Multilinear Algebra

Multilinear algebra Multilinear algebra D. G. NORTHCOTT F.R.S. Formerly Town Trust Professor of Mathematics at the U...

Author: D. G. Northcott

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Multilinear algebra

Multilinear algebra

D. G. NORTHCOTT F.R.S. Formerly Town Trust Professor of Mathematics at the University of Sheffield

The right of the University of Cambridge to print and sell all manner of books was granted by Henry VIII in 1534. The University has printed and published continuously since 1584.

CAMBRIDGE UNIVERSITY PRESS Cambridge London New York New Rochelle Melbourne Sydney

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521262699 © Cambridge University Press 1984 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1984 This digitally printed version 2008 A catalogue recordfor this publication is available from the British Library Library of Congress Catalogue Card Number: 83-27210 ISBN 978-0-521-26269-9 hardback ISBN 978-0-521-09060-5 paperback

Contents

1 .1 .2 .3 1.4 .5 i.6 .7

Preface

ix

Multilinear mappings General remarks Multilinear mappings The tensor notation Tensor powers of a module Alternating multilinear mappings Symmetric multilinear mappings Comments and exercises Solutions to selected exercises

1 1 1 4 6 6 10 13 15

2 Some properties of tensor products General remarks 2.1 Basic isomorphisms 2.2 Tensor products of homomorphisms 2.3 Tensor products and direct sums 2.4 Additional structure 2.5 Covariant extension 2.6 Comments and exercises 2.7 Solutions to selected exercises

19 19 19 22 25 28 29 31 37

3 Associative algebras General remarks 3.1 Basic definitions 3.2 Tensor products of algebras 3.3 Graded algebras 3.4 A modified graded tensor product 3.5 Anticommutative algebras 3.6 Covariant extension of an algebra 3.7 Derivations and skew derivations

42 42 42 44 47 51 54 56 56

vi

Contents 3.8 3.9 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 5

Comments and exercises Solutions to selected exercises

58 65

The tensor algebra of a module General remarks The tensor algebra Functorial properties The tensor algebra of a free module Covariant extension of a tensor algebra Derivations and skew derivations on a tensor algebra Comments and exercises Solutions to selected exercises

69 69 69 72 74 75 76 78 80

The exterior algebra of a module General remarks The exterior algebra Functorial properties The exterior algebra of a free module The exterior algebra of a direct sum Covariant extension of an exterior algebra Skew derivations on an exterior algebra Pfaffians Comments and exercises Solutions to selected exercises

84 84 84 87 89 93 95 96 100 105 111

The symmetric algebra of a module General remarks The symmetric algebra Functorial properties The symmetric algebra of a free module The symmetric algebra of a direct sum Covariant extension of a symmetric algebra Derivations on a symmetric algebra Differential operators Comments and exercises

117 117 118 120 121 121 122 123 124 126

7 Coalgebras and Hopf algebras General remarks 7.1 A fresh look at algebras 7.2 Coalgebras 7.3 Graded coalgebras 7.4 Tensor products of coalgebras 7.5 Modified tensor products of coalgebras 7.6 Commutative and skew-commutative coalgebras 7.7 Linear forms on a coalgebra 7.8 Hopf algebras

130 130 130 133 134 135 143 150 151 153

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

7.9 7.10 7.11 7.12 7.13 7.14 8 8.1 8.2 8.3 8.4 8.5 8.6

Contents

vii

Tensor products of Hopf algebras E(M) as a (modified) Hopf algebra The Grassmann algebra of a module S(M) as a Hopf algebra Comments and exercises Solutions to selected exercises

156 158 160 163 166 169

Graded duality General remarks Modules of linear forms The graded dual of a graded module Graded duals of algebras and coalgebras Graded duals of Hopf algebras Comments and exercises Solutions to selected exercises

175 175 175 178 184 188 191 194

Index

197

Preface

This account of Multilinear Algebra has developed out of lectures which I gave at the University of Sheffield during the session 1981/2. In its present form it is designed for advanced undergraduates and those about to commence postgraduate studies. At this general level the only special prerequisite for reading the whole book is a familiarity with the notion of a module (over a commutative ring) and with such concepts as submodule, factor module and homomorphism. Multilinear Algebra arises out of Linear Algebra and like its antecedent is a subject which has applications in a great many different fields. Indeed, there are so many reasons why mathematicians may need some knowledge of its concepts and results that any selection of applications is likely to disappoint as many readers as it satisfies. Furthermore, such a selection tends to upset the balance of the subject as well as adding substantially to the required background knowledge. It is my impression that young mathematicians often acquire their knowledge of Multilinear Algebra in a rather haphazard and fragmentary fashion. Here I have attempted to weld the most commonly used fragments together and tofillout the result so as to obtain a theory with an easily recognizable structure. The book begins with the study of multilinear mappings and the tensor, exterior and symmetric powers of a module. Next, the tensor powers are fitted together to produce the tensor algebra of a module, and a similar procedure yields the exterior and symmetric algebras. Multilinear mappings and the three algebras just mentioned form the most widely used parts of the subject and, in this account, occupy the first six chapters. However, at this point we are at the threshold of a richer theory, and it is Chapter 7 that provides the climax of the book. Chapter 7 starts with the observation that if we re-define algebras in terms of certain commutative diagrams, then we are led to a dual concept ix

x

Preface

known as a coalgebra. Now it sometimes happens that, on the same underlying set, there exist simultaneously both an algebra-structure and a coalgebra-structure. When this happens, and provided that the two structures interact suitably, the result is called a Hopf algebra. It turns out that exterior and symmetric algebras are better regarded as Hopf algebras. This approach confers further benefits. By considering linear forms on a coalgebra it is always possible to construct an associated algebra; and, since exterior and symmetric algebras have a coalgebra-structure, this construction may be applied to them. The result in the first case is the algebra of differential forms (the Grassmann algebra) and in the second case it is the algebra of differential operators. The final chapter deals with graded duality. From every graded module we can construct another graded module known as its graded dual. If the components of the original graded module are free and of finite rank, then this process, when applied twice, yields a double dual that is a copy of the graded module with which we started. For similarly restricted graded algebras, coalgebras and Hopf algebras this technique gives rise to a full duality; algebras become coalgebras and vice versa; and Hopf algebras continue to be Hopf algebras. Each chapter has, towards its end, a section with the title 'Comments and exercises'. The comments serve to amplify the main theory and to draw attention to points that require special attention; the exercises give the reader an opportunity to test his or her understanding of the text and a chance to become acquainted with additional results. Some exercises are marked with an asterisk. Usually these exercises are selected on the grounds of being particularly interesting or more than averagely difficult; sometimes they contain results that are used later. Where an asterisk is attached to an exercise a solution is provided in the following section. However, to prevent gaps occurring in the argument, a result contained in an exercise is not used later unless a solution has been supplied. Once the guide-lines for the book had been settled, I found that the subject unfolded very much under its own momentum. Where I had to consult other sources, I found C. Chevalley's Fundamental Concepts of Algebra, even though it was written more than a quarter of a century ago, especially helpful. In particular, the account given here of Pfaffians follows closely that given by Chevalley. Finally I wish to record my thanks to Mrs E. Benson and Mrs J. Williams of the Department of Pure Mathematics at Sheffield University. Between them they typed the whole book; and their cheerful co-operation enabled the exacting task of preparing it for the printers to proceed smoothly and without a hitch. Sheffield, April 1983 D. G. Northcott

Multilinear mappings

General remarks Throughout Chapter 1 the letter R will denote a commutative ring which possesses an identity element. R is called trivial if its zero element and its identity element are the same. Of course if R is trivial, then all its modules are null modules. The standard notation for tensor products is introduced in Section (1.2) and from there on we allow ourselves the freedom (in certain contexts) to omit the suffix which indicates the ring over which the products are formed. More precisely when (in Chapter 1) we are dealing with tensor products of K-modules, we sometimes use ® rather than the more explicit ®R. This is done solely to avoid typographical complications. 1.1

Multilinear mappings Let Mt, M 2 , . . . , Mp (p> 1) and M be JR-modules and let : Mi x M2 x • • • x Mp-+M

(1.1.1)

be a mapping of the cartesian product M t x M2 x • • • x Mp into M. We use mi9 m 2 , . . . , mp to denote typical elements of Mu M 2 , . . . , Mp respectively and r to denote a typical element of R. The mapping cf> is called multilinear if cj)(mu...9m/i

+ m",...

= (t>(mu...,

,mp)

m j , . . . , m p ) + (m 1? ..., m " , . . . , mp)

(1.1.2)

and

(j){mu...,

rmh . . . , mp) = r(j){mu . . . , mi9...,

mp).

(1.1.3)

(Here, of course, i is unrestricted provided it lies between 1 and p.) For example, when p = 1 a multilinear mapping is the same as a homomorphism of K-modules. Suppose now that (1.1.1) is a multilinear mapping. We can derive other 1

2


multilinear mappings from it in the following way. Let h: M-+N be a homomorphism of K-modules. Then h ° (j) is a multilinear mapping of Mx x M 2 x • • • x Mp into JV. This raises the question as to whether it is possible to choose M and (j) so that every multilinear mapping of Mx x M 2 x • • • x M p can be obtained in this way. More precisely we pose Problem 1. To choose M and (j) in such a way that given any multilinear mapping \I/:M1XM2X--XMP->N

there is exactly one homomorphism h: M —• N (of R-modules) such that h ° = i^.

This will be referred to as the universal problem for the multilinear mappings of M x x M 2 x • • • x Mp. We begin by observing that if the pair (M, 0) solves the universal problem, then whenever we have homomorphisms ht: M —• N (i = 1,2) such that hx° (f) = h2o (j), then necessarily h1 = /z2. Now suppose that (M, ) and (M', (/>') both solve our universal problem. In this situation there will exist unique R-homomorphisms X.M-+M' and A': M' —• M such that A ° 0 = ' and X' ° (f)' = (j). It follows that (A' ° >1) ° 0 = 0 or (k' ° X) ° (j) = i ° 0, where i is the identity mapping of M. The observation at the beginning of this paragraph now shows that X' ° X = i and similarly X ° A' is the identity mapping of M'. But this means that X\M-+M' and X'.M'^M are inverse isomorphisms. Thus if we have two solutions of the universal problem, then they will be copies of each other in a very precise sense. More informally we may say that if Problem 1 has a solution, then the solution is essentially unique. Before we consider whether a solution always exists, we make some general observations about free modules. From here on, until we come to the statement of Theorem 1, we shall assume that R is non-trivial. We recall that an ^-module which possesses a linearly independent system of generators is called free and a linearly independent system of generators of a free module is called a base. Now let X be a set and consider homogeneous linear polynomials (with coefficients in R) in the elements of X. These form a free K-module having the elements of X as a base. This is known as the free module generated by X. (If X is empty, then the free module which it generates is the null module.) Any mapping of X into an K-module N has exactly one extension to an Rhomomorphism of this free module into N. We are now ready to solve Problem 1. Let U(Ml, M 2 , . . . , Mp) be the free R-module generated by the cartesian product M x x M 2 x • • • x Mp. Of course, this has the set of sequences (mls m 2 , . . . , mp) as a base. The elements of U(M1, M 2 , . . . , Mp) that have one or other of the forms

Multlinear mappings " , . . . , mp)-(mu

..., m'h...,

mp)

/

- ( m 1 , . . . , m ; , . . . ,mp)

(1.1.4)

and

(ml9...,

rmh . . . , mp)-r(mu

. . . , mi9...,

mp)

(1.1.5)

generate a submodule F(M 1? M 2 , . . . , M p ) say. Put M=(7(M1,M2,...,Mp)/K(M1,M2,...,Mp)

(1.1.6)

and define 0 : M x x M 2 x • • • x Mp-*M

(1.1.7)

so that 0(mx, m 2 , . . . , mp) is the natural image of (ml5 m 2 , . . . , mp), considered as an element of U(Ml9 M 2 , . . . , M p ), in M. Since the elements of M p ) described in (1.1.4) and (1.1.5) become zero in M, 0 U(Ml9M2,..., satisfies (1.1.2) and (1.1.3) and therefore it is multilinear. It will now be shown that M and provide a solution to our universal problem. To this end suppose that ij/: Ml x M 2 x • • • xMp—>N is multilinear. There is an .R-homomorphism in which (ml9m2,..., mp) is mapped into \j/(ml,m2,..., mp). Since \jj is multilinear, the homomorphism maps the elements (1.1.4) and (1.1.5) into zero and therefore it vanishes on P / (M 1 ,M 2 ,... ,MP). Accordingly there is induced a homomorphism h:M—+N which satisfies /i(0(m 1 ,m 2 ,... ,m p )) = ^ ( m 1 , m 2 , . . . ,m p ). Thus /i°(/> = ^. Finally if /i': M —• N is also a homomorphism such that h' ° (/> = ^, then /i and /i' have the same effect on every element of the form 4>(ml,m2,..., mp). However, these elements generate M as an /^-module and therefore h = h'. Thus (M, (/>) solves the universal problem. We sum up our results so far. Theorem 1. Let M l 9 M 2 , . . . , Mp {p > 1) be R-modules. Then the universal problem for multilinear mappings of Mx x M 2 x • • • x Mp has a solution. Furthermore the solution is essentially unique (in the sense explained previously). Corollary. Suppose that (M, c/>) solves the universal problem described above. Then each element ofM can be expressed as a finite sum of elements of the form (f)(mum2,... ,mp). Proof A solution to Problem 1 has just been constructed and it would be easy to check that this particular solution has the property described in the corollary. We could then utilize the result which says that any two solutions are virtually identical. However, it is more interesting to base a proof

4


directly on the fact that (M, (f>) meets the requirements of the universal problem. This is the method employed here. Suppose then that M' is the R-submodule of M generated by elements of the form (ml9 m 2 , . . . , mp). Also let hx: M-+M/M' be the natural homomorphism and h2:M^>M/M' the null homomorphism. Then h1o(j) = h2 ° <j> and therefore ht = h2. However, this implies that M = M'. Let xeM = M'. Then x can be expressed in the form x = rcj)(ml9m2,...,

mp) + r'<j)(m\, m 2 , . . . , m'p) + • • *,

where the sum is finite. However, r(/)(mum2,...,

mp) = (f)(rmu m 2 , . . . , mp)

and similarly in the case of the other terms. The corollary follows.

1.2

The tensor notation Once again Ml9 M 2 , . . . , Mp, where p> 1, denote R-modules and we continue to use ml9 m 2 , . . . , mp to denote typical elements of these modules, and r to denote a typical element of R. Let the pair (M, (f>) provide a solution to the universal problem for multilinear mappings of Mx x M 2 x • • • x Mp. It is customary to write M = M, ®RM2®R--

®RMp

(1.2.1)

and to use m1 ® m2 ® • • • ® mp to designate the element (^(m^ m 2 , . . . , mp) of M. When this notation is employed, Mx ®R M2 ®R • • • ®R Mp is called the tensor product of Ml9 M 2 , . . . , Mp over i^. It will be recalled that the solution to the universal problem is unique only to the extent that any two solutions are copies of each other. Thus we can have different models for the tensor product. However, if we have two such models, then they are isomorphic (as modules) under an isomorphism which matches the element represented by mx ® m2 ® • • • ® mp in the first model with the similarly represented element in the second. On account of this we usually do not need to specify which particular model we are using. Next, because is multilinear, the relations • • • ® mp - - - ®mp + m1 ® • • • ® mf( ® • • • ®mp (1.2.2) and m1®"

- ®rmt®"

- ® mp = r(ml ® • • • ® mx ® • • • ® mp)

(1.2.3)

both hold. Moreover the corollary to Theorem 1 shows that each element of M x ®KM 2 ®R"' ®RMp is a finite sum of elements of the form ml®m2®" • ® mp. Finally the fact that the tensor product provides the

The tensor notation

5

solution to the universal problem for multilinear mappings may be restated as Theorem 2. Given an R-module N and a multilinear mapping i//:MlxM2X'"xMp^N there exists a unique R-module homomorphism h, of Mx ®RM2®R" ®R Mp into N, such that

'

h{m^ ® m2 ® • • • ® mp) = \j/(m1, m 2 , . . . , mp) for all ml9 m 2 , . . . , mp. We interrupt the main argument to observe that when p = 1 the universal problem can be solved by taking M to be M x and to be the identity mapping. This confirms that for p = 1 the tensor product Mx ®R M2 ® R * * • ®RMp is just M x as we should naturally expect. The reader will have noticed that the full tensor notation is rather heavy. To counteract this we shall often use a simplified version. Indeed, because in Chapter 1 we shall only be concerned with a single ring R, it will not cause confusion if we use Mx ® M 2 ® • • • ® Mp in place of the typographically more cumbersome but more explicit M x ®RM2 ®R" ' ®RMp. Nevertheless, in the statement of theorems and other results likely to be referred to later we shall restore the subscript which identifies the relevant ring. So much for matters of notation. Before we leave this section we shall seek to gain insight into the nature of tensor products by examining the result of forming the tensor product of a number of free modules. Suppose then that, for 1 < i N is also a homomorphism satisfying /i'°0 = ^, then h and h! agree on the base Bx x B2 x • • • x Bp of M and therefore /i = /i'. The proof is therefore complete. 1.3

Tensor powers of a module Let M be an ^-module and p > 1 an integer. Put Tp(M) = M ®RM®R--®RM,

(1.3.1)

where there are p factors. The K-module Tp(M) is called the p-th tensor power of M. These powers will later form the components of a graded algebra known as the tensor algebra of M. For the moment we note that 7; (M) = M. Also, if M is a free K-module and B is a base of M, then Tp(M) is also free and it has the elements b1 ®b2®- - - ®bp, where bt e B, as a base. This follows from Theorem 3. 1.4

Alternating multilinear mappings As in the last section M denotes an K-module. In this section we shall have occasion to consider products such as M x M x • • • x M and M ® M ®- - - ® M. Whenever such a product occurs it is to be understood that the number of factors is p, where p > 1. A multilinear mapping rj: M xM x •••

xM-+N

is called alternating if f/(m 1 ,m 2 ,... ,m p ) = 0 whenever the sequence (m1? m 2 , . . . , mp) contains a repetition. (To clarify the position when p= 1 we make it explicit that all linear mappings M-^N are regarded as alternating.) Suppose that n has this property. If lN be a multilinear mapping and suppose that ri(ml9 m2,..., mp) = 0 whenever mi = mi + 1for some i. Then rj is an alternating mapping. Proof The argument just before the statement of Lemma 1 shows that rj(ml9m2, • . . , mp) changes sign if we interchange two adjacent terms in the sequence (m1? m 2 , . . . , mp). Now suppose that (ml,m2,..., mp) contains a repetition. Then either two equal terms occur next to each other or this situation can be brought about by a number of adjacent interchanges. It follows that rj(ml,m2,..., mp) = 0 so the lemma is proved. Consider an alternating multilinear mapping rj, of M x M x • • • x M into an K-module N, and suppose that h: N-+K is a homomorphism of Rmodules. Then h ° rj is an alternating multilinear mapping of M x M x • • • x M into K. This observation leads us to pose the following universal problem. Problem 2. To choose N and Y\ in such a way that given any alternating multilinear mapping there exists exactly one homomorphism h: N-^K (of R-modules) such that C = horj. To solve this problem we consider the tensor power Tp(M) and denote by JP(M) the submodule generated by all elements mx ® m2 ® • • • ® mp, where ( m 1 , m 2 , . . . , mp) contains a repetition. (It is understood that J 1 (M) = 0.) Put N = Tp(M)/Jp(M) and let rj be the mapping of M x M x • • • x M into N which takes {mx,m2,..., mp) into the natural image of mx ® m2 ® • • • ® mp in N. Then rj is an alternating multilinear mapping. If now £:MxMx-xM->K is also an alternating multilinear mapping, then, by Theorem 2, there is a homomorphism of M ® M ® • • • ® M into K which takes mx ® m2 ® • • • ® mp into C(wi1? m 2 , . . . , mp). This homomorphism vanishes on Jp(M) and so it induces a homomorphism /z: AT —• X. It is clear that h ° >/ = £. If we have a second homomorphism, say h': N-^K, and this satisfies h' °rj = £, then /i andft'agree on the elements ^/(rnj, m 2 , . . . , mp) and therefore they agree on a system of generators of N. But this ensures thatft= ft'.

8


It has now been shown that the universal problem for alternating multilinear mappings of M x M x • • • x M has a solution. The solution is unique in the following sense. Suppose that (AT, rj) and (AT, rjf) both solve Problem 2. Then there are inverse isomorphisms X: N —> N' and k': N'-+N such that A°rj = rj' and X' °Y\' = r\. The situation is, in fact, almost identical with that encountered in dealing with uniqueness in the case of Problem 1. Let us suppose that (N, rj) solves the universal problem for alternating multilinear mappings of M x M x • • • x M. Put EP(M) = N (1.4.1) and mx /\m2 A • • • Amp = rj(m1,m2,... ,mp). (1.4.2) Then, because fy is multilinear, we have mx A • • • A (mj + m") A • • • A mp = mx A • • • A m[ A • • • A mp + m1 AAm;' • A • • * A m „ (1.4.3) and ml A • • • Arm; A • • • Amp = r(m1 A • • A• • * • A m p ). (1.4.4) But */ is also alternating. Consequently mx AWI2 AA • • • A m p = 0 whenever (ml,m2,..., mp) contains a repetition and, by Lemma 1, miy Ami2 A • • • A ^ = ±mx Am2 A • • • Amp, (1.4.5) where the plus sign is to be used if (il9 i2,..., ip) is an even permutation of (1,2,...,/?) and the minus sign if the permutation is odd. The module Ep(M) is called the p-th exterior power of M. As we have seen it is unique in much the same sense that Tp(M) is unique. Since when p = 1 we can solve Problem 2 by means of M and its identity mapping, we have The defining property of the exterior power is restated in the next theorem. Theorem 4. Given an R-module K and an alternating multilinear mapping C:MxMx-xM-+K there exists a unique R-homomorphism h: Ep(M)-+K such that for all ml,m2,...

,mpin M.

Corollary. Each element of EP(M) is a finite sum of elements of the form wij A m2 A • • • A mp.

A proof can be obtained by means of a trivial modification of the argument used to establish the corollary to Theorem 1, so details will not be given.

Alternating multilinear mappings

9

The mapping of MxMxxM into Ep(M) which takes (ml,m2,.. •, mp) into ml A m2 A • • • A mp is multilinear and so, by Theorem 2, there is induced a homomorphism TP(M)—>£P(M). This is surjective because the image of mx ® m2 ® ' * * ® w p is mx A m2 A • • • A mp. We refer to Tp(M)—>Ep(M) as the canonical homomorphism of the tensor power onto the exterior power. Note that when p = 1 the canonical homomorphism is the identity mapping of M if we make the identifications T1(M) = M = EX{M). As before we let JP(M) denote the #-submodule of Tp(M) generated by all products mx ® m2 ® • * * ® mp in which there is a repeated factor. Theorem 5. The canonical homomorphism Tp(M) —• Ep(M) is surjective and its kernel is Jp(M). Moreover Jp(M) is generated, as an R-module, by all products mx ® m2 ® • * • ® mp, where mi = mi + 1 for some i. Proof. Let J'P(M) be the submodule of Tp(M) generated by all mx ® m2 ® * * * ® mp with mt = mi + 1 for some i, and let J"P{M) be the kernel of the canonical homomorphism. Then J'P(M) ^JP(M)^ J"P(M). Next, by Lemma 2, the mapping

6: M x M x • • • xM-+Tp{M)/J'p(M), in which 9{mx, m 2 , . . . , mp) = m1 ® m2 ® • • • ® mp + J'p(M), is multilinear and alternating and therefore, by Theorem 4, there is a homomorphism X: Ep(M)-+ Tp(M)/J'p(M) such that X(ml A m2 A • • • A mp) = mx ® m2 ® • • • ® mp 4- J'p(M). But if A is combined with the canonical homomorphism Tp(M)—>£p(M), then the result is the natural homomorphism

Tp(M)^Tp(M)/J'p(M). Thus the natural homomorphism vanishes on J"P{M) and therefore j ; ( M ) c j p ( M ) . Accordingly J'p(M) = Jp(M) = J"p(M) and the theorem is proved. It is useful to know the structure of the exterior powers of a free module. Let us assume therefore that M is a free jR-module and that B is one of its bases. On the set formed by all sequences (bl9 b2,...9 bp) of p distinct elements of B we introduce the equivalence relation in which (bt, ft2,..., bp) and (b'l9 b29..., b'p) are regarded as equivalent if each is a permutation of the other. From each equivalence class we now select a single representative. In the next theorem the set formed by these representatives is denoted by IP(B). Theorem 6. Let M be a free R-module having the set B as a base. Then Ep(M) is a free R-module having the elements bx A b2 A • • • A bp as a base, where (bl9 b2,..., bp) ranges over Ip(B).

