641.6700417.Subsequently at least 30 more Fermat numbers have been proved composite, and to date no further Fermat primes have been found. Constructions for a regular triangle, square, and pentagon were given in Euclid's "Elements." The first explicit construction of the 17-gon was given by Erchinger in about 1800. Richelot and Schwendenwein found constructions for the 257-gon in 1892, and Hermes spent 10 years on the construction of the 65537-gon at Gottingen around the turn of the century. The construction processes seem now to have been computerized (see the article by Bishop [ S ] ) .

- +

Exercise 6.3. (1) Observe that 641 = 54 + Z4 = 5 2' 1. Conclude that 641 I 5 4 . 2 2 8 + 232and 641 I 5 4 . 2 2 8- 1, then show that 641 1 F, without finding the other factor. (2) If 0 I n < m E Zshow that F, 1 F, - 2. Conclude that (F,, F,) = 1, and then conclude that there are infinitely many primes in H.

As a final application of Theorem 6.2 we discuss the quadrature, or squaring, of a circle, which is the construction of a square having the same area as the given circle. An ancient Egyptian solution was to take the side of the square to be times the radius of the circle, which amounts to assuming that TI = a considerably better approximation than that which appears in I Kings 7:23, or in I1 Chronicles 4:2. If the circle is taken to have radius 1, then its area is n,and the problem is to construct a segment of length f i .But TI E a(,/%), and [Q(x):Q] is infinite by Theorem 5.5, so &can not be constructible by Theorem 6.2.

w,

Exercise 6.4. The spiral of Archimedes has equation r = 0 in polar coordinates. Sketch the curve and observe that it first crosses the y axis (for r > 0) at height n/2. If the spiral is given show how to square a circle.

7 Norm and Trace

115

7. NORM AND TRACE The concepts introduced in this section will not be needed in the remainder of the book except for a brief encounter in Section 4 of Chapter VI. The concepts are of considerable importance in the study of algebraic number theory. Suppose F is a field and K is a finite extension of F , say with [ K : F ] = n. If we view K as a vector space over F , then each a E K determines an F-linear transformation t o :K K , defined by t,(u) = au, all u E K . Define the norm NKiFand the truce TrKIF,both maps from K to F , by means of --f

&,,(a)

= det(t,),

TrK,F(u)= trace (fa).

When confusion is unlikely we write simply N for N K / F and Tr for TrK,F. The most readily apparent means for computing values of N and Tr is to choose a basis for K over F , represent each t,, a E K , by a matrix, then compute the determinant and trace as usual in linear algebra. For the usual reasons N and Tr are independent of the basis chosen. For example, take F = Q and K = Q(fi), where I # m E Z*and m is square-free. Then { 1, f i }is a basis for K over F . If a = r + s f i E K , then t, is represented by the matrix [i",", and consequently N ( a ) = r 2 - ms2 (as in Section 5 of Chapter II), and Tr(a) = 2r. Elementary properties of the determinant and trace functions for linear transformations translate immediately to corresponding properties of N and Tr, yielding the following proposition.

Proposition 7.1.

Suppose [ K :F] = n, u,b E K , and u,z: E F. Then

(1) = N(a)N(h), (2) Tr(ua uh) = u Tr(a) + z~Tr(b), (3) N ( u ) = u", Tr(u) = nu.

+

Proposition 7.2. Suppose K is a finite extension of F and F c_ L G K , with [ K : L ] = S. If a E L, then TrK,F(a) = S . TrLiF(a)and NK/F(a) = (NLIF(a))'. Proof. Define t,: K K as above. Choose a basis { h , ,. . . ,b,} for L over F and let t,l L be represented by the matrix A = [aii] relative to (bi},i.e., ab, = ajibj for each i. Choose a basis { c I ,..., cs) for K over L. By Proposition 1.1 { b i c j } is a basis for K over F , which we order as b,c,,b,c,,. . .,b,c,; b,c,,. . . , b r c 2 ;., . . Then we have

cJ={

t,(bicj) = a(bicj)= (uh,)ci=

xiz, a k j ( 6 k C j )

for all i and j.

Consequently the matrix representing t , relative to { b i c j ]is block diagonal of

111 Fields and Galois Theory

1 16

the form

with r As down the diagonal. Thus det(t,) r . trace A, and the proposition follows.

= (det A)'

and trace(t,)

=

Theorem 7.3. Suppose K is a finite Galois extension of F , and F c L G K . Set G = G ( K : F ) and H = F L , and let { $ ~ ~ , 4 ~ , . . . , 4 kbe} a set of coset representatives for H in G . If a E L, then Tr,,p(U) =

x{#i(~): 1 Ii I kj,

NL,p(a) =

n { 4 i ( a ) :1 I i I k}.

and

Proof.

Let m ( x ) = b,

+ b , x + ... + b , - l x ' - l + x r

be the minimal polynomial for a over F. We assume first that L = F(a), so r = k . Then { 1 , a,. . .,a k - ' } is a basis for L over F , and the matrix representing t, on L is

A=

sometimes called the companion matrix of m(x). Thus TrLiF(a)= - b k - and N,,,(a) = (- l ) k b ,(expand the determinant along the first row of A ) . But also - b,- is the sum of the roots (in K ) of m(x),and (- l ) k b , is the product of the roots (see the beginning of Section 5, above). Those roots are precisely (4i(a):1 I i I k), by Proposition 3.3, and the theorem follows in this special case. In general say [L:F(u)]= s, so rs = [ L : F ]= k. Set J = %F(a) = Stab,(a), so [ J : H ] = s. Each left coset of J in G is then a union of s distinct left cosets of H . We may relabel if necessary and assume that 41,.. . ,4r are coset representatives for J in G. The remaining H-coset representatives c$i are distributed evenly among the cosets 1,.. .,# J . Then @,(a),...,@,(a)are the distinctrootsofm(x).Inthelist 4,(a), . . . , q 5 r (U ) , & + ~ ( U ) , . . . , d ka)eachof ( those r distinct roots appears s times due to the s cosets of H lying within each coset of J = Stab,(a). Thus by the first part of the proof and by Proposition 7.2

,

7 Norm and Trace

117

we have

Corollary. If K is a finite Galois extension of F and a E K, then N K / F (= ~ )n{d)(at:$E G(K:F))

and

Proposition 7.4. If K is any field and S is any set of field automorphisms of K , then S is K-linearly independent in the vector space of all functions from K to K. Proof. We may clearly assume that IS1 2 2. Suppose by way of contradiction that S is dependent, and take a dependence relation a141

+ a*& + ... + a&.

= 0,

with all 4i E S , all a, E K * , and k( 2 2 ) minimal. Choose b E K with $ ,(b )# 42(b). Then we have al4l(bc)

+ a,42(bc) + +

ak4,(bc)

=0

+ ak4k(b)4k(C)

=0

"'

or

a141(b)41(c) + a242(b)($2(C) +

' "

for all c E K , and hence another dependence relation

al4l(b)4, + a 2 4 A b M 2 + '.. + ~*4k(bM,= 0. Multiply the original dependence relation by 4,(b)and subtract from the new relation to obtain a2(d)2(b) -

4i(b))42+

' "

f ak(4k(b)

- 41(b))4k

O*

contradicting the minimality of k

Corollary. If L is a finite separable extension of F, then TrL/F maps L onto F. Proof. Since Tr is an F-linear transformation from L to the onedimensional F-space F, its image is either F or 0. If K is a Galois closure of L over F , then by Theorem 7.3 TrL/Fis a sum of automorphisms of K. Those automorphisms are linearly independent, so TrL,Fis not the zero function, and hence Tr(L) = F .

118

111 Fields and Galois Theory

The next theorem appeared in D. Hilbert's famous "Zahlbericht" of 1897. Theorem 7.5 (Hilbert's Satz 90). Suppose K is a finite Galois extension of F with G = G ( K : F ) cyclic, say G = (a) of order n. If a E K , then N,,,(a) = 1 if and only if a = b/a(b) for some b E K . Proof. that

e:

If a

= b/a(b),

then by the corollary to Theorem 7.3 we see

N ( u ) = [ b / ~ b ][ob/oZb] . . . [a"- b / ~ " b ]= 1

since (T"

= 1 E G. Suppose N(a) = 1, a E K . By Proposition 7.4 the automorphisms 1,a,. . .,an- are K-linearly independent, so the function

*:

'

:I( n

4 = a . 1 + (ua(a))a + (aa(a)a2(a))a2 + ... +

)

a'(a) a"-'

from K to K is not 0. Note that the coefficient of a"-' in 4 is N ( a ) = 1, by the corollary to Proposition 7.4. Thus we have b = 4(c) # 0 for some c E K . But then (verify) ab = (l/a)[b - ac]

and hence a

+ aflc = (l/a)[b

-

ac]

+ c = b/a,

= blob.

Theorem 7.5 affords an alternative proof of Proposition 4.8. See Exercise 8.57 below. The map NKiF restricts to a homomorphism of multiplicative groups, K * + F*, and Theorem 7.5 describes its kernel when K is Galois over F with cyclic Galois group. We conclude this section with an analog of Theorem 7.5 for the homomorphism TrKiFof additive groups. The proofs are also analogous. Theorem 7.6. Suppose K is a finite Galois extension of F with G = G ( K : F )cyclic, say G = (a) of order n. If a E K , then Tr(a) = 0 if and only if a = b - a b for some b E K . Proof. e: Compute, using Theorem 7.3. +: Suppose Tr(a) = 0. Choose c E K with Tr(c) = 1 (Corollary, Proposition 7.4). Set n-2

Then

8 Further Exercises

119

and so

b - ob = uc +

n-2

n- 1

1 aa'(c) 1 o'(u)o"-'(c) -

i= I

j= 1

n- 2

=a

n- 1

i=O

o'(c) =

aTr(c) = a

(Theorem 7.3 was used twice). An application of Theorem 7.6 appears in Exercise 8.58.

8. FURTHER EXERCISES Find a splitting field K E C for x 3 - 2 E Q [XI,and determine [K :Q]. 2. Find a splitting field K c C for x 5 - 1 E Q [ x ] , and determine [ K : Q ] . 3. Suppose K is an algebraic extension of F and R is a ring, with F G R E K . Show that R is a field. 4. Find all solutions to x 2 1 = 0 in the ring W of quaternions. Why is your answer not at odds with Proposition 1.7? 5. If f :C + Iw is a (ring) homomorphism show that f ( a ) = 0 for all a E @. 6. If S = { & : p E Z,p prime} show that [Q(S):Q] is infinite. 7. Suppose K is an extension of F , a E K is algebraic over F , and deg rn,(x) is odd. Show that F ( a 2 )= F(a). 8. Show that the field A 2 C of algebraic numbers is algebraically closed. 9. (S. Raffer). Let F be a field that is not algebraically closed, and suppose that the class Y of all algebraic extensions K of F is a set. Say 1 9 ' 1 = a. Construct different algebraic extensions of F corresponding to each element of the power set 2" and obtain a contradiction. 10. If K E C is a splitting field over Q for x 3 - 2 find all subfields of K . 11. Determine the Galois groups (over Q) of the following polynomials: (a) x3 - 1, (b) x 5 - 1, (c) x6 - 1, (d) X' - 2, (e) x4 - 2, (f) x4 + 1, (g) x4 - 7x2 + 10, (h) x 6 3x3 2. 12. Suppose f(x) is irreducible in F[x] and K is a Galois extension of F . Show that all irreducible factors of J ( x ) in K[x] have the same degree. [ H i n t : Show that G ( K :F ) acts as a permutation group on the factors.] 13. Let G = G(R:Q). 1.

+

~

+

120

111 Fields and Galois Theory

+

(1) If E G and a S b in R show that +(a) 5 +(b). [Hint: b - a is a square in R.] (2) Show that G = 1. [Hint: If not choose 4 E G and a E [w such that +(a) # a. Choose b E Q between a and +(a).] 14. Suppose K is a finite Galois extension of F , and F c E c K , F c L c K . Show that the join E v L is Galois over E , that L is Galois over E n L, and that G ( E v L : E ) z G(L:E n L). 15. Let F = @, let K be the field @ ( t ) of rational functions, and let G = G ( K :F ) . If and 6 in G are determined by 4(t)= i t and 6 ( t ) = l/t, where ( E @ is a primitive nth root of unity, show that H = ($,O> I G is isomorphic Show that 9 H = @(t" t - " ) . with the dihedral group 0,. 16. For any f ( x ) E F [ x ] set f"'(x) = f ( x ) , f " ) ( x ) = f ' ( x ) , and in general let f ' " ) ( x ) be the derivative of f ( " - " ( x ) , 1 I n E Z. If f(x), g(x) E F [ x ] set h(x) = f ( x ) g ( x )and show that

+

+

h'"'(x) =

i, ( nk ) f'"-k'(X)g'k)(X)

k=O

(this is Leibniz's rule). 17. If char F = 0, a E F , and f ( x ) has degree n in F [ x ] show that

f ( x )=

n

1 (f'k'(a)/k!)(x

-

u)~.

k=O

18. Give an example of fields F c E E K such that K is normal over E and E is normal over F but K is not normal over F (think about and r 2 ) . 19. Find a primitive element (over Q) for K = Q ( f l , $ ) G @. 20. Let s and t be distinct indeterminates over H,, and let F = Z,(s,t), the field of rational functions in s and t . Let a and b be roots of xP - s and x p - t , respectively, in some extension of F , and let K = F ( a , b). Show that K is not a simple extension of F. 21. Suppose char F = p # 0 and K is an extension of F . An element a E K is called purely inseparable over F if it is a root of a polynomial of the form xPk - b E F [ x ] , 0 I k E Z. (1) Show that if a E K is both separable and purely inseparable over F , then a E F. ( 2 ) Show that the set of all elements of K that are purely inseparable over F constitute a field. Conclude that there is a unique largest "purely inseparable" extension of F within K . 22. A field F is called perfect if either char F = 0 or else char F = p and F = F" = { a " : E~ F } .

4

8 Further Exercises

121

(1) If F is finite show that the map a - a P is a monomorphism and conclude that F is perfect. ( 2 ) Show that the field H J t ) of rational functions in the indeterminate t is not perfect. (3) Show that a field F is perfect if and only if every finite extension K of F is separable over F , and hence every f(x) E F[x] is separable.

23. Suppose F, and F, are Galois fields, with q = p" and r = p", p prime. Show that F, has a subfield (isomorphic with) F, if and only if n I in. 24. List all subfields of F, if q

= 220;if

q = p 3 0 , p prime.

25. Suppose the Galois fields F, and F, are both subfields of a field K . Determine F, n F, and F, v F,.

26. If 0 < k E H show that there exists an irreducible polynomial of degree k over any Galois field F,. 27. If F is a finite field of characteristic p show that every element of F has a unique pth root.

28. If Pis a prime and 1 n E Zshow that f ( x ) = xp" - x is the product of all the monic irreducible polynomials over Z,whose degrees are divisors of n. (Hint: Use Exercise 23.) 29. Let F be a finite field. (1) Show that the product of all elements in F* is - 1. ( 2 ) Conclude Wilson's Theorem: If p E Z is a prime, then ( p - I ) ! = - 1 (mod PI. 30. If F is a finite field show that every element of F is a sum of two squares in F. (Hint: See Exercise 1.12.8, and let S be the set of squares in the additive group F.) 31. Solve x J - 1 = 0 explicitly in R and show that cos(2?~/5)= ( - 1 + ,/3)/4. [Hint: Write

+ x 3 + x 2 + 1 = XZ(X2 + x + 1 + l/x + l/x2) and make the substitution y = x + l/x.] x4

fl$

32. (Artin). Let F be a field, Q c F c A, maximal with respect to F (Why does F exist?). (1) If F E K E A, with K normal and finite over F , and K # F , show that G = G ( K :F ) is a 2-group having a unique subgroup of index 2. Conclude that G is cyclic. (2) If F E L c A and [ L : F ] is finite show that L is normal over F and G ( L : F )is cyclic. Conclude that the set of finite extensions of F (in A) is an ascending chain.

122

111 Fields and Galois Theory

33. Suppose F is a field of characteristic p # 0, n = p‘rn with ( p , m) = 1 and k > 0, and K is a splitting field for xn - 1 over F . Show that K has roots of unity of order rn but not of order n, and in particular K has no primitive nth roots of unity. 34. Show (by factoring) that Q12(x)is reducible in Z,

35.

If n > 1 is odd show that Q2,(x)

36. Let p be a prime and set f(x) = 1

=

,[XI.

a,,( -x).

+ x + x 2 + ... + xp-l.

Show that for any field F the irreducible factors of f(x) in F[x] all have the same degree. 37.

Show that

+

f(x) = X’ - 7x6 - 1 8 9 ~ 1~ 7 0 1 ~~ 7 E Q[x]

is not solvable by radicals. 38. Factor x“ - 1 into a product of polynomials of degrees 1 or 2 in R[xJ. (Hint: The nth roots of unity in U2 have the form rl = p

i i n

= cos(2kn/n)

+ i sin(2knln).

Except for f 1 they occur in complex conjugate pairs.) 39. Suppose F is a field and K = F(x), the field of rational functions in the indeterminate x over F . If u E K\F show that u is transcendental over F . 40.

If F is a field show that the set of monomials

{xk,xk,s... .;”: 0 I ki < i, 2 I i I n ) is a basis for K = F(xl,x2,. . . ,x,) over Fo = F(a,, e 2 , . . ,0,).

+ +

41. Write x: x t ... + x,’ as a polynomial in the elementary symmetric polynomials ( T ~ , C T. .~.,,en. 42. Find the sums of the squares of the roots of each of the polynomials f(x) = x 3 - 3x2 7 and g(x) = 3x4 - x 3 5x 7.

+

+

+

Show that the elementary symmetric polynomials el,.. . ,enare algebraically independent over any commutative ring R with 1, i.e., if

43.

f ( x , , . . . ,x,)

E

RCx,,. . .,x,]

and

f(a,,. . .,(T,)

= 0,

then f(x,, .. . ,x,) = 0. (Hint: Take a counterexample of minimal degree and view it as a polynomial in x, with coefficients in R[x,, . . . ,x,Show that the “constant term” is not 0, substitute 0 for x, and use induction on n.)

8 Further Exercises

123

44. The discriminant of the general polynomial f ( x ) = n { ( x - x i ) :1 I iI nf is

D

=

n{(x,

- x,)’: 1

Ii < j 5 n}.

Write D as a poiynomial in u,, . . .,on in the cases n

=2

and n

=

3.

45. If char F # 3 show that a monic cubic polynomial f ( x ) = a,

+

NIX

+ u2x* + x 3 E F [ x ]

can be written as

.fW = B(Y) = Y 3 + a y + b by means of the substitution x the same Galois group over F .

=y

-

a 2 / 3 .Observe that f ( x ) and g ( y ) have

46. Suppose F is a field,f(x) = .x3 + ax + b E F [ x ] , and K 2 F is a splitting field for f ( x ) over F . Write D for the discriminant of f ( x ) (see Exercise 44, above) and let d be a square root for D. (1) Show that D = -4a3 - 27b2 and that d E K . (2) If f ( x ) is irreducible show that G = G(K :F ) is the alternating group A , if and only if d E F. (3) Conclude, if f ( x ) is irreducible, that G is A , if -4a3 - 27b2 is a square in F, and otherwise G is S , .

47. Use Exercise 46 (and also 44 if necessary) to compute the Galois groups over Q of the following cubics: (1) .x3 - 4.x + 2, ( 3 ) x 3 + 6x2 + 9~ + 3, ( 5 ) x 3 - 3 9 ~ 26,

+

+ + +

( 2 ) x 3 - 12x 8, 3~ - 1 1 , (4) x 3 - 6.x’ (6) x3 - 8 4 ~ 56.

48. Determine whether or not the angles n/4, n/6,5n/6,2n/3, and 2x15 can be trisected with straightedge and compass.

49. Show that an angle of n degrees, n

E

Z,is constructible if and only if 3 1 n.

50. The spiral of Archimedes (see Exercise 6.4) can be used to trisect any angle. Place the angle with vertex at the origin 0 and initial side o n the x axis, and say the other side first cuts the spiral at P. Trisect the segment OP and draw circles about 0 through the trisection points. Show that lines from 0 through the points of intersection of those circles with the spiral trisect the original angle. 51. Solve the Delian problem by intersecting the two parabolas x 2y = XI,

=yz

and

124

111 Fields and Galois Theory

52. If cose = a/b, with a, b E E,(a,b) = 1, b > 1, and b not divisible by a cube (> 1) in E,show that 8 can not be trisected by straightedge and compass. 53. Verify the construction shown in Fig. 6 (Richmond, 1893) of a regular pentagon inscribed in a circle. Let A B be a diameter and OC IAB a radius. Bisect A 0 at D and choose E on A B such that D E = DC. Let FG be the perpendicular bisector of OE. Then BG is an edge of the pentagon (see Exercise 31).

A

D

O

F

€

B

Figure 6

54. A curve called a cissoid was studied by Diocles (ca. 180 B.C.) in connection with the Delian problem. In terms of analytic geometry the cissoid can be taken to have the equation y 2 = x3/(l - x). Sketch the curve. If the cissoid meets the line y = -2x + 2 at the point P show that the line joining the origin to P meets the line x = 1 at height y = p. 55. If F is a finite field and K is a finite extension of F show that both TrK,F and NK/F map K onto F . 56. If K is finite over F but not separable show that TrKIF= 0. 57. Use Hilbert’s Satz 90 (Theorem 7.5) to prove Proposition 4.8. (Hint: Apply Satz 90 to a primitive pth root of unity, and show that the resulting element of K is a primitive element.) 58. Suppose char F = p , K is a Galois extension of F , and [ K :F ] = p . Show that K = F ( a ) for some a with m,Jx) of the form x p - x - b, b E F . [Hint: Let G ( K : F )= (a), use Theorem 7.6 to write 1 = uc - c, and set a = c p - c.]

Chapter IV

1

Modules

1. PRELIMINARIES

Suppose R is a ring. An additive abelian group M is called a (lef) R-module if there is defined a scalar multiplication ( r , x ) ~ rEx M satisfying the requirements

+ +

(i) r ( x y) = rx (ii) ( r s)x = rx (iii) (rs)x = r(sx)

+ ry,

+ sx, and

for all r, s E R, x, y E M. It follows easily that r . 0 = 0 . x = 0 and that ( - r ) x = r ( - x ) = -rx. Right R-modules are defined similarly. Exercise 1.1. Recall that the set End(M) of all endomorphisms of the abelian group M is a ring (Example 5, p. 48). If M is an R-module show that there is a ring homomorphism 4: r Hdr from R to End(M), with 4r(x)= rx, all r E R , x E M . Conversely, if M is an abelian group and 4: R + End(M) is a homomorphism show that M becomes an R-module if we define rx = 4r(x). This provides an alternative definition of R-modules. Perhaps the most obvious source of examples of modules is to take R = F , a field, and M = I/, any vector space over F. In fact the theory of modules can reasonably by viewed as an attempt to generalize linear algebra by allowing more general scalars. Further examples abound; we Iist a few of them.

EXAMPLES 1. Take R = Zand let M be any additive abelian group. Then rx already has a meaning for any r E Z,x E M , and it serves to define a module action. Thus the theory of abelian groups is subsumed by the theory of modules. 125

126

IV Modules

2. If S is a ring and R a subring of S , then S is an R-module with rx for r E R, x E S , the usual product in S . 3. If R is a ring and J is a left ideal in R, then J is an R-module with rx for r E R, x E J , the product in R. 4. For a very important example let R = F[x], F a field, let Y be a vector space over F, and let T : V + V be a linear transformation. If

+ a , x + ... + a,x” E F [ x ] define f(T) to be a,f + a , T + . . . + a,,T”, also a linear transformation on V . It f ( x ) = a,

is easily verified that “substitution” of T for x effects a homomorphism f ( x ) ~ f ( Tfrom ) the polynomial ring R into the ring of all linear transformations of V (see Theorem 11.3.4).If we define f ( x ) . u = f(T)(u),the result of applying the linear transformation f ( T ) to the vector u E V ,then Y becomes an F[x]-module, which we will usually denote by V,. 5. For a rather trivial class of examples let R be any ring, M any additive abelian group, and define rx = 0 for all r E R, x E M . Exercise 1.2. (1) Verify that V,, in Example 4 above, is an R-module. (2) If F = Q, I/ is the Q-space of all column vectors with 2 entries from Q, T :Y -P Y is the result of multiplication by the matrix A = and f ( x ) = xm - x, determine the module action f ( x ) u on an arbitrary vector u = [g] in V,.

[A :I,

If R is a ring with 1, M is an R-module, and 1 . x = x for all x E M , then M is called a unitary R-module (some authors prefer to make that requirement part of the definition of a module). Any vector space (as an F-module) and Examples 1,3, and 4 above provide examples of unitary modules. Example 2 may not, even when both R and S are rings with 1, as an example on page 48 shows. A subgroup N of an R-module M is called a submodule if rx E N for all r E R and x E N . If N is a submodule of M , then the quotient group M I N is also an R-module if we define r(x + N ) = rx + N , all r E R, x E M . Proposition 1.1. If M is an R-module and { M a } is any nonempty collection of submodules of M , then n , M , is also a submodule.

If M is an R-module and S G M is any subset, then by Proposition 1 . 1 there is a unique smallest submodule of M that contains S as a subset, viz., R(S) = n ( N : S E N and N is a submodule of M ) . We call R(S) the submodule of M generated b y S. If M = R ( S ) for some finite set S we say that M is jinitely generated over R, and if M = R ( a ) for some single element a E M we say that M is cyclic over R.

1

Preliminaries

127

Exercise 1.3. Let V, be the module of Exercise 1.2(2).Show that V, is a cyclic module and that N = { [ : ] : a E Q } is a cyclic submodule (neither, of course, is a cyclic group). If M and N are both R-modules, then a homomorphism (or an Rhomomorphism, if we wish to stress the ring R ) from M to N is a (group) homomorphism f : M + N such that f(rx) = rf(x) for all r E R, x E M. The usual language of R-isomorphism, R-automorphism, etc., will be used without further explanation. The kernel and image of an R-homomorphism are defined as usual; they are clearly submodules of M and N, respectively. Denote by Hom,(M, N) the set of all R-homomorphisms from M to N. Exercise 1.4. Show that HomR(M,N) is an abelian group if we define (j’+ g)(x) = f(x) + g(x). If R is a commutative ring show that HomR(M,N) is an R-module if we define (rf)(x) = r . f ( x ) . If R is a commutative ring with 1 and N is a unitary R-module show that Hom,(M,N) is a unitary R-module. It is a very simple matter to extend the basic homomorphism and isomorphism theorems from abelian groups to arbitrary R-modules. We state the results.

Theorem 1.2 (The Fundamental Homomorphism Theorem for Modules). Suppose M and N are R-modules and f E Hom,(M, N ) has kernel K . If q :M -+ M / K denotes the canonical quotient R-homomorphism, then there is a unique R-isomorphism 9: MIK + Imf c N such that the diagram f

is commutative. Proposition 1.3. Suppose A4 and N are R-modules and f : M + N is an epimorphism, with kernel K . Then there is a 1-1 correspondence between the set of all submodules L of N and the set of all those submodules P of M that contain I<, given by L * f - ’ ( L ) = P. In particular each submodule of a quotient module M / K has the form P / K for some submodule P , K E P E M.

Theorem 1.4 (The Freshman Theorem for Modules). Suppose K and N are submodules of an R-module M, with K E N. Then N/K is a submodule of M/K and (M/K)/(N/K) E M/N.

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IV Modu!es

Theorem 1.5 (The Isomorphism Theorem for Modules). If K and N are submodules of an R-module M , then I< N and K A N are submodules of M and (K N ) / K z N / ( K n N ) .

+

+

An R-module M # 0 is called simple if its only submodules are M and 0. A sequence M = M , 2 MI 2 M ,

?*.*

2 Mk = 0

of submodules of an R-module M is called a composition series if each M i +is a maximal (proper) submodule of M i , or equivalently (see Proposition 1.3) if every factor M i / M i + is simple. Theorem 1.6 (Jordan-Holder). position series

Suppose an R-module M has com-

M = Mo 2 M I ? . . . 2 Mk = 0

and M

= No 2

N,

? ... 2

N,,, = 0.

Then k = m and there is a 1-1 correspondence between the sets of factors so that corresponding factors are R-isomorphic. Proof. Essentially the same as the proof of Theorem 1.5.5.

An R-module M is called Noetherian if it satisfies the ascending chain condition (ACC)for submodules, i.e., if M , c M , _C is any ascending chain of submodules of M , then either the chain is finite or there is some k such that M, = M kfor all m 2 k. In either case we say that the chain terminates. (Recall the corresponding notion for commutative rings; see p. 66.) - a -

Proposition 1.7. The following statements regarding an R-module are equivalent. (i) M is Noetherian. (ii) Every submodule of M is finitely generated. (iii) Every nonempty set { M,) of submodules of M has a maximal element with respect to set inclusion. Proof. i ii: If a submodule N is not finitely generated choose x, E N and set M , = R(x,). Then choose x2 E N\M, and set M , = R(x,,x2), etc. The process does not terminate since N is not finitely generated, and the sequence M, c M 2 c ... violates the ACC for M . ii => iii: If there is a set of submodules without a maximal element we may choose a strictly increasing infinite chain M, c M2 c ... from the set. Set N = U M , , a submodule of M . Then N = R(x,,. ..,x,) for some xl,.. . ,xk in N , and hence x i is in some Mj, for each i. If jm is the largest of the ji,then all

2 Direct Sums. Free Modules

129

xi are in Mi_ and hence N = M j m ,contradicting the fact that the sequence is strictly increasing. iii 3 i: Obvious. N

Proposition 1.8. Let K be a submodule of the R-module M and set M / K . Then M is Noetherian if and only if both K and N are Noetherian.

=

Proof. -: All submodules of K are submodules of M , so they are all finitely generated, and hence K is Noetherian by Proposition 1.6. Any ascending chain N , E N , G ... of submodules of N has the form M , / K E M 2 / K E ..., with each Mi a submodule of M and M , E M , c so thk chain must terminate and N is Noetherian. -=: Let M , c M2 c ... be an ascending chain of submodules of M . Then the chain M , n K E M2 n K c . . . terminates, say at Min K , and likewise the chain (MI K ) / K E ( M , + K ) / K c ... terminates, say at ( M , + K ) / K . Set la = maxi j , m f . If k 2 n and x E Mkrthen ( M k + K ) / K = ( M , + K ) / K , so x + K = y + K for some Y E M,. But then x - Y E M , n K = M,,n K , and hence x E M,. Thus the chain M , E M , E ... terminates at M, and M is Noetherian. +..,

+

2.

DIRECT SUMS, FREE MODULES

The definition and construction of a (direct) product for an arbitrary nonempty family { M a } of R-modules is completely analogous to the corresponding definition and construction for the direct product of groups, and will not be repeated. We will write ~ { M , : cEI A } , or just Mu, for the product of a nonempty family of R-modules. The notion of direct sum, in a sense dual to the notion ofdirect product, is also of considerable importance for modules. Suppose (M,:a E A } , A # 0, is a family of R-modules. A direct sum of the modules Mu is an R-module M together with a family i,: M , + M , a E A, of R-homomorphisms with the following universal property. Given any R-module N and R-homomorphisms fa: Mu + N , a E A, there is a unique f E Hom,(M, N ) such that f i , = fa for all CI E A, i.e., the diagrams

n

are all commutative. Note that the definition results from the definition of a product by simply reversing ail the arrows. In the language of category theory the direct sum is a coproduct in the category of R-modules.

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IV Modules

Proposition 2.1. If a direct sum exists for a family {Ma} of R-modules, then it is unique up to R-isomorphism, and each i, is an R-monomorphism. Proof. As usual (see the proof of Proposition 1.6.1).

It is perhaps reasonable by now to be a bit brief in discussing the construction of a direct sum. Theorem 2.2. Any nonempty family {M,:a direct sum M .

E

A } of R-modules has a

n

Proof. Take M to be the submodule of Ma consisting of those m such that ma = 0 for all but finitely many a E A (in particular, if A is finite, then M = Ma).Define i,: Ma + M by agreeing that i,(x) = m if and only if ma = x and ms = 0 for all # a. It is easy to check that i , E Hom,(M,, M). Given any R-module N and family f a :M, -,N of R-homomorphisms define f :M -,N by setting f ( m ) = f,(m,):a E A } (there are only finitely many nonzero summands). Then f E Hom,(M, N ) , f i , = f a for all a, and f is uniquely determined.

n

I{

i,

Exercise 2.1. Prove the last statement in the proof above, and show that M).

E Hom,(M,,

Let us agree to write @(M,:a E A ) , or simply O M , , for the direct sum of the Ma. For a finite collection MI, M,, ..., M, of R-modules we write M , 0 M, 0 ... 0 M, for the direct sum. In general, if {M,:a E A } is any collection of submodules of an R-module M, we shall write I{M,:tx E A } for the submodule consisting of all sums z { x , : a E A } , where x , E M, and x , = 0 for all but finitely many a E A. The next theorem is an analog for modules of Theorem 1.6.3, and describes an R-module as the internal direct sum of submodules. Theorem 2.3. Suppose M is an R-module, {Ma}is a family of submodules, R ( U M , ) = M , a n d M , n { u M p : b # t l , b E A } = O f o r all a. Then M is R-isomorphic with O M , . of submodules we will When M is the internal direct sum of a family {Ma} often write M = @Ma. It will generally be clear from the context whether O M , represents an internal or an external direct sum. Exercise 2.2. Show that M = OM,, an internal direct sum of submodules, if and only if each x E M has a unique expression of the form x = x1 + x2 + ... + x k for some k, with xi E MU3.

M

Proposition 2.4. If M,, M,, . . . ,M, are Noetherian R-modules, then = MI 0 M2 0 . . . @ M, is Noetherian.

2 Direct Sums. Free Modules

131

Proof. It will suffice, by induction, to assume that M = M , 0M , . Then N, = { (x, 0):x E MI}is a submodule of M that is R-isomorphic with MI,so N, is Noetherian, and MINI is R-isomorphic with M , [via ( x , y) N, H y], so M/N, is Noetherian. Thus M is Noetherian by Proposition 1.8.

+

A ring R is called left ( r i g h f )Noetherian if it satisfies the ascending chain condition for left (right) ideals, or equivalently if R is Noetherian when viewed as a left (right) R-module.

Proposition 2.5. Suppose R is a left Noetherian ring with 1 and M is a finitely generated unitary R-module. Then M is a Noetherian R-module. Proof. Say M = R(u, ,..., a,). Let N be the R-module R" = R 0R 0. . . @ R. Then N is Noetherian by Proposition 2.4. Define f : N + M by setting f ( r , , ..., r,) = rial

+ ... + r,u,.

Clearly f E Hom,(N, M ) and ,f is onto since M is unitary. Thus M ?z N/ker f and M is Noetherian by Proposition 1.8. Let R be a ring with 1 and B an arbitrary set. A free R-module based on B is a unitary R-module F together with a function 4 :B + F such that given any unitary R-module M and any function 8 : B + M there is a unique f~ Hom,(F, M) such that f 4 = 01, i.e., the diagram B

yyF 4

M

is commutative. Exercise 2.3. (1) If B = 0 show that F = 0 is a free R-module based on B. (2) If B = {b), a singleton, show that F = R is a free R-module based on B, with 4(6) = 1 E R. As usual we may check that a free R-module F based on a set B is unique (to within R-isomorphism) if it exists, and that 4 is 1-1. In order to construct a free R-module (if B # 0) let M,, = R, viewed as a left R-module, for each B E B and set F = OM, = @ { R : P E B } , a direct sum of IB1 copies of the R-module R. ForeachpEBdefine4(/3)=mEFbymeansofrn,,= I E M , = Randm,=O if B # E B. It is then a routine matter to verify that F is a free R-module based on B. Exercise 2.4. Do so.

132

IV

Modules

Since 4:B --t F is 1-1 we may identify p with 4(p)for each fl E B, and hence rjbj = 0, then it is assume that B E F . If b , , . . .,b, E B, r l , . . . ,rt E R, and immediate from the construction that r = . . . = r k = 0. In general we say that a subset B of an R-module M is R-linearly independent if given b , ,.. . ,bk E B and r l , . . . , r k E R with x r j b j = 0 we may conclude that all r j = 0. A basis for an R-module M is an R-linearly independent subset of M such that M = R(B).

I;=,

Theorem 2.6. If R is a ring with 1, then a unitary R-module is free if and only if it has a basis. Proof. a : See the construction and discussion above. If M = 0, then B = /zI is a basis. +: Let B be a basis for M . If B = 0, then M = 0 and M is free. Suppose then that B # 0. Let 4: B + M be the inclusion map, i.e., 4 ( b ) = b for all b E B. Set Mb = Rb for each b E B ; then r Hrb is an R-isomorphism between R and Mb. The fact that B is a basis allows us to apply Theorem 2.3 and conclude that M = @{M,:b E B } , which is isomorphic with @{R:b E B } , and we may conclude that M is free just as above.

As a particular case we observe that a free abelian group is (to within isomorphism)just a direct sum of some number (possibly transfinite) of copies of the additive group H of integers. Note also that if R = K , a field, and V is a vector space over K , then V has a basis (see the Appendix), and hence V is a free K-module. We know in that case that any two bases have the same cardinality, and we shall see shortly that the same result holds for free abelian groups, but it does not hold true in general (see Exercise 7.6).

Theorem 2.7. Suppose R is a ring with 1 having an ideal I such that R/I is a division ring, and F is a free R-module. Then any two bases of F have the same cardinality. Proof. Set E = I F = { r x : r E I, x E F } , and note that E is a submodule of F. Then F I E is a vector space over K = R/I, with scalar multiplication given by ( r + I ) ( x + E ) = rx + E(Verify!). If B is a basis for F set 6 = b + E for each b E B, and set B = { b : bE B } . Each x E F I E can be written as x

k

= i= 1

ribi + E

=c(ri i

+ I)G,

ri E R, bi E B,

so B spans F I E over K . If

(i.e., = 0 in FIE), then c i r i b iE E , so there are sl,. ..,s, E I such that C i r i b i = C i s i b i(there is no harm in using the same set { b i } of basis elements,

2 Direct Sums. Free Modules

133

some coefficients may be 0). Thus x i ( r i - si)bi = 0, so ri = si, all i, and hence each ri + I = I. Thus B is a K-basis for FIE (in particular the map b H 6 is 1-l), and F I E has K-dimension ( B ( The . same argument applies to any other basis for F , so all bases must have the same cardinality. Corollary. If R is a commutative ring with 1 and F is a free R-module, then any two bases of F have the same cardinality.

Proof. Use Proposition 11.1.8 and the corollary to Proposition 11.1.9.

Because of Theorem 2.7 we may define the rank of a free module F over a ring R with 1 having a division ring as a homomorphic image to be the cardinality of any basis for F. In particular the notion of rank is defined for free abelian groups. Theorem 2.8. If R is any ring with 1 and M is a unitary R-module, then M is a homomorphic image of a free R-module F .

Proof. See the proof of Theorem 1.10.1. Corollary. If R is a commutative ring with 1 and M is a unitary R-module, with M = R ( S ) for some set S E M , then M is a homomorphic image of a free R-module F of rank IS/. The ring R = Z,has an ideal M = 2R, so M is a submodule of the free R-module R. Note, however, that M is not a free R-module, since /MI = 2 and jR\ = 4. Thus submodules of free modules are not necessarily free. Theorem 2.9. Suppose R is a PID, F is a free R-module, and E is a submodule of F. Then E is also free, and rank(E) Irank(F).

Proof. We may assume E # 0. Let B be a basis for F. For any C G B set F, = R (C) and Ec = E n F,. Let Y be the set of all triples (C, C ’ , f ) , where C’ E C G B, E, is free, and f : C’ + E, is a function such that f(C’) is a basis for E,. Then Y # 0 since (@, $3,@) E 9. Partially order Y by agreeing that (C, C ’ , f ) I(D,D‘,g) if C C D, C’ c D‘, and g 1 C‘ = f . Zorn’s Lemma applies and we may choose a maximal element ( A , A’, h ) in Y . The proof will be complete if we can show that A = B, since E = E B . If A # B choose b E B\A and set D = A U { b } . If ED = EA, then ( A , A ’ , h )< (D, A’,h), contradicting maximality. If ED # EA, then there are various elements y + rb E ED, with y E F A , r E R, and in some cases r # 0. Let I be the set of all r E R such that y + rb E E for some y E FA. Then I is an ideal in R, say I = (s), and so w = x + sb E E for some x = FA (note: s # 0). Set D’= A’ { b ) and extend h’: D’+ ED by setting h‘(b)= w. If z E ED, then z = y + rb for some y E FA and r = r’s E I , so

u

z = ( y - r’x)

+ r’w,

and

z - r‘w = y - r‘x E E n FA = EA.

134

IV Modules

It follows that R(h'(D')) = ED. A dependence relation on h'(D') translates easily into a dependence relation on D, so h'(D') is a basis for ED. Thus ( A ,A', h) < (D, D', h'), maximality is violated again, and the theorem is proved.

Corollary. Suppose R is a PID, M is a finitely generated unitary R-module, and N is a submodule of M. Then N is finitely generated. Proof. As shown in the proof of Proposition 2.5 there is an Rhomomorphism f:R" + M for some integer n. Thus f - ( N ) is a submodule of R", so it is free of rank m 5 n by the theorem. Thus N = fs- ' ( N )has a set of rn generators.

Proposition 2.10. Suppose R is a PID, M is an R-module, F is a free R-module, and f E Hom,(M, F ) is onto. Then M has a free submodule E that is R-isomorphic with F such that M = E 0 kerf.

zf=l

Proof. If B is a basis for F choose xb E M such that f ( x b )= b for each b E B. Set E = R ( { x b : bE B } ) . If r,xb,= 0, then 0 = f ( c r i x b , )= C r i b i , so all ri = 0 and { x , : b E B} is linearly independent. Thus { x , : b E B } is a basis for E , so E is free and is clearly R-isomorphic with F. If x E M write f ( x )= 1 ribi,bi E B, and note that then x - c r i x b , E kerf. Thus M = E + kerf. Since E n kerf = 0 we conclude that M = E 0 kerf by Theorem 2.3. 3. FINITELY GENERATED MODULES OVER A PID

Assume throughout this section that R is a PID. The most important examples will be R = F , a field, R = Z, and R = FCx]. All R-modules considered will be assumed unitary. If M is an R-module, then an element x E M is called a torsion element if rx = 0 for some r # 0 in R. This generalizes the idea of an element of finite order in an abelian group. If x E M is a torsion element we define its annihilator, or order ideal, to be A ( x ) = { r E R:rx = O } . Then A ( x ) is an ideal in R, so A ( x ) = (s,) is principal; the generator s,, which is determined up to associates, is called the order of x. We write 1x1 = s, for the order of x . Exercise 3.1. Let R = Q [ x ] and let M = V, as in Exercise 1.2(2). If and u = [ y ] find JuJand Jut.

u=

[A]

An R-module M is a torsion module if every element of M is a torsion element. At the other extreme M is torsian-free if 0 is the only torsion element in M . Proposition 3.1. Let T be the set of all torsion elements in an R-module M . Then T is a submodule and M / T is torsion free.

3 Finitely Generated Modules over a PID

135

Proof. If x , y E T say 1x1 = a and J y J= b. Then ab(x - y) = b . a x a . by = 0, and if r E R, then a ( r x ) = r(ax) = 0, so T is a submodule of M . Suppose r{x + T ) = rx + T = T i n M / T . Then r x E T, so A(rx) # 0. If 0 # s E A ( r x ) ,then sr . x = 0, so 0 # sr E A ( x ) ,and x E T.Thus M / T i s torsion free. The submodule Tof torsion elements in M is called its torsion submodule. In particular every abelian group has a torsion subgroup consisting of all its elements of finite order. Note that the proof of Proposition 3.1 did not actually require that R be a PID, but only an integral domain.

Proposition 3.2. If M is a finitely generated torsion-free R-module, then M is free of finite rank. Proof. Let {xl,. . .,x f l } be a generating set for M and let { y , , . . . , y m ) be a maximal linearly independent subset of {xi,. ..,x,,}. Let N = R(y, ,..., y,), a free submodule of M. For each x ithere exists some si # 0 in R and elements rij E R such that sixi x j r i j y j= 0, by the maximality of { yj}, and so sixi E N . Set s = n{si: 1 I i I n } , so that sx E N for all x E M . Define f : M + N by setting f ( x ) = s x ; then f E Hom,(M, N ) and kerf = 0 since M is torsion-free. Thus M is R-isomorphic with a submodule of N , so M is free and of finite rank by Theorem 2.9.

+

Theorem 3.3. If M is a finitely generated R-module with torsion submodule T, then M has a free submodule F of finite rank such that M = T @ F . The rank of F is uniquely determined by M . Proof. Since M I T is torsion-free and the quotient map y ~ :M + M / T is onto we conclude from Propositions 2.10 and 3.2 that M = T 0F for some free submodule F of finite rank that is R-isomorphic with M / T . Any other free submodule F‘ such that M = T 0F’ is also R-isomorphic with M I T so rank F’ = rank F . As a consequence of Theorem 3.3 we may define the rank of a finitely generated R-module M to be rank ( F ) , where M = T 0 F, T is the torsion submodule of M , and F is free. Suppose that M is a torsion module and that there is some r E R* with rx = O f o r a l l x ~M . S e t l = { s ~ R : s x = O , a l l x ~ M } . T h e n l i s a n i d e a l i n R ; say 1 = (a), so a I r. The generator a of 1, which is determined up to associates, is called the exponent of M . It is the least common multiple of the orders of all elements of M . Of course not every torsion module has an exponent. For example, take M = @ { Z,:n E Z}.However, if M = R(x,, . . .,x,) is finitely generated and torsion, and if lxil = r i , then r = n { r i :1 5 i 5 n ) annihilates every x E M , and M has an exponent.

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IV Modules

Exercise3.2. Suppose M is an R-module, that x and y are torsion elements with orders r and s, respectively, and that r and s are relatively prime in R. Show that x + y has order rs.

Proposition 3.4. Suppose M is a torsion module having exponent r. Then M has an element of order r. Proof. Recall that R, being a PID, is a UFD. Write r = H { p f a :1 I i I k } , a product of distinct prime powers, with each e , > 0, and set ri = r/pi. Then r J' ri and so for each i there is some xi E M with rixi # 0. Set y i = (r/pe')x, and note that pFzyi= 0 but pF1-'yi = rixi # 0, and hence (yil = p f ' for each i. Set x = y , ... y,. Then 1x1 = r by Exercise 3.2. If M is an R-module and s E R define M I S ] = {x E M : s x = 0 } ,the set of all elements in M having order a divisor of s.

+ +

Exercise3.3. Show that M [ s ] and s M modules of M .

=

{sx:x E M } are both sub-

Recall that if r, s E R, not both 0, then the symbol (r, s) is used both to denote the ideal generated by { r, s} and to denote a GCD for r and s,which is in fact a generator for that ideal. Both uses appear in the next proposition. Proposition 3.5. Suppose M

= R (y

) is cyclic of order r, and s E R. Then

(i) M [ s l = R < ( r / ( r , s ) ) y z ) R / ( r , s ) ,and (ii) sM = R ( ( r , s ) y ) z R/(r/(r,s)), so M [s] is cyclic of order (r,s) and sM is cyclic of order r/(r,s). Proof. Note that M [ s ] = {uy:u E R and suy = 0}, and that suy = 0 if and only if u is a multiple of r/(r,s).Thus M [ s ] = R ( ( r / ( r ,s))y). Define a map 4 :R + M [ s ] by setting +(u) = (ur/(r,s))y.Then C#I is an epimorphism and k e r 4 = (r,s), so (i) is proved. The proof of (ii) is similar and is left as an exercise.

Corollary. If (r,s) = 1, then M [ s ] = 0 and s M

=

M.

Proposition 3.6. Suppose C = R ( z ) is a cyclic R-module of order ro, M is a finitely generated torsion R-module whose exponent divides r o , N is a submodule of M , and f E Hom,(N, C). Then f can be extended to g E Hom,(M, C ) . Proof. If M = R(x,, . . . ,xk) set N, =N

+ R(x,),

N, = N,

+ R(x,),

..., Nk = N k - l + R(x,) = M.

By induction it will suffice to show that f can be extended to an Rhomomorphism from N , to C . If s is the order of x1 + N in N , / N , then s 1 ro (since roxl + N = N ) , say ro = sf. Thus s x , E N , so f ( s x , ) C~ , and t f ( s x l )= f ( t s x , ) = f(roxl) = 0, so f ( s x l ) has order dividing t in C. On

3 Finitely Generated Modules over a PID

137

the other hand f(sxl) = uz for some u E R, so tuz = 0 and ro I tu, or t s I tu. Thus s I u, say u = su, and so f(sxl) = s(uz). Set zo = uz E C and define f ’ E Hom,(N @ R, C) by setting f ’ ( y , r ) = f(y) + rzo. Next define h E Hom,(N 0 R, N,) by setting h( y, r) = y rx,. Let us see that ker h E kerf’. If (y,r) E kerh, thenrx, = - Y E N,sor(x, + N) = Oin N,/N,and henceslr in R, say r = - ws. Then y = swx,, (y, I ) = (swx,, -sw), and we have

+

f ’ ( y , r ) = f’(swx,, -sw)

= f(swxl) - swzo = f(swxl) -

wf(sx,) = 0.

Since h maps N 0R onto N, we may define f, on N, by setting f,(h(y, r ) ) = f’(y, I ) (f,is well defined since ker h c kerf’). Note that if y E N, then

f(Y)= f’(Y7O) = fl(h(Y70)) = f l ( Y ) + 0 . x1 = fl(Y), so f, extends f to N, and the proof is complete. C

Theorem 3.7. If M is a finitely generated torsion R-module, then = M , 0M , 0 . . . 0M,, where each Mi is cyclic of order r i , with ri 1 r i - 2 I i 5 k, and r , the exponent of M .

M

,,

Proof. By Proposition 3.4 we may choose an element x, E M whose order is the exponent r , of M . Set M , = R(x,). By Proposition 3.6 the identity map i: M , -,M , extends to an R-homomorphism 9,: M -,M , . Set N2 = kerg,. If x E M , then

Sl(X - g,(x)) = g,(x) - 91(x) = 0 since g1 extends the identity map, so x - g,(x) E N, and

+ (x - Sl (4)E M , + N,. If x E M , n N,, then x = g,(x) = 0, so M = M , 0N2 by Theorem 2.3. If N, x

= g,(x)

has exponent r , , then r2 I r , ; we may choose x2 E N2 of order r2 and set M2 = R(x,). Then N2 = M2 0N, as above, hence M = M , 0M 2 0N,, and we may continue. The process must terminate in a finite number of steps with some Nk = 0 since M is Noetherian (Proposition 2.5). Corollary. M z R/(r,) 0R/(r2)0. . .0 R/(rk). Proof. The map r Hrxi from R to Mi in the proof above has kernel (ri), so Mi E R/(ri) by the FHTM.

