Foundations of Galois Theory (Dover Books on Mathematics)

FOUNDATIONS OF

GALOIS THEORY M. M. POSTNIKOV TRANSLATED BY

ANN SWINFEN

DOVER PUBLICATIONS, INC. MINEOLA. NEW YORK

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Bibliographical Note This Dover edition, first published in 2004, is an unabridged and unaltered

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republication of the first English edition originally published by Pergamon Press,

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London, and The Macmillan Company, New York, in 1962. It is the English translation of the work first published in Russian by Fizmatgiz, Moscow, in 1960. b.0

Library of Congress Cataloging-in-Publication Data .t:

Postnikov, M. M. (Mikhail Mikhailovich) [Osnovy teorii Galua. English] Foundations of Galois theory / M.M. Postnikov ; translated by Ann Swinfen.

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p. cm. Originally published: New York : Pergamon Press, 1962. Includes bibliographical references and index. ISBN 0-486-43518-0 (pbk.) 1. Galois theory. I. Title. own

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QA211.P613 2004

512'.3-dc22 2003067440

Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

CONTENTS PAGE

Foreword Preface

vii viii (7r

1. THE ELEMENTS OF GALOIS THEORY 1. THE ELEMENTS OF FIELD THEORY

I 1

2

2. NECESSARY FACTS FROM THE THEORY OF GROUPS O0-

1. The definition of a group 2. Subgroups, normal divisors and factor groups 3. Homomorphic mappings 3. GALOIS THEORY cwt

1. Normal extensions 2. Automorphisms of fields. The Galois group 3. The order of the Galois group 4. The Galois correspondence 5. A theorem about conjugate elements 6. The Galois group of a normal subfield 7. The Galois group of the composition of two fields

4 6 7 9 12 14 14

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C3.

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1. Preliminary remarks 2. Some important types of extensions 3. The minimal polynomial. The structure of simple algebraic extensions 4. The algebraic nature of finite extensions 5. The structure of composite algebraic extensions 6. Composite finite extensions 7. The theorem that a composite algebraic extension is simple 8. The field,of algebraic numbers 9. The composition of fields

16 16

18 21

25 25 28 31

35 38 39 41

II. THE SOLUTION OF EQUATIONS BY RADICALS 1. ADDITIONAL FACTS FROM THE GENERAL THEORY OF GROUPS

1. A generalization of the homomorphism theorem 2. Normal series 3. Cyclic groups 4. Solvable and Abelian groups 2. EQUATIONS SOLVABLE BY RADICALS

1. Simple radical extensions 2. Cyclic extensions 3. Radical extensions

45 45 46 49 52 58 58

60 65 V

CONTENTS PAGE

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71

73 73 75 79 81 0000

1. The Galois group of an equation as a group of permutations 2. The factorization of permutations into the product of cycles 3. Even permutations. The alternating group 4. The structure of the alternating and symmetric groups 5. An example of an equation with Galois group the symmetric group 6. A discussion of the results obtained

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3. THE CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

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4. Normal fields with solvable Galois group 5. Equations solvable by radicals

85 89

4. THE UNSOLVABILITY BY RADICALS OF THE GENERAL EQUATION OF

DEGREE n 3 5 bin

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1. The field of formal power series 2. The field of fractional power series 3. The Galois group of the general equation of degree n 4. The solution of equations of low degree

92 92 97 101 105

FOREWORD THE GALOIS theory of equations is simultaneously the source of modern abstract algebra and one of the most concrete applications of abstract algebra. For the problem of solving equations becomes familiar at a very early stage of a school course in mathematics; and the question, answered in the negative by Galois, whether all polynomial equations with integer coefficients can be solved by processes

involving only the ordinary operations of arithmetic together with the extraction of n`b roots must seem a natural and unavoidable one to anyone well versed in school algebra. Thus Galois theory constitutes an excellent introduction to group theory and provides a strong (and historically significant) motivation .r.

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for the introduction of the basic concepts of abstract algebra. The present work, by a leading Russian mathematician, sets itself the limited aim of explaining the basic theory and describing how it is "a)

applied to prove the unsolvability `by radicals' of equations of degree n > 5. On the other hand the treatment is almost entirely

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vii vii

self-contained (the necessary group theory, in particular, is given in its entirety), and it is therefore hoped that this book will commend itself to university students at the stage at which they first begin to grapple with modern algebra.

P. J. HILTON

Birmingham

Vii

PREFACE Txis BooK is intended in the first instance for students in their second "C,

or third year at university, who are starting to learn Galois theory. CAD

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For this reason the reader is assumed to possess only the fundamentals of higher algebra to the extent of the programme of a first year course

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at university. On the other hand, the theoretical material presented in the book is not accompanied by examples, as it is assumed that these will be provided in the course of lectures or seminars. The problems included in the text are of a quite trivial character and are intended solely for the self-discipline of the reader. It should be noted that the order of presentation adopted in this book differs from the order in which Galois theory should be presented in lectures (for instance, groups of permutations should appear considerably earlier in a course of lectures). Galois theory is presented in this book for fields contained in some unique `universal', algebraically closed field of characteristic 0 (more specifically, the field of complex numbers). This makes it ram

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possible to avoid the difficulty of founding an abstract theorem on the existence and uniqueness (to within isomorphism) of the decomposi-

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tion field of a given polynomial. On the other hand, no actual loss of generality occurs in such a method of presentation, because, as is well known, any field can be embedded in an algebraically closed field. ,-r

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Another, less essential, peculiarity of the method of presentation adopted in this book lies in the fact that we carefully avoid the use of Vii

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the theorem on the extension of an isomorphism, replacing it by possibly less elegant, but certainly more accessible results in the theory of symmetric functions. Further, we investigate, more CAD CAD

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pedantically than is usually done, the relations between the different definitions of a finite extension, and we base the presentation of the group-theoretical material on the concept of a homomorphism (we note in passing that for homomorphisms `onto' and isomorphisms

`into' we use special terms, which have recently appeared in mathematical literature and are rapidly superseding the customary terminology). Viii

PREFACE

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Because the theory of groups plays only a subsidiary part in Galois theory, it is presented only in so far as it is necessary for the development of general Galois theory and its application to the problem of '.7

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solving equations by radicals. For instance, although we present the concepts of a normal series and its refinement, yet there are no theorems in this book like Schreier's theorem or the Jordan-Holder theorem. 'L3

In presenting the theory of permutations, the theorem on the factorization of permutations into the product of independent cycles is proved in detail, and the concept of an even permutation emerges

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from the consideration of the factorization of a permutation into the product of transpositions. Although we do not insist on the superiority of this method of introducing the concept of an even permutation (as compared with the standard method, based on the consideration of inversion in permutations), we nevertheless consider ""'

that this method is worth attention. We prove the simplicity of the ,'y

alternating group as it was proved by Redei in a recent paper. Redei's proof, as given by us, is simpler than the usual proof of Bauer. L..

In considering the solution of equations by radicals, we restrict ourselves to the problem of the solution of equations by arbitrary (possibly reducible) radicals. Thus the cyclotomic equations are solvable by radicals by definition, and this, of course, essentially simplifies the theory. Although with such an approach to the soluran

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tion of equations by radicals the results obtained cannot, for instance, be applied to the problem of constructing regular polygons with the ear

help of ruler and compasses (as the theory of Gaussian periods fn'

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remains entirely outside the framework of our exposition), nevertheless the simplification of the theory achieved in this way is so con-

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siderable that, for a preliminary introduction to the fundamental ideas on which the application of Galois theory to the problem of solution by radicals is based, the consideration of only irreducible radicals is not appropriate.

In the last chapter of the book general equations (i.e. those (7)

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having literal coefficients) are considered. Because the field of coefficients of these equations is the field of rational functions, if we continue to treat the question from the point of view indicated above, we are obliged in particular to prove that this field can be embedded in an algebraically closed field (namely, in the field of fractional power .'t.

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series). The fact that the field of fractional powers is algebraically closed has been proved by Ostrowsky with the help of Hensel's

PREFACE

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lemma. This proof, although not effective, is considerably simpler than the constructive proof, based on Newton's polygon and frequently reproduced in Russian. As a reference book for the material of a preliminary course we use A. G. Kurosh's book A Course in Higher Algebra, which in the text is called simply `The Course'. The page numbers given here are those of the fourth and fifth editions. The author takes this opportunity to thank V. G. Boltyanskii and D. K. Faddeyev, who read this book in manuscript and made many valuable suggestions. THE AUTHOR

I. THE ELEMENTS OF GALOIS THEORY

CHAPTER 1

THE ELEMENTS OF FIELD THEORY 1. Preliminary remarks

We define a field as a non-empty set P of complex numbers, possessing the following properties :

(1) if aaPandbeP, then a + bePandabaP; (2) if a e P, then - a e P and a -1 e P (when a 0). Examples of fields are: the field of rational numbers R, the field of real numbers D and the field of complex numbers C. The field P is called a subfield of the field K, and the field K an extension of the field P, if every element of the field P belongs to the

field K, i.e. ift P c K. Any field (in our sense) is a subfield of the field of complex numbers.

It is easy to see that every field contains unity, and hence also

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the whole field of rational numbers R, i.e. any field is an extension of the field of rational numbers. In modern algebra one takes the abstract definition of a field as a set with two algebraic operations satisfying definite axioms (see The Course, p. 28). As opposed to such " abstract " fields, a field in our sense is called a number field. The theory expounded in this book can without difficulty also be extended to the case of arbitrary fields. The transition from number fields to arbitrary fields involves, fundamentally, only purely technical difficulties. These difficulties are

connected with the fact that in an arbitrary field some multiple of

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unity can be equal to zero, and an irreducible polynomial can possess multiple roots. Fields in which this difficulty does not arise are called fields of characteristic 0 (see The Course, pp. 32 and 213). Besides the number fields, the field of rational functions, for instance, belongs to this class. Another more essential difficulty, arising in the transi-

tion from number to arbitrary fields, lies, in particular, in the fact that different fields, generally speaking, are in no way connected

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t The notation P e K does not exclude the case when P coincides with K. $ [Editor's footnote: This does not follow from (1) and (2); one needs to postulate that P contains at least two elements.] 1

2

FOUNDATIONS OF GALOIS THEORY

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with one another: for instance, it is impossible to speak of the sum of elements of two different fields. It is more convenient from every point of view to overcome this difficulty by restricting the class of fields considered to subfields of some sufficiently wide `universal' field. It is indeed in this way, by choosing as the universal field the field of complex numbers, that we again arrive at the number fields. In the general case it is sufficient to require that the universal field be algebraically closed, i.e. to require that any polynomial over this field is decomposable in it into linear factors. It is easy to verify that all the theory expounded below remains valid without any alteration, if by a field one understands a subfield of some fixed, but otherwise arbitrary algebraically closed field of characteristic 0. cow

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2. Some important types of extensions

An extension K of a field P is called finite, if in the field K there exist elements ai, ... , such that any element ft e K can be written ...

in a unique way as a linear combination of these elements with coefficients from the field P:

b1,...,bnnP.

..fl

fl =bla1+ ...

A system of elements al, ... , a possessing this property is called a basis of the field K over the field P.

One can also approach the concept of a finite extension from another angle, by noting that any extension K of a field P can be L."

considered as a linear space over the field P. In fact, the elements of the field K can be added together and multiplied by elements of the per,

field P, and the two operations (addition, and multiplication by

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elements of the field P) clearly possess all the necessary properties. From this point of view, the extension K is finite if and only if it has finite dimension (as a linear space over the field P), and a system of elements is a basis for it (in the sense defined) if and only if it is a basis for it in the sense of the theory of linear spaces. Because all bases of a finite-dimensional linear space have the same number of vectors, then, in particular, all bases of the field K over the field P have the same number of elements. This number is called the degree of the field K over the field P and is denoted by [K: P] (from the point of view of the theory of linear spaces, the degree of the field K is its dimension as a linear space over the field P). !1,

3

ELEMENTS OF FIELD THEORY

Problem. To prove that the degree [K: P] is equal to unity if and only if K = P. Let P be an arbitrary (number) field and al, ... , a arbitrary numbers (i.e. elements of the field C). We consider all possible fields which are extensions of the field P and contain the numbers al, ... , Such fields exist, since, for example, we may take the field C of all complex numbers. It is easy to see that the intersection of all these fields is also a field (in general, it is proved without difficulty that the intersection of any family of fields is itself a field). This intersection is,

clearly, the minimal extension of the field P containing the

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numbers al, ... , a (the fact that it is minimal means that this intersection is a subfield of any other extension of P containing the numbers al, ... , a.). This minimal extension is denoted by P(al, ... , a,J and is called the extension generated by the numbers al, ... , It is clear that P(al, ... , P if and only if al, ... , a e P. Problem. To prove that the field P(al, ... , can be defined as '-'

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the set of all numbers obtained as the result of applying to the numbers of the field P and the numbers al, ... , a all possible

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combinations of the four arithmetical operations. The number a is called algebraic over the field P, if it is the root of some polynomial with coefficients from the field P (not identically equal to zero). Any element of the field P, clearly, is algebraic over this field (if the converse is also true, i.e. if any number algebraic over the field P belongs to this field, then P is called an algebraically closed field; see para. 1). It is clear, moreover, that any number L."

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algebraic over the field P is also algebraic over any extension of the field P. We emphasize that the converse assertion, generally speaking, is not true. For instance, any complex number is algebraic over the field D of real numbers (because it is the root of a quadratic trinomial with real coefficients), whilst there exist numbers (indeed,

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real numbers), not algebraic over the field R of rational numbers. As an example of numbers non-algebraic over the field R one can mention the well-known numbers e and n, whose nonalgebraic nature is proved in complete courses on the theory of numbers. An extension K of the field P is called algebraically generated if it 0

is generated by some finite system of numbers algebraic over the field P, i.e. if there exist numbers al, ... , aS algebraic over the field P such that K = P(al, ... , c ). If, in particular, s = 1, then the field K = P(al) is called a simple algebraic extension of the field P.

4

FOUNDATIONS OF GALOIS THEORY cep

An extension K of the field P is called a composite algebraic extension if there exists a chain of subfields

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P=LocL1 c ... cLs_1 cLs=K,

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beginning with the field P and ending with the field K, such that for any i = 1, ... , s the field L, is a simple algebraic extension of the field Li_1. If Li = Li_1(aj), i = 1, ... , s, then the field Kis denoted by P(a1)(a2) ... (as). We emphasize that in this definition the algebraic nature of the numbers aZ, ... , as over the field P is not assumed. Finally, an extension K of the field P is called algebraic if each of its elements is algebraic over the field P. Thus we have introduced the following five types of extensions: (1) finite extensions; (2) algebraically generated extensions; (3) composite algebraic extensions; (4) simple algebraic extensions; (5) algebraic extensions. In this chapter we will study the relations holding between these types of extensions, and also the structure of the extensions of each of these types (except, however, the last type).

3. The minimal polynomial.

The structure of simple algebraic

extensions.

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Let P be an arbitrary field and a a number algebraic over the field P. By definition, the number a is the root of some polynomial over the field P. The polynomial f (x) having the least degree of all the polynomials with this property is called the minimal polynomial of the algebraic number a. This polynomial is irreducible, because otherwise the number a would be the root of at least one of its factors of smaller degree, which is impossible by hypothesis. Any polynomial, whose root is the number a, is not relatively prime to the minimal polynomial f (x) and, consequently, is divisible by this polynomial. In particular, an irreducible polynomial with root a can differ from the minimal polynomial only by a constant factor. In other words, an irreducible polynomial with root a is defined uniquely (to within a constant factor). The degree n of this polynomial is called the degree of the algebraic number a over the field P. The degree n is equal to unity if and only if a e P. Let a be a number algebraic over the field P, f (x) its minimal

ELEMENTS OF FIELD THEORY

5

polynomial and n its degree. We consider the set K of all numbers fl for which there exists a polynomial g(x) over the field P such that

fl = g(a). It is obvious that

K c P(a). We will prove that K is a field. Because the sum, difference and pro-

duct of elements of K clearly belong to K, it is only necessary to prove that for any number /3 e K distinct from zero the number fl-' also belongs to K. By definition, 9 = g(a),

where g(x) is some polynomial over the field P. Since g(a) # 0, the polynomial g(x) is not divisible by f (x) and, consequently (by virtue of the irreducibility of the polynomial f (x)), the polynomials g(x) and f (x) are relatively prime. Therefore, according to a well-known theorem (see The Course, p. 197), there exist polynomials u(x) and v(x) over the field P such that

f(x)u(x) + g(x)v(x) = 1.

Putting x = a in this equation, we obtain: /iv(a) = 1,

i.e. l-1 = v(a), so that /3-' e K. Thus the set K is in fact a field. Because, by definition, P e K °CD

and a e K, K is an extension of the field P, generated by the number a. Therefore by virtue of the minimal nature of the field P(a):

P(a) c K. Comparing this inclusion with the inclusion K e P(a), we see that

K = P(a). Thus we have proved that for any element l3 of the field P(a) there can be found a polynomial g(x) over the field P such that I = g(a). This polynomial is not defined uniquely, because one can add to it any polynomial divisible by the polynomial f (x). In other words, if the difference g(x) - g, (x) is divisible by the polynomial f (x), then g(a) = gl(a). Conversely, if g(a) = gl(a), then the polynomials g(x) - gl(x) and f (x) are not relatively prime (because they have

6

FOUNDATIONS OF GALOIS THEORY

the common root a) and, hence, the polynomial g(x) - g1(x) is divisible by the polynomial f (x). Thus g(a) = gi(a)

if and only if the difference g(x) - gl(x) is divisible by the polynomial .f W.

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In particular, if r(x) is the remainder after division of the polynomial g(x) by the polynomial f (x), then g(a) = r(a). Hence, any element of the field P(a) can be written in the form r(a), where the degree of the polynomial r(x) is smaller than n (i.e. smaller than the degree of the polynomial f (x)). In other words, for any element /3 e P(a) there exist elements bo, b1, ... , bn_ e P (coefficients of the polynomial r(x)) such that

(1) /3 = bo + bta + ... + bn_lan-1. Because the difference r(x) - r1(x), where r(x) and r1(x) are polynomials of degree less than n, is divisible by the polynomial f (x) of

degree n if and only if r(x) = r1(x), then this representation is unique. Thus any element /3 of the field P(a) can be written uniquely in the form (1). In other words, the elements

1,a,...,a"-1 generate a basis of the field P(a) over the field P. Hence, the simple algebraic extension P(a) is a finite extension and its degree [P(a) : P] is equal to the degree of the number a. In other words, the class of exten-

sions of type (4) is contained in the class of extensions of type (1). 4. The algebraic nature of finite extensions

Let /3 be an arbitrary element of a finite extension K of a field P and let [K: P] = n. Because in an n-dimensional linear space any n + 1 vectors are linearly dependent, then, in particular, the elements 1,p, ... 'fin

are linearly dependent over the field P, i.e. in P there exist numbers co, c1, ... , cn, at least one of which is not equal to zero, such that CO+C1/3+ ... +Cn/3"=0.

This means that the number # must be a root of the polynomial

CO+Clx+ ... +Cnx"

ELEMENTS OF FIELD THEORY

7

and, hence, is an algebraic number (over the field P). Thus it has been proved that any finite extension is algebraic, i.e. the class of extensions of type (1) is contained in the class of extensions of type (5).

Moreover, we see that the degree (over P) of any element of a finite extension K of the field P does not exceed the degree n of this extension. Now let al, ... , a be a basis of the field K over the field P. Because

the numbers al, ... , a are, by what has been proved, algebraic (over P), then the extension P(al, ... , generated by them is an algebraically generated extension. By virtue of the minimal nature of this extension it is contained in the field K: K.

On the other hand, because from al, ... , a e P(al, ... , it follows for any numbers b1, ... b e P, E P(al, ... , then any element of the field K is contained in the field P(al, ... ,

that blal + ... + i.e.

Hence,

K = P(al, ...

,

Thus it has been proved that any finite extension is algebraically generated, i.e. the class of extensions of type (1) is contained in the class of extensions of type (2). 5. The structure of composite algebraic extensions

a°.

Let K = P(al)(a2) ... (aS) be a composite algebraic extension of the field P. It will be shown that any element of the field K can be expressed in the form of a polynomial (over P) in al, a2, ... , aS, i.e. that for any element fi e K there exists a polynomial g(xl, ... , xS) over the field P (in s variables) such that

P=g(a1,...,a"). We will prove this assertion by induction with respect to s. If s = 1, then K = P(al), and, hence, in this case the theorem is valid (see para. 3). Now supposing that the theorem is already proved for

the field L = P(al) ...

we consider an arbitrary element

8

FOUNDATIONS OF GALOIS THEORY

/3 E K. Because K = L(as), then over the field L there exists a polynomial h(x) such that /3 = h(as). Let h(x) = Yo + Y1x1 + ... + ynxn, where yo, Y1,

,

yneL.

By the principle of induction, for any i = 0, 1, ... , n there can be found a polynomial hi(x1, ... , xs_1) (in s - 1 variables) such that Yi = hi(a1, ... , as-1)

Hence, letting g(xl, ... , xs) = h0(x1, ... , xS-1) + h1(x1, ... , xs-1)xS + ... n

we obtain fl

=g(z1,...,as)

Thus our assertion is completely proved. Now we consider an arbitrary algebraically generated extension P(a1, ... , as) of the field P and we define by induction the fields Lo, L1, ... , Ls, letting

Lo=P,L1 =L0(al),...,Li=Li-1(ai),...,Ls= L.,-1(as) Because for any i = 1, ... , s the number ai, algebraic over the mar'

field P, is also algebraic over its extension L1_1, then the field Li is a simple algebraic extension of the field Li-1 and, hence, the field LS is a composite algebraic extension P(a1)(a2) ... (as) of the field P. Therefore, simply by the assertion proved, any element of the field LS can be expressed in the form of a polynomial (over P) in a1, ... , as and, hence, belongs to the field P(a1, ... , as). In other words,

LSCP(a1,...,as). On the other hand, the field LS contains all the numbers a1, ... , as and, by virtue of the minimal nature of the extension P(a1, ... , as) :

P(al, ... , as) c L. Hence (because LS = P(a1)(a2) ... (as)), P(a1, ... , as) = P(a1)(a2) ... (as).

Thus, any algebraically generated extension is a composite algebraic extension, i.e. the class of extensions of type (2) is contained in the class of extensions of type (3).

9

ELEMENTS OF FIELD THEORY

In particular, it is thus proved that any element of an algebraically generated extension P(a1, ... , as) can be expressed in the form of a polynomial over the field P in the elements al, ... , as.

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6. Composite finite extensions .y.

Let L be a finite extension of the field P, K a finite extension of the field L:

P= Lc K, al, ... , am a basis of the field L over the field P and fl, ... , /3 a .fl

basis of the field K over the field L. Thus,

m = [L: P],

n = [K: L].

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It can be shown that the mn elements aipj, i = 1, ... , m; j =

1,

, n, form a basis of the field K over the field P, i.e. in the first place,

...

any element of the field K is a linear combination of the elements ai/3j with coefficients from the field P and, secondly, the elements aifj are linearly independent (over the field P). In fact, any element /3 of the field K is, by definition, a linear combination of the elements Nl, ... , /3n with coefficients from the field L: /3 = YA + ... + Yn/3n, where Yl, ... , YnEL, n

E YA. j=1

(1)

On the other hand, for any j = 1, ... , n the element yj is a linear combination of the elements al, ... a, with coefficients from the field P:

y j = cl jal + ... + c,,, jan where c1j, ...