10


Proof. Let N be the free K-module generated by Ip(B). Define a mapping n.BxBx

••• xB-^N

as follows. W h e n (bl9 b2,..., bp) c o n t a i n s a repetition n{b1, b2,..., bp) is t o be zero. W h e n , h o w e v e r , bi,b2,-..9bp a r e all different t h e r e is a u n i q u e bp), such t h a t (b\, b'2,..., b'p) p e r m u t a t i o n (b[, b'2,..., b'p\ of (bi9 b2,..., belongs to Ip(B); in this case we put where the plus sign is taken if the permutation is even and the minus sign if it is odd. The mapping n has a unique extension (denoted by the same letter) to a multilinear mapping of M x M x • • • x M into AT, and the construction ensures that the extension is alternating as well as multilinear. We claim that (N,rj) solves Problem 2. Clearly once this has been established the theorem will follow. Suppose then that £:MxMx'xM-+K is an alternating multilinear mapping. Since N is free, there is a homomorphism h:N—>K such that h ( b 1 , b 2 , . . . , b p ) = C ( b l 9 b 2 , . . . , b p ) whenever (bl9b2,...9bp) is in Ip(B). Thus h°n and ( are alternating multilinear mappings which agree on Ip(B) and therefore on B x B x • • • x B as well. If follows that h°rj = C. Moreover it is evident that h is the only homomorphism of N into K with this property. The proof is therefore complete. 1.5

Symmetric multilinear mappings Once again M denotes an ^-module and all products M xM x • • • x M and M ® M ® ' " ® M are understood to have p (p > 1) factors. Let N be an /^-module. A multilinear mapping 9:MxMx-"xM-^N

is called symmetric if 0(ml9m2, • • •, mp) = 0(mlV mlV . . . , mip\

(1.5.1)

whenever mx,m2,...,mp belong to M and (il ,i2,..., ip) is a permutation of ( 1 , 2 , . . . , p). Clearly 6 is symmetric provided 6(ml9m2,..., mp) remains unaltered whenever two adjacent terms are interchanged. If 6 is a symmetric multilinear mapping and h: N—+K is a homomorphism of JR-modules, then h ° 6 is a symmetric multilinear mapping of M x M x • • • x M into K. This inevitably prompts the next universal problem. Problem 3. To choose N and 6 so that given any symmetric multilinear mapping

Symmetric multlinear mappings

11

co: M x M x • - x M —+K h: N-^K such that h°6 = oj. there exists a unique R-homomorphism As Problem 3 can be treated in very much the same way as Problem 2 there will be no need to go into the same amount of detail. Denote by Hp(M) the submodule of Tp(M) generated by all differences w 1 ®m 2 ® *** ®mp — mii®mi2® • • • ®m,-, where (il9i29...,

(1.5.2)

ip) is a permutation of ( 1 , 2 , . . . , p). Put

N=Tp(M)/Hp(M) and defined: MxMxxM—•Nsothat 0(m 1 ,m 2 ,... ,m p ) is the natural image of mx ® m2 ® • • • ® mp in N. Then 6 is multilinear and symmetric. Moreover considerations, closely similar to those encountered in solving Problem 2, show that N and 6 solve the present problem. Of course the solution is unique in the same sense and for much the same reasons as apply in the case of the other universal problems. Now let (JV, 6) be any solution to Problem 3. We put SP(M) = N

(1.5.3)

m1m2

(1.5.4)

and . . . mp = 0(ml9 m 2 , . . . , mp).

Then m±m2 . . . (wij + w i " ) . . . mpz=m^m2

. . . m[...

mp-\-mim2

. . . m'l...

mp

(1.5.5) and

m1m2 . . . (rmt)...

mp = r{mlm2

...mi...mp)

(1.5.6)

because 6 is multilinear, and mlm2 . . . mp = miimi2... mip (1.5.7) because it is symmetric. (Once again (il9 i 2 , . . . , ip) denotes an arbitrary permutation of ( 1 , 2 , . . . ,p).) The module Sp(M) is called the p-th symmetric power of M. As in earlier instances, when p = 1 our universal problem is solved by M and its identity mapping. Consequently S1(M) = M. The next theorem records the characteristic property of the general symmetric power. Theorem 7. Given an R-module K and a symmetric multilinear mapping there exists a unique R-homomorphism h: Sp(M)^>K h(mlm2 ... mp) = w(ml,m2,..., for all m1,m2,...,

mp.

mp)

such that

12


Corollary. Each element of SP(M) is a finite sum of elements of the form m1m2 . . . mp. As is to be expected a proof of this corollary can be obtained by making minor modifications to the arguments used when establishing the corollary to Theorem 1. We next observe that, by Theorem 2, there is a homomorphism Tp(M) —• SP(M)9 of ^-modules, in which mx ® m2 ® ' ' * ® mp is mapped into mlm2 . . . mp. This is known as the canonical homomorphism of Tp(M) into Sp(M). Evidently it is surjective. If we identify T^M) and S^M) with M, then, for p = 1, the canonical homomorphism is the identity mapping. We recall that Hp(M) is the submodule of TP(M) generated by elements of the form (1.5.2). Theorem 8. The canonical homomorphism Tp(M) —• SP(M) is surjective and its kernel is Hp(M). Furthermore Hp(M) is generated by all differences mx ® •• • ®mi®mi -ml®--'

+1

® • •• ®mp

® m x + 1 ® m < ® ••• ® m p ,

where the second term is obtained from the first by interchanging two adjacent factors. Here the demonstration parallels the proof of Theorem 5. We leave the reader to make the necessary adjustments. Finally let us determine the structure of SP(M) when M is a free K-module with a base B. To do this we consider all sequences (bl9 b29... 9 bp) of p elements of B (on this occasion repetitions are allowed) and we regard the sequences (bl9 bl9. . . 9 bp) and (bfl9 b'l9..., b'p) as equivalent if each can be obtained by reordering the other. From each equivalence class we now select a representative, and we denote by I*(B) the set consisting of these representatives. Theorem 9. Let Mbea free R-module with base B. Then Sp(M) is also a free R-module and it has as a base the elements b1b2 ... bp9 where (bl9 b2,..., bp) ranges over I*(B). Remark. If the elements of B are regarded as commuting indeterminates, then this result says that Sp(M) can be thought of as consisting of all homogeneous polynomials of degree p, with coefficients in R9 in these indeterminates. Proof. Let N be the free .R-module generated by I*(B) and define d:BxBx

••• xB—>N

by putting 6(bl9 b2,..., bp) = {b'l9 b'2,..., b'p\ where (b'l9 b'l9..., b'p) is the rearrangement of (bl9 b2,..., bp) that belongs to I*(B). Naturally 6 extends

Comments and exercises

13

to a multilinear mapping of M x M x • • x M into N and it is evident that the multilinear mapping so obtained is symmetric. It now suffices to show that (N, 0) solves Problem 3. Suppose then that

co'.MxMx-

- - xM —>K

is multilinear and symmetric, and define the jR-homomorphism h: N-+K so that, for (b1,b2,... ,bp) in I*(B), we have h(bl,b2,... ,bp) = co(bl9 b2,..., bp). Then h ° 6 and co are symmetric multilinear mappings which agree on I*(B). It follows that they must agree on B x B x • • x B, and this in turn implies that h°6 = co. This completes the proof because it is clear that there is only one homomorphism M —• K which, when combined with 6, gives co. 1.6

Comments and exercises In this section we shall make some observations to complement what has been said in the main text and we shall provide a number of exercises. A few of these exercises are marked with an asterisk. Those that are so marked are either particularly interesting, or difficult, or used in later chapters. Solutions to the starred questions will be found in Section (1.7). Throughout Section (1.6) R is assumed to be non-trivial. Since we are concerned with multilinear algebra it is fitting that the scope of linear algebra (in the present context) should be made explicit. Often by linear algebra is meant the theory of vector spaces and linear transformations. When the term vector space is used it is frequently understood that the scalars are taken from a field. We, however, will only require our scalars to belong to an arbitrary commutative ring and when we make this change the terminology normally changes as well. What had previously been termed a vector space now becomes a module and what had been described before as a linear transformation is now referred to as a homomorphism. Thus, for us, linear algebra will mean the theory of modules (and their homomorphisms) over a commutative ring. One of the key problems of linear algebra is to determine when a system of homogeneous linear equations has a non-trivial solution. Let us pose this problem in a very general form. Suppose that M is an ^-module and consider the equations a11ml +al2rn2+ • • -+alqmq = 0 a21ml +a22m2 + • • • +a2 m =0 : : ::: aplmx -\-ap2m2 + • • • +apqmq = 0.

(161)

14


Here the au belong to R and we are interested in solving the equations in the module M. Of course one solution is obtained by taking all the mt to be zero. Any other solution is called non-trivial. The first exercise gives a necessary and sufficient condition for the existence of a non-trivial solution. This result is known as McCoy's Theorem. In order to state it we first put

A

=

021

«22

(1.6.2)

so that A is the matrix of coefficients, and we denote by(Uv(A) the ideal that is generated by the v x v minors of A. Then 8loM)2«iM)2«2(i4)2---, where by M0(A)we mean the improper ideal R. Note that % (A) = 0 for v > min(p, q) because, for such a v, there are n o v x v minors. Exercise 1*. Show that the equations (1.6.1) have a non-trivial solution, in the R-module M, if and only if $lq(A) annihilates a non-zero element of M. Deduce that if M ^ O and pO

is also exact.

Proof. The corollary is derived from the theorem by taking all the fi9 with the exception of ffl, to be identity mappings. This ensures that N( = 0 for all The property described in the corollary is usually referred to by saying that tensor products are right exact.

Tensor products and direct sums

25

2.3

Tensor products and direct sums Let {Nt}ieI be a family of submodules of an ^-module N. Then N is the direct sum of these submodules if each n e N has a unique representation of the form

n = YJ n h

(2.3.1)

16/

where n( e Nt and only finitely many summands are non-zero. When this is the case we shall usually write AT = 5 > ,

(As.).

(2.3.2)

16/

However, when N is the direct sum of a finite number of submodules, say of Nl9 N2,..., Ns, we shall use N = Nl®N2@---®Ns

(2.3.3)

as an alternative. Suppose that we have the situation envisaged in (2.3.2). Then for each i e / we have an inclusion mapping oÂ^—»iV and a projection mapping nt: N —> Nt. (For n e N the latter picks out from the representation (2.3.1) the summand indexed by i.) Both a{ and n{ are #-homomorphisms. They have the properties listed below: (A) 7ij ° (Tt is a null homomorphism if i ^j and it is the identity mapping of Ntifi=j; (B) for each neN, 7c,(n) is non-zero for only finitely many different values of i; (C) for each neN, Y. Let us now make a completely fresh start. Suppose that N is an Rmodule, and that {Nt}ieI is a family of ^-modules which, however, are no longer assumed to be submodules ofN. Suppose too that, for each iel, we are given homomorphisms a^. Nt—>N9 7rf: N—>Nt and that these satisfy conditions (A), (B) and (C). Since nt °Gt is the identity mapping of Ni9 o{ is an injection and n{ is a surjection. In particular o{ maps Nt isomorphically onto M' —• 0fee^n exact sequence of R-modules and let P be a projective R-module. Then the induced sequences 0^>P®RM"^P(g)RM-+P®RM'->0 and 0->M" ®RP-+M

®RP-+Mf

®RP-+0

are exact. Proof Of course we need only consider the first sequence. Let / : M" —• M be an injective homomorphism. By Theorem 4 Cor., it will suffice to show that P ® / is also injective. We begin with the case where P is free. Suppose that F is a free /^-module and that {£i}ieI is one of its bases. Let x G F ® M". By Theorem 5 Cor., x = £ (f, ® m'/), where the mj' are in M" and only finitely many of them are non-zero. Assume that (F ® / ) ( x ) = 0. Then Z«i®/«)) = 0 and therefore /(mj') = 0 for all f G /, again by the corollary to Theorem 5. But / is an injection. Consequently all the m" are zero and hence x = 0. This proves that F ® / is an injection. Finally assume that F = P@Q, where F is free and P and Q are submodules. There exist homomorphisms a: P —• F and TT : F —• P such that 7c ° o- is the identity mapping of P. Now (7i ® M") ° (a ® M") = (TT ° cr) ® M" and this is the identity mapping of P ® M". It follows that a ® M" is an

28

Some properties of tensor products

injection. But we already know that F ® / is an injection so (F®f)°((j®M") = (T(g)f is an injection as well. Since cr®/= (a ® M) ° (P ® / ) , this implies that P ® / is an injection, which is what we were aiming to prove.

2.4

Additional structure In Section (2.4) S, as well as R, denotes a commutative ring with an identity element. If M is both an R-module and an S-module, then we shall say it is an (R, S)-module provided (i) the sum of two elements of M is the same in both structures, and (ii) s(rm) = r(sm) whenever reR, seS and msM. Let M and M' be (R, 5)-modules. A mapping / : M —• M' is called an (R, Syhomomorphism if it is a homomorphism of K-modules and also a homomorphism of S-modules. A bijective (R, S)-homomorphism is, of course, referred to as an (R, S)-isomorphism. Suppose that M x , . . . , M;i _ x, M ;| + x,..., Mp are JR-modules and that M;1 is an (R, S)-module. For s e S define fs: M/4 —> M;< by /s(m/4) = sm/r Then / s is where fs an R-homomorphism and therefore Ml®'-'®fs®"'(g)Mp9 occurs in the /i-th position, is an endomorphism of the K-module M x ® • • • ® M^ ® • • • (g) M p . For x in M x ® • • • ® M/4 ® • • • ® Mp define sx by 5X = (MX ® • • • ® / s ® • • • ® M p )(x).

(2.4.1)

Then, because fs + a = fs+fa and fsa = fs°f(T for s,aeS, it is easily verified that M x ® • • • ® M/4 ® • • • ® Mp is an S-module. Indeed we can go further and say that it is actually an (R, 5)-module with s(m1 ® • • • ® m,, ® • • • ® mp) = mx ® • • • ® sm^ ® • • • ® mp. (2.4.2) Finally, if gt: Mf—>Mj is an R-homomorphism for i= 1 , . . . , fi— 1, fi +1, . . . , p and ^;1: M/f —• MJ4 is an (K, S)-homomorphism, then ^ 1 ® * - ® 0/< ® ' * * ® ^ p is itself an (jR, S)-homomorphism. After these preliminaries let A be an K-module, C an S-module, and B an (R, 5)-module. The considerations set out in the earlier paragraphs show that A ®RB is an (R, S)-module and therefore (A ®RB) ® S C is an (R, S)module. Similar considerations show that A ®R(B®SC) is an (R, S)-module as well. The next theorem, which asserts that (Ai ® 1). Corollary. The covariant extension of a protective R-module is a protective S-module. Proof Suppose that F is a free ^-module and that P, Q are #-submodules with F = P ®Q. By Theorem 5, F ®RS = (P ® R S) 0 (Q ®R5),

(2.5.2)

this being, in the first instance, a direct sum of/^-modules. However, F ®R S, P ®RS and Q ®R S are all of them S-modules, and now (2.5.2) is seen to be a direct sum of S-modules. But it has just been shown that F ®R S is S-free. It therefore follows that P ®RS and Q ®RS are 5-projective. Our final result in this chapter shows that tensor products and covariant extensions commute. Theorem 9. Let M1,M29.-.,Mp phism

be R-modules. Then there is an isomor-

(M, ®R S) ®s (M 2 ®R S) ®s - - • ®s (Mp ®R S) *(M! ®RM2 ®R-- ®RMp) ®RS of S-modules in which (mx ® sx) ® (m2 ®s2)®"'® to (m1®m2®- ® mp) ® sls2 ... s .

(mp ® sp) corresponds


31

Proof. We begin with the case p = 2. By Theorem 7, there is an (R, S)isomorphism (Mx ®R S) ®s (Af2 ®R S)«Mx ®* (5 ® s (M 2 ®* 5))

(2.5.3)

which matches (mx ® sx) ® (m2 ® s2) with mx ® (sx ® (m2 ® s2)). Next, Theorem 3 provides an S-isomorphism S ®s(M2 ®RS)*M2®RS

(2.5.4)

which pairs sx ® (m2 ® s2) with m2 ® s ^ . Indeed this is also an (R, S)isomorphism. Thus (2.5.3) and (2.5.4) together provide an (R, S)isomorphism (M, ®RS) ®S(M2 ®RS)*M1

®R (M2 ®RS)

in which (mx ® sx) ® (m2 ® 52) goes into m1 ® (m2 ®sls2). corollary to Theorem 1 provides an K-isomorphism M x ®^ (Af2 ®R S)«(Af! ®* M 2 ) ®R 5,

Finally, the (2.5.5)

where ml®(m2®s) is associated with (m1®m2)®s. The case p = 2 follows as soon as it is noted that (2.5.5) is an S-isomorphism as well as an Risomorphism. When p = 1 the theorem becomes a tautology. The general result follows by induction if we make use of Theorem 1 and the case where p = 2. 2.6

Comments and exercises This section will be used to clarify certain aspects of the theory of tensor products by means of comments and exercises. As in the last chapter solutions are provided to the more interesting and difficult exercises, and the exercises which come into this category are marked by an asterisk. Let Af' be a submodule of an K-module M and N' a submodule of an Rmodule N. The purpose of the first exercise is to show that usually Af' ® Nf cannot be regarded as a submodule of Af ® N. More precisely, the inclusion mappings M-+M and N'^N induce a homomorphism M' ®N'—>M ®N,

(2.6.1)

and now the point being made is covered by Exercise 1. Show by means of an example that the homomorphism Af' ® N''—• Af ® N of (2.6.1) need not be an injection. (Hint. Take R = Z, Af = Af' = Z/2Z, N = Z, and AT = 2Z. As always Z denotes the ring of integers.) By contrast Exercise 2 embodies an important constructive result which partially offsets the purely negative assertion of Exercise 1. Exercise 2*. Let xl,x2,...,xs

be elements of an R-module

Af, and let

32


yi» ^2? • • • •> y$ belong to an R-module N. Suppose thatxl ® yx + x2 ® y2 + ' ' ' + xs® ys = Oin M ® N. Show that there exist finitely generated submodules M' and AT, of M and N respectively, such that (i) the x( are in M' and the yt in AT, and (ii) x1®y1+x2®y2

+ --+xs®ys

= 0 in M'

Exercise 3*. Let I be an ideal of R and let M be an R-module. Establish an isomorphism M ®R (R/I)&M/IM of R-modules. The next exercise provides a version of a very useful result known as Nakayamas Lemma. It appears here because it helps to solve Exercise 5. Exercise 4*. Let M be a finitely generated R-module. Show that if M = JM for every maximal ideal J, then M = 0. The condition that M be finitely generated cannot be dropped. This may be seen by taking R to be Z and letting M be the field of rational numbers considered as a Z-module. Exercise 5*. Let M be an R-module. Show that the following two statements are equivalent: (i) for a finitely generated R-module N we have M ® N = 0 only when N = 0; (ii) for every maximal ideal J, of R, JM is different from M. We next consider a matter that concerns tensor products of homomorphisms. To this end let / : M —• N and / ' : M' —• N' be homomorphisms of K-modules. Then, as we saw in Section (2.2), these induce a homomorphism f®f:

M®M'^N ®N'.

Now the homomorphisms of M into N can be added and they can be multiplied by elements of R to produce new homomorphisms; in fact the homomorphisms of M into N form an .R-module. This module is denoted by Hom R (M, N) and this is sometimes simplified to Hom(M, N) if there is no uncertainty as to the ring of scalars. Since f ® f belongs to Hom(M ® M', N ® AT'), we have a mapping Hom(M, N) x Hom(M', AT)-> Hom(M ®M',N

® N')

in which (/, / ' ) is taken into / ® / ' , and this mapping is bilinear (see (2.2.7) and (2.2.8)). It therefore induces a homomorphism Hom(M, N) ® Hom(M', AT) - • Hom(M ®M\N®N')

(2.6.2)

of .R-modules. At this point we encounter an ambiguity in our notation because f ® f


33

could denote either an element of Hom(M, N) ® Hom(M', N') or an element of Hom(M ® M ' , N ® AT'); indeed what the mapping (2.6.2) accomplishes is to change the former interpretation into the latter. In practice this double meaning does not cause problems. Usually / ® / ' means the homomorphism of M ® M' into N ® JV' and in any event the element of doubt can be removed by examining the context. The next exercise has been included to emphasize the difference between the two meanings of / ® / ' . Exercise 6. Let R be the ring Z/4Z and let I = 2Z/4Z so that I is an ideal of R. Put M = M' = R/I and N = N' = R. Show that in this case the homomorphism Hom(M, N) ® Hom(M', N') - • Hom(M ®M\N

® AT)

of (2.6.2) is neither an injection nor a surjection. We know, from Theorem 4 Cor., that tensor products are right exact. Thus if 0 —• E —• E —> E" —• 0 is an exact sequence of K-modules and M is an arbitrary ^-module, then the derived sequence M®F->M ®£-»M®£"->0 is exact. However, for some choices of M it can happen that the five-term sequence 0->M®F-»M®£->M®F'->0 is not exact. This observation leads on to the definition of an importance class of /^-modules. Definition. The R-module M is said to be 'flat' or 'R-flaf if whenever 0—• E' —-• E —• E" —> 0 is an exact sequence of R-modules the resulting sequences 0-^M®F-^M®£-^M®£'/-^0

(2.6.3)

and 0-»£'®M —£®M->F'®M->0 (2.6.4) are. exact. The reader should note that two sequences have been mentioned in the interests of symmetry. It is very easy to show, using Theorem 2, that if either one of (2.6.3) and (2.6.4) is exact, then so too is the other. An alternative (but equivalent) way of defining flat jR-modules is the following: M is flat if and only if M ® / and f ® M are injections whenever f is an injective homomorphism. Of course, if either / ® M or M ® / is an injection, then the other is an injection as well. This again is a consequence of Theorem 2. From Theorem 6 we see that all projective modules, and hence all free modules, are flat. It will appear presently that the class of flat modules can be strictly larger than the class of projective modules, but first we exhibit a non-flat module.

34


Exercise 7. Show that if the ideal I contains a non-zero divisor and I is different from R, then R/I is a non-flat R-module. The next two exercises concern straightforward properties of flat modules. Exercise 8. Show that an arbitrary direct sum of flat R-modules is a flat Rmodule. Exercise 9. Show that if M1,M2,...,Mq Ml®M2®'"®Mqisa flat R-module. A deeper result is contained in

are flat R-modules, then

Exercise 10*. Let M be an R-module with the property that every finitely generated submodule is contained in a flat submodule. Show that M itself is a flat R-module. It is now a simple matter to give an example of a flat module which is not projective. Such an example is provided by Exercise 11 *. Show that the field Q of rational numbers, when considered as a Z-module, is fiat but not projective. The remaining comments on Chapter 2 all have to do with covariant extension. We already know from Theorem 8 and its corollary that projective modules stay projective and free modules stay free when they undergo such an extension. As the next exercise shows, a similar observation applies to flat modules. Exercise 12. Let R^>Sbea homomorphism of commutative rings and let M be a flat R-module. Show that M ®R S is a flat S-module. The main text makes a brief mention of different ways in which covariant extensions can arise, but so far we have not examined in detail any special cases. This omission will now be put right. The simplest situation occurs when we have an ideal / of R and we use the natural ring-homomorphism R—+R/I. For an .R-module M the corresponding covariant extension is M ®R (R/I) and this we know is naturally isomorphic to the J?//-module M/IM (see Exercise 3). The reader will find it an instructive exercise to check that the isomorphism described in Chapter 1, Exercise 5 can be regarded as a special case of Theorem 9. Next let Xl9 X2,..., Xn be indeterminates. Then R is a subring of the Xn~\ and covariant extension by means of polynomial ring R[X1, X2,..., the inclusion homomorphism is referred to as adjunction of the indeterminates Xl9 X2,..., Xn. This can be looked at in a more down-to-earth manner.


