Elements of Mathematics: Algebra II: Chapters 4-7 9783540007067, 9783642616983, 3540007067, 3642616984

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ELEMENTS OF MATHEMATICS

Springer Berlin Heidelberg New York Hong Kong London Milan Paris Tokyo

NICOLAS BOURBAKI

Algebra II Chapters 4-7 Translated by P.M. Cohn & J. Howie

,

Springer

Originally published as

ALGEBRE, CHAPITRES 4 A 7 Masson, Paris, 1981

Mathematics Subject Classification (2000): 06-XX, 1OXX, 20XX, 1SXX, 18XX

Cataloging-in-Publication Data applied for A catalog record for this book is available from the Library of Congress. Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at http://dnb.ddb.de

ISBN-13: 978-3-540-00706-7 DOl: 10.1 007/978-3-642-61698-3

e-ISBN-13: 978-3-642-61698-3

This work is subject to copyright. All rights are reserved. whether the whole or part of the material is concerned. specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on

microfilm or in any other way. and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9. 1965. in its current version. and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. Springer-Verlag Berlin Heidelberg New York a member of Springer Science+ Business Media http:/www.springer.de © Springer- Verlag Berlin Heidelberg 2003 The use of general descriptive names. registered names. trademarks. etc. in this publication does not imply. even in the absence of a specific statement, that such names are exempt from the relevant pro-

tective laws and regulations and therefore free for general use. Cover Design: Design & Production GmbH. Heidelberg Printed on acid-free paper 41/3111 db 54321 SPIN 11419846

To the reader

1. The Elements of Mathematics Series takes up mathematics at the beginning, and gives complete proofs. In principle, it requires no particular knowledge of mathematics on the readers' part, but only a certain familiarity with mathematical reasoning and a certain capacity for abstract thought. Nevertheless, it is directed especially to those who have a good knowledge of at least the content of the first year or two of a university mathematics course.

2. The method of exposition we have chosen is axiomatic, and normally' proceeds from the general to the particular. The demands of proof impose a rigorously fixed order on the subject matter. It follows that the utility of certain considerations will not be immediately apparent to the reader unless he has already a fairly extended knowledge of mathematics. 3. The series is divided into Books and each Book into chapters. The Books

already published, either in whole or in part, in the French edition, are listed below. When an English translation is available, the corresponding English title is mentioned between parentheses. Throughout the volume a reference indicates the English edition, when available, and the French edition otherwise. (Set Theory) designated by E Theorie des Ensembles (Theory of Sets) Algebre (Algebra(1» A (Alg) Topologie Generale (General Topology) TG (Gen. Top.) FVR Fonctions d'une Variable Reelle Espaces Vectoriels Topologiques (Topological EVT (Top. Vect. Sp.) Vector Sp!lces) INT Integration Algebre Commutative (Commutative Algebra(2» AC (Comm. Alg.) VAR Varietes Differentielles et Analytiq ues Groupes et Algebres de Lie (Lie Groups and Lie Algebras(3» LIE (Lie) TS Theories Spectrales

e)

e) e)

So far, chapters I to VII only have been translated. So far, chapters I to VII only have been translated. So far, chapters I to III only have been translated.

VI

ALGEBRA

In the first six books (according to the above order), every statement in the text assumes as known only those results which have already been discussed in the same chapter, or in the previous chapters ordered as follows: E ; A, chapters I to III ; TG, chapters I to III ; A, from chapters IV on ; TG, from chapter IV on ; FVR ; EVT; INT. From the seventh Book on, the reader will usually find a precise indication of its logical relationship to the other Books (the first six Books being always assumed to be known). 4. However we have sometimes inserted examples in the text which refer to facts the reader may already know but which have not yet been discussed in the series. Such examples are placed between two asterisks: *... *. Most readers will undoubtedly find that these examples will help them to understand the text. In other cases, the passages between *... * refer to results which are discussed elsewhere in the Series. We hope the reader will be able to verify the absence of any vicious circle. 5. The logical framework of each chapter consists of the definitions, the axioms, and the theorems of the chapter. These are the parts that have mainly to be borne in mind for subsequent use. Less important results and those which can easily be deduced from the theorems are labelled as « propositions », « lemmas », « corollaries », « remarks », etc. Those which may be omitted at a first reading are printed in small type. A commentary on a particularly important theorem appears occasionally under the name of « scholium ». To avoid tedious repetitions it is sometimes convenient to introduce notations or abbreviations which are in force only within a certain chapter or a certain section of a chapter (for example, in a chapter which is concerned only with commutative rings, the word« ring» would always signify «commutative ring »). Such conventions are always explicitly mentioned, generally at the beginning of the chapter in which they occur.

Z

6. Some passages in the text are designed to forewarn the reader against serious errors. These passages are signposted in the margin with the sign (- p » ; the set of polynomials u such that deg u ~ p is thus an A-submodule of A [(Xi)i E d, equal to Po + PI + ... + P p with the above notations. Let E be an A-module. The family (E ® P n)n E N is a graduation of type N of the module E [(X;); E 11 = E ® A A [(X;); E 11 of polynomials with coefficients in E. We extend the conventions adopted above for the degree of inhomogeneous polynomials to this case.

PROPOSITION 1. - Let u and v be two polynomials. (i) If deg u '# deg v, we have u+v'#O

and

deg(u+v)=sup(degu,degv).

If deg u = deg v we have deg(u + v) ~ deg u. (ii) We have deg(uv) ~ deg u + deg v. The proof is immediate.

Let J c I and B = A [ (Xi )i E I _ J] ; we shall identify A [ (Xi )i E I] with B [ (Xi )i EJ ] (III, p. 453 f.). The degree of u E A[(Xi)iEd, qua element of B[(Xi)iEJ], is called the degree of u with respect to the Xi of index i E J (III, p. 454). n

Let u =

L

akXk E A [X] be a non-zero polynomial of degree n in a single

k~O

indeterminate. The coefficient an which is '# 0 by hypothesis, is called the leading coefficient of u. A polynomial u '# 0 whose leading coefficient is equal to 1 is called a monic polynomial. In A [Xl' X 2 , ... , Xq] the number of monomials of total degree p is equal to the number of elements (nk)l -+ f (x) of T [ into E is called the rational function associated with f (and E); we sometimes denote it by 1. If 9 E K ( (Xj )j EI) we have T [ n T geT [ + g' T [ n T geT [g' hence the rational function associated with f + 9 (resp. fg) is defined on T [ n T 9 and has the same value on this set as 1 + 9 (resp. 19). Let T, be the set of x E T [ such that f(x) is invertible; if x E T" x is substitutable in II f and the rational function associated with 1I f takes at x the value f (x I.

t

If K is an infinite commutative field, f E K «Xj)j EI)' 9 E K «Xj)j E I) and 1, 9 are the rational functions associated with f, 9 (and K), and if l(x) = 9 (x) for all x E T [ n Tg then f = g. For if f = ulv and 9 = ul/v l , where u, v, UI' VI are polynomials, we have u(x) VI (x) = UI (x) V (x) for all x such that v(x)vI(x) oF 0, hence UVI = ulv (IV, p.18, Th.2). Therefore the mapping f >-+ 1 is injective and we shall often identify f and 1.

No.4

RATIONAL FRACTIONS

A.IV.23

* Using the factoriality of K[(Xi)iEI] (Comm. Alg., VII, § 3, No.2 p. 502 and Cor. of Th. 2 p. 506), one easily shows the following: for every f E K «Xi )i EI) there exist u, v E K [(Xi)i Er1 such that: 1) f = u/v ; 2) for x E KI to be substitutable in f it is necessary and sufficient that v(x)#,O.* 4.

Differentials and derivations

Let K be a commutative field. By III, p. 558, Prop. 5, every derivation D of K[(Xi)iEd extends in a unique fashion to a derivation D of K«Xi)iEd. If D, DI are permutable derivations of K [(Xi)i Ed, then the bracket [D, D' 1 = DD' - D'D is zero, hence

[D, D'l which is a derivation of K( (Xi)i E I)

extending [D, D' 1is zero; in other words, D and D' are permutable. In particular the derivations Di (IV, p. 6) extend to derivations of K ( (Xi )i E I) again denoted by Di and which are pairwise permutable. If f E K «Xi)i EI), Di f is also written D x/ or

aa1.

or f'x;. When there is only a single indeterminate X one uses the

I

. Df , dX df ' f' . notatIOn Let B = K[(Xi)iEd, C = K«Xi)iEI). By III, p. 574, Prop. 23, the canonical mapping

is an isomorphism of vector C-spaces. Bearing in mind III, p. 570, we see that the vector C-space OK (C) admits as a basis the family (dXi)i E I of the differentials of the Xi. Let i be the coordinate form of index ion 0K(C) relative to that basis. Then the mapping u >--+ -+ u(x) of A[[KJ] into E; we obtain (4)

A.IV.30

§4

POLYNOMIALS AND RATIONAL FRACTIONS

Let f= (fi)iEI E (A[[J]])I and 9 = (9j)jEJ E AJ,K. We denote by f(g) or fog the element (fi «9j)j EJ ))i E I of (A[[K]])I. If f E Au, we have fog E AI,K because the mapping f~ f«9j)jEJ) of A[[I]] into A[[K]] is continuous. Let fE (A[[J]])I, g E AJ,K' hE AK,L. Then go h E AJ,L and by (4), we have

(5)

(f 0 g)

4.

0

h= f

0

(g

0

h) .

Invertible formal power series

PROPOSITION 5. -

In the ring A[[T]] of formal power series in one indeterminate

the polynomial 1 - T is invertible, and we have (1 -

Tt 1 =

L r. O. Now there is a ring homomorphism 'P: A [[T]] -+ A [[I]] such that 'P (T) = t, and

1 - T is invertible in A[[T]] (Prop. 5); consequently 1 - t is invertible in A[[I]], and hence so is u. Remark. - Let A be the set of all formal power series with constant term 1. By Prop. 6, A is a commutative group under multiplication; the multiplicative group of A [[I]) is thus the direct product of A and the multiplicative group of A. We shall equip A with the topology induced from that of A [[I]). For each ~ E N(I) we have in IV, p.26 defined the ideal a~ of A[[I]] ; then 1 + a~ is a subgroup of A and the family (1 + a~) is a fundamental system of neighbourhoods of 1 in A. Since the multiplication in A is continuous, we see that A is a topological group (Gen. Top., III, p. 223) ; in other words, the mapping I t-+ 1-1 is continuous in A.

Let K be a commutative field and .0 the subring of the field of rational fractions K ( (Xi )i E I) formed of rational fractions in which the element 0 of K I is substitutable. If

f

E .0, we have

f

=

'!. , where u and v are polynomials such that v

the constant term of v is =Ie- 0, hence v is invertible in K[[I]]. We can verify at once that the element uv- 1 of K[[I]] depends only on f; we say that the formal power series uv- I is the expansion at the origin of the rational fraction'!. . The mapping

f

v

~

uv- 1 is an injective homomorphism of.o into K[[I]] ; we shall often identify

.0 with its image under this mapping.

No.5

5.

A.IV.31

FORMAL POWER SERIES

Taylor's formula for formal power series

Let X = (Xi)i E I and Y = (Yi)i E I be two families of indeterminates relative to the same index set I. We denote by X + Y the family (Xi + Yi)i E I of formal power series in A [[X, Y]]. It is clear that we can substitute Xi + Y i for Xi in a formal power series U E A [[X]], the result being written u (X + Y). For each v E N(I) we denote by dVu the coefficient of yv in the formal power series u(X + Y) considered as belonging to A[[XJJ[[YJJ (III, p. 456). In other words, we have

(6)

u(X

+ Y) =

L dVu(X). yv

(u E A [[X 1])

.

Substituting (0, X) for (X, Y) we obtain (7)

In other words, the constant term of dVu is the coefficient of Xv in u. Since the mapping u ~ u(X + Y) of A[[X]] into A [[X, V]] is continuous, the mappings u ~ d Vu of A [[X 11 into itself are again continuous. As in the case of polynomials (IV, p. 7) we can prove the formulae (8) APA(1 L.>. L.>. U

(9)

(p + Minto M and f the composite mapping

TS(M) ® TS(M)": TS(M Et> M)

~

x + y of

~ TS(M)

where h is the canonical homomorphism. If z, z' E TS (M), then f(z ® z') = zz'. For let i be the mapping x ~ (x, 0) of Minto MEt> M. We have u 0 i = Id M, hence TS (u) 0 TS (i) = IdTs(M) ; therefore f(z ® 1) = TS(u) (h(z ® 1» = TS(u) (TS(i)(z» = z. Likewise f(l ® z') = z', whence f(z ® z') = f(z ® 1) f(l ® z') = zz'. 7.

Coproduct for symmetric tensors

Let M be a free A-module, and LlM = Ll the diagonal homomorphism x ~ (x, x) of Minto M Et> M. Let c M = c be the unital homomorphism of graded A-algebras composed of the homomorphisms: TS(M) ~ TS(M Et> M)': TS(M) ® TS(M) where IT is the canonical isomorphism. Equipped with c, TS (M) is a graded Acogebra. For all x E M and each integer p ~ 0 we have TS(Ll) ("Yp(x» = 'Yp«x, x», hence by (8), (10) r+s=p

In particular c(x)

(11)

Let (Xi)i

E

I

=

x ® 1+1® x .

be a family of elements of M, and for v

E N(I)

put x. =

fl 'Y. (Xi)' j

i

Then (12)

This follows from (10) because c is an algebra homomorphism.

E

I

No.7

A.IV.S1

SYMMETRIC TENSORS AND POLYNOMIAL MAPPINGS

PROPOSITION 8. - Let M be a free A-module, then with its algebra and cogebra structures, TS (M) is a graded commutative and cocommutative bigebra. The counit is the A-linear mapping E : TS (M) --. TS o(M) = A which is zero on TsP (M) for p >- 0 and such that E (1) = l. We know that the A-algebra TS (M) is associative, commutative and unital. On the other hand, the coproduct is by construction a homomorphism of graded algebras; now the fact that the cogebra TS (M) is coassociative and cocommutative follows by an easy calculation from the Formula (10). The mapping E of TS (M) into A is a homomorphism of graded algebras such that E (1) = 1. Finally for every x EM we have E{-yp (x» = 0 ifp >- 0, E{-yO(X» = 1 ; if we bear in mind (10), this shows that (E 1) 0 c = (1 E) 0 C = IdTS(M) ; thus E is the counit of

TS(M).

Let M and N be free A-modules and u an A-homomorphism of Minto N ; then TS (u) is a bigebra homomorphism. For we have aN 0 u = (u, u) 0 aM' hence the diagram PROPOSITION 9. -

TS(M)

TS(ll.,,)

• TS(MEB M)

TS,.,j

cr

• TS(M) ® TS(M)

)TS'.' • TS,.)

JTS'..)

TS(N)

TS(ll.N)

~

TS(N EEl N)

TS(N) ® TS(N) ,

~

where IT and T are canonical isomorphisms, is commutative (IV, p.49). Thus cNoTS(u)= (TS(u) TS(u»oc M · PROPOSITION 10. - Let M be a free A-module, then the primitive elements (III, p. 602) of the bigebra TS (M) are the elements of M. Let (e;); E I be a basis of M, and for v E N(I) put ev = "Yv,(e;). Let

n

L

z=

j

E I

Avev be an element of TS (M), then by (12) we have

vE Nil)

c(z) =

L Av p,uEN(I),p+cr=lI

p, ..

hence

c(z)-lz-zl=

L

Ap+ (i) and (iv) : let h satisfy the conditions of (ii). By Prop. 4, (ii) (IV, p. 47) there exists a linear mapping 9 / of Tq (M) into N such that h = 9 / I TSq (M). Let 9 be the q-linear mapping of Minto N associated with 9 /, then for any X EM we have f (x)

=

h (-y q (x ))

=

9 / (x ® x ® ... ® x )

=

9 (x, x, ... , x) ,

No.9

whence (i). On the other hand, if (e;); (IV, p. 46)

on writing e v =

E

E

I

f(LA;e;) = i

L Iv I

AVh(ev)· ~

q

(iii) is clear. (ii): given (e;), (u v ) satisfying the conditions of (iii), let us write "VVi(e;) and recall that (ev)lvl =q is a basis of TSq(M). Let h be the

~

~

n

ev =

is a basis of M, we have, by formula (6)

I

n "VVi (e;) ; we thus have i

(iv) (iii)

A.IV.55

SYMMETRIC TENSORS AND POLYNOMIAL MAPPINGS

i E I

homomorphism of TSq(M) into N defined by h(e v ) = u v

L A;e;

x = i

E

;

then for each

in M we have

I

f(x)=f(LA;e;) =

L Ivl

i

AVUv=h( ~q

AVev ) =h("Vq(x».

L Ivl

~q

DEFINITION 3. - Let M and N be A-modules and q an integer ~ O. Suppose that M is free, and denote by Pol~ (M, N) or simply Polq (M, N) the set of mappings ofM into N satisfying the conditions of Prop. 13. The elements ofPolq (M, N) are called homogeneous polynomial mappings of degree q of Minto N.

Prop. 13 (i) defines a homomorphism of A-modules:

Prop. 13 (ii) defines a homomorphism of A-modules:

These homomorphisms are called canonical. They are surjective. Examples. - 1) The homogeneous polynomial mappings of degree 1 of Minto N are the A-linear mappings of Minto N. 2) Let (N;); E I be a family of A-modules, f; a mapping of Minto N;, i E I, and f: M - t Ni the mapping with components fi. For f to be a

n

j E

I

homogeneous polynomial mapping of degree q it is necessary and sufficient that each f; be a homogeneous polynomial mapping of degree q. 3) Let (Mj)j E J be a finite family of free A-modules and u:

nM

j

multilinear mapping; then u is polynomial of degree Card(J).

E

J

j -t

N a

A.IV.56

§5

POLYNOMIALS AND RATIONAL FRACTIONS

4) Let (Xi)i E I be a family of indeterminates, N an A-module and u E N [(Xi )i E I] a homogeneous polynomial of degree q. The mapping (Xi)i E I ~ U ( (Xi )i E I) of A (l) into N is a homogeneous polynomial mapping of degree q : this is seen at once by condition (iii) of Prop. 13. If I is finite, every homogeneous polynomial mapping of degree q of A (I) = Al into N is of that form. 5) The mapping (Xi)i E N ~ X~ + xl + ... + X~ + ... of A (N) into A is a homogeneous polynomial mapping of degree 2. If A = Z/2Z, it coincides with the linear mapping (Xi)i E I ~ Xo + Xl + ... + Xn + ... 6) Let f E Polj" (M, N), let B be a commutative ring, p a homomorphism of B into A and M' and N' the B-modules derived from M and N by means of p. Assume that M' is free; then f E Polfi (M', N'): this follows at once from condition (i) of Prop. 13. PROPOSITION 14. - Let M, N, P be A-modules, q and r integers;:: 0, and assume that M and N are free. If f E Polq(M, N), f' E polr(N, P), then f' 0 f E polqr(M, P). There exists a q-linear mapping 9 of Mq into N and an r-linear mapping g' of N r into P such that

f (x) = 9 (x, X, ... , x ) f'(y) = g'(y,y, ... ,y)

for all x EM, for all YEN.

Hence for every x E M we have f'(f(X)) = g'(f(x), f(x), ... , f(x)) = g'(g(x,x, ... ,x), ... , g(x,x, ... , x))

and the mapping (Xl' Mqr into P is qr-linear.

... ,Xqr)~g'(g(Xl'

... ,Xq), ... ,g(Xq(r-I)+I' ... ,x qr )) of

PROPOSITION 15. - Let M be a free A-module, N an A-module and q an integer ;:: O. We suppose that the mapping y ~ q!y is an automorphism of N. Let f E Polq (M, N), then there exists one and only one symmetric q-linear mapping h ofMq into M such that f(x) = hex, x, ... , x) for all x E M. For any Xl' ... , Xq E M we have

(16)

(- l)q h(X l ,X2, ... ,xq)=-q-'-

L He {1.2 •...• q}

•

(-1)

CardH

f

(

)

LXi' i

E

H

a) There exists a q-linear mapping 9 of Mq into N such that f(x) = 9 (x, x, ... , x) for all x E M. Let us define a q-linear mapping h of Minto N by h(xl' X2' ... , x q)

=, 1

q.

L orE

g(Xa(I)' Xa(2), ... , xa(q»).

6q

Then h is symmetric and f(x) = h (x, x, ... , x) for all x E M. b) Let h be a symmetric q-linear mapping of Mq into N such that

No. 10

SYMMETRIC TENSORS AND POLYNOMIAL MAPPINGS

A.IV.57

f (x) = 9 (x, X, ... , x). Let I be the linear mapping of Tq (M) into N such that h(Xl' ... , Xq) = I (Xl ® ... ® Xq) for any Xl' ... , Xq E M. We have

= (-l)ql(s(Xl®···®Xq))=

He {t."q} (_l)CardH/("Yq(~HX;))

by Prop. 3, (v) (IV, p. 45). Now

and this proves formula (16) and the uniqueness of h. PROPOSITION 16. Let M be a free A-module, N an A-module, q a poslt!ve integer and u the canonical homomorphism of Hom (TSq (M), N) into Polq (M, N). (i) If A is an infinite integral domain and N is torsion-free, then u is an isomorphism. (ii) If the mapping y >--'» q!y in N is injective, u is an isomorphism. In the two cases of the proposition we have to prove that u is injective, that is,. every linear mapping f of TSq(M) into N which is zero on "Yq(M), vanishes. Assume A to be an infinite integral domain and N torsion-free. For every z E TSq (M) there exists a E A - {O} such that az is an A-linear combination of elements of "Yq(M) (IV, p. 47, Prop. 5). Hence af(z) = f(az) = 0, and so f(z)=O. Suppose next that the mapping y >--'» q!y in N is injective, then by IV, p. 45, Prop. 3, (v), fvanishes on s. Tq(M). Hence if z E TSq(M), we have q!f(z) = f(sz) =0, and so f(z)=O.

Let M be a free A-module, N an A-module, q a positive integer, Polq (M, N) and (e;)i E I a basis ofM. In the two cases of Prop. 16 there exists a

COROLLARY. -

h

E

unique family (UV)VEN(I),lvl = q of elements ofN such that h

(I. Aiei) I E

for all (Ai)

10.

E

A

I

I.

Iv I

=

AVu v q

(I).

Polynomial mappings

DEFINITION 4. Let M and N be A-modules and assume that M is free. Let Map(M, N) be the A-module of all mappings of Minto N. The submodule PoIHM, N) of Map (M, N) is denoted by PoIA(M, N) or simply Pol(M, N) ;

I.

q>O

its elements are called polynomial mappings of Minto N.

A.lV.58

POLYNOMIALS AND RATIONAL FRACfIONS

§5

Let (ei)i E I be a basis of M and suppose that I is finite; by Prop. 13 (IV, p. 54) a mapping f of Minto N is polynomial if and only if there exists a polynomial F in the indeterminates Xi with coefficients in N such that f

(I Xi ei ) I E

=

F(x)

I

for every family x = (Xi)i E I in A (I). This property is independent of the basis chosen for M and it justifies the terminology « polynomial mapping ». PROPOSITION 17. - Let M be a free A-module and B an associative, commutative and unital A-algebra. Then Pol A (M, B) is a sub-B-algebra of the algebra Map(M, B). This follows from Def. 4 and Prop. 13, (iv) (IV, p. 54). PROPOSITION 18. - Let M, N, P be A-modules, and assume that M and N are free. If fEPol(M,N), gEPol(N,P), then gofEPol(M,P). We can reduce at once to the case where there exists an integer q such that 9 E Polq (N, P) ; then there exists a q-linear mapping h of Nq into P such that 9 (y) = h (y, y, ... , y ) for all YEN. Writing f as a sum of homogeneous polynomial mappings we are thus reduced to proving that the mapping

of Minto P, where fi E Pol q; (M, N), is polynomial. For i = 1, ... , q there exists a qi-linear mapping li of M q ; into N such that fi(X) X E M. Hence

=

li(X,x, ... ,x) for all

from which our assertion follows. Lemma 2. -

Let N be an A-module, n an integer "'" 0 and

Suppose that there exist ao, aI' ... , an E A such that f(a o) = ... = f(a n ) = 0, and such that for i #= j the homothety of ratio a i - aj in N is injective, then f = O. (This lemma generalizes the Cor. of IV, p. 16.) The lemma clearly holds for n = 0 ; we shall prove it by induction on n. We have

I

f(X) = f(X) - f(ao) = i

= 1

m;(Xi - a~) = (X - ao) g(X)

No. 11

SYMMETRIC TENSORS AND POLYNOMIAL MAPPINGS

A.IV.59

where g is an element of N [X] of the form ma + mj X + ... + m~ _ 1 xn - 1. The hypotheses of the lemma imply that g (0.1) = ... = g (an) = 0, hence g = 0 by the induction hypothesis, and so f = O. PROPOSITION 19. Let M be a free A-module, N an A-module, G an infinite additive subgroup of A, and suppose that the homotheties ofN defined by the nonzero elements of G are injective. Then Pol (M, N) is the direct sum of the Polq(M, N). Let fa, fl' ... , fn be such that fi E Pol i (M, N) and suppose that we have the relation fa + ... + fn = O. Let x E M, then for all A E G we have

By Lemma 2, applied to the polynomial

L fi (x) Xi

we have

i = 0

fo(x) = ... = fn(x) = O. COROLLARY. - Assume that A is an infinite integral domain; let M be a free Amodule and N a torsion-free A-module. (i) We have Pol(M, N) = EEl Polq(M, N) and each Polq(M, N) may be q .. O

identified canonically with Hom (TSq (M), N). (ii) Let f E Pol (M, N) and (ei)i E I a basis ofM. There exists one and only one family (u.). E NO) of elements of

Nsuch

that f (

L AA)

i E I

L

= l.I

A·u. for all

e N(I)

(Ai) E A (I). The assertion (i) follows from Prop. 16 and 19, and (ii) follows from (i) and the Cor. of Prop. 16.

11.

Relations between S(M*), TS(M)*gr and Pol(M, A)

Let M be a free A-module, we shall equip the graded dual TS(M)*gr with the structure of a commutative, associative and unital graded 1 algebra, derived from the graded cogebra structure of TS(M) (III, p. 580). By III, p. 497 there exists a unique homomorphism of graded A-algebras

e : S (M *) ..... TS (M) *gr inducing in degree 1 the identity mapping of M*.

1 A graded homomorphism of degree - k of TS (M) into A is here considered as an element of degree k of TS (M) *gr (II, p. 377).

A.IV.60

POLYNOMIALS AND RATIONAL FRACTIONS

§5

PROPOSITION 20. - If the A-module M is free and finitely generated, then e is an isomorphism of graded algebras. Let (ei)i E I be a basis of M and (et)i E I the dual basis of M*. For v E N 1 put

eV

=

TI 'YVi(e i

E

i) E

TS(M).

I

By Prop. 4 (IV, p. 47) the family (ev)vEN 1 is a basis of TS(M); let (en be the basis of TS(M)*gr dual to (eJ. In the light of III, p. 505, Th. 1 it is enough to show that for any v E N 1 we have eV* =

TI (etfi , j

or also that for p, IT E N 1 we have e p* . from IV, p. 50, Formula (12).

eI

e: = e *

p+a ;

but this last assertion follows

Remark 1. - In the same way we see that if M is a finitely generated free Amodule, then the graded algebra S(M)*gr defined in III, p. 593, may be identified with TS(M*). PROPOSITION

21. -

The canonical homomorphism of A-modules (IV, p. 55) u: TS(M)*gr -+ PolA(M, A)

is an algebra homomorphism. Let a E TSq(M)*, bE Tsr(M)*, x EM; we have

= (a ® b, c(-Yq+r(X») = = (a ® b, 'Yq(X) ® 'Yr(X» = (a, 'Yq(x» (b, 'Yr(x» = u(a)(x). u(b)(x) ,

u(ab)(x) = (ab, 'Yq+r(x»

whence the result. Remarks. 2) The composite homomorphism Am = u 0 e : S (M *) -+ Pol A(M, A) is the unique unital homomorphism of algebras inducing the inclusion of

M* = POII(M, A) in Pol (M, A). If M is finitely generated free and A is an infinite integral domain, then AM is bijective (Prop. 20 and Cor. of Prop. 19). In particular if A is an infinite integral domain, then the canonical homomorphism f ~ 1 of A [Xl' ... , Xn] into Pol (An, A) (IV, p. 4) is an isomorphism. 3) Consider the coproduct cs: S(M*) -+ S(M* x M*) (III, p. 575, Example 6). For any v E S(M*), x, Y EM, the polynomial mapping AMxM(CS(v»: M x M -+ A maps (x, y) to AM(V)(X + y). For the two algebra homomorphisms

No.1

A.IY.61

SYMMETRIC FUNCTIONS

of 5 (M *) into Map (M x M, A) defined in this way agree on M*, in virtue of the relation (v ® 1 + 1 ® v ) (x, y) = v (x + y)

(v

E

M

*) .

§ 6. SYMMETRIC FUNCTIONS

1. Symmetric polynomials

Let n be a positive integer. For every permutation rr E 6 n let 'P" be the automorphism of the A-algebra A[X}, ... , X n] which maps Xi to X"(i) for 1 ~ i ~ n. It is clear that rr >--+ 'P" is a homomorphism of 6 n into the group of automorphisms of A[X}, ... , Xnl. We shall put rrf = 'P,,(f) for rr E 6 n and f E A[X}, ... , Xn]. The polynomial f is said to be symmetric if rrf = f for all rr E 6 n ; the symmetric polynomials form a unital graded sub algebra of A[X}, ... , Xn]; we shall denote it by A[X}, ... , xn]sym in the rest of this paragraph. For every positive integer k we denote by ~k the set of k-element subsets of the set {I, 2, ... , n} and put (1)

Sk

=

L

n

Xi'

He'llk ieH

When we wish to specify the integer n we shall write We have in particular

sk,n

instead of

Sk'

So

= 1

S}

=

L Xi L XiX

l=s:;..;;n S2

=

j

l:Ei- n. It is clear that Sk is a homogeneous symmetric polynomial of degree k ; we shall call it the elementary symmetric polynomial of degree k. In the ring A[X}, ... , X n , U, Y] we have the relation

(2)

. (U +

f1

YX i )

=

.