138

IV Modules

Theorem 3.8 (Uniqueness). Suppose M is a finitely generated torsion R-module and that M

=

M , @ - “ @ M k = N , @ . . . @ N,.

,

with Mi cyclic of order r i , N j cyclic of order s j , ri I r i p for each i, sj I s!- for each j , and neither rk nor s, is a unit in R . Then k = m, ri and si are associates in R, and Mi z Ni for each i.

I

Proof. We may assume k 2 m.Choose a prime p E R such that p r k ,and hence p 1 ri, 1 Ii k. Since M has both r , and s1 as exponent r l and s1 must be associates ( r , sl), so p s,. Suppose p 1 s j for 1 I j S n, where n m, but p $ s,+ (if it exists). Since p is a prime R / ( p ) = K is a field. By Proposition 3.5 we see that

,

-

I

M [ p ] = M , [ p ] O . . . O M k [ p ] ~ @ { K S: li l k } , a vector space of dimension k over K , and also M[P]=N,Cp7O.*.ON,[pl = Ni[p]O..*ON,[p] =@{K:l l j r n } ,

-

-

a vector space of dimension n over K . It follows, since n Im I k, that m = k. We have already observed that r l s,, and we wish to show that ri si for all i. If not suppose that ri si for 1 I i I n - 1, but that r, s, say because sn,t‘rn. Set M’ = r , M , then

-

M’ = rnMl @ ... @ r,M,-

+

, = rnNl @ . .. @ rnNn-

@ r,N, @

with r,N, # 0, violating the conclusion drawn above. Thus ri furthermore we have Mi 2 R / ( r i )= R / ( s i )z Ni

-

. ‘.,

si for all i , and

for all i.

The orders r l ,..., rk of the cyclic submodules M , ,..., Mk are called the invariant factors of the torsion module M . In case R = Z,so M = A is an abelian group, we take the orders ri = ni to be positive integers, as usual. In so A is finite. that case we see by Theorem 3.7 that ] A ]= n,n2**.ntr Theorems 3.7 and 3.8 thus tell us that all finite abelian groups are determined up to isomorphism by their invariant factors. For example, if n = [A1 = 100 we may write 100 = 5 0 . 2 = 2 0 . 5 = 10. 10, and Z, Z,,@ Z,,Z,,0 E,, Z,,@ Z,,are all the abelian groups of order 100. If n = 48 we have 48 = 2 4 . 2 = 1 2 . 4 = 1 2 . 2 . 2 = 6 . 2 . 2 . 2 , and a corresponding list of five groups. We may combine Theorems 3.3, 3.7, and 3.8 for a complete description of the structure of finitely generated modules over a PID.

3 Finitely Generated Modules over a PID

139

Theorem 3.9. If M is a finitely generated R-module, then there exist a nonnegative integer m and nonunits r , , . . . ,rk in R, with r, I r i - for 2 2 i I k, such that M = M I 0... @ Mk 0F , where Mi is cyclic of order ri and F is free of rank m. Equivalently

,

M

Z

R / ( r , )@ . . . @ R/(rk)@ R".

The module M is determined up to R-isomorphism by m and the sequence r l , . . . ,rk of invariant factors.

Corollary. If A is a finitely generated abelian group, then there exist a nonnegative integer m and integers n , , . ..,n k ,all larger than I , with n, 1 n i _ for 2 I i I k, such that A

2

Z,,@ . . . @ Z,,@ 12".

The group A is determined up to isomorphism by m and the sequence n , ,. . . ,nk of invariant factors, and the order of the torsion subgroup of A is n l n 2 ... n k . A torsion R-module M is called primary, or p-primary, if its exponent is some power p' of a prime p E R.

Proposition 3.10. Suppose M is a torsion module with exponent r E R. If with ( s , t ) = 1, then M = M [ s ] @ M [ t ] . Consequently if r = p'l'p;'. . . &k, where the pi are distinct primes in R , then M = M [p ; I ] 0. . . O M [p ? ] , a direct sum of primary submodules. r

= st,

Proof'. We may write 1 For any x E M we have

= as

x = t . bx

+

+ bt for some a, b E R by Proposition 11.5.2. S ' UXE

If x E M [ s ] n M [ t ] , then x = asx Theorem 2.3.

M[s]

+ M[t].

+ b t x = 0. Thus M

=

M [ s ] @ M [ t ] by

Note that the primary submodules in the direct sum decomposition in Proposition 3.10 are unique: MCpj'] is the submodule consisting of all elements of M having order a power of the prime divisor p i of r. We may now apply the Invariant Factor Theorem 3.7 to each primary submodule of a torsion module M , the result being the next theorem.

Theorem 3.11. Suppose M is a finitely generated torsion module of exponent r = { pp': 1 I i I k } , where the piare distinct primes in R and each e, 2 1. Then M = @{Mi: 1 I i I k ) , where Mi is pi-primary, and Mi = @{Mij: 1 2 j I k , } , where M i j is cyclic of order pFi1, with 1 5 eij I eicj-1 ) I e, for all i and j . M is determined up to isomorphism by the set { p:": 1 I j I k,, 1 I i I k } of prime powers, which are called the elementary divisors of M .

n

140

IV Modules

The statement of Theorem 3.1 1 applies directly to finite abelian groups; it will not be restated as a corollary. As an application of Theorem 3.1 1 let us describe all abelian groups A of order 72 = 23 * 3' by means of elementary divisors. If p1 = 2 and p 2 = 3 we require that z j e l j = 3 a r ~ d z ~=e ,2.~Thus the possible sets of elementary divisors are {23,32}, (2 3,3,3}, {22,2,32}, {2,, 2,3,3}, {2,2,2,32}, and (2,2,2,3,3). Thus up to isomorphism an abelian group of order 72 is one of

Z,OZ,,

Z,@Z,OZ,,

Z,0Z,0 Z,0 Z9,

Z40Z20Z9, Z,OZ,@~,OZ,, and Z2 0 Z,0Z,0Z,0 Z,.

4. APPLlCATlONS TO LINEAR ALGEBRA We concentrate now on the modules M = V', where V is a finitedimensional vector space over a field F and T : V + V is a linear transformation (see Example 4, p. 126). Since the set of all linear transformations from V to V is itself a finitedimensional vector space over F the set {Z, T, T 2 , . .} is linearly dependent, and hence u,Z + a , T -k - - '+ ukTk= f ( T )= 0 for some nonzero f ( x ) E F [ x ] . Just as in the proof of Proposition 111.1.2 we see that there is a unique monk polynomial mT(x)of minimal positive degree such that m,(T) = 0, and that if f ( x ) is any polynomial such that f(T)= 0, then mT(x)I f ( x ) in F [ x ] . We call mT(x)the minimal polynomial of the linear transformation T. For example, if F = Q and T is multiplication by i], then mT(x)= x 2 - 2x + 1; if T is multiplication by [ 1': iy], then mT(x)= x 2 - x - 2. We recall from linear algebra that if T: V -,V is represented by the matrix A , then the characteristic polynomial of T is f ( x ) = det(A - xZ).

[A

Exercise 4.1. Suppose F

=Q

and

T

4 - 6 A= 4 - 7 [ 6 -12

iq

multiplication by

d 3

4.

Show that T has minimal polynomial mT(x)= x 2 - 3x polynomial f ( x ) = - x 3 + 4 x 2 - 5x + 2.

+ 2 and characteristic

It is clear from the very definition of the minimal polynomial m,(x) that Vr is a torsion module and that m,(x) is its exponent. Any basis for V as an F-

4 Applications to Linear Algebra

141

vector space also generates V, as a module, so V, is finitely generated. Note that if W is any subspace of V , then T W E W if and only if f ( T )W E W for all f ( x ) E F [ x ] , so the submodules of V, are just the T-invariant subspaces of I/. Exercise 4.2. Suppose W = R ( v ) is a cyclic submodule of V,, and that W has order f ( x ) E F [ x ] , where degf(x) = k > 0. Show that the set { v , Tu, T 2 u , ., .,T k - ' o ) is a (vector space) basis for W. We call v a cyclic vector for W .

Suppose F is a field and

f ( x ) = a,

+ a,x + ... + a,-,xn-l + x n

is a monic polynomial in F [ x ] . Define the companion matrix of f ( x ) to be the n x n matrix 0 1 0 C ( f )= 0

0 0 1 0

0 0 . . ' 0 --a0 0 0 ... 0 --a1 0 0 ... 0 - a 2 1 0 ... 0 -a3

.. .. .. .. . . . .

.. .

.. .

0 0 0 0 ... 1 - a n - l

For example, the companion matrix of f ( x ) = 2 - 3 x

+ x 3 is

0 0-2 1 03 ] .

c ( f ) = 0[

Proposition 4.1. If F is a field, f ( x ) E F [ x ] is monic of degree n, and T is the linear transformation of F n represented by the companion matrix C ( f ) , then the minimal polynomial m T ( x )if f ( x ) .

Proof. Let e , , e , , . . .,enbe the usual basis elements for F", i.e.,

etc. Then Te, = e i + lfor 1 5 i 5 n - 1, and T e n = - a 0 e1 - a 1 e 2 - ... - an- 1en. Thus e, = T e , , e3 = T 2 e l , . . ,en = T n - l e l ,and hence Ten = T"el = -a,le, - a , T e l - ... - a n - l T " - l e l ,

142

1V Modules

which says that f ( T ) e , = 0. But then also and hence f ( T ) u = 0 for all u E V . If with k < n and g ( T ) = 0, then g(T)e,=

k

6,T'ei =

j=O

k

b j e j + l = 0,

j=O

so all bj = 0 and g ( x ) = 0. Thus mT(x)= f ( x ) . Exercise 4.3. Show that the characteristic polynomial of a companion matrix C(f)is + f ( x ) .

Proposition 4.2. Suppose W = R ( u ) is a cyclic submodule of V,, and that the order of W is the monic polynomial f ( x ) E F [ x ] of degree k. Then the restriction of T to W is represented by the companion matrix C ( f ) with respect to the basis { u, Tu, ...,T k- u ) for W. Proof. We saw in Exercise 4.2 that { u, To,. . ., T k - u } is indeed a basis for W. If 0 _< i I k - 2, then T(T'u)= T i + l u ,and if f ( x ) = a0

+ a1x + + "'

+ Xk,

uk-lXk-'

then

T(Tk-'u) = T k v = -a,lu

- U,

TU- ... - ak-l T k - ' o

since u has order f ( x ) . The proposition follows.

Theorem 4.3. If V is a finite-dimensional vector space over a field F and T : V 4 V is a linear transformation, then there is a basis for V with respect to which the matrix that represents T has the block diagonal form

rc(fl 1

1 C(X)]

where the f i(x) are the (rnonic) invariant factors of V,, so that fi(x)lL-,(x), 2I i I k. Furthermore w,(x) = fl(x) and the characteristic polynomial of T is f ( x ) = kn{fi(x): 1 I i I k}. Proof. By Theorem 3.7 we know that V, = @{M/;:lI i I k), where = F [ x ] ( u i ) is cyclic of order fi(x), the ith invariant factor of VT,chosen

4 Applications to Linear Algebra

143

m o n k That T has the matrix A is thus a consequence of Proposition 4.2. We have m,(x) = fl(x)since both are the exponent of V,, and the statement about the characteristic polynomial follows from Exercise 4.3. Corollary (The Cayley-Hamilton Theorem). The minimal polynomial m T ( x )is a divisor of the characteristic polynomial f ( x ) , and hence f ( T )= 0.

The matrix A in Theorem 4.3 is called a rational canonical matrix for T. The word "rational" refers to the fact that the entries of A lie in F itself and no extension field is required. Recall from linear algebra that two matrices A and B represent the same linear transformation T :V + V (with respect to different bases for V )if and only if there is an invertible matrix C with entries in F (the change-of-basis matrix) such that B = C - ' A C . In that case we say that A and B are similar matrices over F . In fact, suppose A represents T with respect to the basis {ui}and B represents T with respect to the basis { wi}, and let C be the matrix representing the identity transformation I : V + V with respect to {wi} in the domain and {ui} in the range, i.e., wi = Cjcjiuj for all i. Then we have

v-v

B

and so A and B are similar over F . The next theorem is thus a direct consequence of Theorem 4.3 and the uniqueness Theorem 3.8. Theorem 4.4. If F is a field and A is an n x n matrix with entries in F , then A is similar over F to a uniqe rational canonical matrix, called its rational canonicalform. Two matrices A and B with entries in Fare similar over F if and only if they have the same rational canonical form. Corollary. Suppose K is an extension field of F , A and B are n x n matrices with entries in F , and A and B are similar over K . Then A and B are similar over F.

Proof. A and B are similar over F to rational canonical matrices P and Q, respectively, say C ' A C = P and D-'BD = Q, all the matrices havingentries in F. But A and B have entries in K , so we must have P = Q by the theorem, and hence A = ( D C - ' ) - ' B ( D C - ' ) . Let us next interpret primary decomposition and elementary divisors in the case M = V, (see Proposition 3.10 and The,orem 3.11).

144

IV Modules

If the minimal polynomial mT(x) of the linear transformation T :V -+ V factors as m,(x) = p,(x)ei:1 I iI k},

n{

where the p,(x)are distinct, monic, and irreducible, and each e, 2 1, then V, is the direct sum of the primary submodules VT[pi(x)ei].Each primary submodule decomposes according to its (monic) invariant factors pi(x)ei,.If the appropriate basis is chosen (via a cyclic vector) in each cyclic submodule, then the union of those bases is a basis for V,, and with respect to that basis the matrix representing T is the block diagonal matrix with the companion matrices C(p : ' ~ )as the diagonal blocks. The uniquely determined set { p , ( x ) e , ~of) monic polynomials is called the set of elementary divisors of T, or of any matrix A that represents T. If follows that two n x n matrices A and B with entries in a field F are similar over F if and only if they have the same set of elementary divisors. Suppose now that one of the irreducible factors p i ( x )of mT(x)is linear, say p i ( x )= x - ai. For example, that is true of all pi(x)if F is algebraically closed. I J ( x - ai)eL'is one of the elementary divisors of T and let Suppose P ~ ( X ) ~= be a cyclic submodule of order (x - ai)ei, with cyclic vector u. Note that T restricts to a linear transformation on and that the minimal polynomial of It follows (Why?) that the set that restriction is pi(x)eij= (x -

vj

vj

{u, (7- - a,l)u,(T- ail)%,. . .,(T - U i Z y i J -

is a basis for

vj.We then have

4

+

TU= U ~ V ( T - ~ i l ) ~ , T ( T - a , l ) v = a,(T - a , l ) u + ( T - U i l ) 2 U , T(T - ~ ~ l ) e ' ~ = - ' ai(T v - u , ~ ) ~ , J -+ ' u(T - a , l ) e ' J - ' ~ , T ( T - ail)e'J-'v = a,(T - a i l y l J -l u

(verify these equations). Thus the matrix of the restriction of T to eij x eij matrix 0 0 ai 0 1 ...

ai

0 0

0 0

...

...

0

... 0 ... 0

0 0 0

-

. .

... ai 0 1

K j is the

4'

-

The matrix A i j is called a Jordan block; a block diagonal matrix whose diagonal blocks are Jordan blocks is called a Jordan matrix. If T :V + V can be

5

Computations

145

represented by a Jordan matrix J, then J is called the Jordan canonical form for T , or for any matrix that represents T. Since the Jordan matrix (when it exists) is completely determined by the elementary divisors of T i t is unique, except that the linear factors p,(x) may be relabeled and the diagonal blocks rearranged accordingly. It is customary to say that the matrices with the blocks rearranged in that way are the “same” Jordan canonical form. Theorem 3.1 1 and the discussion above yield the next theorem. Theorem 4.5. If T : V + V is a linear transformation such that all irreducible factors of mT(x)are linear, then V has a basis relative to which T is represented by a Jordan matrix. If F is algebraically closed and A, B are n x n matrices with entries in F , then A and B are similar over F if and only if they have the same Jordan canonical form. Theorem 4.6. If F is any field, V is a finite-dimensional vector space over F , and T: V + V is a linear transformation, then T can be represented by a diagonal matrix if and only if the minimal polynomial m T ( x ) splits as a product of distinct linear factors in F [ x ] .

Proof. It is necessary and sufficient that T be represented by a Jordan matrix all of whose Jordan blocks are 1 x 1, which holds if and only if every elementary divisor is of degree 1. Thus the exponent of each primary submodule must be linear and the exponent of V,, which is m,(x), is just the product of those distinct linear factors.

5. COMPUTATIONS Suppose again that R is a PID, and assume that all R-modules discussed are unitary. Suppose M is a free R-module of rank n, with { x l , x 2 , . . .,x , } as a basis, N is a free R-module of rank m, with ( y l , y 2 , . .., y,] as a basis, and f E Hom,(M, N ) . For each x i we may write f ( x i ) = x ; = l a j i y j ,with aji E R. Just as in linear algebra we say that f is represented by the m x n matrix A = ( a i j ) relative to the bases { x i ) for M and { yj} for N ; the map f~ A is an R-isomorphism from Hom,(M, N ) onto the R-module of all m x it matrices with entries from R. It should be stressed that in discussing the matrix representing an Rhomomorphism it is important that the bases { x i }and { y j } are ordered; clearly a different matrix might result if basis elements are permuted. Theorem 5.1. Suppose R is a PID, M and N are free R-modules of finite rank, and f E Hom,(M, N ) . Set E = Im(f). Suppose there are bases { x l , .. .,x,} for M and { y,, . . .,y,,,} for N such that the matrix representing f

146

IV Modules

[:: I:

relative to {xi) and (yj) has the block diagonal form

D 0 ,

B= 0

01

where U1

'..

0

D=[:

and

'.. d ] ,

with ui E U(R), all i, 0 # d j 4 U(R), all j , and d j I d j + l for 1 I j I k - 1. Then the quotient module N / E is the direct sum of the cyclic submodules R(y, + E), s + 1 I i Im, its torsion submodule has invariant factors d , , d k - . ., d l , and its rank is m - s - k. Proof.

Note that

if l ~ i ~ s , if s + 1 I i I s + k, if s + k + l l i I n ,

UiYi

f ( x i )= d i - x y i [O

and that yi E E if 1 I i Is since ui E U(R). Thus { yi + E : s + 1 I i I m} generates N I E as an R-module. If we write q.= R(yi + E ) , s + 1 I iI m, wehave thenthat N / E = W , + l + . . . + W r n . S u p p o s e x ~ W , n x { W ; . : j # i } , say x = -r. ,yi E = c { r j y j E : j # i).

+

+

Then I : rjyj + E E and we may write rn

x

s+k

S

C1 rjyj =

s+

Sjujyj + j = s + 1 Sjdj-syj, j =1

with all sj E R. But then sj = 0 if 1 5 j I s, rj = 0 if s + k + 1 j m, and r j = s j d j - , if s 1 I j I s + k since {yj} is a basis for N . Thus rjyj E E for s + 1 5 j Is + k and we conclude that x = 0. Thus N / E = @{@:s + 1 I i I m}. Suppose now that ri(yi + E ) = E in N I E , ri E R. As above we have

+

ITzs+ rn

C rJYJ. . -- 1 s+k

sjdj-,yj,

jZS+1

s+ 1

where sj E R, so rj = s j d j - , if s + 1 j I s + k and rj = 0 if s + k + 1 I j I m. It follows that each yi E, s + 1 I i I s + k, has order d i - , in N I E and the set

+

{yi+E:s+k+ l l i l m }

5

Computations

147

is R-linearly independent. Thus @{w:s+l
is the torsion submodule of N / E , and the invariant factors are dk,.. . ,d , (see Theorem 3.9). If f E Hom,(M,N) is represented by the matrix A relative to (ordered) bases ( x i } and { y j ) , then changes in the bases result in changes in A . Let us consider some very specific sorts of changes and their results. If x i and xi are interchanged then the ith and j t h columns of A are interchanged. The same change in A can be effected by a matrix multiplication A H AC,,, where Cij is the matrix that results from interchanging the ith and jth columns in the n x n identity matrix I. For example

Note that the matrix Cij is invertible; in fact C; = 1. Likewise if y i and y j are interchanged the effect on A is to interchange the ith and jth rows. The same change is effected by the matrix multiplication A H RijA, where R i j results from interchanging the ith andjth rows of I . Note that R: = I . If x i is replaced by x i + r x j for some r E R and j # i the resulting set is still a basis, and the effect on A is to add r times the j t h column to the ith column (verify this). Again the same effect can be achieved by a matrix multiplication A H ACij,, where Cij,is the result of adding r times the jth column of I to the ith column. The matrix Cij, is also invertible, with C,' = C i j ( - , ) .Similar remarks (with a slight twist) apply to the invertible matrix Rijr obtained by adding r times the jth row of I to the ith row. The result of multiplying R, . A is to add r times thejth row of A to the ith row, but that row operation corresponds to replacing y j by y j - ryi in the basis for N . The changes in A discussed above are called elementary row and column operations. The net effect of several row and/or column operations in succession is to replace A by B = P A Q , where P and Q are invertible matrices with entries in R, obtained as products of various Rijs and Rij,s (for P ) and various Cijs and Cij,s (for Q). In general we say that rn x n matrices B and A with entries in R are equiualenr (over R ) if B = P A Q for invertible matrices P and Q with entries in R. It is sometimes convenient, although we will not find it strictly necessary, to admit a third type of elementary operation, viz., multiplying a row or column by a unit u from R. This reduces to multiplication by Rij, or Cij, if we allow i = j a n d t a k e r = u - 1.

148

IV Modules

Theorem 5.2. Suppose R is a Euclidean ring with function d: R * + E , and A = (aij)is an m x n matrix with entries from R. Then there are invertible matrices P and Q with entries in R such that PAQ = B has the block diagonal form

I:: 1 11

B= 0

D 0 ,

as in Theorem 5.1. The diagonal entries di are unique up to associates. Proof. By the discussion above it will suffice to show that A can be transformed to B by a succession of elementary row and column operations. We may of course assume that A # 0. Note first that we may apply elementary operations until the nonzero matrix entry with minimal d-value divides every other matrix entry (Why?), and that the situation will then persist when any further elementary operations are applied. We may further arrange, by interchanging rows and/or columns, that an entry with minimal d-value is a , in the upper left-hand corner. Then, since a, divides each a l i in the first row and each ail in the first column, further elementary operations will reduce the matrix to the form

,

[;l

3

with al dividing each entry of C. The procedure can now be applied to C and continued until the desired diagonal form is reached. As for uniqueness we may view A as the matrix representing an R-homomorphism f from R" to R". Then B also represents f and by Theorem 5.1 the di are the invariant factors of the torsion submodule of R"/f(R"),hence are unique by Theorem 3.8. Theorem 5.2 actually holds true when R is a PID (see Exercise 7.21) but the proof is a bit more complicated and is not in general constructive. Elements with minimal d-value must be replaced by elements with a minimal number of prime factors (counting multiplicities), and elementary row and column operations are not generally sufficient. Matrices in block diagonal form

[s :],

where

C = rc

1'

d '

with ad - bc E U ( R ) ,can also be necessary; they correspond to changes in bases that result in what are sometimes called secondary row and column operations on the representing matrix. Let us look in detail at a small example. Suppose f : E 2 + Z z is given by

5 Computations

149

so f has matrix A relative to the bases {el , e 2 } in both M = H 2 and N = 2'. We wish to perform elementary operations on A to transform it to diagonal form. It will be convenient to compute the matrices P and Q in the process in order to find the new bases for M and N relative to which f is represented by the diagonal matrix. To that end we need only perform successively on identity matrices the same row operations (for P ) and column operations (for Q ) as are performed on A . One possible sequence ofelementary operations that succeeds for A is given (in order) by R z l l , R l Z 1 ,R z l l , C , , , - , , , . Accordingly we have A=[

"I%[ "I%[

-6

8

-2

-2

20]% 14

14

and

Observe that PAQ = D, as desired, and hence AQ = F ' D . Since D is diagonal the effect of multiplying P - times D is to multiply the ith column of P-' by the ith diagonal entry of D for each i. Thus in general the bases for M and N relative to which f is represented by D are determined by the columns of Q and of P - ' , respectively. For the present example we have 1 0 R2WlI 1]-[-1

[o

[-; -;I=

1 0 R121-11 1]-[4

,]A[=

R*1,-1,

11-

P-l,

so the bases consist of x1

2 -1

x2

=[-;'I

for M

and

If we set E = Im(f) E N and G = N / E , then, by the proof of Theorem 5.1, G = ( y l E ) 0 ( y 2 E ) , G has invariant factors 2 and 34, and (GI = 68.

+

+

150

IV

Modules

The matrix B = PAQ that is equivalent to A in Theorem 5.2 is called the Smith normal form of A . The thrust of Theorem 5.2 is that any Rhomomorphism f: M + N , where R is Euclidean and M and N are free of finite rank, can be represented by a matrix in Smith normal form. Exercise 7.21 indicates the same result for any PID. An important application is the solution of a system of linear equations with coefficients in E (or in any Euclidean ring). The system UllX,

+

UlZX2

+ ... + u,,x,

= c1,

UZlXl

+ 422x2 + ... +

U2,Xfl

= c2,

Um1Xl

+ um2x2 + ... +

UmnX,

= c,

can be written as A X = C, where A = (aij)is the coefficient matrix, X = (x,, x 2 , . . . ,x,,)' is the column vector of unknowns, and C = (clrc2,. . . ,c,)'.

We may change variables via X = QY, with Q invertible over H, so that the system becomes AQY = C, which is equivalent with PAQY = PC if P is also invertible. For appropriate P and Q we have PAQ = B in Smith normal form, so the system is BY = PC, which is easily solved if solutions exist, and the solutions X to the original system are obtained as X = Q Y . For example, let us find all integral solutions of the system -33x1

+ 42x2 - 20x3 = -26, + 13x3 = 16.

21x1 - 27x2

This can be written as A X

=

C with

A=[

X

= (x1,xZ,x3)',

Change-of-basis matrices p=[:

:I

1

-33 42 -20 21 -27 13 '

and

C = (-26,16)'. 0

and

Q=[:

1

i]

2

transform A to Smith normal form

(verify). If we change variables X = QY we obtain PAQY = BY = PC = [-:I. Thus the system becomes 1 . y, = - 4 , 3 . y, = 6, so that y, = -4, y 2 = 2, and y, is unrestricted. We may set y , = a, an arbitrary integer, and

5 Computations

[it] ['5 1' [;j [ ,:,!

151

we find that

=

=

-2

+ 3a

=

2 3 3

with a E I arbitrary, is the full set of solutions to the system A X = C . Consider now the ring R = F [ x ] , F a field, and an R-module V,. Choose a basis { u l , u 2 , . . .,u , } for I/ so that T is represented by a matrix A. Let N = R", the free R-module consisting of all column vectors with n entries from R. In this setting we have the following theorem.

Theorem 5.3. Let E be the submodule of N generated by the columns of the matrix A - x l . Then Vr is R-isomorphic with N / E . Proof. If we set e l = (l,O, . . . ,O)', e2 = (0, 1,0,. . .,O)', { e l ,e 2 , .. . , e n } is an R-basis for N . The mapping 4: N -+ 4 ( e i ) = ui is an R-homomorphism from N onto V,, so it FHTM, to show that ker 4 = E. For 1 Ii I n let zi be A - XI, i.e., zi =

c ajiej n

as usual, then V, determined by will suffice, by the the ith column of

...,

- xei E N .

j= 1

Then j= 1

j= 1

for all i, since A represents T relative to { u l , . . . , o n } . Thus E E ker 4. On the other hand let W be the F-subspace of N / E spanned by { e i + E : 1 I i IH}. Observe that x(ei

+ E ) = xe, + E = xi a j i e j = c a j i e jj

zi

-

)+

x a j i e j - xe,

( i

E

+ E = xaji(ej + E ) E W i

since zi E E , so W is in fact a submodule of N I E . But N = R(el, ..., e n ) , so W = N / E , and if u E N we may write u = x i c i e i z, where ci E F and z E E. Consequently

+

b , ( ~=) C c i $ ( e i ) + 4(z) i

= Cciui i

since E E ker 4, and hence u E kerb, if and only if all ci= 0,i.e., if and only if U=ZEE.

152

IV Modules

Theorems 5.1, 5.2, and 5.3 make possible the computation of invariant factors for any linear transformation T o n a finite-dimensional vector space I/. If T is represented by a matrix A relative to a basis { u l , . . .,u,} for V , set M = N = R", where R = F [ x ] , and let zi E N be the ith column of A - X I . Define an R-homomorphism f: M -, N by setting f ( e i )= zir 1 Ii In; set E = I r n ( f ) = R(z, ,...,z,), and we have V, z N / E by Theorem 5.3. The matrix representing f relative to the basis {el,. . ., e n } for both M and N is just A - X I .By Theorem 5.2 we may choose new bases so that f is represented by a diagonal matrix B with each diagonal entry dividing the next, and then see by Theorem 5.1 that the nonzero nonunit diagonal entries are the invariant factors of N / E , and hence of V, (the diagonal entries will in fact all be nonzero since V, is a torsion module). If we write P ( A - x l ) Q = B, diagonal, then ( A - x I ) Q = P - ' B , and since B is diagonal the ith column of P - ' B is just the ith column of P - ' multiplied by the ith diagonal entry of B. Thus the bases for M and N relative to which f is represented by the diagonal matrix Bare, respectively, the column vectors of Q and of P - ' . If we write y , , y 2 , . . ., y , for the columns of P - then, by Theorem 5.1,

',

N/E = O{R(y,

+ E ) : s + 1 Ii S n ) ,

where the first s diagonal entries of B are units. The isomorphism, call it 0, given in Theorem 5.3 (and its proof) from N / E to V, is determined by O(ei + E ) = ui, so if the column y i is written as yi = xjyjiej, then O(yi + E ) = y,pj = u i , say. Note here that each y j i E R = F [ x ] and uj E V , so yjiuj means the module action of polynomials on vectors. We conclude that V, = @ { R ( u i ) : s + 1 Ii I n } . With this information and the aid of Proposition 4.2 and Theorem 4.3 we may obtain the F-basis for Y relative to which T is represented by its rational canonical form. Note that the F-dimension of R(ui) is equal to the polynomial degree of its order as an R-module, i.e., the polynomial degree of the corresponding invariant factor. Once cyclic vectors are obtained for the rational canonical form it is a straightforward matter to obtain a basis for the Jordan canonical form (provided, of course, that it txists). Each cyclic submodule is first decomposed as a direct sum of primary submodules as indicated in Proposition 3.10. If the order of each is a power of a linear factor, then a basis is chosen as in the discussion preceding Theorem 4.5. A few examples should make the procedures clear. First take V = Q3 and let T be represented by the matrix

cj

A=

5 -8 6 -11 [6 -12

1

4 6

5 Computations

153

relative to the basis {el,e2,e3}.The sequence C13, CZl2,C31(x-5),4rR21(-3,2), R31(x-71/4,C3213/2), R3,(-,, of row and column operations yields = R32(-2)R31(x-7)/4R21(-3/2)1

=

[ ["

(x

Q = c32(3/2~c31(x-S)/4c212c131

=

-3/2 5)/4 -2

+

:]

]

1

9

0 1 3f2 1 2 (x 7)/4

for which 4 P ( A - x1)Q = 0 0

1

+

:]

0 1-x

0

(1 - x2)/4

(verify). Thus the invariant factors of T are fl(x)

=

mT(x) = x2 - 1 and

f 2 ( x ) = x - 1.

The rational canonical form for T is thus 0 1 0

Generators for the cyclic submodules are obtained by means of the columns of P - ' as discussed above. Since p-'

= R21(3/2)R31(7-x)/4R3221 =

[

1

0 0

3/2 (7 - x)/4 2 O] 1

we have

u1 = e,

3 7-x + -e2 +2 4

e3

3 1 + -e2 + -(71 - A)e3 = 0, 2 4 u2 = e2 + 2e3, = el

and u3 = e3.

Applying Proposition 4.2 to the cyclic vectors u2 and u3 we obtain a basis w1 = u 3 , w2 = Tw, = (4,6,7)',and w3 = u2. The change-of-basis matrix

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IV Modules

(from (el,e2,e3}to { w l ,w 2 ,w3})is then

[P 1: 1

L=O

6

1 ,

and L - ' A L = C, the rational canonical form. Since mT(x)= (x - l)(x + 1 ) has distinct linear factors the Jordan canonical form for T is diagonal, by Theorem 4.6. The primary submodules of R < u 3 ) have generators u1 = (x - l)u3 = ( A - f)u3 = (4,6,6j'

and u2 = (x

+ l)u3 = ( A + I)u3 = (4,6,8)',

eigenvectors for the eigenvalues - 1,1, respectively. Since u3 = w3 is also an eigenvector for eigenvalue 1 the Jordan form for T,relative to {ul, u 2 , u , } , is -1

J=[

0 0

y].

The matrix M with columns u l , u 2 , u3 serves as a change-of-basis matrix. That end result would probably have been reached more quickly and more directly by means of the standard techniques of linear algebra. The basis vectors u l , u2, u3 are simply linearly independent solutions to the homogeneous systems of equations (A - IZ)X = 0,3, = - 1 , l . Next take V = Q4, and suppose that T: V + Vis represented by

.=I 3 -1

1-1

2-1 3 -4 3 - 33 - 13 -4I 1

relative to {el,e2,e3,e4}.Then

5 Computations

155

Thus mT(x) = f,(x)

= (x

+ l)(x

-

2)'

=

-8

+ 4x + 6x2 - 5x3 + x4

is the only invariant factor of T , and the rational canonical form of T is 8

0 0 0

In this case P-

- R2,( - 1,R31,3- x , R 4 1 3 R 3 2 ( - x 2 +

=I

1

-1

3-X 3

I:

7 x - 9,R42(3x-6,R431

0 0 1 0 - x 2 + 7 ~ - 9 1 -(.~-2),/27 3~ - 6 1 - 2(x - 2),/27

R341 R43[(x-2p, - 2711

1

2

Thus u1 = p - le4 = (O,O, 1,2)' is a cyclic vector for the rational canonical form of T , and u , , together with u2 = Tu, = (- 1,1, -5, - 5 ) ' , u, = TU,= (-4,3,2, - l)', and u4 = Tu, = ( - 12,6, - 1, - 11)' provides a basis relative to which T is represented by the rational canonical form C. The minimal polynomial mT(x) factors as p,(x)p,(x), with pl(x) = x + 1, p2(x)= (x - 2),. The corresponding primary submodules have cyclic vectors (x - 2),u, = (O,O, -81, -81)' and (x + l)ul = (-1,1, -4, -3)L. For arithmetical convenience we may replace (O,O, - 81, - 81)' by u1 = (O,O, 1, 1)L. Then u , , together with 02 = (x

+ l)U1 = ( -

1,1, -4, -3)',

~3

= (X -

2)uZ = ( - 3,2,5, O)',

u4

= (x -

2)u3 = (0, - 3, - 3,O)'

and

1

provides a basis for the Jordan canonical form 0 0 2 0 1 0 0

-1

J =

0 0 2 1

0' 0 0 2

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IV Modules

For a final example take V = F4,Q c F E @, and suppose that T: V + Vis represented by the matrix

relative to {e,,e2,e3,e4). Then

P(A - xZ)Q

=

0 -1

0

0

0 0

(x2

+ 1)2

where = R34R34xR43(-1~R42(1+3x)R32(1+x-x2)R24(-1) R41(-3)R31(x-2)R21(- 3)R121,

Q = C43

(x3

+ xz + 3)C34(- 1 ) c42(1 - x) C3~ ( -x1)c4I - 3)

c312c21(-1-x)c13~~

Thus m,(x) = (x2 + 1)2 = x4 + 2x2 + 1. The vector u , = P-’e4= (l,O, 1,l)’ is a cyclic vector for V,, and u l r together with I42 = Au, = (1,0,0, l)’, ~3 = A u = ~ (-2, -1, -2, -2)‘ and u4

= A U 3 = (1,1,1,0)’

provides a basis relative to which T is represented by its rational canonical form 0 0 0-1 1 0 0 0 0 10-2 0 0 1 0

.-[

If F E [w, then x2 + 1 is irreducible in F[x] and T has no Jordan canonical form. If, on the other hand, x2 + 1 splits in FCx], then mT(x)= (x - i)2 ( x + i)2, where i 2 + 1 = 0. As cyclic vectors for the primary submodules V,[(x i)2] and V,[(x - i ) 2 ] we may take

+

u1 = ( x - ZJ’U, = (-3

- 2i, -1, -3, - 3

- 2)’

5

Computations

157

and w 1 = (x

respectively. Then u1,u2 = (x 03 =

+ i)2ul = ( - 3 + 2i, - 1, - 3, - 3 + X)',

+ i ) u , = (2 + i, 1 + i, 1 + i, 1 + i)',

w1,

and u4

= (x -

i ) w l = (2 - i, 1 - i, 1 - i, 1 - i)'

provide a basis for the Jordan canonical form

I ; ;1 pl

o0 0o 001

r-i

1 - i

J=

0 0

0 0 O i O O

l

for T .

i

The Smith normal form can also be used to compute invariants of finitely generated abelian groups given by presentations. Much as in Section 10 of Chapter I we may write an additive presentation

whereby we view G as the quotient of the free abelian group N with basis {a,, . . , ,a,} by the subgroup

If we set M

=

Z" and define f

E

Hom,(M, N ) by setting

then Im(f) = E and f is represented by the matrix A = [ r i j ] . Theorems 5.1 and 5.2 apply directly, so we may determine the invariant factors and the torsion-free rank of G. For example, take G to have generators al,a2,a3,a4, subject to the relations

+ 6 ~ 1 8 ~ -2 12a3 + u4 = 0, 4a, IOU, - 12a3 + u4 = 0. 4 ~1 4 U 2 - 6 ~ 3 a4 = 0, -

-

158

IV Modules

Then the representing matrix

I-; -.-[“ ; :] -4

A=

is equivalent with

-8 -10

-1;

1 0 0

0 0 0

by Of the sequence R14, R41(-4)? R 3 1 6 , R214, c21(-1)7 c31(-1)> c23(-l)i R,,(-,,, R,,(-,,, R231, and R24. Thus the invariant factors of G are 2 and 6, and the torsion-free rank of G is 1, by Theorem 5.1. The torsion subgroup is isomorphic with E 2 0 Z6, of order 12, and G Z,0 Z6 0 Z.

6. TENSOR PRODUCTS Suppose Vis a finite-dimensional vector space over the real field R and that

T: V - + V is a linear transformation whose minimal polynomial is mT(x)=

+ 1. Then T has no eigenvalues in R, but the equation mT(x)= 0 can of course be solved in the extension field @.It is in fact customary to say that T has & i as complex eigenvalues. Corresponding eigenvectors would necessarily allow scalar multiplication by nonreal complex numbers, so it would seem desirable to extend V to a vector space over C.In particular it should then be possible to multiply elements of the original vector space V by elements of the extension field C. Phrased somewhat more abstractly we wish to be able to “multiply” together elements of two different IW-modules M = C and N = V. The tensor product of modules is a useful device that makes such multiplication possible in great generality. Suppose R is a ring, M is a right R-module, and N is a left R-module. If A is an abelian group, then a function f:M x N -+ A is called a balanced map if x2

0) f ( x + u, Y ) = f(x9 Y ) + f(%Y), (ii) f ( x , y + 0 ) = f(x, A + f ( x , 4, and (iii) f ( x r , Y) = f ( x , r y ) for all x, u E M, all y, u E N, and all r E R. A tensor product of a right R-module M and a left R-module N is an abelian group T together with a balanced map t: M x N -+ T such that given any

6 Tensor Products

159

abelian group A and any balanced map f :M x N + A there is a unique homomorphism g: T + A such that f = gt, i.e., the diagram

M x N

+A ,f

is commutative.

Proposition 6.1.

If a tensor product exists it is unique up to isomorphism.

Exercise 6.1. Prove Proposition 6.1. Exercise 6.2. If M = gP and N = E,, both viewed as abelian groups (i.e., Z-modules), show that T = 0 is a tensor product for M and N .

Theorem 6.2. If R is a ring, M is a right R-module, and N is a left Rmodule, then a tensor product of M and N over R exists.

xi

Proof. Let F be the free abelian group based on the Cartesian product M x N ; a typical element of F is thus a finite sum ni(xi, y i ) , niE Z, x i E M, y i E N . Let H be the subgroup of F generated by all elements of the forms

0) ( x + u, Y ) - (x. Y ) - (u, Y), (ii) ( x , Y 4 - (x, y ) - (x, 4, and (iii) (XI, Y ) - ( x , r y )

+

for all x , u E M , all y , v E N , and all r E R . Set T = F / H and define t: M x N + T by setting t ( x , y ) = ( x , y ) + H . It is clear from the definition of H that t is a balanced map. Suppose A is an abelian group and f : M x N + A is a balanced map. If we define h: F + A by setting

then h E Hom(F, A ) and H Iker h since f is balanced. Thus we may define g from F / H = T to A by setting g(a + H)= h(a) for each a E F. Then g E Hom( T, A ) and gt(x, y ) = g ( ( x ,y )

+ H ) = Mx, Y ) = f(x9 Y ) ,

so gr = f . For the uniqueness of g note that t ( M x N ) generates T , so if g’ E Hom( T,A ) and g‘t = f , then g’t = gt, and hence g’ = g, Consequently T is a tensor product for M and N .

Because of Proposition 6.1 we may take the group T constructed in the proof of Theorem 6.2 and call it the tensor product of M and N over R . For a

160

IV Modules

more convenient notation let us write x O y for (x, y) + H E T, and consequently write C i x i@ y i , a finite sum, for a typical element of T. We write M ORN for T itself (as constructed). The notation x 0 y suggests that we have indeed “multiplied” together the elements x E M and y E N . Observe that

+

+ u 0 y, + XO 0,

( x u)O y = x 0 y x O ( y u) = x @ y

+

and xr @ y = x 0 ry

for all x, u E M , all y , v E N, and all r E R, giving the basic properties of the multiplication. Exercise 6.3. Show that x @ 0 = 0 @ y

= 0 for

all x E M, all y E N.

For example, take R = Z, M = Q, and N = h,for some integer n > 0. Then for any x E M , y E N we may write

x 0y

= ( x / n ). n O

y = ( x / n )0 ny = ( x / n )0 0 = 0.

Thus M ORN = 0 in this case. Exercise 6.4. Suppose R is a ring, M , and M , are right R-modules, N, and N , are left R-modules, f E Hom,(M,, M,), and g E Hom,(N,, N,).

(1) Show that there exists a unique h E Hom,(M, ORN,, M2 ORN2) such that h(x 0 y ) = f ( x ) @ g( y ) for all x E M , , y E N, . (Hint: define a (balanced) map from MI x N, to M, ORN2 via (x, y ) H f ( x ) O g ( y )and see the definition of a tensor product.) The unique homomorphism h is denoted by f Q g. (2) Suppose further that f ‘ E Hom,(M,, M3) and g’ E HOm,(N,, N3). Show that ( f ’ 0 g ’ ) ( f @ g ) = f ’ f O g’g. Suppose S and R are rings. An abelian group M is called an S-R-bimodule if it is simultaneously a left S-module and a right R-module, satisfying the further requirement that s(xr) = (sx)r for all s E S , x E M, and r E R. Suppose now that M is an S-R-bimodule and N is a left R-module. We may make a left S-module of the tensor product M ORN , as follows. For each s E S define f,:M x N + M ORN via f,(x, y ) = sx 0 y . Then f, is a balanced map (verify), and there is a unique homomorphism gs: M O R N+ M O R N such that g,(x 0 y ) = sx O y for all x E M, y E N . As a consequence we may define a module action on M ORN by setting

for each s E S.

6 Tensor Products

161

Exercise 6.5. Verify that M ORN is a left S-module as indicated above. If M is a unitary left S-module show that M ORN is also a unitary S-module.

Proposition 6.3. Suppose R is a ring with 1, N is a unitary left R-module, and R is viewed as an R-R-bimodule with ordinary multiplication in R as the module actions. Then R ORN is R-isomorphic with N . Proof. The map f:R x N -+ N defined by f (r, y ) = ry is easily seen to be balanced. Thus there is a homomorphism g : R ORN -+ N such that g(r 0 y ) = ry for all r E R and y E M . It is easy to check that g is an Rhomomorphism, and g is onto since N is unitary. If ri 0 y i is in ker g, then x i r i y i = 0. But then

xi

i

i

i

so ker g = 0 and the proposition is proved. Proposition 6.4. Suppose M is a right R-module and N , and N2 are left Rmodules. Then M OR(N10N2)

( M ORN1) 0 ( M ORNz),

Proof. Set N = N , 0 N 2 and define homomorphisms p1 and p2 from N to N via p l ( y l ,y 2 ) = ( y 1 , O ) and P A Y , ,y2) = (0, ~ 2 l Clearly . Pi(N) z Ni- Set fi = 1 @ p i :M ORN + M ORN [see Exercise (6.4)]. It is easy to check that f, + f2 = 1, the identity map, that f1f2 = f2f1 = 0, and that f; = 1;. for each i. As a consequence, if we set K i = Im(J), then M ORN = K , 0 K2 (verify). Thus it will suffice to show that Ki S M ORNi,i = 1,2. Define 1 , : M x N , + K , by setting

,

t,(X,Y,) = x O ( Y , ~ o ) = f , ( x 0 ( Y , ~ o ) ) .

Then t is balanced. Suppose that f : M x N , -+ A is a balanced map for some abelian group A . Define h: M x N A by setting h ( x , ( y l ,y z ) ) = f ( x , y l ) , then h is also balanced. Thus there exists a unique homomorphism k from M ORN to A such that h = kt. Let g = k K,, so g E Hom(K,, A), and observe that -+

I

gt,(x,y,) = S ( m ( Y l , o ) ) = k(XO(Y,?O)) = kt(X%(Y,,O)) = h(X,(Y,,O)) = f ( x , Y l ) ? so g t , = f.That g is unique follows from the fact that Im(t,) generates Kl.

Thus K , z M O R N , . A completely analogous argument shows that K , z M ORN , , and the proposition is proved.

162

IV Modules

The diagram below may help in following the proof just completed.

M x

h

Remark: Clearly a result analogous to Proposition 6.4 holds if M

=

M,Q M,,and if the Mi are S-R-bimodules, then the isomorphism is an S-

isomorphism.

We may now solve the problem alluded to in the first paragraph of this section, viz., extending the ground field of a vector space. Suppose V is a vector space over a field F and K is an extension field of F. We shall assume that V is finite dimensional, although the assumption is not at all necessary. Let { x l , .. .,x , } be an F-basis for V , so V = F x , 0 .. . O F x , z F". We may view K as a unitary K-F-bimodule. Then V x = K 8, Vis a unitary K-module, i.e., a K-vector space. Note that

VK=K

@IF V Z K @,(F") Z ( K @IFF)"Z K "

x=

by (analogs to) Propositions 6.3 and 6.4. Thus dim,( V") = dim, V . a, 8 v , E V K we may write vi = bjixj,with bji E F, Given x = and hence

,

Consequently (1 @ xl,. .., 1 0 x,) is a K-basis for V K .The function u H 1 8 v is a 1-1 map from V into V K ,so we may identify V with the image and view V as a subset (not a K-subspace) of V x . Exercise 6.6. If K is an extension field of F, V is an F-vector space, and V x = K 0, V, then V K is also an F-vector space. Show that dim,(VK) = [K :F] dim, V .

If F is a field and V, Ware F-vector spaces, then arguments analogous to those above show that dim,( V 0, W) = (dim, V)(dim, W).

6 Tensor Products

163

If{x,, . . .,x,} and { y,,. . . , y,} are bases for V and W , respectively, then {xi @ y j : 1 I i I n, 1 I j I m}

is a basis for V OFW. Suppose that S: V + V and T: W + W are linear transformations, and that they are represented by the matrices A = ( a i j )and B = ( b i j ) relative to the bases {xi} are { y j } . Recall from Exercise 6.4 that S 0 T: V OFW -+ V OFW is characterized by the fact that (S Q T) ( u 0 w) = S(u) Q T(w)for all u E V , w E W. If the basis {xi 0 y j } is ordered

0 YI,XI 0 ~

XI

0

2 7 * . * * ~ 1Y m ,

x2

0~

0 .~m>.,**

l r . . . r ~ 2

XnO ~lr...,xnO Y,, then the matrix A Q B that represents S 0 T has the block form (aijB),which is commonly called the Kronecker product of A and B. This construction is of considerable importance in the theory of group representations, and thereby in the study of quantum mechanics.

Exercise6.7. Verify that S 0 T, as above, is represented by the Kronecker product of A and B. If R is a commutative ring then a ring A is called an R-algebra if A is a left Rmodule such that r(ab) = (ra)b = a(rb) for all r E R and all u, b E A. Examples abound. A Z-algebra is nothing but an arbitrary ring A. If R is a commutative ring with 1, then any polynomial ring R[x,, . . . ,x,] is an Ralgebra. If F is a field and K is an extension of F, then K is an F-algebra. If V is an F-vector space, then the ring A = HornF(V , V) is an F-algebra. If R is a commutative ring, A and B are R-algebras, and A is an R-Rbimodule, then A ORB is an R-module. Let us show that it can be made into an R-algebra as well. Fix a E A, b E B, and define f a b : A x B + A ORB by setting &(C, d ) = ac 0bd. Then f a b is a balanced map and hence there is a unique homomorphism gab: A ORB + A Q R B such that g&(C Q d) = ac 0 bd for all c E A, d E B. We define

xi

Since ga&, is an R-homomorphism it is a routine matter to check that the multiplication, so defined on A ORB, makes A ORBan R-algebra. It is unitary if R is a ring with 1 and A is unitary as a left R-module.

Theorem 6.5. Suppose R is a commutative ring with 1, and I and J are ideals in R. Then (R/I) Q R ( R / J )and R / ( I + J) are isomorphic R-algebras.

Proof. Define t : R / I x R / J + R / ( I + J) by setting t(r + I, s + J ) = rs + + J. Then t is a balanced map. If A is an abelian group and f : R / I x R / J A is a balanced map define g : R / ( I + J) -,A by setting

I

+

g(r

+ 1 + J ) = f ( r + I, 1 + J).