35

Let M be an K-module and consider formal polynomials Z mvlv2...vHX\lXV22 ' ' ' ^n"'

where the coefficients mVlv2...Vn are taken from M. It is clear that we can add such polynomials and that in this way we obtain an abelian group. Indeed if Xn~\ can we denote the group by M[X l 9 X2,. • •, Xn~\, then M\_Xl9 X2,..., X n ]-module in which be regarded as an R[XU X2,..., (Here, of course, r e R and m e M.) It is now a straightforward exercise to verify that there is an isomorphism M[X19 X29...9Xn-]vM

®R R[X19 X29...,

XJ

2

Xvn» is matched with of R[Xl9 X29...9 Xj-modules in which mX\'X\ ... 2 m ® XyX^ ... XI*. Note that R[Xx, Xl9..., Xn~\ is a free and therefore flat ^-module. The final example is taken from the general theory of fractions. First we recall how fractions are formed from the ring R. Let Z be a non-empty multiplicatively closed subset of R, that is to say Z has the property that if ax, a2,..., an (n >0) belong to it, then axa2 ... an is in Z as well. Note that in allowing n = 0 as a possibility we are in fact assuming that the identity element of R is a member of the multiplicatively closed set. We now consider formal fractions r/a, where the numerator r is in R and the denominator a is in Z. Two such fractions, say rl/al and r2/a2, are regarded as the same and we write rxjax =r2/a2 if (and only if) aa2rx = aaxr2 for some <jeZ; in other words we introduce an equivalence relation but without using a special notation for the equivalence classes. It is then possible to define addition and multiplication of fractions (compatible with the definition of equality) so that r a

r' a'

a'r-\-ar' aa'

and r r' rr' a a' aa'' after which it is readily checked that the fractions form a commutative ring. This ring is known as the ring of fractions of R with respect to Z and in what follows it will be denoted by Z -1 (jR). Note that the identity element of Z " 1 ^ ) is 1/1, and that the fraction r/a is the zero element of Z " 1 ^ ) if and only if a'r = 0 for some a' e Z; also Z ~1 (R) is a trivial ring only when 0 is a member of Z. In general R is not a subring of Z " * (R) but we do have the canonical ring-

36


homomorphism K—• JL

\K)

{Z.O.J)

in which r ofR is mapped into the fraction r/1. Let us examine what happens when we use (2.6.5) to pass from modules over R to modules over Z "* (R). To this end suppose that M is an .R-module and let xeM ® /? Z~1(.R). Then x = mx ® Xx + m2 ® A2 + • * * + ms ® As, where mf- e M and A,- G Z " 1 (R). Now we can write the fractions Xt with a common denominator in Z, say ^. = r./cr, and then r

V

*

Thus each element of M ®R Z " l (R) can be written in the form m ® (1/cr), where m e M and a e Z . Exercise 13*. Le£ m belong to the R-module M and let o belong to the multiplicatively closed subset Z of R. Show that m®(l/<j) = 0 in M ® R Z ~ l (R) if and only if o'm = 0 for some o' e Z. The construction which produced the ring Z~ X (JR) from R can also be applied directly to an .R-module M. Thus we may consider fractions of the form m/cr, where meM and oeZ; and naturally we treat m1/Gl and m1JG2 as the same if oo2ml =aalm2 for some AX ®A2®-

- ®Ap

f

in which (al9 a2,..., ap, a\,a 2,..., a'p) is mapped into a±a\ ® a2a!2 ® • • • ®apa'p. This (Chapter 1, Theorem 2) induces a homomorphism Al®A2®'"®Ap®Al®A2®"-®Ap -+Al®A2®--®Ap

(3.2.1)

of ^-modules. Moreover (Chapter 2, Theorem 1) we have an Risomorphism ® Ap) p

® - ' ® A

p

(3.2.2)

in which, with a self-explanatory notation, (a1®--- ®ap)®(a'l® • • • ®a'p) corresponds to ax ® • • • ®ap®a\ ® • • • ®a'p. If therefore we combine (3.2.1) and (3.2.2) the result is a homomorphism (Ax ® - • • ® Ap) ® (Ax ®'' - ® Ap)-+ Ax ® A2 ® • • • ® Ap (3.2.3) in which (ax ® • • • ® ap) ® (a[ ® • • • ® a'p) is carried over into axa\ ® a2d2

®"-®apa'p.

We now define a mapping /*: {Ax ® - • • ® Ap) x {A^ ® - • • ® Ap) -+Al®A2®--®Ap

(3.2.4)

by requiring jx(x, x% where x and x' belong to Ax ® A2 ® • • • ® Ap, to be the image of x ® x' under the mapping (3.2.3). Evidently \i is bilinear and

Tensor products of algebras

45

= ala'l ® a2a'2 ® • • • ® apa'p. It follows that if x = al®-®afp, then

(3.2.5)

®ap, x' = a\®''

n

* ®a'p, and x = a'{®-'

rivix, x'\ x") = fi(x, ii{x\ x")).

(3.2.6)

But \i is bilinear. Consequently (3.2.6) holds when x, x' and x" are any three elements of Xx ® A 2 ® * * • ® Ap. We are now ready to turn Ax ® A2 ® • • • ® Ap into an ^-algebra. To do this we define the product of two of its elements, say x and x\ to be fi(x, x'). It then follows from (3.2.6) that multiplication is associative; and the bilinearity of \i ensures that it is distributive with respect to addition. Let et be the identity element of At. By (3.2.5) we have '" ®ep,a1®--

®ap)

and therefore for all x in Ax ® A2 ® • * • ® Ap. Thus A^ ® A2 ®' * * ® Ap is a ring with identity. Finally for r e R we have fi(rx, x') = rfi(x, xf) = fi(x, rx') and therefore A±® A2®- - - ® Ap\s &n /^-algebra. We summarize our conclusions so far. Theorem 1. Let Al9 A2,.. •, Ap be R-algebras. Then Al®RA2

®R'-®RAp

is an R-algebra, where the R-module structure is the usual one and the product - • ®ap and a\ ® a!2 ® • • • ® a'p is axa\ ®a2a'2 ® • • • ®apdp. of al®a2®It will now be shown that the results of Section (2.1) yield certain isomorphisms between algebras. We can deal with these fairly rapidly. Theorem 2. Let Au A2,...,AP there is an isomorphism

and BUB2, ...,Bq

be R-algebras. Then

A1®R'"®RAp®Bl®R"-®RBq « ( A t ®R• • • ®R Ap) ®R (Bx of R-algebras in which al®---®ap®bl®--'®bq (a1®--®ap)®(b1®'-®bq).

®R'"®RBq) is associated with

Proof By Theorem 1 of Chapter 2, there is an isomorphism /, of R-modules, that maps Ax® • • • ®Ap®B1® • • • ®Bq onto (Ax® • • • ®Ap)®(B1® • • • ®B ) so that

46

Associative algebras f(a1®'-®ap®bl®--®bq) = K ® • • • ® ap) ® (bx ® • •

Let x = ax ® • • • ® a p ®xbx ® • • • ® bq and x' = b'q. Then, by Theorem 1, xx' = ala'l ® • and therefore /(**') = ( a ^ i ® • • • ® apa'p) ® ( b ^ ; ® • • • ® bgb;) But / is R-linear. Hence if y and / a r e any two elements of Ax ® • • • ® 4 p ® #i ® • • • ® Bq, then f(yy') = f(y)f(y')- The theorem follows because / is bijective. Before we leave this result we note that the argument used to deduce the corollary to Theorem 1 of Chapter 2 now shows that there is an isomorphism (Al®RA2)®RA3*A1®R(A2®RA3)

(3.2.7)

of K-algebras in which (al ® a2) ® a3 corresponds to al ® (a2 ® a3). The next theorem is derived, in much the same way, from Theorem 2 of Chapter 2. The details are left to the reader. Ap be R-algebras and let (i l5 i 2 , . . . , ip) be a Theorem 3. Let Al9A2,..., permutation of ( 1 , 2 , . . . , p). Then there is an isomorphism Al®RA2®R--®R

Ap&Atl

®R Ah ®R"- ®R Aip

of R-algebras in which ax ® a2 ® • • • ® ap corresponds to aix ® aiz ® • • •

v It will be recalled that # itself is an K-algebra. Theorem 4. L^f A be an R-algebra. Then there is an isomorphism R ®RA Â of R-algebras which matches r ®a with ra. There is a similar algebra-isomorphism A ®RRzzA, where a ®r is matched with ra. Proof We need only consider the first assertion. By Theorem 3 of Chapter 2, there is an isomorphism f:R®RA—+A of K-modules for which f(r®a) = ra. That / is compatible with multiplication is clear because (r1a1)(r2a2)=(r1r2)(a1a2). We next examine tensor products in relation to homomorphisms of algebras. To this end let A1,A2,..., Ap and Bx, B2,..., Bp be .R-algebras and suppose that we are given algebra-homomorphisms f: At—>Bi for i = 1,2,...,/?. From Section (2.2) we know that / i ® fi ® ' " ' ® fP- Ax ®R A2 ®R • • • ®R Ap -^Bl®RB2®R-"®RBp (3.2.8)

Graded algebras

47

is a homomorphism of R-modules. We claim that in the present instance it is actually a homomorphism of R-algebras. For it is clear that fx ® fi ® ' ' * ® fp takes identity element into identity element. Now suppose that x and xf belong to A1 ® A2 ® • • • ® Ap. It suffices to show that the image of xx' is the product of the separate images of x and x'. Indeed it is enough to prove that this is so when x and x' have the special forms al ® a2 ® • * * ® ap and a\ ® a'2 ® • • • ® a'p. But in these circumstances what we wish to prove is obvious. 3.3

Graded algebras Let A be an R-algebra and {An}nel a family of R-submodules of A indexed by the integers. The family is said to constitute a grading on A provided that (i) A=Ân

(As.),

neZ

and (ii) areAr and aseAs together imply that

araseAr+s.

An K-algebra with a grading is known as a graded R-algebra. The elements of An are said to be homogeneous of degree n. Thus condition (ii) says that the product of two homogeneous elements is again homogeneous, and the degree of the product is the sum of the degrees of the two factors. We shall frequently meet with situations where An=0 for all n\, each element of Ap is a sum of products of p elements of Ax. Furthermore Ao is generated, as an R-module, by the identity element 1A and therefore Ao is contained in the centre of A. Proof All the assertions become obvious as soon as it is noted that (i) A is generated, as an R-module, by products of elements of Ax (including the empty product), and (ii) such a product is homogeneous of degree p, where p is the number of factors. We need some further terminology. Let A and B be graded jR-algebras. A mapping f:A—>B is called a homomorphism of graded algebras if it is a homomorphism of /^-algebras which preserves degrees, i.e. if it satisfies f(An)^Bn for every n. (Here {An}neI and {Bn}neZ are the respective gradings.) By an isomorphism of graded algebras we naturally mean a bijective homomorphism of graded algebras. If f:A—>B is such an isomorphism, then, for each n, f maps An isomorphically onto Bn. Now let if be a two-sided ideal of a graded R-algebra A. Definition. The two-sided ideal H is called 'homogeneous' if whenever aeH all the homogeneous components of a belong to H as well. Lemma 2. Let H be a two-sided ideal of a graded R-algebra A. Then H is homogeneous if and only if it can be generated by homogeneous elements. Proof. If H is homogeneous, then it is certainly generated by the homogeneous components of the various members of H. Thus H has a homogeneous system of generators. Conversely suppose that Q is a set of homogeneous elements and that Q generates H. Then each element of H is a sum of elements of the form axb, where a, be A and x e Q. Indeed we may suppose that a and b are themselves homogeneous. Thus if a e H, then a is a sum of homogeneous elements of H and therefore the homogeneous components of a belong to H. This completes the proof. Now suppose that H is a homogeneous two-sided ideal of the graded Ralgebra A. Put Hn = HnAn.

(3.3.1)

Then H=YjHn

(As.).

(3.3.2)

Graded algebras

49

We already know that AjH — B say is an K-algebra. Suppose that we put BH = (AH+H)/H.

(3.3.3)

Then Bn is an R-submodule of B and

We claim that the sum (3.3.4) is direct. To see this suppose that £„ bn = 0, where foneBn and there are only finitely many non-zero summands. For each n e / w e choose an e An so that bn = an + //, taking care to arrange that an = 0 whenever bn = 0. Then £„ an belongs to / / and therefore, because H is homogeneous, all the an are in H. But this means that all the bn are zero. Accordingly our claim is established. Clearly the product of an element of Br and an element of Bs is an element of Br+S. Hence the R-algebra A/H is graded by its submodules {(An +H)/H}neI and the natural mapping A —+A/H is a homomorphism of graded algebras. This mapping induces, for each n, a homomorphism of An onto Bn whose kernel is An • • • JP) is a second sequence of integers, then for XEAJ and y e i j w e have where / + J = (ix +jx, i2 +j2,..., ip +jp).

50

Associative algebras

It will be convenient to put \I\ = ii+i2 + ~' + ip

(3.3.10)

and

? \l\ = n

R"-

®RA\PK

(3.3.11)

\l\ = n

Then as a consequence of (3.3.9) we have A=YaAn

(d.s.)

(3.3.12)

nel

and it is clear that the product of an element of An and element of Ak belongs toAn+k. Let us sum up. It has now been shown that if A(1\ A{2\..., A{p) are {1) i2) {p) graded K-algebras, then A ®RA ®R- • • ®RA is also a graded Ralgebra, where the module of homogeneous elements of degree n is

We call {An}weZ the total grading on the tensor product. Evidently if the gradings on A(1\ A{2\ . . . , Aip) are non-negative, then the total grading on A{i) ® A{2) ® • • • (x) Aip) is non-negative as well. We next review some of our basic isomorphisms from the standpoint of the theory of graded algebras. Suppose then that A{1\ A{2\ . . . , A(p) and £ (1) , B{2\ . . . , Biq) are graded K-algebras. It will be recalled that Theorem 2 provides an explicit isomorphism ^(A(1)

®R-"

®RA(P)) ®R (£ (1) ®R- • • ®RB™)

(3.3.13)

of ordinary, that is ungraded, /^-algebras. However, on this occasion each side of (3.3.13) is a graded algebra and it is clear (from the way the isomorphism operates and the way the gradings are defined) that the degrees of homogeneous elements are preserved. Accordingly what we have is an isomorphism of graded algebras. Let us consider Theorem 3 in a similar way. To this end suppose that (A*i> M2> • • •» Vp) i s a permutation of ( 1 , 2 , . . . , p). Then the isomorphism A(1)®RAi2)®R'"®RAip) «i4 0AI+J,

w h e r e / + J = (ix -\-j1, i2 +j2, . . . , i p + ; ' ) . P u t (3.4.2)

JV(J,J)=£y. and

r>s

(3.4.3) We are now ready to describe the new or modified multiplication.

52


Suppose that

I

J

are elements of A. Then the modified product of a and a' is defined to be 5>(/,JWa,,a;).

(3.4.4)

IJ

Thus if ait ® ai2 ® • • • ® aip belongs to Aj and a^ ®fl}2® • • • ® a}p belongs to Xj, then, whereas their ordinary product is aixa'h ® «l2«}2 ® * • * ® af a) , their modified product is e(/, J ) ^ ^ ® al2fl}2 ® • • • ® aipa'jp.

(3.4.5)

Clearly modified multiplication is distributive with respect to addition on both sides, and e{1) ® e(2) ® • • • ® e(p\ where e(r) is the identity element of A(r\ is still neutral for the new multiplication. Also the property that corresponds to (3.1.1) continues to hold. Hence in order to check that we still have an R-algebra it suffices to verify that modified multiplication is associative. Let

I = (il9i2>->>ip)>

J = UiJ2>-

• • JP)

a n d

K = (kl,k2,...

,kp) b e

sequences of integers and suppose that xeAj, yeAj and zeAK. It will suffice to verify the associative law in the case of x, y and z. Indeed we may go further and assume that x = aix ® • • • ® at , y = a'jx ® • • • ® a) and z = 0*, ® ''' ®a'k • ^ u t n o w 5 i n y i e w °f (3.4.5) what we wish to establish will follow if we show that 9

K).

However, this is clear because e(/ + J, K) = s(I, K)e(J, K) and e(/, J + K) = e(/, J)£(/, K). The new K-algebra will be called the modified tensor product of A{1\ A{2\ . . . , A{p\ It will be denoted by Ail) ®K A{2) ®R'"®R A{p) (3.4.6) {1) i2) ip) in order to distinguish it from A ® A ® • • • ® A . Note that the total grading on A{1) ® A{2) ® • • • ® Aip) also serves as a grading on A{1) ® A(2) ® • • • ® Aip\ Finally we recall that the two tensor products have the same identity element; and that for x e At, y e A3 the modified product of x and y is e(/, J) times the ordinary product. The new tensor product is important in the theory of exterior algebras. In the remainder of this section we shall prepare the way for its application. Theorem 6. Let A{1\ A(2\ . . . , Aip) and B(1\ B(2\ . . . , B(q) be graded Ralgebras. Then R

A™ ® R Bw ® R • • • ® R B(«»

(3.4.7)

A modified graded tensor product

53

and (A^

®R'--®R

A^) ®R (B(1) ®R--®RB™)

(3.4.8)

are isomorphic graded R-algebras under an isomorphism which matches a{1)® • • • ®aip)®b{1)® • • • ®biq) with (a{1)® • • • ®aip))®

Proof. We already know that there is an isomorphism / , of graded algebras, which maps A{1)® • • • ®Aip)®B{1)® • • • ®Biq) onto il) {p) {1) (A ® • • • ® A ) ® (B ®" • ® Biq)) and matches elements in the way described. It is therefore enough to show that / behaves properly with respect to modified products. Suppose then that / = (/1,f2,... ,/p), V=(vl9v2,...9vp), J= an( O'i»7*2» • • • Jq) i W = (wls w 2 , . . . , wq) are sequences of integers. We put IJ= (il9..., ip,jl9... Jq) and define VW similarly. We also put \J\=Ji + J2 + '"+Jq a n d \V\=Vi+V2 + '-+Vp. Let x = aii®--®aip®bji®---®bjq and y = ocVi® • • • ®ocVp® PWl®'-'® PWq belong "to A^ ®---® A\p) ®Bfi)®---® Bf and A™ ®'"®A[p)p® B{^1 ®--®Biq)q respectively. The modified product of x and y is e(/J, VW)aticcVi ® • • • ® aipocVp

q q

and the image of this under / is e(IJ,VW)(aiixVi®---®aip(xVp) ®(bjxvi®'"®bjqpwq).

(3.4.9)

Thus the theorem will follow if we can show that (3.4.9) is the product o f (ati ®-- ®aip)®(bji ®-- ®bjq) a n d (a y .(x) • • • ®<xVp)® (PWl ®- " ®PWq) in (3.4.8). However, this latter product is where { is the product of ati ®- • • ® a{ and ocVi ® • • • ® ocVp in A{1) ® A{2) ® • - - ® A{p\ and n is the product of bh ® - • • ® bjq and PWl®" ' ® Pw in B{1) ® B(2) ®" ' ® B{q). Accordingly is just Vx®--

'®aipav)

and therefore the proof will be complete if we show that e(/J, VW) = (-l)lJme(I, VW, W). However, this follows readily from (3.4.2) and (3.4.3).

54


We make one final observation concerning modified products. Suppose that ft:A®^B®

(* = l , 2 , . . . , p )

are homomorphisms of graded algebras. Then of course / x ® f2 ® • • * ® fp is also a homomorphism of graded algebras. But modified and unmodified tensor products have the same ^-module structure. Hence the same mapping fx ® f2 ® * • • ® fp is a homomorphism ®RB{2) ®R'"

®RBip)

(3.4.10)

of K-modules. This certainly preserves degrees and identity elements. We claim that in fact (3.4.10) is a homomorphism of graded algebras. To see this let x and y belong to A(1) ® A(2) ® • • • ® Aip). We have to show that the image of their modified product is the modified product of their images. For this we may assume that x = aix ® at2 ® • • • ® at and y = aji ® ocJ2 ® • • • ® (Xj belong to A^ ® • • • ® A\p) and Af^ ® • • • ® A^p) respectively, where we have made use of our usual notation. However, in this case the desired result follows from the formula for the modified product (see (3.4.5)). 3.5

Anticommutative algebras Let {An} neZ be a grading on an K-algebra A. We say that the graded algebra is anticommutative provided anam = ( - i P a m a B

(3.5.1)

whenever am and an are homogeneous elements of degrees m and n respectively, and provided also that a2m = 0

(3.5.2)

whenever the degree of am is odd. Thus we see that in an anticommutative algebra the square of a homogeneous element of degree one is zero. As the next lemma shows there is an important special situation where the converse holds. Lemma 3. Let the graded R-algebra A be generated (as an R-algebra) by its module AY of homogeneous elements of degree one. If now a 2 = 0 for every aêA^ then A is anticommutative. Proof From Lemma 1 we see that An = 0 for all n 1 and n > 1. Note that if x, y e A x, then from (x + y)2 = 0,

Anticommutative algebras

55

= 0 and y2 = 0 it follows that 0.

(3.5.3)

Now suppose that u = xxx2 . . . xm, ^ = ^1^2 • • • >*> where every xt and ^ belongs to Av Then, by (3.5.3), vu =

yly2...ynx1x2...xm

= ( - i r x ! X 2 . . . x m y ^ 2 . . . yn = (- l)mnuv. But every element of Am (respectively An) is a sum of elements such as u (respectively v) so the formula (3.5.1) holds quite generally. From here on we suppose that m> 1 and is odd. Let am belong to Am. Then am = ux + u2 + * • • + us, where each ut is a product of m elements of A x, and w? = 0 as is clear from (3.5.3) and the second hypothesis of the lemma. Again, by what has just been proved,

(-l)m2uiUj=-uiuj

ujui =

and now it follows that a^ = 0. Thus the lemma is proved. Theorem 7. The modified tensor product of a finite number of anticommutative R-algebras is itself an anticommutative R-algebra. Proof Let A and B be anticommutative .R-algebras. If we can show that A ® B is anticommutative, then the general result will follow by virtue of Theorem 6. Let x = ar®bs,y = ocp®Pa, where ar, ap are homogeneous elements of A and bs, fia are homogeneous elements of B, and where the degree of each element is indicated by its suffix. If x * y denotes the modified product of x and y, then On the other hand so that But in A ® B the element x has degree r + 5 = m say whereas y has degree p + o = n say. Consequently y * x = (— \)mnx * y. Now suppose that m is odd. Then one of r and s is odd and therefore the modified square of x namely {-\)sr

+

r2+s2 2

a r®b2s

is zero. Thus for A ® B we have verified the anticommutativity conditions for homogeneous elements of a special form. That these conditions hold for general homogeneous elements follows by linearity.

56


Covariant extension of an algebra The process of covariant extension as applied to modules was explained in Section (2.5). Here we shall examine it in relation to algebras. Let R, 5 be commutative rings and assume that we are given a ringhomomorphism 3.6

a):R-+S

(3.6.1)

that preserves identity elements. Assume further that A is an R-algebra. From Section (2.5) we know that A ®R 5 is an 5-module. On the other hand 5 is an R-algebra (with w as its structural homomorphism) so that A ®R 5 is an K-algebra and in particular it is a ring. If we examine the 5-module structure and the ring structure on A ®R 5 we find that they interact in such a way that we have an 5-algebra. Thus the covariant extension of the Ralgebra A is the S-algebra A ®R 5. Note that the structural homomorphism 5—• A ®RS maps s into lA ®s. Now suppose that {An}nel is a grading on A. By Theorem 5 of Chapter 2, A®RS=YJ(An®RS)

(As.).