,

a P,

M

yj =

i=1

Cijai

Substituting these expressions in formula (1), we obtain: n+ m = L G cijai/ j j=1 i=1

Thus, any element of the field K is a linear combination of elements of the form aifj with coefficients from the field P.

10

FOUNDATIONS OF GALOIS THEORY

Now we suppose that in the field P there exist elements kip such

that m

n

E E kj.;a,fj = 0.

j=1 i=1

For any j = 1, ... , n we shall set Y;_

ke,ai

The elements yl, ... , yn belong to the field L and satisfy the relation Y1p1 -I-

... + Ynpn = 0.

Because the elements fl,, ... , /3n form a basis of the field K over the field L, then from this relation it follows that

Y1=...=Yn=O. Thus, for any j = 1, ... n m

kijai = 0.

Hence, since the elements al, ... , a. form a basis of the field L over the field P, then kt, = 0 for all i and j. Thus it is proved that the

system of elements at/; is linearly independent.

From the assertion proved it follows that the field K is a finite extension of the field P and its degree is equal to mn, i.e.

[K: P] = [K: L][L : P]. It is easy to generalize this relation: if

P=L0

L,c: ... cLi_1cLic ... cLS=K,

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where for any i = 1, ... , s the field Li is a finite extension of the field L;_1, then the field K will be a finite extension of the field P and

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[K: P] _ [K: Ls-1] ... ELI: Lt-1] ... [L1 : P]. Oaf

For the proof it is sufficient to apply induction with respect to s. This theorem is applicable, in particular, to any composite algebraic extension, because, as we know, any simple algebraic extension a,.

is a finite extension. Thus we see that any composite algebraic

may

extension is a finite extension, i.e. the class of extensions of type (3) is contained in the class of extensions of type (1).

Because all the elements of a finite extension of a field P are

11

ELEMENTS OF FIELD THEORY

C.'

algebraic (over P), then, in particular, for any composite algebraic extension P(al)(a2) ... (as) the elements al, 062, ... , as are algebraic over P. Therefore the extension P(al, ... , as) is an algebraically generated extension. Hence, according to what was proved in para. 5: P(a1)(a2) ... (00 = AU 1, ... , as).

..'

Thus, any composite algebraic extension is an algebraically r-.

V'1

...

"!`

b0.5

generated extension, i.e. the class of extensions of type (3) is contained in the class of extensions of type (2). Comparing this remark with the results of the preceding paragraph, we see that the class of composite algebraic extensions coincides with the class of algebraically generated extensions. Moreover, if K = P(al, ... , as), then K = P(a1)(a2) ... (a), and conversely. 'TI

Further, as was proved in para. 4, the class of finite extensions (i.e. type (1)) is contained in the class of extensions of type (3), i.e. by what has been proved, also in the class of extensions of type (2). 'v)

chi

.ti

...

Hence, the class of finite extensions coincides with the class of

,'b

composite algebraic extensions. Comparing both these theorems, we see that the following three assertions are equivalent: (a) the field K is a finite extension of the field P; (b) the field K is a composite algebraic extension of the field P; (c) the field K is an algebraically generated extension of the field P.

t''

;=,

Thus, all three terms " finite ", " composite algebraic " and " algebraically generated " mean (when applied to extensions) one and the same thing. We conclude this paragraph with some remarks relating to subfields of finite extensions. '.h

Let K be an arbitrary finite extension of the field P and let L be a subfield of it, containing the field P:

P c L c K.

rA'

It is evident that L is finite over P (because it cannot contain an infinite system of elements linearly independent over the field P), and K is finite over L (because any linear combination over P is automatically a linear combination over L). Hence, we arrive at the conditions of applicability of the theorem proved at the beginning of this paragraph. Therefore

[K:P]=[K:L] L:P].

12

FOUNDATIONS OF GALOIS THEORY

5Z'

Cep

""S

ti'

All

Thus, any subfield L (containing the field P) of a finite extension K of the field P is a finite extension, and its degree [L : P] is a factor of the degree [K: P] of the field K. The corresponding quotient[K : P]/[L: P] is equal to the degree [K: L] of the field K over the field L. Because the simple algebraic extension P(a), generated by some element a of the finite extension K, lies in K, and its degree is equal to the degree of the number a, then, consequently, the degree (over P) of any element of a finite extension K of the field P divides the degree [K: P] of the field K over the field P. This is a refinement of the inequality proved in para. 4. Problem. To prove that a finite extension of degree n is a simple algebraic extension if and only if there exists in it an element having degree n. ran

7. The theorem that a composite algebraic extension is simple

Vii

coo

In this section we prove the following theorem: any composite algebraic extension K = P(a1)(a2) ... (;) is simple, i.e. there exists a number 0 such that

K = P(8). First we consider the case s = 2, when K = P(a1)(a2). Let f1(x) and f2(x) be the minimal polynomials (over P) of the numbers a1 and a2 respectively (as we know, these numbers are algebraic over P) and let QQ

f'1)

fin , /3n

01 = al) (01

(1)

be the roots of the polynomial f1(x) and Y1, "' , Y,n

(Y1 = a2)

(2)

be the roots of the polynomial f2(x). Because the polynomials fl(x) andf2(x) are irreducible, then amongst the roots (1), as also amongst the roots (2), none are identical. We consider the elements

A - fl,

(3)

Y1 - Y; vii

where i = 1, 2, ... , n, and j = 2, ..., m (thus j 0 1). The number of these elements is equal to n(m - 1) and, hence, is finite. Therefore

13

ELEMENTS OF FIELD THEORY

in the field P (even in the field R of rational numbers) one can find a number c, not equal to any of the numbers (3). We set

8 = al + cat

(i.e. 0 = /'t + cyl). Because the number c is not equal to any of the numbers (3), then (4)

9 # Qc + cy j II.

""'

foreach i = 1,2, ...,n andj = 2, ...,m. The number 0 belongs to the field K and, hence, is algebraic. The simple algebraic extension P(O) generated by it is contained in K:

P(9) c K.

(5)

We consider the polynomial

gl(x) =fl(6 - cx).

r.°o

via

.ti

This a polynomial over the field P(6), having a root a2 in common with the polynomial f2(x) (which also can be considered as a polynomial over the field P(6)). From inequality (4) it follows that the polynomials gl(x) and f2(x) have no other common roots (because .-.

if gl (y j) = 0, then the number 6 - cy j will be a root of the polynomial

O-'

..y

.fl

.-'

40.

fl(x), i.e. 0 - cyj = li for some i, which by construction is possible only for j = 1). Hence the greatest common divisor of these polynomials is the binomial x - 062. But, as is well known (The Course, p. 194), the greatest common divisor of two polynomials over some field (in our case over the field P(O)) is also a polynomial over the same

field. Therefore a2 a P(O)

and, hence,

al = 6 - cat a P(0). By virtue of the minimal nature of the extension P(al, a2) it follows from this that P(al, a2) c P(9).

Comparing this inclusion with the inclusion (5) and taking into account that P(al, a2) = P(al)(a2), we obtain: P(al)(a2) = P(4).

Thus for s = 2 the theorem is proved. The case of an arbitrary s is reduced to the case s = 2 by a trivial application of the method of complete induction.

14

FOUNDATIONS OF GALOIS THEORY

The theorem proved means that to the list of equivalent properties of extensions mentioned in the preceding paragraph we can add the following property: (d) the field K is a simple algebraic extension of the field P. In other words, the finite (i.e. composite algebraic, i.e. algebraically generated) extensions are exhausted by the simple algebraic extensions.

8. The field of algebraic numbers

In the preceding paragraphs it was proved that the classes of extensions of types (1), (2), (3),and (4) coincide. It remains to clarify

the connexion between these extensions and the extensions of type (5) (i.e. algebraic extensions). As was proved in para. 4, any finite extension is algebraic. We will prove at once that the converse coo

'-.

is not true, i.e. that the class of algebraic extensions, generally speaking, is an essentially wider class than the class of finite

pad

G]+

vii

.r'..

extensions. In what follows this result is not used; it is presented by us only to clarify the complete system of relations between the classes of extensions introduced. Let P be an arbitrary field. We consider the set K of all numbers algebraic over the field P. Let a e K and l3 e K. Then the extension P(a, f) is algebraically generated and, hence, is a finite extension. Therefore all its elements, and this means, in particular, the elements a + f, c$, -a and a-1 (if a 0), are algebraic over P, i.e. belong

to K. Hence the set K is a field. By definition, it is an algebraic

(ND

extension of the field P. We suppose that over the field P there exist irreducible polynomials of as high a degree as desired (in particular, the field R of rational numbers satisfies this condition; see The Course, p. 347). Then the field K will contain elements of as high a degree as desired, and therefore its degree cannot be finite, i.e. the field K will be an infinite extension. Thus in fact there do exist infinite algebraic extensions (at least over the field of rational numbers). Problem. To prove that the field K of all algebraic numbers over the field P is algebraically closed. 9. The composition of fields

Let K1 and K2 be arbitrary fields. Their composition K is the minimal field containing both the field K1 and the field K2. The

i--+

ELEMENTS OF FIELD THEORY

15

existence of the field K follows from the fact that it can be defined as the intersection of all the fields containing both K1 and K2. An example of a composition is the extension P(a1, a2) generated by the numbers al and a2. This extension will be the composition of the extensions P(al) and P(a2). A simple method, suitable in all the interesting cases, of constructing a composition is described in the following theorem: If the fields K1 and K2 are extensions of some field P, and if there exist numbers 81, ... , 0 such that

K2=P(01,...,e), then

K = K1(01, ...

, Os).

In fact, because P c K1, then the field K1(01, ... , 0) contains the field K2 = P(01i ... , B) (and clearly, moreover, the field K1). Therefore, by virtue of the minimal nature of the composition:

KcK1(01,...,Os). On the other hand,

K1(01,...,0)K, because

K1cK and 01,...,83eK. '-r

We apply this theorem to the case when the numbers 01, ... , 0,, are algebraic over P, i.e. to the case when the field K2 is a finite extension of the field P. The numbers 01, ... , 9s algebraic over the field P are also algebraic over the field K1. Therefore any element of the field K = K1(01, ... , BS) can be expressed in the form of a polynomial in B1, ... , BS with coefficients from the field K1 (see para. 5). Hence it follows that any element of the field K can be represented in the form

001 + ... + ;fJ,, where a1, ... , ar E K1, N1, ... , Nr e K2 (in fact

(1)

maybe taken

as mononomials in 01, ... 0j. Thus: if at least one of the extensions K1, K2 of the field P is finite, then any element of their composition K has the form (1).

Problem. To prove that the composition of finite extensions is a finite extension.

CHAPTER 2

NECESSARY FACTS FROM THE THEORY OF GROUPS 1. The definition of a group

C1.

One says that in a non-empty set G there is defined an algebraic operation, if there is given a rule by which to any two elements a e G, b e G there corresponds some uniquely defined element c e G. The

..r

coa

',5

element c is usually denoted by ab, in conformity with which the algebraic operation considered is called multiplication. Sometimes the element c is denoted by a + b, and then the algebraic operation is called addition. As a rule we will use the first, multiplicative, notation. The set G with an algebraic operation is called a group, if (1) for any elements a, b, c e G (ab)c = a(bc);

(2) there exists an element e e G such that III

ae = ea = a coo

for any element a e G; (3) for any element a e G there exists an element a-' e G such that

aa-'=a-'a=e. ".sue'

Condition (1) (the rule of associativity) permits one to define in a

unique way the product of any finite number of elements of the group, i.e. it permits one to prove that the product of any n elements .+.

.fly

is independent of the distribution of the brackets. For a detailed

Lam".

proof see The Course, p. 25. In particular, one can speak of the product of n elements equal to each other, i.e. one can introduce the concept of the power a" of the element a with positive integral exponent. The element e, mentioned in condition (2), is called the identity of the group and is sometimes denoted by 1. It is easy to prove (see The Course, p. 360) that the identity of a group is defined uniquely. 16

NECESSARY FACTS FROM THE THEORY OF GROUPS

17

The element a-1, mentioned in condition (3), is called the inverse of the element a. It can be proved (see The Course, p. 362) that for any element a the inverse element a-1 is defined uniquely. Moreover, for any element a e G and any positive integral n (an)-1 = (a- )n (see The Course, p. 362).

We introduce the power of the element a with negative integral exponent by setting a-" _ (an)-1

(i.e. a-n = (a-1) ).

Moreover, we set

a° = e. It is easy to verify that all the usual rules of operation with powers remain valid in any group. Let g be an arbitrary element of the group G. We consider all possible powers of it ...g-2,9-1,9°=e, 91=9,92,...

"jam

If all these powers are distinct, then the element g is called an element of infinite order; otherwise it is called an element of finite order. Let g be an element of finite order, i.e. g"' = g"2 for some integers n1 and n2. Without loss of generality, we can suppose that n1 > n2, i.e. that the number N = n1 - n2 is positive. Because gN = g"1(g"2)-1, then g' = e. Thus for any element of finite order there exist positive

numbers N such that g' = e. The least of these numbers is called the order of the element g. Let n be the order of the element g and let gm = e, where m is some whole (not necessarily positive) number. We divide (with remainder) the number m by n :

m=nq+r, 0,
We note that only the identity e of the group G has order 1.

18

FOUNDATIONS OF GALOIS THEORY

If the operation given in the group is denoted by the sign +, i.e. if, as one says, the group is given in additive notation, then the element

e is called the zero and is usually denoted by the symbol 0. Analogously, the element a-1 is denoted in this case by -a and is called the additive inverse of the element, and the element an is denoted by na and is called the n-multiple of the element a. We emphasize that in a group the validity of the commutativity -..

rule (ab = ba) is not, generally speaking, assumed. Groups whose operation satisfies this rule are called commutative or Abelian. 2. Subgroups, normal divisors and factor groups

A non-empty subset H of the group G is called a subgroup if ,!m

(1) the product h1h2 of any elements h1 e H, h2 e H belongs to H. (2) for any element h e H the inverse element h-1 belongs to H.

It is obvious that any subgroup is automatically a group. The .'C

identity of the group, as it is easy to see, belongs to any subgroup. C3'

Problem. To prove that the intersection of two subgroups is a r'.

subgroup. We note that the subset of the group G consisting of its identity e, and also the group G itself, are subgroups. These subgroups we will call trivial subgroups.

.cam

true'

Let G be an arbitrary group and H some subgroup of it. The subset of the group G consisting of all elements of the form hg, where h is an arbitrary element of the subgroup H, and g is some fixed element of the group G, is called the coset containing the element g, with respect to the subgroup H, and is denoted by Hg. It is obvious that g e Hg (because e e H). Let g' be any arbitrary element of the coset Hg. By definition:

g'=h'9 where h' is some element of the subgroup H. We consider the coset

Hg' containing the element g'. Any element of this coset has the form hg', i.e. the form hh'g, where h e H. Hence, because We H, then any element of the coset Hg' belongs to the coset Hg, i.e.

Hg'cHg. On the other hand, any element hg e Hg can be represented in the form h(h')-'h'g = h(h')-1g'. Because h(h') a H, then, consequently, hg e Hg'. Thus,

Hg c Hg'.

NECESSARY FACTS FROM THE THEORY OF GROUPS

19

Thus it is proved that

Hg' = Hg, i.e. the coset Hg' containing any element g' from the coset Hg coincides with the coset Hg. Hence it follows that if two cosets intersect, then they coincide.

;wt)

In fact, the coset Hg containing an element g lying in the intersection of the given cosets coincides with each of these cosets. Now it is easy to prove that two elements g1 and g2 of the group G belong to the same coset of the subgroup H if and only if g1g21 e H. In fact, if g1g21 e H, i.e. g1g21 = h, where h e H, then g1 = hg21 i.e. g1 e Hg2, and, consequently, the element g1 belongs to the same coset Hg2 as that to which the element g2 belongs. Conversely, if there exists a coset Hg such that g1 e Hg and g2 e Hg, then Hg1 = Hg2 and, consequently, g1 e Hg2, i.e. g1 = hg2, where h e H. Therefore 91921 e H. Finally, the coset Hg coincides with the subgroup H if and only

ifgeH.

For the proof it is sufficient to note that the subgroup H can be considered as the coset containing the identity e.

Let Hg be an arbitrary coset of the subgroup H. We define a mapping co of the subgroup H on the coset Hg, setting, for any element h e H, w(h) = hg.

It is obvious that this mapping is one-one (because if h1g = h2g, then, multiplying on the right by g-1, we obtain h1 = h2). Thus for any coset of the subgroup H there exists a (generally speaking, not unique) one-one mapping of the subgroup H on this coset. In particular, if the subgroup H is finite (i.e. consists of a finite number of

elements), then all the cosets of H contain the same number of elements.

We apply this result to the case when the group is finite. Let n be

the number of elements of the group G. Any subgroup H of a finite group G is, clearly, finite, and the number m of its elements does not exceed n. Let k be the number of distinct cosets of the subgroup H. (This number is finite by virtue of the finiteness of the group G.) By the proof above, these classes, in the first place, do not intersect, and in the second place, each of them consists of m elements.

Therefore all these classes together contain km distinct elements. Hence, noting that any element g of the group G necessarily belongs

20

FOUNDATIONS OF GALOIS THEORY

10'

"'O

to some coset (namely, the coset Hg), we see that the total number n of elements in the group is equal to km. The number of elements of a finite group is usually called its order.

cep

Thus the number n introduced above is the order of the group G, and the number m is the order of the subgroup H. The number k of residue classes with respect to the subgroup H is called the index of the subgroup H (in the group G). The equation n = km proved above establishes the validity of the following theorem: 4-.

The order of a finite group is divisible by the order of any of its subgroups. The corresponding quotient is equal to the index of the subgroup. .fir'

This theorem is known as Lagrange's Theorem.

Now we will return to the consideration of arbitrary (possibly infinite) groups. The subgroup H of the group G is called a normal divisor if for any element h e H and any element g e G the element ghg-1 belongs to H.

It is obvious that in any Abelian group any subgroup is a normal divisor.

Problem. To prove that the intersection of normal divisors is a normal divisor.

It is obvious that if the subgroup N is a normal divisor of the group G, then it will also be a normal divisor of any subgroup H c G containing it. It is necessary to bear in mind that the converse, generally speaking, is not true: if

NcHcG asp

and N is a normal divisor of H, then N need not also be a normal divisor of G.

Problem. To prove that the intersection N n H of the normal F.;

divisor N and an arbitrary subgroup His a normal divisor of H. The trivial subgroups (i.e. G and e) are, obviously, normal divisors. A group not having any other normal divisors is called simple. Let H be an arbitrary normal divisor of the group G and let Hg1, Hg2 be cosets of the normal divisor H. Choosing arbitrary elements h1g1, h2g2 in the cosets Hg1, HgZ, we consider the product h1g1h292 of these elements. Because h191h292 = h1(91h2911)9192

and, by normality, g1h2g11 a H, which implies also h1(g1h2g11) e H, then this product lies in the coset Hg1g2, and hence its coset coincides

v3,

NECESSARY FACTS FROM THE THEORY OF GROUPS

21

with the coset Hglg2. Thus, with any choice of elements from given cosets the coset of their product is determined uniquely. This uniquely defined coset is called the product of the cosets Hg1 and Hg2 and is denoted by Hg1 Hg2. By what has been proved : H91'Hg2 = H9192

DCa

_E;'°

Thus we have defined in the set of all cosets of the normal divisor H an algebraic operation. It is easy to verify that with respect to this operation the set of cosets is a group (the identity is the coset H = He, .3'

and the coset inverse to the coset Hg is defined by the formula (Hg)-1 = Hg-1). This group is called the factor group of the (IQ

,ti

off-

group G with respect to the normal divisor H and is denoted by G/H. For a finite group G the factor group G/H is finite and its order is equal to the index of the subgroup H.

If H = e, then the factor group G/H coincides, clearly, with the group G, and if H = G, then the factor group G/H consists merely of one element (the identity). 3. Homomorphic mappings

Let G and G' be arbitrary groups. A mapping : G --* G' of the group G into the group G' is called a homomorphism (or homomorphic mapping), if it carries a product into a product, i.e. if for any elements g1, g2 of the group G q5(9192) = 0(91)0(92)e''

Setting in this equation g1 = g2 = e, we obtain 4.(e) = O(e)o(e), whence it follows that O(e) is equal to the identity of the group G'.

Further, setting g1 = g, g2 = g-1, we obtain 4(e) = 0(g)0(g1), i.e. 0(g-1) = 4(g)-1. Thus a homomorphism carries the identity into the identity and inverses into inverses. A homomorphic one-one correspondence is called an isomorphism (or isomorphic mapping). Two groups are called isomorphic if there exists at least one isomorphic mapping of one group onto the other. Isomorphic groups possess identical algebraic properties and in the general theory of groups they are regarded as identical (in connexion with this see The Course, p. 39). Let 0 : G - G' be an arbitrary homomorphism. It is obvious that 4°,

for any subgroup H of the group G the set of all elements of the group G' having the form q5(h), where h e H, is a subgroup of the

22

FOUNDATIONS OF GALOIS THEORY

...

group G'. This subgroup is called the image of the subgroup H under the homomorphism 0 and is generally denoted by 4)(H).

'p,

p".

°-h

0q'

Cam,

'8-

In particular, the subgroup 4)(G) is defined as the image of the group G under the homomorphism 0. This subgroup is also called the image of the homomorphism 0 and it is sometimes denoted by Im 0. If 4)(G) = G', i.e. if for any element g' e G' there exists an element g (generally speaking, not uniquely defined) such that 4)(g) = g', then 0 is called an epimorphic mapping (or simply an epimorphism) of the group G onto the group G'. The special significance of epimorphisms is due to the fact that any homomorphism 0 : G --* G' can be considered as an epimorphic mapping of the group G onto the subgroup q5(G). This remark is easily generalized: for any subgroup H of the group G the homomorphism 0 can be considered as an epimorphic mapping of the .CD

+-'

subgroup H onto the subgroup 4)(H). More exactly, the homoCAD

"t3

In,

morphism 0 determines an epimorphic mapping 0': H -+ 4)(H) (the mappings 0 and 0' coincide on the elements of the subgroup H and differ only in the fact that 0 is defined on the whole group G, and 0' only on the subgroup Hj). Again let 0 : G -+ G' be an arbitrary homomorphism and let H and H' be normal divisors of the groups G and G' respectively such that qS(H) c H'. We consider in the group G an arbitrary coset Hg. An element hg of this coset is mapped by the homomorphism ¢ into the element ¢(h)4)(g), belonging to the coset H'qS(g). Thus all the elements of the coset Hg go into elements of one and the same coset H'¢(g). We denote this uniquely defined coset by (Hg) : .-:

a''

cad

a-+

fn'

.may

'7r

.fl

+-+

NCO

.fl

(Hg) = H'q5(9) Thus we have defined a mapping : G/H -3 G'/H'. (IQ

This mapping is a homomorphism, because

(Hg1H92) = (H9192) = H'4)(9192) = H'4)(91)4)(92)

= H'0(9i)H'0(92) = RH91AH92) t Editor's footnote: This remark is inaccurate since ¢ and 0' also differ in their range.

NECESSARY FACTS FROM THE THEORY OF GROUPS

23

We will say that the homomorphism is induced by the homomorphism 0. We emphasize that it is defined only when O(H) c H'. In the special case when the normal divisor H' consists merely of

the identity e' of the group G', we see that if *b(H) = e', then the homomorphism 0 : G -* G' induces a homomorphism : G/H -* G'. It is obvious that the homomorphism induced by an epimorphism is an epimorphism. Now we consider the set N of all elements mapped by the homo-

Ca.