I

Un-kykSk ;

i:= 1

by appropriate substitutions we deduce the relations

•

f1

(3) i

=

1

(1 +TXJ =

•

I

skTk ,

A.IV.62

POLYNOMIALS AND RATIONAL FRACTIONS n

f1

(4)

(X - Xi) =

n

L

§6

(_l)n-k sn _ kX k .

i = 1

Let us put E = A [Xl' ... , Xn] and S = A [Xl' ... , xn]sym. A-algebra S of symmetric polynomials is generated by Sl' ... , Sn. elements Sl' ... , sn ofE are algebraically independent over A (IV, p. 4). family of monomials Xv = X;-(l) ... x~(n) such that 0 ~ v(i) -+ I\Ia is an injective homomorphism of 6 n into the automorphism group of K(Xl' ... , Xn). For each f E K(Xl' ... , Xn) we have (I\Ia f) (Xl' ... , Xn) = f(Xa(I)' ... , Xa(n»' The rational fractions f such that I\IaU) = f for all CI E 6 n are called symmetric rational fractions. The set of al! symmetric rational fractions in Xl' ... , Xn is a subfield of K(Xl' ... , Xn)· CI E

PROPOSITION 2. - The field of symmetric rational fractions in Xl>"" Xn is the field of fractions of the ring of symmetric polynomials in Xl' ... , X n· Let two V

=

f

K(Xl' ... , Xn) be a symmetric rational fraction, and let Ul' elements of K[Xl' ... , Xn] such that f = u l ". We E

n I\Ia(vd E K[Xl' ... , Xn]

ae

VI

VI

and

= fv E K [Xl' ... , X n ]; then

U

V

be put

is sym-

3,.

metric, hence u is symmetric, because

f

is, and

f

=

~ , whence the result. V

COROLLARY. - Let Sl' S2, ... , sn be the elementary symmetric polynomials in Xl' ... , X n· For every rational fraction 9 E K(SI' S2' ... , Sn) the sequence (Sl' S2' ... , sn) is substitutable in 9 and the mapping 9>-+ 9(SI' s2' ... , sn) is an isomorphism ofK(SI' S2' ... , Sn) onto the field of symmetric rational fractions in Xl' ... , X n • This follows from Prop. 2 and Th. 1 of IV, p. 62. 3.

Symmetric formal power series

Let I be a set, X = (Xi)i e , a family of indeterminates and A [[X]] the algebra of formal power series in the Xi' For every permutation CI E 6[ there exists a unique continuous automorphism 'Pa of the algebra A [[X]] which maps Xi to Xa(i) for each i E I (IV, p. 28, Prop. 4) ; it is clear that CI >-+ 'Pa is a homomorphism of 6, into the group of continuous automorphisms of the algebra A[[X]]. Let f E A [[X II be a formal power series; we put CI f = 'Pa (f) and we shall say that the formal power series f is symmetric if CI f = f for each CI E 6 [. The symmetric formal power series form a closed subalgebra of A[[X]] which is denoted by A [[X llsym and is equipped with the topology induced by that of A [[X]]. Let T be an indeterminate. In the ring of formal power series A [[X, T]] the

A.IV.68

POLYNOMIALS AND RATIONAL FRACfIONS

§6

family (Xj T)j E I is summable, hence the family (1 + X jT)j E I is multipliable (IV, p. 27, Prop. 2) ; moreover we have (14) where the formal power series sk (15)

Sk =

E

A[[X]] is defined by

L (TI HE'llk

Xj

)

(k

~ 1)

iEH

(we denote by ~k the set of all k-element subsets of I). In particular we have Sl = Xj. When I is finite with n elements we have Sk = 0 for k>- n ; more

L

ieI

precisely, when I = {I, ... , n} then the formal power series sk is nothing other than the elementary symmetric polynomial of degree k in XI' ... , X n . Let S = (Sk)k "" I be a sequence of indeterminates. Since the formal power series sk is of order ~ k, and belongs to A [[X]]sym, conditions a) and b) of Prop. 4 of IV, p. 28 are satisfied with E = A [[X ]]sym ; there exists thus a unique continuous Aalgebra homomorphism 'PI : A [[S]]

such that 'PI (Sk)

=

--->

sk for each integer k

A [[X ]]sym

~

1.

2. - a) If I is a finite set of n elements, then 'PI induces a bicontinuous isomorphism of A[[SI' ... , Snll onto A [[x]]sym. b) IfI is infinite, 'PI is a bicontinuous isomorphism of A [[S]] onto A [[X ]]sym. In case a) we put B = A[[SI' ... , Snll and equip this algebra with the topology induced by that of A [[S ]] ; we also denote by 1/11 the restriction of 'PI to B. In case b) we put B = A [[S]] and 1/11 = 'PI. We shall equip the polynomial algebra A [S] with the graduation of type N for which Sk is of weight k for every integer THEOREM

k~1.

Lemma 1. - Let 1 be a finite subset of I, r an integer such that Card 1 ~ rand fa symmetric formal power series which is homogeneous of degree r in the X j (i E I). Let 1 be the formal power series obtained by substituting 0 for X j for each i in 1-1. If

Let us put

f =

L

Ia I

0. =

~

1 = 0,

then we have f = O.

aa Xa (where r

10.1

is the length

L o.j i E1

of the multiindex

(o.j)j E I). If 0. is a multiindex of length r, and l' is the support of 0. (the set of I such that o.j -=1= 0), then Card l' "'" r, hence there exists a permutation a E 6 1 such that a(l') c 1. Put I3 j = o.o--I(i) for i E I, then the monomial XJ3 = X:i(j) depends only on the indeterminates Xj (j E 1), whence aJ3 = 0 by i

E

TI

iEI

No.3

A.IV.69

SYMMETRIC ruNCTIONS

the hypothesis 1 = O. Since f is symmetric, we have a" = alJ and since a. was arbitrary, we conclude that f = O. Lemma 2. - Let f be a symmetric formal power series of degree r in the Xi (i E I). There exists a unique polynomial P E B n A [S], isobaric of weight r, such that f = "'I(P). The case when I is finite follows from Th. 1 (IV, p. 62). Assume that I is infinite and choose a finite subset J of I, containing r elements. We keep the notation of Lemma 1 ; we remark that every isobaric polynomial of weight r in the Sn (n ... 1) depends only on SI' ... , S, and that of!> ... , of, are the elementary symmetric polynomials in the r indeterminates Xj (j E J). If P is an isobaric polynomial of weight r in the Sn and h = f - "'I(P), then ii = 1 - P (of 1, ... , of,) and Lemma 1 shows the relation f = "'I (P) to be equivalent

to

1=

P (of 1,

... ,

of,). By Th. 1 (IV, p. 62) there exists a unique polynomial

PEA [S], isobaric of weight r and such that

1=

P (of l'

... ,

of,), whence the result.

Lemma 3. - For every integer m ... 0 let cm be the ideal of the algebra A[[X]]sym consisting of all symmetric formal power series of order ... m. The sequence (cm)m~O is a fundamental system of neighbourhoods of 0 in A[[x]]sym. The lemma clearly holds when I is finite, so let us assume that I is infinite. For every finite subset J of I, consisting of m elements, denote by j the set of elements of N(I) of length -< m and with support in J. Further denote by a; the set of formal

power series containing no term aX" with a. E 1. Since j is a finite subset of and every finite subset of N(I) is contained in a set of the form j, the family (aJ) is a base of neighbourhoods of 0 in A[[X]] (IV, p. 26). Now Lemma 1 implies the relation a; n A [[X ]]sym = cm for every subset J with m elements, and this proves Lemma 3. N(I)

Since there are only a finite number of monomials of a given weight in the Sk' every formal power series fEB may be written in a unique way as f = P" where P, is an isobaric polynomial of weight r in B n A [S]. For every

L

, .. 0

integer m ... 0 let b m be the ideal of B consisting of all formal power series of the above type such that P, = 0 for O:os; r -< m. The sequence (bm)m~o is a base of neighbourhoods of 0 in B. With the above notation (P,) is a symmetric formal power series in the Xi' homogeneous of degree r, and so this is the homogeneous component of degree r of (I). Lemma 2 shows that is an algebra homomorphism of B onto A [[X] ym , mapping bm to cm for every integer m ... 0; now Lemma 3 shows to be bicontinuous, and this completes the proof of Theorem 2.

"'I

"'I

r

"'I

"'I

A.IV.70

4.

§6

POLYNOMIALS AND RATIONAL FRACTIONS

Sums of powers

We again write X = (Xi)i E I for a family of indeterminates. The symmetric formal power series sk are defined as before by (16)

Sk =

TI

')

H~k

(k "'" 1 ) ,

Xi

i eH

where ~k is the set of all k-element subsets of I. We also write (17)

L X7

Pk =

(k "'" 1 ) .

i E I

This is a symmetric formal power series which is homogeneous of degree k.

For every integer d "'" 1 we have

Lemma 4 ( n. Then Newton's relations may be written pz = sJPI - 2s z P3 = sJPz - SzPI .+ 3s 3

+ ... + (_l)n-l sn _ zPI + (-1 t(n -1)Sn_1 szPn - Z + ... + (- 1 tSn -!PI + (- 1 t + Insn

Pn-I = sJPn-2 - szPn-3 = sJPn - I -

Pn

and (22)

Pk = sJPk-1 - szPk-Z

+ ... + (- 1 t+ ISnPk _ n (for k:> n) .

The first n of the above relations hold for any I ; we find for example

More generally, let S = (Sn)n;;. I be a family of indeterminates. Let us define the polynomials Pd E Z[SI, ... , Sd] recursively by PI = SI and d-l

Pd =

I

k

~

(-l/-ISkPd_k+(-l)d+ldS d

(d~2).

1

Then we have the« universal formulae» Pd = Pd(SI, ... , Sd) holding over any ring A and family X of indeterminates. Let P = (Pk)kool be a sequence of indeterminates. Since Pk is homogeneous of degree k in A [[X]], there exists precisely one continuous A-algebra homomorphism AI: A [[P]]

->

A [[x]]sym

such that AI(Pd = Pk for every integer k ~ 1 (IV, p. 28). If we assign to Pk the weight k, then AI transforms an isobaric polynomial of weight n in the Pk into a formal power series which is homogeneous of degree n in the Xi' a) If I is a finite set with n elements and n! . 1 is invertible in A, then AI induces a bicontinuous isomorphism of A [ [P I, ... , P n]] onto A [[X ]]sym. b) If I is infinite and A is a Q-algebra, then AI is a bicontinuous isomorphism of A [[P]] onto A [[X]]sym. PROPOSITION

3. -

We shall treat the case a) only, the case b) being quite similar.

A.IV.72

§6

POLYNOMIALS AND RATIONAL FRACfIONS

By Th. 2 (IV, p. 68) we can identify A [[X] ]sym with the algebra of formal power series A[[SI' ... , Sn]], with Sk corresponding to sk. By Lemma 4 of IV, p. 70, there exist formal power series 91' ... , 9 n of order ~ 2 in the indeterminates sl' ... , Sn such that

Since k! • 1 is invertible in A, Lemma 2 of IV, p. 35 proves the existence of an automorphism T of the topological A-algebra A[[X]]sym which maps Sk to Pk for 1 ~ k ~ n. Now Prop. 3, a) is an immediate consequence. COROLLARY. Let ~I' ... , ~n' TIl' ... , TIn be elements of A and suppose that A is an integral domain. a) Ifsk(~I' ... , ~n) = Sk(TlI' ... , TIn) for 1 ~ k ~ n, then there exists a permutation (J" E 6 n such that Tli = ~- deg (qu - q'v); show that there exists q" E L such that deg(u - q"v) -< deg v. d) Given u, v, u', v' E L, homogeneous and such that uv = u'v' and deg v "'" deg v' "'" 0 ; there exists tEL such that u' = ut, v = tv'. e) Given u, vEL, homogeneous, such that uv = vu, there exists WE L, a, b E K and m, n E K such that u = aw m , v = bw n• f) Let u, vEL be such that there exist a, bEL, non-zero, with au = bv ; there exist m, dEL such that the left multiples (resp. right divisors) common to u and v are left multiples (resp. right divisors) of m (resp. d).

§ 2

1) Suppose A is an integral domain; let fbe a polynomial of the ring A [Xl' X z, ... , Xn 1of degree "" k; in X; (for 1 "" i "" n). For each value of i (1 "" i "" n) let H; be a set of k; + 1 elements of A. Show that if f

f

=

(Xl' X Z, ... ,

Xn)

=

0 for each element (X;) E

n" H;, then

O.

2) Assume that A is an infinite integral domain; let be a set of polynomials oF 0 in the ring A [Xl' X z, ... , Xnl. Show that if the power of is strictly less than that of A, then there exists a subset H of An equipotent with A such that for each X = (x;) E Hand for each f E we have f(x) oF O. 3) Suppose that A is an integral domain and let B be an infinite subset of A. Show that if the polynomial f E A [X 1 has degree >- 0, then the image of B under the polynomial function x >-+ f (x) is equipotent with B. 4) Suppose that the additive group of A contains an infinite subgroup G whose non-zero elements are not divisors of zero in A. Show that the mapping f>-+ of A [Xl' ... , Xp 1into the algebra of mappings of AP into A is injective (observe that a polynomial of degree n in one indeterminate cannot have more than n distinct roots lying in G). This holds in particular when there is in A an element x o, not a divisor of 0 and of infinite order in the additive group of A.

1

5) Let K be a field of infinitely many elements and let Q be the quaternion algebra over K, of type (- 1,0, - 1) (III, p. 445). Show that the polynomial XZ + 1 has infinitely many roots in Q . .,. 6) Let K be a finite field with q elements. a) In the ring K [Xl' X z, ... , Xn] let Q be the ideal generated by the n polynomials

A.IV.88

§2

POLYNOMIALS AND RATIONAL FRACTIONS

Xi - Xi (1 "" i "" n). Show that if f E a, then f(x l , x2, ... , xn) = 0 for all (Xi) E K n (observe that the multiplicative group K* is of order q - 1). b) If f is any polynomial of K [XI' X 2 , ••• , Xn], show that there exists precisely one polynomial

1

such that deg i

1"" q -

1 for 1 "" i "" n and such that

f == 1 (mod

We thus have deg J"" deg f; if f is such that f(x l , x 2, ... , xn) = 0 for each (Xi) E K n, f belongs to the ideal a, which is thus the inverse image of 0 under the

a).

homomorphism f ~ J (use Exercise 1). c) Let fl' f2' ... , fm be non-zero polynomials of K[XI' X 2 , ••• , Xn] such that fi (0, 0, ... ,0) = 0 (1 "" i "" m) and the sum of the total degrees of the fi is 0 such that k • 1 = 0 in A. Let h be an integer ::> 0 and f tJIe Then 0. is a root of order k of f and a root of order polynomial (X - a )k+ h + (X - 0. "'" k + h - 1 of D f.

i.

* 10) Let fbe a non-zero element of Z[XI' ... , Xn]. Let N(q) be the number of zeros of fin the cell IXII ""q, ... , IXnl ""q. Then N(q)/qntends to 0 as q_+oo. * 11) Let P be the set of all prime numbers, P' an infinite subset of P and fEZ [(Xi)]' For each pEP' let fp be the element of (Z/pZ)[ (Xi)] obtained by applying to the coefficients of fthe canonical homomorphism of Z onto Z/pZ. Suppose that for each PEP' and each X E (Z/pzt we have fp(x) = 0; then f = O. 12) Let M be an A-module, and let 0.0 , ... , an E A be such that for i ~ j the homothety with respect to o.i - o.j in M is bijective. For any mo, ... , mn E M there exists one and only one f E M[X] such that /(o.i) = mi for 0"" i "" nand deg / "" n. 13) Let o.i (1 "" i "" n) be n distinct elements of a commutative field K, ~i (1 "" i "" n) and 'Yi (1 "" i "" n) 2n arbitrary elements of K. Show that there exists one and only one polynomial f E K[X] of degree"" 2n - 1 such that f(o.;) = ~i and /'(o.i) = 'Yi for 1 "" i "" n (Hermite interpolation formula) (to begin with consider the case where 2n - 1 of the 2n elements ~i' 'Yi are zero).

§3

A.IV.89

EXERCISES

§ 3

1) Let E be an associative commutative and unital algebra over a commutative field K ; let x = (Xi)i e I be a family of pairwise permutable elements of E. Let U be the subring of K (Xi )i e I consisting of all f E K (Xi )i e I such that x is substitutable in f Show that if in E the set of non-invertible elements is an ideal, then the same is true of U. Show that the ideal of non-invertible elements in U is then maximal. 2) a) Let K be a commutative field and u = amXm + am + IXm + I + ... + anXn a polynomial of K[X] such that am #= 0, an #= 0 (0"" m "" n). Show that if 9 is a rational fraction of degree d #= 0 in K(X), then u(g) is non-zero and is of degree nd if d:> 0, and of degree md if d""O. b) Deduce that a non-constant rational fraction 9 of K (X) is substitutable in every rational fraction of K(X) (observe that if 9 is of degree 0, there exists a E K such that 9 - a has degree"" 0). 3) A rational fraction f E K (XI' X z, ... , Xn) is said to be homogeneous if it is equal to the quotient of two homogeneous polynomials (the denominator being #= 0). Show that for fto be homogeneous it is necessary and sufficient that

where d is the degree of

f

4) Show that the Lagrange interpolation formula (IV, p. 16, Prop. 6) may be written

f(X)

l3 i

=

w(X) i~1 w'(ai)(X - a i )

where w(X) = (X - al)(X - a z ) ... (X - an)' Deduce that if 9 is a polynomial of K [X] of degree "" n - 2, then

(consider the polynomial f(X)

=

Xg (X».

* 5) Let K be a commutative field of characteristic O. Show that if a rational fraction u

E

K(X) is such that Du

=

0, then u is a constant.

6) Generalize Euler's identity to homogeneous rational fractions (Exercise 3) over a field of characteristic 0 (consider the rational fraction Z and use Exercise 5).

*

~ f(ZX I , ZXz, ... , ZX n) with respect to

zm

7) Let K be a commutative field. Show that [K(T): K]

=

Sup (Card (K), Card(N».

A.lV.90

POLYNOMIALS AND RATIONAL FRACTIONS

(Use the linear independence of the l/(T - a), a

E

§4

K, to show that

[K(T): K] "'" Card(K).)

§

1) Let u (X)

=

L

4

a.X· be a formal power series over a commutative field K.

"=0

a) For u to be a rational fraction of K (X) it is necessary and sufficient that there should exist a finite sequence of elements (A;)I "'; '" q of elements of K, not all zero, and an integer d "'" 0 such that for all n "'" d we have

b) Put

H~k) =

a.+ k _ 1

a.

an +

«Xn+l

Cl n + 2

Q.n+k

+2

0."+3

Cln

a.+k_1

a.+k

a. +2k -

Cln

«(

1

Hankel determinant») show that if H~q+j I) j "'" 0, then u(X) is a rational fraction (use a». c) Prove the identity

=

+k +

1

2

0 and H~ql j ~ 0 for every integer

(cf. III, p.639, Exercise 10). Deduce that if Hf::/)=O for O"",j"",r-l, then the r determinants Hf:l j where 1 "'" j "'" r are either all zero or all non-zero. d) Deduce from b) and c) that for u(X) to be a rational fraction it is necessary and sufficient that there exist two integers d and q such that H~\j I) = 0 for all integers j "'" O. e) Let k be a subfield of K, and identify K(X), k(X) and k[[X]] with sub rings of K«X». Prove that k[[X]] n K(X) = k[[X]] n k(X).

2) Let aI' a 2 , .•• , ap be integers >- 0; denote by a. the number of finite sequences (x; )1 "'; '" p of integers "'" 0 satisfying the equation

0, and let ~" be the number of finite sequences (Xj) of at most n terms, all of whose terms lie in F and such that Xj = n. Show that the

L j

formal power series

L

~"X"

over Q is the expansion of the rational fraction

n=O

1

LX·

1-

.eF

4) Let E be a vector space with an infinite basis over a field K of characteristic 2 ; let B be the exterior algebra AE on this space, which is a commutative ring. Give an example of a formal power series U E B [[X]] such that u 2 = 0 but such that there exists no element 'Y # 0 of B for which 'YU = O. 5) a) b) c) d) e)

Let E = A[X 1 , ..• , Xp], F = A[[X 1 , ... , Xp]]. If D is a Z-derivation of E into F, we have w (Du ) "'" w (u) - 1 for every non-zero u in E. If D' is a Z-derivation of F, then w (D'v) "'" w (v') - 1 for every non-zero v in F. If E is equipped with the topology induced by that of F, then D and D' are continuous. Every Z-derivation of E into F extends in a unique fashion to a Z-derivation of F. Let !?} be the F-module of A-derivations of F. If DE!?} and DXj = Uj, then

D

=

p

L ujD j. i

=

1

f) (D 1 ,

••. ,

Dp) is a basis of the F-module !?}.

6) Let (uJx e L be a family of elements of A [ [Xj multipliable, the family (ux)x e L is summable.

n

e I'

If the family (1 + uJx eLlS

* 7) If A is reduced, then so is A [[ (X;)j e d]. * 8) We denote by (1 + X? the element exp(T log (1 + X» of Q[[X, T]]. a) Show that (1 + X? = T(T - 1) ... ,(T - n + 1) X". (The coefficient of X" is a

L

n.

";;110

polynomial in T whose value is known for T = n EN.) b) Show that (1 +X?(1 +X)T = (1 +X?+T. Write down explicitly the identity between binomial coefficients so obtained. Show that (1 + X + Y + Xy? = (1 + X? (1 + y? §

5

I) (Algebra of sequences of exponential type) Given two integers p, q E N, we denote by

«P, q»

the binomial coefficient (p + q )!. Let k be a commutative ring and B an p! q! associative commutative unital k-algebra. By a sequence of exponential type of elements of B we understand a sequence a = (ap)peN such that ao = 18 and apa q = «P, q» ap + q for any p, q E N. The set of all sequences of exponential type of B is written C (B). a) Verify that there is on C (B) a unique k-algebra structure such that

(a + b )p

=

L

a,b,.

r+s =p

(ab)p (Aa)p

= =

p! apbp , APap ,

A.IV.92

POLYNOMIALS AND RATIONAL FRACTIONS

§5

for any sequence a, bE tf (B) and pEN, ?. E k. The mapping a ....... a l is a homomorphism of the algebra tf (B) into B. b) Suppose that k contains a subring isomorphic to Q. Show that for every x E B the sequence f (x) defined by f (x)p = ~ x P is of exponential type. Show that f is an

p.

isomorphism of B onto tf (B ). c) Let B' be an associative commutative unital k-algebra and let h be a homomorphism of B into B'. For each a E tf (B) the sequence a' defined by a~ = h (a p ) for any pEN belongs to tf (B'). Show that the mapping tf (h) : a ....... a' is a homomorphism of If (B) into If (B'). Verify that 8 (g 0 h) = If (g) 0 If (h) when 9 and hare composable kalgebra homomorphisms. d) Let E be a k-module and TS (E) the algebra of symmetric tensors on E. Show that for each x E E, ("Yp(x)) is a sequence of exponential type in the algebra TS(E) (IV, p. 45). 2) (Functor r) Denote by k a commutative ring and by E a k-module. By the gamma algebra of E we understand the associative commutative unital algebra defined by the generating system N x E and by the relators (cf. III, p. 450) (0, x) - 1 , (p,?.x) - ?.P(p, x) ,

(p,x+y)-

L

(r,x)(s,y) ,

r+s=p

(p, x)(q, x) -

«P, q))(P + q, x)

,

where p, q range over N, x, y range over E and?. over k. This algebra is denoted by fCE). For each pEN we denote by "Yp the mapping of E into r(E) composed of the injection x ....... (p, x) and the canonical homomorphism of the free commutative algebra on N x E into fCE). a) Verify that for each x E E, the sequence "Yp(X) is a sequence of exponential type of elements of r (E) (Exercise 1) and that the mapping "Y : x ....... ("Yp(x))p e N is a homomorphism of the k-module E into the k-algebra If (fCE)). b) (Universal property of fCE)). Let B be an associative commutative unital k-algebra and 'P a homomorphism of the k-module E into If (B). Show that there exists a unique algebra homomorphism h: reE) -+ B such that 'P = If(h) o"y. c) Show that there exists a unique unital homomorphism of algebras E : r (E) -+ k such that E("yp(X)) = 0 for all p:> 0 and all x E E. d) For each pEN let rp(E) be the submodule of r(E) generated by the elements "YVI (XI) ... "Yv,(x,)

where sEN, v E N' and

Iv I =

LVi = p. Verify that the submodules rp(E) form a

graduation of the algebra fCE). Show that "YI is an isomorphism of E onto r I (E) (to verify that "YI is injective, consider the algebra B = k x E with the product defined by the formula (?., x) (f.L, y) = (?f.L,?..Y + f.LX) and show that there exists a homomorphism of reE) into B which maps "YI (x) to (0, x) for any x E E). e) Let E and F be two k-modules and h a homomorphism of E into F. Show that there exists a unique homomorphism of graded algebras fCh): fCE) -+ reF) such that "Yp 0 h = r (h) 0 "Yp for all pEN. Verify that if 9 and h are homomorphisms of k-modules such that 9 0 h is defined, then fCg 0 h) = reg) 0 fCh).

§5

A.IV.93

EXERCISES

f) Let h : E --.. F be a surjective homomorphism of k-modules. Show that r(h) is surjective and that its kernel is the ideal of f(E) generated by the elements 'Yp(x) with p:> 0 and x E Ker(h). g) Let E and F be two k-modules. Show that there exists a unique homomorphism 'P of the algebra f (E x F) into the algebra f (E) ® k f (F) such that

for any pEN, x E E and Y E F. Show that this homomorphism is an isomorphism compatible with the graduations. 3) Let k be a commutative ring and E a k-module. Show that there exists a unique homomorphism fl. of the algebra f(E) (Exercise 2) into the algebra f(E) ®k r(E) such that

for all x E E and all pEN. Show that the coproduct fl. is coassociative, cocommutative (III, p. 580-581) and defines on f(E) a graded bigebra structure (III, p. 585). Show that the homomorphism E defined in Exercise 2, c) is a counit. 4) Let E1 and E2 be two modules over a commutative ring k and let h be a homomorphism of El into E 2. Verify that r(h) (Exercise 2, is a homomorphism of graded bigebras (cf. Exercise 3).

e»

5) Let feZ) be the gamma algebra of the Z-module Z (Exercise 2). For each pEN put T(P) = 'Yp(1).

a) Show that (T(P»p E N is a basis of the Z-module r(Z). Show that there exists a unique homomorphism 6 of the algebra feZ) into the algebra feZ) ® r(Z) such that 6(T(P»

=

p! T(P) ® T(P)

for all pEN. b) Let B be a commutative ring and H the set of homomorphisms of the ring feZ) into the ring B. Show that for every f E H the sequence f(T(P»,p E N is a sequence of exponential type of elements of B. Show that the mapping 'P: f>-+ (f(T(P»)pEN is a bijection of H onto ~ (B) (Exercise 1). c) For f, 9 E H denote by f * 9 the homomorphism of r(Z) ® feZ) into B defined by (f * g)(u ® v) = feu) g(v) for any u, v E feZ). Verify that for any f, 9 E H we have

'P(f) + 'P(g) 'P(f)q>(g)

= =

'P«(f * g) fl.) , 'P«(f*g)o6). 0

6) Let n be an integer:> 0 and let r(Z/nZ) be the gamma algebra of the Z-module Z/nZ (Exercise 2). Show that for every p:> 0, fp(Z/nZ) is isomorphic to the quotient of Z by the subgroup generated by the integers nk«k,p - k», where 1 ~ k ~p. Show that ifp and n are distinct primes, then fp(Z/nZ) is'cyclic of order n. Show that if n is even, then f 2(Z / nZ) is cyclic of order 2n.

A.IV.94

POLYNOMIALS AND RATIONAL FRACTIONS

§5

7) (Extension of scalars) Let k be a commutative ring, L an associative commutative unital k-algebra and E a k-module. a) Let r (L ®k E) be the gamma algebra of the L-module L ®k E (Exercise 2). Show that there exists a unique homomorphism e of L ® k r (E) into r (L ® k E) compatible with the algebra structure on L and such that

for all x E E and pEN. b) Show that e is an isomorphism of graded bigebras over L. 8) Let k be a commutative ring and E a k-module. a) Show that there exists a unique homomorphism g of the algebra r(E) (Exercise 2) into the algebra TS (E) of symmetric tensors over E (IV, p. 43) such that

(g

0

"Yp)(x) = x ® x ® ... ® x

(p factors x) ,

for any x E E, pEN. Show that if E is a free k-module with basis (ej)j E I' then g is an isomorphism of graded bigebras (cf. Exercise 3 and IV, p. 51) and the elements "Ya = "Ya(j)(ej ), where ex ENOl, form a basis of r(E) .

IT

b) Let ~ be a mapping of (1, n) into N and let p

=

.

I

Hi). Show that for every sequence

i = 1

XI'

I

X2,

X~(I)

•.• ,

x. of elements of E, the image of "Y~(I)(XI) ... "Y~(.)(x.) under g is x~(P) where 'Il runs over the set of all mappings of (1, p) into (1, n) such that

® ... ®

~

Card 'Il-I(i) = ~(i) for each i E (1, n) (cf. IV, p. 45, Prop. 3). c) Let fbe the mapping of ®P E into rp(E) defined by f(x i ® ... ® xp) = "YI (XI) .•. "YI (xp) for any XI, ... , xp E E. Show that for each tensor t E ®P E, (g 0 f)(t) is the symmetrization of t. Show that for each U E rp(E) we have (f g)(u) = p! u . 0

.,. 9) (Polynomial laws) Denote by E and F two modules over the commutative ring k. By a polynomial law ofE in F we understand, for each (associative commutative and unital) kalgebra L, a mapping 'PL of L ® k E into L ® k F satisfying the condition: If Land R are two k-algebras and f is a unital homomorphism of L into R, then

a) Let 'P be a polynomial law of E into F, let L = k [(Tj )j E d be the polynomial algebra in a family of indeterminates T j over k and let (Xj)j E I be a family of finite support of elements of E. Show that there exists a unique family of finite support (y~)~ E N(!) of elements of F such that 'PL

(I T

j

®

xj

)

=

I T~ ® y~ . ~e N(I)

j

Show that if R is an algebra over k, then for every family (aj)j E I of elements of R we have (1)

'PR (

I I

aj

® xj

)

=

I a~ ® y~ . ~e N(I)

§5

A.IV.95

EXERCISES

b) A family (crJ)j E J of polynomial laws of E into F is called summable if for every algebra L over k and every u E L ® E, the family crJL (u) has finite support. Show that if (crJ)j E J is a summable family of polynomial laws of E into F, then there exists a unique polynomial law 'P of E into F such that for every k-algebra L and every U E L ® E we have L crJL(U) = 'PL(U). This polynomial law is called the sum of the laws ~ and is denoted by j

L crJ. j

c) Let E, F, G be three k-modules, 'P a polynomial law of E into F and", a polynomial law of F into G. Show that there exists a unique polynomial law "'I of E into G such that for every k-algebra L we have "'IL = "'L 0 'PL. This law "'I is called the composition of 'P and '" and is denoted by '" 0 'P. d) Let p be an integer '-'" O. A polynomial law 'P of E into F is said to be homogeneous of degree p if for every algebra Lover k we have 'PL (au) = aP'PL (u) for any a ELand U E L ® E. Show that if 'P is homogeneous of degree p, then in formula (1) we have YE = 0 for I ~ I = Hi) oF p. Determine all homogeneous polynomial laws of degrees 0

L

and 1. e) With the same notation as in c), show that if 'P and", are homogeneous laws of degrees p and q respectively, then '" 0 'P is a homogeneous law of degree pq. f) Let p be an integer '-'" O. For every commutative k-algebra L we denote by EL the injection a >-+ aT of L into the polynomial algebra L[T] and by Pf. the homomorphism of the L-module L[T] into L which associates with each element L akTk E L[T] the coefficient ap • Let 'P be a polynomial law of E into F. Show that the k

mappings cpf. = (Pr ® Id F ) 0 'PL[TI 0 (EL ® IdE) form a polynomial law of E into F which is homogeneous of degree p. This polynomial law is called the homogeneous component of.. degree p of the law 'P. Show that the homogeneous components of 'P form a summable family with sum 'P. 10) Denote by k a commutative ring and by E a k-module. For each integer ® E into L ® rp(E) composed of "Yp : L ® E ~ r p (L ® E) and the canonical isomorphism of rp (L ® E) onto L ® rp(E) (cf. Exercises 2 and 7). a) Verify that the mappings "Yp.L form a polynomial law (Exercise 9) of E into rp(E), which is homogeneous of degree p ; this law will in what follows be denoted by 'Yp. Show that for every k-module F and every cr E Hom (rp(E), F) the mappings 'PL = (Id L ® cr) 0 'Yp.L form a polynomial law of E into F which is homogeneous of degree p. This law is called the law associated with cr. Let (Xi)i E I be a family of elements of E and (ai)i E I a family with finite support of elements of L. Show that

p '-'" 0 and every k-algebra L we denote by "Yp.L the mapping of L

'PL(Lai®Xi ) i

where

I~ I =

L Hi), a E =

=

L

aE®cr(-YE(x))

EeN(I).IEI ~p

fl aiE(i) and "YE(x) = fl "YE(i)(x;).

b) Let 'P be a polynomial law of E into a k-module F, homogeneous of degree p. Show that there exists a unique linear mapping cr of rp (E) into F such that 'P is the law associated with

A.IV.96

POLYNOMIALS AND RATIONAL FRACTIONS

§5

a. (Suppose first that E is free on a basis (Xi)i E I ; using Exercise 9, show that in this case there exists a unique mapping a E Hom (fp(E), F) such that Relation (1) of Exercise 9 holds for every algebra L and every family (ai)i E I with finite support of elements of L.