(3.6.2)

nel

Of course according to the theorem just quoted the right-hand side is to be understood as a direct sum of ^-modules. However, A ®R 5 and An ®R 5 are 5-modules so that (3.6.2) is also a direct sum of 5-modules. Moreover the product of x in Am®RS and y in An®RS belongs to Am+n®RS. Accordingly the 5-algebra A ®RS is graded by {An ®RS}neZ. Let us turn our attention to the tensor product, over R, of R-algebras Al9A2,..., Ap. (For the moment these algebras need not be graded.) By Theorem 9 of Chapter 2, there is an isomorphism (Ax ®R 5) ®s {A2 ®RS)®S--' *{Al®RA2®R'-®R

®s (Ap ®R 5) Ap) ®R 5

(3.6.3)

of 5-modules in which (ax ® s t ) ® (a2 ® s2) ® • • * ® (ap ® sp) corresponds to (al ® a2 ® • • * ® ap) ® s1s2 . . . sp. This isomorphism is compatible with multiplication and therefore (3.6.3) is actually an isomorphism of 5algebras. Thus, for algebras, covariant extension commutes with the formation of tensor products. Finally suppose that each of the K-algebras Al9 A2, . . . , Ap is graded. Then the two sides of (3.6.3) are graded 5-algebras and it is easy to see that the isomorphism preserves degrees. Hence when Al9A2,. • . , Ap are graded R-algebras (3.6.3) is an isomorphism of graded S-algebras. 3.7

Derivations and skew derivations Let {An} neZ be a grading on an i^-algebra A and let k be an integer.

Derivations and skew derivations

57

A mapping h: A —• A is said to be of degree k if it raises the degree of each homogeneous element by this amount. By a derivation on A we shall understand an #-homomorphism D: A-+A, of degree — 1, which satisfies D(aP) = (Doc)P + (x(DI3)

(3.7.1)

for all a, j8 in A. Note that if D is a derivation, then n

D(a x a 2 . . . a j = £ a x . . . ai _1(Z)a i)ai + 1 . . . an.

(3.7.2)

i=l

Suppose that £>, D' are derivations on A and that reR.lt is easily checked that the endomorphisms D + D' and rD of A are themselves derivations. In fact the derivations on A form an tf-module. Of course D°Df is an endomorphism of A9 but (usually) it is not a derivation. Lemma 4. Suppose that the graded R-algebra A is generated (as an Ralgebra) by its homogeneous elements of degree one, and let D, D' be derivations on A. Then D' ° D = D° D'.If D and D' agree on all homogeneous elements of degree one, then D = D'. Proof Let a1? a 2 , . . . , an, where n>2, belong to Av Then D<xt and D'OL{ belong to Ao and are therefore in the centre of A. Also (Df ° D)OL{ = 0 because As = 0 for all s < 0 . Next, from (3.7.2) we have (Ly°D)(<x1ai2...aH) 7)

+ D(a j )D'(a l .))a 1 ... a f . . . a , . . . aw,

where the symbol A indicates that the term over which it is placed is to be omitted. It follows that

... aJ r

so that D °D = D°D'as stated. The final assertion of the lemma also follows from (3.7.2). Derivations play an important role in the theory of commutative algebras. For anticommutative algebras the concept needs to be modified. By a skew derivation on A we shall mean an .R-homomorphism A: A —> A, of degree — 1, which satisfies A(a/J) = (Aa)j8 + ( - l)ma(A£)

(3.7.3)

for all a e Am and ft e A. The skew derivations on A also form an R-module. Lemma 5. Let A be a skew derivation on the graded R-algebra A and let o^, a2, . . . , 0Ln be homogeneous elements of degree one. Then

A ( a i a 2 . . . a n ) = £ ( - l )i +1ai...(Aaf)...aB.

(3.7.4)

58


Proof. Since

. an) = (A(a1a2 ... a^^K + t - l)" the lemma follows by induction. Lemma 6. Let the graded R-algebra A be generated (as an R-algebra) by its homogeneous elements of degree one, and let A, A' be skew derivations on A. Then A ° A = 0 and A' ° A + A ° A' = 0. / / A and A' coincide on the module of homogeneous elements of degree one, then A = A'. Proof Let a l5 a 2 , . . . , an be homogeneous elements of degree one. Then Aa^ belongs to the centre of A. Two applications of (3.7.4) now show that (A ° A)(axa2 . . . an) = 0 whence A ° A = 0. This proves the first assertion and the second follows by expanding (A + A') ° (A + A') = 0. The final assertion is clear from (3.7.4). Derivations and skew derivations both behave well in relation to the process of covariant extension. To be precise let co: R—+S be a homomorphism of commutative rings and let D (respectively A) be a derivation (respectively skew derivation) on a graded K-algebra A. Then D ®S (respectively A®S) is a derivation (respectively skew derivation) on the graded S-algebra A ®R S. 3.8

Comments and exercises Perhaps the first comment to be made concerning the notion of an algebra is that it subsumes more than appears at first sight. Thus if Q is any ring with an identity element, then Q is an algebra over the integers. Here the structural homomorphism Z—>Q maps the integer n into nln. Additional comments, supplemented by exercises, are to be found below, and, as in Chapters 1 and 2, the exercises for which solutions are provided are marked with an asterisk. Xn are indeterminates, then the We begin by noting that if Xl9 X2,..., polynomial ring K[X l 9 X2,..., Xn~] is a commutative .R-algebra. Exercise 1. Show that an R-algebra A can be generated (as an algebra) by a single element if and only if it is a homomorphic image of the algebra R[X~], where X is an indeterminate. Observe that this exercise shows that an algebra which can be generated by a single element is necessarily commutative. Now let £ be a multiplicatively closed subset of R. We explained in Section (2.6) how the ring E " 1 ^ ) of fractions is constructed and we noted that there is a ring-homomorphism R —• Z " * (R) in which r of R is mapped into the fraction r/1. As a result of this homomorphism £ " x (R) becomes a commutative .R-algebra.


59

This said, let S x and E 2 be two multiplicatively closed subsets of R. Then S f 1 ^ ) and Z^" 1 ^) are ^-algebras so we can form their tensor product (3-8.1) and this too will be a commutative R-algebra. On the other hand, the set of all products CXG2, where (T1enL1 and 0 let An be the K-submodule generated by all words with n letters, and put Aq = 0 when q 1. However, this implies that jS = 0 because p is a sum of homogeneous elements of strictly positive degrees. Thus a = a 0 and the solution is complete. Exercise 5. Let A and B be graded R-algebras. Show that the twisting isomorphism T: A®B^*B®A produces an isomorphism T:A®B

-^->B®A

of graded algebras. Solution. Let x and y belong to A (g) B. Since T preserves degrees it will be enough to show that T(x *y)= T(x) * T{y), where an asterisk has been used to indicate a modified (as distinct from an ordinary) product. But T is an isomorphism of ^-modules. We may therefore confine our attention to the case where x = ar®bs and y = ap(x)j8ff; here ar9 ccp denote homogeneous elements of A and bs, Pa homogeneous elements of B, the degrees of these elements being indicated by their suffixes. On this understanding x*y = ( - l)sparocp ® bspa and therefore

p

On the other hand T(x) = ( - l)rsbs ® ar and T(y) = {- \)paPa ® ocp. Consequently

= T(x * y) which is what we were seeking to prove. Exercise 6. Let A(1\ A(2\ . . . , Aip) be graded R-algebras (sl9 s2,.. •, sp) be a permutation of (1, 2 , . . . , p). Show that

and let

Solutions to selected exercises

67 (3.9.1)

and A{Sl) ® A^ ® •• • ® A{SJ

(3.9.2)

are isomorphic graded algebras. Solution. Suppose that \A' and k'\A'—>A such that X o (f) = (j)' and Xf °cf)f = (j). Thus the problem has (essentially) at most one solution. Theorem 1. The universal problem possesses a solution. If (A,(f)) is any solution, then (i) (ii) (iii) (iv)

\M—*A is an injection; A is generated, as an R-algebra, by (j){M); there is a grading {An}neI on A with Ax =4>(M); for p>l, Ap is the p-th tensor power of M with ml ® m2 ® • • •

(v) the structural homomorphism R—>A maps R isomorphically onto Ao. Remarks. Before starting the proof it may be helpful if certain points are clarified. For instance it should be noted that, once (ii) has been established, Lemma 1 of Chapter 3 ensures that there is at most one grading on A that has the property described in (iii). To amplify (iv) we observe that the mapping of the p-fold product MxMx- xM into Ap which takes (m 1 ,m 2 ,... ,m p ) into (m1)(mp) is multilinear. The theorem asserts that Ap and this mapping together solve the universal problem for multilinear mappings of M x M x • • • x M (see Problem 1 of Chapter 1). Proof. It suffices to exhibit one solution to the problem which satisfies (i)-(v). To this end, for p > 1, put Ap = M ®M ® • • • ® M where there are p factors. Thus Ax =M. We also put A0 = R and define an K-module A by ZP Our immediate concern will be to turn A into an R-algebra. For p > 1 and q > 1, Theorem 1 of Chapter 2 provides an isomorphism Ap(x) AqzzAp+q. Next, from Theorem 3 of the same chapter we obtain isomorphisms Ao ® Aq&Aq and Ap® A0&Ap. (Note that when p = q = 0 the latter isomorphisms coincide.) Thus for every p > 0 and q > 0 we have an explicit isomorphism Ap ® Aq % Ap+q and hence there is a bilinear mapping Vpq:ApxAq-+Ap+q

(4.1.1)

in which iipq(xp, yq) is the image of xp ® yq under the isomorphism in question. Accordingly if m1,m2,..., mp and m\, m' 2 ,..., m'q belong to M, then p®m'l®---®mq.

On the other hand if reA0 = R, then we have

(4.1.2)

The tensor algebra yq) = ryq

71 (4.1.3)

and ^po(xp,r) = rxp,

(4.1.4)

where now xpeAp, yqeAq and p>0, q>0. Suppose next that p>0, g > 0 , t>0 and let xp, yq, zt belong to Ap, Aq, At respectively. Using (4.1.2), (4.1.3) and (4.1.4) it is a straightforward matter to verify that Hp+qAftpqiXp* y zt) = fip,q + t(xP, t*qt(yP> zt))-

(4.1-5)

We are now ready to define multiplication on A. Let x,yeA. These elements have unique representations in the form x = x0 + xx + x2 + • • • and y = yo + y1+y2 + " ', where xn, yn belong to An and the sums are essentially finite. The required product of x and y is then fi(x, y), where Kx,y)=

Z

»pq(xP,yq).

p>0,q>0

It is readily checked that multiplication is distributive (on both sides) with respect to addition, and from (4.1.5) it follows that it is associative as well. Next IK belongs to Ao and fi(lR, y) = y, [i(x, lR) = x by virtue of (4.1.3) and (4.1.4). Finally if reR, then fi(rx9 y) = rfi(x, y) = /i(x, ry).

These observations taken together show that A is an (associative) Ralgebra, and that 1^, considered as an element of Ao, is its identity element. For n < 0 p u t An = 0. Then {An}nel is a grading on A, and, since M = Al9 M is a submodule of A. Let (/>: M —• A be the inclusion mapping. Evidently A and (/> satisfy conditions (i)-(v). The proof will therefore be complete if we show that A and 4> solve the problem which was enunciated at the beginning of the section. Assume then that \I/:M—+B is an K-linear mapping of M into an ^-algebra B. For p>\ there is a multilinear mapping of the p-fold product MxMx-'xM into B that takes (m 1 ,m 2 ,... ,mp) into ^ ( m i ) ^ ^ ) . . . i//(mp), and this will induce a homomorphism hp: Ap—+B, where hp{m1 ®m2 ®'''

®m p ) = ^(m 1 )^(m 2 )... i//(mp).

Since A0 = R, we may take h0: Ao—>B to be the structural homomorphism R—>B. But A is the direct sum of all the An with n > 0. Consequently there is an K-homomorphism h: A-^B which agrees with hn on An. Certainly h preserves identity elements. Also if x and y are homogeneous elements, then, using whichever of (4.1.2), (4.1.3) or (4.1.4) is appropriate, we can verify that h(xy) = h{x)h(y). It follows that h: A-+B is a homomorphism of Ralgebras.

72

The tensor algebra of a module

Finally for XGM = AX we have h(x) = \j/(x) and therefore h°(ft = \l/. But M = Ai generates A as an R-algebra and so there can only be one algebrahomomorphism (of A into B) which when combined with (ft gives \ft. Consequently the proof is complete. These conclusions will now be cast into a different form. Suppose that (A, (ft) solves our problem. Put T(M) = A

(4.1.6)

and denote by {Tn(M)}neZ the grading on T(M) that results from Theorem 1. Since (ft maps M isomorphically onto Tt (M) we can use this isomorphism to identify the two modules. Thus TX(M) = M

(4.1.7)

so that T(M) contains M as a submodule. Indeed by this device the notation is greatly simplified. We know that, as an K-algebra, T(M) is generated by Tl(M) = M so that the grading is non-negative. Also the structural homomorphism maps R isomorphically onto T0(M). Hence, when it suits us, we can identify T0(M) with R. Finally, if ® is used to denote multiplication on T(M), then, for p > 1, Tp(M) is the p-th tensor power of M as defined in Section (1.3). It is because of this last property that T(M) is called the tensor algebra of M. Our next theorem merely restates, but now in the new terminology, that the tensor algebra solves the universal problem with which we started. Theorem 2. Let T(M) be the tensor algebra of the R-module M, and let \//: M—> B be an R-linear mapping of M into an R-algebra B. Then \\i has a unique extension to an algebra-homomorphism of T(M) intoB. 4.2

Functorial properties Let /:M—>iV be a homomorphism of K-modules and let T(M), T(N) be the tensor algebras of M and N respectively. Since N is a submodule of T(N), Theorem 2 shows that / has a unique extension to a homomorphism T(f):T(M)^T(N) of K-algebras. Note that if ml, m 2 , . . . , mp belong to M, then TifUnti ® m2 ® • • • ® mp) = f(m1)®f(m2)

® • • • ®f(mp). (4.2.1)

Clearly if N = M and / is the identity mapping of M, then T(f) is the identity mapping of T(M). We next observe, as a consequence of (4.2.1), that a homogeneous element of degree p > 1 keeps its degree when T(f) is applied. Moreover T(f) maps T0(M) isomorphically onto T0(N). Consequently T(f) is actually a homomorphism of graded algebras and therefore, for each seZ, it induces a homomorphism

Functorial properties

73

Ts(f):Ts(M)^Ts(N)

(4.2.2)

of JR-modules. In fact, as is shown by (4.2.1), when p> 1 we have Tp(f) = f®f®'--®f

(p factors)

(4.2.3)

and to this we may add the observation that T o (/): T0(M)-+ T0(N) is the identity mapping when we make the identifications T0(M) = R and T0(N) = R. Next suppose that we have #-homomorphisms f:M—>N and g:K-^M. Evidently T(f)oT(g)=T(fog) (4.2.4) and hence, for each s e Z, Ts(f)°Ts(g) = Ts(f°g).

(4.2.5)

Thus in the language of the Theory of Categories T(M) is a covariant functor from /^-modules to graded K-algebras. Note that if / : M —• N is an isomorphism, then T(f) is also an isomorphism and T ( / - 1 ) is its inverse. This is because T{f~l)°T{f) and T{f)°T{f~l) are identity mappings. Now let K be a submodule of the K-module M. Since T1(M) = M, K consists of homogeneous elements of T(M) of degree one. Accordingly K generates a homogeneous two-sided ideal of T(M). Theorem 3. Let f\M^>N bea surjective homomorphism of R-modules and let K be its kernel. Then T(f): T(M)-+T(N) is surjective and its kernel is the two-sided ideal that K generates in T(M). Proof. It is clear from (4.2.1) that T(f) is surjective. Let I(K) be the twosided ideal generated by K and put Ip(K) = I(K) n Tp(M). Then, for p > 1, Ip(K) is the submodule of Tp(M) generated by all products mx (x) m2 ® • * • ® mp, where mi, m 2 , . . . , mp belong to M and at least one m{ is in K. Since Tp(f) = f ® / ® • • • ® /, it follows (Chapter 2, Theorem 4) that Ip(K) is the kernel of Tp(f). Consequently the kernel of T(f) is as required. We shall now give a second proof that the kernel of T(f) is I(K). This alternative proof has the advantage that it can be adapted to deal with similar situations that arise in the study of other algebras. be the natural homomorphism of graded Let h: T(M)^T(M)/I(K) algebras. Since T1(M) = M and 71(K) = X, the submodule of T(M)/I{K) formed by the homogeneous elements of degree one can be identified with M/K, and then, in degree one, h induces the natural mapping M-^M/K. On the other hand, / induces an isomorphism of M/K onto N. Let g:N ^> M/K be the inverse of this isomorphism. We now have an K-module

74


homomorphism N-+T(M)/I(K) and this, by Theorem 2, extends to an algebra-homomorphism T(N)-+ T(M)/I(K). Let us combine the algebrahomomorphism T(N)^> T(M)/I(K) with T(f): T(M)-> T(N). The result is an algebra-homomorphism T(M)-+ T(M)/I(K) which maps m of 7i(M) into h(m). But Tx (M) generates T(M) as an R-algebra, so in combining the two homomorphisms we have recovered h: T(M)—+ T(M)/I(K). It follows that the kernel of T(f) is contained in the kernel of h, that is to say it is contained in I(K). The proof is now complete because the opposite inclusion is trivial.

4.3

The tensor algebra of a free module In Section (4.3) we shall assume that R is non-trivial. (This is to avoid certain tiresome and unimportant complications.) Let M be a free Rmodule and B one of its bases. We wish to describe the structure of T(M). We know that T0(M) is a free R-module having the identity element of T(M) as a base. Also, by Theorem 3 of Chapter 1, if p > 1, then Tp(M) is a free R-module and it has as a base the set formed by all products bx ® where bt e B. Accordingly T(M) is itself a free R-module and b2®---®bp, the totality of products b1 ® b2 ® • • • ® bn, where b(eB and n > 0 , constitutes a base. To complete the description of the algebra all we have to do is explain how two elements in our base are to be multiplied. But here there is no problem because the product of bx ® b2 ® •' * ® bn and b\ ® b'2 ® ''' ® K is simply the basis element bx ® • • • ® bn ® b\ ® • • • ® b'x. Thus T(M) is what is called the free algebra generated by the set B. The next theorem gives another characterization of T(M) when M is free. Theorem 4. Let M be a free R-module with a base B and let (ft: £ —> A be a mapping of B into an R-algebra A. Then has a unique extension to a homomorphism T(M)—*A of R-algebras. Remark. This result shows that T(M) solves a certain universal problem involving mappings of B into an R-algebra. We leave the reader to make this precise. Proof The mapping 0 : B—>A can be extended to an R-linear mapping (j>: M^>A. Let h: T(M)—+A be a homomorphism of R-algebras. Then h extends if and only if it extends (/>. However, we know (Theorem 2) that there is precisely one h with the latter property. Theorem 5. Let P be a projective R-module. Then the tensor algebra T(P) is projective as an R-module. Consequently (for all n) Tn(P), since it is a direct summand of T(P), is also projective.

Covariant extension of a tensor algebra

75

Proof. Choose an ^-module Q so that P ®Q = F (say) is a free K-module. Let G\ P—>F be the inclusion mapping and n: F—>P the projection onto the first summand. Then T(F) is a free K-module (because F is free) and the mappings T(o)\ T(P)-+T(F) and T(n): T(F)-+T{P) are IMinear. But n ° a is the identity mapping of P and therefore T(n) ° T(M and v: M—+N be homomorphisms of R-modules such that v°u is the identity mapping of N. Then u is an injection, v is a surjection, and M is the direct sum of u(N) and the kernel, Ker v, of v. In particular N is isomorphic to the direct summand u(N) of M. Proof It is obvious that u is an injection and v a surjection. Let meM. Since m=(m — u(v{m))) + u(v(m)) and m — u(v(m)) is in Ker v, it follows that M = Ker v + u(N). Now let x e u(N) nKeri;. To complete the proof it suffices to show that x = 0. But x = u(n) for some neN and now n = (v°u)(n) = v(x) = 0. Accordingly x = 0 and the proof is complete. 4.4

Covariant extension of a tensor algebra Let M be an ^-module and co:R-+S

(4.4.1)

a homomorphism of commutative rings. Our aim is to study the effect of the covariant extension associated with (4.4.1) on the tensor algebra of M. Since here we shall be concerned with more than one commutative ring, we shall embellish the symbol for the tensor algebra by writing TR (M) rather than just T(M). This will help us to avoid certain ambiguities. We know that TR(M) is a graded K-algebra and therefore (see Section (3.6)) its covariant extension 7^ (M) ® R S is an 5-algebra that is graded by the family {Tn(M) ®R S}neI of 5-submodules. The elements of degree one form the 5-module M ®RS and this module generates TR (M) ®R S as an 5algebra. By Theorem 2 there is a homomorphism X:TS(M®RS)^TR(M)®RS

(4.4.2)

of 5-algebras which extends the inclusion mapping of M ®R S in TR (M) ®R S. (Here Ts (M ®R S) denotes the tensor algebra of the 5-module M (x)R S.) Evidently k preserves degrees and therefore it is a homomorphism

76


of graded S-algebras. However, as the following theorem shows, much more is true. Theorem 6. The mapping X:TS(M®RS)-+TR(M)®RS {described above) is an isomorphism of graded S-algebras. Proof Any S-algebra, and in particular TS(M ®R S), can be regarded as an ^-algebra by leaving the ring structure unchanged and using (4.4.1) to turn it into an K-module. Now the mapping M —> TS(M ®RS) which takes m into m ® 1 is K-linear. Accordingly, by Theorem 2, there is a homomorphism

O:TR(M)-+TS(M®RS) of R-algebras such that 0{m) = m ® 1 for all m e M. Consider the mapping TR(M) x S—• TS(M ®RS) in which the image of (x, s) is s6(x). This is a bilinear mapping of /^-modules, and therefore it induces a homomorphism

fx:TR(M)®RS->Ts(M®RS) of K-modules which is such that fi(x ® s) = s6(x) for all xeTR (M) and seS. It is now a simple matter to verify that \i is actually a homomorphism of Salgebras and that \iiyn ® 5) = m ® s for meM and seS. Consider /x ° k and X ° fi. Each of these is a homomorphism of S-algebras and each leaves M ®RS elementwise fixed. It follows from this (and the fact that the S-algebras concerned are generated by M ®R S) that \i ° X and X ° \i are identity mappings and hence that X is an isomorphism. The proof is now complete. In view of Theorem 6 we may write Ts(M ®RS) =TR(M)®RS

(4.4.3)

and Tn (M ®R S) = Tn (M) ®R S,

(4.4.4)

and we may note, in passing, that (4.4.4) is a special case of Theorem 9 of Chapter 2. 4.5

Derivations and skew derivations on a tensor algebra Let M be an ^-module. An /^-linear mapping of M into R is called a linear form on M. If now D is a derivation on T(M), then, since it is of degree —1, it induces an R-homomorphism Tl(M)—^T0(M). But 7; (M) = M and we know that we can identify T0(M) with R. Thus D extends a linear form on M. Moreover it follows from Lemma 4 of Chapter 3 that two different derivations cannot extend the same linear form.

Derivations and skew derivations on a tensor algebra

11

Similar considerations, this time based on Lemma 6 of Chapter 3, show that each skew derivation extends a linear form and that distinct skew derivations give rise to distinct linear forms. In the next two theorems it will be proved that T(M) is fully endowed with both derivations and skew derivations. Theorem 7. Let f be a linear form on the R-module M. Then there is exactly one derivation on T(M) which extends f Proof. Suppose that p>l. There is a multilinear mapping of the p-fold product M x M x • • x M i n t o Tp_ ^ M ) in which (ml5 m 2 , . . . ,rap ) becomes p

X / K M ^ i ® • •' ® fht ® ••' ® mp).