G]+

P'+

Col

`°-

'ti

morphism 0 : G -> G' into the identity e' of the group G'. If a e N, b e N, i.e. ¢(a) = e', 4(b) = e', then 4i(ab) = e'e' = e', i.e. ab e N. Analogously, if a e N, then a-1 e N. Moreover, if a e N and g is any element of the group G, then 4(gag-1) = q (g)q5(g)-1 = e', i.e. gag-1 e N. Thus N is a normal divisor of the group G. This normal divisor is called the kernel of the homomorphism 0 and is denoted by Ker 0. If 4(a) = ¢(b), then /(ab-1) = e', i.e. ab-1 e N. Conversely, if

ab-1 e N, then 4(a) = fi(b). But ab-1 e N if and only if a and b

VII

-0-

belong to the same coset with respect to the subgroup N. Thus, in the homomorphism 0 : G -* G' elements of the group G go into one and the same element of the group G' if and only if they belong to the same coset of the kernel of the homomorphism 0. If Ker ¢ = e, then the homomorphism ¢ is called a monomorphism. From what has just been proved it follows that the monomorphism

0 : G -+ G' carries distinct elements of the group G into distinct elements of the group G', i.e. it is an isomorphic mapping of the group G onto the subgroup q5(G) of the group G'. In particular, a monomorphic and epimorphic mapping is an isomorphism, and conversely.

Because for any homomorphism 0 : G --* G' with kernel N q5(N) = e',

then the homomorphism 0 induces a homomorphism : G/N -+ G'.

The homomorphism

is defined by the formula

(Ng) = q5(g).

If (Ng) = e', i.e. q5(g) = e', then g e N and, hence, Ng = N. In

24

FOUNDATIONS OF GALOIS THEORY

other words, the kernel of the homomorphism

consists of only one

coset N-the identity of the group GIN, i.e. the homomorphism is a monomorphism. Thus: any homomorphism 0 : G -+ G' induces a monomorphism ..+

..r

Imo-

ii:

y..

: GIN -> G', where N = Ker 4). This assertion is known as the homomorphism theorem. If the homomorphism 0 is an epimorphism, then, as we know, the homomorphism will also be an epimorphism, and this means also an isomorphism. Thus, any epimorphism 0 : G -+ G' induces an isomorphism : GIN -+ G',

+.+

where N = Ker 0. The group G' is called a homomorphic image of the group G if there exists at least one epimorphic mapping of the group G onto the group G' (it is the custom to speak thus of a " homomorphic image ", although, of course, it would be more systematic to speak of an " epimorphic image "). From the proposition proved it follows immediately that any homomorphic image of a group is +-'

isomorphic to some factor group of it. We note that the converse assertion is also true: any factor group GIN of the group G is a homomorphic image of the group G.

For the proof it is sufficient to construct just one epimorphic mapping 0 of the group G onto the factor group GIN. Such a mapping can, for instance, be defined by the formula 4)(g) = Ng.

We note that the mapping ¢ so defined is none other than the mapping induced by the identity mapping of the group G on itself (in

the general definition it is necessary to take as H the identity subgroup, and as H' the normal divisor N).

CHAPTER 3

GALOIS THEORY 1. Normal extensions

G.:;

v".

In the whole of this chapter it is assumed that there is given some fixed field P. We will call this field the fundamental field. All other fields are supposed to be extensions of this fundamental field. We emphasize that the fundamental field can be chosen quite arbitrarily. Let f (x) be an arbitrary (generally speaking, reducible) polynomial over the field P. The extension P(al, ... , of the field P, generated .,m

by all the roots al, ... , a of the polynomial f (x), is called the decomposition field of this polynomial (we note that this definition coo

differs from the definition taken in The Course, p. 212, where a cep

decomposition field is any, not necessarily minimal, extension of the

field P containing the roots al, ... ,

e-'

w..

".sue'

According to chapter 1, para. 5 any element of the field P(al, ... , can be expressed in the form of a polynomial in al, ... , a with coefficients in the field P. A finite extension K of the field P is called a normal extension if any polynomial irreducible over P, having at least one root in K, decomposes into linear factors in K. In other words, an extension K of the field P is normal if it satisfies the following two conditions: (1) K is finite over P; (2) if a polynomial irreducible over P has at least one root in K, then K contains the decomposition field of this polynomial. Normal extensions of the fundamental field P we will also call normal fields.

Two numbers algebraic (over P) are called conjugate (over P) if their minimal polynomials (over P) coincide (more exactly, differ by a

constant factor). In other words algebraic numbers are conjugate if they are roots of one and the same polynomial irreducible over P. The concept of conjugate numbers permits the following method of reformulating the definition of a normal extension: an extension K of the field P is normal if (1) K is finite over P; (2) any number conjugate to some number in K also belongs to K. 25

26

FOUNDATIONS OF GALOIS THEORY

This form of the definition of a normal extension is often the most convenient. Let K be an arbitrary normal extension of the field P. Because the

field K, by definition, is finite over P, there exist elements al, ... , as e K such that K = P(al, ... , as). Let fi(x) be the minimal polynomial of the number ai, i = 1, ... , s,

over the field P. Because the field K is normal (i.e. is a normal .n'

.CD

extension of the field P), then the polynomials fi(x), having roots in it, decompose into linear factors in K. Hence the product

f(x) =fi(x) ... fs(x) CC"

coo

of the polynomials fi(x), ... , fs(x) also decomposes into linear factors in K, i.e. the field K contains the decomposition field Q of the polynomial f (x). On the other hand, the numbers al, ... , as are cad

roots (not all the roots!) of the polynomial f (x), and therefore the field K is contained in the field Q. Consequently, K = Q. Thus, any normal field is the decomposition field of some polynomial. p.,

Problem. To prove that any normal field is the decomposition JR.

field of an irreducible polynomial. It will be shown that all the decomposition fields are exhausted by the normal fields, i.e. any field which is the decomposition field of some polynomial (over the field P), will be a normal extension of the field P. For the proof of this important assertion we need some facts from the theory of polynomials in n variables, which also have an independent interest. _ (1 2 ... n) Let a it i2 .. in finy

,..,,

41.

..,

be an arbitrary permutation of order n (see The Course, p. 74, and also below, Part II, ch. 3, para. 1). To any polynomial g(xl, ... , xa) in n variables over the field P we relate a polynomial ga(xl, ... , with the help of the permutation a, defining it by the formula ga(x1, ... , xn) = g(xi,, ... xj.

It is obvious that ge = g

and

(ga)b = gab

GALOIS THEORY

27

We note that ga = g for all permutations a if and only if the polynomial g is a symmetric polynomial. Now let

a1=e, a2,...,ani be all the permutations of order n, indexed in an arbitrary way by the numbers from 1 to n!. We consider the polynomials gal = g, gal, ... , ga,,,,

(1)

where g is an arbitrary polynomial in n unknowns x1i ... , xn. Operating on these polynomials by an arbitrary permutation a of order n, we obtain the polynomials al'

gala = ga,

ga2a, ... , ga,,,a

(2)

Because the permutations a1'a, a2a, ... , anla,

ax'

clearly, exhaust all the permutations of order n (there are n! of them and they are all different), then the polynomials (2) coincide with the polynomials (1) to within order of sequence. It follows from this that any symmetric polynomial in gal, gal, ... , gal, is also a symmetric polynomial in x1, ... , x,,, i.e. if F(y1, ... , yn!) is a symmetric polynomial in the n! variables yl, ... , Yn1, then, substituting for yi the polynomial gat(x1, ... , xa), we obtain a symmetric polynomial in x1, ... , xn. In particular, all the coefficients of the polynomial

G(x;x1,...,xn)= i=1 fi (x-ga,(x1,...,xn))

(3)

(considered as a polynomial in the unknown x) are symmetric

CAD

polynomials in x1, ... , xn and, hence (see The Course, p. 241), can be expressed in the form of polynomials (with coefficients in the field P) in the elementary symmetric polynomials. Now we will return to the proof of the assertion formulated above. Let K be the decomposition field of some polynomial f (x) over the field P. Then, as already noted above, any element /3 of the field K can be written in the form of a polynomial in the roots al, ... , an of the polynomial f (x) (generally speaking, in many different ways), i.e. there exists a polynomial g(x1, ... , xn) in the n unknowns x1, ... , xn such that

..:

9 =g(a1,...,a.).

28

FOUNDATIONS OF GALOIS THEORY

We consider the polynomial G(x) = G(x; a,, ... ,

an),

where G(x; x,, ... , x,) is the polynomial (3), constructed for the polynomial g(x,, ... , xn). By definition: _

n1

G(x) = rj (x i=1

where fla = ga,(a,, ... , an) a K.

According to what was said above, the coefficients of the polynomial G(x) can be expressed in the form of polynomials (over P) in the elementary symmetric polynomials in a,, ... , an, i.e. can be expressed in the form of polynomials in the coefficients of the polynomial f (x). Hence, G(x) is a polynomial over the field P. The minimal polynomial h(x) of the number f3 (over the field P) has the root fi = /, in common with the polynomial G(x) and therefore divides the polynomial G(x). Hence, all the roots of the polynomial h(x), i.e. all the numbers conjugate to the number IB, are contained amongst the numbers and therefore belong to the field K. Thus we have proved that all the numbers conjugate to any element of the extension K (finite, as we know) belong to K. Hence the field K is normal. 2. Automorphisms of fields. The Galois group (7)

."s.

'-'

A one-one mapping S of some field K onto itself is called an automorphism if it carries sums into sums and products into products, i.e. if for any elements a, fl of the field K (a + Q)S = as + as, l

(1)

(ams = asps,

,a3

coo

a'0

(the element into which the element a goes under the automorphism S we will denote by as). We emphasize that an automorphism must be a one-one mapping (transformation), i.e., besides conditions (1), it must also satisfy the following requirements: (a) for any element a e K the element as is uniquely defined and belongs to K;

GALOIS THEORY

29

(b) if a 0 /3, then as 0 Ps; (c) for any element /3 e K there exists an element a e K such that

as=P. From condition (b) it follows that the element a mentioned in condition (c) is defined uniquely. Hence, denoting this element by 9-1

a=/3s we obtain a (clearly one-one) transformation S-1 (i.e. the inverse transformation). This transformation is uniquely characterized by the fact that for any element a e K

(as )s = a

(2)

It will be shown that the transformation S -1 is also an automorphism. In fact, for any elements a e K and /3 e K (as-1 + /3s-1)S (as )s + (/3s )s = a + /3

=

and, hence, by definition: (a +

fl)S-1

= as-1 +

fls

Analogously it is proved that (cYa)s-I = as-1$s-1.

The product ST of two automorphisms S and T is the transformation obtained as a result of successively performing first the transformation S, and then the transformation T; for any element or e K the element aST is defined by the formula cST = (ccS)T.

It can be verified immediately that the transformation ST is also an automorphism.

Problem. To prove that multiplication of automorphisms is associative.

Multiplication of automorphisms, clearly, possesses an identitythis role is played by the identity automorphism E, which leaves all elements of the field K invariant:

aE=a. By definition (see formula (2)):

S-1S = E.

(3)

30

FOUNDATIONS OF GALOIS THEORY

Now we consider the automorphism (S-1)-1, inverse to the automorphism S-1. By definition: (S-1)-1S-1 = E.

(4)

Multiplying this equation on the right by S and using formula (3), we obtain: (S-1)-1 = S. Substituting this expression in formula (4), we obtain:

SS-1=E. And so, S-1S = SS-1 = E.

Thus we see that with respect to the operation of multiplication of automorphisms the set of all automorphisms is a group. This group is called the automorphism group of the field K. Problem. To prove that any automorphism leaves invariant all rational numbers (in particular, the numbers 0 and 1). Now we suppose that the field K is a normal extension of a field P. In this case an automorphism S of the field K is called an automorphism

over the field P, if it leaves all elements of the field P invariant, i.e. if for any element c e P CS = C.

It is obvious that the set of all automorphisms over P is a subgroup 'a$

of the group of automorphisms of the field K. This subgroup is called the Galois group of the field K over the field P and is denoted by G(K, P). We emphasize that the Galois group is considered only for normal fields.

Let

f(x) =Co+C1x+

... +C"x"

be an arbitrary polynomial over the field P, having at least one root a in K:

Co+Cla+ ... +C"a"=0.

(5)

Applying an automorphism S in the Galois group G(K, P) to equation (5), we obtain, as is easy to see, CO + Clots + ... + C"(CCs)" = 0

GALOIS THEORY

31

(because cs = c1 for any i = 0, 1, ... , n), i.e. f(as) = 0.

Thus, any automorphism in the Galois group G(K, P) carries every root of a polynomial over the field P again into a root of this same polynomial.

From this it follows in particular that for any number a E K and any automorphism S E G(K, P) the number as is conjugate over the field P to the number a.

Remark. If one considers the familiar concept of a linear trans-

formation and the fact that a linear transformation of a finitedimensional space is one-one if and only if it does not carry any P'+

vector distinct from zero into the zero vector (the proof of this fact can be found in any sufficiently complete course in linear algebra), then it can be shown that in defining the concept of an automorphism of the field K over the field P the condition of being one-one can be omitted, i.e.:

`ti

any mapping Sofa normal field K into itself possessing the properties (1) and leaving invariant all elements of the field P is one-one, i.e. is an automorphism of the field K over the field P. In fact, if c E P and oc E K, then (cot)S = csas = cots.

Moreover, for any elements a and f3 of the field K

(a+f3)S=as+/3S. This means that the mapping S is a linear transformation of the

""Y

'k..

...

field K, considered as a linear (finite-dimensional) space over the field P (see Ch. 1, para. 2). Therefore, by virtue of the fact in the theory of linear transformations mentioned above, it is sufficient, to prove the assertion stated, to show that if a 0, then also as 0. But if a 0 0, then in the field K there exists an element P such that a,6 = 1 and, hence, as f3s = 1. Thus, in fact, as 0 0. 3. The order of the Galois group

Let K be an arbitrary normal extension of the field P. According to Ch. 1, para. 7 the extension K is a simple algebraic extension, i.e. in K there exists an element 0 such that

K = P(0).

32

FOUNDATIONS OF GALOIS THEORY

The degree n of the minimal polynomial f (x) of the element 0 is equal to the degree [K: P] of the field K over the field P. Any element a of the field K can be written uniquely in the form a = CO + c19 + ... + Cn_10n-1,

where co, c1, ... , C,,-IC-P. (1)

As was proved in the preceding paragraph, any automorphism S from the Galois group G(K, P) carries the root 0 again into a root of

the polynomial f (x). In other words, to every automorphism S E G(K, P) corresponds some root of the polynomial f (x) (with a chosen root 0). We will study this correspondence in more detail.

Let 0' be an arbitrary root of the polynomial f (x). Because the field K is normal and 0 E K, then 0' E K. We define a transformation S of the field K into itself, setting for any element (1) from this field as = co + C1O' +

... +

Cn-kiln-1

(2)

Because the representation of the element a in the form (1) is unique, formula (2) defines the element as uniquely. The definition of the transformation S can, clearly, be formulated in the following way : if

a=g(0), where g(x) is a polynomial over the field P, having degree less than n, then as = 9(0')

Now we consider a polynomial g(x) over the field P of arbitrary degree, and let a = 9(e) We divide (with remainder) the polynomial g(x) by the polynomial .f (x):

g(x) = f (x)q(x) + r(x).

(3)

Setting x = 0 in this equation, we obtain (because f (0) = 0)

a = r(0). .s:

Because the degree of the polynomial r(x) is less than n, then it follows from this that as = r(0').

GALOIS THEORY

33

On the other hand, setting x = 0' in formula (3), we obtain

g(8) = r(8'). Hence as = 9(B').

Thus 9(9)S = 9(8')

independently of the degree of the polynomial g(x). Now let at = 910), az = 92(8)

be arbitrary elements of the field K. Then al + ccz = 91(9) + 92(8), a1a2 = 91(e)92(8)

and, hence, Y/]

=a(al + az)5=91(B)+9z(6')a+, (alaz)S = 91(8')92(8') _42.

..y

Thus the transformation S preserves sums and products, i.e. possesses properties (1) para. 2. Moreover, this transformation, clearly, leaves invariant all the elements of the field P. Therefore (see the remark in para. 2) the transformation S is an automorphism ...

of the field K over the field P, i.e. it belongs to the Galois group G(K, P). i-+

The fact that the transformation S is an automorphism, i.e., t-+

.»,

besides the properties (1) para. 2, it also possesses the property of being one-one, can also be proved without using the remark in para. 2. In fact, we consider the field P(8'). Because 0' e K, then vi'

P(8') c K. On the other hand, the degree of the field P(8') over the field P is equal to the degree of the polynomial f (x), i.e. is equal to the degree of the field K. Hence,

P(8') = K.

From this it follows that together with the description (1) any element of the field K has a unique description of the form

a=c, +c0+... where c', ci, ... Cn_1 a P.

+C'_le'n-1,

(4)

34

FOUNDATIONS OF GALOIS THEORY

.,,

Now we define a transformation S' of the field K into itself, setting for any element (4) from this field

aS'=c'+ci6+

... +cn_18"-1

Because, clearly, C/)

,.%

C/)

S'S=SS'=E (i.e. S' = S-1), then the transformation S is, as asserted, a one-one transformation of the field K on itself (because from as = Jas it _ ass i.e. that a = 9 and for any element a e K follows that there exists an element fi, namely 13 = as', such that $s = a). ass'

'hr

The automorphism S constructed carries the root 0 into the root 0': Bs = 9',

i.e. this automorphism corresponds to the root 0' in the sense indicated above. Thus it has been proved that for any root of the polynomial f (x) there exists in the Galois group G(K, P) an automorphism to which this root corresponds. It will be shown that the automorphism is uniquely defined by the corresponding root, i.e. if es=0T, then

S=T. In fact, if OS = 9T, then OST-1 = 0, i.e. the automorphism ST-1 leaves the root 0 invariant and, hence, leaves invariant any expression of the form ..F.

co + CIO + ... + cn_19n-1,

where co, ... , cn_1eP,

i.e. leaves invariant any element of the field K. Thus, ST-' = E and therefore S = T.

a.)

Thus the elements of the Galois group G(K, P) (i.e. the automorphisms of the field K over the field P) stand in one-one correspondence with the roots of the polynomial f (x), and, hence, the number of them, i.e. the order of the group G(K, P), is equal to the number of roots of the polynomial f (x), i.e. is equal to n (all the roots of the polynomial f (x) are distinct, because this polynomial is irreducible). Thus we have proved that the order of the Galois group G(K, P) is equal to the degree of the field K over the field P.

35

GALOIS THEORY

4. The Galols correspondence

obi

As above, let K = P(6) be an arbitrary normal extension of the fundamental field P and G(K, P) its Galois group over the field P. In this paragraph we will consider an extension L of the field P, contained in the field K:

PcLcK.

CPU.

E.,

Such an extension we will call an intermediate field. The polynomial f (x) over the field P, whose root is the number 0, can also be considered as a polynomial over any intermediate field L.

It is obvious that its decomposition field over L is the field L(9) (why?). Hence, the field L(B) is normal over the field L. On the other hand, because P c L, then P(9) c L(B), i.e. K c L(B), and because L c K and 0 e K, then L(9) c K. Hence, K = L(O). Thus, the field K is normal over any intermediate field L. Therefore one can speak of the Galois group G(K, L) of the field .,,

,_,

r-+

K over the field L. According to what was proved in the preceding paragraph the order of the group G(K, L) is equal to the degree of the field K over the field L. yam.

The elements of the group G(K, L) are, by definition, automorphisms of the field K, leaving invariant any element of the field L.

Because P c L, then these automorphisms also leave invariant any element of the field P, i.e. they are elements of the Galois group G(K, P) of the field K over the field P. Thus, G(K, L) c G(K, P), i.e. the Galois group of the field K over the field L is a subgroup of the Galois group of the field K over the field P. Its order is equal to the degree [K: L] of the field K over the field L. Now let H be an arbitrary subgroup of the Galois group G(K, P). It is obvious that the set of all elements of the field K, left invariant

by any automorphism from the subgroup H, is a subfield of the field K. This subfield contains the field P, i.e. is an intermediate field. We will denote it by K(G, H). Let

Ti=E, T2,...,T. be all the elements of the subgroup H (thus m is the order of the subgroup H). We consider the polynomial

h(x) = fl (x - 0T'). i=1

36

FOUNDATIONS OF GALOIS THEORY

Its roots are the numbers OT' = 0,

0T2,

...

,

OT-.

(1)

Under an automorphism T e H these numbers go into the numbers 0T,T = OT, 6T2T, ... , OT-T. (2) But the elements

T,T=T, TZT,...,T.T, per.'

clearly, exhaust all the elements of the subgroup H (there are m of them and they are all distinct). Hence the numbers (2) coincide with the numbers (1) to within order of sequence. In other words, in any

automorphism T e H the roots of the polynomial h(x) are only transposed. Therefore any symmetric polynomial in these roots, in particular any coefficient of the polynomial h(x), is left invariant by the automorphism T and, hence (in so far as T is any automorphism from the subgroup H), belongs to the field K(G, H). Thus the polyCD.

nomial h(x) is a polynomial over the field K(G, H). Hence, the minimal polynomial of the element 0 over the field K(G, H) is a factor of the polynomial h(x), and therefore its degree (i.e. the degree of the number 0 over the field K(G, H)) is less than or equal to m. But, as we saw above, the field K is a simple algebraic extension of CAD

CAD

any intermediate field (and this means, in particular, of the field K(G, H)), generated by the number 0. Therefore the degree of the field K over the field K(G, H) is equal to the degree of the minimal polynomial (over K(G, H)) of the number 0, i.e. by what has been proved, less than or equal to m. Now we consider the Galois group G(K, L) of the field K over the

field L = K(G, H). According to para. 3 the order of this group is equal to the degree of the field K over the field K(G, H) and therefore

".s

°-O

is less than or equal to m. On the other hand, the group G(K, L) consists, by definition, of all automorphisms of the field K leaving invariant the elements of the field L = K(G, H) and therefore contains the subgroup H. Hence, its order cannot be less than m. From this it follows that the order of the group G(K, L) is equal to m and therefore it coincides with the subgroup H. Thus,

if L = K(G, H), then G(K, L) = H. Now let L be an arbitrary intermediate field and let H = G(K, L). We consider the field K(G, H). It is obvious that L c K(G, H).

37

GALOIS THEORY

According to Ch. 1, para. 6

[K(G, H) : L] -

[K: L] [K : K(G, H)] a?.

On the other hand, by what has just been proved, the degree

..,,

coo

[K: K(G, H)] of the field K over the field K(G, H) is equal to the order of the group H = G(K, L), i.e. is equal to the degree [K: L] of the field K over the field L. Hence, [K(G, H) : L] = 1, i.e. L = K(G, H). Thus if H = G(K, L), then K(G, H) = L. We see, thus, that to any intermediate field L there corresponds some subgroup of the group G(K, P) (namely, the group G(K, L)), and for any subgroup H of the group G(K, P) there exists an intermediate field L (namely, the field K(G, H)), to which this subgroup f74

v°,

.s~

vii

.fl

corresponds, and to different intermediate fields correspond different subgroups (because if G(K, L1) = G(K, L2), then L1 = K(G, G(K, L1))

= K(G, G(K, L2)) = L2). In other words, we have constructed a one-one correspondence between the set of all intermediate fields and the set of all subgroups of the Galois group. This correspondence is called the Galois correspondence.