IV

164

Modules

It is perhaps not obvious that g is well defined. If r + I + J = I’ + I + J , then r’ = r + a + b, with a E I and b E J . Thus r’+ I = r + b + I, f(r’ I , 1+ J ) = f ( r b I, 1 J ) =f ( r + I, 1 J ) f ( b I, 1 J), and , f ( b I, 1 J ) = f(1 I,b J ) = f(1 1,O) = 0, so g is well defined. It is clear that g is a homomorphism (of abelian groups), and gt(r + 1 , s + J ) = g(rs I J ) = f ( r s + I , 1 J ) = f ( ( r I)s, 1 J ) = f ( r I,s(l J ) ) = f(r 1,s J ) since f is balanced. Thus gt = f,and g is uniquely determined since t maps R/I x R I J onto R/(I + J ) . Thus Rl(1 + J) and (R/I)@,(R/J) are isomorphic abelian groups by Proposition 6.1. We remark that r + I J (r + I ) @ (1 + J ) is an explicit isomorphism between them, and that it is an Ralgebra isomorphism.

+

+

+ +

+

+

+

+

+ + + +

+ +

+

+

+

+ +

+

+

+

+

Corollary. h, Oah, 2 &,,.

Proof. Z, = Z/(m), Z,= h/(n),and (m) + (n) = (m, n).

7. FURTHER EXERCISES 1. If R is a ring with 1 and M is an R-module that is not unitary show that Rm = 0 for some nonzero m E M . 2. Give an example of an R-module M having R-isomorphic submodules N , and N , such that M / N , and M I N , are not isomorphic. 3. Suppose Vis a finite-dimensional vector space over a field F , viewed as an F-module. Describe a composition series for Y and determine its factors. 4. A sequence KL M 3 N of R-homomorphisms of R-modules is exact at M if Im(f) = ker g . A short exact sequence 0 -+ K + M N -+ 0 is exact at K , M , and N . If 0 -+ K -+ M -+ N + 0 is short exact show that M is Noetherian if and only if K and N are both Noetherian. 5. Suppose M , , M , and N are submodules of an R-module M , with M , G M 2 . Show that there is an exact sequence -+

0 ( M , n N ) / ( M , n N ) M , / M , -+ ( M , + N ) / ( M , + N ) 0. 6. Let F be a free abelian group with countably infinite basis { u l , u 2 , u 3 , .. .}, and let R = End(F). Show that R, as a free R-module, has one basis B , = (l,), but also another basis B , = (41,42), where 4,(u2,) = a,, 4,(a2,-,) = 0,4,(a2,) = 0, and 4,(u2”-,) = a,, n = 1,2 ,... . -+

-+

7 Further Exercises

165

7. Give an example of an R-module M over a commutative ring R where the set T(M) of torsion elements of M is not a submodule. 8. If F is a field set R = F [x,, x2,x3,...], the ring of polynomials in a countably infinite set of distinct indeterminates. Let I be the ideal (xl,x2,. . .) in R. If M = R and N = I show that M is a finitely generated R-module but N is a submodule that is not finitely generated. Is N free? 9. Suppose R is a PID and M = R(a) is a cyclic R-module of order r, 0 # r E R. Show that if N is a submodule of M, then N is cyclic of order s for some divisor s of r. Conversely, M has a cyclic submodule N of order s for each divisor s of r in R. 10. Suppose R is a commutative ring and M is an R-module. A submodule N is called pure if rN = rM n N for all r E R. (i) Show that any direct summand of M is pure. (ii) If M is torsion free and N is a pure submodule show that M/N is torsion free. (iii) If M / N is torsion free show that N is pure. 11. Suppose L, M, and N are R-modules and f : M + N is an R-homomorphism. Define f*: Hom,(N, L ) + Hom,(M, L ) via f*(4):rn I+ $(f(rn)) for all 4 E Hom,(N, L),rn E M. (i) Show that f* is a Z-homomorphism. (ii) If R is commutative show that f* is an R-homomorphism. (iii) If K , M , and N are R-modules show that K + M + N + 0 is an exact sequence of R-homomorphisms if and only if 0 + Hom,(N, L ) + Hom,(M, L ) -+ Hom,(K, L )

is an exact sequence of Z-homomorphisms for all R-modules L. 12. (i) If R is an integral domain show that free R-modules are torsion free. (ii) If R is an integral domain with 1 that is not a field exhibit a torsion free R-module that is not free. 13. Suppose R is a commutative ring and M is an R-module. Then the Rmodule M * = Hom,(M, R) is called the dual module of M. The elements of M* are commonly called R-linear functionah on M. If M is free of finite rank, with basis {xl, x 2 , . . .,x,,}, show that M* is also free, with basis {fi,fi,. . .,f,}, where

(the dual basis). Conclude that M and M * are R-isomorphic in that case. 14. If R is a commutative ring with 1 and M is an R-module define a function 4: x + 2 from M to its double dual M ** = (M*)* by setting 2(f ) = f ( x ) , all

166

IV Modules

f E M * (see Exercise 13). Show that I#I is an R-homomorphism. Under what circumstances is 4 a monomorphism? 15. Use invariant factors to describe all abelian groups of orders 144 and 168. 16. Use elementary divisors to describe all abelian groups of orders 144 and 168. 17. If p and q are distinct primes use invariant factors to describe all abelian groups of order (i) p 2 q 2 ,

(ii) p44,

and

(iii) p",

1I nI 6.

18. If p and q are distinct primes use elementary divisors to describe all abelian groups of order p3q2. 19. Suppose R is a PID and M is a finitely generated torsion R-module. Show that the elementary divisors of M are just the prime power factors (with multiplicities) that result when the invariant factors of M are factored into prime powers. Show on the other hand that the invariant factors can be recovered from the list of elementary divisors as follows: the last invariant factor is the LCM of all the elementary divisors. If the elementary divisors whose product is the last invariant factor are removed from the list, then the LCM of those that remain is the next invariant factor, etc. 20. If R is a PID and M is a finitely generated torsion R-module show that the decomposition of M by means of its invariant factors expresses M as a direct sum of the smallest possible number of cyclic submodules, and that its decomposition by means of its elementary divisors expresses M as the direct sum of the largest possible number of (nonzero) cyclic submodules. 21. Generalize Theorem 5.2 to modules over any PID, using secondary row and column operations when necessary. 22. Find all solutions X E Z3to the system of equations AX = 0 if A is

(i)

- 1 11, 1 0 2

(ii)

[

0 2-1 1 - 1 01, 2 0 -1

(iii)

[

-3 -4 0 0

3

1 01.

4 -1

23. Find all integral solutions to the following systems AX = B of equations:

0

(ii) . = [ I - -2'

2 -1 0 - 101,

7 Further Exercises

(iv) A = [ 6 8

19 14 22]. 30

167

=[3

24. What are the invariant factors and the elementary divisors of a diagonal matrix over a field F? 25. If a matrix A over a field F has minimal polynomial m ( x ) and characteristic polynomial f(x) show that f(x) is a divisor of m ( ~in)F[x] ~ for some positive integer k. 26. Suppose V = { ( u ,b)':a, b E Q}, T :V -+ V is multiplication by -10 18 -6 111'

'=[

and M is the Q[x]-module V T .Find the elementary divisors of M and write M as a direct sum of cyclic submodules. 27. Determine whether or not 3 0 2 5 -8 4 A= 0 1 -11 and B = [6 -11 6 1 -4 0 3 6 -12 7

[

are similar over Q. 28. Find an explicit similarity transform relating

1

0 -2 -4 0 0 0 A = 2 41 and A ' = [-2 2 -11. 0 -1 -2 -4 4 -2 (Hint: They have the same rational canonical form.) 29. If A is an n x n matrix over a field F show that A is similar (over F ) to its transpose A' (see Exercise 28). 30. Find the characteristic polynomial, invariant factors, elementary divisors, rational canonical form, and Jordan canonical form (when possible) over Q for each of the following matrices:

168

(9

[

IV Modules

0 -2 2 6 81, -1 - 3 -4 1

(vi)

[

1

-4 -8

0 0 7 -4 12 -7

31. An n x n matrix A over a field F is called idempotent if A 2 = A. (i) What are the possible minimal polynomials for an idempotent matrix? (ii) Show that an idempotent matrix is similar over F to a diagonal matrix. (iii) Show that idempotent n x n matrices A and B are similar over F if and only if they have the same rank. 32. An n x n matrix A over a field F is called nilpotent if A" = 0 for some positive integer k.

(i) If A is nilpotent and A # 0 show that A is not similar to a diagonal matrix. (ii) Show that a nilpotent matrix A has a Jordan canonical form over F , and list the possible Jordan forms for A . 33. Suppose A and B are n x n matrices over a field F and that :] and [! g] are similar over F . Show that A and B are similar over F .

[t

34. Suppose A and B are finitely generated abelian groups and that A @ A z B e B. Show that A 2 B. 35. Compute the invariants of the abelian groups with generators a,, . . . ,a,, subject to the following relations:

(i)

n = 2, a, - 5a2 = 0,

3a,

+ 7a, = 0,

(ii)

3, - 3a3 = 0, 2a, + 5a3 = 0, n

a1

a,

+

=

7 Further Exercises

(iii) 2a, -

a2

a, - 3U2 a,

+

a,

n = 3, = 0,

(iv) 2a,

= 0,

+ a3 = 0.

-

a2 a2

3aI

169

n = 4, 5a3 = 0, - 3a3 = 0, - 7a3 = 0.

+

36. If R is a commutative ring and M, N are R-bimodules show that M ORN and N ORM are isomorphic as R-modules. 37. If R is a commutative ring with 1 and x1,x2are distinct indeterminates show that R[x,,x,] and R[xl] ORR[xJ are isomorphic as R-algebras. 38. If A and B are finitely generated abelian groups show that A OzB is a finitely generated abelian group. Find its rank and elementary divisors in terms of those of A and B. 39. Suppose A is a finitely generated abelian group. (i) Compute A OnQ. (ii) Define f : A + A OaQ by setting f ( a ) = a 0 1 for all a E A . Show that f is a homomorphism. Under what circumstances is f a monomorphism? 40. If A is an abelian group show that Z,OHA 2 A / n A . 41. If F is a field and K is an extension field show that M,(K) z K @,M,(F) (as F-algebras). 42. Suppose that F is a field, K is a finite normal extension, and that G = G ( K : F )is a direct product of two subgroups G , and G , . Set L , = F G , and L, = F G , . Show that the ring L, OFL, is a field, with a subfield isomorphic with F, and that if L , O F L 2 is viewed as an extension of F , then K and L , OFL , are F-isomorphic. 43. Suppose R is a commutative ring with 1 and E, F are free R-modules, with respective bases { x , : ~E A } and { y p : P E B } . Show that E ORF is free with basis {xo0 y p : a E A , E B}. 44. Suppose R is a commutative ring and L, M , N are R-modules. Show that Hom,(L, Hom,(M, N ) ) and Hom,(L QR M, N ) are isomorphic R-modules. Conclude in particular that ( L 0, M)* E Hom,(L, M * ) (see Exercise 13). (Hinti If f~ Hom,(L, HomR(M,N ) ) and a E L write f(a) as f,.Try mapping f + S, where f(a 0 b) = f,(b) for a E L, b E M.) 45. If K + M + N + 0 is an exact sequence of (left) R-modules and Lis a right R-module show that L ORK + L ORM + L ORN + 0 is an exact sequence of abelian groups (see Exercise 6.4 for the homomorphisms). 46. Suppose R is a ring with 1. A unitary R-module P is called projective if given an exact sequence M 4 N 40 of R-modules and an R-homomorphism f : P + N , then there is an R-homomorphism h: P + M such that f = gh, i.e.,

170

IV Modules

the diagram P

M-N-0

is commutative. (i) Show that free modules are projective. (ii) If P = PI 0 Pz show that P is projective if and only if both PI and P2 are projective. Generalize to arbitrary direct sums. 47. Show that an R-module P is projective if and only if P is a direct summand of some free module F (see Theorem 2.8). 48. Show that an R-module P is projective if and only if given any short exact sequence 0 + K M + P -+ 0 we may conclude that M z K 0 P (we say that the sequence splits). 49. (i) If R is a PID or a division ring show that every projective R-module is free. (ii) Give an example of a projective module that is not free. 50. If P is a projective R-module show that there is a free R-module F such that F E F 0 P (see Exercise 47). 51. Let R be a ring with 1. A unitary R-module Q is called injective if given any exact sequence 0 -+ K 3 M of R-modules and a homomorphism f: K + Q, then there is a homomorphism h: M + Q such that f = hg, i.e., the diagram O-K-M --f

is commutative. If Q = Q 1 x Q 2show that Q is injective if and only if both Q I and Qz are injective. Generalize to arbitrary direct products. 52. Show that if an R-module Q is injective, then given any short exact sequence 0 -+ Q + M + N - 0 of R-modules we may conclude that M z Q 0 N (the sequence splits) or, equivalently, Q is a direct summand of every R-module that contains it as a submodule. 53. An (additive) abelian group A is called divisible if nA = A for all nonzero n E Z. (i) Show that A = Q is divisible. (ii) Show that any homomorphic image of a divisible group is divisible.

7 Further Exercises

171

Thus, for example, Q / Z is divisible. (iii) Show that no finitely generated abelian group A( # 0) can be divisible. 54. If A is an abelian group show that A has a unique largest divisible subgroup D, and that A = D O B, where B has no nonzero divisible subgroups. (Hint: Take D to be the subgroup generated by the union of all divisible subgroups of A.) 55. If A is any abelian group there is a divisible group D with a subgroup isomorphic with A . (Hint: There is a free abelian group F with a subgroup K such that A F / K ; F is a direct sum of copies of Z, and Z IQ.) 56. If an abelian group A is injective as a Z-module show that A is divisible. (Use Exercise 55.)

Chapter V

Structure of Rings and Algebras

1. PRELIMINARIES We begin by describing an important class of examples. Suppose G is a finite group and R is a commutative ring with 1. Denote by RG the set of all functions a: G -+ R. With the usual addition of functions and multiplication of functions by elements of R it is easy to see that RG is a unitary R-module. By a mild abuse of notation we may view each x E G as an element of RG, viz., we may take x to be the function from G to R such that x ( x ) = 1 and x ( y ) = 0 if x # y E G. Then clearly each a E RG can be written uniquely as a = c { a x. x : x E GI,with each a, E R, from which it follows that the elements of G constitute a basis for RG, and RG is a free R-module of rank [GI. The basis G is closed under (group) multiplication, and that multiplication extends naturally to a multiplication on all of RG by means of ( C a x x ) ( C b y Y= ) C{a,b,.

XY:X,

y E GI = C ( ( C ( a , b , : x y = u ) ) . U : U E G ) .

Exercise 1 . 1 . Verify that R G is a ring, and in fact an R-algebra, and that the multiplication in RG is given by ab(x) = C { a ( x y - ' ) b ( y ) :y E G}

for all a, b E RG, x E G. (This multiplication is sometimes called the convolution product.) The R-algebra RG is called the group algebra of G over R. Exercise 1.2. Let G = { l , x , x * } be a. cyclic group of order 3 and let R = Z,. Write out addition and multiplication tables for the group algebra RG. 172

I

173

Preliminaries

If R is any ring and M is a left R-module we define the annihilator of M in R to be AR(M) = A(M) = { r E R : r x = 0, all x

E

M}.

Then A(M) is an ideal (two sided) in R. We say that M is afaithful R-module if A(M) = 0. Since A(M) * M = 0 there is an obvious R/A(M)-module action on M , viz., [ r + A(M)] . x = rx for all r E R, x E M . The annihilator of M in R/A(M) is just the zero element A(M), so M is always a faithful R/A(M)module. Recall (Exercise IV.l.l)that an abelian group M is an R-module if and only if there is a ring homomorphism f:R + End(M). The annihilator A(M) is simply kerf, and M is faithful if and only if f is a monomorphism. The fact that M is a faithful R/A(M)-module is thus a consequence of the FHT for rings. If M is a left R-module we define the centralizer of M to be

C ( M ) = HomR(M,M ) = { g E E n d ( M ) : g ( r x )= rg(x),all r E R, x

E

M}.

Note that C ( M )is a subring of End(M). If we recall again that there is a homomorphism f : R --+ End(M), say with f : r H f , and f,(x) = rx, then g E C ( M ) if and only if gf,(x) = g(rx) = rg(x) = 1;g(x) for all r and x, i.e., if and only if g commutes with all 5 . Hence the name “centralizer.” An R-module M is called simple, or irreducible, if A(M) # R and the only submodules of M are 0 and M.Note that the condition A(M) # R serves only to rule out the trivial module action rx = 0, all r E R. Recall that a division ring is a ring R with 1 in which every nonzero element is invertible, i.e., U(R) = R*. A division ring clearly can have no zero divisors. Proposition 1.1 (Schur’s Lemma), If R is a ring and M is a simple Rmodule, then C(M) is a division ring. If N is another simple R-module, then either M and N are R-isomorphic or else HomR(M,N ) = 0. Proof. If 0 # f E C ( M ) ,then kerf and Im(f) are submodules of M, and kerf # M so kerf = 0. Also Im(f) # 0 so Im(f) = M, and hence f is invertible in End(M). It is easy to check that in fact f - E C ( M ) , so C(M) is a division ring. The proof of the second statement is similar; if 0 # f E HomR(M,N ) then kerf = 0 and Im(f) = N , so f is an R-isomorphism.

Suppose that F is a field, that the ring R is a unitary F-algebra with 1, and that M is a unitary R-module. Then M is automatically also an F-vector space, with scalar multiplication given by a . x = ( a . 1). x for all a E F and all X E M. Proposition 1.2. Suppose F is an algebraically closed field, R is a unitary F-algebra with 1, and M is a simple unitary R-module that is finite

174

V Structure of Rings and Algebras

dimensional as an F-vector space. Then C ( M )z F. In fact, C ( M ) = { a . I : a E F}, I being the identity map on M . Proof. If g E C ( M ) ,then g is an F-linear transformation from M to M . Let f ( x ) = m,(x) be its minimal polynomial over F. Since F is algebraically closed we may write

f(x) = n { ( x - ai):1 I i I k } ,

with aiE F.

By definition of the minimal polynomial f(g) = n { ( g - a i l ) :I I i Ik ) = 0.

But each g - ail E C ( M ) ,and C ( M )is a division ring by Schur’s Lemma. Thus in fact f ( x ) must be linear, and g = a1 for some a E F. An R-module M is called semisimple (or completely reducible) if it is a direct sum of simple submodules, or if M = 0. Exercise 1.3. If M is a semisimple R-module and 0 # x R x # 0. (Hint: First assume that M is simple.)

E

M show that

Proposition 1.3. Suppose L and N are submodules of M with L c N , and that there are submodules K and P of M such that M = L 0 K = N 0 P . If weset Q = N n K , then N = L O Q .

+

Proof. It is clear that L Q E N and that L n Q = 0, so L + Q = L @ Q G N . If X E N write x = y + z, with Y E L and Z E K . Then z = x Y E K n N = Q,so N G L O Q . Corollary. If M is a module in which every submodule is a direct summand, then every submodule has the same property.

Proposition 1.4. Suppose M # 0 is an R-module such that if 0 # x E M, then R x # 0. Then the following are equivalent. (i) M is semisimple, (ii) M is a sum (not necessarily direct) of simple submodules, and (iii) every submodule of M is a direct summand.

Proof. (i) (ii) is obvious. (ii) => (iii): Say M = C ( M a : aE A ) , with each M, simple, and let N be a submodule of M . By Zorn’s Lemma there is a maximal collection { M b : b E B E A } such that N n x { M b : bE B) = 0. Set K = c { M , : b E B}; then N + K = N 0 K . If N 0 K # M , then some Ma, a E A\B, is not contained in N 0 K , and consequently ( N 0 K ) n Ma = 0 since M, is simple. But then N n (Ma + K ) = 0 (Why?), contradicting the maximality of { M b : bE B } . Thus M = N 0 K and N is a direct summand of M .

I

175

Preliminaries

(iii) 3 (i) By Zorn’s Lemma there is a submodule P of M that is maximal with respect to being a direct sum of simple submodules of M (conceivably P = 0). If P # M write M = P 0N and choose x E N , x # 0. By another application of Zorn’s Lemma there is a submodule Lof N that is maximal with respect to x $ L. If we write M = L 0K and set Q = N n K then N = L 0 Q by Proposition 1.3. Let us show that Q is simple. Clearly Q # 0, or else N = L, contradicting x E N \ L. If Q is not simple it has a proper nonzero submodule Q1, and we may write Q = Q1 O Q, by the corollary to Proposition 1.3. But now the element x cannot be in both L O Q , and L O Q , , since ( L @ Q , ) n ( L O Q , ) = L a n d . u $ L . T h u s o n e o f L @ Q , and L O Q , would violate the maximality of L, and so Q is simple. (We have incidentally shown that M has simple submodules, and hence P # 0.) But now the submodule P 0Q violates the maximality of P, so P = M and the proof is complete. Question: Where was the hypothesis used that Rx # 0 if 0 # x

E

M?

Corollary. If M is a semisimple module then submodules and homomorphic images of M are also semisimple. Proof. For submodules apply the corollary to Proposition 1.3. If K is a homomorphic image of M , say K = f ( M ) , set N = kerf. We may write M = N 0L, and then K % L, which is semisimple, so K is semisimple.

Proposition 1.5. Suppose M is a simple R-module and I is a left ideal in R that is not contained in A ( M ) .Then Ix = M for some x E M . If I is a two-sided ideal not contained in A ( M ) ,then l y = M for all nonzero y E M . Proof. Choose x E M such that I x # 0. Then Ix = M since Ix is a submodule. If I is two-sided and 0 # y E M then l y is a submodule so l y = 0 or l y = M . If l y = 0 set N = { z E M : l z = 03. Then N is a submodule and N # 0 since x E N , so N = M and hence I G A ( M ) , a contradiction. Thus l y =M.

Corollary. If M is simple, then Rx

=

M for all nonzero x E M .

A left ideal I in a ring R is called modular if there is an element e E R such that r - re E I for all r E R. Equivalently, r I = ( r l ) ( e I) for all r E R. We call e a right relatioe unit for the modular left ideal R. Similarly a right ideal J is called modular if there is a corresponding left relative unit. Note that if R is a ring with 1 then every proper ideal is modular, with e = 1. Note also that if I is a modular (left or right) ideal with relative unit e and I # R, then e $ l .

+

+

+

Proposition 1.6. An R-module M is simple if and only if M z R/I for some maximal modular left ideal I of R . Proof. *: Choose x E M , x # 0, and define f : R -+ M by setting f ( r ) = rx. Then Im(f) = M by the corollary to Proposition 1.5, so M Z R/ker f by the FHT for modules. Set I = ker f and note that I , as a submodule

176

V Structure of Rings and Algebras

of R, is a left ideal. Note also that I is maximal since M is simple (see Proposition IV.1.3). Choose e E R such that f ( e ) = x, or ex = x. If r E R, then f ( r - re) = rx - rex = 0, so r - re E I , and I is modular. -=: We must show that M = R/I is a simple R-module if I is a maximal modular left ideal. If e is a relative right unit corresponding to I then e # I, and e(e

+ I) = e2 + I = e + (e2 - e) + I = e + I ,

so e 4 A ( M ) and A ( M ) # R. The only submodules of M = R/I are 0 and M since I is maximal (again see Proposition IV. 1.3), and so M is simple. Exercise 1.4. Show that every proper modular (left or right) ideal in a ring R is contained in a maximal modular (left or right) ideal.

Theorem 1.7 (Maschke). Suppose G is a finite group, F is a field, and either char F = 0 or char F = p with p $ IG(.If R is the group algebra FG and M is a unitary R-module with finite F-dimension then M is semisimple. Proof. Suppose N # 0 is a submodule of M . Choose a vector subspace I/ of M such that M = N 0 V (as a vector space over F). Let ,f:M -,N be the projection of M onto N along I/, i.e., if x E M and x = y z, with y E N and z E I/ then f ( x ) = y . Define g : M -, M by means of

+

g(m) = IGI-'C{xf(x-'m):x E C} for all m E M , and observe that g is an F-linear transformation from M to M . If y E G, then yg(y-'m)

=

IGI-'C(yxf((yx)-'m):x

E

G) = g(m),

all m E M , since y x ranges over all of G as x ranges over G with y fixed. We may conclude that g E CR(M),since G is a basis for R = FG. If m E M and x E G then xf(x-'rn ) E xN c N , so g ( N ) 5 N . If n E N and x E G,thenx-'n E N , s o f ( x - ' n ) = x-'n,and thereforexf(x-'n) = x x - l n = n, so y(n) = n, and therefore g 2 = g. If we set K = (I - g ) M , then K is an R-module since I - g E CR(M). If m E N n K , then m = g(m), and also m = ( I - g)(m') for some m' E M , and hence m = g ( l - g)(m')= ( g - g 2 ) ( m ' )= 0. Thus M = N 0 K , and M is semisimple by Proposition 1.4(iii). Maschke's theorem is of considerable importance in the representation theory of finite groups. 2. THE JACOBSON RADICAL

If R is a ring, then the Jacobson radical of R is the ideal J(R) = n { l : I= A ( M ) for some simple R-module M } .

2 The Jacobson Radical

177

If there are no simple R- modules we define J ( R ) = R and say that R is a radical ring. If J ( R ) = 0 we say that R is a semisimple ring. A rather basic point of view in the structure theory of rings is that semisimple rings are “manageable,” and hence that the Jacobson radical J ( R ) is a measure of deviation from manageability. It will be convenient to have several alternate descriptions of J(R). If I is a left ideal in a ring R set ( I : R ) = { r E R : r R c I ) . This is sometimes called the quotient of I by R . It should not, of course, be confused with the notion of a quotient ring. Proposition 2.1. If I is a left ideal in a ring R and M is the R-module R / I , then A , ( M ) = ( I : R), and in particular ( I :R ) is an ideal in R . Proof. If r E R we have r E A ( M ) if and only if r ( R / I )= I if and only if rR E I if and only if r E ( I : R ) . Exercise 2.1. Show that if I is a modular left ideal in a ring R then ( I : R) c I , and if J is a two-sided ideal in R with J C I then J E ( I : R).

Theorem 2.2. If R is not a radical ring, then J ( R ) is the intersection of all quotients ( I :R ) of maximal modular left ideals I in R . Proof. Propositions 1.6 and 2.1.

If R is a ring define a new binary operation * on R by setting r * s = r + s rs for all r, s E R . An element s E R is called leji quasi-regular if there exists r E R such that r * s = 0, in which case r is called a left adverse for s; r is then right quasi-regular and s is a right adverse for r. An element of R that is both left and right quasi-regular is said to be quasi-regular. A left (right) ideal in R is called left (right)-quasi-regularif each of its elements is left (right) quasiregular. If R is a ring with 1 note that (1 + r)(1 + s) = 1 if and only if r + s + rs = 0, and hence s E R is left (right) quasi-regular if and only if 1 s has a left (right) inverse. Thus s is quasi-regular if and only if 1 + s E U(R).

+

+

Exercise 2.2. (1) Show that ( R ,*) is a semigroup with identity 0. (2) If R,, denotes the set of quasi-regular elements in the ring R , show that (Rqr,*) is a group. Proposition 2.3. Let J denote the intersection of all the maximal modular left ideals in a ring R, or J = R if R has no maximal modular left ideals. Then J is left quasi-regular. Proof. If s E J set I ‘ = { r + rs:r E R ) . Then I ‘ is a modular left ideal with right relative unit e = - s. If I ’ # R, then I ’ is contained in a maximal modular left ideal I (see Exercise 1.4), and s E J G I , so - rs E I , for all r E R . But then r = - r s + (r + rs) E I for all r E R, a contradiction. Thus I’ = R, and in particular there is an element r E R such that --s = r + rs, i.e., r * s = 0.

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V Structure of Rings and Algebras

Theorem 2.4. If R is not a radical ring, then J ( R ) is the intersection of all the maximal modular left ideals I in R . Proof. As in the proof of Proposition 2.3 let J be the intersection of all the maximal modular left ideals I of R . Since ( I :R ) c I for each I we have J ( R ) E J by Theorem 2.2. If r E R\J(R), then rM # 0 for some simple R-module M , and hence rx # 0 for some x E M . But then I = {s E R : s x = 0 } is a maximal modular left ideal in R (see the proof of Proposition 1.6), and r $ I so r $ J . Thus J ( R ) = J .

Corollary. J ( R ) is left quasi-regular. Theorem 2.5. If I is any left quasi-regular left ideal in R then I E J ( R ) . Consequently J ( R ) is the unique maximal element (with respect to set inclusion) in the set of all left quasi-regular ideals in R . Proof. If not, then I M # 0 for some simple R-module M , and hence I x = M for some x E M by Proposition 1.5. Choose s E I such that sx = - x , and let r E R be a left adverse for s. Then

0 = (r * s)x = (r + s + rs)x = rx

+ sx + rsx = rx - x - rx = - x ,

and hence x = 0, a contradiction since M # 0. Note that if s E J ( R ) then r * s = 0 for some r E R, and then r = - s rs E J(R). Thus r is also left quasi-regular, so in fact r is quasi-regular with (unique) adverse s [see Exercise 2.2(2)], and the same applies to s. Thus in fact J ( R ) is a quasi-regular ideal in R, and ( J ( R ) ,*) is a group. Furthermore, J ( R ) is the unique largest ideal of R with that property, by Theorem 2.5. If I and J are left ideals in R denote by I J the left ideal in R generated by all elements ab, a E I , b E J. Thus IJ is the set of all finite sums aibi,where 1 5 i E E , ai E I , and bi E J . Thus IJ c J . Completely analogous definitions apply if I and J are right ideals (then IJ e I ) , or I and J are two-sided ideals (then I J G I n J ) . In particular we may define I' c_ I , and by induction I" c I for any ideal I (left, right, or two sided). An element r in a ring R is called nilpotent if rk = 0 for some integer k 2 1. For example

is a nilpotent element in the ring of 2 x 2 matrices over Z, since r'

= 0.

Exercise 2.3. If rand s are nilpotent elements of a ring R and rs = sr show that r + s is nilpotent. Show by example that the result may fail if rs # sr. If every element of a left (or right) ideal I is nilpotent, then I is called a nil left (or right) ideal. For example, the principal ideal ( 2 ) is a nil ideal in Z, .

2 The Jacobson Radical

179

A left (or right) ideal I in R is called a nilpotent left (or right) ideal if I" = 0 for some integer n 2 1.

Proposition 2.6. A nilpotent ideal I (left, right, or two-sided) is nil. Proof. If I" = 0, then a l a 2. - . a , = 0 for all choices of a, E I , and in particular a" = 0 for all a E I . Proposition 2.7. If r is a nilpotent element in a ring R, then r is quasiregular. Consequently every left, right, and two-sided nil ideal in R is contained in J ( R ) . Proof. If r" = 0 set s = - r + r 2 - r 3 * . . k I n - ' and verify that s is both a left and right adverse for r. Apply Theorem 2.5.

+

Theorem 2.8. If R is a ring, then J ( R ) consists of the set of all eiements s E R such that rs is left quasi-regular for all r E R . Proof. Set J

=

{s E R : r s is left quasi-regular for all r E R } .

Then J ( R ) E J is clear since J ( R ) is a left quasi-regular ideal. If J # J ( R ) choose r E J\J(R) such that rM # 0 for some simple R-module M . Choose x E M such that rx # 0. Then Rrx = M by Proposition 1.4, and we may choose a E R for which arx = - x . Then ar has a left adverse b and we have 0 = 0 . x = (b * ar)x

= (b

+ ar + bar)x = bx - x - bx = - x ,

a contradiction.

Theorem 2.9. If R is any ring, then J ( J ( R ) )= J(R), i.e., J ( R ) is a radical ring. Proof. J ( J ( R ) )E J ( R ) by definition. If s E J ( R ) , then rs is left quasiregular for all r E J ( R ) by Theorem 2.8, and the left adverse of rs is in J ( R ) [recall that ( J ( R ) ,*) is a group]. Thus s E J ( J ( R ) ) ,again by Theorem 2.8.

Theorem 2.10. If R is a ring, then J(R/1(R)) ring.

= 0, so R/1(R) is a semisimple

Proof. We may assume that J ( R ) # R. If M is any simple R-module, then J(R) G A ( M ) , so M is also a simple R/J(R)-module. Conversely if M is a simple R/J(R)-module,then M is also an R-module with the action defined by rx = [ r + J ( R ) ] x ,and as such M is a simple R-module. The annihilator in R / J ( R ) of M is just A , ( M ) / J ( R ) , and J ( R / J ( R ) )= ~ { A , ( M ) / J ( R ) : simple} M = ( n { A R ( M )M : simple))/J(R) = J(R)/J(R) = 0 in R/J(R).

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V Structure of Rings and Algebras

Theorem 2.11.

If R is a ring and I is an ideal in R, then J ( 1 ) = I n J(R).

Proof. If s E I n J ( R )and r E I , then rs is left quasi-regular (in R). If a E R is a left adverse, then a = -rs - ars E I , so rs is in fact left quasi-regular in I . Thus I n J ( R ) E J ( 1 ) by Theorem 2.8. Suppose then that r E J ( I ) , and consider the left ideal Rr. Note that (Rr)2 = RrRr E R l r E Ir, and that Iris a left quasi-regular left ideal in I , since r E J ( I ) . Thus (Rr)2is a left quasi-regular left ideal in R, and hence (Rr)2 G J(R). Let us show that in fact Rr E J(R). If not consider [Rr + J ( R ) ] / J ( R ) ;it is a nonzero left ideal in R / J ( R ) whose square is 0. Since [Rr + J ( R ) ] / J ( R )is nilpotent it is nil, by Proposition 2.6; hence it is contained in J ( R / J ( R ) )by Proposition 2.7. That is a contradiction, since R / J ( R ) is semisimple by Theorem 2.10. Thus Rr E J(R), so sr is left quasi-regular for every s E R , and hence r E J ( R )by Theorem 2.8. The proof is complete. Corollary 1. If R is a radical ring and I is an ideal in R, then I is a radical ring. Corollary 2. If R is a semisimple ring and I is an ideal in R, then I is a semisimple ring. For example, let R = Z,, and suppose n = p;*p;'...pF is the prime factorization of n. Set m = p 1 p 2 . . pk. Since Z,= Z/(n) the ideals of Z,all have the form (a)/(n),where (a)is a principal ideal in Zwith (a) 2 (n), i.e., a I n in Z. Clearly then the maximal ideals in h,are (pl)/(n), .. .,(pk)/(n),and hence

~(h,) = n { ( p i ) / ( n ) : i I i I k} = ( n { ( p i ) : l Ii Ik } ) / ( n )= (m)/(n). Consequently

Z"/J(Z,) = ( W w ( ( m ) / ( n ) ) GI by the Freshman Theorem for rings. We conclude that Z,is semisimple if and only if n is square-free. 3. THE DENSITY THEOREM

A ring R is called simple if R 2 # 0 and the only (two-sided) ideals in R are 0 and R . Note that R 2 = 0 for a ring R if and only if rs = 0 for all r and s in R, i.e., the multiplication is trivial. Every additive subgroup of R is an ideal when the multiplication is trivial. Thus the restriction R 2 # 0 simply rules out the additive groups h,,p prime, with trivial multiplication, as simple rings. Note that R 2 = R for a simple ring, since R 2 is an ideal. Clearly every simple ring must be either semisimple or radical, since J ( R ) is an ideal. An example of a radical simple ring has been given by Sasiada and Cohn [33].

3 The Density Theorem

18 1

Suppose D is a division ring and let R = MJD), the set of all n x n matrices with entries from D. With the usual matrix operations R is a ring and a unitary left D-module, even when D is not commutative. If we denote by Eij E R the matrix having 1 E D as its ijth entry and 0s elsewhere it is easy to see that R has { E i j }as a D-basis, so R is a free D-module of rank n2, i.e., a (left) D-vector space of dimension n2. If D is a field, then R is a D-algebra. Let us see that R = M,(D) is a simple ring. Clearly R 2 # 0, since R is a ring with 1. Suppose J # 0 is an ideal in R. If 0 # A E J write A = (aij)and suppose in particular that aijis a nonzero entry. Then EiiAEjj = aijEijE J , and hence a,; '(aijEij)= Eij E J . But then for any k and rn we have EkiEijEj, = E,, E J , and so J = R. A ring R is called primitive if there exists a faithful simple R-module M . An ideal P in a ring R is called a primitive ideal if R I P is a primitive ring. Observe that a primitive ring R is semisimple, for if M is a faithful simple R-module, A ( M ) = 0 and hence J ( R ) = 0 since J(R) c A ( M ) . Proposition 3.1. If a ring R is both simple and semisimple, then R is primitive.

Proof. Since R is semisimple there is a simple R-module M and the ideal A ( M ) is not all of R. Thus A ( M ) = 0 since R is simple, and so M is faithful. Proposition 3.2. An ideal P in a ring R is primitive if and only if P = ( I : R) for some maximal modular left ideal I of R, hence if and only if P = A ( M )for some simple R-module M . Proof. The two conclusions are equivalent by Propositions 1.6 and 2.1. If Pis a primitive ideal let M be a faithful simple RIP-module. Then M is a simple R-module, with P c A,(M). In fact P = A,(M), for otherwise M would not be faithful as an RIP-module. Conversely, if P = A ( M ) for some simple Rmodule M then M is a faithful simple R/P-module, so P is a primitive ideal.

Corollary. If R is not a radical ring then J(R) is the intersection of all the primitive ideals in R. Recall that if M is a simple R-module, then by Schur's Lemma D = C ( M )is a division ring. We may endow M with the structure of a unitary left D-module (i.e., a left vector space over D) by defining f . rn = f ( m )for all m E M , f E D. If r E R define T,: M --* M by setting T,(rn) = rm, all rn E M . Note then that if f E D = C ( M )we have

W m ) = r . f ( m ) = f(rm) = f . Urn), and so each T, is a D-linear transformation on M . The map r w T, is a ring homomorphism from R into Hom,(M, M ) ,with kernel A ( M ) .If R is primitive

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V

Structure of Rings and Algebras

and M is a faithful simple R-module, then A(M) = 0, so r H T, is a monomorphism and we may (and shall) view R as a subring of Hom,(M, M ) by identifying r with T,. Suppose M is a simple R-module and D = C(M). We say that R acts densely, or that R is dense, on M if given any finite D-linearly independent set { m , , .. . ,mk}c M, and any other elements n,, . . . ,nk (not necessarily distinct) in M, there is some r E R such that T,(mi)= rmi = ni, 1 I i I k. Exercise 3.1. Suppose M is a simple R-module and D = C(M). If dim,(M) is finite and R is dense on M show that r H T,is an epimorphism. If R is primitive, with A(M) = 0, conclude that R = Hom,(M, M). Proposition 3.3. Suppose M is a simple R-module, D = C(M), Vis a finitedimensional D-subspace of M, and m E M\V. Then there is an element r E A R ( V )such that rm # 0. Proof. Induction on n = dim(V). The result is true by the definition of simplicity if n = 0, so suppose n 0 and assume the result for subspaces of dimension n - 1. Choose a D-subspace W of V , with dim,(W) = n - 1, and choose u E V\W, so that V = W 0 Du. Suppose by way of contradiction that A( V ) m = 0. By induction there is some r E R such that rW = 0 but ru # 0, and hence A ( W ) u # 0. But A(W)u is a submodule of M, so A(W)u = M. Define f : M -+ M by setting f(su) = sm for all s E A( W). (Why is f well defined?)Note that if r E R, then

=-

f(r(su))= f((rs)u)= (rs)m = r(sm) = rf(su), so f E C(M) = D. If r E A ( W ) ,then f ( r u ) = rm = rf(u),so r(m - fu) = 0 for all r E A( W ) .It follows from the induction hypothesis that m - fu E W. But then m E W + fi c W + Do = V, a contradiction, and the proposition is proved.

Theorem 3.4 (The Jacobson-Chevalley Density Theorem). If M is a simple R-module, then R acts densely on M. In particular, if R is a primitive ring, M is a faithful simple R-module, and D = C(M), then R is (isomorphic with) a dense ring of D-linear transformations on M. Proof. Suppose { m , , .. .,mk} is a D-linearly independent subset of M and n,,..., n , E M . F o r e a c h i , l I i I k , s e t

6 = @ { D m , : i # j , 1 I j Ik}. Then mi 4 6 , so by Proposition 3.3 we may choose ri E A ( J 0 such that r,m, # 0. Consequently, by the corollary to Proposition 1.5, Rrim, = M, and so s,r,m, = n, for some s, E R (note, though, that siriK = 0, and hence s,rimj = 0 if i # j ) . If we set r = sjr, it follows that rmi = ni, 1 Ii I k, as desired.

,

4

Artinian Rings

183

Theorem 3.5. Suppose R is a primitive ring. Then there is a division ring D such that either (i) R is isomorphic with the ring 0, = M,,(D) of all n x n matrices with entries from D for some n, or (ii) there is an ascending chain R , G R, E R , E ... of subrings of R and for each k an isomorphism fk: R, -+ Dk = M,(D). Proof. Since R is primitive there is a faithful simple R-module M. Set D = C ( M ) ,a division ring by Schur’s Lemma. If dim,(M) = n, finite, then by the Density Theorem and Exercise 3.1 we see that R z Hom,(M,M), which is isomorphic (as a ring and as a D-vector space) with D,, by the usual argument from linear algebra. If dim,(M) is not finite let { m , , m , , . . . } be an infinite D-linearly independent subset of M , and for each k 2 1 set M, = 0 { D m i :1 5 i I k). Set R, = { r E R : r M , S Mk}for each k. It follows from the Density Theorem that every D-linear transformation T: M k4 M k has the form T = T, for some r E R with T,(m)= rm for all m E M,, and so rt+ T, is an isomorphism from R , to HOm,(M,, Mk),which is isomorphic with Dk. 4. ARTINIAN RINGS

A descending chain I, 2 I, 2 I , 2 . * . of left ideals in a ring R is said to terminate if it is finite or if there is some index k such that 1, = 1, for all n 2 k. If every descending chain of left ideals of R terminates we say that R satisfies the descending chain condition, or the DCC. A ring that satisfies the DCC is also called an Artinian ring. For example, R = E is not an Artinian ring since the chain (2) 2 (4) 3 (8) 2 (16) 2 ... does not terminate. On the other hand, if R = M,(D), the ring of n x n matrices over a division ring D, then any left ideal I of R is a D-subspace. If I, 2 I , 2 ... is a descending chain then dlmD(Ik+ I dim,,(Ik) I n2 for all k, so the chain must terminate. Thus R = M,,(D) is Artinian. The argument just given shows that any ring that is a finite-dimensional unitary F-algebra for some field F must be Artinian. In particular, every group algebra FG, for a finite group G, is Artinian. Strictly speaking, a ring satisfying the DCC for left ideals should be called leji Artinian, with a corresponding definition for right Artinian rings. The theory of right Artinian rings is completely analogous to the theory of left Artinian rings. It is not the case, however, that the two concepts are equivalent (see Exercise 5.32 for an example). A ring R is said to satisfy the minimal condition (for left ideals) if every nonempty set of left ideals of R contains a left ideal that is minimal with respect to set inclusion.

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V Structure of Rings and Algebras

Proposition 4.1. A ring R is Artinian if and only if it satisfies the minimal condition. Exercise 4.1. Prove Proposition 4.1. Theorem 4.2. If R is Artinian, then its Jacobson radical J ( R ) is nilpotent, and hence nil. Consequently J ( R ) is the unique largest nilpotent ideal in R . Proof. Write J ( R ) = J. The chain J 2 J 2 2 J 3 2 ... must terminate, so = J" for some positive integer n and all integers k 2 0. If J" # 0 let Y be the set of all left ideals I in R such that I E J" and J"I # 0. Then Y # 0since J" E 9, so there is a minimal element I, in 9 'by Proposition 4.1. Choose a E I , such that J"a # 0. Then J"a is a left ideal, J"a E J", and J" . J"a = J2"a # 0, so J"a E Y . Since J"a E I,, which is minimal in 9, we have J"a = I,. Choose b E J" for which ba = -a. If c is the adverse of b, then

Jn+k

0 =a =a

+ ba = a + ba + c(a + ba) + ( b + c + cb)a = a + (c*b)a = a,

a contradiction. Thus J" = 0 and J is nilpotent. For the second statement see Propositions 2.6 and 2.1.

Corollary. If R is Artinian, then every nil ideal in R is nilpotent. Proposition 4.3. If R is a simple Artinian ring, then R is semisimple. Proof. Since J ( R ) is nilpotent by Theorem 4.2 and R 2 = R we have J ( R ) # R. But J ( R )is an ideal, so J ( R ) = 0. Theorem 4.4 (Maschke). If G is a finite group and F is a field, with char(F),+'IGJ, then the group algebra FG is a semisimple ring. Proof. Since FG has F-dimension IGI and each left ideal is an F-subspace it is clear that FG is Artinian. Set J = J(FG),then J is nil by Theorem 4.2. If we view FG as an FG-module then it is a semisimple module by Theorem 1.7, and J is a submodule, so we may write FG = J 0 I for some left ideal I , by Proposition 1.4(iii).Write 1 = a b, with a E J , b E 1. Then a = a . 1 = a(a + b) = a2 + ab, and a - a2 = ab E J n I = 0, so a = a'. But then a can not be nilpotent unless a = 0, which says 1 = b E I , and hence I = FG, J = 0.

+

Theorem 4.5 (Wedderburn-Artin). following are equivalent:

If R is an Artinian ring, then the

(i) R is simple. (ii) R is primitive. (iii) R is isomorphic with the ring D,, = M,,(D) of all n x n matrices with entries from some division ring D for some positive integer n.

4 Artinian Rings

185

Proof. We saw that (iii) => (i) on page 181, and we know that (i) (ii) by Propositions 4.3 and 3.1. In order to show that (ii) *(iii) we may view R as a dense subring of Hom,(M, M ) by the Density Theorem, where M is a faithful simple R-module and D = C(M). By Theorem 3.5 (and its proof) it will be sufficient to show that dim,(M) is finite. If it is not we may choose an infinite D-linearly independent set { m , , m,, . . . } E M. For each positive integer k set Mk = O ( D m j : 1 I j I k } . Since R is dense on M we may choose rk E R, for each k, such that rkmi = 0 for 1 I i I k but rkmk+l# 0. Thus rk E A(Mk)\A(Mk+l),so A ( M , ) 2 A ( M , ) 2 ... is a descending chain of ideals in R that does not terminate. That is a contradiction since R is Artinian, and the theorem is proved.

Corollary. If R is Artinian, then every primitive ideal P in R is a maximal modular ideal and every maximal modular ideal is primitive. Thus J ( R ) is the intersection of all the maximal modular ideals of R (J(R) = R if there are no maximal modular ideals). Proof. The ring R I P is primitive and Artinian, so it is simple, and hence P is a maximal ideal, modular since RIP is a ring with 1. Apply the corollary to Proposition 3.2. Note that the word “modular” in the corollary to Theorem 4.5is necessary. For example, if R is the subring (0,2,4,6) of Z,,then f = {0,4}is the unique maximal ideal in R (it is not modular) but J ( R ) = R . A nonzero ideal f in a ring R is a minimal ideal if the only ideals J with 0 E J c f are J = 0 and J = f . Note that if 1 is a minimal ideal in R , then either I’ = 0 or I’ = I , and that in the latter case I is a simple ring. (Why?) For example if R = Z,,then I = (2)is a minimal ideal with f = 0; ifR = Z,,then I = (2) is a minimal ideal with f = I. Theorem 4.6 (Wedderburn-Artin). If R is a semisimple Artinian ring, then R has only finitely many minimal ideals f,,Z2,... , f k , and R = I , 0I , 0 .. . 0f k . Each minimal ideal fj is a simple Artinian ring, so lj 2 D::), the ring of nj x nj matrices over a division ring D”’. Proof. Since R is semisimple the intersection of all the primitive ideals in R is 0, by the corollary to Proposition 3.2. Let PI be a primitive ideal and choose (if possible) a primitive ideal P, different from P I .Then choose (again if possible) a primitive ideal P3 that does not contain PI n P,. Inductively choose (as long as possible) a primitive ideal Pn+,that does not contain 1 I i I n ) . The process must terminate in a finite number of steps since { 4: 1 I i I n}},“= would be a descending chain of ideals that otherwise does not terminate. Thus we obtain primitive ideals P , , ..., P, such that { pi: 1 Ii I m} = 0. Relabel if necessary and choose a set { P I , .. .,Pk} of primitive ideals such that 0 [ P i :1 I i I k } = 0 but no proper subset

n{c:

{n

186

V Structure of Rings and Algebras

n

intersects in 0. Set Ij = { 5 :1 I i I k, i # j } for 1 I j 5 k. Since 4 is a maximal ideal (Corollary, Theorem 4.5) and lj is not contained in 5 we have lj + 6 = R for each j , and also r j n 4 = n{S:1I i I k} = 0. Thus R = l j @ e for each j , and lj E R / 4 . Since 5 is maximal lj can have only 0 and Ij as ideals, and 1: # 0 since R is semisimple (see Theorem 4.2). It follows that each rj is a minimal ideal in R and is a simple Artinian ring. Set Qi = 0 { 5:1 I j I i), for 1 I i I k, and note that R = I , @ Q1.By the Isomorphism Theorem (11.1.6) we have Q1/(Q1

+

n 5)Z (Q1 p j ) / p j = RJ€$

if 2 Ij Ik,

so each Q, n 6 is a maximal ideal in Q l , 2 I j Ik. Furthermore,

n{Qln 5 : 2 I j I k} = n{e:15 j I k ) = 0, and no proper subset of { Q1n 5:2 I j I k} intersects in 0. As above we see that

n

Q , = ( { Ql

n 5:3 5 j

I k)) 0 (Q, n P2) = I , @

Q2,

and hence R = I, 0 I, 0 Q , . Inductively R = I, 0 I , 0 ... 0 lj Q Q j , and hence R = I, Q I , 0 . . * 0 1, since Qk = 0. Since lj is simple it is isomorphic with the ring Dli) of nj x nj matrices over a division ring D"' by Theorem 4.5. We note in passing that as a result each 4 is a ring with 1 and hence R is a ring with 1. If I is any minimal ideal in R, then R l = 1 and so I

= (0{ Ij } ) . I = @ {

4 . I :1 Ij I k } .

Each direct summand is an ideal in I so exactly one summand is nonzero and I = ljl for some j . But then I = lj since ljl is a nonzero ideal contained in both of the minimal ideals I and l j .Thus I , , . . . ,1, is the full set of all minimal ideals in R.