(4.5.1)

i= 1

(Here, as usual, the A over m{ indicates that this factor is to be omitted.) Consequently there is induced an K-homomorphism Dp: 7p(M)—• Tp_l(M\ where Dp(m1 ® m2 ® * * • ® mp) is the element (4.5.1). Note that

Dx=f. Next the homomorphisms D l9 Z)2, D 3 , . . . give rise to an Rendomorphism D: T(M)-+ T(M), of degree — 1, which agrees with Dp on Tp(M). To complete the proof we show that D is a derivation. Let ml9 m 2 , . . . , mp and mi, m' 2 ,..., m'q belong to M and put x = ml® m2 ® •' • ® mp, x' = m\®m'2®''' ® m'q. Then D(x ® x') = Dinti ®--®mp®mf1®--®mq) p

= (Dx) ® x' + x ® (Dxf).

(4.5.2)

Since this relation extends by linearity to any two elements of T(M), the proof is complete. Theorem 8. Let f be a linear form on the R-module M. Then there is one and only one skew derivation on T(M) which extends f Proof This parallels the proof of Theorem 7. We first show that for each p> 1, there is an K-homomorphism Ap: Tp(M)^> Tp_1(M), where p

Ap(m! ® m2 ® - - - ® mp)= X (-l)i

+l

f(î)(î ®"-®rhi®'"®

mp).

These combine to yield an endomorphism A: T(M) —• T(M), of degree — 1,

78


which extends /. Moreover, in place of (4.5.2) we obtain A(mx ® • • • ® mp ® mi ® • • • ® m'q) = A(m1 ® • • • ® mp) ® (mi ® • • • ® m'q) + ( - l)p(mx ® • • • ® mp) ® A(mi ® • • • ® m'q). This shows that A is a skew derivation and ends the proof. 4.6

Comments and exercises We make here a few comments on the theory of tensor algebras and provide some exercises. As usual certain of these exercises have been marked with an asterisk; for these particular exercises solutions are provided in the next section. It is clear that if an ^-module M is generated by a given set of elements, then the same set of elements will also generate T(M) as an K-algebra. It follows that if M is a cyclic ^-module, then T(M) can be generated by a single element. Consequently (see Exercise 1 of Chapter 3) the tensor algebra of a cyclic module is a homomorphic image of the polynomial ring &[X] and, in particular, it is commutative. The first exercise builds on this observation. Exercise 1*. Let M be an R-module with the property that every finitely generated submodule is contained in a cyclic submodule. Show that its tensor algebra is commutative. For example, if R is an integral domain and F is its quotient field, then the tensor algebra of F (considered as an .R-module) is commutative. We remarked in Section (4.3) that if M is a free .R-module with a base B, then T(M) is the free R-algebra generated by B. (The notion of the free algebra generated by a set was enlarged upon in Section (3.8).) The next exercise is concerned with such an algebra in the special case where the ground ring R is an integral domain. Exercise 2*. Let R be an integral domain and M a free R-module. Further let x and y be non-zero elements of the tensor algebra T(M). Show that the product of x and y in T(M) is not zero. Once again let M be an K-module. In Section (3.8) we saw how a commutative, graded K-algebra can be formed from the matrices

[o r

m

where reR and meM. Let us denote this algebra by T(M) and let Tn{M) denote the submodule formed by the homogeneous elements of degree n. We recall that T0(M) consists of all the matrices


79

whereas F J M ) is made up of the matrices 0 ml

o oj Finally Tn(M) = 0 if n±0,1. Consider the mapping M—>F(M) in which m of M is mapping into

O ml

o oj This mapping is R-linear and so it extends to a homomorphism X\T{M)^T{M)

(4.6.1)

of /^-algebras. Evidently X preserves degrees and is surjective. Exercise 3*. Let X\ T(M)—>r(M) be the homomorphism of graded Ralgebras described in (4.6.1). Determine its kernel and show that k is an isomorphism if and only if M (g)RM = 0. Exercise 4. Let R be an integral domain and F its quotient field. Use Exercise 3 to give a simplified description of the tensor algebra of the Rmodule F/R. The next exercise is of general interest. Let I^R be an ideal of R and let M be an ^-module. Then IT(M) is a homogeneous two-sided ideal of T(M) and therefore T(M)/IT(M) is a graded K-algebra. But / annihilates this algebra so that T(M)/IT(M) = A (say) can be regarded as a graded R/Ialgebra. Now the submodule Au of A, formed by the homogeneous elements of degree one is the image of Tl(M) = M, and the kernel of Tl(M)-Â1 is IM. Hence the natural mapping T(M)-+T{M)/IT(M) induces a homomorphism M/IM->T(M)/IT(M) of jR//-modules and this in turn induces a homomorphism 0: TR/I(M/IM)-+T(M)/IT(M)

(4.6.2)

of R/7-algebras. (Here by TR/I (M/IM) we mean the tensor algebra of M/IM considered as an K/7-module.) Exercise 5*. Show that the homomorphism :TR/l(M/IM)^T(M)/IT(M) of (4.6.2) is an isomorphism of graded R/l-algebras. This result enables us to make the identification TR/I (M/IM)= T(M)/IT(M).

(4.6.3)

80


Our final comments concern generalized ordinary and skew derivations. These were defined in Section (3.8). First suppose that D is a generalized derivation of degree i on the tensor algebra of the /^-module M. Then D induces an K-linear mapping of Tx(M) = M into 7^ + x (M). Next \Re T0(M) and so that D(lR) = 0. It follows that the mapping To{M)-+ Tt(M) induced by D is a null mapping. Again, if p > 1 and m1? m 2 , . . . ,mp belong to M, then it is easily seen (using induction on p) that p

D(m1 ® m2 ® • • • ® mp) = £

m

i ® ' * * ® ^ m v ® ' ' ' ® "V (4.6.4)

v= 1

It follows that D is fully determined by its degree and its effect on

Tl(M) = M. Next assume that A is a generalized skew derivation on T(M) of degree i. Everything that was said about D in the last paragraph applies to A except that (4.6.4) has to be replaced by A(mx (x) m2 ® • • • ® mp) = £ (-l)(v

+1)l

m1®---®Amv®---®m;,.

(4.6.5)

v = l

The next exercise generalizes Theorems 7 and 8. Exercise 6*. Let M be an R-module, let i be a given integer, and let f: M —+Ti + l(M) be an R-linear mapping. Show that there is exactly one generalized derivation of degree i and exactly one generalized skew derivation of degree i, on T{M), that extend f 4.7


Exercise 1. Let M be an R-module with the property that every finitely generated submodule is contained in a cyclic submodule. Show that its tensor algebra is commutative. Solution. Let £, £' belong to T(M). We wish to prove that f f = £'£ and this will follow with full generality if we can establish it on the assumption that £ = mx ® m2 ® • • * ® mp, £' = m\ ® m2 ® • • • ® m'q, where the mi and m) are in M. Now, by hypothesis, there exists a cyclic submodule N, of M, containing all the mt and the m). Let (/>: T(iV)^T(M) be the algebrahomomorphism that extends the inclusion mapping of N into M. Then there exist x and x' in T(N) such that (j)(x) = ^ and cj)(x') = £)' and therefore cj)(xx') = ££' and (j){x'x) = £'£. But T(N) is commutative because N is cyclic. Hence ££' = £'£ as required.


81

Exercise 2. Let R be an integral domain and M a free R-module. Further let x and y be non-zero elements of the tensor algebra T(M). Show that the product of x and y in T(M) is not zero. Solution. We can write x 2 + • • • +xh

(

where xt e T{{M) and yê 7}(M), and it will suffice to prove that the product of xh and yk is not zero. Equally well we may assume for the rest of the proof that x belongs to Th(M) and y to Tk(M). Before proceeding note that under the canonical isomorphism

Th(M)®Tk(M)*Th+k(M) the element of Th+k(M) that corresponds to x ® y is precisely the product of x and y in T(M). Put U=Th(M) and V= Tk{M). These are free K-modules because M is free. Let wl9 w 2 , . . . , up be a base for U and vl9 v2,..., vq SL base for V. Then (see Chapter 1, Theorem 3) the elements ut ® v} form a base for U ® V. Next, we can write x = alul -\-a2u2 + • • • +apup and y = b1vl +b2v2 + • • • + bqvq, where af, ^ belong to R, and then

But x 7^ 0 and ^ ^ 0. Hence we can find i and j so that af # 0 andfc^# 0 and then we have atbj^0 because R is an integral domain. However, the products ur (x) vs are linearly independent over R and therefore it follows that x ®y^0. This completes the solution. Exercise 3. Let X: r(M)—>r(M) be rfte homomorphism of graded Ralgebras described in (4.6.1). Determine its kernel and show that X is an isomorphism if and only if M ®RM = 0. Solution. Let m1? m 2 , . . . , mp belong to M. Then X{m1 ® m2 ® • • • (g) mp) is the product of the matrices

Lo o j

(i=l,2

p)

and this is zero if p>2. Thus A maps all of T2(M), T3(M), T 4 (M),... into zero. Let x belong to T(M) and let x = x o + x1 + x 2 + • • • be its decomposition into homogeneous components. Then xoeR, and

x1eM

82


x0 It follows that K e r ^ = X Tn(M). n>2

Of course, since k is surjective it is an isomorphism if and only if Ker k = 0. Now if k is an isomorphism, then M ® M =T2(M) = 0. On the other hand, if M ® M = 0, then M ® M ® - - ® M = 0 provided there are at least two factors (see Chapter 2, Theorem 1). Consequently if M ® M = 0, then Tn (M) = 0 for all n > 2 and therefore Ker k = 0. This completes the solution. Exercise 5. Show that the homomorphism :TR/I(M/IM)^T(M)/IT(M) of (4.6.2) is an isomorphism of graded R/I-algebras. Solution. It is clear that (/> preserves the degrees of homogeneous elements so that it is enough to construct an algebra-homomorphism, in the reverse direction, which when combined with (/> (in either order) produces an identity mapping. Let ft: T(M)—• T(M)/IT(M) be the natural homomorphism and for m in M let m denote its natural image in M/1M. Next, for the moment, let us regard the ^//-algebra TR/I(M/IM) as an K-algebra. Then the jR-linear mapping M^»TR/I(M/IM) which takes m into m extends to a homomorphism T{M)^> TR/I(M/IM) of K-algebras. But this vanishes on IT(M) and so it induces a homomorphism This is a homomorphism of K-algebras and of ^//-algebras and it satisfies il/(h(m)) = rh for all m in M. Now (p(m) = h(m) by construction so that \jj{(j)(m)) = rh and 0(^(/z(m))) = /z(m). But the elements m generate the R/Ialgebra TR/I(M/IM) and, because M generates T(M), the elements h(m) generate the K/7-algebra T(M)/IT(M). Thus \jj°(f) and 0 ° ^ must be identity mappings and with this the solution is complete. Exercise 6. Let M be an R-module, let i be a given integer, and let f: M —• Ti + 1(M) be an R-linear mapping. Show that there is exactly one generalized derivation of degree i and exactly one generalized skew derivation of degree i, on T(M), that extend f Solution. We shall only establish the assertion concerning generalized skew derivations because the other assertion can be dealt with similarly, the details being somewhat simpler in the case of ordinary derivations.


83

Suppose that p > 1. The mapping of the p-fold product MxMx into Ti+p(M) which takes (ml9 m 2 , . . . , mp) into

- - xM

is multilinear and therefore it induces an R-linear mapping

Ap:Tp(M)-^Tp+i(M) which is such that p

Ap(ml®m2®--®mp)

= £ (- l ) 0 ^ 1 ^ ® • • • ® f(mll) ® • • • ® mp. A* =

i

Next we define A o: T0(M)-+ T((M) to be the null homomorphism. Then Ao, A 1 , A 2 , . . . can be combined to give an R-linear mapping A: T(M)-+T(M) of degree i which agrees with An on Tn(M) for all n > 0 . Since Ax=f, A extends /. Now to show that A is a generalized skew derivation we have to check that, for x and y in T(M), we have A(x ® y) = (Ax) ® y + Jl (x) ® {Ay), where J is the main involution (see Section (3.8)). However, it suffices to do the checking when x and y are homogeneous elements whose degrees are strictly positive. Suppose then that xeTp(M) and yeTq(M), where p> 1 and q> 1. We have to show that A(x ®y) = (Ax) ® y + ( - l)ipx ® (Ay). But now we can specialize still further and suppose that x = ml ® m2 ® * * * ® mp and y = m\ ® m'2 ® • • • ® m'q, where mi and m) are elements of M. On this understanding we have

A(x®y)= £ n=i

(m'v)

= (Ax) ® y + ( - l) ip x ® (Ay) which is what we require. This establishes that there is a generalized skew derivation of degree i that extends / That there can be at most one such skew derivation was already noted in Section (4.6).

5 The exterior algebra of a module

General remarks The exterior algebra is one of the most interesting and useful of the algebras that can be derived from a module. As we shall see it is an anticommutative algebra having intimate connections with the theory of determinants. Our aim in this chapter will be to establish all its main properties; and in so doing we shall follow a pattern which can be used again, with only small modifications, to study a related algebra, namely the symmetric algebra of a module. As usual R and S are reserved to denote commutative rings with an identity element, and we again allow ourselves the freedom to omit the suffix from the tensor symbol when there is no uncertainty concerning the ground ring. Algebras are understood to be associative and to have an identity element; and homomorphisms of rings and algebras are required to preserve identity elements. Finally T(M) denotes the tensor algebra of an ^-module M. 5.1

The exterior algebra Let M be an /^-module. We propose to define its exterior algebra, and, as in the case of the tensor algebra, we shall do this by means of a universal problem. However, because we are now in a position to make use of the properties of tensor algebras, the details on this occasion will be much simpler. Suppose that 0 : M —> A is an jR-linear mapping of M into an ^-algebra A, and suppose that (c/>(m))2=0 for all meM. If now h:A^>B is a homomorphism of K-algebras, then h ° (/> is an K-linear mapping of M into B with the property that, for every meM, the square of (h °4>)(m) is zero. Problem. To choose A and (j) so that given any R-linear mapping \//: M —• J9 84

The exterior algebra

85

(B is an R-algebra) such that (i/^(m))2 = 0 for all meM, there shall exist a unique homomorphism h: A-^B, of R-algebras, such that / j ° ^ = f

Evidently the problem has at most one solution. More precisely, if (A, (j>) and (A\ (/>') both solve the problem, then there exist inverse isomorphisms k: A—• A'and A': A' -Â,oiK-algebras, such that A ° = (/>' and A'° satisfy the five conditions. To complete the proof we must now verify that they solve the universal problem. Let meM. Then m ®meJ(M) and therefore ((/>(m))2 = 0.

86

The exterior algebra of a module

Finally suppose that \j/: M —• B is an K-linear mapping (of M into an Ralgebra B) which is such that (i//(m))2 = 0. Then (Chapter 4, Theorem 2), \j/ extends to an algebra-homomorphism T(M)—+ B which, it is clear, vanishes on J(M). Accordingly there is induced a homomorphism h: A—+B of Ralgebras which satisfies (ft°(/>)(m) = i/f(m) for all meM. This shows that h°(j) = \l/ and now (ii) ensures that there is only one algebra-homomorphism with this property. Thus the proof is complete. We next introduce some convenient terminology. Let (A, (j>) solve our universal problem. Put E(M) = A

(5.1.1)

and let {En (M)} neI be the grading referred to in Theorem 1. Since maps M isomorphically onto E1(M) we can use this fact to make the identification E1(M) = M.

(5.1.2)

In this way M becomes a submodule of E{M) and, as we know, it generates E(M) as an K-algebra. The grading on E(M) is, of course, non-negative and the structural homomorphism R—>E(M) maps R isomorphically onto EQ(M). It follows that we may make the further identification E0(M) = R

(5.1.3)

when it is convenient to do so. The symbol A will be used for multiplication on E(M). This secures that, for p > 1, Ep(M) is the p-th exterior power of M according to the definition given in Section (1.4). Note that mAm = 0

(5.1.4)

for all min M = E1(M). The algebra E(M) is called the exterior algebra of the module M. The following theorem simply records the fact that it solves the universal problem with which we started. Theorem 2. Let E(M) be the exterior algebra of the R-module M and let \jj\ M—+B be an R-linear mapping, of M into an R-algebra B, such that (\jj(m))2 = 0 for all meM. Then \j/ has a unique extension to a homomorphism E(M)^B of R-algebras. Before we proceed to investigate the properties of the exterior algebra, let us make quite explicit its connection with the tensor algebra. By Theorem 2 of Chapter 4, the inclusion mapping M —• E(M) has a unique extension to a homomorphism T{M)^>E{M) of K-algebras. This algebrahomomorphism is surjective. We call it the canonical homomorphism of T(M) onto E{M). Now both algebras are generated by M and the canonical homomorphism leaves the elements of M fixed. It follows that the degree of

Functorial properties

87

a homogeneous element is preserved and therefore T(M)—+E(M) is a homomorphism of graded algebras. In particular for each p e Z there is induced a homomorphism Tp(M)-+Ep(M)

(5.1.5)

of K-modules. Of course for p = 0 this is an isomorphism and for p = 1 it is the identity mapping. Suppose next that p > 1 and let ml, m 2 , . . . , mp belong to M. Then in (5.1.5) m1 ®m2 ®''' ®mp is mapped into ml /\m2 A • • • /\mp and therefore (5.1.5) is the canonical homomorphism of the p-th tensor power onto the /?-th exterior power according to the definition given in Section (1.4). Theorem 3. The exterior algebra E(M) of the R-module M is anticommutative. Proof. Since m A m = 0 for all m e M, the theorem follows from Lemma 3 of Chapter 3 and the properties of E(M) that have already been noted. 5.2

Functorial properties Let / : M —• N be a homomorphism of K-modules. We can regard N as a submodule of the exterior algebra E(N) and then f(m) A f(m) = 0 for all meM. It follows (Theorem 2) that / has a unique extension to a homomorphism E(f):E(M)-+E(N)

(5.2.1)

of K-algebras. This is such that if m1,m2,. • •, mp belong to M, then £(/)(»»! Am 2 A - - - Amp) = f(ml)Af(m2)A"-

Af(mp).

(5.2.2)

Moreover, since M consists of the elements of degree one in E(M), the homomorphism E(f) preserves degrees. Thus E(f): E(M)-Ê(N) is a homomorphism of graded algebras and therefore, for each peZ, it induces a homomorphism Ep(f):Ep(M)Êp(N)

(5.2.3)

of K-modules. Of course E0(f) is an isomorphism and, when p>l, EP(f)(mi Am2 A'" A ^ P ) = = / ( ^ i ) A / ( m 2 ) A - - - Af(mp). (5.2.4) Naturally when N = M and / is its identity mapping, E(f) is the identity mapping of E(M). Next suppose that in addition to / : M — • N we are given a second homomorphism g: K-+M of K-modules. Then E(f)oE(g) = E(fog)

(5.2.5)

Ep{f)°Ep{g) = Ep{fog)

(5.2.6)

and also

88


for all peZ.ln particular the exterior algebra provides a second example of a covariant functor from K-modules to graded R-algebras. It follows, as in all such situations, that if / is an isomorphism of modules, then E(f) is an isomorphism of algebras and E(f~l) is its inverse. At this point it is convenient to mention a notation which is very commonly used in the theory of exterior algebras and of which the reader should be aware. This is the wedge notation which we have already employed to describe multiplication. When used more extensively we put f\M = E{M\

(5.2.7)

/\pM

= Ep{M\

(5.2.8)

£(A

(5.2.9)

= Ep(f).

(5.2.10)

A/

=

and /\pf

Accordingly (5.2.5) and (5.2.6) become (/\f)°(f\g) = f\(f°g) and (A P / ) ° (A P 9) ~ f\P ( / ° d) respectively. However, since we aim to develop the theories of several algebras along parallel lines, we shall keep to a uniform notation. This will help a little when making comparisons. Theorem 4. Let f:M-+Nbea surjective homomorphism of R-modules and let K be its kernel. Then E(f): E(M) —• E(N) is a surjective homomorphism of graded algebras whose kernel is the two-sided ideal which K generates in E(M). Remark. Since M is a submodule of E(M) so too is K. It is therefore meaningful to speak of the two-sided ideal which K generates in E(M). Proof The corresponding result for tensor algebras is Theorem 3 of Chapter 4. Two proofs were given of that result and, of these, the second one can be readily adapted. The details are left to the reader. We now add a few remarks that are relevant to the study of the exterior algebra of a finitely generated module. Let ex, e2,... ,en belong to the K-module M and let elements w1,M 2,...,Mr of M satisfy relations U = eC

s l ls + e2C2s+ ' ' ' +enCns ( s== U 2, . . . , r). (5.2.11) (Here the coefficients ctj are in R and, for convenience, we have written them on the right-hand side of the module elements which they multiply.) For integers jl9j2, - • • Jr> a 'l between 1 and n, we put

(5.2.12)

The exterior algebra of a free module

89

Lemma 1. Let the situation be as described above. Then ux A u2 A • • • A ur = Y, ieh A eh A " ' ' A ej)Chh...Jr> where the sum is taken over all sequences (jl,j2, • • • Jr) which satisfy 1 <jt < j2 < ''' <jr < n. In particular ul A U2 A • • • A ur = 0 if r>n. Proof We have ux A u2 A • • • A ur={elcl!

+ • • • + encnl) A (excl2 + • • • + encn2)

A-"

A{elclr+---+encnr)

= E K A ^ 2 A • • • A eir)cixlcil2

. . . c l>?

(5.2.13) where the summation is taken over a// sequences (i1, /2, • • • > 0 of r integers lying between 1 and n. However, eix Aeiz A • • • Aeir = 0 if the sequence (il9 i2,..., ir) contains a repetition. Hence from now on we may assume that r0. By Theorem 5, Er(M) = 0 if r>n. On the other hand, if 1 < r < n , then Theorem 6 of Chapter 1 shows that Er(M) has a base consisting of the products uji A uJ2 A • • • AMJP, where UiJn • • • Jr) is a sequence of integers satisfying 1 <jl <j2< ''' <jr(M1 ® • • • ® Mq) generates E{MX) ® • • • ® E(Mq) asanK-algebraand(0(m 1 ,m 2 ,..., m^)) 2 =0for (ml9 m 2 , . . . ,m^)in M x ©

94


Theorem 8. The modified tensor product £(M X ) ®RE(M2) ®R-- ®RE(Mq),

(5.4.3)

together with the R-linear mapping 4> defined in (5.4.2), constitutes the exterior algebra of the direct sum M x © M2 ® '' * © Mq. Furthermore the grading which (5.4.3) possesses by virtue of being a modified tensor product is the same as its grading as the exterior algebra of M x © M2 © * * * © Mq. Proof For the moment let us leave on one side the assertions about gradings. For the rest, the isomorphisms (Mx © • • • © Mq_l) © Mq^M1 ®'"®Mq.1®Mq and (Chapter 3, Theorem 6) ®"-®

E(Mq_l) ® E{Mq)

show that the first part of the theorem will follow if we can establish it when q = 2. Suppose then that U and V are ^-modules, put M=U ®V, and define the K-linear mapping 0 : M —» E(U) ® E(V) by 0(w, v) = u ® 1 + 1 ® v. We have to show that E(U) ® E(V) and 0 constitute the exterior algebra of M. As a first step we note that, since (0(w, v))2 = 0 for all UGU and veV, Theorem 2 shows that 0 extends to a homomorphism X\E(M)-+ E(U) ® E(V) of tf-algebras. Naturally X(u, v) = u ® 1 + 1 ® v. Next the inclusion mappings U-+M and V-+M extend to homomorphisms h: E(U)Ê(M) and k: E(V)Ê(M) of graded K-algebras. Let us consider the mapping E(U) x E(V)—> E(M) in which (x, y) becomes h(x) Ak(y). This is a bilinear mapping of jR-modules. From this, and because E(U) ® E(V) and E(U) ® E(V) coincide as K-modules, it follows that there exists an JR-linear mapping where A'(x ® y) = h(x) A k(y) and X\u ® 1 + 1 ® v) = (u, v) for u in U and v in V. Let x e Er(U), y e ES(V\ f e Ep(U) and n e Ea{V). By (3.4.5), the product of x ® y and f ® n in E(U) ® E(V) is ( - l)sp(x A^)®(yArj) and the image of this under X' is But h and k preserve degrees and E(M) is anticommutative. Consequently the image in question is (h(x) A k(y)) A (fc({) A k(ri)) = X'(x®y)A

X'{^ ® rj).