Treat the general case by introducing a free k-module E' and a surjective homomorphism h ; E' ---> E. The polynomial law \p, composed with the law defined by h is a polynomial law of E' into F which is associated to a linear mapping a'; f p (E' ) ---> F; show that a' is the composite of fp (h) and a mapping a ; fp (E) ---> F such that cp is associated with a (cf. Exercise 2, f».) 11) We keep the notation of Exercise 10. A linear mapping of f(E) into F will be called.a series over E with values in F. The series over E with values in F form a k-module Qenoted by S(E, F). For n E N we say that the series a E S (E, F) is of order ~ n if Ker(a) :::> fp(E) for all p -< n. In what follows we shall equip S (E, F) with the topological group structure having as neighbourhood base of 0 the k-modules consisting of the series of order ~ n (n EN). Let cp be a polynomial law of E into F. Show that there exists a unique series S (E, F) such that; (i) for every k-algebra L and every u E L ® E the family (Id L ® a) "Yp,L (u) (p EN) has finite support ; (ii) CPL(U) = (Id L ® a) 0 "YP,L(U), a

E

L p

(Apply the result of Exercise 10, b)). This series a is said to be associated to the law cpo 12) Let k be a commutative ring, E a k-module and B a k-algebra (not necessarily associative). Let .1 be the coproduct of f(E) (cf. Exercise 3). For any series a, a' E S (E, B) we denote by a • a' the series m 0 (a ® a') 0 .1, where m denotes the linear mapping of B ®k B into B defined by the product in B. a) Show that the mapping (a, a') ~ a • a' defines on S (E, B) a k-algebra structure (cf. Exercise 11). Show that if B is associative (resp. commutative, unital) then S (E, B) is associative (resp. commutative, unital) (cf. III, p.578-584). b) Assume that B is associative and unital. Show that the invertible elements of S(E, B) are the series a such that a(l) is invertible in B. c) Suppose that E is a free k-module and (ei)i E I is a basis of E. For each 0: E N(I) put e (a) = II"Ya(i)(ei ) and denote by f" the series over E with values in k defined by f"(e(~») = 8a,~ (Kronecker index) for each ~ E N(I) (cf. Exercise 8). Show that f"f~ = f" + ~ for any 0:, ~ E N(I). Deduce that when I is finite, the algebra S (E, k) is isomorphic to the algebra of formal power series k[[ (Xi)i Ed]. 13) (Divided powers of a series.) The notation is as in Exercises 10 and 11. Let E and F be two k-modules and a a series over E with values in F. Denote by 0' the homogeneous polynomial law of degree 1 of f(E) into F defined by a. For any integers p, mEN we denote by a!:) the series associated with the polynomial law "Yp 0 0' 0 ("Yo + "YI + ... + "Ym) (cf. Exercise 10). a) Show that as m tends to 00, the sequence a!:) converges in S(E, fp(F)) to a series a(P). This series a(P) is called the divided p-th power of the series a.

§5

A.IV.97

EXERCISES

b) Show that if (x;}j

E

I

is a family of elements of E and" u(P)(-Y'1(X»

=

L

n

E N(I),

then

"'I~(m)(u("'Im(x»),

(E2' .EN(I)

where !f is the set of all mappings with finite support of N(I) into N satisfying the conditions

L ~(a) =p

.

and

L ~(a) a ="

.

(Apply Ex. 10,

a».

c) Show that if u, T E S(E, F), then for each pEN we have (U+T)(P)=

L

u(r)u(s)

r+s=p

(product calculated in the algebra S(E, r(E», cf. Exercise 12). d) Show that for every integer pEN we have uP = p! u(P), where uP is the p-th power of u

in the algebra S (E, r(F». e) Let ~E and ~F be the respective co products in f (E) and f (F). Show that for all pEN we have ~F 0 u(P) = (u(r) ® u(s» 0 ~E •

L

r+s=p

14) With the notation as in Ex. 13, suppose that u E S(E, F) is of order"" l. a) Show that for each pEN, u(P) is of order"" p. Deduce that the family (u(P»p E N is summable. Writing A(u) = L u(P), verify that for every n E N, p

A(u)(f.(E» c fo(F) + ... + f.(F). b) Show that if Ker(u) ~ f m(E) for all m oF 1, then A(u) = f(u 0 "'II). c) Show that A(u) is a homomorphism of the cogebra f(E) into the cogebra f(F), compatible with the counits (cf. Ex. 13, d) Show that if u, T E S(E, F) are two series of order"" 1, then A(u + T) = A(u) A(T), the product A(u) A(T) being defined by the product in f(E) (cf. Ex. 13,

e».

c».

15) With the notation as in Ex. 14, let E, F, G be three k-modules, u E S (E, F), T E S (F, G), and suppose that u is of order"" 1. The series T 0 A (u) E S (E, G) is called the composition of u and T; it is denoted by To u. a) Let 'P be a polynomial law of E into F (Ex. 9) and", a polynomial law of F into G, and denote by u'" and u oll the series associated with 'P and", respectively. Show that if the homogeneous component of degree 0 of 'P is zero, then u'" is of order "" 1 and u oll 0 u'" is the series associated with the polynomial law", 0 'P. b) Show that if u is of order "" p and T is of order "" q, then T 0 u is of order ""pq. c) Show that for every integer p, (T 0 U )(p) = T(P) 0 A (u). Deduce that if T is of order "" 1 and if H is a k-module, then for any p E S(G, H) we have (p 0 T) 0 U = po (T 0 u). d) Suppose that G is a k-algebra. Show that the mapping T 1-+ T 0 A (u ) is a homomorphism of the algebra S (F, G) into the algebra S (E, G). 16) Let E and F be two modules over a commutative ring k and let p be an integer

"" o.

A.IV.98

§6

POLYNOMIALS AND RATIONAL FRACfIONS

a) Show that if p "'" 2, the mapping Ap of Hom(rp(E), F) into FE which associates with -.(R) of Dk as follows: if U I , ... , U/ are the equivalence classes under R, numbered so that Card(U I ) ~ ... ~ Card (Ud, put >-.(R) = (ai' ... , ak) with a i = Card (U i ) if 1"", i "'" I and a i = 0 if 1 1 and b:> 1, this would mean a rf. p, b rf. P and ab E p in contradiction with the fact that p is prime. The number p is therefore prime and by passage to quotients f defines an isomorphism of Fp = Zip onto a subfield P of K'. In both cases P is a subfield of A contained in the centre C of A, and it is a prime field. Let L be a subfield of A ; then P n L is a subfield of P, and since P is prime, we have P n L = P, whence PeL. If P' is a subfield of A and is a prime field, then by what has been said, PcP', whence P = P' because P' is a prime field. COROLLARY 1. - Let K be a field. There exists a unique subfield ofK which is a prime field, and this is the least subfield of K. COROLLARY 2. - For a field to be prime it is necessary and sufficient that it should contain no subfield other than itself 2.

Characteristic of a ring and of a field

We shall define the characteristic of a ring A only when A has a subfield. When this is so, let f be the unique ring homomorphism of Z into A, and let n be the unique positive integer generating the ideal of Z which is the kernel of f (I, p. 111) ; then the integer n is called the characteristic of A. Let A be a ring for which the characteristic is defined ; then A does not reduce to O. By Th. 1 there exists a unique subfield P of A which is a prime field; we shall call it the prime subfield of A. By the proof of Th. 1 there are the following two possibilities : a) the characteristic of A is 0, P is isomorphic to Q, b) the characteristic of A is a prime number p, P is isomorphic to Fpo

If the characteristic of A is zero, there exists a unique ring homomorphism of Q into A; its image is the prime subfield of A, contained in the centre of A. Therefore there exists a unique Q-algebra structure of A compatible with the ring structure. When the characteristic of A is a prime number p, we have the corresponding properties on replacing the field Q by the field Fpo

PROPOSITION 1. - Let A be a ring not reduced to O. a) For A to be of characteristic 0 it is necessary and sufficient that the mapping x ~ n . x of A into itself should be bijective, for every integer n ",

No.3

PRIME FIELDS. CHARACTERISTICS

A.V.3

b) Let p be a prime number. For A to be of characteristic p it is necessary and sufficient that p • x = 0 for all x E A. Let f be the unique homomorphism of Z into A ; we have n • x = f (n) x for every integer n and every x in A. For A to be of characteristic 0, it is necessary and sufficient that f should extend to a homomorphism of Q into A, that is, f(n) should be invertible in A for every n#-O (1, p. 113) ; this proves a). Similarly for A to be of characteristic p it is necessary and sufficient that f should annihilate pZ, that is, f(P) = 0, or also that p . x = 0 for all x E A ; this proves b).

Let us take for A a not necessarily commutative field. The centre of A is a (commutative) field; therefore the characteristic and the prime subfield of A are defined. Remarks. - 1) Let A and A' be two rings not reduced to O. Suppose that the characteristic of A is defined and that there is a homomorphism u of A into A'. The image under u of the prime subfield of A is a subfield P' of A', isomorphic to P, and hence prime. It follows that the characteristic of A' is defined and is equal to that of A. If A and A' are of characteristic 0 (resp. p -# 0), the mapping u is a homomorphism of algebras over Q (resp. Fp). 2) Remark 1 shows that if A is a ring of characteristic 0 (resp. P -# 0), the same holds of any ring A' containing A as subring, or of any quotient of A by a two-sided ideal a -# A. In particular, if K is a field, every subfield of K and every extension field of K have the same characteristic as K. 3) Let A be an algebra not reduced to 0 over a field K. Since the mapping A >-+ A • 1 of K into A is a ring homomorphism, Remark 1 shows that the characteristic of A is defined and equal to that of K. 4) Since the field Q is infinite, every ring of characteristic 0 is infinite; it follows that every finite field has non-zero characteristic. 5) Let A be a ring not reduced to 0, whose additive group is a torsion-free Zmodule, and put B = Q ®z A. The mapping x >-+ 1 ® x of A into B is injective (II, p. 314), hence A is isomorphic to a subring of a ring of characteristic O.

3. Commutative rings of characteristic p In this No. and the following one p denotes a prime number.

2. - Let A be a commutative ring of characteristic p. The mapping a P is an endomorphism of the ring A, that is, we have the relations

THEOREM

a

~

(1) (2)

r

(a + b = a P + b P (abr = aPb P

for a, b in A. Formula (2) follows from the commutativity of A. To prove (1) we use the

binomial formula (a + b r = a P + b P +

L

p-l

(

~)

. aib P- i ; since p . x = 0

i ==1

X E

A, it suffices to prove the following lemma:

for all

A.V.4

§1

COMMUTATIVE FIELDS

Lemma 1. -

Let p be a prime number and i an integer in the range from 1 to

(~) is an integer divisible by p.

p - 1, then the binomial coefficient

We argue by induction on i, the case i = 1 being immediate from the formula

(~)

= p.

integer i i

rt pZ,

Suppose that 2 "'"

(~)

i "'" P -

= (p - i + 1 ) (i

we have

(~)

1 and that

~ 1)

(i ~ 1) is divisible by p. Then the

belongs to the prime ideal pZ of Z ; since

E pZ and the lemma follows.

Let A be a commutative ring of characteristic p and f an integer"", O. From Th. 2 we deduce by induction on f that the mapping a ~ a P ' is an endomorphism of the ring A. In particular we have the relation (3)

"

(al + ... + anY =

af + ... + a~ ,

for any ai' ... , an in A. The mapping a ~ a P is sometimes called the Frobenius endomorphism of A. Taking A = Fp and ai = 1, we obtain from (3) the relation:

(4)

,== n

nP

mod. p

(n E Z, fEN) .

,

For each subset S of A we denote by sP the set of elements of A of the form p x ' with XES I. In particular, if K is a sub ring of A, the set KP'is a subring of A. If K is a subring of A and S a subset of A we denote by K [S] the subring of A generated by K US; when A is a field, we denote by K (S) the field of fractions of K [S], that is, the subfield of A generated by K U S.

2. - Let A be a commutative ring of characteristic p, K a subring of A, S a subset of A and f a positive integer. a) We have K[Sr' = KP'[sP'], and if A is a field, K(SY' = KP'(sP'). b) If the K-module K[S] is generated by the family (ai)i E I of elements of A, then the K-module K[sP'] is generated by the family (ar\ E I. PROPOSITION

r'

Since K [S] is the sub ring of A generated by K US, its image K [S under the P endomorphism 'IT: a ~ a ' of the ring A is the subring of A generated by the image KP' U SP' of K U S under 'IT, whence K [S r' = KP' [sP']. The case of fields is treated similarly; this proves a). It is clear that the family (ar\ E I generates the KP'-module K [S The K-

r'.

module K [sP'] is generated by products of the form xf' ... x~' = arbitrary in S, hence also by the set K [S from this.

XI' ... , Xn

(XI ... Xn

y' with

r'. Assertion b) follows directly

1 Of course the set Sp'should not be confused with the set product of pI sets equal to S, nor with the set of products of p' elements belonging to S.

No.4

4.

PRIME FIELDS. CHARACTERISTICS

A.V.S

Perfect rings of characteristic p

DEFINITION 2. - A ring A of characteristic p ~ 0 is said to be perfect if it is commutative and the mapping a ~ a P is bijective. If the ring A is perfect of characteristic p, the mapping a ~ apt is an automorphism of the ring A for every integer f ~ 0 ; the inverse automorphism is denoted by a ~ a 11pt or a ~ a P- t and the image of a subset S of A under this automorphism is written Sllpt or SF- t . It is clear that (aP',t = ap'+t for all

a E A and any integers e and f (of whatever sign). Let A be a commutative ring of characteristic p. For every integer f ~ 0 let us write n 1 for the kernel of the endomorphism a ~ apt of the ring A. Then (n I) I", 0 is an increasing sequence of ideals of A ; since every positive integer is majorized by a power of p, the ideal n = U nl consists of all the 1",0

nilpotent elements of A. In particular, if A is perfect, every nilpotent element of A is zero. 3. - Let A be a commutative ring of characteristic p ~ O. By a perfect closure of A we understand a pair (A, u) where A is a perfect ring of characteristic p and u is a homomorphism of A into A satisfying the following universal property: (PC) IfB is a perfect ring of characteristic p and v is a homomorphism of A into B, then there exists a unique homomorphism h of A into B such that v = h 0 u. The universal property (PC) implies at once the uniqueness of the perfect closure, in the following sense: if (A, u) and (A', u') are two perfect closures of A, then there exists a unique isomorphism h of A onto A' such that u' = h 0 u (cf. E, IV, p. 23). We shall now establish the existence of the perfect closure: DEFINITION

3. - Let A be a commutative ring of characteristic p ~ O. There exists a perfect closure (A, u) of A. Moreover, the kernel of u is the set of all nilpotent elements of A and for each x E A there exists an integer n ~ 0 such that x p ' E u(A). For each integer n ~ 0, put An = A ; when m ~ n we define a homomorphism 'lTm,n of An into Am by 'lTm,n(a) = a pm -". We thus obtain a direct system of rings THEOREM

(An' 'lTm,n) (I, p. 120) ; let

A be the direct limit of this system and Un the canonical

homomorphism of An = A into A ; we also put u = uo. By construction of the direct limit the kernel n of u is the union of the kernels of the homomorphisms

A.V.6 '7Tn

,o : a

§1

COMMUTATIVE FIELDS f-+

a pn of A into A, thus it consists of all the nilpotent elements of A. The

ring A is commutative of characteristic p by Remark 1 of V, p.3. The ring A is the union of the ascending sequence (un (A»n ~ 0 of subrings. We have un (A y" = A ; hence for each x

E

A there

exists an integer n .... 0 such that

Ap = A. Let x E A be such that x P = 0; choose an integer n .... 1 and an element a E A such that x = un (a). Then we have un _ I (a) = un (a y = 0 ; by definition of the direct limit there exists an integer m .... n such that '7Tm _ I,n - I (a) = 0, that is, apm -" = O. We thus have '7Tm,n(a) = 0, whence un(a) = 0, that is, x = O. Therefore the ring A is perfect of characteristic p. Let v be a homomorphism of A into a perfect ring B of characteristic p. For every integer n .... 0, the mapping b f-+ b P" is an automorphism of B and so there exists a homomorphism vn of An = A into B characterized by v(a) = vn(ay". We then have vm 0 '7Tm,n = vn for m .... n .... 0; by definition of the direct limit there exists a homomorphism h ~f A into B such that vn = h 0 un for all n .... 0; in particular we have v = Vo = h 0 Uo = h 0 u. Finally let h' be a homomorphism of A into B such that h' 0 U = v. Let x E A ; as we have seen, there exist an integer n .... 0 and an element a E A such that x p " = u (a). Then we have

x p" E u(A). We also have un(A) = u n +I (AY, whence

h(xY" = h(u(a» = v(a) = h'(u(a» = h'(xY" ,

and since B is perfect, we find h (x) = h' (x). Thus we have h' completes the proof that (A, u) is a perfect closure of A.

=

h, and this

3. - Let B be a perfect ring of characteristic p and A a subring ofB. Write AP- oo = U Ap- I and denote by j the canonical injection of A in

PROPOSITION

,~o

N-

oo

•

(AP- oo ,

Then AP- oo is the smallest perfect sub ring of B containing A and j) is a perfect closure of A.

For each integer feZ denote by '7T I the automorphism b f-+ bpi of B. The sequence of subrings '7T_ I(A) of B (for f .... 0) is ascending and its union N- oo is thus a subring of B. We have '7TI (AP- oo ) = U '7T_ (f _l)(A) = AP- oo , hence ,~O

AP-

oo

is a perfect sub ring of B. Finally let Bo be a perfect sub ring of B containing

A; for every integer f .... 0 we have '7T_ I(A) C '7T_/(B o) = Bo, whence AP- oo c Bo. If v is a homomorphism of A into a perfect ring B' of characteristic p, then for every integer f .... 0 we can define a homomorphism h, of '7T_ ,(A) into . -I B' by h , ('7T_/(a» = v(ay for all a E A. We see at once that h'+1 agrees with h I on '7T _ I (A) ; thus there exists a homomorphism h of AP- 00 into B' which induces h I on '7T _ I (A) for all f .... 0 and in particular, h extends ho = v. If h' is another

No.6

PRIME FIELDS. CHARACfERISTICS

A.V.7

extension of v to a homomorphism of AP- 0, we have (IV, p. 83, Prop. 11), dis(f)

=

Am -

2

n

(aj - aY

* o.

i a): Let c be the leading coefficient and D the discriminant of f; the resultant of f' and fis equal to ± cD (IV, p. 84, Formula (54», hence is non-zero ; therefore (IV, p. 78, Cor. 2) the polynomials f and f' are relatively prime in K[X]. a) => e): Let A be the K-algebra K [X]I (f) and x the image of X in A ; by III, p. 573, Prop. 22, the A-module OK (A) is generated by the elements dx, subject to the single relation f' (x) dx = O. By V, p. 33, Th. 3, the K-algebra A is thus etale if and only if f' (x) is an invertible element in A, which means that f and f' are relatively prime in K [X]. DEFINITION 2. - A polynomial f E K [X] is said to be separable if it is non-zero and satisfies the equivalent conditions a), b), c), d) and e) of Prop. 3.

Remarks. - 1) Let L be an extension of K and f a non-constant polynomial in K[X]. Bye) of Prop. 3 and V, p. 32, Cor. 2 it comes to the same to suppose that fis separable, whether considered as element of K[X] or of L(X]. On the other hand, it may well happen that f is irreducible in K[X] but not in L(X].

A.V.38

COMMUTATIVE FIELDS

§7

2) Let IE K[X]; we know (IV, p.13, Prop. 13) that there exist irreducible polynomials 11' ... , 1m in K [X] such that I = 11 ... 1m. Let n be an algebraic closure of K ; since an irreducible polynomial g E K [X] is the minimal polynomial over K of each of its roots in n, two distinct irreducible polynomials in K [X] have no common root in n. Condition d) of Prop. 3 now shows that I is separable if and only if the polynomials 11' ... , I m are separable and pairwise distinct. PROPOSITION 4. - Let f be an irreducible polynomial in K [X]. Then the following conditions are equivalent,' a) f is separable. b) There exists an extension L of K in which f has a simple root. c) The derivative f' of f is not zero. d) The field K is of characteristic 0, or it is of characteristic p "# 0 and f;' K[XP). We note first that an irreducible polynomial in K[X) is not constant. It is clear that a) implies b) (take an algebraic closure of K for L). If x is a simple root of fin an extension L of K, we have f' (x) "# 0 (IV, p. 17, Prop. 7), so b) implies c), and the equivalence of c) and d) follows from V, p. 9, Cor. Suppose finally that f' "# 0; let x be a root of f in an algebraically closed extension 0 of K. Since f is the minimal polynomial of x over K and deg f' -< deg f, we have f'(x)"# 0, and so x is a simple root of f (IV, p. 17, Prop. 7). Therefore f is separable and we have shown that c) implies a). COROLLARY 1. - For a field K to be perfect it is necessary and sufficient that every irreducible polynomial of K[X) should be separable. If the field K is of characteristic 0, K is perfect and every irreducible polynomial of K [X) is separable, by d) above. Suppose then that K has characteristic P"#O. Suppose first that K is perfect. We have K[XP) = K[X)P, hence there exists no irreducible polynomial of K[X) belonging to K[XP). By Prop. 4, every irreducible polynomial of K [X) is then separable. Suppose next that K is imperfect, whence K "# KP. Let a be an element of Knot belonging to J.(P; the polynomial XP - a is irreducible in K [X) (V, p. 24, Lemma 1), and it belongs to K[XP), and so is not separable. COROLLARY 2. - Let f E K [X) be a non-zero polynomial. For f to be separable it is necessary and sufficient that there should exist an extension L of K which is a perfect field and such that f has no repeated factor in L [X). Let 0 be an algebraic closure of K ; if fis separable, fhas no repeated factors in O[X] (Prop. 3, Conversely, if L is a perfect extension of K such that fhas no repeated factor in L[X), then fis separable in L[X) (Cor. 1 and Remark 2), hence in K[X) (Remark 1).

d».

No.3

3.

SEPARABLE ALGEBRIC EXTENSIONS

A.V.39

Separable algebraic elements

DEFINITION 3. - Let E be an extension ofK. An element x ofE which is algebraic over K is said to be separable over K if the algebraic extension K (x) of K is separable. PROPOSITION 5. - Let E be an extension ofK, x an element ofE algebraic over K and f the minimal polynomial of x over K. Then the following conditions are equivalent: a) x is separable over K ; b) the polynomial f is separable; c) x is a simple root of f The equivalence of a) and b) follows from Prop. 3, that of b) and c) from Prop. 3 and 4 (cf. V, p. 37 and 38).

If an element x of E is a simple root of a polynomial 9 of K [X], it is separable over K. For the minimal polynomial fof x over K divides gin K[X] (V, p. 16, Th. 1), so x is a simple root of f COROLLARY 1. -

COROLLARY 2. If an element x of E is algebraic and separable over K, it is algebraic and separable over every extension K' of K contained in E. Let f be the minimal polynomial of x over K. Then x is a simple root of f by Prop. 5, and since fbelongs to K' [X], the element x of E is separable over K' by Cor. 1.

3. - Suppose that K has characteristic p '1= O. For an element x ofE to belong to K it is necessary and sufficient for it to be both separable algebraic and p-radical over K. The stated condition is clearly necessary. Conversely suppose that x is separable algebraic over K and p-radical of height e over K. Since x is separable over K, the minimal polynomial f of x over K does not belong to K [XP] (Prop. 4 and 5) ; since x is p-radical of height e over K, we have f (X) = XP' - x p ' (V, p. 24, Prop. 1) ; we conclude that e = 0, so X E K. COROLLARY

PROPOSITION 6. -

Let E be an extension of K. a) If E is algebraic and separable over K, every element of E is algebraic and separable over K. b) Conversely, let A be a set of elements ofE, algebraic and separable over K and such that E = K(A) ; then E is algebraic and separable over K. If E is algebraic and separable over K, the same is true of the extension K(x) of K for every x E E, whence a). Under the hypothesis b), the extension E is algebraic over K (V, p. 18, Cor. 1).

A.V.40

COMMUTATIVE FIELDS

§7

Let F be a subextension of E of finite degree over K. By V, p. ] 1, Cor., there exist elements Xl' ... , Xm of A such that Fe K(Xl' ... , xm) and we have K(Xl' ... , xm)

=

K[xl' ... , xm]

(V, p. 18, Cor. 1) .

By the hypothesis on A, the algebras K[xJl, ... , K[xm] are etale over K; hence the same is true of K[xJl ® ... ® K [xm] (V, p. 32, Cor. 1). Now F is isomorphic to a sub algebra of a quotient algebra of K[xJl ® ... ® K [xm], and so is etale (V, p. 30, Prop. 3). For an algebraic extension E to be separable over K it is necessary and sufficient that every element of E should be a simple root of its minimal polynomial over K. It is enough to apply Prop. 5 and 6. COROLLARY. -

4.

The theorem of the primitive element

Let E be an extension of K ; an element X of E is said to be primitive if E = K[x]. For the extension E to possess a primitive element it is necessary for [E : K] to be finite. THEOREM 1. -

Let E be an extension of K. Then the following conditions are

equivalent,' a) E possesses a primitive element ; b) there exist only a finite number of subextensions of E. These conditions are satisfied when E is a separable extension of finite degree. Suppose first that E possesses a primitive element x, and let fbe the minimal polynomial of X over K. For each monic polynomial gEE [X] dividing f in E [X] denote by Eg the subextension of E generated by the coefficients of g. Since the possible polynomials 9 are finite in number (if f splits in E [X] into a product of r monic irreducible polynomials, the number is bounded by 2~), the subextensions Eg are finite in number. To prove b) it is therefore enough to' show that every subextension L of E is one of the E g • Now if L is a subextension of E, then we have L[x] = E ; if 9 is the minimal polynomial of x over L, we have [E: L] = deg(g). Besides, 9 is a divisor offin L[X], hence in E[X]; so we have Eg eLand E = Eg[x]. Since g(x) = 0, we have [E: Eg] ~ deg(g), therefore [E: Eg] ~ [E: L] and so L = Eg as we wished to show. Next we observe that Condition b) implies that the extension E is of finite degree: by Remark 2 of V, p. 18 it suffices to prove that it is algebraic; now if z is an element of E transcendental over K then the subextensions K(zn), n E N are pairwise distinct. To show that b) =? a) we now distinguish two cases: A) If the field K is finite, the field E is a vector space of finite dimension over K

No.5

SEPARABLE ALGEBRIC EXTENSIONS

A.V.4l

and hence is a finite set. Therefore I (V, p. 78, Lemma 1) there exists an element x of E generating the multiplicative group of E, and we have E = K [x]. B) Suppose now that the field K is infinite. If b) holds, the extension E is of finite degree, so b) can also be expressed by saying that E possesses only a finite number of subalgebras. This being so, the implication b) =;> a) is a consequence of the following more general proposition (for which the hypothesis that the field K be infinite is indispensable, cf. V, p. 153, Ex. 5 of § 7) : PROPOSITION 7. - Suppose that K is infinite; let A be a commutative K-algebra possessing only a finite number of sub-algebras ( for example an etale K-algebra, V, p. 30, Prop. 3) and let V be a vector subspace of A generating A. Then there exists x E V such that A = K [x]. Let AI' ... , An be the subalgebras of A distinct from A. If x rf. Al U ... U An' then the sub-algebra K [x] cannot equal any of the Ai and so must coincide with A. Further, since V generates A, it is not contained in any of the subspaces Ai' Prop. 7 is therefore a consequence of the following lemma: Lemma 1. - Let A be a vector K-space, V, AI' ... , An subspaces of A. If Card(K) ~ nand if V is not contained in any of the Ai' then V is not contained in Al U ... U An' Arguing by induction on n, we need only prove that if V ¢: An and V cAl U ... U An' then V c Al U ... U An -I' Let x E V, x rf. An' and let y be arbitrary in V. If y E Kx, we have y E Al U ... U An _I ; if not, then the elements x and y + Ax, A E K are strictly greater than n in number and belong to Al U ... U An' so two of them belong to the same Ai' Thus there exists i, 1 ~ i ~ n such that either x E Ai and y + Ax E Ai for some A E K, or y + Ax E Ai and y + j-lX E Ai for two distinct scalars A, fA. E K. In both cases we conclude that x E Ai and y E Ai ; but this implies that i 01= n, hence y E AI U ... U An -I as we wished to show.

This completes the proof of the equivalence of a) and b) in Th. 1. Finally if the extension E is separable and of finite degree, condition b) holds, by V, p. 30, Prop. 3. S.