Corollary. A semisimple Artinian ring is a ring with 1. Theorem 4.7. Suppose F is an algebraically closed field and A is a finitedimensional F-algebra that is semisimple as a ring. Then A is isomorphic with F,,, 0 F , , , @ . . . @ F,,,,whereFn,istheringofnjx njmatriceswithentries from F. Proof. By Theorem 4.6 it will suffice to prove the theorem under the further assumption that A is simple, and in fact that A = On,where D = C ( M ) for some simple A-module M. By Proposition 1.2 we have D = F . 1 E F and soAz&.

The final step in describing the structure of semisimple Artinian rings would be a description of all division rings. Unfortunately no description is

4 Artininn Rings

187

known. The next theorem does serve as a complete description of all finite division rings, in view of Theorem 111.3.10.

Theorem 4.8 (Wedderburn). A finite division ring D is a field. Proof. (Witt [40]). Let F = {n . 1 :n E B,1 6 D). Then, as on page 80, F is a prime field, so F z Z, for some prime p . If dim,(D) = r, then ID1 = p'. Let 2 = {x E D:xy = yx for all y E D), the center of D. Then Z is a field, F c Z , and Z* is the center Z(D*) of the group D*. Say IZI = p" = 4. If dim,(D) = t, then st = r. For each x E D define the centralizer of x in D to be C,(X) = { y E D:xy = yx}. Then C,(X) is a subdivision ring of D, Z E C,(x), and if x # 0, then C,(X)* = C,,(X) the group-theoretic centralizer. Set d , = dimzCD(x). Then d , 1 t (Proposition 111.1.1 applies even if D is not commutative) and if x # 0, then the number of elements in the conjugacy class of x in D* is

[D*: c&(x)] = (4' - l)/(qdx- 1) by Proposition 1.2.3. Suppose {xl,.. . ,xk} is a set of representatives for the conjugacy classes of noncentral elements in D*. Then by the class equation (page 14) we have q' - 1 = 4 - 1 +

I{(4' - I)/(qdxl

-

1): 1 5 i

k). Set f(x) = x' - 1 and let g(x) = @,(x)be the tth cyclotomic polynomial. Then g(x) E Z[x], g(x) 1 ffx), and if o E C is a primitive tth root of unity, then g ( x ) = n { ( x - o j ) : l s j s t a n d ( j , t ) = 1) (see page 97). If d (t and d < t , then xd - 1 1 x' - 1 (by high school algebra) and (xd - 1,g(x)) = 1 since the roots of xd - 1 are not primitive tth roots of unity, so we may write x' - 1 = (xd - l)g(x)h(x),and h(x) E B[x] by the division algorithm. Substitutingx = q we see that (4' - l)/(qd - 1)is in Z.Observe also that - d ) : l Ij g(q) = n((4

s t and ( j , t ) = l ) ,

and that 1q - oil > 4 - 1 (see Fig. l), and hence Ig(q)1 > q - 1.

FIGURE 1

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V Structure of Rings and Algebras

Suppose now that t > 1. Then g(q))(q' - l)/(qdxe- l), 1 5 i I k, and g(q) q' - 1, so by the class equation g(q) q - 1. That is in conflict with the fact that Ig(q)1 > q - 1, so t = 1 and D = 2 is a field.

I

I

If F is an algebraic field extension of the real field R, then either F = R or F z C by the Fundamental Theorem of Algebra (Theorem 111.3.10). We are familiar with the division ring W of quaternions which has R as a subfield (in fact R is the center of W), with dim#) = 4. If a E W then left multiplication by a determines an R-linear transformation T of W.If f ( x ) E R [ x ] , then f ( T )is left multiplication by f ( a ) , so a is a root in W of the minimal polynomial of T over R, and hence a is algebraic over R. Consequently W is algebraic as a ring over R. The next theorem tells us that there are no more such examples. Theorem 4.9 (Frobenius). Suppose D is a division ring containing R in its center, with D algebraic over R. Then D is isomorphic either with R, 6,or W. Proof (Palais [29]). As noted above it is sufficient, by the Fundamental Theorem of Algebra, to assume that D is not commutative. Choose d E D\R and set K = R + Rd, a two-dimensional R-subspace of D. Note that elements of K commute with one another when multiplied, since R is central. Choose a subspace F of D that contains K and is maximal with respect to the property that its elements commute when multiplied. By the maximality of F any a E D that commutes with all elements of F must lie in F , for otherwise F + Ra would be a larger commuting subspace. In particular, if a,b E F , then ab E F , and if a # 0, then a-' E F, so F is a field. But then F is an algebraic extension of R, and F # R, so F E C,and in fact F = K = R Rd. Since F z 6 wemay choose i E F such that i2 = - 1. Then F = R + Ri, and we may identify F with C. Thus we may also view D as a (left) vector space over @. Define a C-linear transformation T: D + D by setting T(a)= ai for all a E D. Then T 2 = - I , but T # k I , so T has minimal polynomial x 2 + 1 and eigenvalues i and -i. Write D and D - for the corresponding eigenspaces, so D + = { a E D : a i = ia} and D - = { a E D : a i = - i a } . Clearly C E D + and, by the maximality of F = C, D + c C,so D+ = C. If a E D, then

+

+

a = (a - iai)/2

+ ( a + iai)/2 E D + + D -

since

(a - iai)i = ai and (a

+ ia = i( - iai + a)

+ iaifi = ai - ia = -i(iai + a).

But also D + n D - = 0, so D = D+ 0 D - . Note that, by the definition of D , ab E D+ if a, b E D-.Choose a nonzero element b E D - and define another Clinear transformation S : D D by setting S(a) = ab for all a E D. Then S is --f

5

Further Exercises

189

invertible (its inverse is right multiplication by b - l ) , and S ( D - ) = D'. Thus dim,(D-) = dim,(D') = 1, hence dim,(D) = 2 and dim,(D) = 4. Since 0 # b E D - we see by the observations made at the beginning of the proof that R + Rb is a field isomorphic with @. Note that

b2 E D+ n (R + Rb) = C n (R + Rb)

=

R.

If b2 were positive it would have two square roots in R, but also the two square roots b in Rb, hence x2 - b2 would have four roots in the field R + Rb, contradicting Proposition 111.1.7.Thus b2 < 0 in R.Replace b by j = b / J f l J to obtain j E D - with j 2 = - 1 and j i = -ij, and set k = ij. Then k E D - and k is not an R-multiple ofj, so { j , k } is an R-basis for D - . Thus { 1, i, j , k } is an Rbasis for D. The usual further facts about multiplication of the basis elements i, j , and k in the ring W of quaternions follow easily, so D z W.

5.

FURTHER EXERCISES

1. Suppose F is a field and let R = M , ( F ) , the ring of 2 x 2 matrices over F. Exhibit a left ideal in R that is not a semisimple ring. 2. If R and S are rings and f : R + S is a homomorphism show that f ( J ( R ) )c J ( f ( R ) ) . 3. If {R,:cIE A } is any nonempty family of rings show that (i) .!(OR,) = OJ(R,), and (ii) R,) = J(R,). Conclude that OR, is semisimple if and only if each R , is semisimple, and likewise for R , . 4. Suppose F is a field and f ( x )E F [ x ] , and set R = F [ x ] / ( f ( x ) ) . Determine J ( R ) (see the example on p. 180). Under what conditions is R semisimple? 5. If F is a field let R be the ring of n x n upper triangular matrices with entries from F , i.e., if A = (aij)E R , then aij = 0 if i < j. Find J ( R ) and show that R / J ( R )is commutative (in fact R / J ( R )z F"). 6. If R is any ring show that J ( M , ( R ) ) = M,(J(R)). 7. An element a = c { u x. x:x E G ) in a group algebra FG is called a class function (cf) if it is constant on each conjugacy class in G, i.e., a( y - ' x y ) = a ( x ) for all x, y E G. Write cf(G) for the set of all class functions on G.

J(n n n

(i) Show that cf(G) is the center of FG. (ii) For each conjugacy class K in G define + K E FG by setting 4 K ( = ~ )1 if x E K,4 K ( = ~ 0) otherwise, i.e., 4K= .x:x E K ). Show that the elements 4 K

c{

190

V Structure of Rings and Algebras

constitute a basis for cf(G). Conclude that the dimension of the center of F G is the class number of G.

8. Suppose R is a commutative ring and I is an ideal in R. Define $to be { a E R: ak E I for some integer k 2 0} (this is sometimes called the radical of the

ideal I ) .

(i) Show that f l is an ideal in R . (ii) If I and J are ideals in R show that (iii) Show that =fl+

Jm

g.

f l = Jm= f l n $,

9. Suppose G is a finite group, F is a field, and K an extension field. Show that the K-algebras K G and K O FFG are K-isomorphic.

10 (RieffelC321). Suppose R is a simple ring with 1 and I is a nonzero left ideal in R, viewed as a left R-module. Write R' for C&). Then R' is a ring and I is also a left R'-module. Set R" = C,.(I) and define 8: R -+ End(1) by means of 8,(x) = rx, all r E R, x E I. Also define $: I -+ End(1) by means of $x(y) = YX, all x, Y E I . (i) Show that 8, E R", all r E R, and $x E R', all x E I . (ii) Show that B is a (ring) monomorphism. (iii) If @ E R" and x E I show that #dx= 8+,(x),and conclude that B(1) = (0,:E ~ I } is a left ideal in R". (iv) Use the fact that I R = I to show that Zm(8) is a left ideal in R", and note that lx,, E Im(8). (v) Conclude that R" and R are isomorphic rings (this is a version of the "Double Centralizer" property for simple rings). 1 1. Suppose p E Z is a prime dividing the order of a finite group G and F is a field of characteristic p . Show that F G is not semisimple. (Hint: Show that a = c { x : x E G} is a nonzero nilpotent element in the center of FG and conclude that a E J(FG).)

12. Suppose R is a ring with 1 and that I that J(R) = I .

=

R \ U ( R ) is an ideal in R. Show

13. Suppose that R is a radical ring that is nilpotent (i.e., R" = 0 for some k, 0 < k E E). Show inductively that R' contains the ith term in the descending central series of the group (R,*),and conclude that (R,*) is a nilpotent group. 14. Show that a primitive commutative ring is a field.

15. Suppose V # 0 is a (left) vector space over a division ring D,and R is a ring of D-linear transformations of V. Suppose further that R is transitive on V ,meaning that if u, u E V , with u, # 0, then ru = u for some r E R . Show that R is a primitive ring, with M = V as a faithful simple R-module.

5 Further Exercises

191

16. Let R be the ring of all matrices of the form

a, b E Q, and let M be the R-module Q 2 of column vectors.

(i) Show that R is a division ring (in fact a field). (ii) Show that M is a transitive R-module (see Exercise 15). (iii) Show that M is a simple R-module. (iv) Show that C ( M )z R. 17. Suppose V is a (left)vector space over a division ring D, and R is a doubly transitiue ring of D-linear transformations of V , meaning that if {ul, u2> is Dlinearly independent in V and { u 1 , u 2 ) G V, then rul = ul and ru2 = u2 for some r E R. Show that R is a primitive ring and that C( V) z D. Conclude that R acts densely on V . 18. Suppose M is a simple R-module and D = C ( M ) . Endow M with the discrete topology, the space M M of all functions from M to M with the product topology, and Hom,(M, M) with the topology it inherits as a subspace of MM. Show that R acts densely on M if and only if R is topologically dense when viewed as a subring of Hom,(M, M). 19. If R is an Artinian ring and f : R -+ S is a ring epimorphism show that S is also Artinian. 20. If R is an integral domain with 1 and R is Artinian show that R is a field. 21. If R is acommutative ring with 1 and R is Artinian show that every prime ideal of R is maximal. (You may wish to apply Exercises 19 and 20.) 22. Suppose R is an Artinian ring with 1 and M is a unitary left R-module. Show that M is semisimple if and only if J(R)M = 0. 23. Describe all semisimple rings having 1296 elements. 24. If p and q are distinct primes describe all semisimple rings having the following numbers of elements: (a) p4, (b) p4q4, (c) pl0, (d) p 8 q 5 . 25. If V is a finite-dimensional vector space over a field F and T: V 4 Vis a linear transformation show that the ring F [ T ]of all polynomials (over F) in T is isomorphic with F [ x ] / ( m T ( x ) )What . is J ( F [ T ] ) ? Determine when F [ T ] is simple and when it is semisimple. (See Exercise 4, above.) 26 (Wedderburn). Suppose that A is a finite-dimensional algebra over a field F and that A has an F-basis consisting of nilpotent elements. Show that A is nilpotent. (Show first that F can be assumed algebraically closed by considering A K = K OFA, K an algebraic closure of F. Then suppose the result false and use induction on dim,(A). Argue that A is semisimple, then that A is simple, hence a matrix algebra. Then use the basis of nilpotent elements to see that all elements have trace 0 and arrive at a contradiction.)

192

V Structure of Rings and Algebras

27. (Jennings [18]). Suppose G is a finite p-group and F is a field of characteristic p. Show that J = J ( F G )has {x - 1 : 1 # x E G } as an F-basis, so dim,(J) = IGI - 1. (Hint: Let W be the F-subspace spanned by all x - 1. Show that W is an ideal and each x - 1 is nilpotent. Apply Exercise 26.) 28. If G is a finite p-group and R = J(Z,G) show that the map x H 1 - x is an isomorphism between G and a subgroup of ( R ,*) (see Exercise 27). 29. Show that a finite nilpotent group is isomorphic with a subgroup of ( R ,*) for some finite nilpotent radical ring R (see Exercises 13 and 28). 30. Let F = Z,,let GI be cyclic of order 4, and let C2 be Klein’s 4-group. Show that the group algebras FG, and FG, are not isomorphic. (Hint: Investigate the “degree of nilpotence” of their Jacobson radicals. See Exercise 27.) 31. If GI is cyclic of order 4 and G2 is Klein’s 4-group show that QG, and QG, are isomorphic Q-algebras. 32. Let

be a fixed nonzero column vector with entries from Q and let R denote the additive group of all row vectors a = [ a , , a,] with entries from Q, endowed with a multiplication as follows. If a = [ a , , ~ , ] , b = [b,,bz] E R , then a b = amb (ordinary matrix multiplication). Note that if A(a) is defined to be alml a2m2 then a b = A(a)b (ordinary scalar multiplication). Set n = [-m,,m,] E R and let N = (n), the principal ideal generated by n. 0

+

0

(i) Show that R is a ring having left (multiplicative)identities but no right identity. (ii) Show that every left ideal of R is a Q-vector subspace, and conclude that R is (left) Artinian. (iii) Show that R does not satisfy the DCC for right ideals (consider additive subgroups of N ) . (iv) If e = [ e , , e , ] E R is chosen so that A(e) = 1 show that e is a right relative unit for the modular left ideal N . (v) Show that 0, N and R are the only (two-sided) ideals in R . Conclude that N = J ( R ) . 33. Suppose F is an algebraically closed field and D is a division ring that is algebraic over F , with F contained in the center of D. Show that D = F . 34. Suppose F is an algebraically closed field and A is a finite-dimensional semisimple F-algebra. Show that the dimension of the center of A is equal to the number of simple direct summands of A .

5 Further Exercises

193

35. Suppose R is an Artinian ring with 1 whose set of left ideals is linearly ordered, i.e., if I , and I , are left ideals then either I , c I , or I , G I , . (For an example see Exercise 11.8.34.) Write J for J ( R ) . (i) Show that J is a maximal ideal and a maximal left ideal in R . (ii) Show that R / J is a division ring. (iii) Show that every proper left ideal in R is of the form J" for some n > 0 in Z,and conclude that every left ideal is an ideal. 36. If I is a minimal ideal in a ring R and I 2 # 0 show that I is a simple ring.

Further Topics

Chapter VI

1. INFINITE ABELIAN GROUPS All groups in this section will be abelian; they will be written additively unless it is explicitly noted otherwise. Recall from Section 3 of Chapter IV that an abelian group A has a torsion subgroup T = T ( A )consisting of all of its elements of finite order, and that A / T is torsion free. If T is a torsion abelian group and p E Z is prime let Tp be the set of all elements in T whose orders are powers of p , Tp= { x E T : l x l = pk, some k}.

Clearly Tp 5.T. If 0 # x E T, then ( x ) is the direct sum of its Sylow subgroups, and in particular x can be written uniquely as x = x 1 + x2 + ... + xk, with 0 # x i E q,for various distinct primes p l , . . ., p k . An application of Exercise IV.2.2 yields the following theorem.

Theorem 1.1 (Primary Decomposition). If T is an abelian torsion group and T, = { x E T : l x l = p k , somek}

for each prime p

E

Z,then T = 0,Tp.

For an important example take T = Q/Z. Then T,

=

+

{ ( a / p b ) Z:a,b E Z,b 2 O}.

It is easy to see that we may choose representative elements of the cosets (alpb)+ Z of the form c/pb, where 0 Ic < pb, b 2 0, and ( p ,c) = 1 . If we let 194

1 Infinite Abetian Groups

195

,Z, denote that set of representative elements and define addition in E,, to be ordinary addition modulo 1, then clearly T, z H,-.

Exercise 1.1. (a) If p is a prime let E, denote the multiplicative group of all pkth roots of unity in @, for various k, i.e.,

E P = (e2"iJ/pk E @:O Ik

E h,j E

E).

Show that E, E ZPm. (b) Show that every proper subgroup of H,, is cyclic and that its subgroups are linearly ordered by inclusion. In fact the proper subgroups may m , =VAL. be listed as 0 = A, IA , IA, 5 ..', with lAkl = p k , and Z Recall from Chapter IV, Section 2, that an abelian group F is free if it is isomorphic with a direct sum of copies of the additive group Z,or equivalently if F has a basis (as a H-module). The cardinality of any basis is called the rank of F. Any function from a basis for F to an abelian group A extends uniquely to a homomorphism from F to A. As a consequence every abelian group A is a homomorphic image of a free abelian group. An abelian group G is called projective if for every exact sequence A 5 B -+ 0 of abelian groups and every g E Hom(G, B ) there is some h E Hom(G, A) for which the diagram

is commutative, i.e., g = fh.

Theorem 1.2. An abelian group G is projective if and only if it is free. Proof. e: Suppose G is free with S as a basis. If A B -+ 0 is exact and g E Hom(G,B) choose a, E A for each s E S such that f(a,) = g(s). Then the

function h: S + A defined by h(s) = a, extends to h E Hom(G, A), and f h = 9, so G is projective. a: There is a free abelian group F and an epimorphism 8: F + 6. If K = kere there is an isomorphism 9:G + F / K . If q: F + F / K is the quotient map, then, since G is projective, there is some h E Hom(G, F) such that the diagram

196

VI Further Topics

is commutative, i.e., g = qh. But then h is 1-1 since g is 1-1, so G is isomorphic with a subgroup of F , and hence G is free by Theorem IV.2.9. An abelian group A is called dioisible if nA = A for every nonzero integer n. It is of course sufficient to require that p A = A for every prime p. If A is divisible, x E A , and 0 # n E Z,then n y = x for some y E A , and we may think of y as the result of “dividing x by n.” The “quotient” y is not necessarily uniquely determined, however. Perhaps the simplest example of a divisible group is the additive group Q of rational numbers. Proposition 1.3. Suppose A is a divisible group and f : A epimorphism. Then B is divisible.

--*

B is an

Proof. If 0 # n E E,then nB = n f ( A ) = f ( n A ) = f(A) = B. As an immediate consequence of Proposition 1.3 we see that Q/Z is divisible, and then that Z,, is divisible. Proposition 1.4. If A is torsion free and divisible, then A each A , is isomorphic with Q.

=

where

Proof. If 0 # n E H and x E A , then x = ny for some y E A . In this case y is unique, for if also nz = x , z E A, then n(z - y) = 0 and hence z = y since A is torsion-free. Thus we may write y = (l/n)x. But then for any rn/n E Q we may write (rn/n)x = rn . (l/n)x, and it is easy to verify that A is consequently a Qvector space. As such it has a basis and the proposition is proved.

An abelian group G is called injective if for any abelian groups A, B in an exact sequence 0 --* B A, and for any g E Hom(B, G), there is some h E Hom(A, G) such that g = hf, i.e., the diagram

0-B-A

I

is commutative. Note that this definition “dualizes” (i.e., reverses the arrows in) the definition of a projective abelian group. In particular, B might be a subgroup of A and f the inclusion mapping, in which case the requirement is that any homomorphism from B to G can be extended to the larger group A. Theorem 1.5. An abelian group G is injective if and only if it is divisible. Proof. =: If 0 # n E Z and x E G define a homomorphism g from B = ( n ) IZ = A to G by setting g(kn) = kx. Extend g to h E Hom(Z, G) and set

1

Infinite Abelian Groups

197

y = h(1). Then

ny

= nh( 1) =

h(n)= g(n) = x

and G is divisible. e: Assume C is divisible and suppose we wish to complete the diagram G 7

O-B---,A

/

Let Y = { ( C , j ) : jE Hom(C, G), f ( B ) I CI A , j f = g},

the collection of all “partial completions” of the diagram. Then Y #

0since

(fB,gf-’)E 9, and 9’ is partially ordered if we agree that (C,, j l )2 (Cz,j2) provided C, 2 C, and j , I C, = j,. By Zorn’s Lemma there is a maximal element ( D , h )E Y , and it will suffice to show that D = A . If not, choose x E A\D and set E = ( D , x ) = D + ( x ) . If D n ( x ) = 0 we may define h’ E Hom(E, C) by setting h’(d + m x ) = h(d), d E D, and check that ( E ,h’) > (D, h), violating maximality. If D n ( x ) # 0, choose k minimal and positive in E for which kx E D. Observe then that elements of E have unique representations as d mx, where d E D and 0 Im < k. Set a = h(kx) E C. Since G is divisible we may write a = kb for some b E G. Define h” E Hom(E, G ) by setting h”(d + mx) = g(d) + mb, d E D and 0 I m < k in Z. Again it is easy to verify that ( E , h”) > ( D ,h), violating maximality, so D = A .

+

Corollary. If A is an abelian group and D is a divisible subgroup of A , then D is a direct summand of A . Proof. If i : D + A denotes inclusion, then 0 + D +!- A is exact, and we have a commutative diagram

D

O--+D-,A for some h E Hom(A,D). Set K = ker h. If x E A, then x = h(x) + [x h(x)] E D + K since h2 = h. If y E D n K , then y = h ( y ) = 0, so A = D 0 K .

Exercise 1.2. Suppose D is an abelian group that is a direct summand of every abelian group A that contains D as a subgroup. Show that D is divisible.

198

VI Further Topics

An abelian group A is called reduced if it has no nonzero divisible subgroups. Clearly Z is reduced, for example.

Proposition 1.6. An abelian group A has a unique maximal divisible D, so in fact subgroup D. If B IA and B is divisible, then B I is divisible}.

=

=

(u{BI A:Bis divisible)).

Proof. Set

D

u{ B

D

I A :B

Then clearly D is divisible and the rest follows. A

Corollary. If A is abelian and D is its maximal divisible subgroup, then 0 K , with K reduced.

=D

Proof. See the corollary to Theorem 1.5.

If p E Zis a prime and A is an abelian p-group set A[p] = {x E A : p x = 0). Clearly A[p] I A , and it is easy to see that A[p] is quite naturally a vector space over E,. Consequently A[p] is a direct sum, with each summand isomorphic with Z,.

Proposition 1.7. Suppose 4 and B are divisible abelian p-groups for some prime p. If A [ p ] z B [ p ] , then A 2 B. Proof. Let g: A [ p ] -,B[p] be an isomorphism, but view g as an element of Hom(A[p], B), and let i: A [ p ] -,A be inclusion. By Theorem 1.5 there exists h E Hom(A, B ) such that the diagram

I

commutes, i.e., h A[p] = g. We show by induction that ker h = 0. Take x E kerh. If 1x1 I p, then x E ALP], so h(x) = g(x) = 0 and hence x = 0. Suppose then that 1x1 = p", n > 1, and assume that the only y E kerh with Jyl 5 p"-' is y = 0. Set y = p x . Then y E kerh and lyl = p " - ' , so n = 1, a contradiction. Thus ker h = 0. Next take any y E B,say I yl = p". Again we shall use induction to see that h is onto. If n = 1, then y E B[pf = h ( g ) I Im(h). Suppose n > 1 and that all elements of B having order p"-' or less are in Im(h). Since pn-'y E B t p ] we have p n - ' y = g(x) = h(x) for: some x E A [ p ] . Choose z E A such that p"- lz = x. Then p " - ' [ y - h(z)] = p0-1y

-

h ( p " - ' z ) = p " - ' y - h(x) = 0,

1 Infinite Abelinn Groups

199

so by induction there is some u E A with h(u) = y - h(z). Thus h(u + z ) = y E Im(h), and the proof is complete. An abelian group A with torsion subgroup T is said to split if T is a direct summand of A , in which case A = T 0 B for some torsion-free subgroup B of A . Note, for example, that every finitely generated abelian group splits, by Theorem IV.3.3. Exercise 1.3. Set A

= n{Z,:aIl

primes p

E

Z}.

0,

Show that the torsion subgroup of A is T = H,, that A is reduced, and that A / T is divisible. Conclude that A does not split. Proposition 1.8. If A is divisible and has torsion subgroup T , then T is divisible, and consequently A splits. Proof. If 0 # n E h and x E T, then ny = x for some x E A . But then (yl Inlxl so y E T and T is divisible. Apply the corollary to Theorem 1.5.

Theorem 1.9. Suppose A is a divisible abelian group. Then there are index sets r and A, for each prime p so that A

= (@{A,:Y E

r))o (@{T,:P prime)),

where each A , z Q and T, z @{Z,,:6(p) E Ap} for all p . Furthermore, A is determined up to isomorphism by the cardinalities (rland IApl, all p . Proof. By Proposition 1.8 A splits as A = G 0 T , where T is torsion and divisible, G is torsion-free and divisible. By Proposition 1.4 G = @ { A y : ? E r},with each A , 2 Q, and Irl is uniquely determined as the dimension of G as a Q-vector space. By Theorem 1 . 1 we have T = OPT,, and T, is divisible for each prime p. For each p choose a set A, for which (A,( is the (uniquely determined) Z,-dimension of T,[ p ] , and set

H, = @{Zp,:6(p)

E

A,).

Then clearly H,[p] 2 Tp[p]for all p , and so H, z T, by Proposition 1.7. The theorem follows. Exercise 1.4. Let C* be the (multiplicative) group of nonzero complex numbers and let

c = { z E @*:I21 = 13, also a multiplicative group. Show that C 2 @*.(Hint: Apply Theorem 1.9to both groups.) Exercise 1.5. Show that every abelian group A is a subgroup of a divisible group. (Hint: We may assume that A = F / K , F free abelian, and that F = @=Z. Use the fact then that Z IQ.)

200

VI Further Topics

2. POLYA-REDFIELD ENUMERATION Suppose G is a finite group acting on a finite set S . The character 8 of the permutation representation is defined by #(x) = I{S E S:xs = s}I

for all x E G. Thus 8 is a function from G to the nonnegative integers; it counts the $xed points in S for each x E G .

Proposition 2.1. The character 8 of a permutation representation of G on S is a class function, i.e., 8 is constant on conjugacy classes in G. Proof. If x,y E G and s E S, then xs = s if and only if x y ( y - ' s ) = y - l s , so s H y - ' s determines a 1-1 correspondence between the sets of fixed points for x and its conjugate x y .

Theorem 2.2 (Burnside's Orbit Formula). If a finite group G acts on a finite set S with character 8, then the number of G-orbits in S is

Proof. Set Y = { ( x , s )E G x S : x s = s } . For fixed x E G there are 8 ( x ) such ordered pairs; for fixed s E S there are (Stab,(x)( of them. Thus

1 9 ' 1= z { B ( x ) : xE G}

= x{IStab,(s)l:s E S } .

Let S,, S,, . . .,S, be the distinct G-orbits in S and choose si E Si for each i. Then C{IStab,(s)l:s

C ISi[IStab,(si)( i=l k

E

S} =

1[G:Stab(si)]lStab(si)l = klG1, k

=

i= I

SO

k

x

Exercise 2.1. If G is transitive on S and IS1 > 1 show that there is some G having no fixed points, i.e., 8(x) = 0.

=

I G J - ' C { # ( x ) : xE G } .

E

As an example of how Burnside's Orbit Formula can be used, imagine coloring a regular tetrahedron with two colors, say red (r) and blue (b), each face being a solid color. There are 24 = 16 actually different colorings, since there are 2 choices for each of the 4 faces, but they are not all distinguishable. For example, the 4 different colorings with 1 red face and 3 blue faces are indistinguishable if we allow the tetrahedron to move about. Clearly two colorings are indistinguishable precisely when one can be rotated to the other. Note that we are viewing the rotation group G of the tetrahedron as acting not

2 Polya-Redfield Enumeration

201

on the set of faces but on the set S of 16 distinct colorings of the faces. Thus the number of distinguishable colorings is the number of G-orbits in S. Recall (Exercise 7.12.19) that G is isomorphic with the alternatinggroup A,. It is easy to verify that (12)(34) E A , corresponds to a rotation of 180"about an axisjoining the midpoints of opposite edges, and that the 3 elements of order 2 are all conjugate to (12)(34). There are 2 conjugacy classes of elements of order 3, each having 4 elements. One contains (123), which corresponds to a rotation through 120" about an axis through a vertex and the center of the opposite face; the other class contains its inverse, (132). In the action on the set of 16 colorings it is easy to see that (12)(34) fixes 4 colorings, since faces adjacent at the axis edges must be the same for a fixed coloring. Similarly (123) and (1 32) each fix 4 colorings, and of course the identity 1 fixes all 16 colorings. Let us tabulate the information obtained thus far. (r

(12)(34) (123) (132) 3 4 4 16 4 4 4 1

Icl(4l I O(a)

Because of Proposition 1.1 the information in the table is sufficient for computation of the number k of orbits by means of Burnside's Orbit Formula. In fact, k = h ( 1 6 + 3 . 4 4 . 4 + 4 . 4 ) = 5,

+

and there are 5 distinguishable colorings. They are, of course, rrrr, rrrb, rrbb, rbbb, and bbbb.

Exercise2.2. Suppose three colors, red (r), blue (b),and yellow (y),are used to color a tetrahedron. Show that there are 15 distinguishable colorings, and list them. The main result of this section, the Redfield-Polya Theorem, is essentially a refinement of Burnside's Orbit Formula, which makes it possible to count orbits in a variety of situations. One further bit of technical apparatus, the cycle index, will be needed. If the finite group G acts on a set S with n elements, each x E G corresponds to a permutation r~ of S, which can be written uniquely as a product of disjoint cycles. If r~ has cycle type (j l ,j,, . . . ,j,,), we will say that x E G has cycle type ( j l ,j z , ... ,j,,). Recall from Chapter I that conjugate elements have the same cycle type. Recall also that if ( j , ,j,, . . . ,j,) is a cycle type, then i .ji = n. If G acts on S , IS1 = n, and x E G has cycle type ( j l j, z , .. . ,J,,) we define the monomial of x to be mon(x) = ci1r{z t',.,

cS=

where t,, t,, . . . ,t , are n distinct (commuting) indeterminates. Then define the cycle index of the action of G on S to be the polynomial (say over the rational

202

VI FurtherTopin

field Q) in t 1 , c 2 , . . .,t, given by

. . , t n ) = 1GI-l x{mon(x):x E G } . Note that if G has conjugacy classes K , , K,, . . . ,K , , with x i E Ki for all i, then 9' = 9',S(tl,.

m

Let us compute several examples of cycle indices. They will prove useful later. It will be convenient to tabulate information regarding conjugacy classes and cycle types. EXAMPLES 1. Let G = S, and S

=

{ 1,2,3}. (12) (123) 1 3 2 Type (3,0,0) (1,1,0) (O,O, 1) CT

1

Icl(4l

Thus 9' = :(t: + 3t1t, + 2t3). 2. Let G = A,, S = {1,2,3,4}. U

Icl(a)l

Type

1 (12)(34) (123) (132) 1 3 4 4 (4,0,0,0) (0,2,0,0) ( L O , 1,O) ( L O , L O )

Thus 2'= &(t? + 3tg + 4t1t, + 4tlt3). Note that this is the same as the cycle index for the action of the rotation group of the tetrahedron acting on the set of 4 faces. It would be instructive at this point for the reader to compute the cycle index for the action of that group on the set of 6 edges of the tetrahedron. The next few examples involve geometric symmetry groups. The reader is urged to construct cardboard models of the appropriate geometric objects and verify the tabulated information. 3. Let G be the rotation group of a cube (see Exercise 1.1.3). We shall compute its cycle indices for 3 different permutation representations: (i) on the set F of 6 faces, (ii) *3nthe set V of 8 vertices, and (iii) on the set E of 12 edges. Recall (Exercise 1.12.28) that G z S,, so we may list elements of S, as representatives of the conjugacy classes. Observe that (1234) corresponds to a 90" rotation and (13)(24) to its square, whereas (12) corresponds to the other type of 180"rotation, whose axis joins midpoints of opposing edges. CT

1

(12)

(123)

Icl(x)l 1 6 8 F-type (6,O,...) (0,3,0,... ) (0,0,2,0,...) V-type (8,O,... ) (0,4,0,...) (2,0,2,0,... ) E-type (12.0,... ) (2,S.O,...) (0,0,4,0,...)

(1234)

(13)(24)

6 3 (2,0,0,1,...) (2,2,0,...) (0,0,0,2,...) (0,4,0,...) (0,0,0,3,...) (0,6,0,... )

2 Polya-Redfield Enumeration

203

Thus we have

+

+

(i) 9& = h(tt + 6t; + 8t: 6 t f t 4 3t:t:), (ii) 2G,v = &(t; + 9t: 8t:t: + 6t:), and (iii) 2G,E = &(ti2 + 6t:t: + 8t; + 6t: + 3tz).

+

4. Let G = C, = (a I a" = l), a cyclic group of order n, acting as a plane rotation group on the set S of n vertices of a regular n-gon. If we label the vertices consecutively as 1,2,3,. . . ,n we may assume that a = (123.. .n) E S,,. We know from Section 1, Chapter 1 that C,, has a unique cyclic subgroup of order d for each divisor d of n, and that a cyclic group of order d has 4(d) generators, i.e., C,, has 4(d) elements of order d. Furthermore, each a kE C,,has order n/(n, k), and in fact akis a product of (n,k) disjoint n/(n, k)-cycles.Thus if 1 I d 5 n and d [ n there are 4 ( d ) elements of order d, each having cycle type (0,. . . ,n/d,. . .,O). Consequently

Thus, for example, if n = 12 then 2 = &(tiZ + t z

+ 2t: + 2t: + 2t: + 4 t , , ) .

5. Let G = D,,, the dihedral group of order 2n, acting on the set of vertices of a regular n-gon. In the presentation

G = (a,b\a" = b2 = 1, b-'ab = a-')

we may assume that a is a generating rotation as in Example 4 above, so the cycle structure of each power akhas already been determined. Write A = (a). Take b to be a reflection whose mirror passes through one of the vertices. Since a-kbak= ba2'wehavecl(b) = {ba2k:0I k < n}. If nisodd,then(a) = (a2), so in that case cl(b) is the entire coset bA (all reflections are conjugate). Each reflection then fixes the vertex on the mirror and interchanges the remaining vertices in pairs, so the cycle type is (1, (n - 1)/2,0,. ..). If n is even there are two classes of reflections, viz., cl(b) and cl(ba). The reflections in cl(b) have mirrors through 2 opposite vertices and cycle type (2,(n - 2)/2,0,. ..). Those in cl(ba) have mirrors through midpoints of opposite edges and cycle type (O,n/2,0,. . .). Thus if n is odd, then 2 = (1/2n)[x{$(d)t!/d: 1 5 d 5 n, d 1 n}

+ nt, t!")"];

if n is even, then %" = (1/2n)[~{4(d)t",d:1I d In, d In)

+ (n/2)t:t$-')/2 + (n/2)r;I2.

In our discussion above of colorings of a tetrahedron we passed from a permutation action on the set of 4 faces to an action on the set of 24 = 16

204

VI Further Topics

distinct colorings. Suppose the faces to be numbered 1,2,3,4 and write D = { 1,2,3,4} for the set of faces. Let R = {r, b}, the set of colorings. Then the set of colorings can be identified with the set of all functions f:D + R, since a coloring is simply an assignment of a color for each of the faces. In general suppose that D (the domain) and R (the range) are two finite sets, and write RDfor the set of all functions from D to R. Note that lRDl = (RIIDI.If G is a finite group that acts on D,then G also acts on R D as follows: if D E G, f E R D ,and d E D,then (cf)(d) = f ( c - ' d ) , giving of E R".

Exercise 2.3. Verify that the action described above of G on R D is a permutation representation. The G-orbits in the set R D of functions are commonly called patterns. Suppose S is a commutative ring with 1 containing the rational field Q as a subring, with 1, = 1, (S will often be a polynomial ring, for example). A function w:R -,S is called a weight assignment on R. For each r E R the ring element w(r) is called the weight of r. A weight assignment w on R induces a weight assignment W on R D by means of for each f:D + R. D} Suppose w is a weight assignment on R, and f l , f 2 E R D are in the same W ( f )= n{w(f(d)):d

pattern, say with fl = of2,D

E

E

G. Then

Wfi) = ncw(f*(d)):dE 0) = n w f 2 ( d ) ) : d E Dl = n(w(f,(~-'d)):d E D}=

W(f2X

since D - acts as a permutation of D.Thus the induced weight assignment W is constant over patterns in RD.If F is a pattern in R Ddefine W ( F )= W ( f )for any f E F. If U is any subset of R define the inventory of U to be Inv(U) = x{w(r): r e U } . Similarly if T is a subset of RD define its inventory to be Inv(T) = C { W ( f ) : fE T } .We agree in general that I n v ( 0 ) = 0. In our earlier example (coloring the tetrahedron) we had D = { 1,2,3,4} and R = (r,b). Take S to be the ring Q[tl,t,] of polynomials in two indeterminates, and define a weight assignment w: R + S by means of w(r) = t , , w(b) = t 2 , As we saw, there are five patterns F,, .. .,F,, represented by f, = rrrr E F,, f2 = rrrb E F,, f3 = rrbb E F3, f4 = rbbb E F4, and fs = bbbb E F,. It is easy to check that IF,I = lFsl = 1, lF21 = IF4[ = 4, and lF31 = 6, and it is clear that W(F,) = W(fl) = tf, W(F2)= t : t , , W(F,) = [:ti, W(F4)= tit;, and W(Fs)= t:. Consequently Inv(R") = t': 4t:t2 6t:t: 4t1t: t: = ( t , + t2)4 = [ I ~ v ( R ) ] ~ ~ ' . The example is a special case of the next result.

+

+

+

+

2 Pblya-Redfield Enumeration

205

Proposition 2.3. Suppose R and D are finite sets and w: R + S is a weight assignment on R. Suppose that D is partitioned into a union of disjoint sets D,,D2,. . .,Dmand let T be the set of functionsf E R” that are constant on each of the subsets Di. Then

n x { ~ ( r ) ~ ” ~ Rl :)r. m

Inv(T) =

E

i= I

Proof. Define $: D -+ {1,2,.. .,m}by agreeing that $ ( d ) = j if and only if d~ Dj. For each J ” E T define 4 = 4,-: {1,2, ..., m} + R by means of 4(j ) = f ( d ) if and only if d E Dj.Then each f E T clearly factors as f = 4 ,-$. If R = { r l , r 2 , . .. , r k } ,then the right hand side of the equation to be proved is the product m

To multiply we choose one term from each factor, in all possible ways, multiply the terms together, and then add the resulting products. To choose a term from each factor is to choose a function 4: { 1,2,. . .,m} + R, then multiplying yields {w(4(i))IDi1},after which we add the results for all such functions 4. But

n?’

w(4(i))’”zt =

H { w ( ~ ( $ ( ~ IDi} I ) :=~n { w ( . f ( d ) ) : dE Di}, E

wheref = 4$ E T. The choice of all possible 4 results in all f of adding is

Corollary. Inv(R”)

=

E

T, so the result

[1nv(R)]lD1.

Proof. If D = { d , ,..., d,) take Di = {di}, 1 5 i i n. Suppose now that D and R are finite sets, w:R + S is a weight assignment, and G is a finite group acting on D, and hence on R”. Recall that i f f € RD,then the weight of f is W ( f )= n { w ( f ( d ) ) : dE D}, and if F is a pattern (i.e., a Gorbit in R”), then the weight of F is W ( F )= W ( f )for any f E F . Define the pattern inventory of G to be PI

=

x{

W(F):all patterns F E R ” ) .

Suppose, for example, that we define w(r) = 1 E Q for all r E R . Then W ( f )= W ( F )= 1 for all f E R ” and all patterns F , and consequently the pattern inventory PI is just the number of patterns. Thus PI, if it can be computed, contains at least as much information as is afforded by the Burnside Orbit Formula applied to the action of G on R D .

206

VI

Further Topics

Theorem 2.4 (Redfield-Polya). Suppose G is a finite group acting on a finite set D, ID( = k, with cycle index 2, and R is a finite set with a weight assignment w: R -,S. Then the pattern inventory for the action of G on R D is

Proof. Let W , , W,, .. . be the distinct weights of patterns in R D , and set = { f R~D :W ( f )= H!}, i = 1,2,. . .. Clearly q.is a union of patterns; say there are mi different patterns in Y i .Thus PI = Cimiy.Note that G acts on each Fi,and that the G-orbits in Fi are just the mi patterns whose union is If Oi is the character of the action of G on Pithen, by the Burnside Orbit Formula, mi = \GI-'C{O,(o):a E G } ,and hence

e

e..

PI = x m i W = 1GI-l x{CO,(o)Iq:o E G } .

xi

i

i

Fix a E G. Then Oi(a)Vis the inventory of {f E RD:af = f},since Oi(o)is just the number of f E Fifixed by a, and so PI = IGI-'

1 (c{W ( f ) : fE R D and af = f } ) .

asC

If of = f and d E D, then f(d)

= ~ - ' f ( d ) = f(ad) = o-'f(ad) = f ( 0 2 d ) = ...,

and f is constant on the o-cycles in D. Conversely, iff E R Dis constant on the 0-cycles in D, then af = f. (Why?) Thus if the a-cycles partition D into disjoint sets D,,D,,.. ., then {f:af = f } is just the set of all f E R D that are constant on each of the subsets Di, i.e., the set T E R D in Proposition 2.3. By that proposition

C{w(f):af = f} = n(x{w(r)lDgI:r E R } ) i

for each fixed a E G. If the cycle type of cr is (jl, j 2 , ...,jJ, then among the sets Di there are j , for which IDi! = l , j , for which IDi[ = 2, etc. Thus

for each o, and consequently

2

Polya-Redfield Enumeration

207

Corollary. The number of different patterns in R D is T ( l R 1 , IRI,. . .,I RI). Proof.

Set w(r) = 1 for all r E R.

For an example let D = F be the set of 6 faces of a cube and let R = {red, blue, yellow). Let S = Q(r, b, y), where r, b, and y are distinct indeterminates, and assign weights by means of w(red) = r, w(b1ue) = b, w(yel1ow) = y. If G is the rotation group of the cube, then the cycle index Z?”G,F is T = &(t:

+ 6r: + 8t: + 6tft4 + 3t:t:)

(see Example 3, above). By the Redfield-Polya Theorem

PI = &(r + b + Y ) + ~ S(r2 + b2 + y 2 ) 3 + 8(r3 + b3 + y3)2 + 6(r b + y)’(r4 b4 + y4) 3(r b + y)2(r2 b2 y2)2] = r6 + b6 + y 6 + r5b + r5y + rb5 + ry5 + b5y + by5 + 2(r3b3+ r3y3 + b3y3 + r4b2 + r4y2 + b4y2 + b2y4 + r2b4 + r2y4 + r4by + rb4y + rby4) + 3(r3b2y+ r3by2 + r2b3y + r2by3 + rb3y2+ rb2y3)+ 6r2b2y2.

+

+

+ +

+ +

(Verify.) The total number of distinguishable colorings (i.e., patterns) is obtained by setting r = b = y = 1; there are 57 of them. Note that the individual terms in the polynomial PI give the number of patterns of various types. For example, the term r6 represents the unique pattern with all faces red; the term 3r3b2yrepresents 3 distinct patterns each having 3 red faces, 2 blue faces, and 1 yellow face, etc. For a second example let us determine the number of distinguishable necklaces that can be made by using 9 red or blue beads. We assume that the beads of the same color are indistinguishable, and that 2 necklaces are indistinguishable if one can be made to look like the other by rotating it or turning it over. In effect, then, we imagine each necklace laid out so the beads are at the vertices of a regular g-gon, and that the beads are acted on by the dihedral group 4 . Take D to be the set of 9 beads, R = {red, blue}, S = Q(r, b) as above, and w(red) = r, w(b1ue) = b. The cycle index for D9 is S = &(t: + 2t: + 6t9 + 9t,t:), so the pattern inventory is

PI = &[(r + b)9 + 2(r3 + b3)3+ 6(r9 + b9) + 9(r + b)(r2 + b2)“] = r9 + r8b + 4r7bZ+ 7r6b3 + 10r5b4 + 10r4b5+ 7r3b6 + 4r2b7+ rb8 + b9. Set r = b = 1 to see that the number of distinguishable necklaces is 46, and observe, for example, that there are 7 patterns each using 3 red beads and 6 blue beads.

208

VI Further Topics

Exercise 2.4. Let p be an odd prime. Discuss the possible necklaces with p beads (i) if red and blue beads are used, or (ii) if red, blue, and yellow beads are used. In particular determine the number of distinguishable necklaces. Exercise 2.5. In how many distinguishable ways can the faces of a tetrahedron be colored if n different colors are available? A different sort of weight assignment is often useful in combinatorics. Suppose D,R, and G are as usual, let x be an indeterminate, and set S = Q [ x ] . Let k: R + Zbe a function, with k(r) 2 0 for ail r E R , and call k(r) the content of R . For f~ RD define the content o f f to be K ( f ) = x { k ( f ( d ) ) :d E D } . Assign weights on R by means of w(r) = xk(')for all r E R . Note that then, for f E RD,

W ( f )= n { ~ ( f ( d ) ) : dE D } = n { ~ ~ ( j ( ~E )D)}: =d x

xK(j).

~ ~ ~= ( ~ ( ~ ) )

For each nonnegative integer m let am= I{r E R:k(r) = m}l, and define the generating function for R (and k) to be a ( x ) = C { a m x m : OI mE

H},

a polynomial whose coefficients count the numbers of elements in R having various contents. If f i and f2 are in the same pattern F in R D it is easily verified that K ( f J = K ( f 2 ) ,so we may define K ( F ) = K ( f )for any f E F . If 0 I rn E Z set A , = I { F : F is a pattern and K ( F ) = m}l,

and define the pattern counting function to be A(x)=

c A,xrn

E i z [x ] .

m

Theorem 2.5. Suppose R and D are finite sets and G is a finite group acting on D, with cycle index 3,generating function a(x), and pattern counting function A(x). Then A ( x ) = .z(a(x), a ( x 2 ) ,. . . ,a(xk)).

Proof. Since PI = PI

c{W(F):all patterns F } and W ( F )

=

= x ~ ( ~we) ,have

A,

+ A 1 x + A 2 x 2 + ... = A(x).

But by the Redfield-Polya Theorem

PI = 3

L+

)

w(r), C w(r)', . . . r

+

a l x + a 2 x 2 ..',a, = 3 ( a ( x ) , a ( x 2 ) ,.. . , U ( X k ) ) . = 3(a0

+ a , x Z + u2x4 + .-.;-.)

2 Polya-Redfield Enumeration

209

If (DI= k, then for each integer m, 0 5 m I k, define D('")to be the set of all m-sets in D, i.e., the set of all subsets of size m in D.Thus ID("')) = (L).If a group G acts on D, then it acts naturally on each D'"')by means of

q{dl,d2,. . .,dm} = {ad,,ad,, . . .,ad,}

for each a E G.

Theorem 2.6. Suppose G acts on D,ID1 = k, with cycle index 3.If x is an indeterminate and 0 5 m I k, then the number of G-orbits in D'") is the coefficient of x"' in 6(1+ x, 1

+ x 2,..., 1 + Xk).

Proof. Take R = {0,1),define a content k : R + Z via k ( 0 ) = 0, k ( 1 ) = 1, and hence assign weights via w(0) = 1, w( 1 ) = x. If f E R D ,then

Thus W ( f )= x"' if and only if precisely m elements of D are mapped by f to 1 , i.e., if and only if f - '( 1) E D("'). There is thus a 1- 1 correspondence between Gorbits of m-sets in D'"' and patterns of content m in R D . But the number of patterns of content m is the coefficient A , in the pattern-counting function A ( x ) . Since the generating function is a(x) = 1 + x the result follows from Theorem 2.5. Corollary 1. The number of G-orbits in D is the coefficient of x in 6 ( 1 x,1 + x 2,...) 1 Xk).

+

+

Identify D with D'".

Proof.

Corollary 2. The total number of orbits of G acting simultaneously on all < k, is

D("'),0 5 m

A,

D

+ A , + A , . . . + A , = S(1+ 1 , l + l 2 ) . . . ) = 2 ( 2 , 2)...)2).

For an example, let G be the rotation group of a cube, acting on the set V of vertices. The cycle index is 9 = $[ty 9 t j + 8tit: 6tif. Thus

=

+

+

+ X, 1 + x', ..., 1 + x') = &[(1 + x ) + ~ 9(1 + x ' ) ~+ 8(1 + x)'(1 + x3), + 6(1 + x4),] = 1 + x + 3x2 + 3x3 + 7x4 + 3x5 + 3x6 + x 7 + x8.

A(x) = 6 ( 1

(Verify.) Thus, for example, there are 3 orbits of 2-sets: one orbit, of size 12, consists of pairs of vertices that share an edge; another, of size 12, consists of pairs of vertices at opposite corners of faces; and the third orbit, of size 4, consists of pairs of vertices at opposite corners of the cube. Exercise2.6. Describe the 3 orbits of 3-sets and the 7 orbits of 4-sets in the example above.

210

VI Further Topics

Exercise 2.7. If the rotation group G of the cube acts.on the set E of edges compute the numbers of orbits of m-sets, 0 I m s 12, and describe the orbits of 2-sets explicitly. Exercise 2.8. Suppose a hollow regular tetrahedron has thin walls, and both the inside and outside of each face is to be colored red or blue. How many distinguishable colorings are there, and how many of them have 3 or fewer of the faces red?