It is now a simple matter to check that k' is a homomorphism of K-algebras.

Covariant extension of an exterior algebra

95

Consider X' ° X. This is an algebra-homomorphism of E(M) into itself and it induces the identity mapping on M. Consequently X' ° X is the identity mapping of E(M). Next X'°X is the identity mapping of E(U)®E(V) because it induces the identity mapping on (j)(M) and we know that, as an algebra, E(U) ® E(V) is generated by 0(M). Hence X is an isomorphism of algebras and therefore E(U) ® E(V), together with (f>, is the exterior algebra of M. As already observed, the first part of the theorem follows in full generality. Finally, let us consider the two gradings on E(M1)®E(M2)®'" ® E(Mq) that are mentioned in the statement of the theorem. In both cases {Ml © • • • © Mq) is the module of elements of degree one. But (j)(M1 © • • • © Mq) generates the algebra and therefore the two gradings must coincide. 5.5

Covariant extension of an exterior algebra Let M be an ^-module and a>: R—>S a homomorphism of commutative rings. Since we are here concerned with more than one commutative ring, we shall, for greater definiteness, use ER(M) to denote the exterior algebra of M. Of course ER(M) is a graded /^-algebra, and therefore (see Section (3.6)) ER(M) ®RS is an S-algebra graded by the Smodules {En(M) ®R S}nel. In particular M ®R S is the module of elements of degree one. It is easily verified that, as an S-algebra, ER(M) ®RS is generated by M ®R S and that, in ER (M) ®R 5, the square of an element of M (g)R S is zero. It follows, from Lemma 3 of Chapter 3, that ER (M) E0(M). But El(M) = M and E0(M) = R. Consequently A gives rise to a linear form on M. Furthermore, by Lemma 6 of Chapter 3, different skew derivations determine different linear forms. Theorem 10. Let f be a linear form on the R-module M. Then there is one and only one skew derivation on E(M) which extends f Proof In view of what has already been said it will suffice to produce a skew derivation that agrees with f on E1(M) = M. Suppose then that p > 1 is an integer and consider the mapping of M x M x • • • x M (p factors) into which takes (ml, m 2 , . . . , mp) into p

£ (—\)l*lf(mi)ml

A • • • Arhi A • • • /\mp.

This mapping is multilinear and alternating. Consequently there is induced a homomorphism Ap: Ep(M)^>Ep__l(M) of K-modules with the property that A ^

A

t

m2 A • • •

A

mp)

(5.6.3)

Note that when p = 1, Ap coincides with /. Next, the various Ap (p = 1,2, 3,...) are the restrictions of a single Rendomorphism A: £(M)—>£(M) that has degree — 1 ; and now it only remains for us to verify that A is a skew derivation. Let xeEr(M) and yeEs(M). We wish to show that A(x Ay) is equal to

Skew derivations on an exterior algebra

97

A(x) A y + (— l)rx A A(y) and for this we may assume that r>\ and s>\. Indeed it is enough to establish the relation in question when x = m1 A m2 A • • • A mr and y = fix A \I2 A • • • A \IS, where the mt and ^ are elements of M. But then, by (5.6.3), r

A(x Ay)= £ (—l)l

+1

f(mi)m1 A • • • A ^ A • • • Amr A / ^ A • • • A/HS A • • • AfljA- ' ' Afis

and with this the proof is complete. Suppose now that / is a linear form on M. By Theorem 10, it has a unique extension, Ay say, to a skew derivation on E(M). We have seen that the skew derivations on E(M) form an /^-module and it is evident that and Arf = rAf

(5.6.5)

for / i , / 2 e M * and reR. Furthermore, by Lemma 6 of Chapter 3, A / ° A / = 0. (5.6.6) Before we start to examine the consequences of these relations it will be convenient to broaden the basis of the discussion. To this end we note first that if yo:M*xM-+R is defined by yo(f, m) = f(m), then y0 is bilinear and the mapping M^>R in which m becomes yo(f, m) is just /. Suppose now that U is an K-module and that we have a bilinear mapping y:UxM-*R,

(5.6.7)

i.e. suppose that we have a bilinear form on U x M . If we fix u in U, then there is a linear form on M that maps m into y(u, m). This linear form will have a unique extension, Au say, to a skew derivation on E(M). Thus Au(m) = y(u,m)

(5.6.8)

for all meM. Furthermore (5.6.4), (5.6.5) and (5.6.6) generalize to give AUl + M2 = A U i +A U 2 ,

(5.6.9)

Aru = rA u ,

(5.6.10)

A u °A u = 0,

(5.6.11)

where the notation is self-explanatory. It follows that there is an i^-linear mapping

98


U->EndR(E(M))

(5.6.12)

in which u has image AM. Note that if q > 1 and m1,m2,..., then, by (3.7.4) and (5.6.8), Au(m!

A

m2

A

q

•••

= £ (—l)l i=

A +i

mq are in M,

mq)

y(u,mi)m1 A - • - Arhi A - • - Amq

(5.6.13)

1

for all u in U. But AU°AU = O. Accordingly (5.6.12) extends to a homomorphism F: E(U)-+ End^ (E(M)) of K-algebras. Note that if ul9 u2,...,

(5.6.14) u belong to U, then

and therefore F^

AU2 A

- • • AW/,) =

AMI°AM2°-

•2 F(w1 A u2 A • - • A up) = AUi ° T(u2 A u3 A - - - A up).

(5.6.16)

We now wish to make explicit the way in which F(w1 A U2 A • • • A up) operates on E(M) and for this it is convenient to introduce some temporary notation. ,tp) to denote a Suppose that \ 1 and that the relation in question has been established for all smaller values of the inductive variable. Then, by the inductive hypothesis, {T(u2 A M3 A • • • A up))(mx A m2 A - - - Amq)

= (-Dp-1I(-l)"c|^(m1Am2A---Am,)K.

(5.6.20)

K

Here the sequence K = (/c2, k3,...,

kp) is required to satisfy 1 < k2 < k3 < • • •

1 . Let wl5 u2,..., mp to M. Then

y(up9 m x )

upbelongto

U

andmi,m2,...,

y(ul9m2)

•••

y(ul9mp)

y(up,m2)

•••

y(up,mp)

Proof. If we take account of (5.6.15) this is simply what (5.6.19) becomes when p = q. 5.7

Pfaffians Let M be an K-module and let y:MxM-+R

(5.7.1)

be an alternating bilinear mapping o f M x M into the ring R (i.e. y is an alternating bilinear form). Furthermore for meM let Am denote the skew derivation on E(M) which satisfies Am(/*) = y(m, ju)

(5.7.2)

when pi belongs to M. (Note that we are continuing to identify E0(M) with R.) Then if m1? m2 e M and r e K, we have Ami +m2 = Ami + Am2 and Arm = rAm. Next for m G M we define an endomorphism Lm:E{M)-+E(M)

(5.7.3)

Lm(x) = mAx.

(5.7.4)

by Note that Lmi +m2 = Lmi + Lm2, Lrm = rLm and, whereas Am has degree - 1, the degree of Lm is + 1 . We now put Am = Lm + Am.

(5.7.5)

Of course Am is also an endomorphism of the R-module E(M). Lemma 3. The mapping M->End R (£(M))

(5.7.6)

in which m of M becomes Am is R-linear. Furthermore A m ° Am = 0 for all meM. Proof The first assertion is clear. Now let xeE(M). Then (Lm°Lm)(x) = m Am Ax = 0,so that Lm°Lm = 0. We also have Am° Am = 0 by (5.6.11). Next

Pfaffians

101

But y(m, m) = 0 because y is an alternating bilinear form. Consequently and therefore A m °L m + L m ° Am = 0. The lemma follows. By Theorem 2, the mapping (5.7.6) extends to a homomorphism £(M) -»End,, (£(M))

(5.7.7)

of/^-algebras. Let us denote the image of an element x of E(M) by Ax. Then A m i . m 2 ..... m p =(L m ,+A m i )o(L m 2 + A m2 )o---o(L mp + Amp)

(5.7.8)

whenever ml5 m 2 , . . . , mp belong to M. We next define Q:E(M)->E(M)

(5.7.9)

Q(x) = A x (l £(M) ).

(5.7.10)

by Clearly Q is an endomorphism of the ^-module E(M) and Q(1£(M)) = 1£(M). It follows that Q induces the identity mapping on E0(M). Furthermore if and ueE(M), then and hence Q(w)).

(5.7.11)

Theorem 11. Let Q be defined as above. Then for every skew derivation A on E(M) we have Q ° A = A ° Q. Proof. Suppose that n > 0. It will suffice to prove that if xeEn(M) and A is a skew derivation, then Q(A(x)) = A(Q(x)). This will be established by induction on n. First we note that if x e £ 0 ( M ) , then Q(A(x)) and A(Q(x)) are both zero. Thus all is well when n = 0. From here on we assume that n > 1 and that the assertion in question has been established for all smaller non-negative values of the inductive variable. Accordingly we have Q(A'(z)) = A'(O(z)) if A' is a skew derivation and zeEn_l(M). Now let x GEn(M) and let A be a skew derivation. We wish to show that Q(A(x)) and A(Q(x)) are the same and for this step we may assume that x = niAy, where meM and yeEn_l(M). Now A(m) belongs to E0(M) = R. Consequently Q(A(x)) = Q(A(m)j;-mAA(j;)) = A(m)Q(y) - m A Q(A(y)) - (Am by (5.7.11). On the other hand, again by (5.7.11),

102


= A(mAQ(y)+Am(Q(>;))) = A(m)fi(y) - m A A(Q(y)) + (A ° A J (Q(y)). But A(Q(y)) = Q(A(y)) by the inductive hypothesis and A ° Am = — Am ° A by Lemma 6 of Chapter 3. Accordingly = Q(A(x)). The lemma follows. We now introduce a second alternating bilinear form y:MxM-+R

(5.7.12)

by putting y(mum2)= -y(mi,m2) (5.7.13) and using this we define Am, Ax and Q just as we defined Aw, Ax and Q using y. By (5.7.13) Am=-Am.

(5.7.14)

Of course, like Q, Q is an endomorphism of E(M) and it commutes with all skew derivations. Suppose that msM and ueE(M). Then, in view of (5.7.14) and (5.7.11), we have Q(mAu) = mAQ(u)-A m (Q(u)).

(5.7.15)

We also have, this time from (5.7.8), A. l A . 2 A ... A m p =(L m i -A m i )o(L m 2 -A m 2 )o---o(L m p -A m p )

(5.7.16)

for m l5 m 2 , . . . , mp in M. Theorem 12. Q and Q ar^ inverse automorphisms of the R-module E(M). Proof. We shall establish the following statement by induction on n: for all n>0,ifxeEn(M), then Q(Q(x)) = x andQ(Q(x)) = x. This will clearly suffice. The statement is true when n = 0 because both Q and Q induce the identity mapping on E0{M). From here on we shall suppose that n > 1 and make the natural inductive hypothesis. Let xeEn(M). In showing that Q(Q(x)) = x we may suppose that x = mAy, where meM and yeEn_1(M). But then, by (5.7.11) and (5.7.15), Q(Q(x)) = Q{m A Q(y) + AJQ(y))} = m A Q(Q(>;)) - Am(Q(Q(>;))) + Q(Am(Q(>;))) = m Ay =x

Pfaffians

103

because Q(Sl(y)) = y by the inductive hypothesis and because Q commutes with Am. Since fi(Q(x)) = x for entirely similar reasons, the proof is complete. Theorem 13. Let ml, m 2 , . . . , mp (p > 1) belong to M. Then y(mum2)

•••

y(m2, mx)

y(m2, m2)

-m

y(mump)

y(mp, mx)

y(mp, m2)

• • • y(mp, mp)

y(m2, mp)

= co\

(5.7.17)

where w is the component of degree zero of the result of applying (Lmi +A m ,)° (Lm2 + A m2 )°- • (Lm+\m) to \E(MY V P *s °dd, then the determinant and co are both zero. Remark. Theorem 13 is of interest not only because it shows that the determinant is a perfect square but also because it identifies its square root. Proof. For u in E(M) let no(u) denote its homogeneous component of degree zero. Let x = ml Am2 A • • • Amp. Then the corollary to Lemma 2 shows that the determinant in (5.7.17) is equal to

Put x = Q(x). By Theorem 12, x = Q(x) and therefore Ax(x) = Ax(Q(x)) = Ax(Ax-(l£(M)) = (AX°AX-)(1£(M)) = A X A X -(1 £ ( M ) ) = Q ( X A X ) .

But x = Q(x) is the result of applying (L m i -A m i )°(L m 2 -A m 2 )o---°(L m p -A m p ) to 1£(M) (see (5.7.16)) so that x belongs to the ^-module spanned by all products mix A mi2 A • • • A mih. However, x A mti A mi2 A • • • A mih = 0 if h >0 and thus we see that x AX = TZO(X)X. Consequently Ax(x) = n0(x)Q(x) and therefore the determinant in (5.7.17) is equal to ( - l) p ( p + 3)/27r0(x)7T0(Q(x)). On the other hand a> = no(Q(x)) so that it follows that the relation (5.7.17) will be established if we show that (_ l)P(r + Wn0(x) = n0(Q(x)). Next, by (5.7.8), Ax = (Lmi +A Mi )o ( L p

=I

+A m ,)o.. .o ( L

(5.7.18)

+ A

104


where A p _ 2s , x is an endomorphism of E(M) of degree p — 2s, and now we may conclude, from (5.7.16), that _

p

A x = X (- 1 ) SA P -2s,xs=0

Accordingly ) = A X ( 1 £ ( M ) ) — YJ Ap_2 s ,x(l£(M)) s=0

and

Thus if /? is odd, then both 7io(x) = 0 and 7io(Q(x)) = 0 and therefore, in this case, both the determinant and co are zero. On the other hand, ifp = 2k then

= (-l) f c and therefore

But (~

as required. We shall now extract from Theorem 13 some information which is directly relevant to the theory of matrices. To this end let

A=

«11

fl

C?2i

^22

12

fli

a PI be a p x p matrix with entries in R. We call A an alternating matrix provided that (i) au = 0 for all i, and (ii) ajt= —atj for all i and j . Suppose then that A is an alternating matrix. We can construct a free Rof p elements and afterwards an module M with a base el9e2,-..,ep alternating bilinear form y: M xM^>R with y(ei,eJ) = aij. This enables us to introduce the Pfaffian of A, which will be denoted by Pf(^4). In fact the Pfaffian is defined by

where 7r0: E(M)-+E0(M) that

is the natural projection. It follows (Theorem 13) (5.7.19)

and when p is odd P((A) is zero. B y w a y of e x a m p l e c o n s i d e r a g e n e r a l 4 x 4 a l t e r n a t i n g m a t r i x

105


12

A. =

— a12

"13

"14

023

024

0

fl34

"034

0

In this case Pf (A) is the component of degree zero of But Le has degree + 1 and Ae has degree — 1. Consequently

= 023014-024013+012034-

Thus the relation Det(A) = (Pf(/4))2 becomes 024

-0i2

0

a23

034

-013 -014

-023 -024

0 -034

0

= (012034-013024+014023) 2

as may be verified directly. 5.8

Comments and exercises The theory of exterior algebras is of particular interest because of its connection with the theory of determinants, and indeed most of what will be said here arises out of this connection. However, there are two comments on the general theory that come to mind. These will be dealt with first. As usual, the examples for which solutions are provided are marked with an asterisk. If M is an R-module, then E(M) - or ER (M) when it is desired to remind the reader of the ground ring in question - always denotes its exterior algebra. First suppose that / is a proper ideal of R and that M is an R-module. Then IE(M) is a homogeneous two-sided ideal of E(M) and therefore E(M)/IE(M) inherits the structure of a graded K-algebra. But / annihilates this algebra, so E(M)/IE(M) can also be regarded as an K/7-algebra. Next, the natural mapping £(M)—• E(M)/IE(M) induces a homomorphism of El{M) = M into E(M)/IE(M) which vanishes on IM. Thus there is induced a homomorphism M/IM-+E(M)/IE(M).

(5.8.1)

To be precise this is a homomorphism both of ^-modules and of R/Imodules. Furthermore the image of every element of M/IM has its square equal to zero. It follows that (5.8.1) can be extended to a homomorphism

106

The exterior algebra of a module >E{M)/IE(M)

(5.8.2)

of ^//-algebras. Exercise 1. Show that the homomorphism (5.8.2) is an isomorphism of graded R/I-algebras. There is a solution to this exercise which is very similar to the solution provided for Exercise 5 of Chapter 4. We shall therefore leave the reader to make the minor adjustments that are required. Note that this result enables us to write ER/l(M/IM) =

ER(M)/IER(M).

(5.8.3)

Our next comment has to do with generalized skew derivations. We already know, from Theorem 10, that, for any R-module M, E(M) is fully endowed with skew derivations. The next exercise extends this result. Exercise 2. Let M be an R-module, let i be a given integer, and let f: M —• Ei + l(M) be an R-linear mapping. Show that there is one and only one generalized skew derivation of degree i, on £(M), that extends f This time we can adapt the solution to Exercise 6 of Chapter 4. The only point to be noted is that if/? > 1, then the mapping of the p-fold product M x M x • • • x M into Ei + p(M) which takes (ml9 m2 , mp) into

is an alternating multilinear mapping. This is because E(M) is an anticommutative algebra (see Theorem 3). Our next comments have to do with Laplace's expansion of a determinant and we begin by recalling what has been established already. Suppose that

C =

c22

(5.8.4)

is an n x n matrix with entries in R and let p be an integer satisfying 1 < p < n. In what follows J = (JiJ2, • • • Jp) respectively K = (kl9 / c 2 , . . . , kp) will denote a sequence of p integers with 1 <j\ < • • • <jp < n respectively 1 < kx8 13)

-

The next exercise provides a relation that is similar to (5.8.13). To solve the exercise we have merely to repeat the above argument, but this time using Laplace's expansion in terms of columns rather than in terms of rows. Exercise 4. / / FKS is defined as in (5.8.12) show that

In the preceding discussion J, K and S denoted increasing sequences of p integers all lying between 1 and n. The total number of such sequences is (p). Let us put these (p) sequences in some order - the particular way in which this is done does not matter. Then the ring elements CJK can be regarded as


109

the entries in an (np) x (np) matrix Cip) say. (This is the p-th exterior power of C.) Likewise the TJK can be regarded as the entries in another (J) x (np) matrix Tip) say. On this understanding (5.8.13) and (5.8.14) can be written in matrix notation as C{p)r{p) = (Det(C))I = r{p)Cip\ (5.8.15) where / denotes the identity matrix with (J) rows and columns. Exercise 5*. Let C =11^-11 be an nxn matrix with entries in R and for l2)alternating matrix and suppose that 1 R' be a homomorphism of commutative rings and let M be an K-module. In this section the symmetric algebra of M will be

6.5

Derivations on a symmetric algebra

123

denoted by SR(M). We note that SR(M) ®RRf is a commutative K'-algebra and that it is graded by the family {Sn(M) ®R R'}neI of K'-submodules. The homogeneous elements of degree one in SR(M) ®RRf form the covariant extension M ®RR' of M, and M ®RR' generates SR(M) ®RR' as an Ralgebra. Let SR> (M ®R R) denote the symmetric algebra of M ®R Rf considered as an ^'-module. Then, because SR(M) ®RRf is commutative, the inclusion mapping M ®RR'—>SR(M) ®RR' extends to a homomorphism X: SAM ®RR')^SR(M)

®RR'

(6.5.1)

of .R'-algebras which evidently preserves degrees. Theorem 6. With the above notation X\ SAM ®RR')^SR(M)

®RR'

is an isomorphism of graded R'-algebras. We leave to the reader the straightforward task of adapting the proof of Theorem 6 of Chapter 4. Finally we observe that Theorem 6 allows us to make the identifications SR. (M ®RR') = SR (M) ®R R'

(6.5.2)

and (for all neZ) RR'.

(6.5.3)

Derivations on a symmetric algebra Let M be an K-module and let us identify the submodule S0(M) of the symmetric algebra with R. If now D is a derivation on 5(M), then D induces a homomorphism of Sx (M) = M into S0(M) = R, so that in effect D extends a linear form on M.

6.6

Theorem 7. Let f be a linear form on M. Then there is one and only one derivation on S(M) that extends f Proof Lemma 4 of Chapter 3 shows that at most one derivation extends / Now suppose that p > 1. There is a symmetric multilinear mapping, of the pfold product M x M x • • • x M into Sp-l(M), in which (ml9 m 2 , . . . , mp) becomes p

£

f{mi)ml...mi...mp,

the A over mt indicating, as usual, that this term is to be omitted. It follows that there exists an K-homomorphism Dp: Sp(M)-^Sp_1(M) that satisfies p

Dp(m1m2 ... mp)= £ fim^m^^...

mt... mp.

124

The symmetric algebra of a module

we obtain an K-endomorphism Next, by combining DX,D2,D3,... D: S(M)^S(M), of degree - 1 , that agrees with Dp on SP(M). A simple verification now shows that D is a derivation. It extends / because Dx = / . 6.7

Differential operators The notion of a differential operator is one that we have not encountered before. It occurs naturally in connection with polynomial rings and it can be extended readily to symmetric algebras. We shall therefore investigate it here. Let M be an K-module and h SL non-negative integer. A differential operator of degree h, on 5(M), is an K-linear mapping cj): S(M)—• S(M) of degree — h with a certain additional property. To describe the property it will be useful to introduce some notation. Suppose that p >h and that mu m 2 , . . . , mp belong to M. Let / = (j1? i 2, . . . , ih) be a sequence of h integers with 1 h and arbitrary elements m1? m 2 , . . . , mp in M. It is clear that the differential operators of a given degree h form an K-submodule of End*(S(M)). This submodule will be denoted by VhS(M). Of course, V0S(M) consists of the endomorphisms of S(M) produced by multiplication by the various elements of R. An explanation of the name 'differential operator' will be given in Section (6.8). Note that a differential operator of degree 1 is identical with a derivation as defined in Section (3.7). There is a second form of (6.7.2) to which it is convenient to draw attention. Suppose that xi9 x2,..., xq are elements of M and that s l5 s 2 , . . . , sq are non-negative integers with s1+s2 + m" +sq>h. If we replace mlm2.. .mp by xj x x 2 2 ... x j , then (6.7.2) becomes

= 1 (l1) • • • M*! 1 "" 1 *?"" 2 • • • Kq~aq>

( 6-7 -3)

where the summation is over all sequences (a1? a 2 , . . . , aq) of integers with Sf

(i=l,2,...,g)

(6.7.4)

Differential operators

125

and - " - + a , = /i.