We repeat once again that, in the Galois correspondence, to the intermediate field L of the normal field K corresponds the Galois group G(K, L) of the field K over the field L, and to the subgroup H of the group G(K, P) the subfield K(G, H), consisting of all elements of

the field K left invariant by every automorphism in H. The order of the group G(K, L) is equal to the degree of the field K over the field L,

b10

and the degree of the field K over the field K(G, H) is equal to the order of the group H. In particular, to the whole group G(K, P) corresponds the field P. Hence, the field P consists of all elements of the field K left invariant by every automorphism of the group G(K, P). v,'

To the identity subgroup E, i.e. the subgroup consisting only of the identity automorphism E, corresponds, clearly, the whole field K.

The Galois correspondence allows the theory of subfields of a 43.

given normal field in some sense to be " mapped " into the theory of the subgroups of its Galois group and thus to be studied by grouptheoretical methods. For instance, from the finiteness of the number of subgroups of a finite group it follows immediately that the number of intermediate subfields of any normal field is finite. To prove this fact without using the Galois correspondence is quite difficult.

38

FOUNDATIONS OF GALOIS THEORY

`.4

When applying the Galois correspondence, it is always necessary to bear in mind that it " reverses the inclusion signs ", i.e. if to the subfields L1 and L2 of the field K correspond the subgroups H1 and H2 of its Galois group, then from L1 c L2

(3)

H1 D H2,

(4)

it follows that and conversely, (3) follows from (4). 5. A theorem about conjugate elements

Let a be an arbitrary element of a normal field K. We consider the elements as' = a,

cS2,

... , as",

(1)

where

S1=E, S2,... , Sn ,dam

are all the automorphisms in the Galois group G(K, P) of the field K Con

over the field P. In any automorphism S of the field K over the coo

a..

field P the numbers (1) go into the numbers S,

$1S

S2S

sns

0C

i.e. they are just subjected to a permutation. Therefore all the coefficients of the polynomial

9(x) = i=1 H (x - as') are left invariant by any automorphism S, i.e. belong to the field P.

coo

.k.

r/'

S1]

p;'

via

In so far as a = as', then the polynomial g(x) and the minimal polynomial f (x) of the element a have a common root and, consequently, the polynomial g(x) is divisible by the polynomial f (x) (because the polynomial f (x) is irreducible). On the other hand, we know (see para. 2) that all the numbers as', ... , as" (among these numbers, generally speaking, some can be equal) are conjugate to the number a, i.e. they are roots of the polynomial f (x). Thus, every root of the polynomial g(x) is a root of the polynomial f (x). Let 9(x) = P1(x)k'P2(x)k2 ...

p,(x)k,

be a decomposition of the polynomial g(x) into the product of powers of distinct irreducible polynomials (which have leading coefficients

GALOIS THEORY

39

equal to one). Because the polynomial g(x) is divisible by the polynomial f (x) and the polynomial f (x) is irreducible, then the polynomial f (x) must coincide with one of the polynomials pl(x), ... , p1(x) (we suppose that the leading coefficient of the polynomial f (x) is equal to one). In order to be specific let f (x) = pl(x), so that 9(x) =f(x)k'P2(x)k2 ... p,(x)k,.

Because all the roots of the polynomial g(x) are roots of the polynomial f (x), and not one of the roots of the polynomials p2(x), ... , pi(x) (by virtue of the irreducibility of these polynomials) can be a root of the polynomial f (x), then the polynomials p2(x), ... , pl(x) cannot have roots, i.e. PZ(x) = ... = PA(x) = 1-

Thus,

9W =f(x)k` From this, in particular, it follows that the numbers as', ... , oes^ exhaust (generally speaking, with repetitions) all the numbers conjugate to the number a. Thus it has been proved that two elements of the field K are conjugate (over the field P) if and only if there exists an automorphism of the field K over the field P, transforming one element into the other.

6. The Galois group of a normal subfield

Let an intermediate field L be a normal extension of the fundamental field P. Then for any element a e L and any automorphism S e G(K, P) the element as also belongs to the field L (because it is conjugate to a; see para. 2). Therefore the formula as.

= as,

a e L,

defines some transformation S' of the field L into itself. It is easy to

see that the transformation S' is an automorphism of the field L over the field P, i.e. an element of the Galois group G(L, P) of the field L over the field P. (The automorphisms S and S' act in the same way in the field L; the difference between them lies in the fact that the automorphism S is defined in the whole field K, and the automorphism S' only in the field L.) It is obvious that (ST)' = S'T',

40

FOUNDATIONS OF GALOIS THEORY '"3

i.e. that the correspondence S -> S'

(1)

tai

is a homomorphism of the group G(K, P) into the group G(L, P). The kernel of this mapping consists of the automorphisms S leaving invariant every element of the field L, i.e. the kernel is the Galois group G(K, L) of the field K over the field L. Because the kernel of any homomorphism is a normal divisor, then, consequently, the subgroup of the Galois group G(K, P) corresponding to the normal intermediate field L (i.e. the Galois group G(K, L) of the .+.

field K over the field L), is a normal divisor of the group G(K, P). Now let L be an intermediate field corresponding to an arbitrary

0

Z/2

...

0c°

normal divisor H of the group G(K, P), i.e. L = K(G, H). Because for any automorphism T e H and any automorphism S E G(K, P) the automorphism STS" 1 belongs to H, then for any number a e L aSTS-I

=o

aST = IS.

Because T is an arbitrary automorphism in H, then from this it follows that as a L. Thus all elements conjugate to every element a e L belong to L, i.e. L is normal over P. Thus it has been proved that in the Galois correspondence to normal subfields correspond normal divisors, and conversely.

'-h

~''

Now we shall return to the consideration of the homomorphism (1). Let G' be its image, i.e. the subgroup of G(L, P) consisting of automorphisms of the form S'. According to the homomorphism theorem (see Ch. 2, para. 3) the homomorphism (1) induces an isomorphism of the factor group G(K, P)/G(K, L) onto the group G'.

coo

Hence, the order of the group G' is equal to the index of the subgroup G(K, L) in the group G(K, P). But this index is equal (why?) to the degree of the field L over the field P, i.e. is equal to the order of the

group G(L, P). Thus the order of the subgroup G' is equal to the order of the whole group G(L, P). Thus the order of the subgroup G' is equal to the order of the whole group G(L, P), whence it follows

that G' = G(L, P). Thus it has been proved that the mapping (1) is epimorphic. The mapping induced by the homomorphism is, hence, an isomorphism of the factor group G(K, P)/G(K, L) onto the group G(L, P). Thus :

GALOIS THEORY

41

the Galois group of a normal intermediate field L over the field P is isomorphic to the factor group of the Galois group of the field K over the field P by the Galois group of the field K over the field L.

7. The Galois group of the composition of two fields

Let the normal extension K of the field P be the composition of the cad

extensions K1 and K2. In the Galois group G(K, P) the subgroup G(K, K1) corresponds to the subfield K1, and the subgroup G(K, K2)

to the subfield K2. The automorphisms in the subgroup G(K, K1) 0

leave invariant all the elements of the field K1, and the automorphisms

in the subgroup G(K, K2) leave invariant all the elements of the field K2. Hence, any automorphism in the intersection G(K, K1) n G(K, K2) leaves invariant any element of the form aa a1N1 + ... +

arQ/l'r,

(1)

where a1, ... , a, a K1, i'1, ... , fl, a K2. But, according to Ch. 1, para. 9, all the elements of the composition K are exhausted by the elements of the form (1) (the results of Ch. 1, para. 9 are applicable because the fields K1 and K2 are finite over P). Hence, the intersection considered contains only the identity automorphism. Thus: if a normal extension K of the field P is the composition of the extensions K1 and K2, then

G(K, K1) n G(K, K2) = E.

(2)

Problem. To prove the converse assertion, i.e. to prove that if the normal field K contains the subfields K1 and K2, satisfying condition (2), then K is the composition of the fields K1 and K2. Now we suppose that the field K1 is normal over the field P. Then

its Galois group G(K1, P) is a homomorphic image of the Galois group G(K, P), and the kernel of the corresponding epimorphism is

the group G(K, K1) (see para. 6). From formula (2) it follows immediately that this epimorphism, when restricted to the subgroup G(K, K2), is a monomorphism. In other words, the group G(K, K2) is isomorphic to some subgroup of the group G(K1, P). Thus: if the normal extension K of the field P is the composition of the normal extension K1 and the (generally speaking, arbitrary) extension K2, then the Galois group G(K, K2) is isomorphic to some subgroup of the group G(K1, P).

II. THE SOLUTION OF EQUATIONS BY RADICALS

CHAPTER 1

ADDITIONAL FACTS FROM THE GENERAL

THEORY OF GROUPS 1. A generalization of the homomorphism theorem

Let 0 : G --). G' be an arbitrary homomorphism and H' some subgroup of the group G'. We will consider the set H of all elements

w^o

of the group G mapped by the homomorphism ¢ into elements of the subgroup H' : g e H if and only if 4)(g) e H'. It is obvious that the subset H of the group G is a subgroup of it. This subgroup is denoted by 4-1(H') and is called the complete inverse image of the subgroup H' under the homomorphism 0. In this terminology, the kernel of the homomorphism 0 is none other than the complete inverse image of the identity e' of the group G'. It is easy to see that the complete inverse image H = 0 - '(H') of a normal divisor H' of the group G' is a normal divisor of the group G. Because 4)(H) c H', then the induced homomorphism is defined : G/H -* G'/H'.

'ZS

It is easy to verify (see Ch. 2, para. 3, of I for the case H' = e') that the homomorphism is a monomorphism. Thus: for any homomorphism 0 : G - G' and any normal divisor H' c G' the induced homomorphism G/H --* G'/H',

where H = 4-1(H'), is a monomorphism; if the mapping 0 is epimorphic, then the mapping

is isomorphic.

If H' = e', then this assertion is reduced to the homomorphism

ono

theorem proved earlier. Problem. To prove that (1) the subgroup H of the group G is the complete inverse image of some subgroup of the group G' in the homomorphism ¢ : G -+ G' if and only if

Ker 0 c H; 45

46

FOUNDATIONS OF GALOIS THEORY

(2) if Ker 0 c H1 and Ker 0 c H2, then ci(H1) = O(H2) if and only if H1 =H2; (3) if the homomorphism 0 is an epimorphism, then 4 -1(Hi) _ 0-1(H2) if and only if Hi = H. To deduce from this that the epimorphism 4) : G -+ G' defines a one-one correspondence between the set of all subgroups of the group G' and the set of those subgroups of the group G which contain the kernel of the epimorphism 0. 2. Normal series

Let G be an arbitrary group and G1, G2 subgroups of it, of which the second is a subgroup of the first: G1

G2-

A chain of subgroups, each contained in the preceding one,

G1=Hoz) Hl=) ... z) H,-1=) Hi=) ... CHs=G2i

(1)

.fl

beginning with the subgroup G1 and ending with the subgroup G2, is called a normal series if for any i = 1, ... , s the subgroup Hi is a normal divisor of the subgroup H,_1 (the subgroup Hi need not also L."

.fl

be a normal divisor of the whole group G). The corresponding factor groups H,_1/H, are called the factors of the normal series (1). We emphasize that, generally speaking, we do not require that the normal series (1) should not contain repetitions: it can easily be that

for some i the subgroup H, coincides with the subgroup H,_1. However, if one desires, one can remove from a normal series all the repeated groups. Normal series

G=Ho=) H1=) ... =) H,-1=) Hi=) ... CHs=e,

(2)

beginning with the group G and ending with the identity subgroup e, have a special significance. Such normal series we will call normal series of the group G. It is obvious that, if the group G is finite, then for any normal series (2) all the factors Hi-1/H, are also finite and n = n1n2 ... n3,

(3)

where n is the order of the group G, and n,, i = 1, ... , s, is the order of the group H,_1/H,. Conversely, if the group G possesses a normal series with finite factors, then the group G itself is also finite and its

ADDITIONAL FACTS FROM THE THEORY OF GROUPS

47

order n is expressed in terms of the orders n1, ... , ns of the factors of the normal series by formula (3). Now let 0 : G -+ G' be an arbitrary homomorphism. It is obvious that if the subgroup H1 of the group G is contained in the subgroup H2: H1 = H2, then the subgroup cb(H1) of the group G' is contained in the subgroup O(H2)

0(H1) _ O(H2).

Moreover, if the subgroup H, is a normal divisor of the subgroup H2, then the subgroup cb(H1) is a normal divisor of the subgroup 4)(H2) (prove it!). Hence, for any normal series

... =) H,-1z) Hiz) ... z) H5=G2

G1=Ho=H1

(4)

the chain ...

q5(G1) = O(Ho) _ O(H1)

4)(H,-1) _ qS(H1)

...

... - N(HS) = 4)(G2) (5)

is a normal series. If, in particular, G1 = G and G2 = e (i.e. if the series (4) is a normal series of the group G), and the mapping 0 is epimorphic, then O(G1) = G' and 4)(G2) = e' (i.e. the series (5) will be a normal series of the group G'). Thus: an arbitrary epimorphism 4) : G -+ G' carries any normal series

G=Ho=) HI=) ...

H1-1H1

...CHs=e

(6)

of the group G into a normal series G' = O(Ho)

O(H1)

...

O(Hi-1) = O(H,)

...

C(Hs) = e' (7)

of the group G'.

We note that for any i = 1, ... , s the epimorphism ¢ induces an epimorphism

H,-1/Hj - 0(Hi-1)l0(Hj) (because the conditions under which the induced homomorphism is defined are clearly satisfied here). Hence: the factors of the series (7) are homomorphic images of the factors of the series (6).

48

FOUNDATIONS OF GALOIS THEORY

Again let 0 : G -+ G' be an arbitrary homomorphism. It is obvious that if the subgroup Hi of the group G' is contained in the subgroup H?: H1C

z,

then the subgroup 4-1(H2) of the group G is contained in the subgroup 0-1(H2):

0-1(Hi) c 0-1(H2 Moreover, if the subgroup Hl is a normal divisor of the subgroup HZ, then the subgroup 0-1(Hi) is a normal divisor of the subgroup 0-1(H2) (prove it!). Hence, for any normal series (8)

the chain

0-1(G1) = 4-1(H'0) - 0-1(Hi) ... - q-1(H!-1)

...

-

4-1(HS) = 0-1(G2)

(9)

'+.

is also a normal series. If, in particular, Gi = G' and G? = e' (i.e. if the series (8) is a normal series of the group G'), and the mapping 0 is monomorphic, then 0-1(G'1) = G and 0-1(G') = e (i.e. the series (9) will be a normal series of the group G). Thus: a monomorphism 0 : G -* G' associates with a normal series

G'=Ho=) Hi=) ... =) H;-1=) H; ... CHs=e'

(10)

of the group G' the normal series

G = c-'(HO')

O-1(Hi)

...

...

_ 4-1(H'-1)-1(Hi)

...-1(H;)

= e (11)

of the group G.

We note that for any i = 1, ... , s the monomorphism ¢ induces a monomorphism Hence: the factors of the series (11) are isomorphic to subgroups of the

factors of the series (10). Now let G be an arbitrary group and

G=Ho=) H1

... =Hi-1=) Hj

...

H,a=e

(12)

49

ADDITIONAL FACTS FROM THE THEORY OF GROUPS

one of its normal series. We suppose that for each i = 1, ... , s a normal series for the corresponding factor group Hi-1/Hi is given: Kit

n

Hi-1/Hi = Kio

...

Ki, We will consider the natural epimorphism Ki.i - '

... zD Kit, = e.

(13)

Oi : Hi- 1 -+ H1- IIHi

(defined by the formula qi(h) = Hih). This epimorphism associates with the series (13) the normal series

1^r

Hi-1 = 0i 1(Kio) _ q5i 1(Ki1) - ... 0-1(Ki;-') _ (p-'(Ki,) ...

Hi. (14)

...

Inserting, for each i = 1, ... , s, the series (14) between the terms Hi-1 and Hi of the series (12), clearly we again obtain a normal series of the group G. This normal series is called the refinement of the series (12) by means of the series (13). Any factor of this series has the form

Oi 1(Ki;-1)/4'i 1(Ki;). According to the general homomorphism theorem (see para. 1) this factor group is isomorphic to the factor group Ki, - I /Kij. +..

Thus, the factors of the refined series are isomorphic to the factors of the refining series (13).

Because any non-simple group possesses non-trivial normal series (i.e. containing non-trivial subgroups), then any normal series, having

^U'

..'

vii

at least one non-simple factor, possesses non-trivial refinements (i.e. not reducible to repetitions). On the contrary, if all the factors of a normal series are simple groups, then all refinements of this cad

normal series reduce to repetitions. (7)

..fl

In conclusion we draw attention to the parallelism between the theorems proved in this paragraph, relating to epimorphisms, and the theorems relating to monomorphisms. We cannot here explain the very deep foundations of this parallelism. 4-.

3. Cyclic groups

A group G is called cyclic if all its elements are powers of a fixed element go. This element go is called a generator of the cyclic group G. Any cyclic group, clearly, is Abelian.

50

FOUNDATIONS OF GALOIS THEORY

An example of a cyclic group is the group of whole numbers with

respect to addition. This group we will denote by the symbol Z It is generated by the number 1 (and also the number - 1). Another rte'

e=.

cyclic group is the group consisting of only one element (the identity). In an arbitrary group G the powers g° of any element g constitute te,

a cyclic subgroup with generator g. The order of this subgroup, clearly, coincides with the order of the element g. Hence by virtue of Lagrange's theorem it follows that the order of any element of a to'

obi

''a

O0'

group divides the order of the group (we note that all the elements of a finite group are elements of finite order). Therefore for any element g of a finite group of order n the equation 9°=e

r''

r7'

"C3

holds. This simple remark is often useful. We note further that a finite group G of order n is a cyclic group if' and only if it possesses an element of order n. This element is a generator.

co-

s1.

.1.'

't3

In fact, if the group G is cyclic and go is a generator, then the order of the element go is equal to n. Conversely, if the group G possesses an element of order n, then among the powers of this element there are n distinct ones, and therefore these powers exhaust

O''

"r1

r-'

.'S

the whole group G. We see thus that a cyclic group can have several distinct generators (namely, any element of order n is a generator). Problem. To prove that any group of prime order is a cyclic group. Let G be a cyclic group with generator go and H a subgroup of it. Because any element of the subgroup H is an element of the group G, r-'

'.3

.ti

.-!

then it can be represented in the form go, where d is a positive or negative whole number (generally speaking, not uniquely defined). C3'

We consider the set of all positive numbers d, for which the element go belongs to the subgroup H. Because this set is non-empty (why?t),

Cdr

.'s

L:.

then in it there exists a smallest number do. It will be shown that any element h of the subgroup H is a power of the element go °. In fact, by definition, there exists a number d such that h = go (the number d may also be negative). We divide (with remainder) the number d by the number do :

d=doq+r,

0
t Editor's footnote: The set is empty if G is infinite and H = e, but the conclusion is obvious in this trivial exceptional case.

51

ADDITIONAL FACTS FROM THE THEORY OF GROUPS (11

Because go = go (go )_9 e H, then by virtue of the minimal nature per

of the number do the remainder r must be equal to zero. Thus, d = doq and h = (go°)9. Thus it has been proved that the element g00 is a generator of the group H, i.e. that the group H is cyclic. Thus : a subgroup of a cyclic group is cyclic.

Problem. To prove that the number do is equal to the index of the subgroup H. We note further that any factor group G/H (and, hence, any homomorphic image) of a cyclic group G is cyclic.

For the proof it is sufficient to note that the residue class Hgo, containing the generator go of the group G, serves as a generator of the group G/H.

In particular, any factor group of the group of whole numbers Z is a cyclic group. We will study these cyclic groups in more detail.

P'+

Because the group Z is Abelian, then any subgroup H of it is a normal divisor. On the other hand, according to what was proved above, the subgroup H is a cyclic group. As the factor groups with respect to the trivial subgroups are well known to us we can assume that the subgroup His non-trivial. Let the number n be a generator of the subgroup H. We can regard this number as positive (why?) and, hence, greater than 1. The subgroup H consists, clearly, of all whole numbers divisible by n. Therefore two numbers belong to the same coset of the subgroup H if and only if their difference is divisible by n, i.e. if 't7

co"

'off

a)-

'L3

they are congruent modulo n (see The Course, p. 29). Thus the cosets

a4.

(or

h,,

"t3'

with respect to the subgroup H are none other than the residue classes of numbers congruent to each other modulo n. In other

co"

s,'

(~D

(1q

1:4

words, the factor group of the group Z with respect to the subgroup His the group (with respect to addition) of residue classes of numbers congruent to each other modulo n. We will denote this group by Z,,. Its generator is the residue class containing the number 1. It will be shown that any cyclic group G e is isomorphic either to the group Z (if it is infinite), or to one of the groups Z (if its order is finite).

In fact, let go be a generator of the group G. We define a mapping 0 of the group Z into the group G, setting 4(a) = go,

a e Z.

52

FOUNDATIONS OF GALOIS THEORY

From the rules of operations with powers it follows that for any a, b e Z: q5(a + b) = &)O(b),

i.e. that the mapping 0 is homomorphic. Its image 4(Z) consists of all elements of the group G which are powers of the element go, i.e. it coincides with the group G. Thus the group G is a homomorphic image of the group Z and, hence, is isomorphic to some factor group of it, i.e. either to the group Z itself or one of the groups Zn (we suppose that the group G does not consist only of the identity). Which of the groups Z. is isomorphic to the group G is uniquely determined by the fact that isomorphic groups must have identical order. Thus the structure of cyclic groups is completely described. An important example of cyclic groups is obtained on the basis of the following considerations. Let n be an arbitrary positive whole number. As is well known (see The Course, p. 55), there exist exactly n distinct complex numbers ao = 1, a1, ... , an-1

which are n-th roots of unity:

a"=l, i=0,1,...,n-1. v,'

The product of any two n-th roots of unity, clearly, is an n-th root of unity, and the number inverse to an n-th root of unity will also be an n-th root of unity. Hence, the set of all n-th roots of unity is a group (of order n). As is shown in The Course, p. 57, any n-th root of unity is a power

of a so-called primitive root. Hence, the group of all n-th roots of unity is a cyclic group of order n. The primitive roots (and only they) serve as generators of it.

Problem. To construct an isomorphic mapping of the group of n-th roots of unity onto the group Zn. 4. Solvable and Abelian groups

The normal series

HL-1:n Hin...=) H., =e

(1)

of the group G is called a solvable series if for each i = 1, ... , s the factor group Hi-1/Hi is a cyclic group. A group possessing at least

53 CS'

("1

ADDITIONAL FACTS FROM THE THEORY OF GROUPS P'+

one solvable series is called solvable; a group not having any solvable series is called non-solvable. As an example of a solvable group one .,1

o-1

.+,

two

...

`L3

"'1

o-1

can clearly take any cyclic group (a solvable series consists of the group G and the identity subgroup e). A non-solvable group is, for example, any simple group, provided it is not cyclic (i.e. if its order is not a prime number). As we know, any epimorphism 0: G -* G' carries the normal series (1) of the group G into a normal series of the group G', whose factors are homomorphic images of the factors of the series (1). Because a homomorphic image of a cyclic group is a cyclic group, then, consequently, any epimorphism 0: G -- G' carries a solvable series of the group G into a solvable series of the group G'. Thus, a homomorphic image of a solvable group is solvable.

Let H be an arbitrary subgroup of the solvable group G. We define a mapping 0 of the group H into the group G, setting

&)=h, heH

C3.

`-'

o-+

,.y

o-1

,-.

CD'

P'+

Cep

80.