Stability properties of separable algebraic extensions

PROPOSITION 8. -

Let E be an extension of K and (Ei)i E I a family of

subextensions of E such that E = K

(u I E

I

E i ). If each extension Ei is algebraic and

separable over K, the same is true of E. This follows at once from Prop. 6 (V, p. 39). 1 The reader may convince himself that Th. 1 is used nowhere before the proof of Lemma 1 of V, p. 78.

A.V.42

COMMUTATIVE FIELDS

§7

9. - Let F be an algebraic extension ofK and E a subextension ofF. For F to be separable over K it is necessary and sufficient that F should be separable over E and E separable over K. Suppose first that F is separable over K ; then E is separable over K by Prop. 1 (V, p. 36). Moreover, every element of F is separable over K (V, p. 39, Prop. 6) hence over E (V, p. 39, Cor. 2) and so F is separable over E (V, p. 39, Prop. 6). Conversely, suppose that F is separable over E and E separable over K. Denote by x an element of F and by fEE [X] the minimal polynomial of x over E. Since E is algebraic over K, Th. 2 (V, p. 18) shows that there exists a subextension E' of E of finite degree over K such that fEE' [X] ; then f is at the same time the minimal polynomial of x over E and over E', and since x is separable over E (V, p. 39, Prop. 6) it is so also over E' (V, p. 39, Prop. 5). Write F' = E' (x) ; then F' is separable and of finite degree over E', and since E is separable over K, E' is separable and of finite degree over K (V, p. 36, Prop. 1). Hence F' is separable and of finite degree over K, by V, p. 32, Cor. 2. Therefore (V, p. 39, Prop. 6) x is separable over K. We have now shown that every element of F is separable over K, hence F is separable over K (V, p. 39, Prop. 6). PROPOSITION

PROPOSITION 10. - Let E and K' be two subextensions of the same extension ofK and let E' = K' (E). Suppose that E is algebraic over K, hence E' algebraic over K' (V, p. 18, Cor. 2). a) If E is separable over K, then E' is separable over K'. b) Conversely if E' is separable over K' and E and K' are linearly disjoint over K, then E is separable over K. Assertion a) follows directly from Prop. 6 (V, p. 39). Under the hypothesis b), let F be a subextension of E of finite degree over K. Then F and K' are linearly disjoint over K, hence the K'-algebra F(K') = K' ® K F is isomorphic to K' (F). Since K' (F) is a subextension of E' of finite degree over K' and E' is algebraic and separable over K', the K'-algebra K' (F) is etale. In other words, the K'-algebra F(K') is etale, and now Cor. 2 of Prop. 4 (V, p. 32) shows that F is etale over K. Thus we have shown E to be separable over K.

6.

A separability criterion

PROPOSITION 11. - Suppose that K is of characteristic exponent p, and let E be an algebraic extension of K, generated by a set s. If E is separable over K, then E = K (Sf'n) for all integers n ~ 0 ; conversely if E is of finite degree over K and E = K (Sf'), then E is separable over K. The case p = 1 is trivial by the Cor. of V, p. 37. Suppose from now on that p-#l. By hypothesis E is algebraic over K and we have E = K (S), hence

K(SP) = K(EP) = K[EP]

b Y V , p. 18 , C or. 1 .

SEPARABLE ALGEBRIC EXTENSIONS

No.7

A.V.43

If E is of finite degree over K, it is a separable extension of K if and only if it is an

etale algebra over K ; the Cor. of V, p. 35 shows that this happens if and only if E = K[EP]. Suppose now that E is separable and of infinite degree over K. Then K [EP] is the union of the subrings K [E'P] where E' ranges over the set of subextensions of E of finite degree over K ; but such an extension E' is separable over K (V, p. 36, Prop. 1), whence E' = K [E'P] c K [EP] by what has been said; finally we have E = K [EP]. By induction on n ~ 0, the relation E = K [EP] implies that E = K [EP"]. COROLLARY 1. - Every algebraic extension of a perfect field Let K be a perfect field of characteristic exponent p, and let extension of K. Then E is separable over K (V, p.36, E = K(EP) by Prop. 11 ; but we have K = KP c P, hence E = so E is perfect. COROLLARY 2 (Mac Lane). -

is a perfect field.

E be an algebraic Prop. 2) whence K(EP) = EP, and

Let K be an algebraic closure of K and

KP~ Bn mapping XI ® ... ® Xn to the function (ai' ... , an_I) ~ aj(xI)'" an-J(xn-J) x n.

S.

Galois cohomology

Let N be a field, r a finite group of automorphisms of Nand K the field of invariants of r. For every integer n"", 1 we denote by GL (n, N) the group of square matrices of order n with coefficients in N and non-zero determinant (II; p.349). We let the group r operate on the group GL(n, N) by the rule a(A) = (a(aij)) for A = (aij)' PROPOSITION 9. - Let (Vcr)crEr be a family of elements ofGL(n, N). For A to exist in GL (n, N) such that Vcr = A-I. a (A) for all a E r it is necessary and sufficient that VcrT = Vcr' a(V for a, T in r. The condition is necessary,' if Vcr = A-I. a (A), then we have T )

GALOIS EXTENSIONS

No.6

A.V.65

The condition is sufficient: we identify the elements of N n with matrices of n rows and one column with coefficients in N. We let the groups r act on N n by For each a E r we denote by u" the mapping x ~ V a (x) of N n into itself. The verification of Formulae (1) to (3) of V, p. 62 is immediate. Moreover we have U 0 U, = u, and since u, is bijective, we have u, = IdNn. Let Vo be the set of vectors x E N n such that u" (x) = x for all a E r. By Prop. 7 (V, p. 63), Vo is a K-structure on N n ; in particular there exist in Vo vectors b l , ... , b n forming a basis of N n over N. The matrix B with columns b l , ... , b n is therefore invertible and the relation u,,(b i ) = b i for 1 ~ i ~ n is equivalent to VO". a(B) = B. Writing A = B- I, we obtain V" = A -la(A) for all a E r. 0" •

E

COROLLARY 1. - Let (C")"Ef be a family of non-zero elements of N. For a -:/= 0 to exist in N such that CrT = a (a) . a- I for all a E r it is necessary and sufficient that C"T = C". a(cT ) for a, T in r. COROLLARY 2. - Let (C")"Ef be a family of elements ofN. For b to exist in N such that aO" = a (b) - b for all a E r it is nesssary and sufficient that aO"T = a" + a(aT) for a, T in r. We have aT(b) - b = [a(b) - b] + a[T(b) - b] for all b in N and a, Tin r, whence the necessity. Conversely suppose that a"T = aO" + a(a T) for any a and T in r. Put

VO"=

(~ ~,,)

for aEr; then we have V"T= V".a(VT) for a, Tin r; by

Prop. 9, there exists thus a matrix A that a (A) = A V" for all a

( a(x) a(z)

E

=

(;

r ; writing

a(y)) = a (t)

(X z

~) with non-zero determinant such

down the relation a (A) = AVO" we find

xa" + y ) za" + t

(a

E

r) .

In particular, x and z belong to K and we have

If x -:/= 0 we have a" = a(b) - b with b = x-Iy; if z -:/= 0, we have the same

relation with b = z-It . Now x and z cannot both be zero because

xt - yz 6.

=

det A -:/= 0 .

Artin's theorem

THEOREM 2 (Artin). - Let N be a field, r a group of automorphisms ofN and K the field of in variants of r. Let V be a vector sub- K -space of N of finite dimension

A.V.66

§1O

COMMUTATIVE FIELDS

over K. Then every K-linear mapping u of V into N is a linear combination with coefficients in N of the restrictions to V of elements of r. Let u be a K-linear mapping of V into N and let V (N) = N KV be the vector N-space derived from V by extension of scalars; denote by u the N-linear form on YeN) such that u(x y) = x. u(y) for x E Nand y E V. For each O'er there exists an N-linear form h" on V (N) such that h" (x y) = xa (y) for x E Nand y E V. The canonical mapping of YeN) = N K V into N K N is injective. Now the Cor. of Prop. 8 (V, p. 64) show that the intersection of the kernels of the linear forms ha on V (N) is reduced to O. Therefore (II, p. 302, Cor. 1) there exist ai' ... , an in

r

n

i

L aiai(x) for all x E V. j

L aih,,;

and ai' ... , an in N such that u =

whence u(x)

=

=1

=1

Let us equip the set NN of all mappings of N into N with the product topology of the discrete topologies of the factors. Th. 2 means that the set of linear combinations with coefficients in N of elements of r is dense in the set of K-linear mappings of N into itself.

Let N be a field, r a finite group of automorphisms 0 fN and K the field of invariants of r. Let n be the cardinal of r. a) We have [N: K] = nand N is a Galois extension ofK with Galois group r. b) Let ai' ... , an be the elements ofr and (Xl' ... , Xn) a basis ofN over K, then det (ai(Xj» i= o. c) Let u be a K-linear mapping of N into N. There exists a unique family aaa(x) for all x EN. (aa)" E r of elements of N such that u(x) =

THEOREM 3. -

L

I1Ef

We equip the ring N K N with the N-algebra structure whose external law is given by A(x y) = x Ay for A, x, y in N. Then the dimension of the vector Nspace N ® K N is [N: K]. The dimension of the product vector N-space N r is equal to n. The mapping \(I defined in the Cor. of Prop. 8 (V, p. 64) is an Nisomorphism of N K N onto N r , whence [N: K] = n. Let.:1 be the group of Kautomorphisms of N. We have r c .:1, hence K is the field of invariants of .:1, and N is a Galois extension of K. Further, the order of .:1 is at most equal to [N: K] by Dedekind's theorem (V, p. 27, Cor. 2) and since the order of r equals [N : K], we have r = Ll. Hence r is the Galois group of N over K, and this proves a).

With the notation of b) put fi = \(I (Xi 1 ); we have fi (a) = (J" (Xi) for 1 "'" i "'" n and O'er. Since \(I is an isomorphism of vector N-spaces, the sequence (fl' ... , f n) is a basis of N r over N, whence det (fj (a i » i= 0, that is,

This proves b).

No.7

GALOIS EXTENSIONS

A.V.67

Finally, c) follows from Th. 2 (V, p. 65) which proves the existence of a family (a"),, E r such that u (x) = a"IT (x) (for all x E N) and from Dedekind's theorem

L

aef

(V, p. 27, Cor. 2) which proves the uniqueness of (a"),, E r.

7.

The fundamental theorem of Galois theory

THEOREM 4. - Let N be a Galois extension ofK and r its Galois group. Let:f be the set of subextensions of N and ~ the set of closed subgroups of r. For every subgroup Ll E ~ we denote by k(Ll) the field of invariants of Ll and for every sub field E E:f we denote by 9 (E) the group of E-automorphisms of N .. Then Ll ~ k (Ll) is a bijection of ~ onto :f, and E ~ 9 (E) is the inverse bijection. A) The relation E = k (g (E)) (for E E :f) is a consequence of the following more precise lemma :

Lemma 1. - Let E be a subextension ofN. Then N is a Galois extension ofE and Gal(N/E) is a closed subgroup of Gal(N/K) with the induced topology. Let x EN; the minimal polynomial f of x over E divides in E [X] the minimal polynomial 9 of x over K (V, p. 17, Cor. 2). Since N is Galois over K, the polynomial 9 is a product in N [X] of distinct factors of degree 1 ; hence the same is true of f and so N is Galois over E. Let r be the Galois group of N over K and Ll that of N over E. By definition Ll is the subgroup of r consisting of all IT such that IT (x) = x for all x E E. Now for each x E E the mapping IT ~ IT (x) of r into the discrete space N is continuous, hence D is closed in r. Let IT E r; for Xl' ... , xn in N let U(Xl' ... , xn) be the set of all T E r such that T (Xi) = IT (Xi) for 1 ~ i ~ n ; put

Then the family of sets U(Xl' ... , xn) (resp. V(Xl' ... , xn)) is a base of neighbourhoods of IT in r (resp. Ll). Hence the topology on Ll is that induced by r. B) The relation Ll = g(k(Ll)) (for Ll E g) is a consequence of the following more precise lemma : Lemma 2. - Let Ll be a subgroup ofr. Let E be the field of invariants of Ll ; then the Galois group of N over E is the closure of Ll in r. The Galois group of N over E is closed in r (Lemma 1) and contains Ll, hence it contains the closure Li of Ll. Let IT be an E-automorphism of N and let Xl, ••• , X n be in N. Since N is Galois over E (Lemma 1) there exist (V, p. 57, Prop. 2) a subextension No of N, Galois of finite degree over E and containing Xl' ... , x n • Let Llo be the image of the subgroup Ll of Gal(N/E) under the restriction homomorphism of Gal (N/E) into Gal (No/E). Since [No: E] is finite, Dedekind's theorem (V, p. 27, Cor. 2) shows that Gal (No/E) is finite. Hence Llo is finite, and since E is the field of invariants of Llo, we have Llo =

COMMUTATIVE FIELDS

A.V.68

§ 10

Gal(No/E) (V, p. 66, Th. 3). In particular, a o contains the restriction of a to No. Therefore there exists TEa such that a and T have the same restriction to No, whence a(xI) = T(XI), ... , a(xn) = T(Xn). It follows that a is a limit point of a in f, and hence Gal (N /E) c:: ~. COROLLARY 1. - Let E and E' be two subfields of N contammg K; then E c:: E' if and only if 9 (E ) ::::J 9 (E'). If a and a' are two closed subgroups off, then a c:: a' if and only if k(a)::::J k(a'). For the two inverse bijections E ~ 9 (E) and a ~ k (a) are inclusion-reversing. COROLLARY 2. L =

Let (Ei)i E I be a family of subfields of N containing K ; put

n Ei and M = K (u

E i ). Then 9 (L) is the smallest closed subgroup of f

i E l iEI

containing U 9 (Ei) and we have 9 (M) i

E

I

n 9 (Ei).

= i

E

I

The first assertion follows from Cor. 1 and the second is immediate. COROLLARY 3. - For i = 1, 2 let Ei be a subfield of N containing K and let a i = 9 (Ei). For any a E f the relations a(E I ) = E2 and aala- I = a 2 are equivalent. For we have T E 9 (a(E I )) if and only if m(x) = a(x), that is, a-ITa(x) = x, for all x EEl; this amounts to saying 'that a-ITa E ai' whence g(a(E I )) = aala- I. COROLLARY 4. - Let E be a subfield ofN containing K and let a = 9 (E). For E to be Galois over K it is necessary and sufficient that a should be a normal subgroup of f. When this is so, the restriction homomorphism of f into Gal (E/K) defines by passage to quotients a topological group isomorphism of f/a onto Gal(E/K). Since N is separable over K, the same is true of E (V, p. 36, Prop. 1). Therefore E is Galois over K if and only if it is quasi-Galois over K ; this also means that a (E) = E for every K-automorphism a of N (V, p. 52, Prop. 1 and p. 54, Prop. 3). By Cor. 3 this is equivalent to aaa- I = a for all a E f. The restriction homomorphism 'P: Gal (N/K) -+ Gal (E/K) is continuous and surjective (V, p. 60, Prop. 3) and its kernel is clearly equal to a = Gal (N /E). Since f is compact, the homomorphism of f / a onto Gal (E/K) derived from 'P by passage to quotients is an isomorphism of topological groups (Gen. Top., I, p. 87, Cor. 2). COROLLARY 5. - Let E be a subfield of N containing K. For E to have finite degree over K it is necessary and sufficient that 9 (E) should be open in f. When this is so, the index (f: 9 (E)) is finite and equal to [E: K]. For 9 (E) to be open it is necessary and sufficient that there should exist a sub extension F of N, of finite degree over K, such that in the notation of V, p. 60,

No.8

GALOIS EXTENSIONS

A.V.69

g(E) contains UF(Id N ) = g(F). The relation g(E):J g(F) is equivalent to E c F by Cor. 1 (V, p. 68), whence the first assertion of Cor. 5. Suppose that [E: K] is finite. Let 0 be an algebraic closure of K containing N as subextension (V, p. 23, Th. 2) and let Jf be the set of K-homomorphisms of E into D. Every element of Jf is induced by a K-automorphism of 0 (V, p. 52, Prop. 1), and since N is quasi-Galois over K, the mapping a ~ a I E of [ into Jf is surjective. For a and a' in [ to have the same restriction to E it is necessary and sufficient that a- la' E 9 (E), whence Card Jf = ([: 9 (E)). Finally since E is an etale algebra over K, we have Card Jf = [E: K] (V, p. 32, Prop. 4), so in conclusion we have ([: 9 (E)) = [E: K]. COROLLARY 6. - For i = 1,2 let E j be a subextension of Nand [ j the Galois group of N over E j • The following conditions are equivalent: a) The group [ is the direct product of the subgroups [I and [2. b) The extensions EI and E2 are Galois over K, we have EI n E2 = K and

For [ to be the direct product of the subgroups [1 and [2 it is necessary and sufficient that the following conditions hold (I, p. 48, Prop. 15) : (i) the subgroups [I and [2 are normal in [ ; (ii) [I n [2 = {E}, where E is the neutral element of [ ; (iii) [ = [I . [2· Now (i) means that EI and E2 are Galois over K (Cor. 4). By Cor. 2, condition (ii) is equivalent to N = K (EI U E 2), and finally if (i) and (ii) hold, [1[2 is the least subgroup of [containing [I U [2 ; it is closed because [I and [2 are compact and the mapping (a, T) ~ aT of [I X [2 into [ is continuous (Gen. Top., I, p. 63, Cor. 1). Cor. 2 now shows (iii) to be equivalent to EI n E2 = K, and this proves the equivalence of a) and b). Remark. - With the notation of Cor. 6 suppose that conditions a) and b) hold. The restriction homomorphisms 'Pi: r -> Gal (E;/K) for i = 1, 2 induce topological group isomorphisms

By a) we see that the mapping u~ ('PI(u), 'Pz(u» is a topological group isomorphism of Gal(N/K) onto Gal(EI/K) x Gal (Ez/K).

8.

Change of base field

Let N be a Galois extension of K and [ the Galois group of N over K ; further let N' be a Galois extension of K' with Galois group ['. We shall identify K (resp. K') with its image in N (resp. N'). Let u be a homomorphism of K into K' and v a homomorphism of N into N' whose restriction to K equals u (cf

A.V.70

§1O

COMMUT A TrVE FIELDS

Fig. 1). Let IT E r' ; since u (K) c K' , IT is a u (K )-automorphism of N' ; moreover v (N) is a Galois extension of u (K), hence IT induces a u (K )-automorphism of v (N) (V, p. 55, Remark 1). In other words, for every IT E r' there exists a unique element v * (a) of r such that

vov*(IT)=aov.

(8)

• N'

N

)

u

K

)

FIG.

I.

• K'

The mapping v * is a homomorphism of Gal (N' /K') into Gal (N /K). For every x E N, the mapping a t--+ v*(a) (x) = v-l(a(v(x») of r' into the discrete space N is continuous, so v * is continuous. Three particular cases are of importance : a) If F is a Galois extension of K and E a subextension of F, we know (V, p. 67, Lemma 1) that F is a Galois extension of E. Let us apply what has been said to the case where N = F, K' = E, N' = F and v = Id F . Then v * is merely the canonical injection j : Gal (F /E) ---. Gal (F /K) .

This is sometimes called the inflation homomorphism.

b) Suppose that in addition E is Galois over K. Let us apply what has been said to the case where N = E, K' = K, N' = F and v is the canonical injection of E into F. Then v * is just the restriction homomorphism 7T:

Gal (F/K) ---. Gal (E/K) .

We know (V, p. 60, Prop. 3) that 7T is surjective, with kernel Gal(F/E) and that by taking quotients it defines a topological group isomorphism of Gal(F/K)/Gal(F/E) onto Gal(E/K) (V, p. 68, Cor. 4). c) Suppose that v- 1 (K') homomorphism

=

K and N'

=

K' (v (N»; let us show that the

v*: Gal(N'/K') ---. Gal(N/K), is a topological group isomorphism, sometimes called translation. For the group Gal(N'/K') is compact, the group Gal(N/K) is separated and v* is continuous; so it is enough (Gen. Top., I, p. 87, Cor. 2) to prove that v* is bijective. Now every element IT of the kernel of v * is an automorphism of N' which induces the identity on K' and on v(N), hence IT = E because N' = K'(v(N»; it follows that v * is injective. Further, the image of v * is a closed subgroup d of Gal (N /K)

No.8

A.V.71

GALOIS EXTENSIONS

(Gen. Top. , I, p. 81, ibid.) and the field of invariants of Ll is equal to

v- 1 (K') = K; therefore we have Ll = Gal(N/K) (V, p. 67, Th. 4) and so v * is surjective. The general case may be reduced to the preceding ones by composition. To begin with we note that K' (v (N » is the field of invariants in N' of the kernel Ll of v*; since Ll is a normal subgroup of Gal(N'/K'), the extension K'(v(N» of K' is Galois (V, p. 68, Cor. 4). Thus v* is composed of the homomorphisms Gal(N'/K') ~ Gal(K'(v(N»/K')! Gal(N/v- 1 (K'» ~ Gal(N/K); in this sequence 1T is the restriction homomorphism associated with the triple K' c K' (v (N ) ) c N /, \)J is the translation isomorphism associated with the central square of the diagram (Fig. 2) and j is the inflation homomorphism associated with the triple K c V-I (K/) c N. The next theorem gives more detailed information about the structure of translation isomorphisms. N

K THEOREM

5. -

---+

I

---+

v- I(K')

K'(v(N»)

---+

1

~

N' FIG.

2.

K'

Let N' be an extension of K generated by two subextensions

K' and N. Suppose that N is Galois over K, with Galois group r and that K' n N = K. Then the extension N' of K' is Galois and the canonical homomorphism 'P of K' ® K N into N' is an isomorphism. Let a E Gal (N /K) and let a' be the element of Gal (N' /K') which corresponds to it under the translation isomorphism; then we have a' o'P = 'P 0 (Id K , ® a). We have N' = K' (N) and N is algebraic and separable over K ; hence (V, p. 42, Prop. 10), the extension N' of K' is algebraic and separable. By Cor. 4 of V, p. 54 the extension N' of K' is quasi-Galois. Therefore the extension N' of K' is Galois. By c) above the mapping a ~ a I N is an homomorphism A of Gal (N //K') onto Gal (N /K). We have N = K' [N] because N is algebraic over K (V, p. 18, Cor. 1), hence 'P is surjective. If a belongs to Gal (N /K), we have

(9)

A-I

(a)

0

'P

= 'P

0

(Id K, ® a) .

Therefore the kernel of 'P is stable under the mappings Id K , ® a, hence of the form K' ®K No with No c N (V, p. 63, Cor.). For x in No we have x = 'P (1 ® x) = 0, hence No = 0 and so 'P is injective. COROLLARY 1. - Let E' be a subfield ofN' containing K'. There exists a unique subfield E ofN containing K and such that E' = K' (E). We have E = E' n N.

A.V.72

§1O

COMMUTATIVE FIELDS

Put E=E'nN, then E'~K'(E). Now put f=Gal(N/K) and d = Gal (N /E ), and define f' and d' similarly. The mapping >.. : a ~ a I N is an isomorphism of f' onto f and also of d' onto d ; in other words, d' consists of those a E f' for which >..(a) belongs to d. If a E f' leaves the elements of K' (E) fixed, we have >..(a) E d, whence a E d' and a leaves the elements of E' fixed; by Cor. I of V, p. 68 we thus have K'(E)~E'. We have proved the equality E' = K'(E), whence 'I'-I(E') = K' ®K E. IfF is a subfield of N containing K and such that E' = K' (F), we have likewise 'I'-I(E') = K' ®K F, whence F = E. 2. - Let N be a Galois extension ofK. Suppose that the Galois group of N over K is the direct product of two closed subgroups fl and f2 and denote by E; the field of invariants off; for i = 1,2. Then the canonical homomorphism of EI ®K E2 into N is an isomorphism. We have EI n E2 = K and N = K(EI U E 2) by Cor. 6 (V, p. 69), and so it is enough to apply Theorem 5. COROLLARY

f

Remark. - Let K and K' be two fields and u a homomorphism of K into K'. Let K, (resp. K;) be a separable closure (V, p. 45, Prop. 14) of K (resp. K') and IT (resp. IT') the Galois group of K, over K (resp. K; over K'). Since K, is a separable algebraic extension of K and the extension (K;, u) of K is separably closed, there exists (V, p. 45, Cor.) a homomorphism v of K, into K; extending u. From v we obtain a continuous homomorphism v * of IT' into IT. Let VI be another extension of u ; since K, is a quasi-Galois extension of K, there exists an element Uo of IT such that v1=vou O• We conclude that Vt(T)=UOIV*(T)U O for all T E IT.

9.

The normal basis theorem

Let N be a Galois extension of K, with Galois group f. We identify f with the canonical basis of the group algebra K(r) (III, p. 446) ; then N may be considered as a left K(r)-module (III, p. 447, Example), so that u .x =

L

a(J" a (x ) for

x

E

Nand

L

u =

a(J" a

In

K (r) .

If N is of finite degree over K, the group f is finite by Dedekind's theorem (V, p. 27, Cor. 2) and we can define the element t = L a in K(r) ; then we have aEr

TrN/K(x) =

L a(x), aEr

that is, TrN/K(x) = t.x for all x

E

N.

Let us define an action on the right by r on N by x a = u- l (x). In a similar way we can consider the multiplicative group N* as a right Z--+x". For example, the notation x 2,,+3H", where , MI = N, M2 = As and L = N. By the Cor. of V, p. 64 there exists a K-isomorphism 'P of N ®K N onto N ® K K(r) which maps x @ y to L xa- I (y) ® a. It is clear that 'P is an isomorphism of N ®K K(r)-modules and so the theorem follows from Lemma 3. * For the second proof we shall use the following proposition: PROPOSITION 10. - Let x EN, then for {x} to form a basis of the K(r)-module N E r "#- O. it is necessary and sufficient that det (aT (x Since K(r) and N have the same dimension over K, to say that {x} is a basis of N over K(r) means that the mapping a ~ ax of K(f) into N is injective. This in turn means that the mapping b ~ b(1 ® x) ofN ®K K(r) into N ®K N is injective (II, p. 306, Prop. 14). Now there is an N ®K K(rtmodule isomorphism of N ®K N onto N ® K K(r) which maps 1 ® x to L a- I (x) @ a. It follows that {x} is a basis

nr"

of N over K(r) if and only if, for every non-zero family of elements (n,), E T of N we have

(L n, ® T) (L a-I (x) ® a)

means that there exists a E r such that

"#- O. But this last relation

L n,a- IT (x) "#- 0, whence the proposition.

A) Suppose that K is infinite; the mapping x ~ det (aT (x )) of N into N is a polynomial mapping over K (IV, p. 54). By extension of scalars from K to N we get a corresponding mapping for the vector N-space N ® K N and we have just seen that the latter is not identically zero (because N ®K N is free of rank lover N ®K K(r»). So there exists x E N such that det(aT(x))"#- 0 (IV, p. 18, Th',2); more generally, from the same reference we have:

A.V.74

§ 10

COMMUTATIVE FIELDS

PROPOSITION 11. - Suppose that K is infinite and let P : N -> K be a polynomial mapping which is non-zero on K. There exists x E N such that P (x) "#- 0 and {x} is a basis of N over K(f).

B) Suppose that K is finite. By Prop. 4 (V, p. 95) I every extension of finite degree over K has a cyclic Galois group. We shall therefore more generally consider the case where the group r is cyclic of order n ; we denote by 'Y a generator of r. The following lemma is a particular case of more general results proved in Chapter VII. The ring A is either the ring Z of rational integers or the ring K [X] of polynomials over the field K. Lemma 4. Let M be a torsion A-module generated by a finite number Xl' ... , xh of elements; then there exists an element x of M whose annihilator (II, p. 219) is equal to the annihilator of M. In both cases A is an integral domain and every ideal of A is principal. When A = Z (resp. A = K[X]), we denote by f!J the set of prime numbers (resp. the set of irreducible monic polynomials in K [X]). For every element a "#- 0 of A there exists then an invertible element u of A and a family (vp (a»p E 9', with finite support, of positive integers such that a = u pvp(a) and u and the integers

fl

PEiJ1

vp(a) are uniquely determined (I, p. 51 and IV, p. 13, Prop. 13). Let ai be the annihilator of Xi (for 1 ~ i ~ h) and a the annihilator of M ; let aI' ... , ah, a be non-zero elements of A such that ai = Aa i and a = Aa ; since a = al n ... n ah, it follows from what has been said that

(10)

vp (a) =

sup Vp (ad

for all p

E

f!J .

l~i~h

.. L et us wnte a m th e form uPInell ... PrnCr) , WI'th PI' ... , Pr d'" Istmct m ;a7J 7, n (1) >- 0, ... , n (r) >- 0 and u an invertible element of A. Let j = 1, ... , r ; by (10) there exists an integer c (j) such that 1 ~ c (j) ~ h and vp/ac(j» = n (j) ; there exists b j in A with ac(J) = Pj(j)b j and the element Yj = bjxc(j) has as annihilator the ideal Apj(J). Let us show that the annihilator b of Y = Yl + ... + Y r is equal to the annihilator a of M. In any case we have a c b, so b is of the form Apr(J) ... p;z(r) with o ~ m (j) ~ n (j) for 1 ~ j ~ r. If we had a "#- b, there would exist an integer j such that 1 ~ j ~ rand m(j) -+ rrx of the group r such that the stabilizer of each point of X is an open subgroup of r. It comes to the same to say that the mapping (rr, x) >-+ rrx of r x X into X is continuous when X is equipped with the discrete topology. Let X be a finite r-set. We define an action of r on the K-algebra K~ of mappings of X into K, by the formula

u,J(x)

(12)

=

rr(f(rr-lx))

for rr E r, f E K~ and x E X. Let @ (X) be the set of invariants of r in K~; this is the sub-K-algebra ofK~ consisting of mappings f: X ---> K, such that f(rrx) = rr(f(x)) for rr E r and x E X.

Let X be a finite r -set and let Xl' rX n form a partition of X. For

Lemma 5. -

be points of X such that the orbits 1 ~ i ~ n let ~j be the stabilizer of Xj in r and let L j be the field of invariants of ~j. Then L l , ... , Ln are separable extensions of finite degree of K, and the mapping f>-+ (f(x l ), ••. , f(xn)) is a Kalgebra isomorphism of@(X) onto Ll x ... x Ln. By hypothesis the subgroups ~l' ... , ~n of r are open and Cor. 5 of V, p. 68, shows that the subextensions L l , ... , Ln of K, are of finite degree over K. Clearly they are separable; now the last assertion of Lemma 5 is immediate. From Lemma 5 and Th. 4 (V, p. 34f.) we obtain immediately the following result.

rx l ,

... ,

... , Xn

A.V.76

COMMUTATIVE FIELDS

§ 10

PROPOSITION 12. For every finite r-set X the algebra 0(X) is etale over K, of degree equal to the cardinal ofX. Moreover, every etale algebra over K is isomorphic to an algebra of the form 0(X).