3. PSL(V) Let V be a vector space of finite dimension n over a field F . Write GL(V) for the group of all invertible linear transformations of V . Any choice of basis in V provides an isomorphism between GL(V )and the group GL(n, F ) of invertible n x n matrices with entries from F. Both are called the general linear group. The special linear group is the subgroup SL(V) of GL(V) [or SL(n,F) of GL(n,F ) ] consisting of those linear transformations (or matrices) having determinant 1 . When F is a finite field with q elements it is customary to denote GL(n, F ) and SL(n,F ) by GL(n,q) and SL(n,q), respectively. Exercise 3.1. Show that the center of GL(V) is the set of scalar transformations, i.e., Z(GL(V))= {al:0 # a E F } ,

and also that Z(SL(V))= {al:a" = 1 E F } . If 0 # u E V write [ u ] for Fu = {au:aE F } , the line through the origin spanned by u, and call [u] a projectiue point. The set of all distinct projective points [u] is called the (n - 1)-dimensional projectiue space based on V ,denoted by Pn-l(V). There is a natural permutation action of GL(V) on Pn-,(V), viz., T [ u ] = [Tu] for T E GL( V ) and [u] E Pn- ( V ) . It is clear from Exercise 3.1 that Z(GL(V ) )acts trivially on Pn-,( V ) ,i.e., that it is in the kernel of the permutation action. Suppose conversely that T E GL(V) acts trivially on every projective point [ u ] . Thus Tu = au for each nonzero u E V , where a E F may depend on u. If v , , v 2 E V are linearly independent suppose Tu, = a,u,, Tu, = a,u,,and T(u, u 2 ) = alz(ul uz). But then, since T(u, + 0,) = Tu, + Tu,, we may conclude that a , = a,, = a 2 , and hence conclude that T = a , 1 E Z(GL(V)). It also follows that Z(SL(V ) )is the kernel of the permutation action of SL(V ) on P, - V ) . Define the projective general linear group PGL( V )to be the quotient group GL(V)/Z[GL( V ) ] , and the projective special linear group PSL( V ) to be

,

+

+

3 PSL(V)

211

SL(V)/Z[SL(V)]. It is a consequence of the remarks above that PGL(V) and PSL(V) act faithfully as permutation groups on the projective space P,- V ) . There are of course isomorphic versions of the projective groups: PGL(n, F ) = GL(n, F)/Z[GL(n, F ) ] and PSL(n, F ) = SL(n, F)/Z[SL(n, F ) ] , which are often denoted PGL(n, q ) and PSL(n, q) when F has q elements. When F is finite the groups are finite. Let us compute their orders. Proposition 3.1. Suppose JFl= q. Then

(i) IGL(n, q)1 = qn("- ' ) I 2 n{qk - 1:1 I k I n } ,

6) \sun,4)) = \PGL(n,q)l = \GL(n,q)\/@?- 11,

and (iii) lPSL(n,q)l = ISL(n,q)l/(n,q - 1).

Proof. (i) The first row of a matrix in GL(n,q) can be any one of q" - 1 nonzero row vectors. There are q" - 4 choices for the second row since it must be independent of the first. Similarly there are q" - q 2choices for the third row in order that the first three are linearly independent, etc. Thus

- qk:OI k 5 n - 1) (GL(n,q)(= n{q" 41 + 2 + ... t ( n 1 ) l J { q k - 1 : l I k I n) ~

= q n ( n - 1)/2 n{qk - 111

s k In}.

(ii) Since IZ[GL(n,q)]J = JF*J= q - 1 it is clear that IPGL(n,q)l = IGL(n,q)J/(q- 1). The determinant map is a homomorphism from GL(n,q) onto F* with kernel SL(n,q), so IGL(n,q)l/ISL(n,q)l = lF*I = 4 - 1. (iii) Since F* is cyclic of order q - 1 we see by Exercise 3.1 that a . 1 E Z(SL(n, q ) )if and only if a" = a4-I = 1, i.e., a ( " ' 4 - ' ) = 1, hence if and only if a is in the unique subgroup of order (n,q - 1) in F*.

Although SL and PGL have the same order they are usually not isomorphic. Exercise 3.2. If F is any field denote by % 9 , ( F ) the group of linear fractional transformations of F , i.e., all rational functions f(x) E F ( x ) of the form f ( x ) = (ax b)/(cx d), where a, 6, c, d E F and ad - bc # 0, with composition of functions as the group operation. Show that % Y 2 ( F ) PGL(2,F). Let 2 2 ( F ) be the subgroup of YY,(F) consisting of those f ( x ) for which ad - bc is a square in F . Show that g 2 ( F ) z PSL(2, F ) .

+

+

212

VI Further Topics

If dim V = n, then a hyperplane in V is any subspace W of dimension n - 1. A linear transformation T # 1 of V is called a transvection if there is a hyperplane W such that TI W = l,, i.e., T fixes each vector in W , and Tv - v E W for all v E V . If T is a transvection of V with fixed hyperplane W choose a basis for V by first choosing v, E V\ W and then choosing a basis (v,,. . .,v,} for W. The resulting matrix representation for T makes it clear that det T = 1, so T E SL( V ) . Exercise 3.3. (1) Show that the inverse of a transvection is a transvection. (2) Suppose V is a subspace of V,, v E Vl\V, and T is a transvection on V with hyperplane W. Show that T extends to a transvection T, on V, whose fixed hyperplane W, contains Wand v.

Proposition 3.2. If v and w are linearly independent in V , then there is a transvection T of V for which Tv = w. Proof. Choose a hyperplane W with v - w E W, v $ W (hence w 4 W ) , anddefine Tbymeansof TI W = I,, To = w. If X E Vwritex = ao + u,with a E F, u E W. Then

Tx - x

= aw

+ u - (av + u) = a(w

-

v)E

W,

so T is a transvection. Proposition 3.3. Suppose W, and W, are distinct hyperplanes in V and v E V\(Wl u W,). Then there is a transvection T of V with T(W,) = W, and Tv = U.

Proof. Note that W, + W, = V , so dim(Wl n W,) = n - 2. Thus W = (W, n W,) + Fv is another hyperplane. Since W, W, = V there are .x E W, and y E W, with x y = v. Note that x 4 W, (or else v E W,), so W, = (W, n W,) + Fx, and similarly W, = (W, n W,) + Fy. Thus V = (W, n W,) + Fx Fy. It follows that x 4 W, for otherwise also y = v - x E W and hence V c W, a contradiction. Define T by means of TI W = 1, and Tx = -y. As in the proof of Proposition 3.2 we see that T is a transvection. Also

+

+

+

T(W,) = T(W, n W,

+ Fx) = W, n W, + F(-y)

= W,,

and Tv = v since v E W

Theorem 3.4.

The set of transvections of V generates SL(V).

Proof. We may assume dim V = n 2 2. Fix R E SL(V); then choose a hyperplane W c V and v E V\ W. If v and Rv are linearly independent then by

213

3 PSL(V)

Proposition 3.2 there is a transvection TI with TIRv = v. If v and Rv are linearly dependent first choose a transvection To such that ToRv and v are linearly independent, then choose a transvection T', with T ; ToRv = v, and set TI = T', To. In either case, then, TIRu = v, with TI a product of transvections. Note that v # T , R W . If T I R W = W set T 2 = l V . If T , R W # W apply Proposition 3.3 to get a transvection T, with T,T,R W = W and T2v = v. Set R , = T2T,R. Since R , E SL( V ) and R,ri = u it follows that R , WE SL(W). Now use induction on n. If n = 2, then R , W = l w , so R , = 1, and R = T F ' T ; If n > 2, then by induction R , 1 W is a product of transvections on W , each of which extends to a transvection fixing 0 on V by Exercise 3.3.2. Thus R , is the product of the extended transvections and so R = T i T ; R,, a product of transvections.

I

I

Theorem 3.5. I f TI and T2 are transvections on V , then TI and T, are conjugate in GL( V ) . If n 2 3 they are conjugate in SL( V ) .

Proof. For i = 1,2 let M/; be the fixed hyperplane of K , choose x i E V\w, and say wi = T x i - x i E M/;. Choose bases { wI,u3,..., u,] for W, and ( w z ,v 3 , . . . ,on} for W, . For each u # 0 in F define S, E GL( V ) by means of Soxl = x,, Sowl = w,, S,ui = oi( 3 5 i I n - l ) , and Sou,, = av, (if n 2 3). Then S,T,Sa-*x, = S,T,x1 = S,(X, + w,)= x2 + W , = T2x2, SoTIS,'w, = S,T,w, = S,W, = w2 = T2w2, S,T,Su-'vi = S,Tlui = Soui = vi = T,vi, 3 < i < n - 1, and

S,T,S,~'v,, = a-'S,T,u,

so SOTIS;

=

= u-'Sou,, = v, =

T,. If n 2 3 we may set b

= (det

T2u,

(if n 2 3),

S , ) - ' and then S,

E

SL( V ) .

Theorem 3.6. Suppose dim V = 2 and let { u , , u 2 ) be any basis for V . Each conjugacy class of transvections in SL(V ) contains one whose matrix relative to { u I , u 2 } is of the form [i 71, a E F*. Proof. If T is a transvection with hyperplane W choose 0 E V\ Wand set w = Tu - u E W . Relative to the basis { u, w } the matrix representing T is then [t :]. If M is the matrix representing T relative t o { v , , v 2 } there is a matrix B E GL(2,F ) with B M B - ' = [t y]. If det B = a-' set A = Then BA

E

SL(2, F )

and

( B A ) . - ' M ( B A )= A - '

Theorem 3.7. If n I 3 and G

=

SL( V ) ,then G'

i:

=

G.

[::I.

;]A=[:

3

2 14

VI Further Topics

Proof. By Theorems 3.4 and 3.5 it is sufficient to exhibit a transvection in G'. Choose a basis { v l , v , , . . . ,v , } for V and define Tl and T2via Tl:VIHV1 - ~ T2:~1 HVI, V ~

if

2 U, ~ H U ~

H U-~ ~ 3 U , ~ H V ~ ,

21i
Clearly TI and T, are themselves transvections, and it is easily checked that T l T , T ~ ' T ~ ' : v l H -uvl3 , v i + v i ,

21iIn,

so Tl T, T ; ' T ; is a transvection in G'.

Corollary. If n 2 3, then PSL( V ) is equal to its derived group. Theorem 3.8. If n

=

2, IF1 > 3, and G

= SL(V), then

G'

=

G.

Proof. Choose a basis { vl, v 2 ) for Vand choose a E F , a 4 (0,k 1 ). Define S E SL(V ) by means of S: v 1 ~ a ' u l-, v 2 H av,, and for each b E F* define & E SL(V) by means of &: v 1 I-+ u1 bv,, u2 Hv 2 . Then S&S-' T b l is represented by the matrix

+

The theorem follows from Theorems 3.4 and 3.6 since b E F* is arbitrary. Corollary. If n

= 2 and

IF1 > 3, then PSL( V )is equal to its derived group.

Exercise 3.4. If n 2 3 or IF1 > 3 show that GL(n,F) has derived group SL(n, F). Determine the derived groups of SL(2,2) and SL(2,3).

The main result of this section is that, with two exceptions, PSL(V) is a simple group. We will need some auxiliary information about permutation groups. Suppose G acts as a permutation group on a set S. A proper subset B L S, with IBI > 1, is called a block for G if either x B = B or x B n B = 0 for all x E G. If G is transitive on S and has no blocks we say that G is primitive on S . For example, the rotation group of a tetrahedron acts primitively on the vertices of the tetrahedron, whereas the dihedral group D4 is transitive but not primitive on the vertices of a square. Exercise 3.5. (1) If IS( = p , a prime, and G is transitive on S, show that G is primitive. (2) If G is transitive on S show that G is primitive if and only if each stabilizer Stab&), s E S, is a maximal subgroup of G. Proposition 3.9. If G is doubly transitive on S, then G is primitive.

3 PSL(V)

215

Proof. Suppose B S S , B # S , and JBJ> 1. Choose a, b E B, a # b, and c E S\B, then choose x E G with xa = a and xb = c. Then a E XB n B and c E xB\B so B is not a block.

Proposition 3.10. Suppose G is faithful and primitive on S and 1 # N 4G. Then N is transitive on S. Proof. Since G is faithful and N # 1 there is an N-orbit B c S with IB( > 1, say B = Orb,(a) = Na. If x E G, then xB = xNa = Nxa = Orb,(xa), so either xB = B or XB n B = 0. Thus B = S , since otherwise B would be a block for G, and so N is transitive.

Proposition 3.1 1. Suppose G acts on S, H I G is transitive on S, and s E S. Then G = H . Stab,(s). Proof. If x E G, then xs = hs for some h E H , so K ' x E Stab&) and thus x E h . Stab&) c H . Stab&).

Theorem 3.12 (Iwasawa). Suppose G is faithful and primitive on S and G = G'. Fix s E Sand set H = Stab,(s). If there is a solvable subgroup K 4H such that G = (U{K":x E G}), then G is simple. Proof. Suppose 1 # N <1 G. By Proposition 3.10 N is transitive on S and by Proposition 3.11 G = N H . Thus NK a N H = G. But then K" I ( N K ) " = NK, all x E G, so U{K":x E G} E NK and G = NK. Since K is solvable K(m)= 1 for some integer m. It is easy to check inductively N . lQk) for all k. Thus that (NK)(k'I G

so N

=

=

G(m)= (NK)'"' _< N . Ktm)= N,

G and G is simple.

Proposition 3.13. If n 2 2, then PSL(V) acts faithfully and doubly transitively on the projective space P, - ( V ) .In particular, it is primitive.

,

Proof. We observed earlier that PSL(V) is faithful on PnPl(V). It will suffice to show that SL(V) is doubly transitive. Take [ul] # [u,] and [wl] # [w2]in P,-,(V), so { u l , u , ) and { w , , w 2 )are linearly independent sets in V . If n = 2 they are bases. If n 2 3 set V, = ( u , , ~ , ) and V, = (wl,w,). Then either V, = V, # V or else V, u V, is not a subspace, so V, u V, # Vin either , uj} are linearly case. If uj E V\(V, u V,), then both { u , , u , , u , } and { w l w,, independent. The argument can be repeated to obtain bases {u,, u , , u 3 , . . .,u,} and { w l w,, , u 3 , . . . ,u,} for V . For any b E F* define E GL(V) by means of u1 Hbw,, u2 Hw 2 ,and U ~ H u i , 3 I i I n. If w i= Cjajiuj, i = 1,2, then det & = b(a1,az2- azlal,). Choose b so that det & = 1. Then & E SL(V) and maps [ul]to [w,]and [u2] to [w,].

2 16

VI Further Topics

Proposition 3.14. If 0 # u E V and A = Stab,,(,,([u]), then A has a normal abelian subgroup B whose conjugates in SL( V) generate all of SL( V). Proof. Choose a hyperplane W with u E V\ W, so V = W 0 Fu. If S E A and w E W write Sw = S'w + a,. u, with S'w E W and a, E F. It is easy to check that S' E GL(W) and that 4: S -,S' is a homomorphism from A into GL(W). Thus B = ker 4 is a normal subgroup of A . Choose a basis {w,,.. . ,w,- for W. For any b E F* define % E A by means of %: w 1H w , bu, w i w w ifor 2 I i S n - 1, and U M U . Then is a transvection (with hyperplane spanned by (w2,. . ., w,,u ) ) , and if w E W , then &w = w + bu, so Tb = 1, and TbE B. If n = 2 the matrix representing relative to (w,,u } is [i71, and b is arbitrary in F*. It follows from Theorems 3.4,3.5, and 3.6 that the conjugates of B generate SL(V ) . If S E B, thenS( W = 1,,soif Sv = ~ ~ - ' c i +wcvi we havedet S = c = 1. Thus if S,, S, E B , then their representing matrices relative to { w , , . ..,w, - 1, u } have the partitioned form

+

[;

r']

and

[;

:'I.

Since

it follows that B is abelian.

Theorem 3.15. simple group.

Except for PSL(2,2) and PSL(2,3) every PSL(V) is a

Proof. Choose u # 0 in V and take A = Stab,,([u]) and B U A as in Proposition 3.14. Write Z = Z(SL(V)) and set H = A/Z = Stab,,,([u]), K = BZ/Z 4H. Then K is solvable (in fact abelian) and the conjugates of K in PSL(V) generate PSL(V) by Proposition 3.14. The theorem follows from Proposition 3.13, the corollaries to Theorems 3.7 and 3.8, and Theorem 3.12. Exercise 3.6. Use the action of PSL on projective space to show that PSL(2,2) z S, and PSL(2,3) E A,, and conclude that the exceptions in Theorem 3.15 are not simple. Show also that PSL(2,4) z A,. Note that IPSL(2, 5)1 = 60, so PSL(2, 5) z A, by Proposition 4.5 of Chapter I, even though IPl(5)(= 6. Observe that IPSL(3,4)1 = 20160 = IA,l, so PSL(3,4) and A, are simple = 3. For groups having the same order. If Z = Z[SL(3,4)], then (Z(= any S E SL(3,4) write for the coset SZ that is its image in PSL. Now let E PSL(3,4) be any element of order 2. Then S2 E 2, so S6 = 1. Replacing S by T = S3 we have = T in PSL and (TI = 2 in SL. Let W c V

s

4 Integral Dependence and Dedekind Domains

2 17

+

be the kernel of the linear transformation 1 T. Thus dim V = 3 = dim W + dim(1 + T ) V .Note that v E W if and only if u + Tv = 0, or Tu = u (char F = 2), i.e., u is fixed by T. Thus W # V , since T # 1. Note also that T(l + T ) v = ( T + T 2 ) u = ( T + I)u,so(l T)uisfixedbyTandhence(l T ) V C W . Thus dim(1 + T)VI dim W and it follows that dim(l T ) V = 1, dim W = 2. Consequently W is a hyperplane and Tv - v E W for all u E V , i.e., T is a transvection. It follows from Theorem 3.5 that all elements of order 2 are conjugate in PSL(3,4). The permutations (12)(34)and (12)(34)(56)(78)are not conjugate in S , , hence certainly not in A , , so A8 has at least two conjugacy classes of elements of order 2. Thus PSL(3,4) and A, are nonisomorphic simple groups having the same order.

+

+

+

Exercise 3.7. Show that PSL(4,2)also has the same order as PSL(3,4) and A , . Show that PSL(4,2) and PSL(3,4) are not isomorphic. ( H i n t : Consider 1 0 0 0 0 0 1 0 0 0 1 1

4.

INTEGRAL DEPENDENCE AND DEDEKIND DOMAINS

All rings considered in this section will be commutative rings with 1 (proper ideals are obvious exceptions), and all modules will be assumed unitary. If R is a subring of S it will be assumed without further mention that 1, = I,. If R is a subring of S , then a E S is integral ouer R if a is a root of some monic f ( x ) E R [ x ] ;the equation f ( a ) = Ois called a relation of integral dependence. If every a E S is integral over R, then S is integral over R , or an integral extension of R. The concept generalizes the notion of algebraic field extensions to commutative rings. Proposition 4.1. Suppose S is integral over R. Then S is a field if and only if R is a field. Proof. *: If 0 # a E R, then a - ' E S is integral over R, so there are bo,...,bk-l E R with b , + b , a - ' + . . . + (a - ' ) k = O

Multiply by ak- ' and observe that U-'

=

-(boak-'

+

"'

+ bk-1) E R.

2 18

VI Further Topics

e:

If 0 # a E S choose f(x) monic of minimal degree in R[x] with

f(4 = 0, say

f ( x ) = a,

+ a,x + ... + xn.

Then a, # 0 (by minimality of degree) and a, =

-a(al

+ a,a + ... + a " - ' ) E R n Sa.

Thus a, = ba for some b E S, and 1 = a; ' a , and S is a field.

= (a;

'b)a,so a has an inverse in S

In order to obtain a useful characterization of integrality we require the following technical proposition.

Proposition 4.2. Suppose that A = (sij) is an n x n matrix with entries from a ring S, that u I , u 2 , .. . ,u, E S, and that

$ sijuj = 0,

I 5 i 5 n.

j= 1

Then (det A)uj = 0,

1 I j I n.

Proof. Fix j and denote by A j the matrix obtained from A by multiplying column j by u j . Thus det A j = (det A ) u j . But det A j is unchanged if we multiply column k of A j by uk and add to columnj for each k # j. The resulting matrix has ij-entry C;!, sikuk= 0 for every i, so its determinant is 0, and hence (det A ) u j = 0.

Proposition 4.3. Suppose R is a subring of S and a E S. The following are equivalent: (a) (b) (c) aM G

a is integral over R, R[a] is a finitely generated R-submodule of S, and there is a finitely generated R-submodule M of S with 1 E M and M.

Proof. (a) (b): If f(x) E R[x] is monic of degree n with f(a) = 0, then a" is an R-linear combination of { 1, a,. . . ,a"- 1, as are all higher powers of a. Thus R [ a ] = R(l,a,.. . , a n - ' ) . (b) =j (c): Take M = R [ a ] . (c) (a): Let { ul,. . .,u,} E S be a set of generators for M as an R-module. Multiply each ui by a and we may write

'

au, = Sl'Ul

+ . ' . + S,"U",

4 Integral Dependence and Dedekind Domains

2 19

for suitable sij E S . Thus

5

(sij -

j= 1

ad,,)uj = 0,

15i

n,

where 6 , is the usual Kronecker delta. Denote by A the n x n matrix (sijasij);by Proposition 4.2 we have (det A)uj = 0 for all j . Write 1 = C j b j u j , bj E R ; then det A = (det A ) . 1 =

c bj(det

A ) u j = 0.

J

But then f ( x ) = (- 1)" det(sij - xSij) is a monic polynomial in R [ x ] with f(a) = 0, so a is integral over R.

Corollary. If a,, a,,. . . ,ak E S are integral over R, then R[a,, a,, . . . , ak]is a finitely generated R-module. Proof. Induction on k. Note that a, is integral over R[a,, .. . ,ai- since it is integral over R.

Exercise 4.1. Suppose R, S , and Tare rings, with S integral over R and T integral over S. Show that T is integral over R. (See Proposition 111.1.5.) If R is a subring of S the inregrat closure of R in S, denoted Ints(R), is the set of all elements a E S that are integral over R. Clearly R E Int,(R). If R = Int,(R) we say that R is integrally closed in S . Proposition 4.4. If R is a subring of S, then Int,(R) is a subring of S and Int,(R) is integrally closed in S . Proof. If a, b E Int,(R), then M = R[a,b] is a finitely generated Rsubmodule of S containing a - b and ub, so a - b and ub are in Int,(R) by Proposition 4.3(c). It follows that Int,(R) is a ring, and Int,(R) is its own integral closure in S by Exercise 4.1. An integral domain R is called integrally closed if it is integrally closed as a subring of its field F of fractions. For example, R = Zis integrally closed. This fact is proved and generalized in the next proposition.

Proposition 4.5. If R is a UFD, then R is integrally closed. Proof. Let F be the field of fractions of R and suppose u E F is integral over R. Write u = a / b with a, b E R. We may assume that (a, b) = 1 in R. Since u is integral we have tin

= an/V =

c riai/bi

n-1

i=O

This Page Intentionally Left Blank

4 Integral Dependence and Dedekind Domains

221

Proof. Define f;: K + F , 1 I i I n, by setting f;.(a) = Tr,,,(uia) for all a E K . Then f;. E Hom,(K, F ) (the dual space of K ) , and it follows from the corollary to Proposition 111.7.4 that f ; # 0. Thus dim,(ker,i.) = n - I. For eachj set V j = n{kerfi: i # j}. I t is easy to verify by a dimension argument that V, # 0 (in fact, dim V j = 1). Choose uj # 0 in V,. Let us show that uj 4 ker fj. If u j E ker fj, then Tr(uiuj) = 0 for all i. Choose y E K with Tr( y) = 1 (the corollary to Proposition 111.7.4) and write yu;' = x i a i u i , aiE F, so y = aiuiuj.Then

xi

1 = Tr(y) = xuiTr(u,uj) = 0,

a contradiction. We may assume, then, that f j ( v j )= 1, and hence Tr(uiujj = hij for all i and j. If x j b j u j = 0, bi E F , then

0

=

( j

so { u j ) is a basis.

)

11 1bjuj

=

1hjTr(uiuj) = bi

for all i,

i

We remark that { u i } is called the dual basis for { u i > relative to the bilinear form (x, y ) Tr,,,(xy) ~ on K .

Theorem 4.9. Suppose R is an integrally closed domain with field F of fractions, K is a finite separable extension of F , and S = Int,(Rj. There is an Fbasis { u l , . . . , u , , } for K such that the free R-module R(u,,.. . , t i n ) contains S. Proof. By the corollary to Proposition 4.7 there is a basis { u l , . . . ,u,,} c S for K over F , and by Proposition 4.8 there is a dual basis { u l , . . . ,u,} with TrK/,(uiuj)= hij for all i and j. If a E S write a = biui, with all bi E F . Then au, E S, and hence Tr,,,(aUi) E R, all i, by the corollary to Proposition 4.6. But

1,

so U E ~ R=UR (~u 1 , ...,u,,). i

Corollary 1.

If R is Noetherian, then S is Noetherian.

Proof. The R-module R(u,, . . . ,u , ) is Noetherian by Proposition IV.2.5, so S is a Noetherian R-module by Proposition IV.1.8. Ideals of S are Rsubmodules, so S satisfies the ACC for ideals, i.e., S is a Noetherian ring.

Corollary 2. If R is a PID, then S is a free R-module of rank n

= [K:F].

Proof. By Theorem IV.2.9 S is a free R-module of rank n or less. It follows from the corollary to Proposition 4.7 that rank (S) = n.

222

VI Further Topics

In the setting of Corollary 2 there is a free R-basis { wl,. . . ,w,} for S . Such a basis is commonly called an integral basis. It will prove useful to generalize the notion of ideal for integral domains. If R is an integral domain with field F of fractions, then a fractional ideal of R is a nonzero R-submodule M of F such that rM E R for some r E R*. Note that then I = r M is an R-submodule of R, i.e., an ideal in R. Basically, M _c F is a fractional ideal if M # 0 and we may simultaneously clear all denominators in M to obtain an ordinary ideal in R. For example, M = { 3 a / 2 : a E Z > G Q is a fractional ideal of Z, with 2 M = (3) E Z. Clearly each ideal in R is a fractional ideal; when it is important to distinguish them from other fractional ideals ordinary ideals will be called integral ideals. If a E F*, then M = Ra is a fractional ideal of R; it is called the principal fractional ideal generated by a. Suppose M and N are fractional ideals of R. Define M

+N

=

{a + b:aE M , bE N } ,

and

1 aibi:aiE M , bi E N , 0 < k E Z

i= 1

just as for integral ideals. Also, define ( M : N )= { a E F : a N E M } ,

often called the quotient of M by N . If M N = R, then M is called invertible. If M I = N for some integral ideal I in R we say that M divides N , and write M I N . Let us write 9 = SRfor the set of all fractional ideals of R, 9 = 3, for the subset of invertible fractional ideals, and 9 = YRfor the subset of principal fractional ideals. The following exercise gathers together a number of elementary facts about fractional ideals.

+

Exercise 4.2. (1) If M , N E 9 show that M n N , M N , M N , and (M:N)E.F. (2) Show that B is a commutative monoid relative to multiplication, with identity element R. (3) Show that 9 is a group, with identity R. If M E 9 show that M - ' = ( R :M ) . (4) Show that 9is a subgroup of 9, with (Ra)-' = Ra-' for each a E F*. ( 5 ) If M , N E B and M I N show that M 2 N . Conclude that if M 1 N and N I M then M = N . Proposition 4.10. If R is an integral domain and M finitely generated R-module.

E YR,then

M is a

44 Integral Integral Dependence Dependence and and Dedekind Dedekind Domains Domains

Proof. R = MM-' Proof. Write Write 11 E E R = MM-' cc E M , then cbi E M M - ' = R for E M ,then cbi E M M - ' = R for cc = =c c .. 11

223 223

as as 11 = =x x yy = = ll aa ii bb ii ,,witha,€ witha,€M M ,, biE biE M M -- '' .. If If all i, so all i, so = = C(cbi)ai C(cbi)ai EE i i

1 1Rai. Rai. i i

Thus Thus M M = = R(al R(al ,,.., ,,..,a,). a,). FR show Exercise If R R is is aa Noetherian Noetherian domain domain and and M M E E FR show that that M M is is aa Exercise 4.3. 4.3. If finitely generated R-module. (Hint: I = r M , an integral ideal, for some Write finitely generated R-module. (Hint: Write I = r M , an integral ideal, for some rr E R* and show that I and M are isomorphic as R-modules). E R* and show that I and M are isomorphic as R-modules). Proposition Proposition 4.11. 4.11. If If every every nonzero nonzero integral integral ideal ideal in in R R is is invertible, invertible, then then 9 = 4, i.e., every fractional ideal is invertible, and F is a group. 9 = 4,i.e., every fractional ideal is invertible, and F is a group. Proof. %, then I = rM is an integral ideal for some r E R*, so Proof. If If M M E E %, then I = rM is an integral ideal for some r E R*, so clearly M = r ' l . Thus clearly M = r - ' l . Thus M M (( rr 11 -- '' )) = = (( rr -- '' II )) (( rr ll -- '' )) = = R, R,

so so M M -- '' = = rr ll -- '' and and M M €9. €9. Proposition Proposition 4.12. 4.12. Suppose Suppose R R is is an an integral integral domain, domain, II ,, and and I, I, are are nonzero nonzero R, and 1'1, = I E &. Then also I , , I, E $R. integral ideals in integral ideals in R, and 1'1, = I E &. Then also I , , I, E $R. Proof. Say I-' I-' = = JJ ,, so so JJ II 11 ll 22 = = R. R. Thus Thus JJ II ,, = = I;' I;' and and JJ II ,, = =1 1 ;; '' .. Proof. Say For For the the remainder remainder of of this this section section we we will will be be concerned concerned with with writing writing ideals ideals as as for products of prime ideals. Such products have been defined only finitely products of prime ideals. Such products have been defined only for finitely many many prime prime ideals, ideals, and and that that finiteness finiteness will will be be understood understood even even when when not not to agree that the product explicitly mentioned. It will also be convenient of the the explicitly mentioned. It will also be convenient to agree that the product of empty R .. Of Of course, course, the the product product of of aa set set empty collection collection of of (nonzero) (nonzero) prime prime ideals ideals is is R {{ P P }} containing containing just just one one prime prime ideal ideal PP is is just just P P itself. itself. Proposition Suppose R R is is an an integral integral domain domain and and Pl,. Pl,... .. ,,P, P, are are Proposition 4.13. 4.13. Suppose R, and set invertible prime ideals in invertible prime ideals in R, and set II = ii ss m =n n{ {C C :: ll I I m }} E E R. R. If ii S If Q1,. Q1,... ..,,Q, Q, are are also also prime prime ideals ideals and and II = =n n {{ Q Q ii :: 11 I I S n}, n}, then then m m= = nn and and {Pi,. . ., Pm) = (Qi,. . ., Q,). {Pi,. ..,Pm) = (Qi,. . ., Q,).

ni

Proof. Proof. Induction Induction on on m. m.We We may may assume assume that that Pl Pl is is minimal minimal in in the the set set {Pi), {Pi), in the sense that if pi G Pl, then PJ = Pl . We have Pl 2 I = Qi, so in the sense that if pi G Pl, then PJ = Pl . We have Pl 2 I = Qi, so Pl Pl 2 2 Qi Qi for since Pl Pl is is prime; prime; we we may may assume assume P, P, 2 2 Q1. Q1. By By the the same same argument argument for some some ii since Q1 Q12 2 pj pj for for somej, somej, so so Pl Pl 2 2 Q1 Q12 2 pj pj and and hence hence Pl Pl = = Q1 Q1 = = pj pj by by the the minimalminimality of Pl. Multiplying by P ; l we obtain ity of Pl. Multiplying by P ; l we obtain nn {{ R ii I ii I nn }} ,, R :: 22 I I Im m }} = =n n {{ Q Q ii :: 22 I I I and and the the proposition proposition follows follows by by convention convention ifif m m= = 1, 1, by by induction induction ifif m m> > 1. 1.

224

V1 FurtherTopics

The point of Proposition 4.13 is that when factorization of an ideal as a product of invertible prime ideals occurs, it is unique. Proposition 4.14. Suppose that R is an integral domain and that every proper ideal in R is a product of prime ideals. Then every invertible prime ideal P in R is maximal. Proof. Take a E R\P and set I = P + Ra. We must show that I = R , so suppose I # R . Set J = P + Ra2 and write I = Pl’.‘Pk, J = Q I . . . Q m for prime ideals 8, Q j . Thus 4,Qj 2 P for all i and j. Write R for RIP, ii = a + P E R, etc. Then 7 = PI . . . Pk and

Since r a n d J a r e principal ideals in R they are invertible (Exercise 4.2.4),so all by Proposition 4.12. By Proposition 4.13 we may conclude that m = 2k, and by relabeling that Q1 = Q2 = pl,.. ., Q2k- = Q z k= pk.Since all contain P it follows that = Qzi = Q z i - for each i , and hence J = 1’. Thus P c J = ( P + R U )E ~ P 2 + Ra. If x E P write x = y + za, with y E P 2 and z E R . Then za = x - y E P, and a q! P, so z E P and x E P 2 + Pa. Thus P E P 2 + Pa E P and P = P 2 + Pa. Multiply by P - to see that R = P + Ra = 1 and the proof is complete.

4 and ej are invertible

Proposition 4.15. If R is a Noetherian domain, then every nonzero integral ideal in R contains a product of nonzero prime ideals. Proof. Let Y denote the set of nonzero integral ideals in R for which the conclusion fails. We must show that Y = 0. If not there is a maximal element I in 9’ (Proposition 11.7.1). Since I cannot be prime there are a, b E R\I with ab E I . By maximality the ideals (a) + I and (b) + I are not in Y so each of them contains a product of prime ideals, and consequently their product does likewise. But then [(a)

+ l ] [ ( b )+ I ] c (ab) + a1 + bl + I E I,

so I contains the same product, a contradiction.

If F G C is a finite extension of the rational field Q then F is called an algebraic number field. The integral closure R of Z in F is called the ring of algebraic integers in F. Thus an algebraic integer in F is any a E F that is a root of some monic f (x) E Z[x]. The rings R , in Chapter I1 provide a class of examples of rings of algebraic integers. Note that in general F is the field of fractions of R by Proposition 4.7. If R is the ring of algebraic integers in an algebraic number field F , then R is Noetherian by Corollary 1 to Theorem 4.9 and R is integrally closed by Proposition 4.4. In the next theorem we establish a third important property

4

Integral Dependence and Dedekind Domains

225

of R, and then use those three properties as defining properties for Dedekind domains. Theorem 4.16. Suppose F is an algebraic number field and R is its ring of algebraic integers. If P is a nonzero prime ideal in R , then P is maximal. Proof. Take a # 0 in P and let m ( x ) = m,,,(x) be its minimal polynomial over Q. Then m ( x )E Z [ x ] by Proposition 4.6. If m ( x ) = a.

+ a,x + . . . + x k ,

then a, # 0, and

a,

= -(ala

+ ... + ak) E P,

so P n Z # 0. It is easy to verify that P n Z is a prime ideal in Z, so P n Z = pZ for some prime p E Z.Thus ZlpZ

=

Z I P n Z 2 (Z + P ) / P 5 R I P .

Since R is integral over Zit is immediate that R I P is integral over (Z + P ) / P . But (Z + P ) / P z Z,is a field, so RIP is a field by Proposition 4.1, and hence P is a maximal ideal in R .

Exercise 4.4. (1) Show that the prime p E P n Z in the proof above is unique. (2) Show that R I P is a finite field (see Corollary 2 to Theorem 4.9). A Dedekind domain is a Noetherian integral domain that is integrally closed, in which every nonzero prime ideal is maximal. Thus, for example, the ring of algebraic integers in an algebraic number field is a Dedekind domain. We shall see shortly that several alternative definitions are possible for Dedekind domains, some of them perhaps aesthetically more appealing than the one above. The definition above, however, tends to be more easily verified in specific cases.

Proposition 4.17. If R is a PID, then R is a Dedekind domain. Proof. Propositions 11.5.9 and 11.5.10, Exercise 11.8.24, and Proposition 4.5. Exercise 4.5. Show that R is a Dedekind domain that is not a PID (see Exercise 11.5.1). Show that the polynomial ring Z [ x ] is a Noetherian domain that is not a Dedekind domain. ~

Exercise 4.6. Suppose R is a Dedekind domain with field F of fractions, K is a finite separable extension of F , and S = Int,(R). Show that S is also a Dedekind domain (see Proposition 4.4, Corollary 1 to Proposition 4.9, and the proof of Proposition 4.16).

226

VI Further Topics

Proposition 4.18. If R is a Dedekind domain and I is a proper integral ideal in R, then ( R : I )# R [clearly ( R : l )2 R ] . Proof. By Proposition 11.1.8 there is a maximal (and therefore prime) ideal P with I S P E R . Choose a E I * and apply Proposition 4.15 to obtain nonzero prime (and therefore maximal) ideals P , , . ..,pk in R with PI. ' . Pk c Ru G P . We may assume that k is minimal in that respect. If P were different from all 4 we could choose a, E e\P for all i (since all 8 are maximal). But then a l a z. . ' a kE P , a contradiction since P is prime. Thus we may asE Ra E P. Since k is sume by relabeling that P = PI, and hence PP, minimal we may choose b E P 2 .- .pk\ Ra. If c = b/a, then c $ R (or else b E Ra), but Cl = (b/U)f E

(b/a)PZ (l/U)PP2.--Pk E R,

so c E (R:l)\R, as required. Proposition 4.19. If R is a Dedekind domain and P prime ideal, then P is invertible, with P - ' = ( R : P ) .

c R is a nonzero

Proqf. Note that P c ( R : P ) PL R by the definition of (R:P).Since P is maximal we have either ( R : P ) P= P or ( R : P ) P= R. Suppose ( R : P ) P= P . If a E ( R : P ) ,then U P 5 P , so a 2 P E U P E P , and by induction a"P c P if 0 < n E Z . Thus b d E P c R for any nonzero b E P and every n, so R [ a ] = R(1,a,a2,. . .) is a fractional ideal of R. By Proposition 4.10 it is finitely generated. Thus a is integral over R by Proposition 4.3, and so a E R since R is integrally closed. But that means ( R : P )c R, which contradicts Proposition 4.18. Thus ( R : P ) P= R , so P E 9; and P - ' = ( R : P ) .

Theorem 4.20.

If R is an integral domain the following are equivalent.

(a) R is a Dedekind domain. (b) Every (integral) ideal I in R is a product of prime ideals. (c) Every nonzero (integral) ideal 1 in R is uniquely a product of prime ideals. (d) Every fractional ideal M of R is invertible, i.e., FRis a group with respect to multiplication.

ni

Proof. (a) (b): If 0 # I # R apply Proposition 4.15 to obtain nonzero prime ideals Pl, . . . ,pk with q E I . We may assume that k is minimal and apply induction on k. If k = 1, then I = Pl since PI is maximal. Suppose then that k > 1 and assume that (b) holds for ideals containing products of fewer than k nonzero prime ideals. Choose a maximal (and hence prime) ideal P with 1 E P , so P, E P . Then some C P , so we may assume p k E P , and hence Pk = P since Pk is maximal. By Proposition 4.19 P is invertible. Multiply by P - ' to see that P ~ " ' P k -E l I F ' . Note that IP-' # R for otherwise I = P , =j

ni

4 Integral Dependence and Dedekind Domains

227

contradicting the minimality of k > 1. By the induction hypothesis there are prime ideals Q1,. . . ,Q , with I P - = Q 1.. . Q,, and hence I = Q 1 .. . Q,P. (b) 3 (c): Apply Proposition 4.13. (c) +(d): It will suffice, by Proposition 4.11 and (c),to show that every nonzero prime ideal P is invertible. Choose a E P , a # 0, and write R a as a product of prime ideals, say Pl ... Pk = R a G P . Then Ra is invertible by Exercise 4.2.4 and each fl. is invertible by Proposition 4.12. Since P is prime P 2 p i , so P = c since 4 is maximal (Proposition 4.14). Thus P is invertible. (d) (a): If I is a nonzero (integral) ideal in R , then I is finitely generated by Proposition 4.10, so R is Noetherian by Proposition 11.7.2. We must show that R is integrally closed (in its field F of fractions). Take a = b / c E F , with b, c E R , integral over R; say a is a root of f ( x ) = a0

+ a l x + ... + x k E R [ x ] .

Let M be the fractional ideal of R generated by { 1, a,. ..,a k -'3. Then Uk

=

-(a0

+

U,U

+ '.'

ak-1Uk-l) E

M.

It follows easily that M2 = ( R + R a

+ ... + R a k - ' ) ( R + R a + ... + R a k - ' ) = M,

so M = MM = R , and hence a E R , as required. Suppose finally that P # 0 is a prime ideal in R , and take a E R\P. Set J = P + R a and set M = J-'P. Then J M = JJ-'P = P, so MI P and hence M 2 P . If b E M , then ah E J M = P , a n d a 4 P , s o b E P . T h u s M = P o r J - ' P = P . Multiplyby P - ' to see that J - ' = R and hence J = R . Thus P is maximal and the proof is complete. There are several other characterizations of Dedekind domains. Two that are well known involve the notions of localization (see Section 2 of Chapter 11) and of projective modules (see Exercise IV.7.45). Exercise 4.7. Suppose R is a Dedekind domain with group 9 of fractional ideals. Show that 9 is a free abelian group with basis the set of nonzero prime ideals of R .

Theorem 4.21.

A Dedekind domain R is a U F D if and only if it is a PID.

Proof. Suppose R is a UFD.It will be sufficient to show that each nonzero prime ideal P in R is principal. Take a E P, a # 0, and write a = n { p p i :1 5 e, E 271,where the p i are distinct primes. Then pi E P for some i, and hence Rpi E P. Since pi is prime Rp, is a prime ideal, so it is maximal, and p = Rp, is principal. Exercise 4.8. If R is a Dedekind domain and M, N if and only if M 2 N .

E 9R show

that M I N

228

VI Further Topics

If R is a Dedekind domain and M , , M ,

E

FR, then a greatest common

I

divisor (GCD) for M, and M , is any M E 9 such that M I M , , M M , , and if N E 9 with N M , , N M , , then N M . A least common multiple ( L C M ) for M , and M , is any M E .F such that M , M , M , M , and if N E 9 with M , 1 N , M , N , then M 1 N . There are obvious extensions to three or more

I

I

1

1

I

I

fractional ideals. If G C D ( M ,,M 2 ) = R we say that M , and M2 are relatively prime. Exercise 4.9. Suppose R is a Dedekind domain. (1) If M , , M 2 € 9 show that G C D ( M , , M , ) = M , + M , and L C M ( M , , M , ) = M , n M2 . Conclude in particular that the GCD and LCM exist and are unique. (2) If Z and J are proper nonzero integral ideals write I

= n{PY':OI ai E

Z, 1


J

= n{P!l:O I bi E

Z,1

I i 5 k},

k}

and

where the Pi are distinct prime ideals and Pp is understood to be R . Set ci = min{ai, bi} and d, = max{ai,bi},for each i. Show that G C D ( 1 , J ) = n { P i n :1 I i Ik} and L C M ( 1 , J ) = n { P f z 1: I i I k } .

+

Conclude that I J = ( I J ) ( l n J ) and that If = I n J if and only if i and J are relatively prime. The quotient group %? = WR = 919'is called the ideal class group of a Dedekind domain R (or of its field F of fractions). Note that l%?l= 1 if and only if R is a PID (and hence a UFD), so /%?I can be viewed as a measure of deviation from unique factorization for elements of R. It can be shown that I%?[= h is finite for algebraic number fields; the class number h is an important numerical invariant in algebraic number theory. Historically the notion of ideal was introduced by Dedekind, following ideas of Kummer, to provide a replacement for unique factorization of elements when rings of algebraic integers are not UFDs. The next proposition will show that, in a rather different sense, ideals in Dedekind domains are never very far from being principal (see the corollary). Proposition 4.22. Suppose R is a Dedekind domain and I , J are proper nonzero integral ideals in R. Then there is an integral ideal L with J L = R (i.e., J and L are relatively prime) and I L E gR.

+

4 Integral Dependence and Dedekind Domains

229

Proof. Write I

=

n(Pf1:OI a, E Z,1 I i I k}

J

=

n { P P o : O2 b, E Z,1 I i Ik }

and with distinct prime ideals 4. For each i choose a, E P:i\P:if' (this is possible because of unique factorization). The ideals Py1+ are pairwise relatively prime so we may apply the Chinese Remainder Theorem (11.6.2) to obtain a E R with a = a,(modP:'+') for all i. Thus a E P;l\P:i+l, all i, so a E P:i = = I . It follows that Pp')Ra for all i. Also Ppi)1J for all i. Since Ppi+l$ Ra it follows from Exercise 4.9 that

ni

+ I J = n P f 8= I . If we write Ra = PFn,then e, 2 a, for all i since I I Ra. Set L = Pfip4', so that L is an integral ideal with 1 L = Ra E 9. Also I = Ra + I J = I L + I J . Multiply by I to see that GCD(L, J ) = L + J = R , so Land J are relatively GCD(Ra,I J ) = Ra

n,

I

-

prime.

Corollary. Each integral ideal I in R requires at most two generators. Proof. We may assume 0 # 1 # R . Choose a E I , a # 0, and apply the proposition to 1 and J = Ra to obtain an integral ideal L with I L = R b for some b E R* and Ra + L = R . Then Ra + Rb = Ra + I L G I . Write 1 = ra + c, with r E R and c E L. If d E I , then d = dra Thus I

+ dc E Ra + I L = Ra + Rb

= (a,b).

We end this section with a detailed look at an easy numerical example. Take R = R p and F = O ( n ) as in Section 5 of Chapter 11. Then R is not a UFD, and 6 = 2 . 3 = (1 + F 5 )(1 - F 5 ) is an example of different factorizations into nonassociated primes. Set l1 = (2), I, = (3), J, = (1 + and J 2 = (1 - F 5 ) , so (6) = Ill, = J , J 2 . Set

F), P,

= GCD(I,, J 1 ) =

I,

+ J,

=

(2,l

+ J--5).

Let us see that Pl is a prime ideal. (Why is Pl # R?) Suppose (x + ) E p1,i.e.,

x (u

uJ-3

(xu - 5yu)

Thus x u

+ (xu + y

u ) p = 2a

+ y u = b and

xu

-

5yu = 2a

+ b + b-

+y

p

for some a, b E h.

+ b = 2a + xu + yu,

)

230

VI Further Topics

so u(x - y ) - u(x

+ 5 y ) = 2a = 0 (mod 2),

and therefore u ( x - y) - u(x - y ) = (u - u)(x - y ) = 0 (mod 2),

so we may assume u = u(mod2). Writing u = u 2c u(l E P,, and P, is prime.

+

+ J-5)

+

Exercise 4.10. Set P2 = I , J 2 , P3 = I , that each is a proper prime ideal.

+ 2c we have u + u

+ J , , P4 = I , + J 2 , and show

Continuing with the example it is clear that I E P, n P2 = Pl P,. If there are ai, b i , c i , d i E Z with =C[2ai i

E (2)

p=

c(

E PIP,

+ b i ( l + P ) ] [ 2 c i+ di(l - J--5)]

+ C bidi(6) E (2) + ( 6 ) E ( 2 ) = I , , 1

so I ,

= P,P2. Similarly, 1, = P3P4, J , =

( 6 )= Ill2 = PlP2P3P4

PIP3,and J2 = P2P4. Thus = JlJ2 = PlP3P,P4,

and uniqueness of factorization is restored at the level of ideals.

5. TRANSCENDENTAL FIELD EXTENSIONS Throughout this section F will denote a field and K 2 F an extension field. A set S E K is algebraically dependent over F if there are distinct elements a,, a 2 , .. . ,ak E S and a nonzero polynomial f ( x 1 , x 2 , . . ., xk) E F [ x ~ , x. .~.,,xk]

with f(a,, a,,. . .,ak)= 0. Otherwise S is algebraically independent over F . Note that a set S = {a> E K with just one element is algebraically dependent if a is algebraic over F; it is algebraically independent if a is transcendental over F. In general an algebraically independent set S G K is called a transcendence set over F. Exercise 5.1. Show that S E K is algebraically dependent over F if and only if there is some a E S that is algebraic over F(S\{a}).

5 Transcendental Field Extensions

23 1

If K is not algebraic over F, then K is called a transcendental extension. If K = F ( S ) for some transcendence set S over F , then K is called a purely transcendental extension of F. The simplest example of a purely transcendental extension is a field K = F ( x ) of rational functions in an indeterminate x over F , or more generally K = F ( x , , x 2 , .. .,x,) with distinct indeterminates x l , . . . ,x, over F . Exercise 5.2. Show that the real field R is a transcendental extension of the rational field Q, but that it is not purely transcendental. A transcendence set B c K over F is called a transcendence basis if it is maximal with respect to set inclusion. By an easy Zorn's Lemma argument any transcendence set S E K is contained in a transcendence basis B. In particular K has a transcendence basis over F. Note that K is algebraic over F if and only if the empty set 0is the only transcendence basis for K over F.

Proposition 5.1. Suppose S C_ K is a transcendence set and a E K\S. Then S LJ { a } is algebraically dependent over F if and only if a is algebraic over F(S). Proof. =.: There is a finite (possibly empty) set { b l , .. . ,b k } G S and a nonzero polynomial f b o , X l , . . %) E ' 1

FCX,,

x, 9 . . . ,X k l

with f(a, b l , . . . ,bk)= 0. Note that xo actually occurs in j ( X ) , since otherwise there would be a nontrivial algebraic dependence relation on S. Define d x o ) = f(x03b,,...,bA E

W)Cxol

and observe that g ( x o ) # 0 but g(a) = 0, so a is algebraic over F(S). G : If g f a ) = 0 for a nonzero polynomial g ( x )over F(S), then we may clear denominators in the coefficients of g(x) and assume that g ( x ) E F[b,,. . . ,b,][x] for some {bl,. . . ,b,} c S. But then the equation g(a) = 0 is in fact a nontrivial algebraic dependence relation over F for {a, 6 1 , . . . ,b k } ,and hence for S u { a } . Corollary. If S is a transcendence set for K over F , then K is a transcendence basis if and only if K is algebraic over F ( S ) . Exercise 5.3. If S G K and K is algebraic over F ( S ) show that there is a transcendence basis B for K over F with B G S.

If S

cK

write a@) = a,,,(S)

= { aE

K : a is algebraic over F ( S ) ) ,

commonly called the algebraic closure of F ( S ) in K . The following

232

V1 Further Topics

observations are easy consequences of the definition of a(S): (i) (ii) (iii) (iv)

S E a@);

if S E T E K , then a(S) G a(T); if a E a(S), then a E a(S’) for some finite set S’ E S ; and a(a(S)) = a(S)

[(iv) follows from Proposition 111.1.51. These four facts, together with the result of the next proposition, show that a determines a “dependence relation” as defined abstractly in Section lof Chapter I of Zariski and Samuel [41]. Proposition 5.2. a E a(S u (6)).