(the mapping 0 is the identity mapping of the group H, considered as a mapping into the larger group G). The mapping 0 is, clearly, a monomorphism. Hence, to the normal series (1) of the group G there corresponds in the subgroup H some normal series (consisting of the subgroups 4-'(Hi) = Hi n H), whose factors are isomorphic to subgroups of factors of the series (1). Because any subgroup of a cyclic group is a cyclic group, then from this it follows that 'CS

a subgroup of a solvable group is solvable. From the proof given it follows that if a solvable group possesses

oil

''S'

''Y'

.'mar

a solvable series of length s (i.e. a series consisting of s - 1 terms), then any subgroup of it also possesses a solvable series of length s (we recall that we allow solvable series with repetitions). From the two theorems proved it follows immediately that all the 'S'

factors of any normal series of a solvable group are solvable groups. It will be shown that the converse assertion is also true:

a group G possessing a normal series with solvable factors is a solvable group. .-!

..ti

In fact, let

Hs=e

G=Ho=)

(1)

C1.

be a normal series of the group G with solvable factors H;_1/H;. °'Y

By definition, the factor group Hi_1/Hi for any i = 1, ... , s possesses a solvable series : Hj- 1/Hj = Kjo

Kt 1

...

Kji -1

Kt;

...

Ki,, = e'.

(2)

54

FOUNDATIONS OF GALOIS THEORY

.T'

0A"

We refine the series (1) by means of the series (2). As we know, the factors of the refined series are isomorphic to the factors of the series (2) and, hence, are cyclic groups. In other words, the refined series is solvable. Thus the group G possesses a solvable series, i.e. it is a solvable group. A special case of the theorem proved is the following assertion: if the group G possesses a solvable normal divisor H, with respect to which the factor group G/H is solvable, then the group G itself is also solvable. L".

In fact, the condition imposed on the group G means that it 'C3

possesses a normal series

Gz) Hz) e

with solvable factors. Hence, by what has just been proved, the group G is solvable. R-0

a.0

This proposition permits one to prove the following assertion, essentially extending the number of examples of solvable groups known to us: any finite Abelian group G is solvable.

We carry out the proof by means of induction with respect to the order n of the group G. If n = 1, then the group G consists only of the identity e and, hence, is solvable. We suppose that the solvability of any finite Abelian group having order less than n has already been proved, and we consider an arbitrary Abelian group G of order n.

'CS

'"!

'.3

0'"

Let g be an arbitrary element of the group G distinct from the identity. Because g 0 e, then the cyclic subgroup H of the group G with generator g has order greater than one, and hence the order of the factor group G/H is less than n (the factor group can be constructed because in an Abelian group any subgroup is a normal coo

pi'

divisor). Because any factor group of an Abelian group is an Abelian group (prove it!), then consequently by the principle of

r"3

CD'

'"'!'

.fl

'CS

)-+

.'sue'

induction the group G/H is solvable. Thus in the group G there is a solvable (and in fact cyclic) normal divisor H, the factor group G/H with respect to which is solvable. Hence, by the theorem proved above, the group G is solvable. At the end of this paragraph we mention two examples of groups which will play a fundamental part in what follows. Let n be an arbitrary positive whole number. We consider the Cam'

set Z of all residue classes of numbers congruent to each other modulo n. The class containing the number a we will denote by [a]

55

ADDITIONAL FACTS FROM THE THEORY OF GROUPS

(in The Course, p. 29, this class is denoted by Ca). In the set Zn, besides addition (with respect to which it is, as we know, a cyclic group), one can also define a multiplication. As with addition, multiplication of classes of numbers congruent to each other is defined by the rule : [a][b] = [ab].

1N.

This multiplication, clearly, is commutative and associative. Moreover, it possesses an identity (namely, the identity of this multiplication is the class [1] containing the number 1). However, with respect to this multiplication the set Zn is not a group, because, for instance, the null class [0] (consisting of the numbers divisible by n) does not have an inverse. We will clarify which classes have inverses. Let [a] be an arbitrary class modulo n, for which there exists an inverse class, i.e. a class [b] such that [a][b] = [1].

a..

Then the number ab - 1 is divisible by n, i.e. there exists a whole number k such that ab + kn = 1. .t]

It is obvious that this equation is possible only when the numbers a and b are relatively prime to the number n. Thus if for the class [a] there exists an inverse class, then the number a is relatively prime to the number n. Now, conversely, let a be relatively prime to the number n. We consider all numbers which can be represented in the form

ax+ny, where x and y are arbitrary whole numbers (positive or negative). Let d be the smallest of all the positive numbers of such a form: (3) d = au + nv. We divide (with remainder) the number a by the number d:

a=dq+r, 0sr
(4)

From formulae (3), (4) it follows that

r=a(l-qu)+n(-qv). Hence by virtue of the minimal nature of the number d it follows that

r = 0, i.e. that a is divisible by d. Analogously it is proved that n is divisible by d. Hence d = 1, because the numbers a and n, by

56

FOUNDATIONS OF GALOIS THEORY

hypothesis, are relatively prime. Thus there exist whole numbers u and v such that

au+nv=1. r,.

Going over to residue classes and making use of the fact that [nv] = [0], we obtain: [a][u] = [1]

Thus the class [u] is inverse to the class [a]. Thus we have proved

gin'

.-'

that the class [a] has an inverse if and only if the number a is relatively prime to the number n. From this theorem it follows immediately that the set of all residue classes modulo n, consisting of numbers relatively prime to n, is a group with respect to multiplication (clearly, Abelian). This group is denoted by Z,, and is called the multiplicative group of residue classes modulo n. Its order is equal to the number fi(n) of positive numbers Ca.

less than n and relatively prime to n. One can show that if, for instance, the number n is divisible by two odd prime numbers, then the group Z is not cyclic. We do not need this fact, and we leave it without proof.

r..

.-!'

Now let n again be an arbitrary positive whole number. We consider the set of all pairs of the form (a, b), where a and b are whole numbers, and where the number a is relatively prime to the number n. We separate this set into classes, placing in one class the pairs (a, b) and (a,, b1) if and only if the number a is congruent to the number a1 modulo n, and the number b is congruent to the number b, modulo n. The class containing the pair (a, b) we will Iv.

denote by [a, b], and the set consisting of all the classes by M. We define in the set M an algebraic operation (" multiplication ") by setting [a, b][c, d] = [ac, be + d].

It is easy to verify that this formula in fact defines in the set M a single-valued operation, i.e. if [a, b] = [a,, b,] and [c, d] = [c,, d,], then [ac, be + d] = [a1c1, b,c, + d1]. Problem. To prove that with respect to the multiplication so defined the set M. is a group. The identity of this group is the class [1, 0], and the inverse element is defined by the formula

[a, b] -1 = [a, - ab], where a is a number such that

[a][a] = [1]

(in the group

ADDITIONAL FACTS FROM THE THEORY OF GROUPS

The group

57

as it is easy to see, is not Abelian (if n > 2). For

example,

[1, 2][a, 0]

[a, 2a],

[a, 0] [1, 2] _ [a, 2].

Problem. To prove that the order of the group M. is equal to n4(n) (where cb(n) is the number of positive numbers less than n and relatively prime to n). acs

From the definition of the group M it follows immediately that the mapping 0 : MM -* defined (as it is easy to see, uniquely) by the formula

0[a, b] = [a], is a homomorphism. This mapping, clearly, is an epimorphism, and therefore the group Z;, is isomorphic to the factor group

M of the mapping Vi. The kernel N consists, clearly, of the elements of the form [1, b].

Because

[1, b][1, b1] _ [1, b + b1], then, mapping the class [1, b] onto the residue class [b], we clearly obtain an isomorphism of the group N onto the group Z. Hence, the kernel N is a cyclic normal divisor of the group M. The factor group of the group M. by the kernel N is isomorphic, as we have seen, to the group Z and therefore is Abelian and, hence, a solvable group.

Thus the group M possesses a solvable (and in fact cyclic) normal divisor, the factor group with respect to which is solvable (and in fact Abelian). Hence the group M is also solvable.

CHAPTER 2

EQUATIONS SOLVABLE BY RADICALS 1. Simple radical extensions

P'+

'-r

A simple radical extension of the field P is the decomposition field K of a binomial equation of the form x" -a=O, where ac-P, a & 0. (1) As is well known (see The Course, p. 56) all the roots of equation (1) are obtained from one of them by multiplication by an n-th root of

unity. On the other hand, any n-th root of unity is a power of a primitive root. Thus if 0 is an arbitrary root of equation (1), and C is a primitive n-th root of unity, then the numbers (2)

0 = 0C°,

exhaust all the roots of equation (1). Hence, the field P(C, 0) contains all the roots of equation (1) and therefore K c P(C, 0).

On the other hand, the field K contains the number 0 and OC and therefore contains the numbers 0 and C = OC/0. Hence, P(C, 0) c K. Thus,

K=P(C,0). It may happen that the field P already contains the root C. In this case the simple radical extension K has the form P(0). At the other extreme arises the case when a = 1. In this case 0 = Ck, where k is a whole number and, hence, K = P(C) (we note that equation (1) is not supposed irreducible). Being a (decomposition) field, the field K = P(C, 0) is normal and therefore we can speak of its Galois group G(K, P). Let S be an arbitrary automorphism in the Galois group G(K, P).

Because the number C is a root of the polynomial x" - 1, then the number CS will also be a root of this polynomial (see part I, Ch. 3, para. 2). Hence, a number a can be found such that CS = C C. 58

(3)

59

EQUATIONS SOLVABLE BY RADICALS

-.)

If the number Ca were a root of the polynomial x' - 1, where m < n, then the number C = 1 would also be a root of the polynomial xm - 1, i.e. it would be an m-th root of unity with m < n. But this is impossible because, by hypothesis, the number C is a primitive n-th root of unity. Hence, the number Ca cannot be an m-th root of unity with m less than n, i.e. it is a primitive root. Therefore the number a is relatively prime to the number n (see The Course, p. 57). coo

1-'

Further, because the number 0 is a root of equation (1), then the number 9's will also be a root of equation (1), i.e. it is possible to find a number b such that OS

(4)

= Sb9.

^v''

.0+,

Thus, to every automorphism S e G(K, P) corresponds a pair of numbers a and b, where the number a is relatively prime to n. This correspondence is not unique because, for instance, the pair (al, b1), where the numbers al and bl differ from the numbers a and b by numbers which are multiples of n, also corresponds to the same automorphism S. We will consider this question in more detail.

Let the pairs (a, b) and (al, b,) correspond to the same automorphism S, i.e. SS

=

ya

C"

oss = Cbbe,

SS = Cal

>

o = rle. gqbStile

In other words, Sa=ya%

Ca-al = 1,

b

Cb =

Cb-b1 = 1. b

Because C is a primitive root of unity, then this equation is possible

if and only if the differences a - ai and b - bl are divisible by n, i.e. if the numbers a and b are congruent modulo n to the numbers al and bl respectively. In other words, the pairs (a, b) and (a1, bl) correspond to the same automorphism S e G(K, P) if and only if these pairs belong to the same class in the sense of Ch. 1, para. 4, i.e. if they define the same element [a, b] of the group Mn. Thus, if

with every automorphism S from the Galois group G(K, P) we cad

associate the element [a, b] of the group M,,, where the numbers a and b are defined from formulae (3) and (4), then we obtain a singlevalued mapping of the group G(K, P) into the group M. We will denote this mapping by the letter 0:

cad

O(S) = [a, b].

60

FOUNDATIONS OF GALOIS THEORY

It will be shown that the mapping 0 is a homomorphism. In fact, if 4)(S) = [a, b] and 4)(T)= [c, d], i.e. CS

= Ca,

ST =Cc,

BS

yy

= bbe, 0T=CdO,

then

yy(ST = (ya)T =

= (ybO)T =

(CT)bOT = ybcyde = Cbc+de arc

OST

(ST)a = yac,

Thus, S(ST) = [ac, be + d], cb(ST) = O(S)O(T).

We will find the kernel of the homomorphism 0. If the automorphism S e G(K, P) belongs to the kernel of the homomorphism 4), then Cs=C, 0s=0' t3,

,..,

.'3''

.r1

i.e. the automorphism S leaves invariant the elements C and 0. Hence, the automorphism S also leaves invariant any polynomial (with coefficients from P) in the elements C and 0. Therefore, since any element of the field K = P(C, 0) can be expressed in the form of a CYO

..O

polynomial in C and 0, the automorphism S leaves invariant any element of the field K, i.e. S = E. Thus, the kernel of the homomorphism 0 consists only of the identity automorphism E, i.e. ¢ is a monomorphism. In other words, ¢ is an isomorphic mapping of the group G(K, P) onto a subgroup of the group M. Because the group M (and therefore also any subgroup of it) is solvable, then it follows from this that .'"

fl.

"S+

the Galois group of a simple radical extension is a solvable group. -co

Cad

a4'

Problem. To prove that if C e P, then the Galois group of the simple radical extension K = P(C, 0) = P(0) is a cyclic group (the order of which divides the number n). 2. Cyclic extensions

OL.

A normal extension K of the field P is called a cyclic extension if its Galois group G(K, P) is cyclic. As an example of a cyclic extension one can take a simple radical extension defined by a binomial equation of degree n, with the condition that the fundamental field P contains a primitive n-th root of unity (see the problem at the end of ;11

61

EQUATIONS SOLVABLE BY RADICALS

`C3

para. 1). The degree m of this equation, generally speaking, is less than n. The equality m = n holds if and only if, equation (1) from para. 1, defining the given simple radical equation, is irreducible.

The purpose of this paragraph is the proof of the following " converse " assertion: If the field P contains a primitive n-th root of unity, then any cyclic extension K of. it of degree n is a simple radical extension, which is defined by an irreducible binomial equation of degree n.

,~a)

Before the proof of this assertion we will set down several preliminary remarks. Let C be a primitive n-th root of unity, and let S be a generator of the group G(K, P) (this group, by hypothesis, is cyclic, i.e. its ele-

ments are exhausted by the powers S° = E, S1, ... , S"). With any element a of the field K we associate an element (SP, a) = a + bpoS + 62QaS2

+ ... +

C(n-i)PcSn_1,

i.e. an element n-1 &-n

(SP, a) = Z SkPa k, k=0

where p is some whole number. The element (p', a) we will call the Lagrange resolvent of the element at, corresponding to the number p.

To begin with we consider the resolvent (S,a)=a+Cccs+S2cS2-{-

... + cn-1aS"

corresponding to the number 1. Because a e K, then P(a) c K.

Hence,

in

the

Galois

correspondence, to the field P(a) corresponds some subgroup H of the group G(K, P) (namely, H = G (K, P(a)). Being a subgroup of a cyclic group, the group H is cyclic. As its generator one can take the element Sd, where d is the index of the subgroup H. Because Sd C H = G(K, P(a)), then acd = a and, hence, cS`d+J = as' for any

i and j. Therefore, denoting by m = n/d the order of the subgroup H, we see that (y

S, a)

n-i

= E y kcSk k=0

m[-1 d-1

= i=0 L j=0 E yid+jaSid+J M-1 d-1

EE

i=0 j=0

Cid+jcSJ =d-1E yjcSJm-1 E S

j=0

S

i=0

b

id.

FOUNDATIONS OF GALOIS THEORY

62

But if d

1, and therefore

n, then Cd

1 - Smd

I'M

m-1

`y C1 = 1 + Cd + ... +

C(m-1)d

1 - rn

= 1 - Cd

1 - yd

O,

~O.

because to = 1. Thus if d 0 n, then (C, a) = 0. Hence, if (C, a)

0, cad

then d = n. On the other hand, the equation d = n means that H = E, i.e. that P(a) = K. Thus it has been proved that if (C, a)

then

0,

P(a) = K.

...

off

In connexion with this assertion the question of the existence of

..'

>v"

elements a e K, such that (C, a) 0, naturally arises. (It is necessary to bear in mind that it need not follow from P(a) = K that ((, a) 0.)

b08= 1.,

For the answer to this question we make use of theorem I Ch. 1, para. 7. According to this theorem in the field K there exists an 1.,

element 0 such that rev

K = P(0).

Because [K: P] = n, then 0 is a root of an irreducible equation of .=i

degree n. We will show that at least one of the resolvents (C, 0),

0n-1)

(C, 02), ... , (C,

(1)

is distinct from zero.

In fact, if all the resolvents (1) are equal to zero, i.e. if Yn-10s"-1

0 + COs + ... + (02)S 02 + C (02)S

= 0,

+ ... + Cn-1(02)S"-1 = 0,

0n-1 + ((en-1)S + ... + Cn-1(Bn-1)S"-1 = 0,

then, since (C, 1)

= 1 +C+ ...

+Cn-1 =0,

the determinant

0

0s

... ...

02

(82)S

...

1

0n-1

1 0S"-1

(02)S"-1 ...

...

1

(On - )S

...

(on-1)S"-1

63

EQUATIONS SOLVABLE BY RADICALS

is equal to zero (its columns are linearly dependent). On the other hand, because the mapping S is an automorphism, then (0`)s' = (0s')` for any i and j. Hence, the determinant written above can be expressed in the following form: 1

1

0 02

Bs

...

1

Bs"-1

(0- 1)2

(05)2

(2)

on-1

(OS)n-1

I

(Os'-) .-I I

...

The determinant obtained is the Vandermonde determinant of the Bs"-' elements 0, OS, ... , (see The Course, p. 92-93). As is well known, it is equal to the product of all possible differences of these elements. But we know that amongst these elements none are Q..

identical (because if 0s' = 0s', then S` = Si; see part I, Ch. 3, para. 3). Therefore the determinant (2) is distinct from zero. However, it was proved above that it is equal to zero. This contradiction proves that not all the resolvents (1) can be identically equal to zero. Comparing the last two assertions, we obtain

in the field K there exists an element a such that (C, a) 0 0, and hence, K = P(a).

Now we consider the resolvent (c', a), corresponding to an arbitrary whole number p. Applying to the resolvent C("-1)pas"-1

(Cp, a) = a + Spa- + C2pas2 + ... +

the automorphism S and making use of the fact that e P), we obtain

Cs

= C (because

+ C(n-1)pa (Cp, a)S = aS + yCpas2 + ... + C(n-2)paS"'1 b

(because S" = E), i.e.

S

(Cp,

a)S

= S-p(bp, a)

(3)

(because Cn -1 = C -1). In particular, (C,

a)S

= C-1(C, a)

0

Raising this equation to the p-th power and taking into account that S is an automorphism, we obtain yy

((b , a)p)S = ay S-p(b, a)p

(4)

64

FOUNDATIONS OF GALOIS THEORY

Dividing equation (3) by equation (4) (on the hypothesis that ((, a) # 0), we obtain a) s _ (p, a) cc)p,

a)p)

(C,

i.e. the automorphism S leaves invariant the number (gyp, a)

C P

(S, a)p.

It is clear that any power of the automorphism S also leaves the number cp invariant. In other words, any element of the group G(K, P) (i.e. any automorphism of the field K over the field P) leaves the number cp invariant (because the whole group G(K, P) is exhausted, Off

by definition, by the powers of the automorphism S), and hence cp e P (see part I, Ch. 3, para. 4). Thus it has been proved that for CAD

any p there exists an element cp in the field P such that (Cr, a) = cp((, a)p.

Hence, all the elements (Cp, a) belong to the field P(a), where a = (C, a). Now we will find the sum of all the Lagrange resolvents (Cp, a) for

p= 0,1,...,n- 1. We have: n-1

y

n--`I n-1

(Ce, a) = Ll E p=0 p=0k=0

y

n-1

n-1

y

E ask p=0 E C" k=0

(5)

But if Ck 0 1, i.e. if k 0 0, then n-1 P=O

rkn-1

y

bkp=

Ck - 1

=0

because Cn = 1. Thus, in the sum (3) only the terms corresponding to k = 0 are distinct from zero, i.e. n-1

P=O

(C", a) = na.

Since (Cp, a) e P(a), it follows from this formula that a e P(a), so that P(a) c P(a), i.e.

K c P(a). Because, on the other hand, P(a) c K (because a e K), then

K = P(a).

EQUATIONS SOLVABLE BY RADICALS

65

Finally, setting p = n in formula (4) and taking into account that C° = 1, we obtain (an)s =

a"'

whence it follows (see above the analogous reasoning for the number

ep) that a" e P. Thus, letting c = a", we see that the number a is a root of the binomial irreducible (why?) equation

x" - c = 0, where c e P. Thus it has been proved that the field K is a simple radical extension, i.e. the assertion formulated at the beginning of this paragraph has been proved.

Problem. To prove that if the field P contains a primitive n-th root of unity then the binomial equation

x" - c = 0, where c e P, is reducible if and only if there exists a divisor m (distinct from 1) of the number n and an element c e P such that c = cm. The structure of cyclic extensions in the general case (when the fundamental field P does not contain the necessary roots of unity) will be studied in para. 4. 3. Radical extensions

An extension K of the fundamental field P is called a radical extension if there exists a chain

P=Lo aL1c ... aLj_1cLjc ... cLs=K

(1)

o''

Coo

>,,a

"''

021

of subfields of the field K embedded each in the next, beginning with the field P and ending with the field K, such that for any i = 1, ... , s the field Lj is a simple radical extension of the field

CAD

.-'

Li_1. The chain (1) here is called a radical series. We emphasize that a radical extension can possess many different radical series. In spite of the fact that in the radical series (1) each field Li is a normal extension of the field L1_1, the whole field K need not be a normal extension of the field P. This is related to the fact that, generally speaking, a normal extension of a normal extension is not a normal extension of the fundamental field. The necessary and sufficient condition for a normal extension of a normal extension to be a normal extension of the fundamental field is indicated in the following lemma.

66

FOUNDATIONS OF GALOIS THEORY

Lemma. Let P be an arbitrary field, L a normal extension of it and K a normal extension of the field L. It turns out that the field K is a normal extension of the field P if and only if there exists a polynomial over the field P whose decomposition field over the field L is the field K. In fact, if the field K is normal over the field P, then there exists a polynomial f (x) with coefficients in the field P such that

K = P(a1, ... , a.), where ai, ... , a are the roots of the polynomial f (x). Then

(because P c L), but on the other hand,

L(ai,...,ajc:K (because L c K and ai, ... , a a K). Hence,

K=L(a1,...,an), i.e. K is the decomposition field of the polynomial f (x) over the field L.

(We note that we have not made use of the fact that the field L is normal in this reasoning.) Conversely, let

where ai, ... , a are the roots of some polynomial f (x) over the field P. Because the field L, by hypothesis, is normal over P, then there exists a polynomial g(x) with coefficients in the field P such that

L=P(/31,...,Ym), where fl , ... , /3m are the roots of the polynomial g(x). Then

;)

K = P(/31, ... , Ym, a ... , (2) Because the numbers Nl, ... , /3,,,, ai, ... , an exhaust the roots of the polynomial g(x) f (x), then equation (2) means that the field K

is the decomposition field of the polynomial &)f (x) with coefficients in the field P and, hence, is a normal extension of the field P. Thus the lemma is completely proved. We will call the field K a normal radical extension of the field P if it is both a normal and a radical extension of this field. The relation

67

EQUATIONS SOLVABLE BY RADICALS

ACS

r.,

vii

between normal radical extensions and arbitrary radical extensions is set out in the following theorem: any radical extension K of a field P is contained in some normal radical extension K. We will prove this theorem by induction with respect to the length s of the radical series (1) which is associated with the radical extension K. Ifs = 1, then K is a simple radical, and therefore also a normal, extension of the field P. Therefore in this case for the field K one can take the field K itself. Now supposing that the theorem has already been proved for all

radical extensions possessing radical series of length s - 1, we consider the radical extension K with the radical series (1) of length s. Because the field L = LS_1 is a radical extension of the field P with a

radical series of length s - 1, then, by the principle of induction, there exists a normal radical extension L, containing the field L: L c L.