Remarks. - a) It is easy to show that for every K-algebra homomorphism 't' of o (X) into K, there exists a unique element x of X such that 't' (f) = f (x) for all f E 0(X). 2) Let X and Y be two finite r-sets. Let 3\(X, Y) be the set of mappings u of X into Y such that u(ux) = uu(x) for all IT E r and all x E X. For u E 3'r(X, Y) we define a K-algebra homomorphism u*: 0(Y) ..... 0(X) by u*(f) = f 0 u. For every homomorphism 'It of 0 (Y) into 0 (X) there exists a unique element u of 3'r(X, Y) such that 'It = u*.

11.

The structure of quasi-Galois extensions PROPOSITION 13. Let N be a quasi-Galois extension of K. We denote by N, the field of invariants_ of the group of all K-automorphisms of N and by N, the relative separable algebraic closure of K in N (V, p. 44). Then: a) N, is the relative p-radical closure of K in N (V, p. 25). b) N, is a Galois extension of K and every K-automorphism of N, extends in a unique fashion to an N,-automorphism of N. c) The fields N, and N, are linearly disjoint over K and we have N = K IN, UN,] ; in other words, the canonical homomorphism ofN, ®K N, into N is an isomorphism. Let n be an algebraic closure of K containing N as subextension (V, p. 23, Th. 2). Every K-automorphism of n induces an automorphism of N because N is quasiGalois. Therefore every element of N, is invariant under the group of K-automorphisms of n, hence p-radical over K (V, p. 53, Cor. 3). Conversely every element of N which is p-radical over K is clearly invariant under every K-automorphism of N and hence belongs to N,. This proves a). Every K-automorphism of n maps N into N, hence N, into N" so N, is a quasiGalois extension of K (V, p. 54, Prop. 3). It follows that N, is a Galois extension of K. Every element of N, n N, is separable algebraic and p-radical over K, hence belongs to K (V, p. Cor. 3) ; we thus have K, n K, = K. Now N is p-radical over N, (V, p. 44, Prop. 13) and separable algebraic over N, (V, p. 56, Th. 1) hence both p-radical and separable over K(N, UN,). Hence we have N = K(N, UN,) (V, p. 39, Cor. 3) and the assertions b), c) follow from Th. 5 (V, p.71).

Let p the characteristic exponent ofK, K an algebraic closure ofK, K, the relative separable closure ofK in K and Kp·z, ct>3' ct>4' ct>s, ct>7' ct>g, ct>g and ct>11 follow directly from (7) ; now we have ct> 1ct> Zct> 3ct>6 = X 6 - 1 and ct>1ct>Zct>3ct>4ct>6ct>lZ = Xl2 - 1, whence ct> 4ct>l2 = XIZ - 1 6 . X6 + 1 4 Z 6 = X + 1 and fmally ct>l2 = -z-- = X - X + 1. The cases n = 6 and X-I X+l n = 10 may be treated similarly.

Remark. - * For every integer n >- 0 a function f-L (n) is defined as follows: if n is divisible by the square of a prime number, we put f-L(n) = 0, otherwise f-L (n) = (- 1 if n is the product of h distinct prime numbers (n (X) =

(8)

TI

(X n/ d

-

1)1l-(d) ,

dl n

or more explicitly (9) PI

n under the unique homomorphism of Z[X] into K[X] which maps X to X.

Lemma 3. - The roots of ct>n in Ks are the primitive n-th roots of unity. Denote by Sn the set of roots of ct>n in Ks. By Formula (6), the set f-Ln (Ks) is the union of the Sd for d dividing n. Every primitive n-th root of unity thus belongs to Sn and the lemma now follows from Prop. 4 (V, p. 81).

A.V.84

§11

COMMUTATIVE FIELDS

PROPOSITION 6. - Let p be the characteristic exponent of K and let n ~ 1 be an integer prime to p. For the polynomial n(X) to be irreducible in K[X] it is necessary and sufficient that the homomorphism Xn: Gal (KslK) -+ (Z/nZ)* should be surjective. By Lemma 3 we have Rn(K) = K(O for each root ~ of n(X) and hence n(X) is irreducible in K[X] if and only if the degree 'P(n) of n(X) is equal to [Rn (K) : K]. Further, the Galois group of Rn (K) over K is of order [Rn (K) : K] and it is isomorphic to the subgroup of (Z/nZ)* which is the image of Xn. Now Prop. 6 follows from the fact that (Z/nZ)* is of order 'P(n). THEOREM 2 (Gauss). - Let Q be an algebraic closure of Q and let n ~ 1 be an integer. a) The cyclotomic polynomial n(X) is irreducible in Q[X]. b) The degree of Rn(Q) over Q is 'P(n). c) The homomorphism Xn of Gal (Q/Q) into (Z/nZ)* is surjective and defines by passage to quotients an isomorphism of Gal (Rn(Q)/Q) onto (Z/nZ)*. Taking account of Prop. 6, we need only prove c). Every integer r prime to n is a product of prime numbers Pl' ... , Ps not dividing n ; so it is enough to show that for every prime number p not dividing n, the mapping x ~ x P of J..l.n (Q) into itself extends to an automorphism of Rn (Q). It suffices to prove that if ~ is a primitive nth root of unity, the minimal polynomial f of ~ over Q is equal to the minimal polynomial 9 of ~P over Q. We shall assume that f =1= 9 and argue by contradiction. The polynomials f and 9 are monic irreducible in Q [X] and divide Xn - 1, so there exists u E Q [X] such that Xn - 1 = fgu (IV, p.13, Prop. 13). Lemma 2 (V, p. 82) shows that f, 9 and u have integer coefficients. Let us denote by fj the polynomial with coefficients in Fp obtained from a polynomial v E Z[X] by reduction modp. We thus have Xn - 1 = Igu in Fp [X]. Further we have g(~P) = 0 and so g(XP) is a multiple of f(X) in Q[X]. By Lemma 2 there exists hE Z[X] such that g(XP) = f(X). h(X). Now we have v(XP) = v(XY for every polynomial v E Fp[X]. By reduction modp we thus obtain

gP = Iii.

If v is an irreducible polynomial in Fp [X] dividing

1, it must then

divide g. Since Ig divides Xn - 1, we conclude that divides Xn - 1 in Fp[X]. This is absurd because the polynomial xn - 1 is separable in Fp[X]. It may be shown that for every abelian extension E of finite degree over Q there exists an integer n ~ 1 such that E is isomorphic to a subextension of Rn (Q). * In other words, the field Q (J..l. oo (C)) is an abelian closure of Q. *(- 0 such that a' E K *n ; then the subfield L of n generated by the roots of the polynomial Xn - a is a cyclic extension of K of degree r. In particular, Xn - a is irreducible if and only if r = n. Remark. - Let a E K * and let r be the least integer >- 0 such that a' E Kn. Let B be the set of roots in K of the polynomial X n /, - a ; then we have Xn - a =

(14)

n (X' - b) ,

be B

by substitution of X' for T in the relation T n /, - a = IT (T - b). By Example 4 each of the polynomials X' - b is irreducible, so that (14) is the decomposition of Xn - a into irreducible polynomials in K [X]. 9.

Artin-Schreier theory

In this No. we shall use p to denote a prime number and assume that K is of characteristic p. We denote by n an algebraic closure of K and by tJ the endomorphism of the additive group of n defined by tJ (x) = x P - x . By V, p. 93 the kernel of tJ is the prime subfield Fp of K. For every subset A of K we denote by K (tJ- 1 (A» the subextension of n generated by all x E n such that tJ(x)EA. Lemma 7. - K(tJ-l(A» is an abelian extension ofK of exponent dividing p. Since the polynomials tJ - a = XP - X - a, a E A are separable over K, the extension L = K(tJ-l(A» is Galois. Let a E Gal (L/K) and x E tJ-l(A); we have tJ(a(x» = tJ(x), hence a(x) -x E Fp' that is, a(x) = x + i, i E Fp. This implies that aP(x) = x + pi = x, hence a P = 1 ; similarly if a' E Gal (L/K) and a' (x) = X + j, then we have a a a' (x) = X + i + j = a' a a (x), so a a a' = a' a a. Lemma 8. - Let L be a Galois extension of K. There exists a unique mapping (a, a) ~ [a, a) of Gal(L/K) x «tJ(L) n K)/tJ(K» into Fp such that for all a E Gal (L/K) and every element x E L such that tJ (x) E K we have, on denoting by tJ(x) the residue class oftJ(x)modtJ(K),

(15)

[a, tJ(x»

=

a(x) - x.

§11

COMMUTATIVE FIELDS

A.V.92

This mapping is Z-bilinear (for a, T E Gal (L/K), a, b E (.P(L) n K)/.P(K), we have [aT, a) = [a, a) + [T, a), [a, a + b) = [a, a) + [a, b) ). For the right-hand side of (15) is an element of Fp which depends only on the residue class of .P (x ) mod .P (K) ; this proves the first assertion; the second may be verified without difficulty. For every Galois extension L of K let us write aL: (.P(L)

n K)/.P(K)

ai.: Gal (L/K)

-->

-->

Hom (Gal (L/K), Fp)

Hom «.P (L) n K)/.P(K), Fp)

for the homomorphisms obtained from the above Z-bilinear mapping (V, p. 87). PROPOSITION 10. - For every Galois extension L of finite degree of K, the homomorphism aL is bijective. Let x E L be such that .P (x) E K and the residue class of .P (x) mod .P (K) lies in the kernel of aL. For every a E Gal (L/K) we have by definition a (x) = x ; hence x E K and .P (x) E .P (K). This proves the injectivity of aL. Now let f: Gal (L/K) --> Fp be a homomorphism; for all a, T E Gal (L/K) we have f(aT) = f(a) + a(f(T»,

f(a)

E

Fp.

By V, p.65, Cor. 2 there exists x E L such that f(a) = a(x) -x for all a E Gal (L/K). Since f(a) E Fp we have .P(a(x» = .P(x), hence a(.P(x» = .P(x) for all a E Gal (L/K) and .P(x) E K. If a is the residue class of .P(x) mod 'p(K), we have f(a) = [a, a), so f = aL(a). COROLLARY. - IfL is a Galois extension ofK, the homomorphism aL is injective and its image is the group Homc(Gal(L/K), Fp) of continuous homomorphisms of the topological group Gal (L/K) into the discrete group Fp. This is proved in the same way as the Cor. of Prop. 9, V, p. 89. THEOREM 5. - a) The mapping A ~ K('p-1(A» is a bijection of the set of subgroups ofK containing .P(K) onto the set of abelian subextensions of exponent dividing p in D. The inverse mapping is L ~ .P (L) n K. b) For every subgroup A of K containing .P(K), the homomorphism a' : Gal (K (.P- 1(A) )/K)

-->

Hom (A/.P (K), Fp)

is bijective, and it is a homeomorphism when Hom (A/.P(K), Fp) is equipped with the topology of simple convergence. c) Let A be a subgroup ofK containing .P(K) and let B be a basis of the vector Fp-space A/.P (K). For each a E B let Xa be an element of D such that .P (x a ) is a representative of a in A. Then the monomials xQ = x;(a) with

n

a

E

B

No.1

FINITE FIELDS

A.Y.93

= (ex (a)) in N(B) such that 0 :!S: ex (a) Gal(K/K) such that TIK(l) = a q. Let

m

:>

r

be the subgroup of Gal (K/K) generated by

0 we have

a; (x) = xqm for

all x

E

a q.

For every integer

K, hence the set of fixed points of

a; is equal to Km. Since Km oF K, we have a; oF 1. Hence there is an isomorphism

TIo of Z onto r which maps 1 to a q • The field of invariants of r consists of the x E K such that x q = x, hence is equal to K. Therefore (V, p. 67, Lemma 2) the group r is dense in Gal (K/K). Since every subextension of K of finite degree over K is one of the fields K m, the subgroups Gal (K/Km) form a fundamental system of neighbourhoods of 1 in Gal (K/K). It is clear that

a;,

r n Gal (K/Km) is the cyclic group generated by

hence equal to TIo(mZ).

From the above remarks about the topology of Z the isomorphism TIo : Z -> r extends in a unique fashion to an isomorphism of topological groups TI K: Z -> Gal (K/K). Let m ~ 1 be an integer; it is clear that the Frobenius automorphism of K relative to Km is Hence we obtain the relation

a;.

4.

Cyclotomic polynomials over a finite field

Let K be a finite field of q elements, n ~ 1 an integer not divisible by the characteristic p of K and Rn a cyclotomic extension of level n of K (V, p. 81). We know that the group IJ.- n(Rn) = IJ.- n of n-th roots of unity in Rn is cyclic of order n, that Rn = K (IJ.- n ) and that there exists an injective homomorphism

'Pn: Gal (Rn/K) such that a(O = ~j Further, if f is Rn over K is cyclic p. 95, Prop. 4). We

->

(Z/nZ)*

for a E Gal (Rn/K), ~ E IJ.- n and j E 'Pn(a). the degree of Rn over K, then the Galois group of of order f, generated by the automorphism a q : x ~ x q (V, have at once:

6. The image under 'Pn of the Frobenius automorphism is the residue class of q mod n. Therefore, taking into account Prop. 6 of V, p. 84 :

PROPOSITION aq

COROLLARY. The degree of Rn over K is the least integer f ~ 1 such that qf == 1 (mod n). For the cyclotomic polynomial a) may be proved in the same way. PROPOSITION

COROLLARY. - Let E and F be algebraically disjoint over K. Let E' be the relative algebraic closure of E in Land F' that of F (Y, p. 19). Then E' and F' are algebraically disjoint over K. Let B be a transcendence basis of E over K ; this is also a transcendence basis of E 'over K. Since E is algebraically disjoint from F over K, B is algebraically free over F, hence over F' (V, p. 108, Prop. 6) ; now we can apply Prop. 12. PROPOSITION 13. - Let L be an extension of a field K and E, F two subextensions of L. a) We have tr. degFF(E) ~ tr. degKE. When E and F are algebraically disjoint over K, then every transcendence basis of E over K is a transcendence basis of

A.V.114

COMMUTATIVE FIELDS

§ 14

F(E) over F and we have tr. degFF(E) = tr. degKE. Conversely, this equality implies that E and F are algebraically disjoint over K when tr· degKE is finite. b) We have tr. degKK(E U F) ~ tr. degKE + tr . degKF. When E and Fare algebraically disjoint over K, we have tr . degKK (E U F) = tr. degKE + tr . degKF. Conversely, this equality implies that E and Fare algebraically disjoint over K when E and F are of finite transcendence degree over K.

a) Let B be a transcendence basis of E over K ; then E is algebraic over K(B), and Cor. 2 of V, p. 18 shows that F(E) is algebraic over F(K(B)) = F(B). By Th. 2 (V, p. 109) B contains a transcendence basis of F (E) over F ; when E is algebraically disjoint from F over K, B is algebraically free over F (Prop. 12) and this is then a transcendence basis of F (E) over F. The three first assertions of a) follow from this. Suppose now that E is of finite transcendence degree over K, equal to that of F (E) over F; since F (E) is algebraic over F (B) and Card B = tr. degFF(E), Cor. 1 of V, p. 110 shows B to be algebraically free over F, so E is algebraically disjoint from F over K (Prop. 12). b) We have K(E U F) = F(E) and so the Cor. of V, p.110 implies the equality:

Now b) follows at once from a) and this equality. PROPOSITION 14. - Let L be an extension of a field K, E and F two subextensions of Land B a transcendence basis of E over K. For E and F to be algebraically disjoint over K it is necessary and sufficient that K(B) and F should be linearly disjoint over K. For E and F to be algebraically disjoint over K it is necessary and sufficient for B to be algebraically free over F (Prop. 12), that is, for the monomials in the elements of B to be linearly independent over F. Since these monomials form a basis of the vector K -space K [B ], it comes to the same to say that K [B ] and Fare linearly disjoint over K. Finally, since K (B) is the field of fractions of K [B ], Prop. 6 of V, p. 14 shows that K [B ] and F are linearly disjoint if and only if this is so for K (B) and F. COROLLARY 1. - If E and F are linearly disjoint, then E is algebraically disjoint from F over K. Conversely, ifE is a pure extension ofK and is algebraically disjoint from F over K, then E and F are linearly disjoint over K. COROLLARY 2. Every pure extension of K is linearly disjoint from every algebraic extension of K ; in particular, K is relatively algebraically closed in every pure extension of K.

No.6

6.

A.Y.115

TRANSCENDENTAL EXTENSIONS

Algebraically free families of extensions

DEFINITION 6. - Let L be an extension of a field K and (Ej)j E I a family of sub extensions of L. The family (Ej)j E I is said to be algebraically free if the following condition is satisfied: '(AF) For each i E I let Aj be a subset ofEj which is algebraically free over K. Then Aj n Aj = 0 for i ", j and U Aj is algebraically free over K. j E

I

Remark. - By Prop. 3 (Y, p. 107) it is enough to verify the condition (AF) for finite subsets A j. We thus obtain the following result: if (Ej)i E I is an algebraically free family, the same is true of (E:)j E I if E: is a sub extension of E j for each i E I; conversely, if every family (E: )j E I' where E: is a finitely generated sub extension of E for each i E I is algebraically free, then (Ej)j E I is algebraically free. On the other hand, for (Ej)j E I to be algebraically free it is necessary and suffic~ent that (Ej)j E J should be algebraically free for every finite subset J of I. Speaking intuitively we may say that the algebraic independence of extensions is a property « of finite character ».

PROPOSITION15. - Let (Ej)j E I be a family of sub extensions ofa given extension L of a field K. Then the following conditions are equivalent: a) The family (Ej)j E I is algebraically free. b) For each i E I the extension E j is algebraically disjoint over K from the extension F j generated by the E j for j ", i . c) There exists a family (Bj)j E I of disjoint subsets of L, such that Bj is a transcendence basis of E j over K for each i' E I, and B = U B j is algebraically free j E

over K.

I

It is clear that a) implies c).

Assuming c), let us choose i in I ; put Cj

=

U B j • For each j ", i every element j >F

j

of E j is algebraic over K(B j ) and a fortiori over K(Cj). By Cor. 1 of Y, p. 18, the field Fj is therefore algebraic over K(Cj). Further, we have Bi n Cj = 0 and B = Bi U C i is algebraically free over K ; therefore Bi is algebraically free over K(C i ) (Y, p.107, Prop. 4), hence also over Fj (which is algebraic over K(C i » by Prop. 6 of Y, p. 108. Thus we have proved that Ei is algebraically disjoint from Fj over K (Y, p. 113, Prop. 12), hence c) implies b). Let us now assume b) and prove a). It is enough to show that if ii' ... , in are distinct elements of I, then the family of extensions (EjI , ... , E,o • ) is algebraically free; we argue by induction on n, the case n = 1 being trivial. Suppose then that n >- 1 and that the family (EjI ' ... , E j• _) is algebraically free; for 1 "" k "" n choose a subset Ak of E jk algebraically free over K and put

A.V.116

§ 14

COMMUTATIVE FIELDS

B = Al U ... U An _ I' By the induction hypothesis the subsets AI' ... , An _ I are pairwise disjoint and B is algebraically free over K ; by b) Ei n is algebraically disjoint from Fin and since B is contained in Fin we have B n An = 0 and B U An = Al U ... U An is algebraically free over K. We have thus shown that the family (Ei 1 , ... , Ei n ) is algebraically free. The following proposition generalizes part b) of Prop. 13 (V, p. 113). PROPOSITION 16. - Let (Ei)i E I be a family of sub extensions of an extension of a field K, and let E be the field generated by U E i . i

a) We have tr . degKE

E

I

L tr. degKEi , with equality when the family (Ei)i

~ i

E

E

I

is

I

algebraically free over K. b) Conversely, assume that tr . degKE =

L tr . degKEi j

and that tr . degKE is

e I

finite; then the family (Ei)i E I is algebraically free over K. For each i E I let Bi be a transcendence basis of Ei over K and put B = UBi' For each i E I, each element of Ei is algebraic over K(B i ), hence over i

E

I

K(B). Now Cor. 1 of V, p. 18 shows that E is algebraic over K(B) ; by V, p. 109, Th. 2, B thus contains a transcendence basis of E over K. If moreover the family (Ei )i E I is algebraically free over K, then the Bi are disjoint and the set B is algebraically free over K. This establishes a) (Set Theory, III, p. 160, Cor. of Prop. 4). Under the hypotheses of b), E is algebraic over K (B) and of finite transcendence degree over K, and we have Card(B) ~ tr. degKE. By Cor. 1 of V, p. 110, B is algebraically free over K and the Bi are disjoint. Now Prop. 15 shows that the family (Ei)i E I is algebraically free over K. Before stating the next theorem let us remark that there exist algebraically closed extensions of K of arbitrary transcendence degree, for example an algebraic closure of an appropriate field of rational fractions. THEOREM 5. - Let (Ei)i E I be a family of extensions of a field K and algebraically closed extension of K. Suppose that the inequality (2)

tr. degK n

n an

L tr. degKEi

"'" i

E

I

holds. Then there exists an algebraically free family (Fi)i E I of subextensions of n such that Fi is K-isomorphic to Ei for all i E 1. For each i E I let Bi be a transcendence basis of Ei over K. Let B be a transcendence basis of n over K. By (2) we have Card B "'" L Card Bi ; hence j E

I

there exists a family (Bi)i E I of pairwise disjoint subsets of B and bijections u i : Bi ---> Bi (for i E I). By Prop. 2 of V, p. 106, U i extends to a K-isomorphism vi of K (Bi) onto K (Bf) ; since n is algebraically closed and Ei algebraic over

No.7

TRANSCENDENTAL EXTENSIONS

A.V.1l7

K (Bi)' the Cor. (V, p. 23) shows that Vi extends to a K-isomorphism of Ei onto a subextension Fi of ll. By construction B: is a transcendence basis of Fi over K, and Prop. 15 (V, p. 115) shows that the family (Fi)i E I of subextensions of II is algebraically free over K. COROLLARY 1. - Let E and II be two extensions of a field K. Suppose that II is algebraically closed, of transcendence degree at least equal to that ofE. Then E is K-isomorphic to a subextension of ll.

2. - Let II be an algebraically closed field of infinite transcendence degree over its prime subfield. Then every field of the same characteristic as II is the ascending directed union of fields isomorphic to subfields of ll. For each field is the ascending directed union of finitely generated subfields over its prime field, and now it is enough to apply Cor. 1. * Example. - This applies particularly in characteristic 0 in taking II = C ( 0 such that 'P (n) ~ [L: Q], which is absurd (V, p. 80, Formulae (2) and (3)). COROLLARY 3. - If E is a finitely generated extension of a field K, then every sub extension E' of E is finitely generated. Let B' be a transcendence basis of E' over K. By V, p. 109, Th. 2, B / is contained in a transcendence basis B of E over K and so is finite, by Prop. 17. Since E / is algebraic over K (B /) and E is a finitely generated extension of K (B /), Cor. 1 shows [E / : K (B /)] to be finite. Now Prop. 17 shows that E' is finitely generated over K.

Prop. 17 may be paraphrased by saying that a finitely generated extension of K is an algebraic extension of finite degree of a purely transcendental extension K(Xl' ... , xn)·

§ 15.

1.

SEPARABLE EXTENSIONS

Characterization of the nilpotent elements of a ring

PROPOSITION 1. - Let A be a commutative ring and x an element of A. For x to be nilpotent it is necessary and sufficient that 1 - xT be invertible in the ring A[T]. We note that A [T] is a subring of the formal power series ring A [[T)) and that

L a): Since every extension of a field of characteristic 0 is separable (V, p. 122, Th. 1) it suffices to examine the case where K is of characteristic p oF O. Let X be the set of all a-algebra homomorphisms of a ® K L into a and f the homomorphism of a ® K L into aX defined by f (u) = (X (u ) )x E x for u E a ® K L. Condition b) means that f is injective and so implies that the ring a ®K L is reduced. Hence condition b) of Th. 2 (V, p.122) is satisfied with K' = a, so L is separable over K. a) => d): Suppose that L is separable over K. Let V be a vector sub-K-space of finite dimension of L, V (0) = a ®K V the vector a-space derived from V by extension of scalars and fa the linear form on V (0) such that fa (x ® y) = xy for x E a, y E V. Denote by G the group of K-automorphisms of a; for a E G we put a v = a ® Id v and geT = a 0 fa 0 a l . For each a E G the mapping g" of V (0) into a is a-linear and maps x ® y to x. a(y) for x E a, y E V. The kernel N" of geT is therefore a vector subspace of V (0)' and the same holds for N = n N". If p is the characteristic

v

"EG

exponent of K, the field of invariants of G in a is equal to KP- d): With the notation of d) we can identify

R with the

relative algebraic

closure of K in [ (V, p. 22, Ex. 2). Suppose that the ring A =

R ®K L

is an

integral domain. Let E be a subextension of R of finite degree over K ; the subring

No.6

REGULAR EXTENSIONS

A.V.143

E ® K L of A is an integral domain and hence is an algebra of finite degree over L ; by the Cor. of V, p. 10 it is a field. Since K is the union the increasing directed set of extensions E of the above type, A is a field (V, p. 11, Prop. 3). The canonical homomorphism of A into L mapping x ® y to xy (for x E K and

K are linearly disjoint over K. d) => a): Under the hypotheses of d) we have L n K = K, so K is relatively

y E L) is therefore injective, so Land

algebraically closed in L ; further if p is the characteristic exponent of K, the field L is linearly disjoint from KP- oo over K, hence L is separable over K (V, p. 123, Cor. 1). COROLLARY

1. -

Let A be an algebra over a field K. For A to be a regular K-

algebra it is necessary and sufficient for the ring K ® K A to be an integral domain. The stated condition is clearly necessary. Conversely, assume that K ®K A is an integral domain and denote by E the field of fractions of A. By Prop. 4 (V, p. 141) the ring K ®K E is an integral domain, hence E is a regular extension of K, by Prop.9; by V, p. 141, Cor., we conclude that A is regular as K-algebra. COROLLARY 2. - Let K be an algebraically closed field. Every K-algebra which is an integral domain is a regular K-algebra. In particular, every extension of K is regular. This follows from Cor. 1. COROLLARY 3. - Let K be an algebraically closed field. If A and B are two Kalgebras which are integral domains, then the same is true of A ® K B. By Cor. 2, A and B are regular K-algebras, and it suffices to apply Prop. 2 (V, p. 141).

6.

Application to composite extensions

10. -- Let Land M be two extensions of a field K and (E, u, v) a composite extension of Land M (V, p.12). Suppose that the ring L ® K M is an integral domain and that the subextensions u (L) and v (M ) of E are algebraically disjoint over K. Then u (L) and v (M) are linearly disjoint over K. Put w = u * v (V, p. 12), denote by F the field of fractions of the integral domain L ® K M and identify L (resp. M) with a subfield of F by means of the mapping x ~ x ® 1 (resp. y ~ 1 ® y) ; then the restriction of w to L (resp. M) is u (resp. v). Let B be a transcendence basis of Mover K (V, p. 109, Th. 1). By hypothesis u (L) and v (M) are algebraically disjoint over K ; therefore (V, p.114, Prop. 14), u(L) and v(K(B)) are linearly disjoint over K. Thus there exists a K-homomorphism u L (B) -> E which agrees with u on L and with v au K (B). By construction Land M are linearly disjoint over K in F ; by Prop. 8 (V, p. 15) the subfields L(B) and M of F are linearly disjoint over K(B). It PROPOSITION

I

:

A.V.144

COMMUTATIVE FIELDS

§ 17

follows that there exists a K-homomorphism w': M [L(B)] ---> E which agrees with u on L (B) and with v on M. But the field F is generated by M U L(B) and M is algebraic over K(B); we thus have M[L(B)] = F (V, p. 18, Cor. 2). Hence we conclude that w' is a K-isomorphism of F onto E whose restriction to L (resp. M) is u (resp. v). Thisshows u(L) and v(M) to be linearly disjoint over K. I

Let fl be an extension of a field K and La subextension of fl which is regular over K. Every subextension M of fl which is algebraically disjoint from Lover K is linearly disjoint. The ring L ® K M is an integral domain by definition of regular extension and it suffices now to apply Prop. 10. COROLLARY 1. -

2. - Let fl be an extension of a field K and L, M two subextensions of fl. Suppose that L is separable over K and that the relative separable closure 0 fK in M is equal to K. If Land M are algebraically disjoint over K, then they are linearly disjoint over K. By Prop. 10 it is enough to remark that the ring L ®K M is an integral domain (V, p. 140, Cor.). COROLLARY

Exercises §1

1) Let A be a not necessarily commutative ring, not zero and without divisors of zero (I, p. 98). Show that if there exists an integer m ~ 1 such that mA = 0, then the set of all integers having this property is an ideal pZ, where p is a prime number, and A is a ring of characteristic p. 2) Let m, n be integers such that 0",; n ",; m. Let p be a prime number and let m = ao + aIP + ... + a"p' and n = ~o + ~1P + ... + ~"p' be the expansions of m and n in base p, where we thus have 0",; a i (Xi

for at least one index i. If on the other hand,

a i for 0",; i ",; r, then

(In (Z/pZ) [X] we have (1 + x)m

(1 + X)"O(l + XP)"'1 ... (1 + XP')"".)

=

3) Let A be a commutative ring. For every polynomial f E A [X] we write in the polynomial ring A [X, Y] of polynomials in two indeterminates over A,

f(X + Y)

L t::.mf(X) ym ,

=

m =0

where t::.m is thus an A-linear mapping of A [X] into itself. a) If Z is an indeterminate and T Z the A-linear mapping of A[X] into A[X, Z] such that Tzf = f (X + Z), show that T Z 0 t::.m = t::.m 0 T Z (where on the right t::.m is the mapping defined above in the ring B [X], where B = A [Z D. Deduce that in A [X] we have

for integers m, n ~ 0 (ct. IV, p. 8). b) Suppose that A is a ring of characteristic p:> 0 and for each integer k

Dk

=

•

t::.pk. Show that If f

Every polynomial f

E

E

A[X] and g

E

k

A[xP ] then Dk(fg)

=

~ 0 put g. Dd + f. Dkg.

A [X] can be uniquely written in the form f (X)

=

I

m =0

where deg (g m)

- 0 with respect to x, then K (x) is an algebraic extension of degree n of K (y ).