Suppose S E K , a, b E K , and b E a(S u {a})\a(S). Then

Proof. Set L = F(S), so b is transcendental over L but algebraic over L(a). Then {a, b} is algebraically dependent over L by Proposition 5.1. Choose S(x1,x2)# 0 in L [ x , , x , ] with f ( a , b ) = 0, and note that x, must occur in f ( x , , x,), for otherwise b would be algebraic over L. Thus with g(a) = 0, so a E a(S u {b)).

Theorem 5.3. If A and B are transcendence bases for K over F , then

I 4 = IBI.

Proof. We may assume that 0 < [ A ]5 IBI. Suppose first that A = { a l , a 2 , . . ,a,,} is finite. Note that B $ a ( A \ { a , } ) , e.g., since a, E a(B) but a, 4 a(A\{a,}). Choose b , E B\a(A\{a,}) and observe that A , = { b 1 , a 2..., , a n }is a transcendence set by Proposition 5.1. Also a, E a(A,) by Proposition 5.2, and it follows that A , is a transcendence basis. Replacing A by A , begins an induction. Suppose now that 1 < k I n and that b,, b,, . . . , bk- E B have been chosen so that 1

= {bl

9 . .

.?

bk-

l,akr.

..)a n }

is a transcendence basis. Then B $ a(&_ l\(ak)) and we may choose bk E B\a(&,\{ak})a It fOllOWS as above that

Ak

=

{ b , , . . ., bk, ak + 1 > . ., an}

is a transcendence basis. Take k = n, then An E B is a transcendence basis, so A, = Band IBI = (AnI = IAl. Suppose then that A is infinite. For each a E A there is a finite set B, G B with a E a(B,). Clearly B = U { B , : a E A}, for if U { B , : a E A } = C # B,then B E K = a ( A ) 5 a(C),contradicting the algebraic independence of B. Thus

IB(= IU{B,:a E A } ( I KoIAl and hence IBI

=

IAl.

= (A/,

5 Transcendental Field Extensions

233

The cardinality of any transcendence basis for K over F is called the transcendence degree of K over F ; it will be denoted by T D ( K : F ) . Exercise 5.4. Show that cise 111.1.4).

Proposition 5.4.

If F

EL

TD(R:Q)

=

T D ( @ : Q= ) lRl (see Exer-

E K , then T D ( K : F )= T D ( K : L )+ TD(L:F).

Prooj. Let A be a transcendence basis for L over F and B a transcendence basis for K over L; clearly A n B = @. Then L is algebraic over F ( A ) ,so L(B)is algebraic over F(A)(B) = F ( A u B); hence K , being algebraic over L(B), is algebraic over F ( A u B) by Proposition 111.1.5. It will suffice, then, to show that A u B is algebraically independent over F . Suppose 0 # f ( X , Y ) E F [ X , Y ] gives a dependence relation f ( u l , . .. ,a,, b,, .. .,b,) = 0, with a, E A , bj E B. Then f ( a , , . . . ,a,, h , , . . . ,h,) can be viewed as a polynomial evaluated at b , , . . . ,b, and having coefficients of the form g(a,,.. . ,a,,,) E F [ a , , . . . ,a,]

G

F(A).

All the coefficients must be 0 since B is a transcendence set over L 2 F ( A ) . But then each such g ( X ) = 0 in F [ X ] since A is a transcendence set over F , and hencef(X, Y ) = 0 in F [ X , Y ] .Thus A u B is a transcendence basis for K over F , and

T D ( K : F )= [ A u BI

= 1.41

+ IBI = T D ( L : F )+ T D ( K : L ) .

Exercise 5.5 (D. Gay). (1) Suppose S, T G K are transcendence sets over F and 6:S -, Tis a 1-1 correspondence. Show that there is a (unique) F-isomorphism T : F ( S ) -+ F ( T ) with 7 1 S = 0. ( 2 ) Let B be a transcendence basis for K = C over F = Q and choose A E B with IAl = IB\AI, For each S c_ A apply (1) to obtain an isomorphism r,:Q(B) -, Q(B\S). Extend ss to an isomorphism between a(B) = C and a(B\S) (see Theorem 111.1.11). (3) Conclude that C has 21'' proper subfields isomorphic with itself, and furthermore 2'' subfields L 2 Iw with [ @ : L ]infinite and L $ R (see Exercise 111.8.13 and note that there is no monomorphism U : R -+ [w except 6' = lR).

If K is a purely transcendental extension of F with T D ( K :F ) = 1 we may as well take K = F ( x ) ,the field of rational functions in an indeterminate x over F . The transcendence basis most ready at hand is B = {x). If a E K * write a = f ( x ) / g ( x ) ,with f ( x ) and g(x) relatively prime in F [ x ] , and define the degree of a to be deg a = max { deg .f(s),deg g ( x ) j .

Proposition5.5. If K and [ K : F ( i . ) ] = dega.

=

F ( x ) and a E K\F, then a is transcendental over F ,

234

V1 Further Topics

Proof. Say deg u

=n

> 0, and write a

f ( x )= a, g(x) = b ,

= f ( x ) / g ( x ) ,with

+ a , x + . . . + a,x", + b , x + ... + b,x"

in F [ x ] , and at least one of a,, b, nonzero. Let y be an indeterminate over K and set h(Y) = h,(Y) = C d Y ) - f ( Y ) E FL-alCyl

KCYI.

Then h( y ) has leading coefficient ub, - a,, so deg h( y ) = n, and h(x) = 0, so x is algebraic of degree n or less over F(a).Thus u must be transcendental over F, for otherwise x would be algebraic over F. If h,(y) were reducible in F [ a ] [ y ] it would also be reducible in F [ u ] [ y ] = F [ a , y ] , by Proposition 11.5.15. Since h,(y) is of degree 1 in a any factorization would take the form h,(y) = u(y)u(u,y), where u ( y ) is of degree 0 in a, i.e., u ( y ) E F [ y ] , and u(u, y ) E F [ u , y ] is of degree 1 in a. If we apply the homomorphism 4: F [ u , y ] + F [ y ] determined by $(a) = 0, $ ( y ) = y , to the equation a g ( y ) - f ( y ) = u(y)u(u,y ) we obtain f ( y ) = -u(y)o(O, y ) , so u ( y ) l f ( y ) in F [ y ] . But then also

4 Y ) I as(Y)= U Y ) + f ( Y h

and consequently u ( y )1 g ( y ) . Since ( f (y ) ,g ( y ) )= 1 in F [ y ] we conclude that degu(y) = 0 and h,(y) is irreducible over F(a). Since K = F(x) = F(u)(x)we conclude further that [ K :F(u)] = n = deg a.

Corollary 1. The minimal polynomial m ( y ) for x over F ( a ) is an F(!x)multiple of ug( y ) - f ( y ) .

Corollary 2. If K deg u = 1, i.e., u = (ax

=

F ( x ) and u E K\F,

then K = F(a) if and only if

+ b ) / ( c x + d), with a, b, c, d E F and ad # bc.

Proof. Since [ K :F ( a ) ] = deg a we have K = F(a) if and only if deg a = 1, hence a = (ax + b ) / ( c x + d ) , as indicated. If ad = bc, then either a = a / c or a = b / d , both in F.

The next theorem requires a bit of information from the beginning of Section 3, above. Theorem 5.6. If K = F(x), with x transcendental over F, then the Galois group G = G ( K : F ) is isomorphic with the projective general linear group PGL(2, F ) . Proof. Any $ E G must carry the primitive element x to another primitive element a, hence ~ ( x=) ( a x + b ) / ( c x + d ) for some a, b, c, d E F with ad # bc, by Corollary 2 above. Conversely, defining $(x) = (ax + b)/(cx d ) completely determines $ E G since K = F(x). Define a map f : GL(2, F ) + G via

+

5 Transcendental Field Extensions

235

+ b ) / ( c x + d ) . It is a routine matter to check that f is an epimorphism. If A = [:I;] E GL(2, F ) and f ( A ) = 1 E G, then ( a x b ) / ( c x d ) = x , or a x b = c x 2 + dx, so b = c = 0 and a = d . Thus f :Uf;]H4, with @ ( x )= ( a x

+

+

+

kerf

=

(a1 : a E F * )

= Z[GL(2,

F)]

and G z GL(2, F)/Z[GL(2, F ) ]

=

PGL(2, F ) .

It will be convenient for the proof of the next theorem to introduce some temporary notation. If f ( x , y ) E F [ x , y ] is viewed as a polynomial in x with coefficientsin F [ y ] we will write f ( x , y ) as f,(x), and similarly f,( y ) will denote f ( x , y ) viewed as polynomial in y with coefficients in F [ x ] . Thus, for example, if f ( x , y ) = xy

-

xy3

+ x3y + x4y2,

then

A,(.) = ( y - y 3 ) x + yx3 + y z x 4 f,(y) = ( x

+ x 3 ) y + x4y2 - x y 3

and and

degf,(x) degf'(y)

= 4; =

3.

Similar remarks apply to rational functions g ( x , y ) E F ( x , y).

Theorem 5.7 (Luroth's Theorem). Suppose K = F(x), with x transcendental over F , and L # F is an intermediate field, F E L G K . Then L = F ( y ) for some y E K that is transcendental over F . Proof. If p E L\F, then x is algebraic over F ( p ) c L by Proposition 5.5, so x is algebraic over L. Let m,(y)

= a,

+ a,y + ... + y"

be the minimal polynomial of x over L, so [ K : L ] = n [since K = L ( x ) ] . At least one of the coefficients ai = a i ( x )is not in F since x is transcendental over F , say ai = y E L\F. Write y = f ( x ) / g ( x )with f ( x ) and g ( x ) relatively prime in F [ x ] , and say degy = k. Then [ K : F ( y ) ] = k by Proposition 5.5, and k 2 n since F ( y ) G L c K . The theorem will be proved if we show that k = n. We may assume that all coefficients aj = a j ( x )are reduced to lowest terms in F(x); then multiply through m,(y) by the least common multiple b, = b,(x) E F [ x ] of their denominators and cancel the denominators. Thus we repIace m,(y) by u,(Y) = bo

+ b , x + ... + b,x",

with each b j = b j ( x ) E F [ x ] and hence u(x, y ) E F [ x , y ] . It is easily verified that u,(y) is primitive over F [ x ] . Since y = ai = b i / b , = f ( x ) / g ( x )it is clear that deg u,,(x)2 k.

236

VI Further Topics

If we set h,(y) = M y ) - f ( y ) E LCyI, then h,(x) LCYl, say

= 0,

so m,(y) 1 h,(y) in

mx(Y)Px(Y)= W ( Y ) - f ( Y ) = C f ( x ) / s ( x ) I g ( y) f(Y) with p , ( y ) E L [ y ] . Set 4x3 Y ) = f W ( Y ) - f ( y ) g ( x )E FCx, Y l and note that deg r,(y) = deg ry(x) = k. Also mx(Y)P,(Y)g(x) = S ( X ) Y ( Y ) - f ’ ( y ) g ( x )= r x ( y ) ,

(*I

so if the left-hand side is viewed as a polynomial in y over F(x), then all denominators of coefficients must cancel and, since u,(y) is primitive, we may rewrite (*) as u(x, y ) q ( x ,y ) = r(x, y ) for some q ( x , y ) E F [ x , y ] . Observe now that

k

= deg r y ( x )= deg u,(x)

+ deg q y ( x )2 k + deg qy(x),

so deg q,,(x)= 0 and q ( x ,y) = q( y ) E F [ y ] (also deg u,,(x)= k). Since q( y ) is primitive (its nonzero coefficients are units in F [ x ] ) we have u,(y)q(y) primitive by Gauss’s Lemma (Theorem II.5.13), and hence r,(y) = u , ( y ) q ( y ) is primitive over F [ x ] . But r(x, y ) = - r ( y , x ) so it follows that r y ( x )= u,(x)q(y) is primitive over F r y ] and consequently q ( y ) must be a constant, q ( y ) = q E F*. Finally, then, n = deg u,( y ) = deg r,( y ) = k . Luroth‘s Theorem is important in the study of algebraic curves. Loosely speaking its significance there is as follows: If an irreducible curve can be parametrized (finitely many points excepted) by rational functions then it can be so parametrized in a 1-1 fashion (see Walker [38, Chapter V, Section 7 1 ) . There is an analog of Luroth’s Theorem for purely transcendental extensions K = F ( x , y ) of transcendence degree 2, due to Castelnuovo and Zariski. It requires, however, that F be algebraically closed and applies only to intermediate fields L such that K is separable over L. Beyond a few counterexamples very little is known for transcendence degree 3 or higher. Suppose F = Fq, a finite field, and set K = F(x), x an indeterminate. Then the Galois group G = G ( K : F ) is the group of linear fractional transformations ( a x + b)/(cx + d ) as in the proof of Theorem 5.6; G z PGL(2,F) and )GI = (q - l)q(q + 1) = q 3 - q by Proposition 3.1. Set L = F G , the fixed field of G in K . Then K is Galois over L by definition, and [ K : L ] = IGI = q 3 - q by the Fundamental Theorem of Galois Theory (111.2.10). By Luroth’s Theorem L is of the form F ( y ) for some Y E K that is transcendental over F. In order to determine a choice for y it will be convenient to describe first a set of generators for G. For each a E F* define p a E G by setting p,(x) = ax, and for each b E F define rb E G by setting rb(x) = x a. Define a E G by means of a ( x ) = I / x .

+

5 Transcendental Field Extensions

237

Exercise 5.6. Suppose 4 E G. (1) (2) where (3)

If $(x)

= ax

+ 6 show that # = 7,,pa.

+

+

If #(x) = (ax 6)/(cx d ) , with c # 0, show that 4 r = a / c and s = (bc - a d ) / ( . Conclude that G = ( ~ ~ , , , T h , C J : U E F * , b E F ) .

=t , p p ~ ~ p ~ ,

Now set y

= ( x q- x)" 'l/(xq2 - x ) ~ " E F(x) =

K.

Recall that x4 - x = nix - a : a E F } ,

and if Fl is a degree 2 extension of F , then IFl/ = q 2 and xq2 - x = n { x a : a E F l } (see Theorem 111.3.11). Thus if we divide xqz- x by xq - x, then the quotient, which is easily checked to be C;=oxi(q-l), has as roots the elements of Fl \F, so it is relatively prime to x 4 - x. Thus we may write

in lowest terms, and degy = q 3 - q = IC(. Thus [ K : L ] = [ K : F ( y ) ] by Proposition 5.5, so it will suffice to show that y E L in order to conclude that L = F(y). If b E F, then

+

rb(x4 - X) = (X

and likewise z b ( x q 2- x)

6 ) 4 - (X

+ 6) = X'

= xqz- x,

-x

+ 6'

-

6 = X'

-

x,

so T b ( Y ) = y. If a E F* then

p,,(xq - x) = a4x4- ax = axq - ax = a(x4 - x),

and po(xq2- x) = a(x4'

-

x),

so p,(y) = y. Finally, a(x4

- x) =

1/x4

- l / x = (x - xq)/xq+

and fJ(X4*

so ~ ( y = ) y. Thus $(y) L = F(y). L

- x) = (x = y for all

-

92 )/x q 2 + 1 ,

4 E G by Exercise 5.6(3),so y E L = 9 G and

Exercise 5.7. Take F = Z,,K

= F(x),

G

=

G ( K : F ) , and L = S G , so

= F(y), with

y

= (x2 - x)5/(x4

-

x)3 = (2 - x)2/(1

+x +x y .

238

VI Further Topics

Exhibit primitive elements (over F ) for each of the six intermediate fields M , LGMEK. 6. VALUATIONS AND p-ADIC NUMBERS

A Diophantine equation is not actually a specific type of equation, it is rather an equation for which only certain types of solutions are of interest. For example, if f ( x l , x2,.

*.

yxk) E Z[XlrX2,.

.., x k ] ,

then the equation f ( xl, . . . ,xk) = 0 is called a Diophantine equation if the only solutions of interest are solutions x 1 = a , , . . . ,x k = a k ,with all ai E 7. For a concrete example consider x 2 - 7y2 - 11 = 0. Viewed as an equation involving real variables x , y its graph is a hyperbola and the equation has infinitely many solutions. The corresponding Diophantine problem asks whether there are any points on that hyperbola having integers for both coordinates. That might appear to be a more difficult question. If there were an integral solution, however, it would also be a solution to the congruence x 2 - 7y2 = 1l(mod 4), or equivalently x 2 + y 2 = 3(mod 4). Since the square of any integer is = 0 or l(mod 4) the congruence has no solution in Z and, a priori, the Diophantine equation has no solutions. The example suggests some necessary conditions for the existence of solutions to certain Diophantine equations: if f ( x l , .. .,x k ) = 0 is to have integral solutions it is necessary that the congruences f(xl, . . .,x k ) = 0 (mod n) have solutions for all moduli n. Another necessary condition is even more obvious: if the equation has integral solutions it has real solutions. For example, x 2 + y 2 + I = 0 has no solutions in integers. Surprisingly, perhaps, there are some equations for which the necessary conditions above are, collectively, also sufficient. An important example is afforded by pure quadratic equations f(x1,.

. .,X k )

=

1

UijXiXj

= 0,

i.j

with all aijE 7. If that equation has a nontrivial real solution and if there is a nontrivial solution to f ( x l ,. . . ,x k ) = 0 (mod p “ ) for all prime powers p” (0 < m E Z), then there is a nontrivial solution in integers. When the necessary conditions are also sufficient it is commonly said that the “Hasse Principle” applies. Around 1900 K. Hensel constructed a field 6, Q for each prime p E Z, the field of p-adic numbers. Each 6,has a subring R, 2 Zof p-adic integers, with the following property. If f ( X ) E Z[xl, . . . , x k ] , then f ( X ) = 0 has a

6 Valuations and p-adic Numbers

239

p-adic integer solution if and only if all the cor_tgruencesf ( X ) = 0 (mod p"), 0 < m E h,have solutions. Furthermore, each Q, is a complete metric space, as is R 2 Q, and methods of analysis are available. It is convenient to define the p-adic fields, and many other important fields, in terms of completions of fields with valuations. A function $ from a field F to the real field [w is called a valuation of F if

(i) $(ab) = $(a)$@) for all a, b E F , (ii) $(O) = 0, $(a) > 0 for all a E F * , and (iii) there exists c 2 1 in R such that if $(a) I 1, a E F, then $(a + 1) I c. Note that 4: F* -,[w* is a homomorphism. If $(a) = 1 for all a E F*, then $ is called the trivial valuation on F. Terminology pertaining to valuations is not universally agreed upon. A valuation as defined above is often call a real, or rank 1, valuation, and sometimes is called an absolute value. There is a more general notion of valuation, in which the positive real numbers are replaced by any ordered group. Exercise 6.1. If $ is a valuation on F and ak = 1 in F , 0 < k E Z,show that F is a finite field, then F can have only the trivial valuation. Probably the most familiar example of a nontrivial valuation is the ordinary absolute value function $=,on F = Q (or on any F G C), i.e., $m(a)= (al, all a E F . For another class of examples take F = Q and let p E Z be a fixed (positive) prime. If r E Q* write r = pVp(*)a/b,with vp(r)E Z,a, b E 7, and ( a , p ) = (b,p)= 1 (note that r=f p " ~ ( ~ ) :primes alI p}). $(a) = 1. Conclude that if

n{

Define $,(r) = p - " p ( ' ) , and, of course, $,(O) the p-adic valuation of Q.

= 0. Then

4,

is a valuation, called

Exercise 6.2. Show that each p-adic valuation 4, on Q is in fact a valuation, and that the constant c in (iii) of the definition can be taken equal to 1. For another example suppose K is a field and let F = K(x), the field of rational functions over K. Suppose a = f(x)/g(x) E F , with f ( x ) , g(x) E K[x]. If degf(x) = rn and degg(x) = n define $(a) = 2"-". Then $ is a valuation on F. Further examples can be obtained by fixing an irreducible (i.e., prime) p(x) E K [XIand proceeding by analogy with the definition of the p-adic valuation on Q. If $2 are nontrivial valuations on F we say that is equivalent with 42, and write $1 $ 2 , if $ 2 ( a ) < 1 for all a E F with $l(a) < 1. Note that if is a

-

240

VI

Further Topics

-

nontrivial valuation and 0 < r E R, and if we define 42:F -, R by means of 42(a)= 4,(a)l, i.e., 42= &, then 42 is also a valuation and 4, 42.We agree that the trivial valuation is equivalent only to itself.

Proposition 6.1. Suppose 4,,& are valuations on F , 4, Then 4,(a) < 1

- d2,

and a E F.

g5z(a)< 1,

4,(a) = 1

if and only if if and only if

4 2 ( a )= 1,

>1

if and only if

#2(a)> 1.

and #,(a)

Proof. We may assume that 41and 42are nontrivial. Note that 4i(a)< 1 (a E F * ) if and only if 4i(a-1)> 1. A moment's thought shows that it will suffice to show that if 4,(a) = 1, then 42(a)= 1. Choose b E F* with 4,(b) < 1. If 0 < n E Z,then +,(a"b) = 4,(b) < 1, so +,(an@ = 42(4n42(b)< 1,

or 42(a)< q52(b)-

Thus

42(a)I By the same argument

lim 4 2 ( b ) - 1 i= n 1. n-r m

42(a-1)5

1, so 42(a)2 1 and hence

42(a)= 1.

Corollary. Equivalence of valuations is an equivalence relation, and the valuations on F are partitioned into equivalence classes. Proposition 6.2. If 41,42are valuations on F and 4 for some positive r E [w.

- 42,

then 42 = 4;

Proof. If both are trivial take r = 1. Otherwise take a E F* with < 1, so also 42(a)< 1. Set r = l ~ g ~ ~ ( a ) / l o g # E, ([wa so ) 42(a)= $,(a)'. Take any b E F* with 4,(b) < 1, and choose m,n positive in Z with m / n > log4,(b)/log4,(4. Then log4,(um)< log4,(b"), so 4 , ( a m )< 4,(b"). It follows that 42(am)< 4,(b") and, reversing the argument, that m / n > log 4,(b)/log 42(a).Since m and n are arbitrary it follows that +,(a)

1% 41(b)/log 4 , ( 4 2 1% 42(b)/log 4 2 ( 4 . The reverse inequality follows symmetrically, so they are equal and hence log42(b)/log4,(b) = log42(a)/log4,(~) = 1.

Thus 42 = 4;. A valuation 4 on F satisfies the triangle inequality if #(a + b) 5 $(a) + 4(b) for all a, b E F .

241

6 Valuations and p-adic Numbers

Proposition 6.3. Suppose 4 is a valuation on F whose constant c [(iii) in the definition] can be taken to be 2, i.e., if 4 ( a ) I 1, then 4(a + 1) S 2. Then 4 satisfies the triangle inequality.

Proof. Take a, b E F , say with 4 ( a ) I 4(b)# 0. Then 4 ( a / b )I 1 so

b ( a / b + 1) = + [ ( a

+ b)/b] I2

or

+ b) I24(W = 2max(&a),4(b)). An easy induction gives

4

(,I1k' )

a, I 2kmax{ @(a,):1 I i I zk}

foralla, E Fif 1 I k E Z.If 1 I n E Zchoosek E Zwithn I 2' < 2n.Thenfor any a,, . . . ,a, E F we may add enough 0's to a, to have 2k summands and conclude that

As a special case we see that @(n . I ) I 2n for all positive n E Z.Furthermore,

Apply that inequality and the Binomial Theorem to obtain

Thus 4(a

Since lim,,,(4n

+ 4)''"

+ b) I (4n + 4)""[4(a) + 4 ( b ) ] . =

1 we conclude that +(a

Corollary. Every valuation satisfies the triangle inequality.

4; #2

4 I is

equivalent with a valuation

Proof. Suppose 41 has constant c > 2, set r Then if 4 2 ( a )I 1 we have b 2 ( a + 1 ) satisfies the triangle inequality.

-

+ h) 5 $(a) + &b).

= log2/logc, =

$l(a

+ 1)'

b2 that

and set 42 = C' = 2, and

I

242

VI Further Topics

If the constant c in the definition of a valuation can be taken to be I, then the triangle inequality can be strengthened considerably. If #(a + 1) 5 1 whenever #(a) 5 1, then 4 is called a nonarchimedean valuation. Otherwise # is archimedean. Note that the ordinary absolute value # m on Q is archimedean. Exercise 6.3. Show that the p-adic valuations archimedean. Proposition 6.4. equivalent:

Suppose

#p

on Q are all non-

# is a valuation on F . Then the following are

(a) # is nonarchimedean; (b) { @ n . 1):n E Z)is bounded in R; (c) #(a b) I max{4(a), # ( b ) ) ,all a, b E F .

+

+

Proof. (a)*(b): Since #(1) = 1 we have #(2. 1) = #(1 1) I 1, and inductively #(n . 1) 5 1 if 1 I n E Z. Since #( - n . 1) = # ( n . 1) we conclude 4 ( n . 1) I 1 for all n E Z. (b) 3 (c): By the corollary to Proposition 6.3 we may assume that # satisfies the triangle inequality. Say # ( n * 1) I M for all n E Zand take a, b E F . If 0 < n E Z,then

and so #(a

+ b) I [ M ( n + l)]""max{~(a),#(b)).

Since

-

lim [ M ( n + l)]""

=

1

fl'W

+

we conclude that #(a b) I max{#(a), # ( b ) ) . (c) (a): If a E F and $(a) I 1, then $(a

+ 1) I max{4(4,$(1)} = 1

so # is nonarchimedean. Corollary. If char F archimedean.

=p

and # is a valuation on F , then

4 is non-

Remark: The proof of Proposition 6.4 shows that if # is nonarchimedean, then $(a . 1) 2 1 for all n E Z. Exercise 6.4. If # is a nonarchimedean valuation on F , a, b E F, and b) = #(b). Conclude that if #(ai) 5 4(a,)

+(a) < #(b) show that #(a

+

243

6 Valuations and p-adic Numbers

for 2 I i I k but $(al i 2 2.

+ a, + ... + a k )< #(al), then $(ai)= $ ( a , ) for some

The next two propositions, due to A, Ostrowski, characterize (up to equivalence) all valuations on Q. Proposition 6.5. If 4 is an archimedean valuation on Q, then $ ordinary absolute value.

-

$ J ~the ,

Proof. (Artin [4]). We may assume that #I satisfies the triangle inequality. Take m, n > 1 in E. It is an easy consequence of the division algorithm that we may write m as a polynomial in n, m = a. + aln + ... + aknk,with a, E H , 0 I a, < n, and ak # 0. Thus

1 ai$(ni) < n 1k $(n'). _< n(k + I)max(l,$(n)k}

+(m) I

k

i=O

i=O

(consider the two possibilities $(a) I 1 and $(n) > 1). Note that k I log m/log n since m 2 nk. Thus $(m)< n(1og m / l o g n + 1) . max { 1, $(n)"gm"ogn1. For any positive integer j we may replace m, above, by m j and conclude that

$(rn)j < n ( j log m/Iog n

+ 1)max{ 1, $(n)j'ogm/'ogn1,

and hence $(m) < [ n ( j l o g m / l o g n

Since limj+,( j a

+

P)'/j

=

+ 1)]'~"n~(l,$(n)'~~~''~~~}.

1 for any a, P E R with a > 0, we conclude that

$(m)Imax { 1 , $ ( n ) l o g m / l o g n } . Since $ is archimedean we may assume that $(m)> 1, so 1 < $(m)I $(n)logm/logn . It follows that $(n) > 1 for all n > 1 in Z. We may thus reverse the roles of m and n above and conclude that $(a)

$(m)logn/logm

- 4w, <

and consequently $(n)l/logn

= $ ( m ) l / l o g m = er

for some positive r E R, independent of the choice of m,n > 1 in H. Taking logarithms we see that log $(n) = r log n = log n', and hence $(n) = n' = I nl'. Since $(-a) = 4(a) and $(a/b) = $(a)/$(b)it follows that $ = $k +m.

Q,

- 4,

+

-

Proposition 6.6. If is a nontrivial nonarchimedean valuation on Q, then for some prime p.

244

VI Further Topics

Proof. Set P and n E H,then

= {aE

E : 4 ( a )< I}, P # ( 0 ) since 4 is nontrivial. If a, b E P

4(a - b) 2 max{4(a),4(b)}< 1

and m a ) = 4(n)4(4

4(a)

< 1,

so Pis an ideal in E.If m,n E Z\P, then 4(mn) = +(m)4(n) 2 1, so mn 4 P and P is a prime ideal. Thus P = ( p ) for some prime p E Z.In particular 4 ( p ) < 1, so $ ( p ) = p - * for some r > 0 in R. If n E Z* write n = pkm, p y m . Then m 4 P , so 4(m)= 1 and 4 ( n ) = p - k r = 4Jn)*. It follows that 4 = 4; 4p. N

If 4 is a valuation on a field F , then 4 determines a topology on F if we take as a neighborhood basis at each a E F the sets B,(a) = { b E F : $ ( b - a) < E } ,

0 < E E R.

It is clear from Proposition 6.2 that equivalent valuations determine the same topology. Since we may assume, by Proposition 6.3, corollary, that 4 satisfies the triangle inequality, it follows easily that F is in fact a metric space, with metric 6 = 6, defined by 6(a,b) = &a - b). As usual a sequence a,, a 2 ,a s , . . . in F converges to a E F if 6(a,, a) = 4 ( a n- a) + 0 as n -+ co.The sequence {a,} is a Cauchy sequence if for any E > 0 in R there is some N E Z such that +(a,,, - a,) < E if m,n 2 N . Clearly a convergent sequence is a Cauchy sequence. If every Cauchy sequence converges in F , then F is called complete (relative to 4). Note for example that F is complete if 4 is trivial. Exercise 6.5. Show that a Cauchy sequence {a,} is bounded, i.e., E R, all n.

$(a,) I M for some M

If 4 is a valuation on F , then a completion of F is a field extension K 2 F with a valuation 8 on K satisfying (i) 8 1 F = 4, (ii) K is complete, and (iii) F is dense in K , i.e., if a E K there is a sequence {a,} in F with O(a, - a) + 0. It is a well-known fact from topology that every metric space has a completion; the construction is in fact a bit easier in the present setting. If 4 is a valuation on F denote by V the set of all Cauchy sequences {a,} E F. If {a,}, {b,} E % define {a,} { b n } = {a, b,} and {a.}{b,] =

+

+

{aflbn}.

Exercise 6.6. Show that % is a commutative ring with 1 relative to the operations above (for multiplication see Exercise 6.5).

Let JY = ((a,} E %?:an + 0). It is easily checked that JV is an ideal since Cauchy sequences are bounded. Suppose I is an ideal of %? with M E I , I # N ,and choose {a,} E I \N. Then {a,}is eventually bounded away from

6

245

Valuations and p-adic Numbers

0, say $(a,) 2 E > 0 for all n 2 N . Define bi = 1 for i < Nand bi = -ai for i 2 N, then set ci = ai+ bi.Clearly {c,} E .,V; let us see that {b,} is a unit in ‘8.If m, n 2 N , then $ ( l / b n - 1 / b m ) = C1/#(aman)Ibfam - an) I (1/E2)$(am - an),

which tends to 0 as rn, n 00. Thus {1/6,} E ‘8and {b,} E U(‘8).Since {c,} E N and {a,} E I we have {b,} E I, so 1 = V and N is a maximal ideal. As a result K = U / M is a field. If a E F the constant sequence {a} is is U, and a- {a} + A” is a monomorphism from F into K , so we may identify F with its image and agree that F E K . If {a,} E V , then {$(a,)} is a Cauchy sequence in R, which is complete, so suppose $(a,) + r E R. If (a,} A”= {b,} + N in K , then {a, - b,} E .1.‘ and it follows that 4(b,) -+ r as well. We may thus define 6( {a,} N )= r and thereby obtain a well-defined function H from K to R. -+

+

+

Exercise 6.7. Verify that 6: K and that 81 F = 4.

-+

R, as defined above, is a valuation on K ,

Theorem 6.7. The field K , above, is a completion of F relative to 4. Proof. It remains to be shown that K is complete and that F is dense in K . We may assume that 6 satisfies the triangle inequality. It will be convenient to establish the denseness first. If Q = {a,} + A’”E K , then for each m we have 6(am- a) = IimkdK,6(am- a,‘), and so

lirn @(a, - ct) = lirn lirn $(am - ak) = 0,

m-a)

m - z k+%

since {a,} E U, i.e., {a,} is a sequence in F that converges to ct E K . Next let {a,} be a 6-Cauchy sequence in K , with CI, = {up’ N } ,a t ) E F , k = 4 2 , . . .. For each n we may choose b, E F with 6(b, - a,) < 2-”, by the denseness of F in K . Then

+

$(bn

- bm)

$(bn - an) + $(an - a m )

which converges to 0 as m, n observe that

-+

cci, so

6(a, - p ) = 6 ( { a p )- b k }

{b,} E %?.Set /.I= {b,}

Thus

n

+ lim d(b, - 8). n

+ Jf E K , and

+ M ) = lirn $ ( a t ) - bk). k

= lim B(cr, - b,)

+ 4 ( E m - bmh

246

VI FurtherTopics

Since d(a, - b,) < 2-" and b, + b by the proof above of denseness, we conclude that a, -+ p and K is complete. Exercise 6.8. Show that the completion K is unique in the following sense. If L, with valuation is another completion of F relative to #, then there is an F-isomorphism f : K + L with 8 = $5 (Hint: For any Cauchy sequence {a,} E F define f(lim, a,) = limna, E L.)

+,

If K is the completion of F relative to #, then the extended valuation (d above) is usually also denoted by 4. We remark that the completion of Q relative to #m is (isomorphic with) [w; in fact, the completion process above is in essence one of the standard methods for constructing R from Q. Suppose now that # is a nonarchimedean valuation on a field F. Define

R = R, P = P,

{ a E F : # ( u ) I11, = { U E F:#(u) < l},

=

and U

=

U, = { a E F:#(u) = 1)

and note that any valuation equivalent with # determines the same sets. Since # is nonarchimedean it is immediate that R is a subring of F ; it is called the ring of integers at #, or the valuation ring at #. If we write R-' = { r - ' : r E R * ) , then F = R u R - l , and in particular F is the field of fractions of R. U

Proposition 6.8. If # is nonarchimedean on F with valuation ring R , then U, is the group of units of R and P = P+ is the unique maximal ideal in R.

=

Proof. If u E U , then #(u) = 1, so # ( # - I ) = 1 and u - l E R, i.e., u E U(R), and conversely. Since # is nonarchimedean it is clear that P i s an ideal in R (see the proof of Proposition 6.6) and P is maximal and unique since R is the disjoint union of P and U .

Since Pis maximal in R the quotient R I P is a field F = F,, called the residue class jield of F at 4. Exercise 6.9. If #, is_ the p-adic valuation on Q describe R, P, and U explicitly and show that F H,.

Proposition 6.9. Suppose # is a nonarchimedean valuation on F and K is a completion of F , with # extended to K. Then F and K have isomorphic residue class fields. R,

Proof. Write R,, R , for the valuation rings in F and in K , etc. Then = R , n F and PF = PK n F. If a E RK, then by denseness #(a - b) < 1

6 Valuations and p-adic Numbers

247

for some b E F , s o a - b E P, and hence b E R , n F = RF. Thus a E b + PK E RF + RK, so RK E RF + PK and hence R , = RF + P,. As a result

K

= RKJPK = (RF

+ PK)JP, 2 RFJRF A PK = RFJPF = F.

For any valuation 4 of a field F the image +(F*) is a multiplicative subgroup of { r E R : r > 0}, called the value group of 4. If the value group is cyclic, then is called a discrete valuation. If 4 is discrete, nonarchimedean, and nontrivial, then one of the two generators of the value group is its largest element less than 1, call it r o . Take n E F with 4(n)= ro, so n E P . If a E P, then &a) I 4(n),so d ( a x - ’ ) 5 1 and a C 1 E R, or a E Rn. It follows that P = R n is a principal ideal.

+

on Q is discrete, and Exercise 6.10. Show that each p-adic valuation is not discrete. that x can be taken to be p . Show that

Proposition 6.10. Suppose 4 is nonarchimedean on F and K is the 4completion of F. Then the value group of K is equal to that of F. In particular remains discrete on K if it is discrete on F. Proof. Take a E K * . Since F is dense +(a - b) < +(a) for some b E F. Thus

4(b) = 4(a + (b - 4)= max { +(a), WJ- a)> = +(a) by Exercise 6.4, so #(a) E 4(F*). Suppose now that F is a field that is complete relative to a nontrivial ak, with each nonarchimedean valuation 4. As usual an infinite series ak E F , is defined to be the sequence { b,} of partial sums, b, = ak. If b, -+ a E F , then a is called the sum of the series, and we write a = ak. Since F is complete ak converges if and only if { b,} is a Cauchy sequence, i.e., +(a, + .. . + a,) + 0 as m, n + m. Since 4 is nonarchimedean

c:=

+(a,

+ + a,) I max(+(ak):rn I k I n } , *.*

so clearly a k converges if and only if 4(ak) ,O as k + 00. Suppose further that 4 is discrete, with n a generator for the maximal ideal P in the valuation ring R . By Proposition 6.2 we may assume that 4(n)= p for some prime p e Z . Choose a representative element a of R from each element (coset) in the residue class field F = R J P ,taking a = 0 in the coset P itself. Denote by A the resulting set of coset representatives so each r E R can be written uniquely as r = a b, with a E A, b E P. Note that b(a)= 1 for every a # 0 in A, since a E R\P = U .

+

Theorem 6.11. In the setting above every element a E F* has a unique representation as a convergent “Laurent” series a = x ~ = , , where , ~ ~rnlE~ 7~ ,

248

VI Further Topics

(possibly rn < 0), each ak E A, am# 0, and $(a) = p - " . Conversely every series of that form converges to an element a E F* with $(a) = p-". We have

and U

1 a k n k : a k E A a n d a o# O m

=

{k=O

I

.

Proof. We prove the converse first. Any series xP=,,,aknk , a, aknk.Then converges since $(aknk)= $(ak)p-k-,0, so write a =

I;=,,,

n-

E

A,

max{d(aknk):rnIk 5 n}

m

= lim max{ p P k : mI n- m

Since a, # 0 we have a = a,nm

n I k } = p-".

+ b, where b = ckm,,+ akn , so k

4(b)Ip-('"+l) < 4(a,nm) = p-", hence $(a) = p - , by Exercise 6.4. Suppose Xkm,,Cknk is another such series, that ck = ak for rn Ik < n, but c, # a,. Then c - a = x p = n ( C k - ak)nk.Each coefficient ck - ak is either 0 (if ck = ak) or else ck - ak E R\P = U , so #(ck - ak) = 1. It follows as above that 4(c - a) = p - " # 0, so c # a and series representations are unique. Now let us see inductively that every a E F* has such a series represo $(an-"') = 1 and sentation. Say $(a) = p-", rn E Z. Then $(a) = +(IT"'), an-"' E U . Choose a, E A with a Y m- a, E P,so that $(an-,

-

a,) = 4(a - a,n"')$(n-'") < 1

or

d ( a - amnm)< &(n"')= p - " . By induction we may write a-amnm =

f

aknk

k=m+ 1

(possibly a,, = 0),and hence a = EF=,,, aknk.The descriptions of R, P , and U follow immediately.

6 Valuations and p-adic Numbers

249

If p > 0 is a prime in H , then the completion 6,, of Q relative to the p-adic valuation 4, is called the field of p-adic numbers. We will write d, for its valuation ring R g , of p-adic integers and write 6,for U ( d , ) . By Exercise 6.9 and Proposition 6.9 the residue class field of Q p is Z,, so it is natural to take A = Ap = (0, l , . . . , p - 1). Thus each a E 6; has a unique expansion as a 4,-convergent series a = xkm_makpk, with m E Z,4(a) = p-”, and ak E A. That series is called the canonical representation of a E 6;. If rn # 0 it is often convenient to “factor out” pm, relabel the coefficients (ak-uk-,,,), and write a = p m C : ~ o u k p kwith , u = C?=,akpkE fip. Sums and products of p-adic numbers in canonical form can be carried out as usual for convergent series, except that coefficients should be reduced (by division) when necessary to integers between 0 and p - 1, the quotients being carried as contributions to coefficients of higher powers of p . In many respects the canonical representation of p-adic numbers is analogous to the decimal representation of real numbers. We may observe, for example, that a p-adic number a is rational if and only if its canonical expansion is ultimately periodic. In order to sketch a proof we may clearly assume that a E fi,, since multiplication by p“ will not alter periodicity. Suppose first that the sequence of coefficients of a E fi, has the form bo,.. .,b j - l,co,. .. , c k the bar denoting periodic repetition. Set b = b,p’ E Q and c = cipi E Q, so a = b + pjc(1 + pk + p Z k + ...). The geometric series converges since A ( p f l k ) = pPflk2 0, so a = b + p’c/(l - p k ) E Q. The proof of the converse amounts basically to representing a E Q * in the form just obtained above. Again take a to be a p-adic unit, so a = d/e, with d , e prime to p in Z and e > 0. Take k to be the order of p when viewed as an element of V(Z,), so pk = l(mod e), and write 1 - p k = e f , f E Z. Thus a = g/(l - p k ) , where g = df is prime to p . Choosej E Z,minimal, for which 191 < p’ (one exception: ifg = - 1 take j = 0). Let c E Z satisfy cpj = g(modpk- 1) and set b = (y - cp’)/(l - p k ) E Z. If a > 0, hence y < 0,choosec with0 I c I p k - 2;ifa < Oandg > Ochoosec with 1 I cI pk - 1. In either case it follows easily that 0 I b < pj. Note that g = (1 - p k ) b + p’c, so a = b p’c/( 1 - p k ) . If we write b = bipi and c = 1;:; cipi, canonically, and l / ( l - p k ) = ~ ~ o pwek obtain i the canonical form of a and see that its sequence of coefficients is b,, . . . ,bj- I r ~ O , .. .,ck; An example may be illuminating. Let us express a = & canonically in Q7. The order of 7mod 15 is k = 4, and 1 - 7 4 = -2400 = 15(- 160), so a = -320/(1 - 74), g = -320. Thus 7’ < 191 < 7 3 = 343 and j = 3. We may solve 343x = 1(mod 2400) (e.g., by using the Euclidean algorithm) and find that x, = 7 is a solution. Thus c = 160 = - 320 . 7 (mod 2400) is a solution to cp’ = g(modpk - l), with 0 I c I pk - 2. We set

c::;

c{LA

+

b

= (-

320 - 160 343)/( -2400)

= 23

250

VI Further Topics

and obtain

+ 7 3 . 160/(1 - 74). Write 23 = 2 + 3 . 7 and 160 = 6 + 1 . 7 + 3 . 7 ' , so - = (2 + 3 . 7) + 73(6 + 1 . 7 + 3 . 72)(1 + 74 + 7 8 + ...) = 2 + 3 . 7 + 0 . 7 ' + 6 . 73 + 1 . 74 + 3 . 7 5 + 0 . 7 6 + 6 . 7 ' + ... u = & = 23

:5

and the sequence of coefficients is 2,3,0,6,1,3,0.

,,

Exercise 6.11. (1) Find the canonical expansion of a = & in 6 6 3 , and 6,. of coefficients. Determine a if p = 7 (2) Suppose a E fi, has sequence and if p = 11. (3) Suppose 0 > a E Z n Choosej minimal in Zwith a + p j > 0, and write

op.

a

+ p'

= a,

+ a l p + ..*+

~ ~ - ~ p j - ' a, , E

Show that the canonical - representation of a in a,,a, )...)a j - l , p - 1.

A,,.

6,has coefficient sequence

Proposition 6.12. The ring R, of p-adic integers is compact (with the topology it inherits as a subspace of the metric space

a,,).

Proof. For 0 5 k E Z set yk = A, = {0,1,.. . , p - 1 ) with the discrete topology, and set X = &:O Ik E Z}, the Cartesian product with the product topology. Then X is compact by Tychonov's Theorem (e.g., see Kelley [21]). Define f: X + k,-by setting f(a,, a , , a 2 , .. .) = akpk.Then f is a 1-1 map from X onto R, by Theorem 6.11, and it will suffice to show that f is continuous. Take o! = (ao,a , , ...) E X , so f(o!) = a E d,. As a basic neighborhood of a we may take N = { b E k,$,(b - a) < p-'], with k 2 0 in Z. If we set

n{

V = { a o } x {al} x ... x { a k )x

n { T : j2 k + I},

then V is open in X , o! E V , and f(V)G N , so f is continuous and R, is compact. We remark that the map f in the proof above is also open, so X and R, are in fact homeomorphic. It follows in particular that R , is totally disconnected. Exercise 6.12. Show that U{pkkp:kEZ}.)

bP

is locally compact. (Hint:

6,=

Theorem 6.13. If f(x) E E[x], then the congruences f(x) 5 O(modp"') have solutions for all m 2 1 if and only if the equation f(x) = 0 has a solution in R,,. A

6 Valuations and p-adic Numbers

25 1

Proof. e: Suppose a = x Y o a k p kE R, is a solution to f ( x ) = 0. For each n set b, = z ; = o a , p k E Z,so lim,b, = a in R,. Thus f(6,) E Z and f ( b , ) + f(a) = 0 (since polynomials are continuous), i.e. 4,(f(6,)) -+ 0, which means that high powers of p divide f(6,) for large n. Now, given m 2 1 choose n large enough so that $,(f(b,)) < p-", i.e., p"(f(b,), or f(6,) = O(mod p"). 3: Choose c, E Z, 1 I rn E Z, with f(c,) = O(modp"). Since all c, E k,, which is compact (Proposition 6.12), there is a subsequence {c,,) that converges, say c,, -+ a E R, as i -+ 00. By continuity f(cmg) + f(a). But p"' I f(c,,); hence 4,(f(cm,)) I p-"', and m i--t a, so f(a) = lim,f(c,,) = 0. A

With not much more effort the polynomial f ( x ) in Theorem 6.13 can be replaced by f ( x l , . . . ,x,) E Z[x,, . . . ,x,], with the same conclusion (see Borevich and Shafarevich [6, p. 411). Exercise 6.13. Show that if f ( x ) = x z - a E Z[x], then f ( x ) has a root in Z if and only if it has a root in R and in A p for every prime p . We conclude this section with a brief discussion of an approximation theorem that has come to play a central role in algebraic number theory. Proposition 6.14. Suppose b,, 4z,.. . ,( b k are inequivalent nontrivial valuations on a field F . Then there is an element a E F with 41(a)> 1, +,(a) < 1 for 2 I i I k. Proof. Induction on k. The case k = 1 is trivial but the induction step will also require the result for k = 2. Since I $ ~and & are inequivalent we may choose 6, c E F with c$,(b) > 1,42(6) I 1, QI1(c)2 1, and 42(c) < 1. Set a = 6c; then @l(a)> 1 and &,(a) < 1. Suppose then that k 2 3 and assume the result for k - 1 or fewer valuations. Choose 6, c E F with $,(6) > 1, 4i(6)< I for 2 Ii I k - 1 and 1 set a, = b"c, 1 I n E Z. Then $l(a,) > 1 and 41(c) > 1, +k(c)< 1. If 4k(6) I 4k(a,) < 1 for all n, and if 2 I i s k - 1, then rbi(a,) = 4,(l1)"4~(c)< 1 for large enough n. Thus we may take a = a, for some n. If ( b k ( 6 ) > 1, then instead set a, = b"c/( 1 + b"), 1 5 n E h. Then 4 (a,) ;f 4 (c) and &(a,) 2 &(C), so 4,(a,) > 1 and @&,) < 1 for large enough n. If 2 I i I k - 1, then $[(a,) ;f 0, and again we may take a = a, for some n.

Corollary. Suppose c$~,. . . ,4kare inequivalent nontrivial valuations on F and 0 < E E R. Then for some a E F we have 41(a - 1) < E , bi(a)< E for 21i1k. Proof. Apply the theorem to get b E F with 4,(6)> 1 and 4,(6)< 1 for 2 I i I k. Set a, = 6"/(l + b"), 1 I n E Z,and take a = a, for sufficiently large n.

252

VI Further Topics

Theorem 6.15 (The Artin-Whaples Approximation Theorem). Suppose that $,,. . .,$ k are inequivalent nontrivial valuations on F , that 0 < E E [w, and that a,, . . . ,ak E F . Then there is an element a E F for which 4i(a - ai) < E, 1 5 i I k .

Proof. We may assume that each bi satisfies the triangle inequality. Set M

= max{$i(aj):

1 I i, j I k}.

Apply the corollary, above, to each 4i in turn and obtain b,, . . . ,b,

#i(bi - 1) < c / k M ,

Set a =

If= ,ajbj. Then

@j(bi)< E/kM

E

F with

if j # i.

Exercise 6.14. If F = Q, a , , . . . ,ak E Z,and 4 p , ,. .. ,dPkare distinct pi-adic valuations show that the conclusion of the Artin- Whaples Theorem follows from the Chinese Remainder Theorem. Exercise 6.15. Let 42,43,$ 5 be the p-adic valuations for p = 2,3,5 on Q and set a2 = a3 = 4, a5 = 6. Find a E Q with 4p(u- u p ) < 0.01, p = 2,3,5. Exercise 6.16. If #,,. .. ,$k are inequivalent nontrivial valuations on F show that n { $ i ( a ) :1 I i S k } = 1 cannot be true for all a E F*. On the other hand let 4,, be the p-adic valuation on Q for each prime p > 0 and show that 4,(a)fl{4p(a):all p } = 1 for all a E Q*.

7. REAL FIELDS AND STURM’S THEOREM The essence of the usual order relationship on R = Z,Q, or R can be captured as follows. There is a subset P of R (the set of positive elements) such that (i) R is the disjoint union of P, (01, and - P = { -x:x (ii) if x, y E P, then x y E P and xy E P.

+

E

P) and

Then x > y if and only if x - y E P . Suppose now that R is any commutative ring. If there is a subset P of R satisfying (i) and (ii) above, then R is said to be ordered (relative to P),the

7

Real Fields and Sturm's Theorem

253

elements of P are called positive and the elements of - P are called negative. We write x > y (or y < x) if x - y E P. Thus P = {x E R : x > 0) and - P = {x E R : x < 0). We write x 2 y (or y I x ) to mean that x > y or x = y, as usual. Note that if x > 0 and y < 0 in an ordered ring R , then x, - y E P , so x( - y ) = - x y E P and hence x y < 0. Similarly if x < 0 and y < 0, then x y > 0. In particular R can have no zero divisors, so every ordered ring is an integral domain. For an example, take R = Q[x] and agree that f ( x ) E P if and only if

f ( x ) = Qg + n , x

+ ... + a,x"

with a, > 0 in Q. With this order we have, for example, 2

+ x > 1 - x - 3x3.