By hypothesis, the field K = LS is a simple radical extension of the field L = Ls_1, i.e. K = L(Q, 0),

where ( is a primitive n-th root of unity for some n, and 0 is an arbitrary root of the equation

x"-(3=0, where$eL. We consider the minimal polynomial g(x) of the number $ over 'd+

the field P. Because the field L is normal and $ e L c L, then L contains all the roots R1 = I, $2, ... , Pr of the polynomial g(x). For any i = 1, ... , r we consider the equation

x"-Q,=0. Let a; be an arbitrary root of this equation (for i = 1 we set a1 = 0) and let K = L(C, av

,

ar)

Because a1 = 0, then the field K contains the field K:

KcK.

68

FOUNDATIONS OF GALOIS THEORY

Further, over the field L the field K possesses the radical series

L=LocLlc ... cLi-1cLic ... cL,.=K, where

(3)

VIA

Li=i = 1, ... , r

because C e L1 c LL-1). (when i > 1 this simplifies to Li = By hypothesis, the field L is a radical extension of the field P, i.e. it possesses a radical series, beginning with the field P and ending with the field L. Extending this series by the series (3), we clearly

obtain a radical series of the field K, beginning with the field P. e..

Thus it has been proved that the field K is a radical extension of the field P. Finally, we consider the polynomial G(x) = g(x"). The coefficients of this polynomial belong to the field P. Because .U.

G(x)=(x"-fl1)...(x"-Pr),

app

Iv.

CAD

then the numbers al, ... , ar are roots of the polynomial G(x). All the remaining roots of this polynomial are obtained from the roots all ... , a,, by multiplication by an n-th root of unity, i.e. by multiplication by a power of the primitive root C. Therefore the field K contains all the roots of the polynomial G(x), i.e. it contains its decomposition field Q over the field L. On the other hand, L c Q, Cl al, ... , a, e Q and therefore K = L(C, al, ... , a,.) c: Q. Hence,

Nip

K = Q, i.e. K is the decomposition field over the field L of the !L,

polynomial G(x). Since the field L is a normal extension of the field

ado

P, and the polynomial G(x) is a polynomial over the field P, then from this, according to the lemma, it follows that the field K is normal over the field P. Thus we have found a field K, containing the field K, which is a normal radical extension of the field P. Thus the theorem formulated above has been completely proved. 4. Normal fields with solvable Galois group

Let K be an arbitrary normal radical extension of the field P. Then to the radical series

P=Lo cL1c... cLi-1cLic... c: L., =K

69

EQUATIONS SOLVABLE BY RADICALS

corresponds the series of subgroups of the Galois group G(K, P)

Hi- InHi

G(K,P)=HoDH1D ...

HS= E,

(1)

where

i = 0, 1, ... , s.

Hi = G(K, L1), ti.

For any i = 1, ... , s we consider the triple of fields

Li_1cL,zK. coo

Because the field Lj is a normal extension of the field Li_1, then the

group H. = G(K, L,) will be a normal divisor of the group ray

Hi_1 = G(K, Li_1). Thus, the series (1) is a normal series.

Further, the factor group H;_1/Hj is isomorphic to the Galois '.S

-0z.

group G(L1, Li_1) of the field L. over the field Lj_1, which, as we know, is solvable (because the field L1 is a simple radical extension of the field Lt-1). Thus, the series (1) is a normal series with solvable factors. The existence of such a series ensures, as we know, (see Ch. 1, para. 4), the solvability of the group G(K, P). Thus: E-"

the Galois group of any normal radical extension is solvable. Now let Q be an arbitrary normal subfield of the field K (as always

bed

Old

it is assumed that Q contains the fundamental field P). Then the Galois group G(Q, P) of the field Q over the field P is isomorphic, as we know, to a factor group of the group G(K, P). Because any

'-r

.-!

'CS

factor group of a solvable group is solvable, then, consequently: the Galois group of any normal subfield of an arbitrary normal radical extension is a solvable group. It will be shown that the converse is also true : any normal field, having a solvable Galois group, is a subfield of some normal radical extension. In other words, the normal fields with solvable Galois group are exhausted by the normal subfields of normal radical extensions. We will prove this assertion first for the cyclic fields, i.e. for normal fields having a cyclic Galois group. Let Q be a normal extension of the field P of degree m with cyclic Galois group G(Q, P). We consider the field K = Q(E),

where s is a primitive m-th root of unity. It is easy to see that the field K is normal over the field P (prove it!). Because the field K is

70

FOUNDATIONS OF GALOIS THEORY

the composite of the normal fields P(c) and Q, then, according to the

theorem in part I, Ch. 3, para. 7, the Galois group G(K, P(s)) is ...

isomorphic to some subgroup of the Galois group G(Q, P). Because the group G(Q, P), by hypothesis, is cyclic, and any subgroup of a cyclic group is cyclic, then, consequently, the group G(K, P(8)) is cyclic. Its order n = [K: P(E)] divides the number m and therefore

the primitive n-th root of unity C is a power of the root s, i.e. it belongs to the field P(s): C e P(E).

cCD

chi

CAD

Thus, the field K is a cyclic extension of degree n of the field P(s), containing a primitive n-th root of unity. Therefore, according to the theorem of para. 2, the field K is a simple radical extension of the field P(E). Since the latter is a simple radical extension of the field P, it has thus been proved that the field K is (by construction, a normal) radical extension of the field P. Thus it has been proved that any cyclic extension Q of the field P is contained in some normal radical extension. Now we go on to the general case. Let Q be a normal extension of the field P, having a solvable Galois group G(Q, P), and let

G(Q,P)=Ho-H1=, ...=, Ht-1=...=H3=E

(2)

be an arbitrary solvable series of the group G(Q, P). Ifs = 1, then

`.sue'

the group G(Q, P) is cyclic and hence, according to what was proved above, the field Q is contained in some normal radical extension of the field P. Now supposing that the theorem has already been proved for normal fields having a Galois group with solvable series of length s - 1, we consider the normal field Q having a solvable Galois group with a solvable series (2) of length s. In this field to the subgroup Hl of the Galois group corresponds some subfield

`--

L = K(G, HI).

t01

t01

'C3

The field L is normal over the field P, and its Galois group G(L, P) is isomorphic to the factor group G(Q, P)/Hl, i.e. it is a cyclic group. Hence, according to what was proved above, the field L is contained in some normal radical extension E of the field P. We consider the composite Q of the fields Q and L. As we know (see part I, Ch. 3,

para. 7), the Galois group G(Q, L) of the composite Q over the coo

field L is isomorphic to some subgroup of the Galois group G(Q, L) of the field Q over the field L (as the fundamental field we take here

EQUATIONS SOLVABLE BY RADICALS

71

the field L). But G(Q, L) = Hl and hence the group G(Q, L), and therefore also any subgroup of it (see Ch. 1, para. 4), possesses a solvable series of length s - 1. Therefore, by the principle of induction, the field Q, and this means also the field Q, is contained in some normal radical extension K of the field L. Because the field L is, by construction, a radical extension of the field P, then the field K will be a radical extension of the field P also. Further, as we know, a radical extension K is contained in some normal radical extension K (it may be that it coincides with K). Thus we have found a normal

radical extension K of the field P, containing the given normal extension Q with solvable Galois group. Thus the theorem formulated above has been completed proved. 5. Equations solvable by radicals

It is said that the root 0 of the equation

f(x) = 0

(1)

over the field P can be expressed in radicals, if there exists a radical extension of the field P, containing the root 0. If all the roots of O.,

equation (1) can be expressed in radicals, then it is said that this

cow

A..

equation is solvable by radicals. It will be shown that if at least one root of an irreducible equation can be expressed in radicals, then the equation is solvable by radicals. In fact, let the root 0 of equation (1) belong to the radical extension K of the field P. As we know, the radical extension K can be extended to a normal radical extension K. Because one root of the irreducible

chi

equation (1) belongs to the normal field K, then all the remaining roots must also belong to it. Thus every root of equation (1) lies in the radical extension K, i.e. can be expressed in radicals. The normal radical extension K, containing all the roots of equation C.'

(1), also contains its decomposition field. Hence, if an irreducible equation is solvable by radicals, then its decomposition field is

contained in some normal radical extension of the field P. The converse is also obvious; if the decomposition field of equation (1) ."Y

is contained in a normal radical extension, then equation (1) is chi

solvable by radicals. But, as we saw in the preceding paragraph, a normal field is contained in some normal radical extension if and only if its Galois group is solvable. Hence,

72

FOUNDATIONS OF GALOIS THEORY

an irreducible equation is solvable by radicals if and only if the Galois group of its decomposition field is solvable.

It is usual to call the Galois group of the decomposition field of f0.

an equation the Galois group of this equation, In this terminology the theorem proved is stated in the following way: an irreducible equation is solvable by radicals if and only if its Galois group is solvable. "C7

Problem. To prove this theorem also for arbitrary equations.

a.+

(Hint: As a preliminary prove that the composite of radical extensions is a radical extension.) We emphasize that the theorems proved in this chapter permit one, given any equation with solvable Galois group, effectively to con-

struct a radical extension containing its roots, i.e. effectively to a0"

express its roots by radicals. see below, Ch. 4, para. 4.)

(For an example of such a construction,

r-)

CHAPTER 3

THE CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS 1. The Galois group of an equation as a group of permutations

We recall (see The Course, p. 74) that a one-one mapping of a finite set M onto itself is called a permutation. The number n of elements of this set is called the degree of the permutation. Because the nature of the elements of the set M plays no part in what follows, then we can assume that the set M consists of the number 1, 2, ... , n. If, in a given permutation a, the number j goes into the number i j, then the permutation is denoted by the symbol

a=(1I1 2i2 .... n) In .

In this notation, the numbers 1, 2, ... , n can be permuted in an arbitrary way (the numbers i1, i2, ... being permuted correspondingly); if j1, j2, ... , jn is an arbitrary permutation of the numbers 1, 2, ... , n, then the symbol .f1 ijt

i2

. . . in

i12 ... ij.

coo

d^"

.-.

denotes the same permutation a. The result of carrying out successively the two permutations a and b (of the same degree) is also, clearly, a permutation. This permutation is called the product of the permutations a and b and is denoted by ab. We emphasize that the permutation ab is obtained by carrying out first the permutation a, and then the permutation b. This remark is .s7

cad

0r.

essential, because when n > 2 multiplication of permutations is '.s

non-commutative. It is easy to see that the multiplication of permutations is associative 73

74

FOUNDATIONS OF GALOIS THEORY

(see The Course, p. 76). In multiplying any permutation a by the identity permutation

_

e

1

2...n

1

2 ... n

the permutation a remains unchanged:

ea = ae = a. Moreover, the product (in any order) of the permutation

a_(1 2 ... n) i2 ... in

l1

by the permutation a _ 1

i2 .. 1n

-(L1

1

2 ... n

(1)

is the identity permutation: a-1a = as-1 = e.

All this means that the set Sn of all permutations of degree n is a group. The permutation e acts as the identity of this group, and the permutation inverse to the permutation a is defined by formula (1). The group S,, is called the symmetric group of degree n. Its order is equal to n!. The subgroups of the group S are called groups of permutations of degree n. In other words, a group of permutations (of degree n) is a group whose elements are permutations of the same degree n, and whose operation is the multiplication of permutations. After these preliminary remarks we return to the Galois groups of equations. Let f (x) be an arbitrary polynomial over the fundamental field P. As we said above, the Galois group of the polynomial f (x) (or of the equation f (x) = 0) is the Galois group G(Q, P) of its decomposition field Q, i.e. the field Q = P(a1i ...

,

an),

".:r

where a1, ... , an are the roots of the polynomial f (x) (numbered in some definite order). We will suppose that the polynomial f (x) does not have multiple roots (which, clearly, does not lessen the generality). As we know (see part I, Ch. 3, para. 2), for any automorphism S e G(Q, P) and any root a; of the polynomial f (x) the

75

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

number as is also a root of this polynomial, i.e. there exists an index k; such that a= ake

O"'

Because the automorphism S is a one-one mapping, and all the roots al, ... , a are distinct, then k1 k; if i j. Hence, the symbol

2

1

-n TV'

( k1 k2 ... k

is the symbol of scme permutation a of degree n. In order to underline the dependence of the permutation a on the automorphism, we will

denote this permutation by 4(S). Thus 0 will be a mapping of the group G(Q, P) into the symmetric group It is obvious that for any two automorphisms S, T n G(Q, P) 4(ST) = q(S)q5(T), :s'

i.e. the mapping 0 is a homomorphism. The kernel of this homomorphism consists of the automorphisms leaving invariant each of the roots al, ... , But if an automorphism of the field Q over the field P leaves invariant all the roots al, ... , a,,, then it also leaves invariant any element expressible in the form of a polynomial (with coefficients in the field P) in al, ... , i.e. it leaves invariant any element of the field Q = P(al, ... , Hence, the kernel of the homomorphism ¢ consists only of the identity automorphism E, i.e. 0 is a monomorphism. In other words, ¢ is an isomorphism of the Galois group G(Q, P) onto a group of permutations. Thus, the '-U'

Galois group of any equation (not having multiple roots) can be con-

sidered as a group of permutations. The degree of this group of permutations is equal to the degree of the equation.

The representation of the Galois group of an equation in the

may

'''

form of a group of permutations is most convenient for an explicit calculation of it in each particular case. In order to be in a position to perform such a calculation, it is necessary, of course, to study in more detail the group of permutations itself. We will go on to this at once. 2. The factorization of permutations into the product of cycles Let

a _-

... n

1

2

kl

k2...k

76

FOUNDATIONS OF GALOIS THEORY

CO"

be an arbitrary permutation of degree n. If for some i the number k; is distinct from i, then it is said that the permutation a properly transposes the number i; in the opposite case it is said that the permutation a leaves the number i invariant. We will consider a cyclic subgroup of the group S,,, consisting of powers of the permutation a. If m is the order of this subgroup, then it consists of the permutations

a =e,a,a ,...,am-1 , 0

2

cad

where all of these permutations are distinct. Let io be an arbitrary number properly transposed under the permutation a. We denote by ik the number into which the number io goes in the permutation ak. It is obvious that the permutation a carries the number ik into the number ik+1 If we had ik = ik+1, then, applying to this equation the permutation a-k, we would obtain io = il, i.e. the permutation a, contrary to hypothesis, leaves the number io invariant. Hence, all the numbers i0, i1, ... are properly transposed by the permutation a. Amongst these numbers no more than m are distinct, because im, clearly, is equal to io. If the numbers cad

10, I1. ... , lm-1

z'~

exhaust all the numbers properly transposed by the permutation a, then the permutation a is called cyclic and is denoted by the symbol (ioil ... im_1). In this case all the numbers i0, il, ... , im-1 are distinct. In fact, if, for instance, ik = Zk+lr where 0 < k + I < in - 1, then, applying to this equation the permutation a-k, we would obtain that io = 1. This equation means that the permutation a' leaves the number io invariant. But for any q the permutation a-4 carries the number iq into the number io, the permutation al leaves the number io invariant and the permutation aq carries the number io into the number iq. Hence, the permutation a' = a-gala" leaves invariant every

number iq, i.e. according to the hypothesis, every number properly transposed by the permutation a. On the other hand, any number left invariant by the permutation a is also left invariant by the per-

mutation a'. Hence, the permutation a' leaves invariant all the numbers, i.e. a' = e, which is impossible, because 1 < m. We note that for any system io, i1, ... , im-i of distinct numbers there exists a cycle (clearly, unique), carrying the number io into the number i1, the number i1 into the number i2, . - . , the number im_2

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

77

into the number im-1 and, finally, the number im-1 into the number io. This cycle is represented by the symbol

(0 ... im-1 ) -

to

it ... 1m-1

11

12 ... 10

1

jn-m

.il ... j.-.),

where j1, ... , j,,-m are those numbers in the series 1, 2, ... , n, distinct

from the numbers io, ... , im-1

We note also that the representation of the cycle in the form .ti

11011

1m-i) is not unique. In fact: (10i1 ... 1m-1) = (11 ... 1m-110) =

= (1m-11011 ... 1m-2),

i.e. the representation of the cycle can begin with any properly transposed number. To within a transformation of such a kind the cycle, as it is easy to see, is unique. The number m of numbers properly transposed by the cycle a is

called its length. From what has been said above it is clear that the length of a cycle is equal to its order.

The least possible length of a cycle is equal to two. Cycles of length two are called transpositions. The transposition (ij) carries the number i into the number j, the number j into the number i and leaves all other numbers invariant. Problem. To prove that a permutation properly transposing only two numbers is a transposition. Any cycle of length m is the product of m - 1 transpositions. In fact, (101112 ... i.-1) = (1011)(1012) ... (101m-1)

Two cycles are called independent, if they do not have properly transposed numbers in common. It is obvious that in the multiplication of independent cycles the order of multiplication plays no part at all (i.e. independent cycles, as one says, commute). It will be shown that any non-identical permutation is the product of independent cycles.

We will prove this assertion by induction on the number s of properly transposed numbers. To this end we note first that the numbers cannot be equal to one. In fact, if the permutation a carries the number i into the number j i, then it cannot leave the number j

invariant, because in the contrary case the two distinct numbers

'ti

78

FOUNDATIONS OF GALOIS THEORY

i and j would be carried by the permutation a into the same number j. /\\

6)b

Therefore s > 2. If s = 2, then the permutation is a transposition, and so the theorem is true for this case. Thus the first stage of the cad

induction is justified. Now we suppose that the theorem has already been proved for all coo

permutations properly transposing fewer than s numbers, and we consider an arbitrary permutation a properly transposing s numbers. C3.

'.2

Let iO be one of the numbers properly transposed by the permutation a. Applying to this the construction expounded above (i.e. operating

on it by the powers of the permutation a), we obtain the numbers iO, il, ... , ik, ... properly transposed by the permutation a (see above). Let iq be the first of these numbers with positive index

r"3

CZ.

00'

'"'

Q,:

'O"

v,'

coinciding with the number iO. Such a number exists because, for instance, the number where m is the order of the permutation a, is equal to the number iO. We will prove that the numbers iO, i1, ... , iq_1 are all distinct. In fact, if, for instance, it = it+p, then, applying to this equation the permutation a-1, we obtain iO = ip, which by virtue of the minimal nature of the number q is impossible. Since the numbers iO, 11, ... , iq_1 are all distinct (and q > 1, because iO i1; see above), then we can form the cycle (i0i1 ... The permutation a(i0i1 ... iq_1) 1 leaves invariant all the numbers left invariant by the permutation a, and, moreover, also all the numbers iO, ... , iq_1. Thus, it in fact transposes no more than C3.

C."

s-q numbers and hence by the hypothesis of the induction it can be factorized into the product of independent cycles. For the completion of the proof it remains- to note that these cycles are also independent of the cycle (iOi1 ... iq_1). 'CS

Because every cycle can be factorized into transpositions, then from the theorem proved it follows that any permutation can be

r7'

coo

v''

coo

C1.

factorized into the product of transpositions (generally speaking, not necessarily independent). The numbers contained in the independent cycles into which a permutation is factorized are the numbers properly transposed by this permutation. Each cycle of the factorization consists of those numbers which are transposed one into the other by the powers of

the given permutation. Thus the number and the structure of the independent cycles into which a permutation is factorized are uniquely

defined by this permutation. In other words, the factorization of a CAD

permutation into the product of independent cycles is unique (to within the order of the factors).

79

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

3. Even permutations. The alternating group

As we saw above, any permutation can be factorized into the product of transpositions. Generally speaking, the same permutation "C7

can be represented in the form of a product of transpositions in many different ways. For instance, it is obvious that

(j k)(i Ic) = (i j)(j k), if i 0 j,

(1)

ifj#k

(2)

and F.'

(i j)(i k) = (i k)(j k),

.ti

.,,

C."

coti

L]. O..

acs

coo

A..

obi

(formulae (1) and (2), as it is easy to see, express the same fact, but in different notations). Lemma. If the product of several transpositions is equal to the identity permutation, then the number of these transpositions is even. We will prove this lemma by induction on the number s of distinct numbers contained in the representation of the given transpositions. The least possible value of the number s is equal, clearly, to 2. If s = 2, then the product under consideration is the power of some transposition and therefore is equal to the identity transposition only when the exponent of the power is even (because any transposition has order 2). Thus, in the case s = 2 the lemma is proved. Now supposing that the lemma has already been proved for any product of transpositions whose representations contain less than s distinct numbers, we consider a product of transpositions equal to the identity permutation (11 i2)(13 14) ... (12q-1 12q) = e,

(3)

:s'

coo

in whose representations exactly s distinct numbers occur. Let i be one of these numbers. Using relation (1) and the fact that independent transpositions commute, we can " move forward " all transpositions in whose representations the number i occurs, i.e. we can pass from the product (3) to an equal product of the form

coo

(4) (ij1) ... (ijp)(k1 k2) ... (k2r-1 k2r), in which all the numbers k1, k2, ... , k2r are distinct from the number i. If p > 1, then, using relation (2) or the relation

(i j)(i j) = e,

(5)

we can pass from the product (4) to a product of the same form, but with smaller p. As a result of a series of such transformations either

80

FOUNDATIONS OF GALOIS THEORY

we completely annihilate all the transpositions in whose representations the number i occurs, or we obtain a product containing only one such transposition: (i j 1)(11 12) ... (12t-1 121)

But this product, clearly, carries the number j1 into the number i and in'

cad

therefore cannot be the identity permutation. Hence, the last case is impossible. Thus, as a result of our transformations, we obtain a product of transpositions, whose representations do not contain the number i, which is equal to the identity permutation. The representations of these permutations clearly do not contain any new numbers. Hence, according to the hypothesis of the induction, an even number of transpositions occurs in this product. It remains to remark that in the transformations described the number of transpositions either is unaltered (when we use relations (1), (2)), or is reduced by two (when we use relation (5)). Therefore the original product (3) also consists of an even number of transpositions. Thus vii

r0.

.-L."

[".

Q..

O-'C

`C3

the lemma is completely proved. Now let a permutation a be factorized in two ways into a product of transpositions: a = (i1 i2) ... (12p-1 12p),

a = (j1 j2) ... (j2q-1 j2q)

(the first factorization contains p transpositions, and the second q). Then

I,,

lit i2) ... (12p-1 12p)(I2q.J2q-1) ... (j2j1) = as -1 = e,

and hence, by the lemma proved, the number p + q is even. Thus, the numbers p and q are either both even or both odd. In other words, in all factorizations of a permutation into the product of transpositions the parity of the numbers of these transpositions will be the same. A permutation is called even if it factorizes into the product of an

even number of transpositions, and odd in the opposite case. According to the theorem proved the parity of a permutation does

not depend on the choice of factorization into a product of ACS

transpositions. Any transposition, or generally any cycle of even length, is an odd cad

G1.

permutation, and any cycle of odd length, in particular any cycle

81

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

of length 3, is an even permutation. The identity permutation is clearly even.

If a = (ii i2)(i3 i4) .:. (is-i is)

is a factorization of the permutation a into a product of transpositions, then a - i = (is is-1) ... (Z4 l3)(i2 ii)

Hence it follows that

a permutation inverse to an even permutation is even, and one inverse to an odd permutation is odd.

Further, if a = (ii i2) ... (is-1 is) b = (J1J2) ... (1t-11t), then

ab = (ii i2) ... (is-1 is i0(J1J2) ... 0:-1Jt) Therefore the product of two even or two odd permutations is an even permuta-

`ti

tion; the product of an even and an odd permutation is an odd permutation. -s~

From this it follows that the set of all even permutations (of a given degree n) is a subgroup of the symmetric group S. This subobi

ova

group is denoted by A and is called the alternating group of degree n. Since for any even permutation a and an arbitrary permutation b

...