§ 3

Let P(X)

A.V.149

EXERCISES

=

o~ i ~ n

I

YiXi be the minimal polynomial of x over K(y). Show that for

;=0

we have either Yi E K or K(Yi) = K(y). b) Deduce from a) that every Y E K (x) such that K (y) = K (x ) is of the form (ax + b)/ (ex + d) where a, b, e, d are elements of K such that ad - be oF 0 ; converse. Find all K-automorphisms of K(x). e) Show that if Y E K (x) is of height n with respect to x, and z E K (y) of height m with respect to Y, then z is of height mn with respect to x. * d) Let F be an extension of K such that KeF c K (x) and F oF K ; let y be an element of F whose height m with respect to x has the least possible value; show that F = K (y ) «< Liiroth's theorem »). (Let P be the minimal polynomial of x in F[T]; show that P (T) = 0 (T, x )/R (x), where 0 is a polynomial of K [T, X] of degree;;.: m in X and is not divisible by any non-constant polynomial of K[X], and R is a polynomial of K[X]. If y = g(x)/h(x), where 9 and h are relatively prime in K[X], observe that by a) the polynomial 9 (T) h (X) - 9 (X) h (T) is not divisible by any non-constant polynomial of K[T] or of K[X] ; deduce that necessarily geT) heX) - g(X) h(T) = e. OCT, X),

where e

E

K,

using Ex. 10.) *

* 12) a) With the notation and hypotheses as in Ex. 11, show that for an extension F of K such that KeF c K (x) to be such that F n K [x] oF K, it is necessary and sufficient that F = K (y) where y = 9 (x) for a non-constant polynomial 9 E K [T]; then we have K(y) n K[x] = K[y]. (Write F = K(y) with y = g(x)/h(x), 9 and h being relatively prime polynomials of K[T]. If there exists a non-constant polynomial P E K[T] such that P (x) E F, we can write P (x) = 0 (y )/R (y), where 0 and R are two relatively prime monic polynomials of K[T], not both constant. Distinguish two cases, according as R = 1 or R is non-constant; in the second case, decompose 0 and R into factors of the first degree in an algebraic closure n of K and observe that if a, b are distinct elements of n, then 9 - ah and 9 - bh are relatively prime in n[T].) b) If F = K(y) with y = g(x), where 9 E K[T] is non-constant, we may suppose that geT) = TgI(T), where gl E K[T]. Show that if K(xg(x» n K[y] oF K, we necessarily have gl (T) = aT" with a E K and n :> O. (Observe that by hypothesis and a) there exist two non-constant polynomials P, 0 in K [T] such that P (xg (x» = 0 (g (x» f- K ; deduce that there are two elements a, b in K and an integer m such that geT) divides a - bTm.) c) Let y, z be two elements of K[x] such that K(y) n K(z) oF K; show that then K[y] n K[z] oF K. d) Find y E K(x) such that K(y) = K(x) but K[y] n K[x] = K. *

* 13) Let K be a field, peT, X) = ao(T) X· + a l (T) X· - I + ... + a.(T) a polynomial in K [T, X]; suppose that: 1° the polynomial a o is not divisible by T; 2° each of the polynomials a l , ... , a. is divisible by T; 3° the polynomial a. is not divisible by T2. Show that under these conditions P, as polynomial of K(T)[X], is irreducible (argue by contradiction, using Ex. 10). * * 1-11 With the notation as in Ex. 11, show that if we take 9 (X) = X and h(X)=X'+X+1 (n;;.:2), the polynomial (g(X)h(x)-h(X)g(x»/(X-x) of K(x)[X] is irreducible (use Ex. 13). *

A.V.IS0

COMMUTATIVE FIELDS

§S

* 15) Give an example of a field K containing two subfields K I, K2 such that K is algebraic over KI and over K z but not over KI n K 2. (Take K = Q(T), KI = Q(T 2), and K2 = Q (T2 - T). Verify that [K: K I ] = 2, [K: K 2] = 2 and KI n K2 = Q. To do this note that an element I (T) of KI n K2 satisfies the conditions 1(- T) = I (T) and 1(1 - T) = f(T).) §

4

1) Let K be a field and (Ei)i e I any family of extensions of K. Show that there exists an extension E of K and for each i E I, a K-isomorphism ui of Ei into E such that E is generated by the union of the ui(Ei ) (argue as in Prop. 4 (V, p. 22)).

* 2) Let K be an algebraically closed field of characteristic 0, F = K «xl/n!)) the field of formal power series in xl/nt, and E = K«X)), where X = (xl/n!)"I, the subfield of F consisting of formal power series in X. For every formal power series I E K «(X)) we denote by w (f) its order. a) Let P(Y) = aoyn + alyn-l + ... + an be a polynomial of E[Y] of degree n, such that an oF O. Show that there exists a strictly increasing sequence (i k)O d "', of integers in the interval (0, n) such that: 1° io = 0, i, = n ; 2° w(a ik ) is finite for 0"" k "" r; 3° for each index j such that 0"" j "" n, distinct from the i k and such that w(a j ) is finite, the point (j, w(aj)) E R2 is above the straight line passing through the points (i k, w(a ik )), (i k _ I'

W

(aik -I)) and strictly above it if j

segments joining the points (i k _ I'

W

W

i k (1 "" k "" r). The union of the (a ik )) for 1 "" k "" r is caned the

Newton polygon of P, the preceding segments are called its sides and the points (ib w(aik)) its vertices. b) Put Pk = i k - i k _ l , 1. Show that if the intersection F n K [(x - 1) (xn - 1)] is distinct from K, then K is of characteristic p :> 0 and n = p', so that K (x) is a p-radical extension of F. (Writing y = x-I and using Ex. 12, b) of V, p. 149, show that (y + 1)" - 1 = ay n for an a E K.) b) Show that if KeF' c K (x) and if K (x) is not a p-radical extension of F' and if K oF F' oF K (x), then there exists z E K [x] such that z f. K and F' n K [z] = K. (Argue by contradiction, using a) and Ex. 12, b) of V, p. 149.)

§7

EXERCISES

A.V.151

c) Under the same hypotheses as in b), suppose that F' = K(y), where Y E K[x] ; show that there exists U E K [x] such that u ¢ K and F' n K (u) = F' n K [u] = K. (Observe that if this were false, we should have F' n K[z] #- K for all z E K[x] such that z ¢ K, using Ex. 12, c) of V, p. 149.) 2) Let K be a field of characteristic p :> 0 and f an irreducible monic polynomial of K [X]. Show that in K [X] the polynomial f (XP) is either irreducible or the p-th power of an irreducible polynomial, depending on whether or not there exists a coefficient of f not belonging to KP (decompose f (XP) into factors of the first degree in n [X], where n is an algebraically closed extension of K). 3) Let K be a field of characteristic p :> 2 and let F be the field of rational fractions K(X, Y). Let E = F(e) be an extension of F generated by a root e of the polynomial

f (Z)

=

Zq, + XZP + Y

of F [Z]. Show that [E: F (eP)] = p, but that E contains no p-radical element over F not in F. (Firstly remark that f is irreducible in F[Z]; if ~ E E existed such that ~ E F, ~ ¢ F, f would be reducible in F (~)[Z]; using Ex. 2, show that then Xl/p and yl/p would belong to E and we would have [E: F] "'" p2.) (See Ex. 1, V, § 7). § 6

1) Let K be a field and A a commutative K-algebra. For A to be etale it is necessary and sufficient that there should exist a finite subset S of A generating the K-algebra A and such that for all XES the subalgebra K [x] generated by x is etale. (If S generates A, the Kalgebra A is isomorphic to a quotient of the tensor product of the K [x], where x runs over S.) 2) Let r be a finite commutative monoid. Consider the following conditions Eo and Ep where p denotes a prime number. (Eo) If x E r and if e, d are integers"", 0, the relation Xe+d = x d implies the relation x·+ 1 =x. (Ep) If x, Y E r, the relation x P = yP implies x = y. Let K be a field of characteristic p ("'" 0) and let A be the K-algebra K(f) of the monoid r. a) For A to be etale it is necessary and sufficient that the condition (Ep) be satisfied. (Use Ex. 1 for the sufficiency. For the necessity observe for p = 0, that x· + 1 - x is nilpotent if Xe+d = x d ; for p:> 0 use the Cor. of V, p. 35.) b) Let X be the set of homomorphisms of r into the multiplicative monoid of K. We have Card X "'" Card r. If K is algebraically closed, then Card X = Card r holds if and only if condition (Ep) is satisfied. c) If r is a group, condition (Eo) always holds; if p :> 0, condition (Ep) is equivalent to saying that each element of r has order prime to p, or also that Card r is not divisible by p. § 7

1) a) Let K be a field of characteristic p :> O. An algebraic extension F of K of finite degree is said to be exceptional if it is not separable over K and if it contains no element

A.V.152

COMMUTATIVE FIELDS

§7

x ¢ K which is p-radical over K, in other words if FP n K = KP (V, p. 151, Ex. 3). Show that if F is exceptional and if E = F" Z E F, Z ¢ E and zP E E, then for each t E K we have zP ¢ EP (t) (observe that otherwise we should have EP (t) = EP (t P) and so t E FP). b) Conversely, let E be a separable algebraic extension of K of finite degree, and let x E E be such that for each t E K we have x ¢ EP(t). Let F = E(x l / P), so that E = F,. Show that F is an exceptional extension of K. (If there existed y E F such that y ¢ E and yP = Z E K, show that we would have z E EP(x) ; this would imply z E EP, hence y E E, which is a contradiction.) A separable extension E of K of finite degree is called strongly separable if the union of the fields EP (t), where t ranges over K, is distinct from E ; it thus comes to the same to say that E = F" where F is a inseparable exceptional extension of K. A strongly separable extension of K cannot be a perfect field. c) Show that if E is a strongly separable extension of K then it is impossible that [E: EP] = p. (Argue by contradiction, showing that [E: EP] = p would imply EP(t) = E for all t E K such that t ¢ KP ; observe on the other hand that if KeEP, E cannot be strongly separable over K unless it is perfect.) d) Show that if [K: KP] = p, then there is no strongly separable extension of K, and so neither is there an exceptional extension of K. 2) Let k be an imperfect field of characteristic p, a E k an element such that a ¢ k P, F the field k(X) of rational fractions in one indeterminate over k; put 2 Y = XP /(XP + a) and K = k(y). 2 a) Show that [F: K] = p2, so that TP - yTP - ay is the minimal polynomial of X in K[T] ; deduce that the relative separable closure of Kin F is E = k(XP) (cf. V, p. 148, Ex. 11). b) Show that F is an exceptional extension (Ex. 1) of K. (Argue by contradiction, assuming bE F such that b ¢ K and b P E K. Observe that K(b) = k(z), where z = r(X)/s(X) with deg r = p, deg s "" p and r, s being relatively prime polynomials in k [X] ; show that zP E K and that the minimal polynomial of X over K is r(TY - zPs(TY up to a factor from K. Conclude, using a) that we would have a E k n FP = k P contrary to the hypothesis.) 3) Let K be a field of characteristic p:> 0, E an extension of finite degree of K, E, and E, the largest p-radical extension and separable extension of K contained in E. Show that E, ®K E, is isomorphic to the relative separable closure of E, in E, with which it will be identified; if it is not equal to E, then E is an exceptional extension of E, and E, ®K E, is a strongly separable extension of E,. Conversely, let E be an exceptional extension of a field L and let K be a subfield of L such that L is p-radical over K and [L: K] -< + 00 ; then L is the largest p-radical extension of K contained in E. 4) Let K be a field, E = K (x) a separable algebraic extension of K and f the minimal polynomial of x in K[X]. a) Let F be an extension of K; in the ring F[X] the polynomial f decomposes into a product fd2 ... f, of irreducible and separable pairwise distinct polynomials. Show that the algebra E ® K F is isomorphic to the product of fields F [X 11 (fj) (1 "" j "" r ). b) Suppose that F = K (y) is a separable algebraic extension of K ; let 9 be the minimal polynomial of yin K[X], and let 9 = 9192 ... 9, be the decomposition of 9 into a product of

§9

EXERCISES

A.V.153

irreducible polynomials in E[X]. Show that r = s and that if m = deg I, n = deg g, I j , nj = deg gj we can, after permuting the gj if necessary, suppose that mj/nj = m/n for 1"" j "" r. mj = deg

5) Let K be a finite field with q elements, and let A be the K-algebra Kn. For A to be generated by a single element, it is necessary and sufficient that n "" q. § 8

1) Let K be a commutative ring, I(X) a monic polynomial in K[X], A the K-algebra K [X]I (f) and x the residue class of X in A. For each a E K we have I (a) = N A/K (a - x). 2) Let K be a field and A a K-algebra of finite degree n. For each integer m relatively prime to n we have A *m n K * = K *m ; in other words, if x E K * and if there exists an element y of A such that x = ym, then there exists an element z in K such that x = zm. § 9

1) The polynomial X2 - 2 is irreducible in Q[X] (cf. I, p. 51, Th. 7) ; let a be one of its roots in an algebraic closure n of Q. Show that the polynomial X2 - a is irreducible over E = Q(a); let ~ be one of the roots of this polynomial in n and let F = E(~) = Q(~). Show that F is not a quasi-Galois extension of Q ; what is the quasi-Galois extension of Q generated by F?

* 2) Let E be the relative algebraic closure of Q in R «< the field of real algebraic numbers ») ; if u is an automorphism of E, then u (x) = x for all x E Q; show that if y E E is such that y:> 0, then also u(y):> O. Deduce that u(y) = y for all y E E. * 3) Show that every algebraic extension of a field K, generated by a set of elements of degree 2 over K is quasi-Galois over K. 4) Let K be a field, I an irreducible polynomial in K[X], separable over K and of degree n and let a i (1 "" i "" n) be its roots in an algebraic closure n of K. Let g be a polynomial in K[X], and h an irreducible factor in K[X] of the polynomial l(g(X». Show that the degree of h is a multiple rn of n and that h has exactly r roots in common with each of the polynomials g(X) - a i (1"" i "" n) of n[X] (consider the conjugates over K of a root of h).

5) a) Let K be a field, n an algebraic closure of K and E, F two subextensions of n. Show that every composite extension of E and F (V, p. 12) is isomorphic to a composite extension of the form (Lw, 1, w), where w denotes a K-isomorphism of F onto a subfield of n, 1 is the identity mapping of E and Lw is the subfield of n generated by E u w(F). b) Deduce from a) that if F is of finite degree n over K, there are at most n composite extensions of E and F that are pairwise non-isomorphic. c) Suppose that E is a quasi-Galois extension of K and F a sub extension of E. Show that every composite extension of E and F is isomorphic to a composite extension of the form (E, 1, v), where v is a K-isomorphism of F onto a subfield of E.

A.V.lS4

§10

COMMUTATIVE FIELDS

6) a) Let K be a field of characteristic exponent p, L an extension of K,

closure of Land R the relative algebraic closure of K in

u-

0, r an integer,,", 3 and nl 1 there exists a cyclic extension of degree n of the field Q. (Reduce to the case where n is a prime number q and use the structure of the multiplicative group (Z/qmz)* (VII, p. 13). * 4) Let K be a field of characteristic p, q a prime number oF p such that K contains the q-th roots of unity and let ~ be a primitive q-th root of unity. a) Let E be a cyclic extension of K of degree q< and CT a K-automorphism of E generating Gal(E/K). Let F be the field of degree q 1), a a Kautomorphism of F generating Gal (F /K). Show that there exist two elements 0., !3 of F such that Tr F / K (!3) = 1 and a(o.) - 0. =!3 P -13 (use Th. 3 of V, p. 85 and the Cor. of Prop. 1 of V, p. 49). Deduce that for all c E K the polynomial XP - X - 0. - c is irreducible in F [X 1; if 0 is a root of this polynomial in an algebraic closure n of F, show that there exists a K-homomorphism 0' of E = F(O) in n extending a and such that 0'(0) = 0 + 13 ; conclude that E is a cyclic extension of degree p' of K, that 0' generates Gal (E/K) and that E = K (0). Finally show that every cyclic extension of degree p' of K containing F is the splitting field of a polynomial XP - X - 0. - c of F [X] for a suitable C E K. 10) a) Let K be a field whose algebraic closure is an extension of prime degree q of K. Such that K is perfect and that q must be distinct from the characteristic of K (use Ex. 9).

§ 11

EXERCISES

A.V.163

b) Show that K contains the q-th roots of unity and that n is the splitting field of an irreducible polynomial xq - a of K [X] ; deduce that q = 2 (in the contrary case deduce 2 from Ex.5 that the polynomial xq - a would be irreducible). Show further that - a is a square in K (lac. cit.), that - 1 is not a square in K, and deduce that n = K (i), where i 2 = - 1. 11) Let K be a field whose algebraic closure n is an extension of finite degree :> 1 of K. Show that n = K (i ) where i Z= - 1. (If we had n c# K (i), show with the help of Galois theory and Sylow's theorem that there would then exist a field E such that K(i) c E c n and n is an extension of prime degree of E; now apply Ex. 10.) 12) Let K be a field of characteristic q, n an algebraically closed extension of K, x an element of n such that x If- K and M a maximal element of the set of sub extensions of n not containing x (V, p. 156, Ex. 9) ; suppose further that M is perfect so that M(x) is a Galois extension of prime degree p of M (lac. cit.). a) Show that either p = 2 and n = M (i) or for each integer r"", 1, there exists a single extension of degree p' of M ; this extension M, is cyclic over M, n is the union of the M, and the M, are the only extensions of finite degree of M. (Note that if a p-group r of order p' has only one subgroup of order p' - I, it is necessarily cyclic, arguing by induction on r and using I, p. 133, Ex. 10, and I, p. 77, Cor. of Prop. 11). b) Suppose that n c# M (i) and that p c# q. Show that M contains the p-th roots of unity; we thus have M (i) = M I = M (u) with uP EM. Show that if u is the p-th power of an element of M I , then MI = M(t), where t is pZ-th root of unity (if u = J3P for J3 E M I, consider the minimal polynomial of J3 in M[X)). c) Under the hypotheses of b) show that if u is not a p-th power of an element of r-l . MI then M, is the splitting field of the polynomial XP - u of MdX] (use b)). d) Under the hypotheses of b), show that if u is the p-th power of an element of M I, there exists "y E MI such that "Y If- M and that M z is the splitting field of XP - "Y; show that "Y is not the p-th power of an element of M z and deduce that M, is the splitting field of the polynomial Xl'

, -I

- "Y of MI [X].

13) A field K is called a Moriya field if every separable algebraic extension of finite degree over K is cyclic. * Every finite field is a Moriya field (V, p. 95, Prop. 4). * The field M of Ex. 12 is a Moriya field. a) For K to be a Moriya field it is necessary and sufficient that for each integer n :> 0 there should exist at most one separable extension of degree n over K. (Use the following fact: if G is a finite group and for each integer m "'" 1 dividing the order of G there exists at most one subgroup of G of order m, then G is cyclic; this follows from the same property for p-groups (Ex. 12) and from I, p. 80, Th. 4). b) If K is a Moriya field, show that every algebraic extension E of K is a Moriya field and that if F is an extension of E of degree mover E, then there exists an extension of K of degree mover K. 14) a) Let K be a field such that there exist algebraic extensions of K of arbitrarily large finite degree, and that all these extensions have as degree a power of the same prime number p. Show that under these conditions, for each power ph (h "'" 0) there exists an extension of K of degree ph (use I, p. 77, Th. 1).

A.V.l64

COMMUT A TIVE FIELDS

§11

b) Let p be a prime number and K a field such that there exists an algebraic extension of K of degree divisible by p if p '" 2 and divisible by 4 if P = 2. Show that there exists an algebraic extension E of K such that the set of degrees of algebraic extensions of E is the set of powers ph of p. (If K is perfect or of characteristic '" p, let KI be the perfect closure of K ; consider a maximal algebraic extension of K 1 , among all those whose elements are of degree not divisible by p over KI ; use a) and Ex. 11.)

15) Let K be a Moriya field (Ex. 13) and P the set of prime numbers dividing the degree of an algebraic extension of finite degree of K. Show that the set of degrees of algebraic extensions of finite degree of K is either equal to the set Np of integers all of whose prime factors are in P or to the subset of N p consisting of integers not divisible by 4. 16) For every prime number p there exists a Moriya field of characteristic p such that the set of degrees of algebraic extensions of finite degree of this field is the set N * of integers >- 0 I. Show that for every set P of prime numbers, there exists a Moriya field K of any characteristic, such that the set of degrees of algebraic extensions of finite degree of K is the set Np (argue as in Ex. 14,

b».

*

17) Let p. be the set of monic polynomials X· + aIX·- 1 + ... + a. of Z[X) whose roots in C all have absolute value "" 1. a) Show that the set p. is finite. Deduce that for every polynomial FE P n' if Fh denotes the polynomial whose roots are the h-th powers of the roots of F, then there exist two distinct integers h, k such that F h = F k. b) Conclude from a) that all the roots of a polynomial FE p. are either zero or roots of unity ( 0 and E an extension of K ; E is a p-radical extension of height"", 1 of K (EP). The cardinal of a p-basis of E over K (EP) (cf. V, p. 103, Th.2) is called the degree of imperfection of E over K. If K(EP) = E, E is said to be relatively perfect over K ; if E is perfect, it is relatively perfect over each of its subfields. The (absolute) degree of imperfection of K is the degree of imperfection of K over its prime subfield Fp, or also the cardinal of a (absolute) p-basis of K. a) If B is a p-basis of E over K(EP), show that for every integer k::> 0 we have E = K(Epk)(B). b) Suppose that E c Kp- n for an integer n ::> O. Show that for the degree [E; K] to be finite it is necessary and sufficient that the degree of imperfection mo of E over K should be finite; then mo is the least cardinal of the sets S such that E = K (S). Show that if k

mk is the degree of imperfection of K (EP ) over K, then mk + 1 "'" m k for all k, and if

f =

L mk,

then [E; K] = pt.

k

2) a) Let E be an extension of a field K of characteristic p ::> 0 and F an extension of E. Show that if B is a p-basis of E over K (EP) and Cap-basis of F over E (FP), then there exists a p-basis of F over K(FP) contained in B U C.

§ 14

A.V.171

EXERCISES

b) Suppose that B is finite and that F is an algebraic extension of K of finite degree. Show that [E: K (EP)] "" [F: K (FP)]. If moreover F is separable over E, show that [E : K (EP)] = [F: K (P')].

3) Let K be an imperfect field and E an algebraic extension of K of finite degree. For an element x E E to exist such that E = K (x) it is necessary and sufficient that the degree of imperfection of E over K (Ex. 1) should be 0 or 1. (To see that the condition is sufficient, note that if E, is the relative separable closure of K in E, we have E = E, (n) and E, = K ([3) and if n rf- E" consider the conjugates of n + c[3 over K, for c E K.) 4) Let K be a field of characteristic p :> 0 and E an algebraic extension of K of finite degree. If r:> 0 is the degree of imperfection of E over K (Ex. 1), show that r is the least cardinal of sets S such that E = K (S). (To see that E can be generated by r of its elements note that if E, is the relative separable closure of K in E, then there exist r elements ai (1"" i "" r) of E such that E = E,(a!, ... , a,), and use Ex. 3.) 5) Let K be a field and E an algebraic extension of K of finite degree. a) Suppose that K is of characteristic p :> O. Show that if the degree of imperfection of E over K (Ex. 1) is :> 1, there exist an infinity of distinct fields F such that KeF c E (reduce to the case where K(EP) = K, and consider the fields K(a + Ab), where the set {a, b} is p-free over K and A E K). b) Conclude that for there to exist only a finite number of fields F such that KeF c E it is necessary and sufficient that the degree of imperfection of E over K be "" 1. 6) Let K be a field of characteristic p :> 0, E an algebraic extension of K of finite degree and N the quasi-Galois extension of K generated by E. a) Let E, and E, (resp. N, and N,) be the largest p-radical extension and the largest separable extension of K contained in E (resp: in N). Show that for E = E,(E,) (in which case E is isomorphic to E, ®K E,) it is necessary and sufficient that E, = N,. b) If the degree of imperfection of E over K is 1, then a necessary and sufficient condition for E = E, (E,) is that the degree of imperfection of N over K be 1. Deduce that if N is a quasi-Galois extension of K, whose degree of imperfection over K is equal to 1, then E = E,(E,) for every sub extension E of N (cf. Ex. 2, c) Conversely if N is a quasi-Galois extension of K of finite degree whose degree of imperfection over K is :> 1, show that if N, ~ N, then there exists tEN such that for the extension E = K(t) of K we have E ~ E,(E,) (if a, b are two elements of a p-basis of N over K(NP) and if x E N is separable over K and x rf- K, take t = a + bx).

b».

§ 14

1) Let E be an extension of a field K and let B be a transcendence basis of E over K. Show that E is equipotent to K x B if one of the sets K, B is infinite, and countable otherwise. * Deduce in particular that every transcendence basis of the field R of real numbers over the field Q of rational numbers has the power of the continuum. * 2) Let K be the field Q (X) of rational fractions in one indeterminate over the field Q of rational numbers. Show that in the ring K[Y] the polynomial y2 + X2 + 1 is irreducible and if E is the extension of K generated by a root of this polynomial (in an algebraic closure of

A.V.l72

COMMUTATIVE FIELDS

§

14

K), then Q is relatively algebraically closed in E but E is not a pure extension of Q. (To see that there is in E no element a rI= Q algebraic over Q note that the existence of such an element would imply the relation E = Q (a)(X), and observe that - (X2 + 1) is not a square in n (X), where n is an algebraic closure of Q.) Show that if i is a root of X2 + 1, then E (i) is a pure transcendental extension of Q (i ).

* 3) Let K be the field C (X) of rational fractions in one indeterminate over the (algebraically closed) field C of complex numbers. Show that in the ring K[Y], the polynomial y3 + X 3 + 1 is irreducible; let E be the extension of K generated by a root of this polynomial (in an algebraic closure of K). Show that E is not a pure extension of C, even though C is algebraically closed. (Show that it is impossible for a relation u 3 + v 3 + w 3 = 0 to exist between three polynomials u, v, w of C [X], pairwise relatively prime and not all three constant. Argue by contradiction: if r is the largest of the degrees of u, v, wand if for example deg (w) = r, deduce from the relation w3 = -

(u + v) (u + jv)(u + j2v)

where j is a pnmltlve cube root of unity, the existence of three polynomials u l , in C[X], pairwise relatively prime, not all three constant, of degrees Hom (G, fLn). Show that 8 is surjective. Put fj = Ker (8). Using the elementary divisor theorem (VII, p. 24), show that B is a free K(fj)-module of rank Card(G ). e) By passing to the field of fractions deduce from d) and a) that pO is the field of fractions of K(f j). Deduce that FG is a pure transcendental extension of K. * § 15

1) Let K be a field of characteristic p :> 0, E an extension of K and B a p-basis of E over K(EP) (V, p. 98). -] a) Assume that E c KP ; let Ko be a subfield of K such that E is separable over Ko. Show that the set BP is a p-independent subset over Ko(KP) (observe that if

A.V.176

§ 15

COMMUT ATIVE FIELDS

(aJ is a basis of the vector Ko-space K, then (aD is a basis of Ko(KP) over Ko). Let C be a subset of K, disjoint from Wand such that W U C is a p-basis of Kover Ko(KP) ; show that B U C is a p-basis of E over Ko(EP). b) Assume that E c KP-' and that E is separable over a subfield Ko of K. Show that if the

degree of imperfection of Kover Ko is finite (V, p. 170. Ex. 1), then it is equal to the degree of imperfection of E over Ko (use

a».

2) Let K be a field of characteristic p > 0, E an extension of K and F a separable extension of E. Show that any two of the following three properties imply the third: a) B is a p-basis of E over K(EP) ; b) C is a p-basis of F over E(FP) ; c) BeE, B U C is a p-basis of 'F over K(FP) and B n C = 0. (Use the fact that if (cl") is a basis of the vector E-space F, then (c~) is a basis of K(EP)[FP] over K(EP) and of E[FP] over E.) 3) Let K be a field of characteristic p > 0, E an extension of K, and F a finitely generated extension of E. Show that the degree of imperfection of F over K is at least equal to that of E over K (reduce to the following two cases: 1" F is separable over E ; 2° F = E (x), where x P E E (cf. V, p. 170, Ex. 2». 4) Let F be a separable extension of a field K of characteristic p > 0. Show that if E is a subextension of F, relatively perfect over K (V, p. 170, Ex. 1), F is separable over E. 5) Let K be a field of characteristic p > 0. Show that if E is an algebraic extension of K or a relatively perfect extension of K, then EP-'" = E(KP-"'). 6) Let K be a field of characteristic p > 0, E a separable extension of K and B a p-basis of E over K(EP). a) Show that B is algebraically free over K (consider an algebraic relation of least degree between the elements of B and put the degrees of the variables which occur in the form kp + h with ~ h ~p - 1). Deduce that if tr. degKE -< + 00, then the degree of imperfection of E over K is at most equal to tr. degKE. b) Show that E is separable and relatively perfect over K(B).

°

°

7) Show that a relatively perfect transcendental extension E of a field K of characteristic is not finitely generated over K. (In the opposite case show that for every transcendence basis B of E over K, E would be algebraic and separable over K(B).)

p >

8) If E is a separable algebraic extension of a field K of characteristic p > 0, then the largest perfect subextension EP'" of E (the intersection of the fields EP' for n"" 0) is algebraic over the largest perfect subfield 1(1'''' of K. (If x E EP"', show that for every integer

°

n> we have x P - ' E K (x) and deduce that the minimal polynomial of x over KP' is the same as its minimal polynomial over K.)

9) Let E be a field, G a group of automorphisms of E and K the subfield of E consisting of the elements invariant under G. a) Show that for every subfield L of E which is stable for the elements of G, Land K are linearly disjoint over L n K. (Consider a linear relation A.jXj = between the elements

L

°

§ 16

EXERCISES

A.V.l77

K which are linearly independent over L n K, with Ai E L not all zero, the number of Ai -:F 0 being the least possible.) b) If L n K is the Jield of invariants of a group of automorphisms of K, show that L is the

Xi E

field of invariants of a group of automorphisms of L(K).

10) Let E be an extension of a field K, let G be a compact subgroup of the topological group Aut K(E) and let F be the field of invariants of G. Show that E is a Galois extension of F and that G is canonically isomorphic to the topological group AutF(E). (Observe that G operates in a continuous fashion on the discrete space E to establish that an element X E E has a finite orbit under the action of G.) 11) Let K be a field of characteristic p :> O. For every commutative K-algebra A, raising to the p-th power makes the ring A into an A-algebra, hence also a K-algebra which is -1 denoted by AP . Let A be a commutative K-algebra. Show that the following conditions are equivalent: (i) A is a separable K-algebra. -1 -1 (ii) The unique K-homomorphism : A (2h KP .... AP such that (a ® A) = aP • A (a E A, A E K) is injective. (iii) For every family (bi)i e I of K which is linearly free over KP and every family (a i )i e 1 of A the equality

L arb i = 0

(It may be verified that (ii)

~

implies that ai = 0 for all i.

(iii) and that (ii) is a consequence of V, p. 123, Th. 2, e).)

12) Let K be a field and (Xi)i e I a family of indeterminates. Show that the algebra of formal power series K [[ (X;)i e I]] is a separable K-algebra. (Apply Ex. 11 ; it may also be verified that for every finite extension K' of K, K [[ (X;)i e I]] ® K K' is isomorphic to K' [[ (Xi )i e I]] and then Th. 2, d) of V, p. 123 applied.) Deduce that the field of fractions of K[[(Xi)ier]] is a separable extension of K (V, p. 120, Prop. 4).