Exercise 7.1. Verify the following facts in an ordered ring R .

(1) If x > y and y > z, then .x > z. (2) If x > y and u 2 u, then x + u > y + u. ( 3 ) If x > y and z > 0, then xz > yz. (4) I f O < x E U ( R ) , t h e n O < x ~ ' . ( 5 ) If x > y > 0 in U(R), then 0 < x - l < y - ' . ( 6 ) If r l r r 2 , .. . , r k E R, then c { r : : 1 5 i 5 k } 2 0. Clearly any subring S of an ordered ring R inherits an order from R if the set of positive elements in S is taken to be P n S . We say that the resulting order on S is induced from the order on R. In the opposite direction suppose an ordered ring R is a subring of a ring K . The order on R is said to be extended to K if K has an order that induces the original order on R . Thus, for example, the usual order on Q induces the usual order on Z,and extends to the usual order on R. Proposition 7.1. Suppose R is an ordered ring and F 2 R is its field of fractions. Then char(F) = 0 and there is a unique extension of the order on R to an order on F.

Proof. Let us first extend the order. Set

i

PF = x E F:x

U

= -with

b

I

a > 0 and b > 0 in R ,

i.e., x > 0 in F if and only if x can be expressed as a quotient of positive elements from R. It is easy to check then that F is ordered, and if a > 0 in R, then a = a 2 / a > 0 in F , so the order on F extends that on R. Uniqueness is clear from the definition of order and Exercise 7.1.4. Finally note that 1 = l 2 > 0 , so n . 1 = 1 2 + f Z + . . . + 1 2 > 0 if n is any positive integer, and hence char(F) = 0.

254

VI

Further Topics

Corollary. The only possible order on Q is the usual order. Proof. The order on Z is unique since its positive elements are just 1, 2=1+1,3=1+1+1,etc.

The order on R = Q [ x ] given as an example above extends uniquely to an order on the field Q ( x ) of rational functions, whereby f ( x ) / g ( x )> 0 in Q ( x ) if and only if f ( a ) / g ( a )> 0 in Q for all sufficiently large positive a E Q. Exercise 7.2. Not all ordered rings have unique orders. For example, R

=

a($) inherits the usual order from R, so that P consists of all a + b.4, a, b E Q, that are positive real numbers. Show that Pl = {a + b$:a b$

E P } determines a different order on Q($).

A field F is called formally real if - 1 is not a sum of squares of elements of F , or equivalently if c ( a f : l I i 5 k ) = 0,

a, E F ,

is possible only if every a, = 0. Note that in any field of characteristic p > 0 we 1’ : 1 I i Ip } = 0, so a formally real field must have characteristic 0. have

c{

xr=

Proposition 7.2. Suppose F is a field and a, b E F are sums of squares in F , say a = af and b = b; . Then ab is a sum of squares in F , as is b- if b # 0.

2 Proof. Clearly ab = C i , j ( a i b j )and b-’ = b(b-’)*

m

=

C (bjbK1)’.

j= 1

A formally real field F is called real closed if the only formally real algebraic extension of F is F itself. Note for example that F = R is real closed since its only nontrivial algebraic extension is (isomorphic with) @, which is not formally real. Proposition 7.3. If F is a formally real field, then F has an algebraic extension K that is real closed. Proof. Zorn’s Lemma. Proposition 7.4. Suppose F is a real closed field and a = Then a = b2 for some b E F.

cf=a : ,

a, E F.

Proof. Suppose not. Then x 2 - a is irreducible in F [ x ] . Let c be a root in a splitting field K = F(c). Since F is real closed K is not formally real, and we may write - 1 = x i ( b i+ mi)’, with bi, ci E F. Thus

7

and we see that

xibici

= 0 since c

Real Fields and Sturm’s Theorem

255

4 F. Thus

a sum of squares by Proposition 7.2, contradicting the formal reality of F. Proposition 7.5. If F is a real closed field and 0 # a E F, then either a or - a is a square in F (but not both). Proof. If a is not a square, then just as in the proof of Proposition 7.4 we may write - 1 = b? a x i c?, with bi, ci E F and c i # 0. Solving for - a we find

xi

+

Xi

so --a is a sum of squares by Proposition 7.2. Thus -a is a square by Proposition 7.4. If both a and - a were squares, then - 1 = -a f a would be a square.

Theorem 7.6. A real closed field F is ordered, with the nonzero squares as positive elements, and the order is unique. Proof. If P denotes the set of nonzero squares in F, then P determines an order by Propositions 7.4 and 7.5. The order is unique since squares must be positive for any order.

Corollary. Every formally real field can be ordered. Proof. Proposition 7.3.

The next theorem generalizes to all real closed fields F an important property of the real field R. In the case of F = R it is a consequence of the Intermediate Value Theorem of calculus. Theorem 7.7. Suppose F is a real closed field and f (x) E F[x] has odd degree. Then f(x) has a root in F. Proof. Suppose the theorem to be false and choose a monic polynomial f(x) E F[x] of minimal odd degree n not having a root in F. Clearly n 2 3. Note that f ( x ) must be irreducible by the minimality of n. Let a be a root of f(x) in an extension field. Then K = F(a)is not formally real and we may write - 1 = xif;.(a)’, where each S,(x) E F[x], degf;.(x) < n = [K:F], and at least onefi(x)is nonconstant. But then g(x) = 1 + f;:(x)’ is a multiple, in F[x], of the minimal polynomial f (x) of a. Write g(x) = f(x)h(x), h(x) E F[x]. Note that degg(x) is even and less than 2n - 1, and

xi

deg g(x) = deg f(x) + deg h(x) = n

+ deg h(x),

256

VI Further Topics

so deg h(x) is odd and less than n - 1. Consequently h(x) has a root b E F , and b is a root of g(x). But then - 1 = f;:(b)’ in F, a contradiction.

xi

Theorem 7.8. If F is a real closed field and i is a root for x 2 + 1 E F[x], then K = F(i)is algebraically closed.

+

Proof. Note that if a bi E K , with a, b E F, then a2 + b2 2 0; hence in F. Say a’ b2 = c2, with c 2 0. Then c2 2 a2,and so c)/2 and ( - a c)/2 are squares in F, say of u and v. If b 2 0 take u 2 0 and v 2 0; if b < 0 take u > 0 and v < 0. Then it is easy to check that (u vi)’ = a bi, so every element of K is a square. It follows from the classical quadratic formula then that there are no irreducible polynomials of degree 2 in K[x], so K has no extension fields of degree 2. That fact, together with Theorem 7.7, allows the proof of the Fundamental Theorem of Algebra (Theorem III.3.9), with R and C replaced by F and K , to be repeated verbatim in the present context.

+

a2 b2 is a square c 2 _+a. Thus (a

+

+

+ +

+

Theorem 7.9 (The Intermediate Value Theorem). Suppose F is a real closed field and f(x) E F[x]. If a, b E F with f(a) < 0 < f ( b ) ,then f(c) = 0 for some c E F between a and b. Proof. By Theorem 7.8 each irreducible factor of f(x) in F[x] has degree 1 or 2. Any irreducible quadratic factor x 2 + rx + s can be written as (x + r/2)2 + (s - r2/4), with s - r 2 / 4 positive (i.e., a square). Thus x 2 + rx + s > 0 for all values assigned to x (from F). If no linear factor of f(x) changed signs from a to b, then f(x) could not change signs, so there is a factor ux + vof f(x) with ua + u < Oand ub + v > 0. If u > 0, then a < - v / u < b, and if u < 0, then b < - v / u < a. In either case c = - v / u is a root of f(x) between a and b.

Suppose F is any field and f(x) E F[x] has positive degree. Apply the Euclidean algorithm (Exercise 11.8.26) to fo(x) = f(x) and fl(x) = f’(x), with one small variation: at each stage subtract the negative of the usual remainder. This process determines the standard sequence fo(x), f l(x),. .. ,fk(x) for f(x). It satisfies the following relations: fdx) = f(x), fib)= f ’ ( 4 ; f d x ) = f1(x)ql(x) - fi(x), degf2(x) < degf,(x); f l ( X ) = f2(x)qz(x) - f3(x), degf3(4 < degf,(x);

7 Real Fields and Sturm's Theorem

257

Next Set g i ( x ) = f,(x)/fk(x), 0 I i I k, and call the sequence go(x), g l ( x ) ,. . . ,g k ( X ) E F [ x ] the Sturm sequence for f ( x ) . Observe that & ( X ) = 1, that G C D ( g , ( x ) , g , ( x ) )= 1, and that for

gi-l(x) = gi(x)qi(x)- g i + l ( x )

1 5 i Ik - 1.

Observe also that f 0 ( x )and go(x) have exactly the same set of roots in F , but that go(x) has no repeated roots (see Proposition 111.3). It is not generally true that g , ( x ) = gb(x). Proposition 7.10. Suppose F is a real closed field, f ( x ) E F [ x ] has positive degree, and go(x),. . . ,g k ( x )= 1 is the Sturm sequence for f ( x ) ,Choose a, b E F with a < b such that f(a) # 0 and f ( b ) # 0. Then b in F and gi(c)= 0 for some i, 0 < i < k, then gi-,(c) and (1) if a I c I

,

gi + (c) differ in sign, i.e., gi - ,(c)gi + I (c) < 0;

(2) if a < c < b and f(c) = 0, then there are c I , c 2E F , with c1 < c < c2, such that go(u)gl(u) < 0 if c, < u < c and go(u)g,(u) > 0 if c < u < c 2 . Proof. (1) Since gi-, ( x ) = g,(x)q,(.w)- g i + l ( x ) and gi(c)= 0 we have g i - ,(c) = -gi+ ](c), so either they differ in sign or both are 0. But gi+,(c) = 0 entails that gj(c) = 0 for all j 2 i, contradicting the fact that ge(c)= 1.

( 2 ) Say that c is a root of f ( x )with multiplicity m, so /(x) = ( x - c)"q(x), q(c) # 0. Thus f l ( x ) = f ' ( x ) = ( x - c)"q'(x)

+ m(x

-

c)"-lq(x)

f k ( x )= ( f ( x ) , f ' ( x )= ) ( x - c)"-ld(x),

with

and d(c) # 0.

We may write q ( x ) = d ( x ) G ( x )and q'(x) = d ( x ) H ( x ) ,with H ( c ) # 0. Consequently go(x) = ( x - c ) G ( x )and yl(x)= ( x - c ) H ( x )+ mG(x),and in particulargl(c) # O.Choosec, < candc, > csuchthatifc, < u < c,,theng,(u) # 0 and G(u) # 0. (Why is this possible?) Note that then, by the Intermediate Value Theorem (7.9), g , ( u ) G ( u )> 0 if c 1 < u < c,, since g,(c) = mG(c) and hence g,(c)G(c) > 0. But then go(u)g,(u) = (u

-

c)g,(u)G(u)

is < 0 if c1 < u < c and is > 0 if c < u < c2. Suppose F is an ordered field and let f o ( x ) f, l ( x ) ,. . . ,f k ( x )be any sequence in F [ x ] . For any a E F define the variation V , ( f ) of the sequence at a to be the number of changes from positive to negative (ignoring 0's) in the sequence fo(a),f l(a),. . . ,.fk(a).Thus V , ( , f )is an integer between 0 and k .

258

VI Further Topics

For example, if F = Q and k = 4 and the sequence {1;.(a)}is 1, - 1,0,1, - 2, then V , ( f )= 3; if { f , ( b ) }is 1,0,1,2, -3, then V,(f) = 1. For the proof of the following theorem it will be convenient to use the usual notation from analysis for intervals in an ordered field F . Thus [ a , b ] = { ~ ~ F : a ~ c ~ b } , (a,b] = (C E F : u < c I b } ,

etc.

Theorem 7.11 (Sturm). Suppose F is a real closed field and f(x) E F [ x ] has positive degree. Suppose a, b E F with a < b and f ( a ) # 0, f ( b ) # 0. If fo(x),fl(x), . . .,f J x ) is the standard sequence for f(x), then V , ( f ) - V,(f) is the number of distinct roots c of f(x) in the interval (a, b).

Proof. Note that fk(a) # 0 and f i ( b ) # 0 since f , ( x )I f ( x ) ,and recall that gi(x) = fi(x)/fk(x) for gi(x) in the Sturm sequeice for ~ ( x ) It. follows that V , ( f )= V,(g) and &(f)= &(g). Thus it will suffice to show that V,(g) - &(g) is the number of roots (automatically distinct) of go(x) between a and b in F. Choose elements a i , 0 I i I m, in [a, b], with u = a0 < a,

< a2 < ... < a,

= b,

so that all roots of all g j ( x ) in the Sturm sequence that lie in [a, b] are included among the a i . Consequently no gj(x) has a root in any of the open intervals (ai,ai+1), 0 I i Im - 1. Take c l E (uo,ul). Each gj(x) remains either always positive or always negative throughout (ao,a,) by the Intermediate Value Theorem, so if no gj(ao) is 0 we have V,,(g) = Kl(g). On the other hand, if gj(ao) = 0 for somej, i < j < k , then g j - l(ao)and gj+ l(ao)differ in sign by (1) of Proposition 7.10; hence gj-l(cl)and g j + l(cl)must also differ in sign by the Intermediate Value Theorem. Thus each of gj,-l(uo), 0, gj+,l(ao)and gj- l(cl), gj(c), g j + l(cl)contribute one sign change in the determination of V,,(g) and &(g), and again K0(g) = K1(g).Similarly, if c, E (a,- 1 , a,), then

K&)

= 1/0,(9).

N e x t c h o o s e ~ ~ E ( a ~ _ ~ , a ~ ) a n~d(ca~ ~ + ,, q + < ~ i) < , lm - 1.Supposefirst that f ( a i ) # 0, and so go(ai) # 0. Then the arguments in the preceding ) place of (ao,ul), and to ( U ~ - ~ , Uin~ )place of paragraph apply to ( a i , a i + l in (a, - 1 , a,), and we conclude that Kz+ ,(g) = V,$(g) = KZ(g).Suppose then that f ( a i ) = gO(ai)= 0. By (2) of Proposition 7.10 (and the Intermediate Value Theorem again) we see that go(ci)and gl(ci)differ in sign, but go(ci+l)and gl(ci+1) have the same sign. Other than that, the arguments used above show the same number of sign changes in g j - l(ci), gj(ci), gj+ l(ci)as in gj- l(ci+ gj(ci+1 ) , g j + l(ci+1) for j > 1. Thus KZ(g)- K, ,(g) = 1 at each i for which f ( a i ) = So(ai) = 0. +

7 Real Fields and Sturm’s Theorem

259

m- 1

Each qz(g) - Kt+l(g) contributes 1 if go(ai)= 0 and contributes 0 otherwise. Thus Va(g)- &(g) is the number of roots of g o ( x )in [a, b]. It is worth noticing that sign changes in the sequence fo(c),fl (c),. . . ,fk(c)are unaffected when some or all of the f;:(x) are replaced by various positive multiples of themselves. Thus at each stage of the Euclidean algorithm we are free to replace f , ( x ) by a positive multiple and proceed to the next stage. This simple observation can simplify the arithmetic considerably. By a slight abuse of language we will continue to call any resulting sequence the standard sequence for f ( x ) . For an example take F = R and f ( x ) = f o ( x ) = x 4 - x - 1. Then f i ( x ) = 4x3 - 1, f 2 ( x ) = 3x + 4,and f3(x) = 1. We have f o b ) fi(4 Sz(4

a --co

-1

+

1 - 5

0

-1 1 - 1 2 11

+ a +

-

f3(4

+ 1

3

1 4 7

31

10

1

-1

+

+

I 1

+

Uf) 2

2 2 1 0 0

It follows that f(x) has just two real roots [neither is repeated since ( f ( x ) , f ’ ( x )= ) 13, one in ( - 1,O) and one in (1,2). For a second example take f ( x ) = 36x2 - 36x + 5. Then fi(x) = 2x - 1 and f2(x) = 1. Thus V,(f) = 2 and V , ( f ) = 0, and f ( x ) has two real roots between 0 and 1. This example shows that Sturm’s Theorem is an improvement over the Intermediate Value Theorem, which would not detect any roots between 0 and 1. EXERCISES

1. Show that f ( x ) = x 3 - 3x - 1 has three real roots and isolate them within intervals of length 1. 2. Show that f ( x ) = x4 + 4 x 3 - 12x + 9 has no real roots. 3. Show that f(x) = x 4 - 6 x 2 - 4x + 2 has four real roots and isolate the roots within four pairwise disjoint intervals.

260

VI FurtherTopics

4. If f ( x ) = x 3 + p x + q E R [ x ] , with p # 0, set D = -4p3 - 27q2. Show that if D < 0, then f ( x ) has one real and two nonreal roots; if D > 0, then f ( x ) has three real roots; and if D = 0, then x = - 3q/2p is a root with multiplicity 2. 5. If

f ( x )= a ,

+ a , x + . . . +u,-lXn-l + X n E R[x]

show that all real roots of f ( x ) lie in the interval [ - b, b ] , where b

= max{l,luol

+ [all + . . . + I U ~ - ~ ~ } .

(Hint: If IuI > b show that If(u)l > 0.) Generalize from R to any ordered field F.

8. REPRESENTATIONS AND CHARACTERS OF FINITE GROUPS We assume throughout this section that G is a finite group and V is a finitedimensional vector space over the complex field C. A representation of G is a homomorphism T from G to the group GL(V )of all invertible linear transformations of V . A basis { u l , . . .,u,,} for V provides an isomorphism between GL( V ) and the group GL(n, C)of all invertible n x n complex matrices, and a corresponding homomorphism ? from G to GL(n,C). The homomorphism ? is called a matrix representation of G. Thus each T(x), x E G , is a linear transformation of V , and f ( x ) is a matrix that represents T ( x )relative to the chosen basis { u l , . , .,u,,}. If dim V = n we say that the representation T(or ?)has degree n, and write deg T = deg ? = n. A representation is called faithful if it is 1-1.

EXAMPLES 1. If G is any finite group and V any complex vector space, then the trivial representation is defined by T ( x ) = I,, all x E G . It is faithful only when G . = 1. 2. Suppose G = (x I x m = l), [ E C is a primitive mth root of unity, V = C, and ?(xi) = (', 0 I i I m - 1. Then T is a faithful representation. 3. Suppose G is a subgroup of the symmetric group S , and set H = G A A,. Set V = C and define ?(x) =

1

-I

if if

X E

H,

XEG\H.

Then T is a representation of G. 4. Suppose G acts as a permutation group on a set S = {s,, . ..,s,}. Let V be a vector space with basis { u l , . . . ,u,,} and define T : G + GL( V ) by means of

8 Representations and Characters of Finite Groups

261

T ( x ) :u i u j if~ and only if xsi = s j . Then T is called a permutation representation of G [note that each ? ( x ) is a permutation matrix, i.e., each row and column of ?(x) has exactly one 1, the other entries are 0's). 5. In the example above suppose S = G and the action is left multiplication. The resulting permutation representation is called the left regular permutation representation of G. It will be denoted in general by R = R G . Exercise 8.1. (1) Write out the permutation matrix representation of the dihedral group D4corresponding to its action on the vectices of a square. (2) Write out the left regular matrix representation if G = Z,,Z,x Z,, S , , or 0,.

Suppose T and S are representations of G on vector spaces V and W , respectively. Then T and S are called equivalent if there is an isomorphism 8: V -+ W such that the diagram

is commutative for all x E G, i.e.,-T(x) = B-'S(x)8 for all x E G. For corresponding matrix representations T and s^ this means there is an invertible matrix M for which ?(x) = M ' g ( x ) M , all x E G, a simultaneous similarity transformation. We write S T (and s^ f )if S and T are equivalent. If T is a representation of G on V and W is a subspace of V we say that W is T-inuariant if T ( x )W c W for all x E G. If the only T-invariant subspaces of V are 0 and V, then T is called irreducible: otherwise T is reducible. Suppose T is reducible, with W a T-invariant subspace, 0 # W # I/. If a basis for W is enlarged to a basis for V, then each ?(x) has the partitioned form ~

r

~

.

r

~

.

It is easily checked that A and B are each matrix representations of G, with respective degrees dim W and dim( V / W ) . If T and S are representations of G on vector spaces V and W define their direct sum T 0 S on V 0 W by means of

( T 0 S)(x): (0, w)++ (T(x)u,W w ) . Clearly T 0 S is also a representation of G. If {ui) is a basis for V and { w j ) is a basis for W , then {(ui, 0)) u { (0, wj)}is a basis for V 0 W and the matrix

262

VI Further Topics

representation (T 0 S)A takes the partitioned form

all x E G. The definition extends in an obvious fashion to the direct sum of any finite number of representations.

Theorem 8.1 (Maschke). Every representation T of a finite group G is equivalent with a direct sum of irreducible representations. Proof. Induction on n = deg T. If n = 1, then Tis irreducible and there is nothing to prove. Assume that n 2 2 and that the theorem holds for representations of degree less than n. For a suitable basis

and so C(xy) = A(x)C(y)

+ C(x)B(y),

all x, y E G.

Thus C(xY)B((xY)- ')W= C(XY)B(Y -

= A(X)C(Y)B(Y - l )

+ C(X),

and so c{c(xY)B((xY)- l ) : Y E G)B(x) = A(x)C{C(Y)B(Y-

l): Y

E G>

+ IGlC(X)

for all x E G. Set

D

=

IG~-'C{C(~)B(~-'):~EG}.

Then the equation above becomes IGIDB(x) = IGIA(x)D

or DB(x) = A(x)D

+ IGIC(x),

+ C(x),

all x E G.

Observe now that

=[

A(x)

It follows that and B.

f'

-

0

A(x)D

+ C(x)

B(x)

A 0 B, and the induction hypothesis can be applied to A

8 Representations and Characters of Finite Groups

263

Proposition 8.2 (Schur's Lemma). Suppose T and S are irreducible representations of G on V and W , respectively. Suppose 8: W + V is a linear transformation with 8 T ( x ) = S ( x ) 8 for all x E G. Then either 8 = 0 or else 0 is an isomorphism and consequently T S.

-

Proof. Suppose 8 # 0. Set W, = ker 8 and V, = Im 8. If w E W, and x E G, then 0 = T(x)Bw = 8S(x)w,so S ( x ) w E ker 8 = W,, i.e., W, is S-invariant. Thus W, = 0 (W, # W since 8 # 0), and so 0 is 1-1. If u E write u = Bu, u E W . If x E G , then T ( x ) u= T ( x ) 8 = ~ 0S(x)u E Im 8 = V,,

el

so V, is T-invariant. Thus V, = V and 8 is onto. If T is a representation of G on V define the centralizer of T to be the algebra C ( T )of all linear transformations A : V + V such that T ( x ) A = A T ( x ) for all x E G. Proposition 8.3. Suppose T is a representation of G. (i) If C ( T )is a division algebra, then T is irreducible. (ii) If T is irreducible, then C ( T )= {a1 : a E

c ) z c.

Proof. (i) If Tis reducible we may assume, by Maschke's Theorem (8.1), that T = T, 0 T2acting on V = Vl 0 V2,with both V;. # 0. Let P be the corresponding projection onto V, . Then P is not invertible, since ker P = V2 # 0. If u E V write u = u , + u 2 , ui E &, Then

P T ( x ) u = P(T,(X)o, + T2(x)u2)= T,(x)u,= T ( x ) P u ,

so P E C ( T )and C ( T )is not a division algebra. (ii) If A E C ( T ) let a E C be an eigenvalue for A . Then ( A - a l ) T ( x ) = T ( x ) ( A - al), all x E G, and A - a1 is singular so it cannot be an isomorphism. Thus A - a1 = 0 by Schur's Lemma (8.2), and A = al. Corollary. If G is abelian and Tis an irreducible representation of G on V , then deg T = 1. Proof. Since G is abelian each T ( x ) ,x E G, is in C(G),and so T ( x ) = a , . 1, a, E @. But then every subspace of V is invariant, so T can be irreducible only

if dim V

=

1.

Proposition 8.4 (Schur). Suppose T and S are inequivalent irreducible representations of G, with deg T = n, and that bases are chosen so that F ( x ) = (tij(x)),s ( x ) = (sij(x)),all x E G. Then

(i) ~ { s i j ( x ) t , ' , ( x - ' ) : x E G} = 0 and (ii) E{tij(X)tkm(X-'):x E G} = 6i,SjkIGl/n, all i,j, k, and rn (as usual d j , = 1 if j = k , = 0 if j # k ) .

264

VI Further Topics

Proof. (i) For any @-matrix M of appropriate size set

GI,

L

= C S l ( x ) M F ' ( x - ' ) : xE

=

1s^( y)$(x)M T(x-')?(y - ') T(y )

and observe that $y)L

X

= C S l ( y x ) M F ( ( y x ) - ' ) T ( y )=

LT(y),

X

all y E G. By Schur's Lemma (8.2) L = 0. Now choose M to be Ejk (see p. 181). Then the irn-entry of L is

1s,j(x)tk,(x

-

') = 0.

X

(ii) This time set

L = E { T ( X ) M T ( X - ~ )E: XG } , and see as above that L E C(f).Thus if M = Ejk we have by Proposition 8.3 that L = ajkl for some ajk E @. The im-entry of L is then ajkdim = Etij(x)tkm(X- ') X

=

=

1tij(x-')tkm(x) X

C tk,(x)lij(x-')

= amidkj.

X

Choose i = m and j # k to see that ajk= 0 if j # k . Next choose i = rn and j = k to see that aii = ajj = a (say) for all i and j. Thus a =1 tij(x)tji(x-')

for all i, j.

X

Sum over j to obtain nu = ~ ~ t i j ( x ) t j i ( x=- 1' ) 1 = [GI, x

j

X

since x j t i j ( x ) ~ j i ( x - is ' ) the ii-entry of T ( x ) F ( x - ' ) = I , and so a Finally, we see that

=

IGl/n.

xtij(x)tkm(X-') = ajkbi, = abjkdi, = di,djklGl/n. X

Exercise 8.2. Suppose Tl, T 2 , .. ., are mutually inequivalent irreducible representations of G , with ?,(x) = (t$')(x)),1 I rn I k. Use Proposition 8.4 to show that the functions tjy), all rn, i, and j, are @-linearly independent. If n, = deg T,, 1 5 rn I k, conclude that n i IlGl, and hence conclude

ck=,

265

8 Representations and Characters of Finite Groups

that there can be no more than I GI mutually inequivalent representations of G. (The key observation is that 1GI is the dimension of the space of all @valued functions on G.) Recall that if A = (aij) is an n x n matrix, then the trace of A is t r A = a i i .It is easy to check that if B is also n x n, then tr(AB) = tr(BA), and consequently if C is n x n and invertible then tr(C- ' A C ) = tr(A).As a result, if T : V + I/ is a linear transformation represented by a matrix A relative to some basis of V , we may define the trace of T t o be tr T = tr A . If T is a representation of G on I/ define its character x = xT, a function ) from G to C, by means of ~ ( x=) tr T ( x ) , all x E G . Observe that ~ ( 1 = tr 1, = dim V = deg T ; ~ ( 1is) called the degree of x. We say that a character x = xT is reducible or irreducible according as T is reducible or irreducible. If T is faithful, then x is called faithful. The character x of a representation T of degree 1 coincides with the (1 x 1) matrix representation 'f.Thus characters of degree 1, which are called h e a r characters, are just homomorphisms from G to the multiplicative group @*. The character of the trivial degree one representation is called the principal character and is often denoted l G . Thus IG(x)= 1, all x E G. Exercise 8.3. Suppose G is a permutation group acting on a set S , let T be the resulting permutation representation of G, and let x = x T . If x E G show that ~ ( xis) the number of fixed points of x in S, ie., of s E S for which xs = s. Proposition 8.5. If T and S are equivalent representations of G, then XT

= XS*

Proof. For appropriate choices of bases we have ?(x) = s(x)for all x

E

G.

Proposition 8.6. Characters are class functions on G, i.e., they are constant on conjugacy classes. Proof. Say x = xT and take x, y E G . Then

~ ( y - l x y )= tr(T(y-'xy)) = tr(T(y)-'T(x)T(y)) = tr T(x) = ~ ( x ) . Proposition8.7. If Tand S are representations of G, then xTes

= zT

+ xS.

Proof. Obvious. Corollary. If T is a representation of G with character x, then there are irreducible characters xl,x 2 , . . . ,xr (not necessarily distinct) such that x =

XI + XZ

+ ... +

Xr.

-

Proof. By Maschke's Theorem (8.1) there are irreducible representations Tl, . . . , T with T Tl 0 Tz @ . . . @ T,. Take traces. Proposition 8.8. If T is a representation of G with character x and x E G , then x ( x - ') = x(x),the complex conjugate of ~ ( x ) .

266

V1 FurtherTopics

Proof. Set H = (x). Then T 1 H is a representation of H . By Maschke's Theorem and the corollary to Proposition 8.3 (applied to T I H)there is a basis relative to which ?(x)

=

[;

... O],

a diagonal matrix. If 1x1 = m, then I

=

f(xm) = ?(x)m

["I1*

=

an

[; "1. =[I -..

a::

so each ai is an mth root of unity, and hence a;' = ifi. But then T(x-1) =

...

... ' 0 1

01 an-

and consequently ~ ( x - ' )= tr ? ( x - ' ) =

'

c;

9

-

a0

Ci =

5).

Exercise 8.4. Suppose ? is a matrix representation of G. Define its contragredient matrix representation ?* by setting F*(x) = ?(x- ')', all x E G . Show that '?* is a representation. If x is the character of ? show that ?* has character Show that '?* is irreducible if and only if ? is irreducible.

x.

If

4,6' are functions from G to C define their inner product to be (4,6')= IGJ-'C{4(x)O(x-'):.x E G}.

Then the inner product is easily seen to be a symmetric bilinear form on the vector space of all functions from G to C. Note that if 4 and 8 are characters of G,then (4,O)is real (use Proposition 8 4 , and for any character x we have

(x,x)= IW' C{lx(x)12:xE G} > 0. Theorem 8.9 (The First Orthogonality Relation). Suppose

xk are all the irreducible characters of G. Then (xi,xj) = dij.

xr, x z , . . .,

Proof. Take i # j and let T and S be representation whose characters are

xi and xj, respectively. Choose bases and say ?(x) = (tij(x)),s(x) = (sij(x)),all x

E

G.Then

8 Representations and Characters of Finite Groups

267

by Proposition 8.4(i), since S and T are inequivalent by Proposition 8.5. On the other hand, (xi,xi)

=

IGI-'CCtii(x)tjj(x-')

=

IGI-lxdijIGI/~i(1)

x i,j

i.j

= IGI- 'xi(1)IGI/xi[1) = 1

by Proposition 8.4(ii). Corollary 1. The characters ill, x 2 , . . . ,x k are linearly independent. Proof. Suppose

c!= aixi

= 0, a, E C. Then

O = (O,xj) = Caixi,xj ( i

)

= Cai(xi,xi)= a j i

for all j.

We observed in the corollary to Proposition 8.7 that a character x of G can be written as a sum of irreducible characters. Thus if xl, xz,. , .,x k are all the irreducible characters of G we may write

x

= nlX1

+ n2X2 +

"'

+ nkxk,

where each ni is a nonnegative integer, the multiplicity of xi as a constituent of x. By Corollary 1 above the coefficients n, are uniquely determined by x. We draw two more corollaries from Theorem 8.9. Corollary 2. If x is a character of G, then x = C:= (x,xi)xi,the coefficient (x, xi) being the multiplicity n, of xi as a constituent of x. Proof. We may write

x = C j n j x j ,and so (x,xi) = C j n j ( x , , x i )= ni.

Corollary 3. If x = C i n i x i and I+/I = C j m j x j are two characters of G, then (x, tj)= nimi.In particular, (x, x) = n z .

xi

zr=

The next result is really just a further corollary of Theorem 8.9, but it is of considerable importance as a test for irreducibility so we list it as a separate theorem.

Theorem 8.10. A character

x of G is irreducible if and only if (x, x ) = 1. Proof. If x is irreducible, then (1,x) = 1 by Theorem 8.9. If x is reducible write x = C i n i x i , with either some n, > 1 or else at least two nonzero ,n: > 1 by Corollary 3 of Theorem 8.9. multiplicities. Thus (x, x ) = 1

-

Exercise 8.5. If T and S are two representations of G having the same character x show that T S.

268

VI Further Topics

Proposition 8.1 1. Denote by p the character of the left regular permutation representation R of G . Then p(1) = IG/ and p ( x ) = 0 if 1 # x E C. If xl,.. .,& are all the irreducible characters of C, then p = Cr= xi(1 ) ~ ~ . Proof. Since R has degree IGI and R(1) = I it is clear that p(1) = [GI. If 1 # x E G, then x y # y for each y E G , and consequently every diagonal entry of R ( x ) is 0. Thus p ( x ) = 0. By Corollary 2 of Theorem 8.9 the multiplicity of xi in p is (

Corollary.

~

3

~

C P ( X ) X ~ (=Xxi(1). -')

=) IGI-'

i

X

xf=xi(

1)2 = I GI.

Proof. p(l) =

Ifzl xi(l)xi(l) = (GI.

Proposition 8.12 (Burnside's Orbit Formula). Suppose G acts on a finite set S and let 8 denote the character of the permutation representation. Then the number of G-orbits in S is (8, lG). Proof. See Theorem 2.2 above in this chapter. Also see Exercise 8.3.

Let x1 = l,, xz,.. . ,x k be the distinct irreducible characters of G, and let = { l}, K 2 , .. .,K , be the conjugacy classes of G. For 1 I i I k and 1 I j I m define oij= lKjlxi(Kj)/xi(l).

K,

Proposition 8.13. Suppose 7;. is an irreducible representation with character xi and K j is a conjugacy class of G. Set Mij = C { 7 ; . ( x ) : xE K j ) . Then Mij = oij1 E C(Ti). Proof. If y E G,then

T(y)-'MijT(y)

=

C{~;.(Y-'XY):X E K j } = Mij,

so M i j E C(7;.).By Proposition 8.3 M, = al, a E @.Taking traces we see tr(Mij) = uxi(l) = C{tr K ( x ) : x E K j } = C { x i ( x ) : xE K j } = IKjlxi(Kj).

Thus u = lKj[xi(Kj)/xi(l) = wij. Proposition8.14. I f x ~ K s s e t n i j s = I { ( y , z ) ~ K x iKj:yz=x)J.Then

(i) nijs depends only on i, j , and s, and not on the choice of x in K , . (ii) IKiI IKjIxt(Ki)xt(Kj)= Xt(l)CsnijsIKsIXt(Ks), and (iii) wtiotj= Is nijsorsfor all i, j , and t. Proof. (i) If y E Ki,z w-lxw = w-'yw * w-lzw.

EKj,

and w E G , then x = yz if and only if

8 Representations and Characters of Finite Groups

269

(ii) With the notation of Proposition 8.13 we see that MtiMtj = C{'&(Y): Y

E

Ki} C('K(2):z E Kj}

= C { T ( ~ z ) : ( y , z E) = C{nijs7;(x):xE

Ki x Kj}

K,} = p l i j s M t s . S

S

Take traces. (iii) Divide by ~ ~ ( 1in) '(ii) and apply Proposition 8.13.

Theorem 8.15 (The Second Orthogonality Relation). If xl,...,zk are the distinct irreducible characters of G and Ki, K j are conjugacy classes of G, then k

C Xt(Ki)Xt(Kjl ) = dijIGI/IKjI. Write K m for the conjugacy class K j '. By t= 1

Proof. we see

Proposition 8.14(ii)

IKiI IKmICXi(Ki)X,(Km)= CnimsIKsI C ~ t ( l ) ~ t ( K s ) f

t

S

= CnimsIKsIP(Ks) =

nimlIGI

S

by Proposition 8.1 1. But niml

=

I{(Y,z) E K i x KJ':yz

=

1}1 = aijlKi1.

Thus, since IKf'I = lKjl, IKiI IKjICXt(Ki)xt(Ki '1 = dijIGI IKiI, f

and the stated result follows.

Theorem 8.16. The number k of irreducible characters of G is equal to the class number rn of G. Proof. Let G act on itself by conjugation, so the G-orbits are the rn conjugacy classes. If 8 is the character of the permutation action, then 8(x) = (C,(x)l for each x E G. By Burnside's Orbit Formula (8.12) m=(O,l,)=

1GI-l

C{B(X):XEG}= I G I - ' C { I C G ( x ) J : x ~ G } k

by the Second Orthogonality Relation (8.15), since ~C,(X)[= IGl/lcl(x)l. Thus

270

VI Further Topics

Corollary 1. The set {xl, x 2 , . . .,x k } of irreducible characters is a basis for the vector space cf(G) of complex-valued class functions on G. Proof. The dimension of cf(G) is equal to the class number m = k, and the k irreducible characters are linearly independent class functions by Corollary 1 to Theorem 8.9.

Corollary 2. Every irreducible character xi is linear [i.e., only if G is abelian.

xi( 1) = I] if and

Proof. If xi(l) = 1 for all i, then by the corollary to Proposition 8.11 we have I GI = Ixi(l)12= k, so every conjugacy class has just one element and G = Z ( G ) is abelian. The converse was the corollary to Proposition 8.3.

xf=

The character table of G is the matrix whose rows are indexed by the irreducible characters xl,. .. ,x k and whose columns are indexed by the conjugacy classes K , , . . .,&. By Theorem 8.15 distinct columns of the character table are orthogonal relative to the usual Hermitian inner product of complex column vectors. For an easy example take G = S,, with K1 = {l}, K , = c1(123), and K , = cl(12). Besides the principal character x1 = 1, there is a linear character x 2 , with x 2 ( K , ) = z 2 ( K 2 )= 1, z 2 ( K , ) = - 1 (see Example 3, page 260). It is immediate from the corollary to Proposition 8.1 1 and the Second Orthogonality ) - 1, and Relation that the remaining character x3 has x3(K1) = 2, ~ 3 ( K 2 = x 3 ( K 3 )= 0. Thus the character table is as indicated.

Exercise 8.6. Compute the character table of the quaternion group Q 2 . For the remainder of this section it will be convenient to write Irr(G) for the set {xl, x 2 , . . .,x k } of irreducible characters of a group G. If T is a representation of G with character x define the kernel of x to be ker x = ker T. It is an easy consequence of Exercise 8.5 that ker x is well defined, i.e., it depends only on x and not on T.

Proposition 8.17. If

x is a character of G , then kerx

=

{ x E G : x ( x )= ~ ( 1 ) ) .

Proof. Let T be a representation with character x. If x E kerz, then T(x) = I so ~ ( x=) tr I = ~ ( 1 )Suppose . then that ~ ( x= ) ~ ( 1 ) As . in the proof of Proposition 8.8 we may assume that ?(x) is a diagonal matrix whose entries

8 Representations and Characters of Finite Groups

271

a,, . . . ,a, are roots of unity in @. Thus X(X) = Cai = ~ I

( 1= ) C IaiI? i

since each [ail = 1. Since equality holds in the triangle inequality we conclude that a, = b, . a,, with 0 < bi E R, for all i, and consequently 1 = /ail= b J a ,1 = bi for all i. But then ~ ( x=) x(l)a, = ~ ( l )so , a, = 1 and hence T ( x ) = I and x E ker x. The ideas in the proof of Proposition 8.17 enable us to describe another useful subgroup of G associated with a character x. Set Z ( x ) = { x E G: Ix(x)j = x( 1)). Thus ker x c Z(x). If T is a representation with character x and if x E Z ( x ) we may assume, as in the proof above, that ?(x) is diagonal, with entries a,, . . . ,a, that are roots of unity in C.As in the proof a, = bial for all i, with 0 < b, E R, and in fact each bi = 1, so ?(x) = a l l . It follows easily that Z ( x )consists precisely of those x E G that are represented by scalar multiples of the identity in any representation with character x. Two immediate consequences are recorded in the next proposition. Proposition 8.18. If x is a character of G, then Z ( x )4G. If x is faithful, then Z ( x ) IZ(G). Suppose K 4G and let T be a representation of G with character x. If K < kerx we may without ambiguity define a representation ?. of G / K by means of ? ( x K ) = T(x),all x K E G / K . Thus the character f o f ‘7.is given by ~ ( x K=) ~ ( x ) all , x E G. Conversely, if ? is a representation of G/K with character f ,and if q: G + G / K is the quotient map, then the composition ‘7.qis ) , it is a representation T of G whose character x = Tu, i.e., ~ ( x=) ~ ( x K and clear that K I ker 1.It is also clear in both cases that f is irreducible if and only if x is irreducible. It is customary to write f = x and let the context determine whether x is being viewed as a character of G or of G I K . In particular, if K a G, then Irr(G/K) consists of those x E Irr(G) such that K 5 kerx. Proposition 8.19. The set of linear (i.e., degree 1) characters of G is Irr(G/G’). Proof. Since G/G’ is abelian each x E Irr(G/G’) is linear by Corollary 2, Theorem 8.16. If x E Irr(G) is linear, then 1:G + @* is a homomorphism. Thus G‘ Iker x, since C* is abelian, and so x E Irr(G/G’). Exercise 8.7. Compute the character tables of the dihedral groups D4 and D,.Note that in both cases the group elements are naturally represented as

2 x 2 matrices.

Recall from Chapter I11 that A denotes the field of algebraic numbers, i.e., all a E C that are algebraic over Q. Say that a E A is an algebraic integer if its

272

V1 Further Topics

(monic) minimal polynomial m,(x) is in Z [ x ] . Thus, for example, algebraic integer but $12 is not.

f l is an

Proposition 8.20. The following are equivalent:

(i) a is an algebraic integer; (ii) Z[ a ] is a finitely generated Z-module; (iii) a is a root of a monic polynomial f ( x ) in Z [ x ] . Proof. (i) =. (ii): If the minimal polynomial of a is m(x) = a,

then ak= -a

,'1 - a ,

+ a , x + ... + X k ,

. a -...-

ak- 1

.ak-' E

a, E

z,

z 1 + +Z '

"

'

'

0'-

',

and it follows easily that { 1, a,. ..,a k - ' } is a generating set for the Z-module

mra1.

(ii) * (iii): Let { fi(a),fi(a), ...,f m ( a ) }be a generating set for Z[a], with L(x) E Z [ x ] , all i. Set

n = max(1 + degA(x): 1 5 i I m).

Since a" E Z [ a ] we may write a" = f ( x )= X"

xr= biA(a),bi

E

Z.Set

m

-

1 bif,(x)E Z [ X ] .

i= 1

Then f ( x ) is monic and f(a) = 0. (iii) =s (i): Let m(x) E Q [ x ] be the minimal polynomial of a. Then f ( x ) = m(x)g(x) for some g ( x ) E Q [ x ] . Write m ( x ) = (a/b)h(x) and g ( x ) = (c/d)k(x), with a, b,c,d E Z and h(x), k(x) primitive polynomials in Z [ x ] . By Gauss's Lemma (Theorem 11.5.13) h(x)k(x)is also primitive. But then, since bdf(x) = ach(x)k(x),we see that bd = QC, and hence f ( x ) = h(x)k(x).It follows that h(x) and k ( x ) are monic, and hence m ( x ) = h(x) E Z[x].

Corollary 1. If a, b E A are algebraic integers, then a algebraic integers.

+ b and ab are

Proof. Let {a,, . . . , a m }and {bl,. . . ,b,) be sets of generators for Z [ a ] and Z [ b ] as Z-modules. Then { a i b j }is a finite set of generators for Z[a, b ] , and Z [ a + b ] , Z [ a b ] are submodules of Z [ a , b ] so they are finitely generated by the corollary to Theorem IV.2.9.

Corollary 2. If a E Q and a is an algebraic integer, then a E Z. Proof. The minimal polynomial over Q for a is m(x) = x - a , and m(x) E Z [ x ] so a E Z.

8 Representationsand Characters of Finite Groups

273

+ + +

If x is a character of G, JGJ= n, and x E G, then ~ ( x = ) a, a, ... a k , where k = x(1) and each a, is an nth root of unity in C (see the proof of Proposition 8.8). But then each ai is a root of x" - 1 E Z[x]. By Proposition 8.20 and its first corollary we see that ~ ( x is) an algebraic integer. Note also that for all x E G.

Ix(x)l I ZIuil = k = ~ ( 1 ) I

Recall that if Irr(G) = (x,,..., xk) and { K , , .. .,&I are the conjugacy classes in G, then wij = [Kjlxi(Kj)/xi(I). Proposition 8.21. For all i and j w i j is an algebraic integer. Proof. By Proposition 8.14(iii) we have w,,mtj = ~ s n , j s wfor , s all f, with nijs E Z. Fix i and t, let u be the column vector whose transpose is u' = ( w t l , .. . ,cot,,), and let N be the k x k matrix whosej-s entry is nijs.Then N u = w,,u, so o,, is an eigenvalue of N and consequently w,, is a root of a monic polynomial in Z[x]. Apply Proposition 8.20 and replace 1, i by i, j .

Theorem 8.22. of ]GI.

{ x l , .. . ,x k } , then each degree xi(l) is a divisor

If Irr(G) =

Proof: Since xi is a class function we have

Thus IGI/Xi(U = (IGI/xi(l))(Xi,Xi) k

1 IKjIXi(Kj)xi(K;')

= (1/~,(1)) J =

1

k

=

1 wijXi(Ki ')>

j=1

which is an algebraic integer by Proposition 8.21 and Corollary 1 to Proposition 8.20. But then IGl/xi( 1 ) is both a rational number and an algebraic integer, so IGI/X,(l) E Z. Proposition 8.23. Suppose xi E Irr G and K j is a conjugacy class of G, and ) lKjl are relatively prime. Then either x i ( K j )= 0 or else that ~ ~ ( 1and Ixi(Kj)l = xi(l), in which case K j E Z ( x j ) . Proof. Choose a, b E Z for which q , ( 1 ) x i ( K j ) / x i1) ( to see that

+ blKjl = 1.

aXi(Kj) + b l K j l ~ i ( K j ) / ~ i (= l ) Xi(Kj)/Xi(l)

or aXi(Kj) + boij

=

xi(Kj)/xi(l).

Multiply by

274

VI Further Topics

Thus xi(Kj)/xi(l) is an algebraic integer, by Proposition 8.21, and lxi(Kj)/xi(l)(I 1 as observed earlier. If the elements of K j have order n let iE C be a primitive nth root of unity and let 93 be the Galois group G(Q(C):Q). Since x i ( K j )is a sum of powers of C the same is true of a(xi(Kj)) for all (T E '3. Furthermore, o ( x i ( K j ) / x i ( l ) is ) an algebraic integer, and Ia(xi(Kj)/xi(l))I 5 1. Set u = n{o(~i(Kj)/~i(l)):g E 9).

Then u is an algebraic integer, Iu( I 1, and, since u is fixed by all (T E 93, u E Q. By Proposition 8.20, Corollary 2, u E Z,and hence u = 0 or u = & 1. If u = 0, then x i ( K j )= 0. If u = f 1 take (T = 1 to see that Ixi(Kj)l= ~ ~ ( 1hence ); K j S Z(xi). Theorem 8.24 (Burnside). If G is a nonabelian simple group, then the only conjugacy class having a prime power number of elements is K , = { 11.

Proof. Suppose K j is a conjugacy class with (Kj( = p" for some prime p , and suppose K j # K , . Note that m 2 1, since Z (G ) = 1. Temporarily relabel the irreducible characters xz,. . . ,xk so that p $ xi(1) if 2 I i I ko but p xi(1) if k , + 1 i k. Since G is simple Z(xi)= 1 for all i > 1. By Proposition 8.23 it follows that x i ( K j )= 0 for 2 I i Ik o . Say xi(l) = pn, for ko + 1 Ii 5 k. Apply the Second Orthogonality relation to columns 1 a n d j of the character table to obtain

I

0=1

+ p ~ { n i x i ( K j ) : k+o 1 < i I k}.

But then O!

+ 1 I i I k}

= C{cixi(Kj):k,

is an algebraic integer, and or = Proposition 8.20.

-

l / p E Q\Z, contradicting Corollary 2 to

The final theorem in this section is a first indication of the great power of character-theoretic methods for finite groups. It should be compared with the application of the Sylow theorems to groups of order p q in Chapter I . Theorem 8.25 (Burnside). If p and q are distinct primes and G is a group of order p8qb, then G is solvable. Proof. If G is not solvable it has a composition factor H that is simple and nonabelian, with IHI = pcqd. Note that c 2 1 and d 2 1 since p-groups are solvable. Let P be p-Sylow in H , choose x # 1 in Z ( P ) ,and let K be the Hconjugacy class of x. Then P I CH(x),so I K ] = [ H : C , ( x ) ] is a divisor of [ H : P ] = qd, and lKl is a prime power. By Theorem 8.24 IKI = 1, so 1 # x E Z ( H ) ,contradicting the simplicity of H .

9 Some Gaiois Group

275

Corollary. A finite nonabelian simple group has at least three distinct primes dividing its order. 9. SOME GALOIS GROUPS

The determination of Galois groups for specific polynomials seems to be more of an art than a science. We present in this section some ad hoc methods that can be useful, and a number of examples. Suppose F is a field, f ( x ) E F [ x ] is monic of degree n, and K 2 F is a splitting field for f ( x ) . If f ( x ) has roots a,,a2, ..., a, in K define the discriminant of f ( x ) to be

D

= Of= n { ( a i - a j ) ’ : 1 I i <j I n}.

Note that D = 0 if and only if f ( x ) has a repeated root in K . If f ( x ) has no repeated roots, then K is a Galois extension of F. If G = G ( K : F ) ,then each g E G permutes the roots { a ; } of f ( x ) , so clearly a(D)= D, and hence D E F G = F . In particular D does not depend on the choice of the splitting field K . In fact, D is left fixed by all possible permutations of a,, a2,...,a, and consequently, by Theorem 111.5.3, D is a polynomial over F in the elementary symmetric polynomials ojevaluated at a,, . . . ,a,, i.e., in the coefficientsof f ( x ) . In order to be able to evaluate some discriminants it will be useful to establish some classical results. For the next three propositions we take R to be a commutative ring with 1; x , , . . . ,x, to be distinct indeterminates over R; and D to be the “generic discriminant” over R; i.e., D = n { ( x i - xi)’: 1 5 i < j I n}. Proposition 9.1 (Vandermonde). Define d

= d ( x , , . . . ,x,)

to be

n { x i - x j : l Ij < i 2 n),

so d 2 = D. Then

d=

1

1

X1

X2

X:

xf

x“- 1 1

...

... ... 1

2

...