,0.

the product bab-1 is an even permutation, the alternating group A. is a normal divisor of the symmetric group S. Since for any two odd permutations a and b the permutation ab-1 is even, i.e. belongs to the group then all odd permutations generate one coset of the subgroup A. Hence, the factor group consists of only two elements, i.e. it has order 2. Therefore the order of the group i.e. the number of even permutations of order n, is equal to in!. Cam"

4. The structure of the alternating and symmetric groups may'

We will study the structure of the group A for different values of n.

For n = 2 the alternating group consists only of the identity permutation e.

FOUNDATIONS OF GALOIS THEORY

82

For n = 3 the alternating group has order 13! = 3 and hence is cyclic. As its generator one can take any even permutation (for instance, the cycle (1, 2, 3)).

For n = 4 the alternating group has order 4-4! = 12 and consists of the following elements : e,

t2 = (13)(24), s2 = (124), s6 = (143), F-+

t1 = (12)(34), Si = (123), s5 = (142),

t3 = (14)(23), s3 = (132),

54 = (134),

s7 = (234),

s8 = (243).

1149

It is easy to verify that

ti = t22 = t3 = e, t2t1 = t1t2 = t3, t3t1 = t1t3 = t2,

Hence, the permutations e, t1, t2, t3 generate a subgroup of the group A4. This subgroup is called the Klein group and is denoted by B. The group B is Abelian and has order 4. Further, it is easy to verify that S1t1S

Slt2S1 1 = tar

S1t3S1 1 = tlr

S2t1S2 1 = t3,

S2t2s2 1 = tlr

S2t3S2 1 = t2,

S3tlS3 1 = t3,

S3t2S3 1 = t1,

S3t3S3 1 = t2r

ay.

.-.

1 = t2,

S4t3S4 1 = t1,

S5i2S5 1 = t3r

S5t3S5 1 - t1r

S6t1S6 l = t3r

S02S6 1 = t1,

S03S6 1 = t2r

ay.

S7t2S7 1 = t1, S8t2S8 1 = t3,

...

ay, r..

S7t1S7 l = tar S8t1S8 1 = t2,

...

S4t2S4 1 = t3,

S5t1S5 1 = t2,

S4t1S4 1 = t2,

S7t3S7 1 = t2,

S8t34 1 = t1.

Hence, the group B is a normal divisor of the group A4. The corresponding factor group A4/B has order 3 and therefore is a cyclic group.

Because the group B is Abelian, then any subgroup of it, for 't3

r'.

w.3

instance the cyclic subgroup C of order 2, consisting of the identity permutation e and the permutation t1, is a normal divisor (of the group B, but not of the whole group A4). The order of the factor group B/C is equal to two, and hence this factor group is a cyclic group. 0

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

83

Thus the chain of subgroups

A4 z) Bz) Cz) e

`t3

C]+ C!1

is a solvable series of the group A4. Thus it has been proved that the group A4 is solvable. The groups A2 and A3 are also, clearly, solvable. Thus, for n < 4 the group A is solvable. Since for n > 3 the group A. is not Abelian, then the groups A3 and A4 are examples of non-Abelian solvable groups. Now we consider the cases n > 5. Let N be an arbitrary normal divisor of the group A, distinct from e. Since N e, then in N there exists at least one permutation t e. A factorization of the

permutation t into a product of independent cycles can be made in one of the following forms:

(1) t = (i0 i1 i2 i3 ...)(...) ... (it has a cycle of length > 4); (2) t = (io it i2)(i3 i4 ...)(...) ... (it has a cycle of length 3 and also other cycles);

(the permutation t is a cycle of

(4) t = (i0 i1)(i2 i3)(...) ...

length 3); (the permutation t factorizes into a e-1

(3) t = (i0 it i2)

.'3

(3.

product of independent transpositions). "'I

(the permutation t is even and therefore cannot be a transposition; the rows of dots denote numbers or cycles, which may equally well not be there). Since N is a normal divisor, then for any even permutation r the permutation rtr-1, and hence also the permutation

rtr-lt-1, belongs to N. For each form of permutation indicated cad

above, we will choose a permutation r of the following form: (1)

r = (i1 i2

(2)

r = (il i2

(3)

r = (il i2 i3),

(4)

r = (ii i2 i 3)

Calculating in each of the four cases the permutation s = rtrt we obtain (1)

S = (i0 12 i3);

(3)

s = (i0 i3)('1 i2);

(4)

S = (i0 i2)(i1 i3)

S= (i0 i3 111214); /

te,

(2)

84

FOUNDATIONS OF GALOIS THEORY

'-.

Thus if in the normal divisor N there exists a permutation t of the form (1), then there also exists a permutation of the form (3). If there exists a permutation of the form (2), then there exists a permutation of the form (1) and hence, by what has just been said, a permutation of the form (3). Finally, if there exists a permutation of the form (3) or (4), then there exists a permutation which is the product of exactly two independent transpositions. Thus, in N there must exist a permutation which is the product of exactly two independent transpositions. Let this permutation be (j112)03 j4) Now let (k1 k2)(k3 k4) be an arbitrary permutation which is the r°.

r..'

t-+

S".

3(~

yam,,

product of two independent transpositions. We consider the permutation

a(klk2k3k4... J1 J2 J 3 J4

where in the place of the dots stand arbitrary numbers (of course, in r`' -14

i.+

the upper row these numbers are distinct from the numbers Cry

k1, k2, k3, k4, and in the lower row from the numbers jl,j2,j3,j4). It is easy to see that a(j1 i2)03 j4)a-1 = (k1 k2)(k3 k4) t3+

SO.

Moreover, denoting the permutation a(j1 j2) by b (for simplification of the formulae), we obtain 'F+

mil!

b(j1j2)(j3j4)b-1 =

a(j1j2)lf1j2)(j3j4)(j1j2)a-

= a(j1 j2)(j3

j4)a-1,

i. e., .v0

= (k1 k2)(k3 k4)

b(j1j2)(j3j4)b-1

.S.

won

4..

The permutations a and b, differing by a transposition, have different parity, i.e. one of them is even and the other is odd. We denote the f..'

even one of the permutations a and b by c, i.e. we set c = a if the permutation a is even and c = b if the permutation b is even. By what has been proved: .fl

C01 j2)(j3

J4c-1

= (k1 k2)(k3 k4) +-'

.Sri'

Because (1 j2)(j3 j4) e N, c e and N is, by hypothesis, a normal divisor in then (kk k2)(k3 k4) e N. Thus we have proved that the

normal divisor N contains all permutations which are products of two independent transpositions.

jay

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

85

o-1

Now we consider a permutation which is the product of two dependent transpositions. Such a permutation has the form Ul J2)(11 J3). Because, by hypothesis, n > 5, then there exist two distinct numbers 11 and 12, not exceeding n and distinct from the numbers j1, j2, and j3. The permutations (j1 j2)(li 12) and (11 12)(j1 j3)

being the products of two independent transpositions, belong, by what has been proved, to the normal divisor N. But (11 j2)(li 12) - (11 12)(1 J3) = 01 J2)(J1 J3)

and, hence, the permutation (jl 32)(il j3) also belongs to N. Thus, to the normal divisor N belongs any permutation which is the product

of two arbitrary transpositions, and hence also any permutation o-'!

which is the product of an arbitrary even number of transpositions, i.e. any even permutation. Therefore the normal divisor N contains all even permutations, i.e. N = An. Thus, if N e, then N = An. In other words, the group A,, has no normal divisors besides the trivial ones, i.e. it is a simple group. Thus we have proved that for n > 5 the alternating group A. is simple, and hence is unsolvable. (because the simple solvable groups are exhausted by the cyclic groups of prime order). 'L3

C".

We note that for n = 2 and n = 3 the group An, clearly, is also simple.

From the results proved with regard to the group A,,, it follows immediately that for n < 4 the symmetric group S is solvable (because it possesses the following solvable series:

S2:ne, ifn=2. S3=)A3=)e, ifn=3.

S4=)A4:nBnC=e, ifn=4), and for n > 5 the group S,, is unsolvable (because it contains the unsolvable group An).

5. An example of an equation with Galois group the symmetric group A group G of permutations of degree n is called transitive if for any

two numbers i, j (of course, it is supposed that I < i, j < n) in the group G contains at least one permutation carrying the number i

86

FOUNDATIONS OF GALOIS THEORY

into the number j. The importance of transitive groups for Galois theory is explained by the following theorem: The Galois group of an irreducible polynomial is transitive.

For the proof it is sufficient to note that if the polynomial f (x) is irreducible, then all its roots al, ... , a are conjugate to each other, and therefore for any pair of roots a1, ai in the field Q = P(al, ... , there exists an automorphism (over P), carrying the root a1 into the root a; (see part I, Ch. 3, para. 5).

Problem. To prove that a polynomial having a transitive Galois group is irreducible.

Not having in view the study of arbitrary transitive groups, we restrict ourselves to the consideration of groups containing at least one transposition.

Let the transitive group G contain the transposition

(i1 i2).

Besides this transposition, the group G may also contain other transpositions of the form (i/1 j). Let (11 12), (11 13),

, (11 1m)

be all the transpositions of the form (i1 j) contained in the group G. Then the group G does not contain any transposition of the form

coo

(jlq), q=1,2,...,m, for which the number j is distinct from the numbers il, i2, ... , (the group G contains transpositions of the form (ipiq), where 1 < p, q < m, because (ip iq) = (i1 ip)(i1 iq)(ii ip)). In fact, for q = 1 this is obvious, and for q > 1 it follows from the relations (j iq) e G and (i1 j) = (i1 iq)(j iq)(ii iq) that, contrary to hypothesis, (i1 j) a G. If now m < n, i.e. if there exists a number j < n, distinct from the numbers il, ... , im, then, since the group G is transitive, it contains at least one permutation a, carrying the number i1 into the number j :

if

a_

(i1

i2 ... im

\31J2...3m.

,

then j1 =i.

From what has been proved above it follows that none of the numbers j1, ... , jm is equal to any one of the numbers it, i2, ... , im, because the permutation a(i1 iq)a -1 = (j1 jq) = (j jq) belongs to the

group G. Hence, 2m < n.

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

87

If 2m < n, then there exists a number k < n, distinct both from the numbers il, ... , and from the numbers jl, ... , By virtue of the transitivity of the group G it contains at least one permutation b, carrying the number it into the number k: if b

k1 k2 .. km

'

then

k1=k. It is proved as before that none of the numbers k1, ... , kis equal to any one of the numbers i2, ... , Moreover, it will be shown that none of the numbers k1, ... , kis equal to any one of the numbers ji, , jm. In fact, if, for instance, kp = jq, then the group G contains the transposition

ab-1(i1 ip)ba-1 = (k' iq), where k' is the number carried by the permutation a into the number k, which is impossible, because the number k', clearly, is distinct from the numbers i1, ... , iq. Hence, 3m < n.

If 3m < n, then by an analogous construction we can find m , 1., distinct from all those previously found, and can thus prove that 4m < n. The process of constructing new numbers ceases only when we exhaust all n numbers 1, 2, ... , n. But because at each step we added precisely m numbers, then such an exhaustion is possible only when m divides n. On the other hand, the process apt,

numbers 11, ...

must necessarily cease, because the number n is finite. Thus we have

proved that the number m divides the number n (the degree of the permutation group G). Because m >, 2, then it follows from this that in the case when n is

a prime number, the number m must coincide with n. Thus in this case the numbers il, ... , i, exhaust all the numbers 1, 2, ... , n, and

therefore the group G contains any transposition (i j) (because (i j) = (i1 i)(i1 j)(il i)). Hence, G = S, because every permutation factorizes into a product of transpositions. Thus it has been proved that a transitive group of prime degree, containing a transposition, coincides with the whole symmetric group.

phi

r-,

We apply this theorem to the problem of finding the Galois group of an irreducible polynomial f (x) of prime degree n. We suppose that all the roots of the polynomial f (x) are real except two. Let,

88

FOUNDATIONS OF GALOIS THEORY

for instance, al, a2 be the unreal roots of the polynomial f (x), and a3, ... , a be its real roots. We suppose further that the fundamental field P consists only of real numbers (for instance, it is the field R of rational numbers). Then the roots al and a2 are, as is well known, conjugate complex numbers : a2 = al.

Any element a of the field Q = P(a1, a2, a3, ... , can be expressed in the form of a polynomial (with coefficients in P) in al, a2, ... , a,,:

a = g(al, a2, a3, ... , an) Because all the coefficients of this polynomial are real numbers by hypothesis, then a = g(al, a2, a3, ... ,

(we recall that the roots a3, ... , hence, a e Q. Therefore, letting

by hypothesis, are real), and a, I'"

as =

we obtain a mapping S of the field Q into itself. From the elementary properties of the operation a -+ a (see The Course, p. 50) it follows easily that the mapping S is an automorphism of the field Q over the fn'

.O.

field P, i.e. S e G(Q, P). The permutation corresponding to the .w!

automorphism S is clearly the transposition (1 2). Thus the Galois ...

.-.

ti,

'.3

group of the polynomial f (x) (considered as a group of permutations) ...

is transitive (because the polynomial f (x) is irreducible) of prime degree n, and contains the transposition (1 2). Therefore this group coincides with the whole group 5,,. Thus, the following theorem has been proved: If: _t/2'0

t".,

...

`C~

(1) the field P consists only of real numbers; (2) the polynomial f (x) is irreducible over the field P; (3) the degree n of the polynomial f (x) is a prime number; (4) the polynomial f (x) has exactly two unreal roots, then the Galois group of the polynomial f (x) is the symmetric group An example of a polynomial over the field R of rational numbers, which satisfies the conditions of this theorem, is the polynomial 'CI

C."

ran

c,,

x5+px+p,

89

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

x5+px+p, 4x4+p, 3px + 4,

-offer

where p is an arbitrary prime number. The irreducibility of this polynomial follows from Eisenstein's criterion (see The Course, p. 347). The Sturm chain for it has the form and hence, according to Sturm's theorem,

1,

the polynomial

x5 + px + p has only three real roots. Thus this polynomial in fact satisfies the conditions of the theorem. This means that its Galois group is the group S5. Because the latter group is unsolvable, then the equation

x5+px+p=0 coo

is not solvable by radicals. Thus, over the field of rational numbers there exist equations of the fifth degree which are not solvable by radicals. Since if all equations of a certain degree n are solvable by radicals

then all equations of smaller degree are also solvable by radicals (why?), then it has been proved that over the field of rational numbers there exist equations of any degree greater than or equal to five which are not solvable by radicals. For the construction of such equations it is sufficient to multiply '.+

the polynomial x5 + px + p by an arbitrary polynomial of the cap

t].

appropriate degree. 6. A discussion of the results obtained

The considerations presented at the end of the preceding paragraph ,°3

allow one to introduce only individual examples of equations not solvable by radicals over the field of rational numbers. Amongst these, for degrees greater than five, reducible equations are certainly cad

'.".+

:°,

.-.

.s'

Lam.

,x.

(pro

to be found. Thus we have still to answer the question of the existence of irreducible equations of degree greater than five, not solvable by radicals. Moreover, there remains open the question of the existence of equations (perhaps reducible) not solvable by radicals over fields Iii

P, other than the field of rational numbers. For every concrete CP"'

field P (at least, if it consists only of real numbers) one can attempt to cad

construct examples of such equations, making use of the theorem proved in the preceding paragraph (here, of course, it is necessary to suppose that the field P is not too large because, for instance, over the field of real numbers any equation is solvable by radicals, since

90

FOUNDATIONS OF GALOIS THEORY

any polynomial factorizes into linear and quadratic factors). The 'C3

fundamental difficulty here lies in proving the irreducibility. Because '"!'

ray

for arbitrary fields there exist no criteria of irreducibility, it is impossible to hope to obtain any general results in this way. In view of these difficulties it is appropriate to treat the question of the solvability by radicals of any equation of a given degree n over

pip

C3'

a given field P on a somewhat different plane, replacing it by the question of the solvability by radicals of the general equation of degree n over the field P. Here by the general equation of degree n over the field P we understand the equation

x"+ax"-'+ ... +an=0,

(1)

,-.

'C3

where al, ... , an are independent variables, which we take to run through all elements of the field P independent of one another. For this purpose it is first necessary to state what is meant by the expression " equation (1) is solvable by radicals ", because the definition

of solvability by radicals which we used above (for equations with numerical coefficients) is not applicable in this case. The first definition, arising in a natural way, of the solvability

by radicals of the general equation (1) can be formulated in the following way: equation (1) is solvable by radicals over the field P if there exists a formula :

`''

.fl

CAD

woo

R(ai, a2, ... , a"), (2) containing, besides the signs of the arithmetical operations, only the sign J, such that for any choice of values a°, a2, ... , a° e P of the coefficients of equation (1) the number R(a°, a2, ... , a°) is a root of the equation (already numerical!):

x"+a°x"-1+...+ann

ate.

(In view of the many-valued nature of the operation / one must here stipulate which values of the root V are being considered.) Formula (2) may, of course, also contain some constant numbers. (7)

It is natural here to demand that these numbers belong to the field P.

With this concept of the solvability by radicals of the general equation it is easy to see that if the general equation of degree n is solvable by radicals over the field P, then any (numerical) equation over the field P is also solvable by radicals (in our previous sense). From this, in particular, it follows that over the field of rational numbers the general equation of degree n > 5 is not solvable by radicals.

91

CONSTRUCTION OF EQUATIONS SOLVABLE BY RADICALS

.:$

The concept presented of the solvability by radicals of the general equation has the defect that it is purely formal and is not related in any essential way to the general concepts of Galois theory. Therefore, if we persist with this point of view, we are not in a position to apply the development of the general theory to the solution of the problem of the solvability by radicals of the general equation over an arbitrary .-.

field. ..+

vii

S".

A more comprehensive point of view lies in considering the general equation (1) over the field P(al, a2, ... , of all rational functions in the variables al, ... , a (having coefficients in the field P). As was said in part I, Ch. 1, para. 1, the whole development of the general theory is applicable not only to number fields, but also to any subfield of an algebraically closed field (of characteristic 0). Therefore if, in considering equation (1) over the field P(al, a2, ... , we wish L""

to apply Galois theory to it, we must prove that the field is contained in some algebraically closed field. If

P(al, a2i ... ,

L3.

this has already been proved, then the concept of solvability by radicals, just as the criterion of solvability found above, will automatically be applicable to the general equation (1). Hence, defining the Galois group of this equation, we immediately solve the problem of its solvability by radicals. A detailed exposition of these questions will be the subject of the following chapter.

CHAPTER 4

THE UNSOLVABILITY BY RADICALS OF THE GENERAL EQUATION OF DEGREE n >, 5 1. The field of formal power series

Let P be an arbitrary field of characteristic 0 (for instance, a number field). A formal power series over the field P in the variable x is an expression of the form a_mx-m +

a-m+lx-m+l

+ ...

... + a_lx-1 + ao + alx + ... + akxk + ... ,

(1)

a)^

where a_m, a_m+1, ... , ao, al, ... , ak, ... are arbitrary elements of

the field P. We emphasize that we consider the series (1) purely a.+

formally, not imposing any restrictions of convergence (it is, in any case, meaningless to speak of the convergence of the series (1) over an arbitrary (non-number) field P). Amongst the coefficients a_m, ... , ao, ... , ak, ... of the series (1) there may be some equal to zero. We will regard the series (1) as moo.

...ti

°+'

.-.

unaltered by the removal (and so also by the addition) of terms 0

having zero coefficients. Power series can be added and multiplied in exactly the same way 'T1

as polynomials. It is easy to verify that with respect to addition and multiplication the set P<x> of all formal power series over the field P in the variable x is a ring. It will be shown that the ring P<x) is a

field, i.e. for any power series f distinct from zero there exists a power series g such that fg = 1. In fact, any power series distinct from zero can be written in the following form:

f=x"(ao+alx+ 92

... +akxk+ ...),

THE GENERAL EQUATION OF DEGREE n > 5

93

where n is some whole number (positive, negative, or zero), and ao 0 0. We define the numbers bo, bl, ..., bk, ... from the equations aobo = 1,

aobl + albo = 0, aob2 + albs + a2bo = 0, aobk + albk_l + ... + akbo = 0,

...............

Because ao 0, these equations successively allow one to define uniquely the numbers bo, bl, ..., bk, .... Now setting

g=x "(bo+b,x+ ... +bkxk+ ...), we clearly obtain

fg=1. ..;

Thus our assertion is completely proved.

We can consider any polynomial ao + alx + ... + apx" (by adding terms with zero coefficients) as a power series. Hence the ring P[x] of all polynomials over the field P in the variable x is a subring of the ring P<x). But because the ring P<x> is a field, then together with any polynomials it also contains all their ratios, i.e. fractional-rational functions (rational functions) in the variable x with coefficients in the field P (this fact is the algebraic equivalent of the well-known fact that any rational function can be expanded as a power series). Thus the field P(x) of all rational functions is a subfield of the field P<x>. Let

F(z)=z"+flz"-1+ ... +f" be an arbitrary polynomial over the field P<x> (with leading coefficient equal to one). The coefficients f1, ... , f" of this polynomial are power series over the field P in the variable x. We will suppose that these coefficients do not contain terms with negative powers of coo

the variable x, i.e.

f;=ato+aj1x+ ... +atkxk+ .. We consider the polynomial (over the field P)

Fo(z) = z" + aloz"-1 + ... + a"o

94

FOUNDATIONS OF GALOIS THEORY

(this polynomial is obtained from the polynomial F(z) by the substitution x = 0). It will be shown that if the polynomial Fo(z) factorizes into the product of two mutually prime polynomials (over the field P) Fo(z) = Go(z)Ho(z),

then there exist polynomials G(z) and H(z) over the field P<x> such that F(z) = G(z)H(z),

where in the substitution x = 0 the polynomial G(z) goes into the -4.

141°

polynomial G0(z), and the polynomial H(z) into the polynomial H0(z). For the proof we write down the polynomial F(z) in the form of a formal series

F(z) = Fo(z) + F1(z)x + ... + Fk(z)xk + ... where

Fk(z)=aIkzn-1+... N''

k=1,2,...,

and we consider the system of equations Go(z)H1(z) + G1(z)Ho(z) = F1(z), Go(z)H2(z) + G1(z)H1(z) + G2(z)Ho(z) = F2(z), .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.........................

GO(z)Hk(z) + Gl(z)Hk-1(z) + ... + Gk(z)Ho(z) = Fk(z),

with respect to the unknown polynomials G1(z), G2(z), ...

Gk(z), ... ,

,

(3)

(4) H1(z), H2(Z), ... Hk(z) .... We will prove that there always exist polynomials (3), (4) such that, in the first place, they satisfy the system of equations (2), and, in the ,

second place, the degree of every polynomial (3) is less than the degree p of the polynomial G0(z), and the degree of every polynomial (4) is less than the degree q of the polynomial Ho(z). Setting for this purpose

B1(z) = F1(z),

B2(z) = F2(z) - Gl(z)Hl(z),

Bk(z) = Fk(z) - Gl(z)Hk-1(z) - Gl(z)Hk-2(z) - ... - Gk-1(z)Hi(z),

THE GENERAL EQUATION OF DEGREE n i 5

95

we can write equations (2) in the following form: Go(z)H1(z) + G1(z)Ho(z) = Bi(z),

...............