(6.7.5)

Note that we have allowed the possibility that some of sx, s 2 , . . . , sq may be zero and that this involves the use of appropriate conventions to cover this situation. Lemma 1. Let h>0 and k>0 be integers, and let and \jj be differential operators, on S(M), of degrees h and k respectively. Then i// ° 4> = ° \f/ and this is a differential operator of degree h + k. Proof Suppose that p>h + k and thatrax,m 2 , . . . , mp belong to M. In what follows / = (il9 i2,..., ih),J = (JiJ2> • • • Jk) and L = (/ l 9 l 2 ,..., lh+k) denote strictly increasing sequences of integers, between 1 and p, whose lengths are h, k and h + k respectively. We shall use (/, J) to denote the sequence obtained from (il9 i2, . . . , j l 9 j 2 , . . . ) by arranging the different integers present so that they form an increasing sequence. We defined /' above. L is defined similarly. It is clear that \\t ° <j> and (j) ° \jj are endomorphisms of degree —(h + k), and from (6.7.2) we see that ^. =L

(6.7.6)

J

This shows that \j/ ° <j> = ° ij/. Again

and now we see that \// ° c/> is a differential operator. A differential operator of degree h {h >0) induces a linear form on Sh(M) and (6.7.2) shows that if we know h and the associated linear form on Sh(M), then the differential operator is fully determined. The next theorem shows that the linear form may be prescribed arbitrarily. Theorem 8. Let h (h >0) be a given integer, let M be an R-module, and let f be a linear form on Sh(M). Then there is exactly one differential operator of degree h, on S(M), which extends f Proof Suppose that p > h and consider the mapping of the p-fold product M x M x • • • x M into Sp_h(M) in which (ml5 m 2 , . . . , mp) becomes

(Here / = (il9 i2,..., ih) satisfies 1 1? (j>2,... so as to obtain an endomorphism c/>:S(M)->S(M) of degree -h which agrees with q on Sq(M). By construction, (/> is a differential operator of degree h and it extends / Finally, we already know that there can be no other differential operator with these properties. Corollary. Let h be a non-negative integer. Then the module WhS(M) of differential operators of degree h is isomorphic to the module of linear forms on Sh{M) under an isomorphism which matches (j), in VhS(M), with its restriction to Sh(M). Of course WhS(M) is an K-submodule of End* (S(M)). Put V S ( M ) = £ V,S(M),

(6.7.7)

h>0

where the sum is taken in EndK (S(M)). It is easy to see that this sum is direct. Indeed in view of Lemma 1 we have Theorem 9. V5(M) is a commutative R-subalgebra of EndR(S(M)) and it is non-negatively graded by its submodules {VhS(M)}h>0. Definition. The graded, commutative algebra VS(M) is called the 'algebra of differential operators' on S(M). In the next chapter we shall meet this algebra in a different guise.

6.8

Comments and exercises We make some miscellaneous comments on the subject of symmetric algebras. As always, if M is an K-module, then S(M) - or SR(M) if we wish to be more explicit - denotes its symmetric algebra. Let us consider what happens to S(M) if we factor out an ideal of R. Suppose then that / is an K-ideal. Then IS(M) is a homogeneous, two-sided ideal of S(M) and hence S(M)/IS(M) is a graded algebra both with respect to R and with respect to the ring R/I. The homomorphism of M = S 1 (M) that is induced by the natural mapping S(M)—• S(M)/IS(M) vanishes on IM and so there results an ^//-linear mapping M/IM-*S(M)/IS(M).

(6.8.1)

But S(M)/IS(M) is a commutative .R/7-algebra and therefore (6.8.1) extends


127

to a homomorphism (j): SR/I (M/IM) -» S(M)/IS(M)

(6.8.2)

of ^//-algebras. The next exercise should be compared with Exercise 1 of Chapter 5. Exercise 1. Show that the homomorphism (6.8.2) is an isomorphism of graded R/I-algebras. This exercise shows that SR/I (M/IM) = SR (M)/ISR (M).

(6.8.3)

Exercise 2. Let M be an R-module, let i be a given integer, and let f: M —• Si + l(M) be an R-linear mapping. Show that there is one and only one generalized derivation of degree i, on S(M), that extends f This, of course, is the counterpart of Exercise 2 of Chapter 5. It can be solved in very much the same way. We turn now to the consideration of differential operators and we begin by indicating the origin of the name. First, however, it will be convenient to introduce some additional notation. Leta = (al9 a 2 , . . . , a n)andj5 = (jS1, /?2, •. •,/?„) be two sequences of n nonnegative integers and let us put

and

(P

(6.8.5)

We shall use a0). By (6.7.3) this will be a differential operator of degree h provided that, whenever \/3\>h, we have

where the sum is taken over all a with aOfor f = l , 2 , . . . , n, then

128

The symmetric algebra of a module g\v\ y

dX S

dXy22...

dXyn»

is a differential operator of degree \y |. It is this fact which accounts for the terminology. Suppose now that/? >0. We know that Sp{M) has {X^} ^| = p asabase. For \(x\=p= \p\, let D(a) be the linear form on Sp(M) which satisfies

Then the D(a) form a base for the linear forms on Sp(M). But, by Theorem 8, D(a) has a unique extension to a differential operator of degree p on S(M); let us use Di<x) also to denote the differential operator. We now see not only that VPS(M) is a free K-module, but also that it has {D(a)} N = p as a base. Note that for every sequence y = (y1,y2,- • •, yn) of non-negative integers, we have (0

ifa

This formula makes it clear that

Exercise 3. With the above notation show that a We turn now to other matters. Let (7 and M be K-modules and suppose that we are given a bilinear form y:UxM-+R.

(6.8.7)

For a fixed M, in £/, the mapping m\-+y(u,m) is a linear form on M and this, by Theorem 7, will have a unique extension, DM say, to a derivation on S(M). Thus, by construction, DB(m) = y(M,m)

(6.8.8)

for all raeM. Next, the mapping U^>EndR(S(M)) given by u\-^Du is Rlinear and, by Lemma 4 of Chapter 3, DUi° DU2 = DU2° DUi for all wx, w2 m U. It follows that there is a homomorphism r: S(U)ÊndR{S(M))

(6.8.9)

of K-algebras which has the property that T{ulu2...up) =DUx°DU2°---°DUp

(6.8.10)

for uu u2,. • •, up in U. But, because a derivation is a differential operator of degree 1, Du° DUi ° • • • ° Du is a differential operator of degree p. Hence


129

(6.8.9) is a degree-preserving, algebra-homomorphism of S(U) into the algebra of differential operators on S(M). Let us examine, in more detail, how this homomorphism operates. First we recall that if 12

C= C

n2

• • * cm

is an n x n matrix, then its permanent, Per(C), is defined by = 2 ^ Cln{l)C2n(2)'

' ' Cnn(n)>

where n is a typical permutation of (1, 2 , . . . , n). The next exercise uses this concept to describe the effect, on m1m2 ... mp, of the differential operator associated with u1u2 ... up. Exercise 4. Let u1,u2,...,up belong to U and m1, m2,... ,mpto M. Show that (DUi ° DU2 ° • • • ° Du )(m1m2 ... mp) is equal to the permanent of the matrix

y{up,mx)

y(u2,m2)

•••

y(u2,mp)

y(up,m2)

•••

y(up,mp)

7 Coalgebras and Hopf algebras

General remarks A coalgebra is a concept which is dual (in a sense that belongs to Category Theory) to that of an associative algebra, and consequently to almost every result that we have concerning algebras there is a corresponding result for coalgebras. Now it sometimes happens that an algebra and a coalgebra are built on the same underlying set. When this occurs, and provided the algebra and the coalgebra interact suitably, the result is called a Hopf algebra. Our concern with these matters stems from the fact that, when M is an i?-module, both E(M) and S(M) are Hopf algebras of particular interest. Whenever we have a coalgebra the linear forms on it can be considered as the elements of an algebra. The algebra which arises in this way from E(M) is known as the Grassmann algebra of M; for S(M) the resulting algebra has very close connections with the algebra of differential operators which was described in Chapter 6. Throughout this chapter we shall follow our usual practice of using R to denote a commutative ring with an identity element; and when the tensor symbol (x) is used it is understood that the underlying ring is always R unless there is an explicit statement indicating the contrary. Finally, it happens to be convenient, before introducing the notion of a coalgebra, to reformulate the definition of an ^-algebra. This reformulation is carried out in Section (7.1). 7.1

A fresh look at algebras Let A be an associative K-algebra with an identity element. Then inter alia A is an jR-module. Further, we have an K-linear mapping \i\ A A —> A which is such that /J.(X ® y) = xy and this, because multiplication on A is associative, makes 130

Afresh look at algebras

131

H®A

A® A® A

• A ®A (7.1.1)

a commutative diagram. Now letn: R—>Abe the structural homomorphism of the algebra A. This too is K-linear and, because rj(l) is the identity element of A, the diagram

(7.1.2) R®A

• A® A
A and n: R—+A which are such that (7.1.1) and (7.1.2) are commutative diagrams. For x, y in A put xy = fi(x ® y). Then it is easily verified that this definition of multiplication turns A into an fl-algebra that has n(l) as its identity element and n: R —> A as its structural homomorphism. When algebras are looked at in this way we shall say that the triple (A, \i, n) constitutes an associative ^-algebra with identity, and (in this context) jx is called the multiplication mapping and n the unit mapping of the algebra. Let us now review the general aspects of the theory of algebras from this new standpoint. Suppose that (A, fiA,nA) and (B, \iB, nB) are (associative) R-algebras and let / be a mapping of A into B. It is readily checked that / is an algebrahomomorphism, in the sense of Section (3.1), if and only if the conditions (i) / is /Minear, (ii) foîA = (f®f)oîB, and (iii) f°nA = nB are all satisfied. Now assume that, for 1 1®1® • • • ®1.

132

Coalgebras and Hopf algebras

Now, by Section (3.2), Ax ® A2 ®''' ®Ap is an K-algebra. In fact its multiplication and unit mappings are given by VAX ®-®AP=(V\ ®^2®-"®^P)°^P

(7.1.5)

and iAl®-®Ap=(rii®ri2®--'®rip)o&iRp)

(7.1.6)

respectively. We turn next to matters involving gradings. Suppose that (A, fi, rj) is an K-algebra and let {An}neI be a family of #-submodules of A such that A=YJAn

(d.s.),

(7.1.7)

MeZ

that is we assume that {An}neZ grades A as an R-module. Then this gives rise to a grading {(A ®A)n}nel on the module A ®A, where (A®A)n=

X Ap®Aq

(d.s.).

(7.1.8)

(Of course (7.1.8) is the usual total grading on A ® A.) This said, if {An}neI happens to be an algebra-grading, then the mappings fi: A ®A—+A and rj: R —> A preserve degrees, it being understood that R is to be graded trivially. Conversely, if \i: A ® A —• A and Y\ : R —* A are degree-preserving, then {An}neI is an algebra-grading on A. Suppose now that A{1\ A{2\ . . . , Aip) are graded K-algebras and let A(i) have fit and rjt as its multiplication and unit mappings. Further, let / = (iu i2,..., ip) and J = (jlJ2, - • • Jp) be sequences of p integers and put (7.1.9) (7.1.10) (see (3.4.2) and (3.4.3)). Then there is an isomorphism of

C41* ® AV ®-"® A\f) ® (Aft ® Ag> ®'' • ® Aff)

(7.1.11)

onto

(Al? ® i4jj>) ® (A\? ®Af2))®-"® (Ajf ® Aff)

(7.1.12)

in which, with a self-explanatory notation, K ®ai2®-"®

aip) ® {a'h ®a'h®-'®

a'jp)

is mapped into e(I, J)(ati ® a'h) ® (ah ®a'h)®--® But (A

{1)

®A

i2)

®-"®

ip)

{1)

A ) ® (A

(aip ® a'jp). i2)

®A

®-"®

A(p)) is the direct sum

of the modules (7.1.11) whereas on the other hand (A(1)®A(1))® i2) i2) ip) ip) (A ® A ) ®-"® (A ® A ) is the direct sum of the modules (7.1.12). Consequently we can combine our various isomorphisms to obtain an isomorphism

Coalgebras

133

(7.1.13) We recall, from Section (3.4), that the modified tensor product A(1) ® A ® • • • ® Aip) is a graded K-algebra. The multiplication and unit mappings of this algebra are given by {2)

AV>®- ® ^ = 0*i ® /*2 ® • ' ' ® /*P) ° AP

(7.1.14)

and (7.1.15) ^ ( 1 > ® - ® ^ = (>7i ® ^2 ® • •' ® r\p)° Ajf> respectively, where Ajf} is the homomorphism previously encountered in (7.1.4). Now that we have described some of the basic features of algebras in terms of mappings (rather than in terms of elements), we are ready to introduce the dual theory which was mentioned briefly in the introduction to this chapter. 7.2

Coalgebras Let A be an R-module and this time suppose that we are given Rlinear mappings A: A —• A ® A and e: A —• R. The triplet (A, A, e) is called an R-coalgebra provided that the diagrams A

A

•

A ®A A®A

(7.2.1)

A®A

A

> A ®A®A

and

(7.2.2) e®A

_

R®A

R ® R

®R

> (R ® - • - ® R) ® (R ® • • • ® R)

commutes, where the horizontal mappings are comultiplications. However, this becomes clear as soon as we examine what happens to the identity

Tensor products of coalgebras

139

element of R when it is transported to the bottom right-hand corner by the two available routes. Now that tensor products of coalgebras have been defined it is natural to ask whether our results on tensor products of algebras have valid analogues. And in fact they do. However, partly because a coalgebra is a relatively unfamiliar object, and partly because of the greater use of mappings made in characterizing coalgebras, it is not an entirely trivial matter to adapt the arguments originally given in Chapter 3. Consequently, until the reader has had a chance to become familiar with the way in which adjustments may be made, we shall expound the dual theory in a fairly leisurely manner. First suppose that Ai9 A2, . . . , Ap and Bl9 B2,..., Bp are jR-coalgebras and let fi'.A^Bi

(7.4.9)

be a coalgebra-homomorphism for i= 1,2,..., p. Then certainly / i ® fi ®''' ® fp is a homomorphism of the module Al®A2®- ®Ap into the module Bl®B2®"'®Bp. However, as the following lemma shows, the homomorphism respects the coalgebra structures.

Lemma 2. With the above assumptions /i ® * * * ®fp'- Ax ®A2 ® • - - ®Ap-+B1 ®B2 ® • • • ®Bp is a homomorphism of coalgebras. Proof Put / = / i ®f2®'"®fp,A

= Al®A2®-"®Ap,B

= B1®

B2®-"® Bp, and let at belong to At (i = 1, 2 , . . . , p). If now, using the abbreviated notation (see the proof of Theorem 1), we set AA,(ai) = X a\ ® a", then &A(al®a2®---®

ap) = Ya {a\®--®

a'p) ® (af[ ®'

and hence

(f®f)(AA(di®"-®ap))

Again, because f is a homomorphism of coalgebras, A f l | (/^) and therefore

= Z (/i a i ® ' *' ® fp;) ® (//; ® • • • ® /g. A c c o r d i n g l y AA<S)B°(f) a p p l i e d t o al®''-®ap®bl®-'-®bq produces the element (7.4.10) and so we have established that ((/> ® 4>) ° A c = A^ ^ ° . Finally when £A®B°(I) and e c are made to act on ax ®- • • ® a p ® î ® * * * ® bq the result in both cases is e^di)...

^ p (0p)£ Bl (M . . . eBq(bq).

It follows that £A0BO(t) = ec

an

d now the proof is complete.

Theorem 3. Let Al9A2,... ,Apbe R-coalgebras and let (il9 i2,..., ip) be a permutation of (1, 2 , . . . , p). Then the canonical isomorphism A1®A2®---®ApÂii

®Ai2®---

®Aip

of R-modules (see Chapter 2, Theorem 2) is in fact an isomorphism of Rcoalgebras. Proof Let 0 : Ax ® A2 ® • • • ® Ap-Âix ® Ai2 ® • • • ® Ai be the moduleisomorphism in question. For 1 B® is a homomorphism of graded coalgebras for i = 1,2,..., p. Then, because • • • ® Aip)->Bil) ® • • • ® Bip) / i ® ® fp: preserves degrees, it is a homomorphism of graded coalgebras on account of Lemma 2. 7.5

Modified tensor products of coalgebras In this section we shall develop ideas that correspond to those described in Section (3.4). Suppose then that A(l\ A{2\ . . . , Aip) are graded ^-coalgebras. We have already seen how the /^-module A=A(1)®Ai2)®'-'®Aip)

(7.5.1)

can be turned into a graded coalgebra (A,AA,eA). We now propose to modify the coalgebra structure on A. The new coalgebra will have the same /^-module structure, the same grading, and the same counit as before; only the comultiplication will be changed and this in a comparatively simple way. We are, as it were, putting a twist into (7.5.1). Let I = (il9 i2, • • •, ip) and J = (j1J2, • • • JP) be sequences of p integers, and, as in (7.1.9) and (7.1.10), put

N(U)=YJUs

(7.5.2)

r>s

and

(7.5.3) We shall use I + J to denote the sequence (^ +jl9 i2 +j2, • • • ? *p+7P)For given / and J, there is an isomorphism of

(A^ ® Atf) ® (422) ® A%) ® • • • ® (Atf ® 4J?)

(7.5.4)

onto in which

K ® a'h) ® (ai2 ®a'h)®'"®

(aip ® a'jp)

is mapped into e(I, J)(ati

® a i 2 ® - - - ® aip) ® (a'h ® a ' h ® - - - ® a ) )

and these various isomorphisms can be combined to give an isomorphism Vp: (A(1) ® A{1)) ® • • • ® (Aip) ® Aip)) - ^ (A{1) ® • • • ® A{p)) ® (A{1) ® • • • ® Aip)) of K-modules.

(7.5.6)

144


Let us compare Vp with the untwisted isomorphism Vp:

which operates as in (7.4.1). If we denote by nu the projection of (A{1) ® • • • ® A{p)) ® (A(1) ® • • • ® Aip)) onto ,4, ® i4 y, that is onto the module (7.5.5), then we find that nu°Vp = e(I9J)7tuoVp.

(7.5.7)

We next define an K-linear, degree-preserving mapping A: A^>A ® A by means of the formula A = Vp°(AAn) ® Ai4(2> ® • • • ® A^(P»). {1)

(2)

(7.5.8) {p)

Now, by Theorem 1, A = A ® y4 ® • • • ® A is a coalgebra. Let us use A and e to denote its comultiplication and counit mappings. Then, by (7.4.4), (7.5.7) and (7.5.8), nIJ°A = e(I,J)nu°A.

(7.5.9)

Lemma 3. The triple (A{1) ® • • • ® A{p), A, e) is a coalgebra and the usual total grading on A(1) ® A{2) ® • • • ® A{p) grades the coalgebra. Proof. It will suffice to prove the first statement and for this a little temporary terminology will be useful. Let us say that an element a of A is pure if it has the form a , , ® ^ ® - - - ® ^ (where ociveA{^) and when this is the case let us use {a} to denote the sequence (zl9 i2,..., ip). Pure elements are, of course, homogeneous and every element of A is a finite sum of pure elements. The idea is to use the fact that (A, A, e) is already known to be a graded coalgebra. Suppose then that a belongs to A and is pure. Then A(a) can be expressed in the form A(a) = £ V ®a", where a', a" are pure and {a'} +{a"} = {a}. Note that nIJ(A((x)) = 0 unless J. Note, too, that

Now suppose that /, J, K are three sequences each consisting of p integers and let nIJK be the projection of A ® A ® A onto At ® A3 ® AK. It is clear that if we identify A ® A ® A with A® (A® A) in the usual way, then nlJK = nI ®nJK. Accordingly

Y

(Aa")

Modified tensor products of coalgebras

145

and K,,K°04®A)°A)(a) = % e({*'}9 {*'})*,(*') ®nJJC(Aa('). But nj,K(A(x") = e(J, K)nj,K(A(x") by (7.5.9) and therefore (ni,j,K° 04 ® A)° A) (a) = e(J, K) £ e({a'}, {a"})7i7(a') ® 7i^(Aa") = e(J, K M / , J + X) X Ti; (a') ® TC^ (Aa") because 7c7 (a') = 0 unless {a'} = /, and nJK (Aa") = 0 unless {a"} =J + K. This shows that (ni,j,K ° (4 ® A) ° A) (a) = e(J, K)e(I, J + K)(nUiK ° (A ® A) ° A)(a) and in a similar manner we can show that (71^^° (A® A)o A)(oi) = e(I,J)e(I + J, K)(nu,Ko

(A® A)° A)(a).

However, it is easy to verify that e(I,J + K)e(J, K) = e(I, J)e{I + J, K) and we know that (A ® A) ° A = (A ® A) ° A because A is a comultiplication. Accordingly KIJ,K ° (A ® A) ° A = n u x

° (A ® A) ° A

for all /, J, /C and therefore (A ® A) ° A = (A ® A) ° A. It remains to be shown that the diagram

(7.5.10) A ®^R
A ®A

>A ®R

-^—+A

is to turn it into £ e({a'}, {a"})e(a")a'. But if 6(a")^0, then {a"} = ( 0 , 0 , . . . , 0) and therefore e({a'}, {a"})= 1. It follows that and this is just a because the mappings A®e

A

A

> A ®A

>A ®R -^—^ A

combine to give the identity mapping of A. This proves that the left-hand triangle in (7.5.10) commutes and the right-hand triangle can be dealt with in the same way. The proof is now complete.

146


The graded coalgebra (A{1) ® • • • ® A{p\ A, e) will be called the modified tensor product or the twisted tensor product of A(1\ A{2\... ,A ( p ) ;andit will be denoted by A{1) ® A{2) ® • • • ® v4(p) to distinguish it from the coalgebra A { 1 ) ®A(2) ®--® A i p ) . We proceed to establish the basic properties of these modified tensor products. Suppose first that f.:A(i)^B(i)

(i=i,2,...,p)

(7.5.11)

is a homomorphism of graded coalgebras. Then fx ® f2 ® • • • ® fp is a homomorphism of the coalgebra A{1) ® A(2) ® • • • ® A{p) into the coalgebra B{1) ® B{2) ® • • • ® Bip\ But changing to modified tensor products does not affect the underlying module structures. Consequently /i®-"®/p:^

( 1 )

®'"

®A{P)->B{1) ® - - - ®Bip).

(7.5.12)

Note that (7.5.12) is JR-linear and preserves degrees. Lemma 4. / / fr ® f2 ® • • • ® fp is considered as a mapping of A{1) ® A{2) ® • • • ® Aip) into B{1) ® B{2) ® • • • ® B{p\ then it is a homomorphism of graded coalgebras. Proof Consider the element ai{ ® a-l2 ® • • • ® a(, where aiveA^\ We can write

where aj, and aj; are homogeneous elements of AUl) the sum of whose degrees is i,,, and then

Put A = A{1) ® • • • ® A{p\ B = B{1) ® • • • ® B{p\ f = / x ® f2 ® • • • ® fp and set a' = ai ® • • • ® a'p, a" = a'[®- - ® a"p. If now we use {a1} to denote the sequence formed by the degrees of a\, a'2, -.. •> a'p and we define {a"} similarly, then we find that AA(ati ®ah®-

-®aip) = % e({a% {a"})ar

Accordingly But and {fa'} = {a'}, {fa"} = {a"} because ffl is a homomorphism of the graded coalgebra A°l) into the graded coalgebra B°°. We therefore have


147

from which it follows that AB ° / = ( / ® / ) ° AA. Thus / is compatible with the appropriate comultiplications. We still have to satisfy ourselves that / preserves counits. However, this follows from the corresponding result for unmodified tensor products because the mappings involved are exactly the same in both cases. The next result corresponds to Theorem 6 of Chapter 3. In order to state it we suppose that A(1\ A{2\ . . . , A{p) and Ba\ B{2\ . . . , B{q) are graded Rcoalgebras. Then, by Theorem 1 of Chapter 2, there is an ^-linear bijection ®A{p)

®B(1) < § • • •

®Biq)

• • • ® A(p)) ® (B(1) ® • • • ® Biq))

(7.5.13)

in which 4>{ah ®--®aip®bh®--®

bjq) = (ati ® • • • ® aip) ® (bji ® • • • ® bjq).