Proof. The result is clear if n = 2, so suppose n 2 3 and assume the result to hold for the n - 1 indeterminates x 2 , .. .,x , . Multiply row n - 1 in the determinant by -xl and add to row n, then multiply row n - 2 by - x l and add to row n - 1,. . ., and finally multiply row 1 by - x1 and add to row 2.

276

VI Further Topics

Then expand along the first column to see that the determinant is equal to

The proposition follows by induction.

For each positive integer k the kth power sum of xl, x2,. . . ,x, is sk = x t

f

xi

+ .' + xf: E R[x1, x27.. . x.1. '

3

Since each sk is clearly a symmetric polynomial it can be expressed, by Theorem 111.5.3, as a polynomial in the elementary symmetric polynomials. The next proposition provides a recursive scheme for expressing sk in terms of 617 ' 0

7

...

9

6,.

Proposition 9.2 (Newton's Identities). The kth power sum sk satisfies sk = sk-,o, - sk-,a,

- Sk-101

- Sk-202

+ ... + (-l)ksl~k-l + (-l)k+lkok + + (-l)n+lSk-nan "*

if if

k I n, k > n.

Proof. We sketch a proof in terms of a generating function, a standard device in combinatorial theory. Let z be an indeterminate over R[xl,x2, ..., x,] and define a(z) = n{l - xiz:l I i I n}.

Observe that a(z) = 1 - ( i l Z

+

O2Z2

+ (-

- ..-

l)"d,Z".

Define s(z) to be the generating function for the sequence formal power series s(z) = Cp=I &Zk. Then n

a'(z) = -

C x i n { 1 - xjz: 1 I j I n,j z i } ,

i= 1

so

(sk},

i.e., s(z) is the

9 Some GaIois Groups

i.e., s(z)o(z) = -za'(z). Thus (setting g o

=

277

1)

Proposition 9.3.

Proof. If A is the matrix with ij-entry xj, 0 5 i 5 n - 1, 1 5 j 5 n - 1, then det A = d by Proposition 9.1, and so

D = d 2 = (det A)(det A ' ) = det(AA'). But AA' is the matrix whose determinant is displayed in the statement of the proposition.

Corollary. The discriminant Dfof any polynomial f ( x )can be obtained as a polynomial in the coefficients of f ( x ) . Proof. The coefficients of f ( x ) are (to within sign) the elementary symmetric polynomials ol, u 2 , .. . , o nevaluated at the roots of f ( x ) . By the proposition Dfcan be evaluated in terms of the power sums sk,and then each sk can be expressed in terms of the oi [i.e., the coefficients of f(x)] by means of Newton's identities.

It should be remarked that the discriminant D, can also be obtained in terms of the resultant R ( f , f ' ) of f ( x ) and its derivative f'(x) (see Van der Waerden [37, p. 871.) It is not necessary to compute the power sums sk in order to compute R ( f , f').However, R ( f ,,f') is usually computed as the determinant of a (2n - 1) x (2n - 1) matrix, so there is no free lunch. Some examples are in order.

278 278

VI Further Further Topics Topics VI

EXAMPLES 1. If f ( x ) = x 2

+ b x + c E F [ x ] , then a1 = -b, ~2

= ~ 1 0 1- 2a2 =

Thus

1

Df = -b

a2 = c, s1 = al,and

- 202 = b2 - 2 ~ .

-b b2 - 2c

1

= b2 - 4c.

Thus the familiar quadratic formula gives the roots of f ( x ) as +( - b f char F # 2.

nf) if

2. Suppose f ( x ) = x 3 + a x 2 + bx + c E FCx], with charF # 3. If we use - a / 3 to change variables then f ( x ) takes the form f ( x ) = g(y) = y 3 + p y + 4 E F [ y ] . (Verify.) Since the roots of g(y) are those of f ( x ) each diminished by a / 3 it is clear that 0, = 0,. For g(y) we have c1 = 0, a2 = p , and c3 = -4. Newton's Identities yield s1 = 0, s2 = - 2 p , s3 = -34, and s4 = 2p2. Thus x=y

3 D=

0

-2p

0 -2p -2p -3q = -4p3 -27q2. - 3 q 2p2

It may be instructive instructive to carry out theclassical solution of thecubicequation thecubicequation g(y) g ( y ) = 0 in order order to see the natural natural role played by the discriminant 0,. Continue Continue to assume that char F # 3, and assume also that char char F # 2. Write y = u uu with u and uu satisfying 3uu = - p . Then

+

y3 = u3

+ u3 + 3uu(u + u ) = u 3 + u3 + 3uuy,

and substituting into g(y) g ( y ) = 0 we obtain u3 + u3 = -4. -4. If z is another (independent) indeterminate we have, since u3u3 = - p 3 / 2 1 , that

+

If we set h(z) = z 2 42 - p 3 / 2 7 , then by Examples 1 and 2 above D,,= 42 + 4p3/27 = -0,127, and by the quadratic formula we may take u3 = -412 +8-J and u3 = -412 If we write u = (-412 + J-)ll3 and v = ( - 4 p - J-)'l3 there are generally three values determined for each of u and u. The values must be paired, however, so that 3uu = - p , giving three values (at most) for y = u + u. If u = u1 and u = u1 are one paired choice for cube roots, and if o # 1 is a cube root of unity, then

Jm.

9 Some Galois Group

279

the solutions to g ( y ) = 0 are Y , = u1

+

u1

= (-4/2

+

J

-

)

1

'

3

+(-q/2

-J

W ) ' 1 3 ,

+ w2u1 = o ( - q q / 2 + & q J i E i ) l / 3

y,

= ou1

y,

= w 2u1

+ w'( -412

+

- J--D,/108)1!3,

mu1

-+ \/--Dg/108)1!3

= 07-412

+ w( - 412 - J

q

p

3

.

These equations are commonly called Cardano's Formulas, although they are the work of Scipio del Ferro and Nicolo Tartaglia. if

Exercise 9.1. Take F

=

Q and use Cardano's Formulas to solve f(x) = 0

(1) f(x) = x 3 - 9~ - 28, ( 2 ) f(x) = x 3 - 1 2 ~ 8 [observe in ( 2 ) that f ( x ) has 3 real roots].

+

If f(x) E F [ x ] has roots a , , a , , . . .,a, in a splitting field K over F set d, = n { a i - a j :1 Ij < i I n } ,

a specialization of the Vandermonde determinant of Proposition 9.1. Thus d: = 0,. Theorem 9.4. Suppose char F # 2, f ( x ) E F [ x ] , and f(x) has distinct roots in a splitting field K over F . View G , = G ( K : F )as a subgroup of S, and set H = G , n A,, the even subgroup. Then B H = F(d,). Proof. If G E S, is a transposition then Ct,(d,) = - d, # d,. Thus if Ct, E G,, then Ct, E H if and only if Ct,(d,) = d,, so F(d,) 5 9 H and YF(d,) 5 H . Consequently F H 5 9 9 F ( d , ) = F(d,), so F(d,) = BH.

d,

Corollary. The Galois group G , of f ( x ) is a subgroup of A , if and only if F , i.e., if and only if the discriminant 0,is a square in F.

E

The corollary above can be quite useful for the determination of Galois groups. Recall that if f ( x ) E F[x] is irreducible, then its Galois group G = G , acts transitively on its set of roots (see Exercise 111.4.1). Proposition 9.5. Suppose char F # 2 and f ( x ) E F[x] is irreducible, separable, and of degree 3. Then its Galois group G = G , is the alternating group A , if the discriminant 0,has a square root in F , otherwise G = S,. Proof. The only transitive subgroups of S, are A , and S,. Thus the proposition is an immediate consequence of the corollary to Theorem 9.4.

280

VI FurtherTopics

Exercise 9.2. Determine the Galois groups of (1) f(x) and (2)f(x) = x3 - 12x + 8.

= x3 - 9x - 28

There is an analog of Proposition 9.5 for quartic (degree 4) polynomials, but a bit of preparation will be needed first. Continue to assume that char F # 2. If

f(x)

= x4

+ b,x3 + b2x2 + b3x + b,

then the change of variables y

=x -

E

F[x],

b,/4transforms f(x) to a polynomial

g ( y ) for which the coefficient of y 3 is 0. Note that the discriminant is

unchanged. Thus we may (and shall) assume that

f(x) = x4 + bx2 + cx + d E F [ x ] . Suppose f(x) has distinct roots a,, a,, a 3 ,a4 in a splitting field K over F. Write G = G , for the Galois group G(K :F) as usual. The sole reason that the alternating group A, is not simple is the existence of the normal subgroup V = (1,(12)(34),(13)(24),(14)(23)}, Klein’s 4-group. In K set

+ u3a4,

+

+

a3 = ~ 1 U 4 a2a3 u2 = a1a3 a2u4, and observe that each ui is fixed by the elements of V. a1 = u1u2

Proposition 9.6. In the above setting 9 ( G n V) = F(al,az,a3).

Proof. We observed above that F(al,a2,a3) c %(G n V). The element a, is fixed by the transposition (12) and hence by the subgroup H , =

( V u {(12)}), a dihedral subgroup of order 8 in S,. The transpositions (13) and (14) and representatives for the other two cosets of H, in S,, and (13)a,= u3 # al,(14)a1 = a2 # a1 so HI is the subgroup of S, that fixes a,. Similarly H2 = ( V u ((13))) fixes a, and H3 = ( V u ((14))) fixes a3. Since HI n H , n H3 = V it follows that if LT E G fixes a,, a,, a3 then CT E G n V , so 9 F ( a , , a,, a 3 )s G n V. Thus %(G n V ) E 9 Y F ( ~ t 1 , a 2 , ~= 3 )F ( a 1 , ~ 2 , ~ 3 ) ,

and the two are equal. It is at least implicit in the remarks and proof above that a,, a,, and a3 are permuted among themselves by all of S,, and hence by G,. It follows that g(x) = (x - a,)(x - Q)(X - u3) is in F[x]. Let us determine its coefficients. Recall that f ( x ) = x4 bx2 + cx + d , so a, + a, + a3 + a, = O , = 0, 0, = b,a, = -c, and c4 = u1a2a3a4= d. The coefficient of x2 in g(x) is

+

-(a1

+ a, + a 3 )= -(a1a2 + a3a4 -t 4 1 4 3 + a2a4 + a1a4 + -

-02

=

-b.

(1243)

9 Some GaIois Group

28 1

Gathering terms judiciously we find that the coefficient of x in g(x) is

aIa2

+

+

+ + a 3 )+ u1u2u4(u1+ a, + u4)

alu2u3(al a,

~ 1 1 ~ 1 3~ ( 2 = ~ 3

+ a 1 a 3 a 4 ( u 1 + a3 + a41

+ u2u3a4(u2+ a3 + a4) - 4alu,a3a, (since o1 = 0), i.e., the coefficient is -40,

= -4d.

Exercise 9.3. Show that the constant term of g(x) is - a I a 2 a 3 2 4O,(r4 = - c 2 4- 4bd (Newton’s Identity for s2 may be helpful).

-03

+

=

Thus we have g(x) = (x - U l N X - a2)(x - a 3 ) = x 3 - bx2 - 4dx - c 2 4bd E F [ x ] ;

+

g(x) is called the resolvent cubic of f(x). Set L = F ( a l , a,, a3),the splitting field for g(x) over F , so L = 9 ( G , n V ) by Proposition 9.6. By the Fundamental

Theorem of Galois Theory

G, = G ( L : F )z G,/gL Observe that a1 - a, = aluz

+ u3a4 -

(1143

=

-~

G,/(G, n V ) .

2

= ~(al 4-

~ 4 ) ( ~-2

a3),

and similarly -~

l = 3

(a1 - ~

3 ) ( ~ -2a,),&,

- a3

= (a1

-~ 2 ) ( ~ 3~ 4 ) .

It follows immediately that 0, = D,. Suppose now that f(x) is irreducible over F , so G, is a transitive subgroup of S,. The possible orders for G, are 24, 12, 8, and 4. Clearly S, and A , are transitive, and they are the unique subgroups of their orders. Subgroups of order 8 are 2-Sylow; there are three of them, all conjugate in S,. Each is dihedral, one being D = ((1234),(13)), and they are transitive. There are two types of subgroups of order 4, both transitive. One is Klein’s 4-group V , which is normal in S, (and unique); the other cyclic, e.g., C = ((1234)). The three cyclic subgroups of order 4 are all conjugate. Except for C and its conjugates every transitive subgroup contains V as a subgroup. Set m = [ L : F ] , L = F(al,ct2,a3)as above. Thus also m = lG,l since G, = G ( L :F ) and L is Galois over F. Since G , z G,/G, n V it follows from the discussion above that G, = S, if and only if m = 6, G = A , if and only if m = 3,andG, = Vifandonlyif m = 1.Since V G DandIC n V ( = 2wehave m = 2 both when G, = C and when G, = D. If G, = C , then [ K : F ] = 4 and [ L :F ] = 2, so [ K : L ] = 2, so f ( x )must be reducible over L in that case. On the other hand if G, = D, then [ K : F ] = 8 and [ L : F ] = 2, so [ K : L ] = 4. In that

,

282

VI Further Topics

case G ( K : L )= S L = V , which is transitive on (a1,u2,u3,a4}, so f(x) is irreducible over L. We summarize in the next proposition. Proposition 9.7. Suppose char F # 2, f(x) E F [x] is separable and irreducible, degf(x) = 4, and g(x) is the resolvent cubic of f(x). Let L E K be the splitting field for g(x) over F and set rn = [ L :F]. If rn = 6, then G = S,; if rn = 3, then G, = A,; if m = 2 and f(x) is irreducible over L, then G, = D ;if rn = 2 and f(x) is reducible over L, then G, = C ; and if rn = 1, then G, = V .

,

The restriction char F # 2 was not used in an essential way in the discussion above, only to change variables. In practice it is convenient to assume that char F # 2 and char F # 3 so that the resolvent cubic is more manageable. Assume for the following examples that F = Q. Thus separability is not in question.

EXAMPLES 1. If f(x) = x4 - x + 1 E Q[x], then reduction of coefficients mod 2 yields f(x) = x4 + x + 1 E Z,[x]. Clearly f(x) has no roots in Z,,and the assumption that f ( x ) factors as a product of two quadratics in Z,[x] leads quickly to a contradiction (verify),so f(x) is ireducible, so f(x) is irreducible in Z[x], hence in Q[x] by Proposition 11.5.15. The resolvent cubic of f(x) is g(x) = x 3 - 4x - 1, which is also irreducible (e.g., since it has no roots in Z, or since it has no roots mod3). The discriminant is 0, = D, = 4, - 27 = 229, which is not a square in Q. Thus G, = S, by Proposition 9.5, and rn = 6. Thus G, = S,. 2. If f(x) = x4 - 6x2 + 8x + 28 E O[x], then f(x + 1) = x4 + 4x3 + 31 which is irreducible in Q [x] by Exercise 11.8.31, so j ( x ) is also irreducible. The resolvent cubic is g(x) = x 3 + 6x2 - 112x - 736. Change variables via x = y - 2 to transform g(x) to h(y) = y 3 - 124y - 496. Then mod 3 we have h(y) = y 3 - y - 1 E Z3[x], which is irreducible since it has no roots in Z,, and hence h( y) is irreducible in Q [y]. Since

Dg= Dh = 4-1243 - 27.496’ we have G, = A , and rn

=

+

3. Thus G

= 984064 = (31.32),

, A,. =

3. If f(x) = x4 - 2x2 4x - 2 E Q[xJ, then f(x) is irreducible by Eisenstein’s Criterion with p = 2. Its resolvent cubic is g(x) = x 3 + 2x2 + 8x = X(X’ so rn = 2 and L

=

+ 2~ + 8),

Q(p). The ring of algebraic integers in Q

R - , = {(a

+ b p ) / 2 : a , b E %, a

_=

b(mod 2))

( g ) is

9 Some GaIois Groups

283

(see Section 11.5). It is easily verified that 2 is a prime in R - 7 , which is a PID (Exercise 11.5.7), so f ( x ) is irreducible in R - , [ x ] , again by Eisenstein’s Criterion, and is irreducible in L [ x ] by Proposition 11.5.15. Thus G, = D, dihedral of order 8, by Proposition 9.7. 4. If f ( x ) = x 4 5 x 2 5 E Q [ x ] , then f ( x ) is irreducible by Eisenstein’s Criterion with p = 5. Thus

+

+

g ( x ) = x 3 - 5 x 2 - 20x

+ 100 = ( x - 5)(x’ - 20),

and L = Q($ ), m = 2. Sincef’(x)is quadratic in x 2 it is easy to find the roots of f ( x ) explicitly, and hence to factor f ( x ) as f(x) = [ x 2

+ ( 5 - f i ) / 2 1 [ x 2 + (5 + 3 1 / 2 1

in L[x]. Thus G, = C, cyclic of order 4 , by Proposition 9.7. 5. If f ( x ) = x 4 1 E Q [ x ] , then f ( x )is the cyclotomic polynomial 0 8 ( x ) ,so f ( x ) is irreducible by Theorem 111.4.2;alternatively f ( x + 1) is irreducible by Eisenstein’s Criterion. Its resolvent cubic is

+

g ( x ) = x 3 - 4x

so m = 1 and G,

=

=

x ( x - 2)(x + 2),

V

Exercise 9.4. Determine the Galois group over Q for each of the following quartics: (a) (b) (c) (d) (e) (f)

f ( x ) = x4 - 2; f ( x ) = x 4 8x 12; f ( x ) = 4x4 x 2 9; f(x) = 0 5 ( x )= x 4 + x 3 + x’ + x f ( x ) = x 4 p x 2 p, p prime; f ( x ) = x 4 4px + 3p, p prime.

+ + + + + + +

+ 1 (see Proposition 111.4.1);

Exercise 9.5. (Solution of quartics). Suppose char F # 2 , 3 and f ( x )= x4

+

+ b x 2 + cx + d E F [ x ] .

To solve f ( x ) = 0 add (ux v ) to ~ both sides and determine u, v E F so that f(x) (ux u)’ = (x’ w)’ for some w E F . Show that u, v, and w satisfy u2 = 2w - b, 2uv = - c , v 2 = w 2 - d , and that 2w is a root of the resolvent cubic g(x). Solve for w (by Cardano’s formulas if necessary; any one of the three roots will do), then for u and v . Then the roots of f ( x )are the four roots of the two quadratics x 2 ux w v and x 2 - ux + w - v. Use this method to solve x4 - 7 x 2 18x - 8 = 0 (use w = 1).

+

+

+

+

+ + +

+

Exercise 9.6. If f ( x ) = x 4 - 4 x 2 show that the Galois group G of f ( x ) over Q is S4. Show that f ( x ) has a root a, E R with 0 < a , < 1, and that if L = Q ( a l ) , then [ L : Q ] = 4, H = Y L 1S,, and H $ A,. Show that H is a

284

VI FurtherTopics

maximal subgroup of S, (see Exercise 1.12.22), and hence there are no fields properly between Q and L. Conclude that a , is not constructible, even though [ Q ( a , ) : Q ] = 2’. Use Exercise 9.5 above to express a, in terms cf radicals. Suppose R is a U F D with field F of fractions, and f ( x ) E R [ x ] is monic of degree n with distinct roots a,, a’,. . .,a, in a splitting field K over F . Let G = G ( K : F ) be the Galois group of f ( x ) . As usual we may view G as a permutation group on { a i } ,hence as a subgroup of S,,, when it is convenient to do so. Let x l , x z , ..., x,, be distinct indeterminates over R [ x ] and set R , = R [ x , , . . . , x , ] , F, = F ( x , , . . . , x , ) , and K , = K ( x , , . . .,x,,).Note that K , is a splitting field for f ( x ) over F,, and that the Galois group G, = G ( K , :F , ) is isomorphic with G. The isomorphism is effected simply by restricting each I$ E G, to K ; in fact, if G, and G are both viewed as permutation groups on { a i }they are identical, so we shall not usually distinguish between them. If M = Cr(x,,x,, . . .,x,,) E K1 and z E S, define a, = C I ( X , - I , , . .., x r - I,,), and note that if also q E S,,,then (a,),, = ar,,.This is essentially the permutation action discussed in Section 111.5, with a slight change in notation. aixi E K1and note that if z E S,,, then Set 8 =

I;=,

e, =

i

~

a = x~u r i x~i .

~

i

-

~

~

Since the roots {ai) are distinct it is clear that if 7 # p in S,,, then 8, # 0,. Define g(x) = n { x - 8,:z E S,,},a polynomial with n! distinct roots in K,. If we view g(x) as a polynomial in x and in x l , . . . ,x,, with coefficients in K it is also clear that those coefficients are polynomials (say over R ; actually over h) in the roots { a , } ,and that the coefficients are invariant under all permutations of (ai}.It follows from Theorem 111.5.3 that the coefficientsare polynomials in the coefficients of f ( x ) ;hence g ( x ) E REX,

9 . .

-

3

x,, X I = Ri[xI.

If 4 E G , we have, of course, another action of 4 on any a E K , , denoted as usual by +u, in which 4 fixes each x i and acts automorphically on the coefficients from K . It is not true in general that 4~ = a4, although

4e = C a , i x i 1

as observed above. If

For fixed

E

T E S,,

S,, define

= C a i x 4 - l i= i

observe that

8,

9 Some Gaiois Group

If tj E G, then tjgo(4 =

so y,(x)

E

n

.(x - $48,:

285

4 E G} = g,(x),

Fl [XI, F, being the fixed field for G. In fact, gu(x) E F1 [ X I n K [ x , , ~

.

2 7 . .t x n t

XI

= F C x , , . . ., xn,

XI.

Observe that [ K l : F l ] = IGI = degg,(x), g,(x) has 8, as a root and G is transitive on the set of roots of g,(x) in K , . It follows that g,(x) is irreducible in F , [ x ] and that K , = F,(O,),with g,(x) the minimal polynomial over F, for 8,. Now let c, = 1, 0 2 , .. . ,ck( k = [S,,:G I ) be a set of (right) coset representatives for G in S,, and for each i write g,(x) = g,,(x). Then

S,

=

{4ai:$E G, 1 5 i 5 n}

and so g ( x ) = n { g i ( x ) 1: I i Ik } ,

representing g ( x ) as a product of irreducible factors in F [ x , , . . . , x , , x ] . Since R 1 [ x ] = R [ x l , ..., x , , , x ] is also a UFD and g ( x ) E R 1 [ x ] it follows from Proposition 111.5.15 and the fact that g ( x ) and all g,(x) are monic that g(x)= g i ( x )must be the prime factorization of g ( x ) in R , [XI. For each t E S,, we have

ni

r

1

with t permuting the irreducible factors of g(x), since { c i s :1 I i I k } is another set of coset representatives for G in S,,. Furthermore, since o, = 1 we see that qslr(x)= g,(x) = g l ( x )if and only if t E G. Thus in the action of S, on the irreducible factors of g ( x ) in R , [ x ] we have Stab,,[g,(x)] = G. We use the ideas developed above to establish a theorem, due to Van der Waerden, that is often useful for determining Galois groups.

Theorem 9.8. Suppose R is a UFD with field F of fractions, P i s a nonzero prime ideal in R , R = R / P , and F is the field of fractions for R. Suppose f ( x ) is monic of degree n in R [ x ] with f ( x ) E R [ x ] the result of reducing coefficients mod P , and suppose that f ( x ) and f ( x ) have no repeated roots in respective splitting fields K and K over F and F. If G = G, = G ( K : F ) and G = G? = G(K:F), then is (isomorphic with) a subgroup of G.

e

,

Proof. Take R ,Fl, K , and g ( x ) as above, and define R , , F, ,El analog i ( x )in R , [ X I we may reduce all coefficients in gously. Since g ( x )factors as R mod P and obtain &x) = g,(x). Each t E S,, permutes the factors g i ( x )of g ( x ) and hence also the factors gi(x)of ij(x), and G has been characterized as

ni ni

286

VI Further Topics

Stab[g,(x)]. The factors g(x) are not necessarily irreducible in R , [XI, they might factor further. The discussion preceding the theorem, as applied to R, G, etc., shows then that G, as a subgroup of S,, is the stabilizer of an irreducible factor of g,(x). But then each $ E G, viewed as an element of S,, permutes the factors gi(x),and gl(x), can only be gl(x) since $ fixes an irreducible factor of g,(x). Thus $ E G and 5 G.

c

Corollary. Keep the hypotheses of the theorem but take R = Z and P = (p) with p prime in Z,so R = F = Z,.Supposef(x) factors in Z,[x] into irreducible factors f;(x)L(x). . . fm(x),with degf;.(x) = n, 2 1. Then G contains a product a1a 2 . . amof disjoint cycles, ai being an n,-cycle. Proof. In this case the Galois group G is cyclic (Theorem 111.3.11) say G = (a). Since G is transitive on the n, roots in K of the irreducible factor fi(x) for each i it is clear that a has the indicated cycle structure, and a € G I G. The next two results also appear in Van der Waerden [37].

Proposition 9.9. If G I S, is transitive on { 1,2,. . .,n> and G contains a transposition a and an (n - 1)-cyclet,then G = S,. Proof: By relabeling we may assume that t = (2 3 .. . n) and then by transitivity take a = (1 k) with 2 I k I n. Then at-' = (1 j ) where 2 I j I n and j = k + l(mod n - 1). Thus G contains all (1 j ) , 2 Ij In. An easy induction on n shows that these generate S,. Theorem 9.10.

If 0 < n E Z there is some f(x) E Q[x] with G,

= S,.

Proof. In view of various examples above we may assume that n 2 4. Choose fi(x), f2(x), f3(x) E ,?![XI, each monic of degree n, so that fl(x) is irreducible in Z2[x], fi(x) splits in Z,[x] as a product of a linear factor and an irreducible factor of degree n - 1, and f,(x) splits in Z,[x] as a product of an irreducible quadratic factor and either one or two (distinct) irreducible factors of odd degree (see Exercise 111.8.26). Then set

f(x) = - 15fi(x) + 10f2(x) + 6f3(x)E ZCxI and set G = G, IS,. Since T(x) = f;(x) in Z,[x] f(x) is irreducible and G, is transitive on the roots of f(x). Since J(x) = L(x) in Z,[x] G contains an (n - 1)-cycle by the corollary to Theorem 9.8. Since f ( x ) = L(x) in ,?!,[XI C contains a product p of a transposition a and one or two cycles of odd length, also by the corollary. A suitable power of p is a, and C = S, by Proposition 9.9. For an example take n = 6. Then fl(x) = x 6 + x + 1 is irreducible mod 2, f2(x) = x6 + 2x2 + x has irreducible factor x s + 2x + 1mod 3, and f3(x) = x6 + x4 + x 3 + x factors as (x2 - 2)x(x3 + 3x + 2), with x 2 - 2

9 Some GnIois Groups

287

+ 2 irreducible mod 5. Thus f(x) = - 15fI(x) + lOf,(x) + 6 f 3 ( ~=) x6 + 6x4 + 6x3 + 20x2 + x - 15

and x 3 + 3x

,

has Galois group G = s6 over Q. The method of Theorem 9.10 is constructive only insofar as it is possible to exhibit the polynomials fi(x), f2(x),and f 3 ( x ) .Unfortunately there seems to be no general constructive procedure available for producing irreducible polynomials of prescribed degrees over Z,.There are some tables; e.g., see Peterson and Weldon [30] for p = 2. Another standard method for showing that every symmetric group S,, occurs as a Galois group over Q uses Hilbert’s Irreducibility Theorem. See Hadlock [13] for an account. The same methods serve to show that every alternating group A, occurs as a Galois group over Q; a nice discussion appears in Paul Feit’s senior thesis [lo]. Exercise 9.7. Use the method of Theorem 9.10 to find polynomials in Q[x] with S, and S, as Galois groups.

We conclude this section with a final example. If f(x)

= X’ -

821+ ~ 3284 E Z[X],

then f ( x )is irreducible by Eisenstein’s Criterion since 821 is a prime in Z.Thus 5 I \G,.l. An application of Proposition 9.3 yields D, = 8214 . 216 . 3, (verify), so G, IA , by the corollary to Theorem 9.4. In Z7[x] we have f(x) = x 5 - 2x

+ 1 = (x - I)(x - 3)(x3 - 3x2

-

x - 2),

and x 3 - 3x2 - x - 2 is irreducible mod 7. Thus there is a 3-cycle in G by the corollary to Theorem 9.8, so 15 I 1 GI and 1 GI = 1530, or 60. If I GI were 15, then G would be cyclic (see Example 2, p. 21), but consideration of possible cycle types shows that A , has no elements of order 15. A subgroup of order 30 in A , would be normal, contradicting simplicity, and so /GI = 60 and G = A , . Exercise 9.8. Show that the following polynomials have Galois group A , over Q. (a) x J + 2Ox + 16, (b) x 5 + 719x - 2876, (c) x 5 + 971x - 3884. Exercise 9.9 (D. Surowski). Choose k E Z so that 3y k and k E l(mod 7), and set m = 5k2 - 1. Show that f ( x ) = x 5 + 5mx + 4m has Galois group A , over Q. (The key steps are to compute 0,and to reduce coefficients mod 3 and mod 7.)

This Page Intentionally Left Blank

Appendix

1

Zorn’s Lemma

A partial ordering on a set Y is a relation I that is (i) reflexive, (ii) antisymmetric, and (iii) transitive, i.e., (i) A I A , (ii) if A I B and B I A, then A = B, and (iii) if A I B and B I C , then A I C for all A, B, and C in Y .If Y has a partial ordering we say that Y is a partially ordered set. The prototypical example of a partially ordered set is a set Y of subsets of some set S, where A I B denotes set inclusion, i.e., A c B. A linear ordering on a set % is a partial ordering I in which any two elements A , B are comparable, i.e., A I B or B I A . A chain in a partially ordered set Y is a nonempty subset V of Y that is linearly ordered by the ordering it inherits from Y by restriction. An upper bound for a chain V in 9is an element B E Y such that A I B for all A E %?(it is not required that B E V). A maximal element in Y is an element M E Y such that if A E Y and M I A, then A = M. We take the point of view that the following statement, Zorn’s Lemma, is an axiom of set theory. Zorn’s Lemma. If Y is a nonempty partially ordered set in which every chain has an upper bound, then ,Y has a maximal element. There are many apparently different but logically equivalent formulations of Zorn’s Lemma. Perhaps the best known is the Axiom of Choice, which asserts that the Cartesian product of any nonempty family of nonempty sets is not empty. An element of that Cartesian product is called a “choice function”; it can be thought of as simultaneously choosing an element from each of the sets in the family. A discussion of various formulations and their equivalences 289

290

Appendix Zttm’s Lemma

can be found in most set theory books. A brief and lucid account appears in Kelley [21]. Let us look briefly at an important application. Suppose V is a (left)vector space over a division ring D.Recall that a subset X of V is linearly dependent if there are distinct elements u , , u,, . , . ,u, E X and (not necessarily distinct) scalars a,, a,, . . .,a, E D,with not all ai = 0, such that a,u,

+ a , u 2 + ..’ + anon= 0.

Otherwise X is linearly independent. The span of a subset X of V is the subspace of V generated by X , obtained by intersecting all subspaces that contain X as a subset. If X # then the span of X consists of all linear combinations a1u1 u2u2 ... anun,where 1 I n E Z, ui E X , and aj E D.A basis for V is a linearly independent subset X of V such that the span of X is I/.

+

Theorem A.l. basis.

a,

+ +

If V is a vector space over a division ring D,then V has a

Proof. Let 9’ be the set of all linearly independent subsets of V . Then 9’ # @ since @ E Y (if V # 0, then also each { u ) E Y if 0 # u E V ) . Partially

order Y by ordinary set inclusion, i.e., A I B if and only if A L B. If % is a thentherearedistinctu,, ..., U , E B chaininYletB = U { A : A€%}.IfB # 9, and scalars a,, . . .,a, E D,not all0,such that a, u1 + ... anon= 0. But each u, is in some A iE %‘and , W is a chain so one of A . . . ,A , contains all the others. By relabeling we may assume that Ai c A , for all i . But then u , , . , . ,u, E A , , which is linearly independent, and we have a contradiction. Thus B E 9, and B is clearly an upper bound for %. By Zorn’s Lemma we may conclude that there is a maximal element M in Y . Let us show that M is a basis for V . To that end let W be the span of M . If W # Vchoose u E V\ W, and set M , = M { u ) . Then M , is linearly indepenM I dent, since u is not a linear combination of elements of M , so MI E 9, M , , and M # MI, contradicting the maximality of M .

,,

+

u

Theorem A.2. Any two bases for a vector space V over a division ring D have the same cardinality. Proof. If V has a finite basis the result is a standard fact from elementary linear algebra. Suppose then that B , and B, are two infinite bases for V. Each u E B , is (uniquely) a linear combination of a finite subset B,(u) c B , . If any x E 8 , were in none of the sets B,(u), then the span of B,\{x) would include B,, and hence would be all of V . But then B , would be linearly dependent, a contradiction, and hence B, = u { B , ( u ) : u E Bl}. Consequently lB21 IK,IB,I = lB,[, where K O is the first infinite cardinal. Similarly lBll I lB21, and hence lBll = lB21 by the Schroder-Bernstein Theorem.

If B is any basis for a vector space V , then ( B (is called the dimension of V .

1. E. Artin, “Geometric Algebra,” Wiley (Interscience), New York, 1957. 2. E. Artin, “Modern Higher Algebra, Galois Theory,” Courant Institute of Mathematical Sciences Lecture Notes, New York Univ. Press, New York, 1957. 3. E. Artin, “Galois Theory,” 2nd ed., Notre Dame Mathematical Lectures, No. 2, Univ. of Notre Dame Press, Notre Dame, Indiana, 1959. 4. E. Artin, “Algebraic Numbers and Algebraic Functions,” Gordon and Breach, New York, 1967. 5. W. Bishop, How to construct a regular polygon, Amer. Math. Monthly 85, 186-188 (1978). 6. Z. I. Borevich and I. R. Shafarevich, “Number Theory,” Academic Press, New York, 1966. 7. H. S. M. Coxeter, “Introduction to Geometry,” 2nd ed., Wiley, New York, 1969. 8. C. Curtis and I. Reiner, “Representation Theory of Finite Groups and Associative Algebras,” Wiley (Interscience), New York, 1962. 9. N. G. DeBruijn, “Polya’s Theory of Counting,” in “Applied Combinatorial Mathematics” (E. F. Beckenbach, ed.), p. 144. Wiley, New York, 1964. 10. P. Feit, The Hilbert irreducibility theorem and Galois groups over Q, Senior Thesis, Harvard College, Cambridge, Massachusetts, 198I . 11. W. Feit, “Characters of Finite Groups,” Benjamin, New York, 1967. 12. S. Haber and A. Rosenfeld, Groups as unions of proper subgroups, Amer. Math. Monthly 66, 491-494 (1959). 13. C. Hadlock, ”Field Theory and Its Classical Problems,” Carus Math. Monographs, No. 19, Mathematical Association of America, Washington, D.C., 1978. 14. G. H. Hardyand E. M. Wright,”An Introduction to theTheoryofNumbers,”4th ed.,Oxford Univ. Press, London and New York. 1960. IS. I. Herstein, “Noncommutative Rings,” Carus Math. Monographs, No. 15, Mathematical Association of America, Washington, D.C., 1968. 16. N. Jacobson, “Basic Algebra 1,” Freeman, San Francisco, California, 1974. 17. N. Jacobson, “Basic Algebra 11.” Freeman, San Francisco, California, 1980. 18. S. A. Jennings, The structure of the group ring of a p-group over a modular field, Trans. Amer. Math. SOC.50, 175-185 (1941). 19. E . Kamke, “The Theory of Sets,” Dover, New York, 1950. 20. 1. Kaplansky, “Fields and Rings,” 2nd ed. Univ. of Chicago Press, Chicago, Illinois, 1972.

29 I

292 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

40.

41.

References J. Kelley, “General Topology,” Van Nostrand-Reinhold, Princeton, New Jersey, 1955. A. G. Kurosh, “The Theory of Groups,” Vol. 2, Chelsea, Bronx, New York, 1960. S. Lang, “Algebra,” Addison-Wesley, Reading, Massachusetts, 1965. C. E. Linderholm, “Mathematics Made Difficult,” World Publ., New York, 1972. S. Lipka, Uber die Irreduzibilitat von Polynomen, Math. Ann. 118,235-245 (1941). J. H. McKay, Another proof of Cauchy’s group theorem, Amer. Math. Monthly 66, 119 (1959). N. C. Meyer, Jr., A new proof of the existence of the free group, Amer. Math. Monthly 77, 870-873 (1970). I. Niven, The transcendence of A, Amer. Math. Monthly 46,469-471 (1939). R. S. Palais, The classification of real division algebras, Amer. Math. Monthly 75, 366-368 (1968). W. Peterson and E. Weldon, Jr., “Error Correcting Codes,” 2nd ed., MIT Press, Cambridge, Massachusetts, 1972. H. Pollard and H. G. Diamond, “The Theory of Algebraic Numbers,” 2nd ed., Carus Math. Monographs, No. 9, Mathematical Association of America, Washington, D.C., 1978. M. A. Rieffel, A general Wedderburn theofem, Proc. N a t . Acad. Sci. U.S.A.54, 1513 (1965). E. Sasiada and P. M. Cohn, An example of a simple radical ring, J. Algebra 5, 373-377 (1 967). I. R. Shafarevich, Constructions of fields of algebraic numbers with given solvable Galois group, Izo. Akad. Nauk SSSR Ser. M a t . 18,515-578 (1954). E. Spitznagel, Jr., Note on the alternating group, Amer. Math. Monthly 75,68-69 (1968). H. Stark, A complete determination of the complex quadratic fields of class-number one, Michigan Math. J . 14, 1-27 (1967). B. L. Van der Waerden, “Modern Algebra,” Vol. 1,2nd ed., Ungar, New York, 1953. R. Walker, “Algebraic Curves,” Dover, New York, 1950. J. C. Wilson, A principal ideal ring that is not a Euclidean ring, Math. Mag. 46,34-38 (1973). E. Witt, Uber die Kommutativitat Endlicher Schiefkorper, Hamb. Abh. 8,413 (1930). 0. Zariski and P. Samuel, “Commutative Algebra,” Vol. 1, Van Nostrand-Reinhold, Princeton, New Jersey, 1958.

I

Index

A

B

Abel, N. H., 99, 103 Abelian, 2 Absolute value, 239 ACC, 66, 128 Adverse, 177 Algebra, 163 Algebraic, 57, 80, 8 I Algebraically closed, 83 dependent, 60,230 independent, 60,23 1 Algebraic closure, 83, 231 Algebraic integer, 224, 27 I Algebraic number, 106 Algebraic number field, 224 Algebraic variety, 73 Alternating group, 17 Angle trisection, 1 1 I Annihilator, 134, 173 Archimedean valuation, 242 Artin, E., 121, 184, 185,252 Artinian ring, 183 Artin- Whaples Approximation Theorem, 252 Ascending central series, 29 Ascending chain condition, 66, 128 Associate, 6 I Automorphism, 8 Axiom of Choice, 289

Balanced map, 158 Basis, 132, 288 Bijection, 2 Bimodule, 160 Binomial Theorem, 75 Bishop, W., I14 Block, 2 14 Burnside, W., 274 Burnside’s Orbit Formula, 200, 268

C Canonical quotient map, 10 Canonical representation, 249 Cardano’s Formulas, 279 Castelnuovo, G., 236 Cauchy sequence, 244 Cauchy’s Theorem, 15 Cayley - Hamilton Theorem, 143 Cayley’s Theorem, 13 Center, 10 Centralizer, 13, 173, 187 Chain, 288 Change-of-basis matrix, 143 Character, 200,265 Characteristic, 80 Characteristic polynomial, 140 Character table, 270

293

294

Index

Chevalley, C., I82 Chinese Remainder Theorem, 7 1 Chronicles 11, 114 Cissoid of Diocles, 124 Class equation, 14 Class number, 228 Closed, 86 Closure, 86 Cohn, P., 180 Common divisor, 62 Commutative, 47 Commutator, 23 Companion matrix, 1 16, 141 Complete, 244 Completely reducible, 174 Completion, 244 Composition series, 25 Congruent, 8 Conjugacy class, 13 Conjugate, 13, 84 Constituent, 267 Constructible, 109 Content, 68,208 Contragredient representation, 266 Convolution, 172 Coset, 8 Coxeter, H. S. M., 2 Cycle, 16 index, 20 1 type, 18 Cyclic group, 7 module, 126 vector, 141 Cyclotomic polynomial, 97

D DCC, 183 Dedekind, R., 63 Dedekind domain, 225 Degree of field extension, 79 of polynomial, 55,59 of representation, 260 del Ferro, S., 99, 279 Delian problem, 112, 123, 124 Dense, 182 Density Theorem, 182 Derivative, 9 1

Derived group, 23 series, 24 Descending central series, 30 Descending chain condition, 183 Dihedral group, 39 Dimension, 289 Diophantive equation, 238 Direct factor, 28 product, 28 sum, 32,70, 129,26 1 Discrete valuation, 247 Discriminant, 275 Divide, 6 I , 222 Divisible group, 170, 196 Division Algorithm, 56 Division ring, 49 Double centralizer, 190 coset, 42 dual, 165 Doubly transitive, 42 Dual basis, 22 1 module, 165 Duplication ofthe cube, 112

E Eisenstein Criterion, 70, 77 Elementary divisors, 144 Elementary row and column operations, I47 Elementary symmetric polynomials, 103 Endomorphism, 8 Endomorphism ring, 48 Epimorphism, 8, 49 Equivalent matrices, 147 representations, 26 1 valuations, 239 Erchinger, 114 Euclidean, 64 Euclidean Algorithm, 77 Euclid's Elements, 114 Euler, L., I14 Even permutation, 16 Exponent, 135 Extension field, 79

Index group, 26 by radicals, 99 External semidirect product, 46

F Factor, 24 Factor group, 10 Factor Theorem, 57 Faithful, 12, 173, 260, 265 F-automorphism, 86 Feit, P., 287 Fermat number, 114 prime, I 14 Ferrari, L., 99 Field, 49 Field of fractions, 53 Finite extension, 79 Finitely generated module, 126 First Orthogonality Relation, 266 Fixed field, 86 Formally real field, 254 Fractional ideal, 222 Free abelian group, 132 generators, 36 group, 35 module, 131 semigroup, 34 Freshman Theorem for groups, 12 for modules, 127 for rings, 5 1 Frobenius, G., 32, 188 Frobenius map, 95 Fundamental Homomorphism Theorem for groups, 1 1 for modules, 127 for rings, 50 Fundamental Theorem of Algebra, 95 of Finite Abelian Groups, 33 of Galois Theory, 90 of Symmetric Polynomials, 105

G Galois closure, 93

extension, 86 field, 96 group, 86 group of a polynomial, 96 Galois, E., 99, 102 Gauss, C. F., I14 Gaussian integers, 64 Gauss’s Lemma, 68 Gay, D. A., 233 GCD, 62 Generalized quaternion group, 45 General linear group, 2 10 General polynomial, 102 Generate ideal, 50 subgroup, 7 submodule, I26 Generating function, 208 Generators and relations, 38 Greatest common divisor, 62, 228 Group, 1 Group algebra, I72

H Haber, S.,42 Hamilton, W. R., 48 Hasse, H., 63 Hasse Principle, 238 Hensel, K., 238 Hermes, J., 114 Hilbert, D., I18 Hilbert Basis Theorem, 74 Hilbert’s Satz 90, I 18, 124 Homogeneous, 59 Homomorphism group, 8 module, 127 ring, 49 Hyperplane, 2 12

I Ideal, 50 Ideal class group, 228 Idempotent, 168 Identity element, 1 Indeterminate, 5 5 Injective, 170, 196 Inner automorphism, 45 Inner product, 266

295

296

Index

Integers mod n, 1 I Integral basis, 222 closure, 2 19 domain, 49 element, 2 I7 extension, 2 17 ideal, 222 Integrally closed, 219 Intermediate Value Theorem, 256 Internal direct product, 29 direct sum, 130 Invariant factors, 138 Invariant Factor Theorem (3.7), 137 Inventory, 204 Inverse, 1 Invertible ideal, 222 Irreducible character, 265 module, 173 representation, 26 1 ring element, 6 I Isomorphism, 8,49 Isomorphism Theorem for groups, 14 for modules, 128 for rings, 51 Iwasawa, K., 2 15

J Jacobson, N., 182 Jacobson radical, 176 Jennings, S., 192 Join, 99 Jordan block, matrix, I44 Jordan Canonical Form, 145 Jordan-Holder Theorem, 25, 128

K Kernel, 8, 270 KingsI, 114 Wein’s 4-group, 5 Kronecker product, 163 Kummer, E., 228 Kurosh, A., 37

L Lagrange interpolation, 76 resolvent, 101 Lagrange’s Theorem, 8 LCM, 228 Least common multiple, 228 Left adverse, 177 coset, 8 ideal, 50 Noetherian ring, 13I quasi-regular, 177 regular representation, 13, 26 1 R-module, 125 zero divisor, 49 Leibniz’s Rule, 120 Length, 24 Linear character, 265 dependence, 288 fractional transformation, 2 1 1 functional, 165 independence, 132,289 ordering, 288 Localization, 54 Luroth’s Theorem, 235

M

Maschke, H., 176, 184,262 Matrix representation, 260 Maximal condition, 73 element, 288 ideal, 5 I McKay, J., 15 Minimal condition, 183 ideal, 185 polynomial, 80, 140 Modular, 175 Module, 125 Monic, 55 Monoid, 1 Monomial, 59, 201 Monomorphism, 8,49 Multiple, 6 1

Index N Negative, 253 Newton’s Identities, 276 Nil, 178 Nilpotent, 168, 178, 179 Nilpotent class, 46 Nilpotent group, 29 Nilpotent ring element, 76 Noetherian module, 128 ring, 66 Nonarchimedean valuation, 242 Norm, 115 Normal closure, 93 field extension, 93 series, 24 subgroup, 10 Normalizer, 14

0 Odd permutation, 17 Orbit, 13 Order, 6, 7, I34 Order ideal, 134 Ordered commutative ring, 252

P Palais, R., 188 padic integers, 238, 249 numbers, 238, 249 valuation, 239 Partial ordering, 289 Partition, 44 Pattern, 204 Pattern counting function, 208 Pattern inventory, 205 Perfect, 120 Permutation, 2 Permutation groups, 12 Permutation representation, 26 1 PgrOUP, 20 PID, 62 Polya, G., 206 Polynomial, 55, 59 Polynomial function, 59

Positive, 253 pprimary, 139 Presentation, 38 Primary, 139, 194 Prime, 6 1 Prime field, 80 Prime ideal, 52 Primitive element, 79 ideal, I8 1 permutation group, 2 I4 polynomial, 68 ring. 181 roots of unity, 97 Principal character, 265 fractional ideal, 222 ideal, 50 ideal domain, 62 Product, 27 Product of ideals, 7 1 Projective abelian group, 195 general linear group, 2 10 module, 169 point, 210 space, 2 I0 special linear group, 2 10 Pure, 165 Purely inseparable, 120 transcendental. 23 I

0 Quasi-regular, 177 Quaternion group, 5 Quaternions, ring of, 48 Quotient, 34, 56 Quotient group, 10 Quotient of ideal, 177,222 Quotient map, 10, 34 Quotient ring, 50

R Radical of an ideal, 190 ring, 177

297

298

Index

Radical, Jacobson, 176 Raffer, S., 1 19 Rank of free abelian group, 195 of free group, 37 of free module, 133 Rank I valuation, 239 Rational canonical form, matrix, 143 Rational functions, 57, 6 I Real closed field, 254 Real valuation, 239 Redfield- P6lya Theorem, 206 Reduced, 198 Reducible character, 265 representation, 261 Reduction of coefficients, 68 Relations, 38 Relative unit, 175 Remainder, 56 Remainder Theorem, 57 Representation, 260 Residue class field, 246 Residue class ring, 50 Resolvent cubic, 281 Richelot, F., 114 Richmond, H., 124 Rieffel, M., 190 Right adverse, 177 coset, 10 ideal, 50 Noetherian ring, 131 quasi-regular, 177 R-module, 125 zero divisor, 49 Ring, 47 Ring of integers, 246 R-linearly independent, 132 R-module, 125 Root, 82 Roots of unity, 97 Rosenfeld, A,, 42 Ruffini, P., 99, 103 S

Sasiada, E., I80 Schur, I., 263 Schur’s Lemma, 173,263

Schwendenwein, 114 Secondary row and column operations, 148 Second Orthogonality Relation, 269 Semidirect product, 46 Semigroup, I Semisimple, 174, 177 Separable element, 92 extension, 92 polynomial, 90, 9 1 Short exact sequence, 44, 164 Similar matrices, 143 Simple extension, 79 group, 18 module, 173 radical extension, 97 ring, 50, 180 Skew field, 49 Smith normal form, I50 Solvable, 24 Solvable by radicals, 99 Span, 288 Special linear group, 2 10 Spiral of Archimedes, 1 14, 123 Spitznagel, E., 17 Split, 82, 170, 199 Split extension, 46 Splitting field, 83 Sporadic group, 27 Square free, 62 Squaring the circle, I14 Stabilizer, 12 Stable, 88 Standard sequence, 256 Stickelberger, L., 32 Sturm, 258 Sturm sequence, 257 Subfield, 79 Subgroup, 6 Submodule, 126 Subnormal, 24 Subring, 47 Substitution, 56,60 Surowski, D., 287 Sylow subgroup, 19 Symmetric group, 3 polynomial, 104 rational function, 103

Index T Tartaglia, N., 99, 279 Tensor product, I58 Terminate, 66, 128, 183 Torsion element, 134 module, 134 subgroup, 135, 194 submodule, I35 Torsion-free, I34 Trace, 1 15 Transcendence basis, 23 1 degree, 233 set, 230 Transcendental, 57, 80, 106 Transcendental extension, 23 I Transitive, I3 Transposition, 16 Transvection, 2 I2 Triangle inequality, 240 Trivial representation, 260 valuation, 239 Tychonov, A,. 250

U Unique factorization domain (UFD), 66

Unit, 49 Unitary module, 126 Universality, 22 Upper bound, 289

V Valuation, 239 Valuation ring, 246 Value group, 247 Vandermonde, 275 Van der Waerden, B. L., 285 Variation, 251

W Wedderburn, J., 63 Weight, 204 Weight assignment, 204 Whaples, G., 252 Wilson, J., 63 Wilson’s Theorem, 121 Witt, E., 187

Zarislu, O., 236 Zero divisor, 49 Zorn’s Lemma, 289

299

Pure and Applied Mathematics A Series of Monographs and Textbooks Editors

Samuel Eilenberg and Hyman Bass Columbia University, New York

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