Go(z)H2(z) + G2(z)Ho(z) = B2(z), Go(z)Hk(z) + Gk(z)Ho(z) = Bk(z),

Now we suppose that, for some k, polynomials G1(z), ... , Gk_1(z), H1(z), ... , H1, _1(z), satisfying the conditions enumerated above, have

already been found. Then we can consider the polynomial Bk(z) as a polynomial known to us. The degree of this polynomial, clearly,

is less than n = p + q. Because the polynomials G0(z) and H0(z) are mutually prime, there exist polynomials Hk(z) andGk(z) such that Go(z)Hk(z) + Gk(z)Ho(z) = 1

(The Course, p. 197). Let Hk(z) be the remainder after division of the polynomial Hk(z)Bk(z) by the polynomial H0(z): Hk(z)Bk(z) = Ho(z)41k(z) + Hk(Z)

Then Go(z)Hk(z) = Go(z)Hk(z)Bk(z) - Go(z)Ho(z)Dk(z) _ (1 - C. k(z)Ho(z))Bk(z) - Go(z)Ho(z)(Dk(z), .a;

Go(z)Hk(z) + Gk(z)Ho(z) = Bk(z),

where Gk(z) = Gk(z)Bk(z) + Go(z)'k(z) .G"

By construction, the degree of the polynomial Hk(z) is less than q and therefore the degree of the polynomial Gk(z)HO(z) wNig

= Bk(z) - Go(z)Hk(z) is less than n = p + q. Hence the degree of the polynomial Gk(z) is less than p = n - q. We note that the polynomials Gk(z) and Hk(z) are defined in a unique way. In fact, if Go(z)Hk(z) + Gk(z)HO(z) = Bk(z) ors

and

Go(z)Hk(z) + Gk(z)H0(z) = Bk(z),

96

FOUNDATIONS OF GALOIS THEORY

where the degrees of the polynomials Gk(z) and Gk (z) are less than p, and the degrees of the polynomials Hk(z) and Hk(z) are less than q, then Go(z)(Hk(z) - Hk(z)) _ (Gk(z) - Gk(z))Ho(z),

whence it follows (by virtue of the fact that the polynomials G0(z) and H0(z) are mutually prime) that the difference Hk(z) - Hk(z) is divisible by the polynomial H0(z). Since the degree of the polynomial Hk(z) - HH(z), by hypothesis, is less than the degree q of the poly-

nomial H0(z), this is possible only if Hk(z) - Hk(z) = 0, i.e. if Hk(z) = Hk(z). Analogously Gk(z) = Gk(z). Thus the existence of the required polynomials Gk(z) and Hk(z) has been proved. We only needed the assumption that the polynomials G,(z), ... ,

Gk-,(z), H,(z), ... , Hk_,(z) had already been found in order to regard the polynomial Bk(z) as known. Because the polynomial .t'

B,(z) is known to us from the beginning (it is equal to the polynomial F,(z)), all the reasoning presented is also applicable to the case k = 1.

Thus, beginning with k = 1, we can successively (and uniquely) define all the polynomials (3) and (4). Now we set

G(z) = G0(z) + G,(z)x + ... + Gk(z)x" + ... ,

H(z) = H0(z) + H,(z)x + ... + Hk(z)xk + ...

(5)

Collecting together terms containing the same powers of the variable z, we see that G(z) and H(z) are polynomials over the field P<x> (of

degrees p and q respectively). On the other hand, multiplying together (formally) their expressions (5) and making use of relation (2), we clearly obtain G(z)H(z) = F(z).

Thus the theorem formulated above has been completely proved. In what follows we need the following assertions, which follow easily from the theorem proved: if the field P is algebraically closed, then the polynomial

F(z) = z" +fiz"-' + ... +f" of degree n > 1 over the field P<x) (having the leading coefficient equal to unity) is reducible if it satisfies the following conditions: (1) none of its coefficients f,, ... , f" contains terms with negative powers of the variable x;

97

THE GENERAL EQUATION OF DEGREE n > 5

(2) at least one of the coefficients fl, ... , f" has a free term, i.e. contains a term with zero power of the variable x; (3) at least one of the coefficients fl, ... , f" does not have a free term, i.e. it begins with a term of positive power with respect to x.

+X4.

In fact, by virtue of condition (1) for the polynomial F(z) there is a definite polynomial Fo(z) (over the field P), obtained from the polynomial F(x) by the substitution x = 0. By virtue of condition (2) the polynomial Fo(z), besides the term z", has at least one term with coefficient distinct from zero. Therefore in the field P there exists at least one root a of the polynomial Fo(z) distinct from zero (we emphasize that the field P is assumed algebraically closed). Let ;s'

Go(z) = (z - a)" be the highest power of the binomial z - a by which the polynomial F(z) is divisible. Thus, Fo(z) = Go(z)Ho(z),

(6)

where the polynomial Ho(z) is relatively prime to the polynomial

This factorization is not trivial, i.e. the degree p of the polynomial Go(z) is not equal to n. In fact, if p = n, then Fo(z) =(z - a)" and, hence, all the coefficients of the polynomial Fo(z) are distinct from zero (we emphasize that we assume the. field P Go(z).

to be of characteristic 0), which contradicts condition (3).

According to the theorem proved above the factorization (6) defines a factorization F(z) = G(z)H(z)

of the polynomial F(z). Thus the reducibility of the polynomial F(z) is proved (because the degree of the polynomial G(z) is equal to p and, hence, is less than n). 2. The field of fractional power series

Suppose, as above, that P is an arbitrary field of characteristic 0.

A fractional power series over the field P in the variable x is an expression of the form aox"o/" + alx"1/" + ... + akx"k/" + ...

,

3.a

where n is an arbitrary positive integer, no, nl, ... increasing integers

no
(1) ,

nk ... are

98

FOUNDATIONS OF GALOIS THEORY

f].

(there may even be negative numbers amongst them, but only finitely many), and ao, a1, ... , ak., ... are elements of the field P. If n = 1, then the fractional power series is nothing other than a formal power series in the sense of para. 1. In exactly the same way as for

power series, we identify fractional power series which differ by terms with zero coefficients. Fractional power series can be added and multiplied by the same rules as formal power series, and, as it is

nee

easy to see, with respect to the operations of addition and multiplication the set P{x} of all fractional power series over the field P in the variable x is a ring. It will be shown that the ring P{x} is afield. For the proof it is sufficient to note that any fractional power series

f = aox"0l" + ... + akx"kln + ...

can be regarded as a formal power series in the variable = xl/n.

Because the ring P of formal power series in the variable is a field, then for the series f, considered as a power series in , there exists (of course, if f 0) a power series g in such that fg = 1. .`s

Replacing the variable conversely by x1/" in the series g, we obtain a (fractional power) series in x, such that fg = 1. CAD

The fundamental property of the field P{x} is described in the

`"'

following theorem:

If the field P is algebraically closed, then the field P{x} is also algebraically closed, i.e. any polynomial F(z) over the field P{x} can be factorized into linear factors. .^.

For the proof of this assertion it is evidently sufficient to prove that any polynomial of degree n > 1 F(z)=Z"+f1Zn-1+,

... +f,,

'17 .-.

over the field P{x} is reducible. Having this in view, we note that without loss of generality we can suppose that in the polynomial 0, then F(z) the coefficient f1 of Z"-1 is equal to zero. In fact, if f1 taking a new unknown

y=z+'/-1, n

we obtain, as it is easy to see, a polynomial with the coefficient f1

THE GENERAL EQUATION OF DEGREE n i 5

99

equal to zero. On the other hand, in such a substitution an irreducible polynomial remains irreducible, and a reducible one reducible. If all the coefficients fi of the polynomial F(z) are equal to zero, i.e. if F(z) = z", then the polynomial F(z) is reducible, so that in this

case the theorem is true. Thus we can suppose that amongst the coefficients fi there are some distinct from zero. Let the expression

for the non-zero coefficient fi as a fractional power series in the variable x begin with the term aix' , where ai 0 0, and ri is some rational number (which may be negative). Let r be the least of the numbers (ri/i). Then for any i (for which fj :A 0)

ri > ir, 0

where equality is achieved for at least one i. Now we make a substitution of the unknown, setting

z=yx Then, as it is easy to see, F(z) = x"G(y), where G(y)=Y"+92Y"-2+

... +9"

(gt = 0, because we suppose that ft = 0), where for any i =.Ax-i"

gi

Hence the expression for the non-zero coefficient gi begins with the term aix"-i"

(i.e. with a term of non-negative degree), and for at least one i the expression for the coefficient gi has a free term. Now let m be the least common denominatort of all the exponents with which the variable x occurs in the series g2, ... , g". Then these ate

&1P

series can be considered as formal power series in the variable = xt i', and hence the polynomial G(y) can be considered as a polynomial over the field P<>. It is obvious that this polynomial satisfies all the conditions of the (considered over the field proposition proved at the end of the preceding paragraph (condition this 3) is fulfilled for it because gt = 0). Hence, over the field polynomial is reducible, i.e. it can be represented in the form of a of positive degree, product of polynomials over the field t (Editor's footnote) that is, the I.c.m. of the denominators.

100

FOUNDATIONS OF GALOIS THEORY

fr'

coy

gar

distinct from n. Setting = x1/m in the coefficients of these polynomials, we clearly obtain a factorization of the polynomial G(y) into a product of polynomials over the field P{x}. Thus it has been proved that the polynomial G(y) is reducible. To complete the proof it remains to note that from the reducibility of the polynomial G(y) the reducibility of the polynomial F(z) follows immediately. Thus

o'0

the fact that the field P{x} is algebraically closed has been completely proved. Together with fractional power series in one unknown x one can define fractional power series in several unknowns x1, ... , xn. The set P{x1, ... , of all the fractional power series over the field P in the unknowns x1, ... , xn can most simply be defined by induction: P{x1, ...

P{x1, ... , xn-1}{xn},

,

i.e. one defines P{x1, ... , as the field of fractional power series over the field P{x1, ... , xn _ 1 } in the variable x,,. One can easily give

a direct (if also somewhat cumbersome) definition of the field P{x1, ... ,

For instance, the elements of the field P{x1, x2}, i.e. cad

the fractional power series in the two unknowns x1 and x2, are expressions of the form

Z L atixlnt/nx mt/m ,

i=o i=o

poi

also algebraically closed, and therefore also the field

ov%V

P{x1 } is

3,0

addition and multiplication of such series are defined by the obvious rules. If the field P is algebraically closed, then, as we know, the field P{x1, x2} (as the field of fractional power series over the algebraically

0.,

closed field P{x1}). By an analogous consideration, the field P{x1i x2, x3} is algebraically closed and, in general, so is any field P{x1, ... , xn}. Thus: if the field P is algebraically closed, then the field P{x1, ... , is also algebraically closed.

In particular, chi

the field C{x1i ... , xn} of fractional power series in n variables with complex coefficients is algebraically closed. As we saw above, the field of rational functions P(x) in the variable

x over the field P is a subfield of the field P<x> of formal power

THE GENERAL EQUATION OF DEGREE n > 5

101

series. On the other hand, since any power series is also a fractional power series, then P<x> c P{x}. Thus,

P(x) c P{x}. From this it follows easily by induction that, in general, for any n

P(xl,... ,x")=P{x1,...,xn}. Hence, in particular: for any number field P the field of rational functions P(xl, ... , xn) is contained in the algebraically closed field C{xl, ... , x"}. Thus we have established the possibility of applying Galois theory to equations over fields of rational functions, and in particular to the general equation of degree n (see Ch. 3, para. 6). 3. The Galois group of the general equation of degree n

We emphasize that by the general equation of degree n we understand an equation of the form

x"+alx"-1+...+a,, = 0,

(1)

where al, ... , a" are independent variables. This equation we consider

(FD

as an equation over the field P = P(al, ... , an) of rational functions in the variables al, ... , an with coefficients in some number field P. Since the field P(al, ... , an) is contained in the algebraically closed field C{al, ... , an}, then as we have already noted repeatedly, all the concepts and methods of Galois theory are applicable to equation (1). In particular, we can speak of its decomposition field (over the field P) : Q = P(tl, ...

, tn),

where t1, ... , to are the roots of equation (1), i.e. fractional power series from C{al, ... , an}. This field contains, in particular, the field P(t1,

... , Q: P(t1i ..

,

tn) C Q.

But in view of the well known formulae of Vieta the coefficients al, ... , an of equation (1) can be expressed rationally in terms of its

102

FOUNDATIONS OF GALOIS THEORY

roots and therefore belong to the field P(tl, ... , t,,). Hence, P C P(t1, ... , tn),

and therefore

P(ti, ... , Q c P(t1, ... , Q. Thus,

Q=P(t,..., tn). P'+

From this it follows that any element of the field Q can be expressed in the form of a rational function in the elements t1, ... , to with coefficients in the field P. In fact, the set of all rational functions in the elements t1, ... , t» with coefficients in the field P is, clearly, a

R,"

0

f11

subfield of the field Q, containing the field P and the elements t1, ... , tn. Therefore by virtue of the minimal nature of the field P(tl, ... , tn) this set coincides with the whole field Q = P(tl, ... , tn). a.>

It will be shown that any element of the field Q can be expressed uniquely in the form of a rational function with coefficients in the field P. In fact, if J 1(tl, ... , t») g1(t1, ... , tn)

f2//(t1, ... , tn)

g2(t1, ... , t»)'

then

g2(t1, ... , tn)f1(tl) ... , tn) - g1(t1, ... , tn)f2(t1, ... , tn) = 0. .+.

But if the fractions f1/g1 and f2/g2 are distinct, then the polynomial 92 f1 - 91 f2 is distinct from zero (i.e. has coefficients distinct from 4.+

zero). Therefore if at least one element of the field Q can be expressed in two different ways in the form of a rational function in t1, ... , tn, then over the field P there exists a polynomial f distinct from zero in the n unknowns x1, x2, ... , xn such that f(t1, ... , tn) = 0.

We consider for this polynomial f all polynomials of the form fa (see part I, Ch. 3, para. 1), where II

2

... n)

'r.

(1

i2 ... l»

is an arbitrary permutation of degree n. By definition: fa(xi, x2, ... , xn) = f (xit, xi2, ... , xi»).

All the polynomials fa are distinct from zero (why?) and hence their

product F is also distinct from the zero polynomial. But, as we

THE GENERAL EQUATION OF DEGREE n i 5

103

know (see part I, Ch. 3, para. 1), this product is a symmetric polynomial. Hence, by the fundamental theorem on symmetric polynomials (see The Course, p. 241) the polynomial F(x1, x2, ... , x-)

can be expressed in the form of a non-zero polynomial (with coefficients in the field P) in the elementary symmetric polynomials in x1, ... , xa. But when x; = t, the latter polynomials coincide to within sign with the coefficients a1, ... , a of equation (1). Hence, over the field P there exists a polynomial g distinct from zero such that g(a1, ... , aa) = F(t1,

,

ta)

On the other hand, the element F(t1, ... , of the field Q is the product of all the elements of the form fa(t1, ... , where a e S. fe (t1, ... , Because this product has the factor f (t1, ... , equal to zero, then F(t1,

,

0.

Hence we have found a polynomial g over the field P distinct from zero such that g(ai, ... , a.) = 0.

''-!'

CO)

But this is impossible, because, by hypothesis, the coefficients a1, ... , a are independent variables and no polynomial in them distinct from zero over the field P can be equal to zero. The contradiction obtained proves that the representation of any element of the field Q in the form of a rational function in t1, ... , t,, is unique. We note that it follows from the proof of the assertion that all the roots t1, ... , t,, are distinct. In fact, if, for instance, t1 = t2, then there exists a polynomial f distinct from zero in the n unknowns

x1, ... , x (namely the polynomial f (x1, ... , x,,) = xl - x2) such that f (t1, ... ,

0. Thus the general equation (1) does not have

multiple roots. Now we consider the Galois group G(Q, P) of the field Q over the

obi

o-+

field P, i.e. the Galois group of equation (1). (We note that the field Q is finite over the field P (because it is the decomposition field of a polynomial) and infinite over the field P (because it contains the elements a1, ... , a,,, which do not satisfy any equation); therefore it is not permissible to speak of the Galois group of the field Q over the field P.)

104

FOUNDATIONS OF GALOIS THEORY

Because equation (1) does not have multiple roots, the Galois group G(Q, P) can be regarded as a group of permutations (see Ch. 3, para. 1). More precisely: there exists a natural monomorphism of the Galois group G(Q, P) into the symmetric group 5,,. The permutation

a= 11Z1

2 ... n i2 ... Z

corresponding in this monomorphism to the automorphism S e G(Q, P), is defined by the relation: tk = tik.

Hence, if the permutation a corresponds to the automorphism S, then for any element A ti, ... , tn)

(2)

g(t1, ... , tn)

of the field Q

(f(tI, ...

,

tn)\S

g(tl, ... , t')

= fa(t1, ...

,

tn) (3)

ga(t1, ... , tn)

Now we shall prove that the monomorphism considered is simultaneously also an epimorphism (and thus also an isomorphism),

i.e. that any permutation a is obtained from an automorphism 10.,

S e G(Q, P).

For this purpose we associate with a permutation a e S a transformation S of the field Q, defining it by formula (3). Because any element of the field Q can be written in a unique way in the form (2), formula (3) in fact defines a unique transformation of the field Q. It is easy to see that the transformation S is one-one (indeed, the inverse mapping is constructed in the same way by means of the vii

permutation a-1) and preserves the operations of addition and multiplication, i.e. it is an automorphism of the field Q. Finally, if the element (2) belongs to the field P, i.e. it can be expressed in terms of a1, ... , then the polynomials f and g are symmetric polynomials and therefore fa = f, g, = g, i.e. the automorphism S leaves the element (2) invariant. Thus S is an automorphism of the field Q over the field P, i.e. S e G(Q, P). It remains to note that the

permutation corresponding to the automorphism S coincides, clearly, with the permutation a.

THE GENERAL EQUATION OF DEGREE n > 5

105 C''

Thus it has been proved that the Galois group G(Q, P) is isotip

morphic to the symmetric group i.e. the Galois group of the general equation of degree n is isomorphic to the symmetric group S. of degree n (whatever the field P is). Hence,

the general equation of degree n where n > 5 is not solvable by radicals (whatever the field P is). On the other hand, if n 5 4, then the general equation of degree n is solvable by radicals (again whatever the field P is). The last result is well known (see, for instance, The Course, p. 282),

but it is not without interest to obtain the known formulae for the solution of equations of degree n < 4 from the general considerations of Galois theory. 4. The solution of equations of low degree

We consider first the quadratic equation

x2+a1x+a2=0.

(1)

CSI

.fl

Let t1, t2 be its roots and Q = C(t1, t2) its decomposition field over the field C = C(al, a2) (because the fundamental field P plays no part, we take it to be the field C of complex numbers). The Galois group of equation (1) is the symmetric group S2. Because this group is a cyclic group of order 2 and, hence, has no subgroups, then the field Q also has no intermediate subfields. Therefore, for example, the field C(tl) coincides with the field Q: Q = C(t1)

The unique non-identical automorphism S of the field Q over the field C carries the root tl into the root t2:

tis =t2 (because otherwise S = E).

According to the general theory we must find the Lagrange resolvents. Because the number -1 is a primitive square root of unity, the Lagrange resolvents in our case have the form

(-1, t1) = tl - t2, (1, tl) = tl + t2.

106

FOUNDATIONS OF GALOIS THEORY

We denote the resolvent (- 1, t1) by the symbol 0:

0=t1-t2. The other resolvent is the elementary symmetric polynomial

t1+t2=-a1. From the equations t1 - t2 = 0 and t1 + t2 = -a it follows that

2t1=-a1+0, 2t2=-a1-0 (which is also consistent with the general theory; see Ch. 2, para. 2), i.e. that

-a1 ± 0 t1,2 =

2

Further,

a2=tl+t2-2t1t2=a2-4a1

and hence

0=.J{a2-4a1}. Thus we have in fact obtained the known formulae for the solution of a quadratic equation. Now we consider the cubic equation

y3+a1y2+a2y+a3=0. Setting

x = y + a3 we can reduce this to the form

x3+px+q=0,

(2)

where 2

3

- a2a1 p= - ai3 +a2, q= 2ai +a3. 27 3 v..

(This transformation is not essential and is performed only in order to simplify the further calculations.) Let t1, t2, t3 be the roots of equation (2) and, hence, Q = C(t1, t2, t3), where C = C(p, q), is its decomposition field.

As we know, the Galois group S3 of this equation possesses the solvable series S3=:) A3=) e.

THE GENERAL EQUATION OF DEGREE n i 5

107

Let L be the intermediate field

C=L=Q, corresponding to the subgroup A3. Then the Galois group G(Q, L) of the field Q over the field L is exactly the group A3. This group is cyclic, of order 3 and a generator of it is, for instance, the permutation (1 2 3). Let S be the automorphism corresponding to this permu-

tation: s s tiS =t2, t2= t3, t3=t1.

Because ti

L. Hence the field L(t1) must coincide

t1, then t1 with the whole field Q :

Q = L(t1)

(why?). According to the general theory we must consider the Lagrange resolvents: (P, t1) = tl + P41 +

P2tiz

= tl + Pt2 +

p2

t3,

(p2, t1) = t1 + p2ti + p4ti = t1 + p2t2 + p4t3,

(p3, t1) = t1 + P3ti + P6ti = t1 + p3t2 + P6 t3, where

_ P

-1 +,/-3 2

is a primitive cube root of unity. Because p2 = p and p3 = 1, then (P, t1) = tl + pt2 + pt3, (P21 t1) = t1 + pt2 + pt3,

(P3,t1)=t1 +t2+t3=0. Adding together all three resolvents, we obtain: 3t1 = (p, t1) + (p2, t1)

(3)

(we do not write down the third resolvent, because it is equal to zero). This result is also consistent with the general theory.

108

FOUNDATIONS OF GALOIS THEORY

According to the general theory the cube of the resolvent (p, t1) must belong to the field L. But (p, t1)3 = ti + t2 + t3 + 3p(t1t2 + t2t3 + t3t1) + N."

+ 3p(t1t2 + t2t3 + t3t1) + 6tit2t3

=t1+t2+ t3- y(tlt2+tyt3+t3t1+t1t2+t2t3+t3t1)+ 3 3

(t1t2 + t2t3 + t3 t1 - t1t2 - t2t3 - t3t1). *..

+ 6t1t2t3 +

""'

2

Expressing the symmetric polynomials in terms of the elementary II`

+..

polynomials (and making use of the fact that t1 + t2 + t3 = 0, tit2 + t1t3 + t2t3 = p, t1t2t3 = -q), we obtain

t;3 + t23 + t3 = - 3q, t1t2 + t2t3 + t3t1 + t1t2 + t2t3 + t3t1 = 3q,

t1t2t3 = -q. Further, it is easy to see that

t12+t22 t3+t3t1-t1t2-t2t3-t3t1=0, where

0 = (t1- t2)(tl - t3)(t2 - t3) Thus

(p,t1)3=-

q+3V-3

227

6.

(4)

2

We see that in fact (p, t1)3 e L, because BS = 0, and therefore 0 e L. Analogously it can be calculated that (P2, t1)3 = -

27

2

q-3

2

3 0.

(5)

Now we find 0. Any transposition carries 0 into - 0, and any even mgr

,U)

permutation leaves 0 invariant. Therefore only the number -0 is conjugate to 0 and, hence, 02 e C. In fact, a simple calculation shows

that 02 = -4p3 - 27q 2.

(6)

109

THE GENERAL EQUATION OF DEGREE n >, 5

Comparing formulae (3), (4), (5) and (6), we find at last the following formula for the solution of a cubic equation:

2+](a+27))

2

+ -J{

3

2-]`4 +27/

i.e. Cardan's well known formula. 7C0

Equations of degree 4 are considered analogously. It is left to the reader as an exercise to work out the corresponding argument.