§ 16

1) Let K be a field of characteristic p :> 0, E a separable extension of K, of finite transcendence degree over K. a) If there exists a transcendence basis Bo of E over K and an integer m"" 0 such that K(EPffl) is separable over K(Bo), show that for every transcendence basis B of E over K there exists an integer n "" 0 such that K(EP") is separable over K(B). b) Deduce that if the condition in a) holds, E admits a separating transcendence basis over K (if S is a p-basis of E over K (EP) and B a transcendence basis of E over K containing S (V, p. 176, Ex. 6), show that B is a separating transcendence basis of E over K, using Ex. 6 of V, p. 176). c) Let K be a field of characteristic p :> 0 and x an element transcendental over K (in an algebraically closed extension n of K). Show that the union E of the extensions K (x p -") of K is a separable extension of K such that tr . degKE = 1 but which does not admit a separating transcendence basis. d) Let K be a perfect field. Show that the perfect closure of the field of rational fractions K(T) is an extension of K of transcendence degree 1 which does not admit a separating transcendence basis.

A.V.178

§ ]6

COMMUTATIVE FIELDS

2) Let K be field of characteristic p >- 0 and E an extension of finite transcendence degree over K. For E to admit a separating transcendence basis over K it is necessary and sufficient that E should be separable over K and the degree of imperfection of E over K equal to its transcendence degree over K. 3) Let E be an extension of a field K and F an extension of E. a) If E admits a separating transcendence basis over K and if F admits a separating transcendence basis over E, then F admits a separating transcendence basis over K. b) If F admits a separating transcendence basis over K and if tr . degKE - 0 and E an extension of K admitting a separating transcendence basis over K. Show that the intersection L of the fields K(EP'), where n runs over N, is an algebraic extension of K. (Consider first the case where tr . degKE (y), in which the ideal (x) is « greater» than the ideal (y). We can keep this « order reversal» in mind by noting that for example 7 has « more multiples» than 91. If we extend the relation x I y to all elements of K, this relation is still equivalent to (x) => (y) in the set fJJ of all principal fractional ideals of K (in which (0) is the smallest element under the relation of inclusion).

As in the previous sections, we will generally be using additive notation in the sequel. However, terminology relating to divisibility will be introduced following the corresponding additive terminology, in paragraphs preceded by the sign (DIV) (in which it is understood that the notation used is that of the present section). In order to make the reader's task easier, certain results will be translated into the language of divisibility, the translation of Prop. 7, for example, being denoted « PROPOSITION 7 (DIV) ».

No.7 6.

ORDERED GROUPS. DIVISIBILITY

A.VI.7

Elementary operations on ordered groups

Let H be a subgroup of an ordered group G ; it is clear that the restriction to H of the ordering of G is compatible with the group structure of H ; we wiII always take H to be ordered in this way, unless otherwise stated. If P is the set of positive elements of G then the set of positive elements of H is H n P. Let (G a ) be a family of ordered groups; according to the definition of the product of ordered sets (Set Theory, III, p. 137) the product group G = G a is

n

equipped with an ordering, the relation « (x a ) ~ (Ya) » between two elements of G being by definition the same as « Xa ~ Ya for all 0. ». It is immediate that this ordering is compatible with the group structure of G ; this ordering makes G an ordered group which we call the product of the ordered groups Ga. The positive elements of G are those elements all of whose components are positive. In the case where all the factors G a are identical to the same ordered group H, then G is the group HI of maps from the index set I into H, the relation « f ~ 9 » between two maps from I into H being the same as « f (0.) ~ 9 (0.) for all 0. E I » ; the positive maps are those which take only positive values. The direct sum of a family (G a ) of ordered groups is defined as an ordered subgroup of their product (II, p.202). Let (G J, E 1 be a family of ordered groups whose index set I is well ordered; recall (Set Theory, III, p. 157) that an order relation, called the lexicographic ordering, is defined on the product set G = G" the relation « (xJ 0 (resp. n, -< 0) ; we have

L n.[J, "'" L (- nJp,.

LE I'

LeI"

In particular, for A E I", this implies that PA

~

L n.[J" and it follows by induction LeI'

from property (P) that we must have PA ~ p, for some LEI'; since p, is irreducible, this would imply PA = p" which is absurd. Hence I" is empty, which proves the proposition. THEOREM 2. Let G be a filtered group. Then the following properties are equivalent: a) G is isomorphic to an ordered group of the form z(I). b) G is lattice ordered and satisfies the following condition: (MIN) Every nonempty set of positive elements of G has a minimal element. c) G satisfies condition (MIN) and every irreducible element of G has property (P). d) G is generated by its irreducible elements, and every irreducible element ofG has property (P). Let us show first that a) implies b). The group Z(I) is lattice ordered, as the direct sum of totally ordered groups. On the other hand let E be a nonempty set of positive elements of Z(I) and let x = n,e, be an element of E (where

L

(eJ denotes the natural basis of

Z(I)) ;

there are a finite number

n (n, + 1) of

elements y of Z(I) such that 0 ~ y ~ x, so the set F of elements of E which are less than or equal to x is a fortiori finite; since it is nonempty, it contains a minimal element (Set Theory, III, p. 170, Cor. 2), which is clearly a minimal element of E. It is clear that b) implies c), by Prop. 14. Let us show that c) implies d). Since G is filtered, it is enough (VI, p.4, Prop. 4) to check that the set F of positive elements of G which are sums of irreducible elements is equal to G+ - {O}. If this were not true, it would follow from (MIN) that the complement of F in G+ - {O} would have a minimal element a ; by definition a is not irreducible, so is the sum of two strictly positive elements x and y; since x -< a and y -< a, these elements belong to F, and so are sums of irreducible elements, and it follows that so is a, which is a contradiction. Finally, d) implies a) by Prop. 15.

No.1

ORDERED FIELDS

A.VI.19

We will apply Th. 2 to the theory of divisibility in principal ideal domains (VII, p. 4) and in unique factorisation domains (AC, VII, § 3), as well as to the study of ideals in a Dedekind ring (AC, VII, § 2).

§ 2. ORDERED FIELDS

1.

Ordered rings

DEFINITION 1. - Given a commutative ring A, we say that an ordering on A is compatible with the ring structure of A if it is compatible with the additive group structure of A, and if it satisfies the following axiom: (OR) The relations x ;>- 0 and y;>- 0 imply xy ;>- O. The ring A, together with such an ordering, is called an ordered ring. Examples. - 1) The rings Q and Z, with the usual orderings, are ordered rings. 2) A product of ordered rings, equipped with the product ordering, is an ordered ring. In particular the ring A E of mappings from a set E to an ordered ring A is an ordered ring. 3) A subring of an ordered ring, with the induced ordering, is an ordered ring.

In an ordered ring, the relations x ;>- y and z;>- 0 imply xz ;>- yz. Indeed these inequalities are equivalent to x - y ;>- 0, z;>- 0 and (x - y) z ;>- 0 respectively. Analogously we can show that the relations x ~ 0 and y ;>- 0 (resp. y ~ 0) imply xy ~ 0 (resp. xy ;>- 0). These results are often invoked under the name of sign rules (two elements are said to have the same sign if they are both ;>- 0 or both ~ 0). They imply that, if A is a totally ordered ring, then every square is positive, and in particular that every idempotent (for example the unit element) is positive.

°

Example. - There is only one totally ordered ring structure on Z: indeed 1 :> 0, whence n :> for every natural number n =F 0, by induction. In contrast there exist ordered ring structures on Z which are not totally ordered (see below).

Let P be the set of positive elements of an ordered ring A. It is known (VI, p. 3, Prop. 3) that P determines the ordering on A. To say that A is an ordered ring is equivalent to saying that P satisfies the following properties: (API) (AP n )

P + PcP PP c P

(AP m )

P n (- P) = {O}

Indeed (API) and (AP m ) state that the additive group of A is an ordered group (VI, p.3, Prop. 3), while (AP II ) is a translation of (OR). Recall that the following condition is necessary and sufficient for the order relation on A to be total: (AP IV )

P U (- P) = A .

A. VI. 20

ORDERED GROUPS AND FIELDS

§2

Example. - In Z, if we take P to be the set of positive (in the usual sense) even integers, we get a ring which is not totally ordered.

Recall also that, in a totally ordered abelian group, the relation n . x = 0 (for a natural number n =fo 0) implies x = 0 (VI, p. 4) ; this gives us the following result. PROPOSITION 1. -

2.

A totally ordered ring is torsion free as a Z-module (II, p. 313).

Ordered fields

DEFINITION 2. - A commutative field, equipped with a total ordering, is called an ordered field if its ordering and its ring structure are compatible. We restrict ourselves to total order relations on fields because· the others are very «pathological» (VI, p. 38, Ex. 6). Examples. - 1) The field Q of rational numbers is an ordered field. 2) A subfield of an ordered field, with the induced ordering, is an ordered field. 3) * The field of real numbers is an ordered field. *

Let K be an ordered field. For all x sgn(x) = 1 sgn(x) = - 1 sgn(x) = 0

E

K we put

if if if

x:> 0,

x { - 1, + I} is a surjective homomorphism whose kernel is closed under addition, then s is the sign map for a unique ordered field structure, where the set of strictly positive elements is the kernel of s. For all x and y in K we have x = sgn (x) Ix I and Ixy I = Ix I Iy I . On the other hand every ordered field is of characteristic zero (Prop. 1). PROPOSITION 2. - Let A be a totally ordered integral domain, and let K be its field of fractions. Then there exists one and only one ordering on K which restricts to the given ordering on A and makes K an ordered field. Every x E K can be expressed in the form x = ab - 1, with a and b in A and b =fo O. If x is positive, then a and b have the same sign, and conversely. Thus we see that, if there exists an ordering on K satisfying the prescribed conditions, then it is unique, and the set P of positive elements is identical to the set of ab- 1, where a and b are elements of A of the same sign, and b =fo O. It remains to show that P satisfies conditions (APr>, (AP n ), (APnr> and (AP rv ). This is obvious for (AP n ) and (AP rv ). For (APr), consider ab- 1 + cd-I, where we may

No.3

ORDERED FIELDS

A.V1.21

assume that a, b, c and d are positive; this sum is (ad + bc )(bdt I, and ad + bc and bd are positive. To show (AP m ), consider an identity of the form ab- I = - cd-I, so that ad + bc = O. If we assume that a and b have the same sign and that c and d have the same sign, then the sign rules show that ad and bc have the same sign; whence ad = bc = 0, so a = c = 0 ; hence P does indeed satisfy (AP m ). Example. - Since Z admits only one totally ordered ring structure (VI, p. 19, example), the field Q admits only one ordering which makes it an ordered field: this is the usual ordering.

3.

Extensions of ordered fields

3. - Let K be an ordered field. An ordered extension of K is a pair (E, u), where E is an ordered field and u is an increasing homomorphism from K to E. Let K be a field, let E be an ordered field and let u: K -+ E be a homomorphism. The relation DEFINITION

x,,;;: y

if u (x) ,,;;: u (y)

is a total order relation on K which gives it an ordered field structure, said to be induced by that of E. If K and E are ordered fields, then a homomorphism u : K -+ E is increasing if and only if the ordered field structure of K is induced by that of E. We will usually identify K with its image in E under u. Examples. - 1) Every ordered field K is an ordered extension of Q. Indeed K is an extension of Q, since it is of characteristic zero, and on the other hand Q can only be ordered in one way, as we have just seen. 2) Let K be an ordered field, and let K (X) be the field of rational functions in one indeterminate over K. Let us define an ordering on the polynomial ring K [X 1by taking the positive elements to be 0 and those polynomials whose leading coefficient is positive. In this way we obtain a totally ordered ring whose ordering extends that of K. By applying Prop. 2 we give K(X) the structure of an ordered extension of K. * For K = R it can be shown that the order relation defined on K(X) in this way is that of growth near + co (ct. VI, p.24, Prop. 4). * THEOREM 1. For an extension E of K to admit the structure of an ordered extension of K, the following condition is necessary and sufficient : (OE) The relation PIxf + ... + p,.x;' = 0 implies

for any finite sequence (Xi,Pi) of pairs of elements Xi ofE and positive elements Pi of K.

A. VI. 22

§2

ORDERED GROUPS AND FIELDS

Condition (OE) is clearly equivalent to : (OE') The element - 1 is not a sum of elements of the form px 2 (x E E, p E K, P ==- 0). Condition (OE) is necessary: if E is an ordered extension of K then the elements Pjx? are positive in E, so zero if their sum is zero. On the other hand Pjx? = 0 is equivalent to pjXj = O. Conversely, suppose condition (OE) is satisfied, then we will define an ordering on E by constructing a subset P of E which satisfies conditions (AP,), (AP,,), (AP",) and (AP,v), and which contains the set K+ of positive elements of K. Such a subset P will certainly make E an ordered extension of K, for we will have K n P = K+ ; indeed, if P were to contain an element - a --+ xy (not necessarily associative), such that, for all a E E, the maps x >--+ ax and x >--+ xa are order-automorphisms of E. Let Xa (resp. "x) denote the element of E defined by (x a ) a = x (resp. a ("x) = x) and suppose that for each a E E the maps x >--+ ax and x >--+ xa are isomorphisms from the ordering of E to the opposite ordering. Show that, for arbitrary x, y, z, (Zinf(x.y»

•

X = (Zy). sup (x, y) .

3) Let G be an ordered group such that the set P of positive elements is not 0 ; show that G is an infinite group, and cannot have a greatest (or a least) element. 4) Let G be an ordered group, let P be the set of positive elements of G, and let fbe the natural map from G onto a quotient group G/H. For f(P) to make G/H an ordered group it is necessary and sufficient that 0 "" y "" x and x E H imply y E H ; then H is called a convex subgroup of G, and G/H is considered as an ordered group in this way. If G is lattice ordered, then a necessary and sufficient condition for G /H to be lattice ordered and for f(sup(x,y»=sup(f(x),f(y» is that the relations Iyl "" Ixl and XEH imply y E H ; show that this says that H is a filtered convex subgroup. Show that a lattice ordered group with no filtered convex subgroups other than itself and {O} is totally ordered (consider filtered convex subgroups generated by two positive elements of G) ; * deduce (Gen. Top., V, p. 25, Ex. 1) that G is then isomorphic to an additive subgroup of the real numbers. * 5) Give an example of an ordered group whose ordering is non-discrete, having nonzero elements of finite order (take the quotient of a suitable ordered group by a subgroup H such that P n H = {O}).

§1

A.V1.31

EXERCISES

6) Let G be an ordered group, let P be its set of positive elements. Show that P - P is the largest filtered subgroup of G, and that it is convex. What is the order relation on the quotient? 7) In the group Z, if P is taken to be the set conslstmg of 0 and the integers "" 2, then the resulting ordered group is filtered but not lattice ordered (show that the set of clements x such that x"" 0 and x"" 1 has two distinct minimal elements). 8) Let x be an element of an ordered group such that y = inf (x, 0) is defined; if n :> 0 is an integer, then nx"" 0 implies x"" 0 (we have ny = inf (nx, (n - 1 ) x, ... ,0) "" inf «n - 1 ) x, ... ,0) = (n - 1 ) y) ;

and hence nx = 0 implies x =

o.

9) In a lattice ordered group G, show that the sum of a family subgroups is a filtered convex subgroup (use VI, p. 12, Cor.). 10) Show that, for every finite sequence group G,

sup(xJ = ~>i -

L inf(xi,xj) + ... +

(Xi)

(H~)

of filtered convex

(1 ~ i ~ n) of elements of a lattice ordered

L

(_1),,+1 Ii O, nXEP implies XEP (conditions (C».

§ 1

EXERCISES

A.V1.33

a) If a is an element of G such that a ~ P, show that there exists a subset P' of G, satisfying conditions (C), such that PcP' and - a E P' (take P' to be the set of x E G such that there exist integers m :> 0 and n ~ 0 and an element YEP with mx = - na + y. b) Deduce from a) that P is the intersection of those subsets T of G such that T + T = T, Tn (- T) = {O}, T U (- T) = G (that is, making G a totally ordered group) and PeT (use Zorn's Lemma). c) In particular if G is an additive group in which every nonzero element has infinite order, then the intersection of all the subsets T of G such that T + T = T, T n (- T) = {O} and TU(-T)=Gis {O} . .,. 21) An ordered group is called lattice orderable if it is isomorphic to a subgroup of a lattice ordered group. Show that it is necessary and sufficient for an ordered group to be lattice orderable that, for every integer n :> 0, the relation nx ~ 0 imply x ~ 0 (to show that the condition is sufficient, use Ex. 20, b) and 19). Show that every lattice orderable group is isomorphic to a subgroup of a product of totally ordered groups. 22) Let G be a lattice orderable group (considered as a Z-module) and E the vector space G (Q) (II, p. 277) ; show that there is a unique ordering on E compatible with the additive group structure of E, inducing the given ordering on G, and such that E is lattice orderable. 23) Let G be the lattice ordered group Z x Z (where Z has the usual ordering) and H the convex subgroup of G generated by (2, - 3) ; show that the ordered group G /H is not lattice orderable (cf. VI, p.30, Ex. 4). 24) Let A be a commutative ring and I the set of all ideals of A. Then a (b + c) = ab + ac for all a, b, c in I, by I, p. 107; in other words I, with the order relation a=> b and the law of composition (a, b) 0-+ ab, is a semi-lattice ordered monoid (VI, p. 31, Ex. 13), the sup of two elements a, b of I being Q n b, and their inf a + b. a) Let K be a field and A = K [X, Y] the ring of polynomials in two indeterminates over K. Consider the principal ideals a = (X), b = (Y), c = (X + Y) in A. Show that (b+c), (a n c) + (b n c) , (a(a + b» n (b(a + b».

(anb)+c~(a+c)n

(a + b) n b (a n b)(a + b)

~

~

b) In the ring A, 1 is a gcd of X and Y, but (X) + (Y) ~ A.

.,. 25) Let I be the semilattice ordered monoid of ideals of a commutative ring A (Ex. 24). For a finite system of congruences x == a j (a j ) to admit a solution whenever any two of them admit a common solution (that is to say whenever a j == aj(a j + aj)

for every pair of indices i, J); it is necessary and sufficient that each of the laws (a, b)o-+a n b, (a, b)o-+a + b, in I be distributive over the other ( 0 implies X-I:> 0 in the ordering defined by P. Show that y 2 3 z2 implies y 3 z, for positive y and z. Deduce that, for arbitrary y:> 0, we have

for every integer n:> 0 (notice that (y + y -

I? 3 m

(2;;:)

for every integer m:> 0).

Deduce that, if the ordered additive group structure defined by P is archimedean (VI, p. 35, Ex. 31), then (y - z )2 3 0 for every pair of elements y, z of P ; if K' is the subfield of K generated by P then X 2 3 0 for all x E K'. * d) * Let K = R (X) be the field of rational functions in one indeterminate over R ; let P be the set consisting of 0 and the rational functions u E K such that u (t) is defined and :> 0 for every real number t. Show that P satisfies the conditions of c) and generates K, but that there exist elements v E K such that v2 ¢ P. * .,. 5) Let A be a lattice ordered ring. A convex ideal (VI, p. 30, Ex. 4) of A is called irreducible if it is not the intersection of two convex ideals distinct from itself. a) Show that the intersection of the irreducible convex ideals of A is 0 (if a E A is nonzero, consider a convex ideal of A maximal among those which do not contain a). Deduce that A is isomorphic to a subring A' of a product

n A"

such that each A, is lattice ordered,

pr, (A') = A, for all L and the ideal {O} is irreducible in A, (as a convex ideal) for all L. b) Show that the following conditions are equivalent in A : ex) IxYI = Ixl . Iyi for all x, y; ~) x. sup(y, z) = sup (xy, xz) for all x 3 0, y, z; -y) x. inf(y, z) = inf(xy, xz) for all x 3 0, y, z; 8) inf(y, z) = 0 implies inf(xy, z) = 0 for all x 3 O. (To see that -y) implies 8), observe that xy "" y • sup (x, 1)). When these conditions are satisfied, then A is said to be strongly lattice ordered. c) Show that a ring A is strongly lattice ordered if and only if it is isomorphic to a subring of a product A, of totally ordered rings. (Using a), reduce to showing that if

n

{D} is irreducible in a strongly lattice ordered ring A, then A is totally ordered; for this, note that if B is a strongly lattice ordered ring then for all b E B the set m of x E B such that inf ( Ix I ' Ib I ) = 0 and the set n of y E B such that Iy I "" Izb I for some z E B are two convex ideals of B such that m n n = {O}.) d) Show that in a strongly lattice ordered ring A the relation inf (x, y) = 0 implies xy = 0; for all z E A we have z2 3 0; for all x, YEA we have xy "" sup (x 2, y2). e) Let A be a lattice ordered ring with no nilpotent element :> 0; show that if inf(x, y) = 0 implies xy = 0 then A is strongly lattice ordered (check condition 8) of

b» .

.,. 6) a) Let K be a lattice ordered field. Show that the following conditions are equivalent: ex) X 2 3 0 for all x E K ; ~) x:> 0 implies x- 1 :> 0 ;

§2

EXERCISES

A.V1.39

'Y) K is strongly lattice ordered (Ex. 5) ; 8) K is totally ordered. (To see that 13) implies 8), notice that 13 implies xy "" x 2 + y2 for x, y E K). b) Let K be the field Q ( .Ji) obtained by adjoining .Ji to the field of rational numbers; as an additive group K is identified with Q x Q via the bijection x + y .Ji >-+ (x, y) ; show that if the product ordering on Q x Q (where Q has the usual ordering) is carried over to K via this bijection, then K is lattice ordered but not strongly lattice ordered. c) The field K = Q(X) of rational functions in one indeterminate X over Q is generated by its set of squares K2. Show that the set P of sums of squares of elements of K defines a nonlattice ordering on K compatible with its ring structure, and such that u:> 0 implies u- 1 :> O. 7) Every element of a commutative field K of characteristic =F 2 is a sum of squares if and only if - 1 is a sum of squares (notice that every element of K is a difference of two squares) . .,. 8) Let A be a nonempty subset of a commutative field K of characteristic =F 2 ; then K is said to be A-orderable if no element of A is a sum of squares in K. We say that K is orderable if there exists a total ordering of K which is compatible with its ring structure. a) Show that if K is A-orderable then it is orderable (Ex. 7), and hence of characteristic O. b) Show that pure extensions and algebraic extensions of odd degree of an A-orderable field K are A-orderable (argue as in Prop. 3 of VI, p.23). c) Let K be an A-orderable field and let b be an element of K not of the form ca - d, where a E A and c and d are sums of squares in K. Show that the field K (,/b") is A-orderable. d) An A-orderable field K is called maximal if no proper algebraic extension of K is Aorderable. Show that a maximal A-orderable field K has the following properties: 1. K is pythagorean, in other words every sum of squares is a square; 2. no element of A is a square; 3. every element of K which is not a square has the form c 2a - d 2, where a E A ; 4. every odd degree polynomial in K[X] has at least one root in K (use b) and c». e) Show that if K satisfies the conditions of d) then every element algebraic over K has degree 2q over K for some q (argue as in Prop. 8 of VI, p. 26, by induction on the exponent of 2 in the degree of the element under consideration). Show that on the other hand no quadratic extension of K is A-orderable. Deduce that K is a maximal A-orderable field (use Galois theory and Prop. 12 of I, p.77). f) Let K be an A-orderable field and n an algebraically closed extension of K. Show that there exists a maximal A-orderable field contained in n and containing K. 9) Let q :> 0 be a non-square natural number; let A denote the set {- 1, Jq} in the field K = Q ( ,iq) ; show that K is A-orderable. Show that there exists an extension E of K such that E is pythagorean, orderable, and that every odd degree polynomial over E has a root in E, but that E does not admit the structure of a maximal ordered field (consider a maximal A-orderable extension of K) . .,. 10) a) Let K be an orderable field and E a Galois extension of K. Show that either E is orderable or there exists an orderable algebraic extension F of K, contained in E, such that E is a quadratic extension of F (use Th. 6 of V, p. 70).

A.VI.40

ORDERED GROUPS AND FIELDS

§2

b) Show that the polynomial X4 + 2 is irreducible over Q, and that the extension K of Q of degree 4, obtained by adjoining a root of this polynomial to Q, contains no orderable subfield other than Q (find all the subfields of K by Galois theory).

11) Let K be an ordered field and G a subfield of K. a) Show that x ~ 0 is infinitely large with respect to G if and only if X-I is infinitely small with respect to G. We say that K is comparable to G if there is no element of K which is infinitely large with respect to G (nor, consequently, any nonzero element of K which is infinitely small with respect to G). A necessary and sufficient condition for K to be comparable to its prime subfield Q is that K be archimedean (VI, p. 35, Ex. 31), * and hence that K be isomorphic to a subfield of R (VI, p. 36, Ex. 33c)). * b) Show, that, in the ring F(G) of elements of K which are not infinitely large with respect to G, the set I (G) of elements infinitely small with respect to G is a maximal ideal; moreover the ordering on the quotient field K(G) = F(G)/I(G) induced from that of F(G) is compatible with the ring structure and is total. c) Show that a class modulo I (G) can contain at most one element of G ; deduce that the natural map from G to K(G) is an isomorphism from the ordered field G onto a sub field G' of K(G), and that K(G) is comparable to G'. 12) Let K be a maximal ordered field, let [be a polynomial over K and let a and b be two roots of F such that a -< b and such that F has no roots between a and b. Show that if g is a rational function over K whose denominator does not vanish for a".; x ".; b then the equation [(x) g (x) + f' (x) = 0 has an odd number of solutions in (a, b) (where each solution is counted with its multiplicity; use Prop. 5 of VI, p. 25). Deduce that if h is a rational function over K having a and b as roots and whose denominator does not vanish in la, b(, then the equation h'(x) = 0 has at least one root in la, b(. 13) Let K be a maximal ordered field, let h be a rational function over K and let (a, b) be a closed interval in which h is defined. Show that there exists c E la, b( such that h (b) - h (a) = (b - a) h' (c) (use Ex. 12). Deduce that h is an increasing function in [a, b) if and only if h' (x) "'" 0 in this interval (to see that this condition is necessary, split the interval by the roots of h' (x) = 0). ,. 14) a) Let K be a maximal ordered field, let E be a subfield of K and let [be a polynomial over E; show that all the roots of [in K belong to F(E) (Ex. 11, b)) (use Prop. 4 of VI, p.24). Deduce that if G is a subfield of K and E c: K is an extension comparable to G (Ex. 11) then the set of elements of K algebraic over Eisa maximal ordered field comparable to G. b) Deduce from a) that, under the hypotheses of a), the field K(G) (Ex. 11, b)) is a maximal ordered field. c) Let [be a polynomial of degree"", 1 over G. Show that [(t) is infinitely large with respect to G if and only if t is infinitely large with respect to G; the element [(t) is infinitely small with respect to G if and only if t is congruent modulo I(G) to a root of [in K (split K into intervals using the roots of [and f' in K, and apply Ex. 13 ; notice that if x -< t and if x E K is not congruent modulo I (G) to t, then there exists Y E K such that x -< Y -< t and y - x E G). ,. 15) Let E be ·an ordered field, let K be a subfield of E such that E is algebraic over K, and let R be a maximal ordered extension of K.

§2

EXERCISES

A.V1.41

a) Among the elements of E - K, let Xo be one whose minimal polynomial fover K has the least possible degree. Show that f'(T) is a product of factors in R[T] of the form (T - a j )2 + el (a j , ej E R), and first order factors T - b j with the b j pairwise distinct elements of K. Deduce that there exists an element Yo in R such that f (Yo) = 0 and such that Xo - z and Yo - z have the same sign, in E and R respectively, for all z E K (cf. VI, p. 43, Ex. 26, e)). b) Deduce from a) that there exists a K-isomorphism from the ordered field E onto a subfield of R. (First show that there exists such an isomorphism for the subfield K(x o ) of E; for any polynomial 9 E K[T] such that deg(g) O. a) Show that if L = K (X) is the subfield of E consisting of rational functions in one indeterminate, then E = L (Ex. 36) up to isomorphism of ordered fields (observe that for every element a> 0 of L there exists an integer n> 0 such that 0 0 ; for each prime number p :> 0, an (abelian) group is said to be p-torsion if all its elements have orders which are powers of p. In this terminology, Th. 1 shows that every torsion abelian group is a direct sum of ptorsion groups. In the case of finite groups, this also follows from I, p. 80, Th. 4.

3.

Applications: I. Canonical decompositions of rational numbers and of rational functions in one indeterminate

THEOREM 2. - Let A be a principal ideal domain, let K be its field offractions and let P be a system of representatives of irreducible elements of A. Given an element x E K, there exist a finite subset H ofP, elements ao E A and ap E A not divisible by p in A (p E H), and integers s (P) :> 0 (p E H) such that

(1)

x = ao +

L app-s(P) , pE fI

where H and the s (P) are uniquely determined by these conditions. Moreover, if Rp denotes a subset of A containing precisely one element of each residue class mod p in A (p E P), then each x E K can be uniquely expressed in the form (2)

where a E A and rph E Rp for all hand p, and all but finitely many of the rph are zero. Consider K as an A-module; then A is the sub module generated by 1. The quotient module K/ A is the quotient of K by the equivalence relation x' - x E A, which is also written, in the notation of VI, p. 6, as x == x' (mod 1) ; let f denote the natural homomorphism from K onto M = K/ A. The quotient module M is a torsion module, for every element of M has the form f(a/b) (a E A, bE A, b =fo 0), whence bf(a/b) = f(a) = O. Hence Th. 1 of VII, p.8 applies. Let Mp (p E P) be the submodule of elements of M annihilated by powers of p ; then f-l(Mp) is the subring Ap of elements of K of the form ap - n where a E A and n ~ 0 is an integer. The module M is the direct sum of the Mp, so every x E K is congruent mod 1 to an element in the sum of the Ap; in other words, formula (1) holds, with the s(P) integers:> 0, and the ap finitely many elements of A such that ap is not a multiple of p. We now show that these conditions on the s (P) and the ap completely determine H and the s (p). Indeed H is then the set of PEP such that the component of f(x) in Mp is =fo O. On the other hand if sand s' are two integers:> 0 such that

No.3

TORSION MODULES OVER A PRINCIPAL IDEAL DOMAIN

A.VII.ll

s ~ s' and if a and a' are elements of A not divisible by p such ap-s == a'p-s' (mod 1), then we deduce that a == a'ps -s' (mod PJ ; if s:> s' a == 0 (mod p), contradicting the hypotheses. This argument also shows that ap is well defined mod pS(P). To complete the proof, we note first of all that in each residue class ,- I

L r"ph

pS in A there exists a unique element of the form

with rh

E

that then each mod

Rp for

h~O

0"", h "'" s - 1. In fact, we proceed by induction on s (the property follows from

the definition of Rp for s = 1): let x of the form s- 2

X -

L

s-2

L r"ph

(rh

E

E

A ; by hypothesis there is a unique element

Rp) in the residue class of x mod pS -

r"ph is a multiple aps -·1 of pS -

1;

1;

then

now there exists a unique element

h~O

rs _ 1 of Rp such that a == rs _ 1 (mod p) ; whence x

==

,- I

L r"ph (mod PJ. To obtain

h~O

formula (2) it is now enough to apply this fact to each ap in formula (1). The uniqueness is clear in view of the above. The following are the most important applications of Th. 2 : I. The ring A is the ring Z of rational integers, and K = Q. Let P be the set of prime numbers:> 0, and for each PEP let Rp be the interval (0, p - 1) in Z. Thus

we have the canonical decomposition

with a

E

Z, eph

E

Z and 0"", eph "'" P - 1.

II. The ring A is the ring E[X] of polynomials in one indeterminate over a commutative field E, and K = E (X). Let P be the set of monic irreducible polynomials in E[X] (VII, p. 5). For PEP we can, by virtue of euclidean division of polynomials (IV, p. 10), take Rp to be the set of polynomials of degree strictly less than that of p. Thus we have the decomposition (called canonical) of a rational function r(X) E E (X) :

where a (X) is a polynomial and vph (X) is a polynomial of degree strictly less than that of p(X), for all p and h. In particular, if the field E is algebraically closed, then the p (X) have the form X - a. with a. E E, and the vph (X) are thus constants. Hence we can say that the vector space E (X) over the field E has a basis consisting of the monomials X" (n "" 0 an integer) and the rational functions of the form xm / (p (X))h, where PEP and hand m are integers with h "" 1 and 0 :;;:: m 1, and let (Z/ aZ)* be the multiplicative group of invertible elements of the ring Z/ aZ. If a = p;(i) is the decomposition of a into

f1

prime factors, then the ring Z/aZ is isomorphic to the product of the rings Z/p;(i'z (VII, p. 3, Prop. 4) and the group (Z/aZ)* is isomorphic to the product of the groups (Z/p;(i'z)*. We are thus reduced to the study of the groups (Z/pnz)*, where p is a prime number; recall (V, p.80) that the order ~(pn) of (Z/pnz)* is pn_ p n-l =pn-l(p -1). Suppose first of all that p :> 2; the natural homomorphism Z/pnz -> Z/pZ restricts to a homomorphism of groups from (Z/pnz)* onto (Z/pZ)*, whose kernel we denote U (pn); by VII, p. 3, Prop. 3 the residue class mod pn of an integer m is invertible if and only if m is coprime to p, that is if and only if the residue class of m mod p is invertible. It follows that U (pn) consists of all the residue classes mod pn of integers congruent to 1 mod p, so has pn - 1 elements, and that there is an exact sequence (3)

{1}

->

U(pn)-> (Z/pnZ)*-> (Z/pZ)*-> {1} .

Similarly, for n "'" 2 let U (2n) denote the kernel of the natural homomorphism from (Z/2nZ)* to (Z/4Z)*; this is a group of order 2n - 2, consisting of all the residue classes mod 2 n of integers congruent to 1 mod 4, and there is an exact sequence (4)

{l}

->

U(2n)

->

(Z/2nZ)*

->

(Z/4Z)*

->

{I} .

Lemma 3. - Let x, y, k be integers with k "'" 0, and let p :> 2 be a prime number. If x"", 1 + py mod p2 then Xpk"", 1 + pk + ly mod pk + 2. If x"'" 1 + 4y mod 8 then X2k"", 1 + 2k+2y mod 2k+3. To prove the first assertion, it is enough to show that, if k "'" 1 and x"", 1 +pky mod pk+l, then x P "", 1 +pk+ly mod pk+2, and then to argue by induction on the integer k. For all a E Z and k "'" 1, it is immediate that (1 + pkay "'" 1 + pk+ la

modpk+2,

hence (1 + P ky + P k+ lz Y = (1 + p key + pz) y "'" "'" 1 + P k + 1 (y + pz ) "'" 1 + P k + ly mod p k + 2

Similarly, for k "'" 1 we have

(1 + 2k + la)2 "'" 1 + 2k + 2a

mod 2k + 3

,

so

(1+2k+ly+2k+2Z)2"",1+2k+2y mod2k+3, whence the second assertion by induction on k.

.

No.4

TORSION MODULES OVER A PRINCIPAL IDEAL DOMAIN

A. VII. 13

PROPOSITION 3. - Let p :> 2 be a prime number and let n :> 0 be an integer; then the group U (pn) is cyclic of order pn - 1; if n"", 2 then the residue class mod pn of an integer x congruent to 1 mod p is a generator ofU (pn) if and only if x is not congruent to 1 mod p2. Let m:> 1 be an integer; then the group U (2m) is cyclic of order 2m - 2 ; if m "'" 3 then the residue class mod 2m of an integer x congruent to 1 mod 4 is a generator ofU (2m) if and only if x is not congruent to 1 mod 8. Since U (pn) has order pn - 1, the order of every element u of U (pn) is a power of n-2 p, and u is a generator of U (pn) if and only if uP oj:. 1. Now if u is the class of x = 1 + py, then u pn - 2 is the class of 1 + p n - ly, by Lemma 3, whence u generates U (pn) if and only if y ¢ 0 mod p, in other words x ¢ 1 mod p2. For example, the class 1 + p generates U (pn). Similarly, the class u of x mod 2n generates U(2n) if and only if U 2n - 3 oj:. 1, which means that x is not congruent to 1 mod 8, by Lemma 3 ; this is satisfied by x = 5.

Lemma 4. - Let A be a principal ideal domain and let 0 -+ N -+ M -+ P -+ 0 be an exact sequence of A-modules. Suppose that there exist coprime elements a, bE A such that aN = 0 and bP = o. Then the exact sequence splits. Ifin addition N and P are both cyclic, then M is cyclic. The module M is torsion, since abM = O. The first assertion follows from Cor. 3 of VII, p. 9. If Nand P are cyclic, then they are finitely generated, and hence so is M (II, p. 17, Cor. 5) ; since each p-primary component of M is isomorphic to a pprimary component either of N or of P, it follows from Remark 2 of VII, p. 10, that M is cyclic. THEOREM

3. -

If a =

n p['(i)

is the prime decomposition of the integer

a:> 1, then the group (Z/aZ)* of invertible elements of the ring Z/aZ is isomorphic to the product of the groups (Z/p[,(i';z)*. If p:> 2 is a prime number and n "'" 1 an integer, then the group (Z / P nZ ) * is cyclic of order p n - 1 (p - 1). The group (Z/2Z)* is trivial; for n "'" 2 the group (Z/2nz)* is the direct product of the cyclic group of order 2n - 2 generated by the residue class of5 mod 2n and the cyclic group of order 2 consisting of the residue classes of 1 and - 1 mod 2n. The orders pn - 1 of U (pn) and p - 1 of (Z/pZ)* are coprime; since U(pn) and (Z/pZ)* are cyclic (Prop. 3 and V, p.78, Lemma 1), the group (Z/pnz)* is cyclic (apply Lemma 4 to the exact sequence (3)). If n "'" 2 then the restriction of the homomorphism v: (Z/2nz)* -+ (Z/4Z)* to the subgroup {I, -I} is bijective; the group (Z/2nz)* is thus the direct product of this subgroup and the kernel U (2n) of v ; the result follows from Prop. 3. Remark. - Let p :> 2 be a prime number and let x be an integer congruent to 1 mod p and not congruent to 1 mod p2; there is an exact sequence (5)

{O}

-+

Z/pn -IZ ~ (Z/pnz)* ~ (Z/pZ)*

-+

{I}

A.VII.14

§3

MODULES OVER PRINCIPAL IDEAL DOMAINS

of groups, where v is the natural projection and where u is induced on the quotients by the map r >---+ xr. Let Zp be the ring of p-adic integers (V, p. 96), and let x be an element of Zp such that x - 1 E pZp and x-I rt p Lzp ; on passing to inverse limits, the exact sequences (5) induce an exact sequence

where v is the natural map, and the continuous map u extends the map n >---+ xn (n E Z). We often put xn = u (x) for n E Zp. Similarly, if x E Zz with x - 1 E 4Z z and x-I rt 8Z z, then there is a split exact sequence {O}

->

Zz ~ Zz* ~ (Z/4Z)*

where u is a continuous extension of the map n

->

>---+

{I} ,

xn.

§ 3. FREE MODULES OVER A PRINCIPAL IDEAL DOMAIN THEOREM 1. -

Let A be a ring such that every left ideal of A is projective (II, p. 231, Def. 1) as an A-module. Then every submodule M of a free left A-module L is a direct sum of modules isomorphic to ideals of A. Let (eJL EI be a basis for L, and let PL be the coordinate functions corresponding to this basis. Choose a well-ordering (Set Theory, III, p. 153) of I and let LL denote the submodule generated by the e). for A ~ L ; put ML = M n Lv The coordinate function PL maps ML onto an ideal OL of A ; since OL is a projective Amodule, there exists (II, p. 231, Prop. 4) a submodule N Lof ML such that the map x >---+ PL (x) from NL into OL is bijective. Let M: be the submodule of L generated by the N). for A ~ L ; we will show that M: = ML for all L, which will imply that M is generated by the family (NJLE I' In fact, suppose M{ = M). for all A -< L ; then PL (x) E OL for all x E M L; thus there exists YEN L such that x - y is a linear combination of finitely many elements e). with A -< L ; in other words x - y is an element of M). for some A -< L; the inductive hypothesis shows that x - y E M{ c:: M:, that is that x E M:, and so M: = M L. It remains to show that the sum of the NL is direct; now suppose there exists a linear relation a L= 0, with

L

a LE N L, where the a L(all but finitely many of which are zero) are not all zero. Let f.L be the greatest index L such that a L~ 0 ; since pfJ-(a).) = 0 for A -< f.L, we have pfJ-(afJ-) = PfJ-

(~aL)

= 0, so afJ- = 0, contradicting the choice of

f.L.

COROLLARY 1. - If every left ideal of A is projective, then every submodule of a projective left A-module is projective. Indeed every projective A-module is a sub module of a free A-module (II, p. 231, Prop. 4), and Th. 1 applies.

No.1

A.VII.I5

FINITELY GENERATED MODULES

COROLLARY 2. - Every submodule of a free module over a principal ideal domain is free. This follows immediately from Th. 1, since every ideal of a principal ideal domain is free. COROLLARY 3. -

Every projective module over a principal ideal domain is free.

Remark. - The proof of Th. 1 shows that every submodule of A (I) is isomorphic to a direct sum EB a where each a is an ideal of A. L,

L

LEI

PROPOSITION 1. - If L is a free module of finite rank n over a principal ideal domain A, then every submodule M of L is a free module of rank ~ n. Indeed M is a free module by Cor. 2 to Th. 1, and it has rank ~ n by the previous remark, or by the following lemma: Lemma 1. - Let L be a module over a commutative ring A, generated by n elements, and let M be a free submodule of L ; then M has rank ~ n. Suppose first that L is free. Let i denote the canonical injection of Minto L. By n+l

III, p. 520, Cor., the homomorphism n+!

n+l

Ai: A

n+l

n+l

M_

A

L is injective; by III,

p. 511, Prop. 6, A L = {O}, so A M = {O} ; it follows that M has rank ~ n (III, p. 518, Cor. 1). Now consider the general case; the module L is a quotient of a free module L' of rank n. There exists a sub module M' of L isomorphic to M (II, p.218, Prop. 21). By the first part of the argument M' has rank ~ n, and the result follows. COROLLARY. - Let E be a module over a principal ideal domain A, generated by n elements. Then any submodule F of E can be generated by at most n elements. Indeed there exists a homomorphism ffrom An onto E (II, p. 216, Cor. 3), and f- !(F), which is a free module of rank m ~ n, is generated by n elements; the images of these elements under f generate F.

§ 4. FINITELY GENERATED MODULES OVER

A PRINCIPAL IDEAL DOMAIN 1.

Direct sums of cyclic modules

Let A be a commutative ring. Recall (II, p. 220, Prop. 22) that a cyclic Amodule is isomorphic to a quotient module A/a, where a is an ideal of A. We will see later (Sect. 4) that every finitely generated module over a principal ideal domain is a direct sum of finitely many cyclic modules. PROPOSITION 1. - Let E be a module over a commutative ring A ; suppose E is a direct sum of n cyclic modules A/ak (1 ~ k ~ n), where the ak are ideals of A ;

A.VII.16

§4

MODULES OVER PRINCIPAL IDEAL DOMAINS p

then, for each integer p :> 0, the A-module A E is isomorphic to the direct sum of the modules A/a H , where for each p-element subset H = {k1' ... , kp } of p

[1, n l, the ideal a H is L ak( j

~

1

Let Xk be the canonical generator of A/ak' that is the image of the unit element of A, so that E is the direct sum of the Ax i (1"" i "" n). Then we know (III, p. 515, Prop. 10) that the exterior algebra A E is isomorphic as an A-module to the n

tensor product ® (A(Ax i )). Now A(Ax i ) is just the direct sum A EEl Ax i , since i

~

I

p

the exterior product of any two elements ofAxi is zero, and so A E is the direct sum of the modules MH

=

(Ax k ) ® ... ® (Ax k ) as H

=

{k l ,

••• ,

k p } runs over

the set of p-element subsets of [1, n 1(with kl -< ... -< k p ) ; now MH is known to be isomorphic to A/a H (II, p. 257, Cor. 4), which completes the proof. We now see that, in the notation of Prop. 1, if the ideals ak form an increasing sequence, they are completely determined by the module E. More precisely: PROPOSITION 2. - Let A be a commutative ring, and let E be a direct sum of n cyclic modules A/ak' where the ak satisfy al C a2 c ... c an. Then, for 1 "" P "" n, p

p

the ideal ap is the annihilator of A E ; if an ¥- A then A E¥-O for 1 "" p "" nand m

A E = 0 for m

:> n. Indeed, in the notation of Prop. 1, we have a H = as(H)' where s (H) denotes the greatest element of the subset H. Since s (H) ~ P for every p-element subset H, and since s (H) = p for H = {I, ... , p}, it follows that ap is the intersection of a H as H varies over the set of p-element subsets of (1, n); the ideal

p

ap is thus indeed the annihilator of A E, by Prop. 1. In the notation of Prop. 2, if an ¥- A, and ifE is also isomorphic to the direct sum of m cyclic modules A/a; with a; c a 2c ... c a;" ¥- A, then m = nand ak = ak for 1"" k "" n ( 1 and n:> 1, then a is even and

18) Let M be the multiplicative monoid generated by a finite number of integers qi :> 1 (1 "" i "" m). Show that, for each integer k:> 0, there exist k consecutive integers which do not belong to M (if hi' ... , h m are m integers:> 1, and k is the greatest number of consecutive integers none of which is divisible by any of the hi' then (k + 1 t 1 "" hll + ... + h;" I ; now notice that, for all r:> 0, every sufficiently large number in M is divisible by at least one qt). Obtain a new proof of Prop. 5 of VII, p. 5 from this result. 19) For every integer n :> 0, show that the exponent of a prime p in the prime factorisation

L OX)

of n! is k

~

[np-kj (which has only finitely many nonzero terms).

I

.,. 20) * Let '!T (x) be the number of primes less than the real number x :> O. For each integer n :> 0, show that

(notice that

n


0). Hence obtain a direct proof of the fact that the binomial coefficients n!1 (m! (n - m)!) are integers. 23) Let

fl

:> 1 be a rational integer; show that the gcd of the

fl -

1 binomial coefficients

( nk ) (1 "'" k "'" n - 1 ) is equal to 1 if n is divisible by two distinct prime numbers, and is

equal to p if n is a power of the prime number p.

24) a) Let n

=

np;k:> 0 be a rational integer expressed in terms of its prime factors. k

Show that the sum of the positive divisors of n is given by the formula

b) We say that n is perfect if 2n =

O. If a and bare:> 2, then there exist integers x and y such that ax + by = 1 and Ix I - 1 there exists an endomorphism Vi E Au which maps Ei -I into EJ. Apply this to the case where A is a principal ideal domain and E is a finitely generated Amodule. 7) Let A be a principal ideal domain with field of fractions K, and let E be a finitely generated, torsion-free A-module of rank n. Let E * be the dual of E ; then E and E* are each isomorphic to An.

A.VII.64

MODULES OVER PRINCIPAL IDEAL DOMAINS

§4

a) Let M be a submodule of E ; show that, if M U is the submodule of E* orthogonal to M, then the module E * IMo is torsion-free, and consequently MO is a direct factor in E* ; the sub module Moo of E orthogonal to MO is equal to En KM (considering E as being canonically embedded in an n-dimensional vector space over K). We can consider E * IM u as being canonically embedded in the dual M* of M ; show that the invariant factors of E * IMo with respect to M* are equal to the invariant factors of M with respect to E (use Th. 1 of VII, p. 18). b) Let F be a second finitely generated, torsion-free A-module, and let u be a linear transformation from E into F. Show that the invariant factors of 'u(F*) with respect to E* are equal to those of u(E) with respect to F (proceed as in Prop. 5 of VII, p.21). 8) Let G be a module of finite length over a principal ideal domain A. For any pair of free modules M and N such that N c M and G is isomorphic to MIN, the invariant factors of N with respect to M which are distinct from A are identical to the invariant factors of G (VII, p. 20, Def. 2) ; their number is called the rank of G. a) Show that the rank r of G is the least number of cyclic submodules of G of which G is the direct sum, and is equal to the greatest of the ranks of the 7T-primary components of G. b) Show that the rank of a submodule or a quotient module of G is at most equal to the rank of G (consider G as a quotient of two free modules of rank r). c) For each X- E A, show that the rank of the submodule X-G is equal to the number of invariant factors of G which do not divide X- (notice that, if E = AI (An), then the module X-E is isomorphic to (AX-)I ((An) n (AX-)). Deduce that the k-th invariant factor of G (in decreasing order) is a gcd of those X- E A such that X-G has rank"" r - k. d) Lct Ank (1 "" k "" r) be the invariant factors of G arranged in decreasing order. Let H be a submodule of G of rank r - q, and let A[3k (1 "" k "" r - q) be its invariant factors, arranged in decreasing order. Deduce from b) and c) that [3k divides nk+ q for 1 "" k "" r - q. Similarly show that, if G/H has rank r - p and invariant factors AYk (1"" k "" r - p) in decreasing order, then "Yk divides nk +p for 1"" k "" r - p. e) Conversely, let L be a torsion module of rank r - q, whose invariant factors AX-k (l "" k "" r - q) are such that X-k divides nk + q for 1 "" k "" r - q. Show that there exist submodules M and N of G such that L is isomorphic to M and to GIN (decompose G into a direct sum of submodules isomorphic to AI Ank). ~ 9) Let A be a principal ideal domain and K its field of fractions; let E be a vector space over K, let M be a finitely generated A-module of rank n contained in E, let N be a finitely generated A-module of rank p contained in KM, and let P be a finitely generated Amodule of rank q contained in KN. a) Let Ani (1 "" i "" p) be the invariant factors of N with respect to M, arranged in decreasing order (Ex. 3). Show that, for 1"" k ""p, the element nk is a gcd of those elements X- E K such that the rank of the quotient module (N + X- (KN n M) )/N is "" P - k (cf. Ex. 8, d)). b) Let A[3 j (1 "" j "" q) be the invariant factors of P with respect to M and A"y j (1 "" j "" q) the invariant factors of P with respect to N, arranged in decreasing order. Show that nl"Yj divides [3j for 1 "" j "" q (cf. Ex. 8). c) Show that "yjn j divides [3j for 1 "" j "" q. (First consider the case n = p = q, and apply Ex. 8. In general, show that we may always assume p = n, and consider a submodule Q of M such that P + pQ has rank p for all pEA; then choose the ideal Ap sufficiently small, and apply Prop. 4 of VII, p. 20.)

§4

A.VII.65

EXERCICES

d) For any submodule H of N of rank k ~ p, let (f.LH) be the fractional ideal consisting of f.L E K such that f.L (M n KH) c H. Deduce from e) that for each 1 ~ k ~ p the element Cik is a gcd of the f.LH as H runs through all submodules of N of rank k .

.,. 10) Let A be a principal ideal domain and let M be a submodule of rank n in E = An; let ACi; (1 ~ i ~ n) be the invariant factors of M with respect to E, arranged in decreasing order. Let N be a submodule of M admitting a complement in M (that is, such that N = M n KN, where K is the field of fractions of A). a) Let P be a complement of N in M. Show that the submodule (P + (E n KN) )/M of ElM is isomorphic to (E n KN)/N. Deduce that, if N has rank p and A~k (1 ~ k ~ p) are the invariant factors of N with respect to E, then ~k divides Cik+n_p for 1 ~ k ~p (use Ex. 8, d». b) Give an example to show that the module ElM is not necessarily isomorphic to the product of the modules (E n KN) IN and E I (P + (E n KN» (take ElM to be indecomposable) .

.,. 11) Let A be a principal ideal domain, let H be a submodule of E = An, and let HO be the submodule of the dual E* of E orthogonal to H (Ex. 7). Let M be a submodule of rank p of E and let ACi; (1 ~ i ~ p) be the invariant factors of M with respect to E, arranged in decreasing order. Let (vB) be the gcd of the ideals f (M) of A as f runs through HO ; show that, for 1 ~ k ~ p, the element Cik is a gcd of the VB for all submodules H of E of rank k - 1 which admit a complement in E (first consider the case k = 1 ; apply the result to the submodule (M + H)/H of E/H, and use Ex. 8, d), applied to the quotient module E/(M + H) of ElM). 12) Let G be a module of finite length over a ring A. Show that a submodule H of G can only be isomorphic to G if it is equal to G (cf. II, p. 212, Prop. 16). 13) Let A be a principal ideal domain with field of fractions K. a) Show that, for every A-module M, the canonical homomorphism eM: M

--> D(D(M» is injective (cf. II, p. 389, Ex. 13). b) If u: M --> N is a homomorphism of A-modules, define natural isomorphisms from D(Im(u» onto Im(D(u», from D(Ker(u» onto Coker(D(u», and from D(Coker(u» onto Ker (D (u». e) Let M be an A-module and N a submodule of M. In the notation of Sect. 9, show that N°O = N (use a) and The module N° has finite length if and only if MIN has finite length. Conversely, for any submodule N' of D (M) of finite length, the module N'o is a submodule of M such that MIND has finite length, and N' = N'oo. d) If M is a torsion A-module, and M" denotes the 'IT-primary component of M, show that D(M) is isomorphic to Hom (M", U,,), where P is a system of representatives of

b».

n

nEP

irreducible elements of A and U" is the 'IT-primary component of KIA for each 'IT E P (VII, p. 54 Ex. 3). e) Deduce from d) that D (KI A) is isomorphic to the product AC")' where

n

nEP

AC") denotes the completion of the ring A(,,) defined in VII, p. 54, Ex. 2, with respect to the topology defined by taking the ideals 'lTn A(,,) to be a basis of open neighbourhoods of 0 (this completion can also be identified with the inverse limit lim (AI A'IT n ».

--

A.VII.66

MODULES OVER PRINCIPAL IDEAL DOMAINS

§4

f) If M is an A-module of finite length, and a is its annihilator, show that D (M) is isomorphic to the dual of the faithful (A/a)-module associated to M.

g) Let M be an A-module of finite length. Show that if N is a submodule of M, then there exists a submodule P of M such that M/N is isomorphic to P and M/P is isomorphic to N (use VII, p.26, Prop. 10 and VII, p.27, Prop. 12) ; two submodules of M with these properties are called reciprocal. Show that, for all a E A, the submodule M (a) of M consisting of all x E M such that ax = 0, and the submodule aM, are reciprocal. .,. 14) Let M be a module of finite length over a principal ideal domain A with field of fractions K. a) Show that if N is a sub module of M then the rank of M is at most equal to the sum of the ranks of Nand M/N (let M = E/H, where E is a free module of rank nand H is a submodu1e of rank n of E ; if N = L/H, where He LeE, notice that, if the rank of N is p, then there exists a submodule R of H, of rank n - p, admitting a complement S in L such that S n H is a complement of R in H ; moreover, (E n KR )/R has rank ~ n - p and is isomorphic to a quotient module of E/L (Ex. 10, complete the proof using Ex. 8, b) Let N be a submodule of M of rank p, and let q be the rank of M/N. Let An; (1 ~ i ~ n) be the invariant factors of M and A~ k (1 ~ k ~ p ) those of N, arranged in decreasing order. Show that ~k is a multiple of ak _ (p + q _ n) (use a) and Ex. 8, c), noticing that (AM)/ (AN) is isomorphic to a quotient module of M/N) .

a» ;

b».

.,. 15) a) Let M be a module of finite length and rank n over a principal ideal domain A, and let Aa; (1 ~ i ~ n) be its invariant factors in decreasing order. Let (~;) (1 ~ i ~ n) be a sequence of elements of A such that ~; divides a; for 1 ~ i ~ n and a; divides ~; , 1 for 1 ~ i ~ n - 1. Let (M;)I";,, n be a sequence of sub modules of M such that M is the direct sum of the M; and each M; is isomorphic to A/An; ; let a; be a generator of M; for 1 ~ i ~ n. Put

Show that the quotient of M by the cyclic submodule Ab is isomorphic to the direct sum of the n modules A/ A~; (1 ~ i ~ n). b) Let M and N be two modules of finite length over A; let An; (1 ~ i ~ n) and A~k (1 ~ k ~ p) be the invariant factors of M and N respectively, arranged in decreasing order; suppose that p ~ n, that ~k divides ak+n_p for 1 ~ k ~p, and that ak divides ~k + (p + q _ n) for 1 ~ k ~ n - q, where q is an integer such that q~n~p+q.

Show that there exists a submodule P of M, isomorphic to N, such that M/P has rank q (decompose each of the two modules M and N as a direct sum of q modules, in such a way as to reduce the problem to the case q = 1 ; then use a) and Ex. 13, g». c) For a module N over A to be isomorphic to a direct factor of a module M of finite length, it is necessary and sufficient that every elementary divisor of N be an elementary divisor of M, and that its multiplicity in N be at most equal to its multiplicity in M. Hence obtain an example of a module M of finite length and a sub module N of M such that the ~

§4

A.VII.67

EXERCISES

rank of M is equal to the sum of the ranks of Nand M/N, and such that N does not admit a complement in M (use a». 16) Let M be a module over a commutative ring A. A submodule N of M is called characteristic if u (N) c N for every endomorphism u of M. For all a E A, the submodule M (a) of elements annihilated by a, and the submodule aM, are characteristic. a) If N is a characteristic submodule of M, and P and Q are two complementary submodules of M, show that N is the direct sum of N n P and N n Q (consider the projections from M onto P and Q). b) Let M be a module of finite length over a principal ideal domain A; let Aa, be the invariant factors of M, arranged in decreasing order (1 "" i "" n); let (M j )\ '" j '" n be a sequence of submodules of M such that M is the direct sum of the M j and each M j is isomorphic to A/ Aaj. Let N be a characteristic submodule of M and let A~, be the annihilator of N n M j (1 "" i "" n); if a j = ~j"Vj, show that ~j divides ~j + \ and "Vj divides "Vj + \ for all 1 "" i "" n - 1 (notice that, for 1 "" i "" n - 1, there exists an endomorphism of M which maps M j + J onto M j , and an endomorphism of M which maps M j onto a sub module of M j + I). c) Conversely, let N be a submodule of M which is a direct sum of submodules N j c M j ; suppose the annihilators A~j of the N j (1 "" i "" n) satisfy the above conditions. Then show that N is a characteristic submodule of M. Deduce that if two characteristic submodules of M have the same invariant factors then they are equal. 17) Let A be a principal ideal domain, let a be a nonzero element of A, and let E be the module A/Aa. Then the product En is isomorphic to An/(aAn). a) Let u be a linear map from En into Em ; show that u can be obtained from a linear map v from An into Am on passing to quotients. b) Let A~j (1"" i ""p "" inf(rn, be the invariant factors of v(An) with respect to Am, arranged in decreasing order; let q "" p be the greatest of the indices i such that a does not divide ~j, and let Bj be a gcd of a and ~j for 1 "" i "" q. Show that the kernel u - \ (0) of u is the direct sum of n - q modules isomorphic to E and q cyclic modules isomorphic to A/ ABj respectively (1 "" i "" q). If a = "VjBj (1 "" i "" q), show that the module u (En) has invariant factors A"Vj.

n»

18) Let C be a principal ideal domain. Consider a system of linear equations

j

L ~

ajj~j = ~j (1 "" i ""

rn )

1

where the coefficients and the right hand sides belong to C, and let A denote the matrix (a j ) with m rows and n columns, and B the matrix obtained by extending A by the (n + 1 )-st column (~j). Show that the system admits at least one solution in C n if and only if the following conditions are satisfied: 1) A and B have the same rank p ; 2) a gcd of the minors of order p of A is equal to a gcd of the minors of order p of B (use Prop. 4 of VII, p. 20 and Ex. 9,

c».

19) Let C be a principal ideal domain. Consider a system of m n

L jd

ajj~j

==

~j (mod

«

linear congruences»

at), where the coefficients, the right hand sides, and the element a all

A.VII.68

MODULES OVER PRINCIPAL IDEAL DOMAINS

§4

belong to e, and where a is nonzero; let A denote the matrix (a ij ), let B denote the matrix obtained by extending A by the column (l3 i ), and let Sk (resp. Sk) be a gcd of the k-th order minors of A (resp. B). Then the system admits at least one solution in e" if and only if: a) If m~n, a gcd of am, am-lSI' ... , aS m _ l , Sm is equal to a gcd of am, am-IS;, (X8~ _ l ' 8~ ;

b) if m>n, a gcd of a"~l, a"Sl' ... , as" is equal_to a gcd of a"+l, a"S;, 8~

+ I"

(Reduce the system of linear congruences to a system of linear equations, and use Ex. 18.) 20) a) Let e be a principal ideal domain, and let [(Xl' .. ,

Xp) =

L ail .. ip~l,it ... ~P,ip (ik)

be a p-linear form on F, where E is the e-module C" (and where ~ij denotes the j-th coordinate of x;). Show that if S is a gcd of the coefficients a it ... ip then there exists an element (ak) E EP such that [(aI' ... , a p ) = S (reduce to the case S = 1, and argue by induction on p, using Th. 1 of VII, p.18). b) Let n> 1 be an integer, and a i (1 ~ i ~ n) be n elements of e with gcd S. Show that there exists a square matrix A of order n over e whose first column is (a i ) and whose determinant is S (use a)). 21) Let e be a principal ideal domain, and let A = (a ij ) be a square matrix of order n over C. a) Show that there exists a square matril\ U of order n over e, with determinant 1, such that the last column of the matrix VA has its first (n - 1) entries zero, and its last entry equal to the gcd of the entries in the n-th column of A (use Ex. 20, b)). b) Deduce from a) that there exists a square matrix V of order n over e, with determinant 1, such that the matrix V A = (l3ij) is lower triangular (