Theorem 5. The bijection (7.5.13) is an isomorphism • • • ® Aip) ® B(1) ® • • • ® Biq) • • • ® A(p)) ® (B(1) ® • • • ® Biq)) of graded coalgebras. Proof It is evident that preserves degrees and that it preserves counits. Consequently we need only show that it is compatible with the appropriate comultiplication mappings. In doing this we shall put

A=A{1)®Ai2)®---®Aip\ B = Bil) ®B{2) ®"- ®B(q\ and C = Aa) ® • • • ® A{p) ® B{1) ® • • • ® Biq) in order to simplify the notation. Now suppose that at eA^ and b; EB\V\ Then

where a'fi, a"H are homogeneous elements of AUl) the sum of whose degrees is itn and bfY, b" are homogeneous elements of B{v) the sum of their degrees being 7,,. Since

a'p ® a"p) ® (X b\ ®b\)

148


it follows that

= 1 *(Ki ® • • • ® *>;}, {a\ ® • • • ® fc;})(ai ® • • • ® vq) ® (*; ® • • • ® &;), where by {ai ® • • • ® b'q) is meant the sequence formed by the degrees of a\, . . . , a'p, b\,..., b'q taken in order. Accordingly

= Z e(Wi ® ' * * ® bi}> K i ® ' * * ® bq})(a' ®fc')® (a" ® &"), where a' = ai ® • • • ® a p , a" = a? ® • • • ® a,, &' = & i ® - - - ® ^ and ft" = Put /' = {a\ ® • • • ® ^p}, J' = {b[ ® • • • ®fo^}and define J" and J" similarly. Also, if /' = (i'i, r 2 , . . . , ip) and J' = (/ l 5 / 2 ,... J'q) let us use / ' J ' to denote the sequence ( i i , . . . , i'p9 fl9... Jq). We then have ((0 ® (j>)o Ac)(ati ® • • • ® aip ® bh ® • • • ® bJq) = X e(I'J\ I"J")(a' ® 6') ® {a" ® b").

(7.5.14)

But, on the other hand, *AK

® *' * ® %) = ! e({af}, {a"}){a'®a") = Y e{I\ F)(af

and e(J\ J"){V ® V) so therefore (A^ ® AB)((atl ® • • • ® afp) ® (bh ® • • • ® ^ ) ) X

® (ft' ® W).

Now a', a" are homogeneous elements of A and b', b" are homogeneous elements of B. Furthermore, if J' = (j'l9... J'q) and /" = ( i j , . . . , *"p), then the degree of b' is | J' | =j\ +f2 + • • • + ; ; and that of a" is |/" | = i'[ + i'i + • • • + fp. Accordingly ® ( ^ ® ' *' ® bjq)) r){-\^J^r'\af®b')®{an®b't).

(7.5.15)

However, it is easy to check that e(l\ I")e(J\ J"){-iyfl]I"l

= e(rj\ I"J")

and so, comparing (7.5.14) with (7.5.15), we see that ( ^ ® 0 ) ° A C and &A®BO(}> produce the same result when applied to a^®- - ®a{ ® bjt ® * * * ® bj. It follows that (0 ® (/>) ° A c = A^^^ ° 0 and with this the theorem is proved.


149

The final result in this section shows that the formation of modified tensor products of coalgebras is a commutative operation. To prepare the way for proving this, first suppose that A and B are graded /^-modules with {An}neZ the grading on A and {Bn}nEl the grading on B. We recall that the twisting isomorphism (see (3.8.5)) T:A®B-^-+B®A

(7.5.16)

is an isomorphism of /^-modules in which ®bn) = (-\rnbn®am.

(7.5.17)

Here, of course, ameAm and bneBn. Now assume that A and B are graded coalgebras. In this case we can regard the twisting isomorphism as mapping the coalgebra A ®B onto the coalgebra B ® A. The next theorem shows that, in this context, T is an isomorphism of coalgebras. Theorem 6. Let A and B be graded coalgebras. Then the R-linear bijection T: A ® B—>B ® A, where T is the twisting isomorphism, is an isomorphism of graded coalgebras. Proof T certainly preserves degrees. Now suppose that ameAm and bneBn. Then

because, since A and B are graded coalgebras, eB(bn)sA(am) is zero except perhaps when m = n = 0. It follows that eB(S)A(T(am

® bn)) = eA(8)B(am ® bn).

Accordingly sB<S)A°T=eA<S)B and we have shown that T preserves counits. We turn now to the question of compatibility with comultiplication. We can write A^ (am) — £ a' ® a", where a\ a" are homogeneous elements of A the sum of whose degrees is m; likewise we have AB(bn) = YJ br ® b", where this time b\ b" are homogeneous elements of B and the sum of their degrees is n. Let us use {#'}, {a"), {b'}, {b") to denote the degrees of these various elements. Then *AQB(am ® bn) = £ ( - l) { *' }{fl V ® V) ® (a" ® b") and therefore (T®T)(AA®B(am®bn)) (~ l){a'm

(br ® a') ® {b" ® a")

150


=

AB(Â(T(am®bn)).

This shows that (T ® T) ° AA^B = AB®A ° T and now the proof is complete.

7.6

Commutative and skew-commutative coalgebras Let A and B be ^-modules, and let us use U:A®B -^>B®A

(7.6.1)

to denote the module-isomorphism in which U(a ® b) = b ®a. If now (A, \i, n) is an K-algebra, then it is a commutative algebra if and only if u A®A >A®A (7.6.2)

is a commutative diagram. Suppose next that (A, ft, rj) is a graded algebra and let T: A ®A^+A ® A be the twisting isomorphism (see (7.5.16)). We shall say that A is a skewcommutative algebra provided that the diagram T

A ®A

• A ®A (7.6.3)

commutes. This is equivalent to requiring that « = (-ir«A (7.6.4) for homogeneous elements am, ocn of degrees m and n respectively. Note that this is weaker than the requirement of being anticommutative; in fact a graded algebra is anticommutative if and only if (i) it is skew-commutative, and (ii) the square of every homogeneous element of odd degree is zero. These observations suggest the following definitions for coalgebras. A coalgebra (A, A, e) will be called commutative if the diagram u A ®A • A ®A

\

A

(7.6.5)

A commutes. Should the coalgebra be graded it will be called skewcommutative provided that

Linear forms on a coalgebra

151

T

A®A

>A®A (7.6.6)

is a commutative diagram. Here, of course, T is the twisting isomorphism. With this definition we have reached the point where the parallelism between algebras and coalgebras has been developed as far as we need. Further results will be found in the exercises at the end of the chapter. Linear forms on a coalgebra It will now be shown that the linear forms on a coalgebra constitute, in a natural way, an R-algebra. To see how this comes about, suppose that (A, A, e) is a given R-coalgebra and put 7.7

A* = HomR(A,R)

(7.7.1)

so that A* is the module of linear forms on A. First we note that the counit e belongs to A* and therefore there exists an ^-linear mapping rj:R-+A*

(7.7.2)

in which rj(r) = re. This mapping will provide the structural homomorphism of our algebra. Next let nR. R ® R —• R be the multiplication mapping of R (considered as an algebra) and let /, g belong to A*. Then ixR ° ( / ® g) ° A also belongs to A* and the mapping A* xA*—+A* in which (/, g) becomes fiR ° (f ® g) ° A is bilinear. It follows that there is a homomorphism fi:A*®A*-+A*

(7.7.3)

of jR-modules which is such that (7.7.4) Theorem 7. Let (A, A, e) be an R-coalgebra. Then (with the above notation) the triplet (A*, /x, rj) is an R-algebra. Proof. Let /, g, h belong to A* and consider the mappings f®g®h

A

> A® A® A

• R®R®R

> R,

(7.7.5)

where AÂ ®A ® A is (A ® A)° A = (A ® A)° A and R®R®R-^R takes rx®r2® r3 into rx r2r3. The total mapping (7.7.5) can be factored into R®h

A

>A ®A

and hence into A

• R®A

HR

>R®R

>R

152


But this is just ii(n(f ®g)® h). In a similar manner we can verify that the complete mapping (7.7.5) is also the same as fi(f ® n(g ® h)). Consequently, if we use \i to define multiplication on A*, then multiplication will be associative. Next, by (7.7.4), and, because (A9A,s) is a coalgebra, (e®A)°A is the canonical isomorphism of A onto R ® A. It follows that fi(e ®f): A-+R takes a into /(a) and therefore fi(e ® / ) = / A similar argument shows that fi(f ® e) = / as well. It is now apparent that A* is an associative K-algebra, with multiplication ^ and identity element e. Furthermore rj:R —>A* is its structural homomorphism. Corollary. / / (A,A,e) commutative algebra.

is a commutative coalgebra, then (A*,/n,rj) is a

Proof. Let /, g belong to A*. If now a, a'eA, then the total mapping u

A® A

f®g

> A ®A

nR

>R®R

>R

takes a ®a' into f(a')g{a). Accordingly v>R°(f ®g)°u=HR°(g® / ) . But A=U ° A because {A, A, e) is commutative. Consequently HR° (f ® g)° A = fiR° (g ® f)° A, i.e. ji(f ®g) = fi{g ® f) which is what we were seeking to prove. We next turn our attention to the case where (A, A, e) is endowed with a coalgebra grading {An} neI. For a given integer k, the linear forms on A that vanish on J]ni:kAn form a submodule of A*; moreover this submodule is isomorphic to HomR(Ak, R) = A* under an isomorphism which maps the relevant linear forms on A into their restrictions on Ak. We may therefore regard A$ as an K-submodule of A*. On this understanding put

9I=S^f

(7-7-6)

kel

so that 21 is a submodule of A*. Note that the sum (7.7.6) is direct. Usually A* and 21 are different, but in the special case where only finitely many Ak are non-zero we always have A* = SU. It is clear that e e Ag because e preserves degrees. Assume now that feA* and g e A*. Then f ® g, which of course maps A ® A into R ® R, vanishes on As ® At unless both s = p and t = q. But A: A —-• A ® A preserves degrees and from this it follows that fi(f ®g)eA*+q. We now see that 21 can be

Hopf algebras

153

regarded as a graded subalgebra of the algebra (A, /x, rj) described in Theorem 7. More precisely we have proved Theorem 8. Let (A, A, e) be a graded R-coalgebra with {An}neZ as its grading. Then (with the above notation) 21 is an R-algebra and it has {A*}neZ as an algebra-grading. (Here A* is the dual of An and it is to be considered as a submodule of the dual, A*, of A.) Corollary. Let the graded R-coalgebra (A, A, e) be skew-commutative. Then 91 is a skew-commutative algebra.

Proof. Let / e A * and g e A*. It is easily checked that ifT: A ® A-Â ®A denotes the twisting isomorphism, then HR°(f®g)°T=(l)pqjiiR° (g ® / ) . But T ° A = A because (A, A, e) is skew-commutative, and now it follows that

7.8

Hopf algebras Let A be an K-module and suppose that we are given R-linear mappings as follows: fi:A®A-+A,

(7.8.1)

n:R-+A,

(7.8.2)

A:A-+A®A,

(7.8.3)

s:A->R.

(7.8.4)

Then A and these various homomorphisms are said to constitute a Hopf algebra provided that the four conditions (i) (ii) (iii) (iv)

(A, \i, n) is an R-algebra; (A, A, e) is an R-coalgebra; A: A-Â ® A and e: A^>R are homomorphisms of algebras; [i\ A ® A-Â and rj: R-+A are homomorphisms of coalgebras;

are all satisfied. When this is the case we shall speak of the Hopf algebra (rj, ft, A, A, e).

The conditions (i)-(iy) are not independent; indeed the precise connections between them are explained in the following lemma. Note that if (i) and (ii) are both satisfied, then A ® A is both an algebra and a coalgebra. Lemma 5. Suppose that (A,fi,rj) is an algebra and that (A,A,e) coalgebra. Then

is a

(a) n'.A®A-^ is compatible with comultiplication if and only if A: A-Â (x) A is compatible with multiplication;

154

Coalgebras and Hopf algebras (b) fi: A ® A—>A preserves counits if and only if e: A^>R is compatible with multiplication; (c) n: R-+A is compatible with comultiplication if and only if A: A—+ A ® A preserves identities; (d) n:R-+A preserves counits if and only if e:A-+R preserves identities.

Proof (a) The mapping fi: A ® A —• A is compatible with comultiplication if and only if A ° ^ = (/i ® fi)° AA(S>A. But

where U: A ® A -^ A ® A is the isomorphism in which U(a ® a') = a' ®a. Consequently our condition becomes A°fi = (/j.®n)°(A®U®A)°(A®A).

(7.8.5)

On the other hand A: A—+ A®A is compatible with multiplication precisely when A° fi = fiA(S)Ao (A ® A) and this is equivalent to (7.8.5) because \iA (S)A — (\i ® \x) ° (A ® (7 ® A). (b) For /i: A ® A^>A to preserve counits it is necessary and sufficient that s ° n = eA (S)A, that is to say we require e ° \i = /nR ° (e ® e). However, this is precisely the condition for a: A—>R to be compatible with multiplication. (c) The structural homomorphism rj: R—+A is compatible with comultiplication if and only if (rj ®rj)°AR = A°rj, and this occurs when and only when A ( l J = l / 1 ® l / 4 . (d) For rj: R^>A to preserve counits we require that eR = e°rj and this occurs precisely when £ ( 1 ^ ) = ^ . We combine some of these observations in the following Corollary. Suppose that (A,fi,rj) is an algebra and that (A9A,e) is a coalgebra. Then fi: A ®A^>A and n.R-+A are homomorphisms of coalgebras if and only if A: A —• A ® A and e: A —» R are homomorphisms of algebras. and (rjB, fiB,B, AB,eB) are Hopf Now suppose that (rjA, fiA,A,AA,sA) algebras. A mapping A^>B is called a homomorphism of Hopf algebras provided it is homomorphism both of algebras and of coalgebras. A bijective homomorphism of Hopf algebras is called an isomorphism of Hopf algebras. Evidently if / : A —• B is such an isomorphism, then / " 1 : B —• A is also an isomorphism of Hopf algebras. Again, if g: A —• B and h: B —• C are homomorphisms of Hopf algebras, then so is their product h ° g. Another concept which we shall need is that of a commutative Hopf algebra. Quite simply, a Hopf algebra is said to be commutative if it is commutative both as an algebra and as a coalgebra. Next assume that (rj, /x, A, A, e) is a Hopf algebra and let {An}nel be a

Hopf algebras

155

family of i^-submodules of A which grades A as an K-module. We then say that {An}neI grades the Hopf algebra provided that it grades the algebra (A, \i, rj) and the coalgebra {A, A, e). This will be the case if \i, rj, A, e all preserve the degrees of homogeneous elements. (Naturally, it is understood that A ® A has the total grading and that R is graded trivially.) Finally, if A and B are graded Hopf algebras and f:A—>B is a degree-preserving homomorphism (of Hopf algebras), then we say that / is a homomorphism of graded Hopf algebras. In the presence of a grading there is an important way in which our definition of a Hopf algebra can be modified. In order to explain how this comes about we shall make a completely fresh start. Assume then that A is an /^-module and that K-linear mappings jtz: A ® A—+ A, rj: R—>A, A: A-+A ®A, and e: A—+R are given. Suppose also that A is graded, as an R-module, by {An}neZ. Should it happen that this makes (A, fi, rj) a graded algebra and (A, A, e) a graded coalgebra, then the twisted product A ® A will exist both as a graded algebra and as a graded coalgebra. Furthermore, in these circumstances we can regard A as a mapping of A into A ® A and \i as a mapping of A ® A into A. With this in mind, we say that the complete system forms a modified Hopf algebra or a twisted Hopf algebra p r o v i d e d t h a t A : A^>A ®A a n d s:A—>R a r e homomorphisms of (graded) ^-algebras and \i\ A ® A-Â and rj: R-+A are homomorphisms of (graded) K-coalgebras. Thus to modify the definition of a Hopf algebra we introduce a grading and then replace A ® A by A ® A. For the new situation the counterpart of Lemma 5 is Lemma 6. Let (A, fi, rj) be an algebra, let (A, A, e) be a coalgebra, and let both the algebra and the coalgebra be graded by {An}n€l. Then (a) fi: A ®A-Â is compatible with comultiplication if and only if A: A-^ A ®A is compatible with multiplication; (b) \i\ A ® A —• A preserves counits if and only if s:A^>R is compatible with multiplication; (c) n:R-Â is compatible with comultiplication if and only if A: A-Â ® A preserves identities; (d) n\R—*A preserves counits if and only if e:A^>R preserves identities. Proof. Because of the close similarity of this result to Lemma 5 we shall only prove (a). For this we observe that /n: A ® A —• A is compatible with comultiplication if and only if A ° /z = (/x ® fi) ° A^ ®A. Now

156


where T: A ®A^*A require is

®A is the twisting isomorphism, so that what we

A o )U = ()U ® fi) o (A ® T ® A) ° (A (g) A).

(7.8.6)

On the other hand, for A: A —• A ® A to be compatible with multiplication it is necessary and sufficient that A°fi = fiAÂ° (A ® A) and this is equivalent to (7.8.6) because fiAÂ = (/x ® //) ° (A ® T ® A). Corollary. Suppose that {A, ft, rj) is an algebra and (A, A, e) is a coalgebra, and let both be graded by {An}neI. Then \i\ A ® A—>A and Y\:R^>A are homomorphisms of (graded) coalgebras if and only if A: A-Â ®A and e: A—>R are homomorphisms of (graded) algebras. For modified Hopf algebras, the definitions of homomorphism and isomorphism are the same as for ordinary (i.e. unmodified) Hopf algebras, except that such mappings are now understood to preserve degrees. A modified Hopf algebra is termed skew-commutative if its algebra and coalgebra components are both skew-commutative. 7.9

Tensor products of Hopf algebras For 1 K is a homomorphism of algebras as well.

Tensor products of Hopf algebras

157

Theorem 9. Let Al,A1,...,Ap be Hopf algebras over R. Put A =Al ® A2®" ' ® Ap and define fiA,rjA, AA,eA as above. Then (nA, fiA,A,AA,eA) is a Hopf algebra. Proof. The theorem follows at once from the fact that A^: A —• A ® A and eA:A-+R are algebra-homomorphisms by applying the corollary to Lemma 5. Definition. The Hopf algebra (rjA, [iA, A, AA, sA) of the theorem is called the 'tensor product' of the Hopf algebras Al9 A2,..., Ap and it is denoted by A1®A2®--®Ap. The corresponding result for modified Hopf algebras presents minor complications. The following lemma will help us to deal with these. Lemma 7. Let A and B be modified Hopf algebras. Then A ®B, which is certainly both a graded algebra and a graded coalgebra, is in fact a modified Hopf algebra. Proof. The comultiplication A ^ ^ o f t h e coalgebra A ®B, can be regarded as a mapping of A ® B into (A ® B) ® (A ® B). Furthermore, it is given by A ^ B = (A ® T® B)o (AA ® AB),

(7.9.6)

where T: A ® B^B ® A is the twisting isomorphism. Next, (7.9.6) can be obtained by piecing together the mappings

A®B -^—^(A ®A)®(B®B)xA ® (A ®B)®B (7.9.7) where the unlabelled isomorphisms are the obvious module-isomorphisms based on Theorem 1 of Chapter 2. However, AA: A—+A ®A and AB: B —> B ® B are algebra-homomorphisms by hypothesis and T. A®B-^*B®A is an algebra-isomorphism by Exercise 5 of Chapter 3. It follows that all the mappings in (7.9.7) are homomorphisms of algebras and therefore

AA®B: A ®B-+(A ®B) ® (A ® B) is a homomorphism of algebras as well. We turn now to eA 0 B . This is the result of combining the mappings (2)

eA®eB

A®B

HR

• R ®R

• R.

Since R ®R and R®R are identical as algebras, we see that A®B: A ®B—+R\s also an algebra-homomorphism. Finally, to complete the proof we have only to apply the corollary to Lemma 6.

£

158


Theorem 10. Let A{1\ A{2\ . . . , A{p) be modified Hopf algebras. Then the algebra and coalgebra structures on A(1) ® A(2) ® • • • ® A(p) interact in such a way as to make it a modified Hopf algebra. Proof. We use induction on p. When p = 1 the statement is tautologous so we shall assume that p > 1 and that the result in question has been proved in the case of p - 1 modified Hopf algebras. This secures that A(1) ® A{2) ® • • • ® A{p~1} is a modified Hopf algebra and therefore, by Lemma 7, the same is true of {A{1) ® A{2) ® • • • ® Ato-V)

®A{p).

Next, Theorem 1 of Chapter 2 provides a degree-preserving moduleisomorphism (7.9.8) However, by Theorem 6 of Chapter 3, this is an isomorphism of graded algebras, and (this time by Theorem 5 of the present chapter) it is an isomorphism of graded coalgebras. Since the right-hand side of (7.9.8) is a modified Hopf algebra, we may conclude that the same holds for the lefthand side. The inductive step is now complete and the theorem follows. 7.10

E(M) as a (modified) Hopf algebra Let M be an ft-module. We already know that E(M) is an anticommutative, graded K-algebra and that S(M) is a commutative, graded K-algebra. But this is far from being the whole story. It will be shown, in this section, that the structure on E(M) can be extended so that E(M) becomes a modified, skew-commutative, Hopf algebra; and later we shall establish that S(M) is a commutative, graded, Hopf algebra. Our reason for dealing with E(M) first is because the details in this case are a little more complicated. When we come to discuss S(M) in a similar manner, the reader will find that there is no difficulty in adapting the treatment provided for exterior algebras. Suppose then that we are given an K-module M. Let us form its exterior algebra E(M) and let us identify E0(M) with #. The multiplication mapping fi: E(M) ® E{M)^>E(M) of the algebra satisfies fi(x®y) = x Ay and the unit mapping rj: R—+ E(M) is simply the identity mapping of R onto E0(M). Now let e:E(M)->R

(7.10.1)

be the projection of E(M) onto E0(M). This mapping will turn out to be the counit of E(M) when we come to consider the latter as a coalgebra. Note that e is a homomorphism of graded R-algebras.

E(M) as a (modified) Hopf algebra

159

The definition of comultiplication requires some preparation. The modified tensor product E(M) ® E(M) of E(M) with itself is an anticommutative R-algebra (see Chapter 3, Theorem 7). Consequently the mapping : M —• E(M) ® E(M) in which (m) = m ® 1 + 1 ® m is not only R-linear, but also satisfies ((j)(m))2 = 0 for all m in M. Consequently can be extended to a homomorphism A: £(M)->£(M) ® £(M)

(7.10.2)

o/ R-algebras which is degree-preserving and which is such that A(m) = m ® 1 + 1 ® m

(7.10.3)

when meM. The homomorphism A is called the diagonalization mapping of £(M). Of course, since £(M) ® E(M) and £(M) ® E(M) coincide as modules (but not as algebras), A can also be considered as an /^-linear mapping of E(M) into E(M) ® E(M). Lemma 8. With the above notation the triple (E(M), A, e) is a coalgebra. This is graded, as a coalgebra, by the exterior powers {En(M)}neZ of the module M. As a graded coalgebra (E(M), A, e) is skew-commutative. Proof To see that the triple constitutes a coalgebra it suffices to verify that the diagrams

E(M)

> E(M)®E(M) E(M)®A A®£(M)

E(M) ® E(M)

(7.10.4)

^

> E(M) ® E(M) ® E(M)

and Ê(M). (7.10.5) £®£(M)

R ® E(M)
E(M) ® R

are commutative. However, all the mappings in (7.10.4) and (7.10.5) are algebra-homomorphisms and we know that the K-algebra E(M) is generated by E1(M) = M. Consequently, when verifying that the diagrams are commutative, we need only examine what happens to an element of M. This reduces the verifications to trivialities. Next, the exterior powers of M grade E(M) as a module and it is obvious that, with respect to this grading, A and e preserve degrees. Accordingly (£(M), A, s) is a graded coalgebra. Finally, to complete the proof we have only to verify that the diagram

160

Coalgebras and Hopfalgebras T

E(M) ® E(M) —> E(M) ® E(M)

E(M) where T is the twisting isomorphism, commutes. But, once again, all the mappings are homomorphisms of algebras so that we need only show that T(A(m)) = A(m) for each element m in M. This, however, is clear. Theorem 11. Let M be an R-module and let the notation be as above. Then (rj, \i, E(M), A, e) is a modified Hopf algebra whose grading consists of the exterior powers of M. As a modified Hopf algebra, E(M) is skewcommutative. Proof. We know from Chapter 5 that (£(M), /i, rj) is a graded algebra and we have just proved (Lemma 8) that (E(M\ A, e) is a graded coalgebra. Furthermore, we noted earlier that A: E(M)-*E{M) ® E(M) and e: E(M) —• R are homomorphisms of algebras. It therefore follows, from the corollary to Lemma 6, that (rj, //, E(M), A, e) is a modified Hopf algebra. This is skew-commutative because (E(M)9 \x, rj) is an anticommutative algebra and, by Lemma 8, (£(M), A, e) is a skew-commutative coalgebra. Thus the theorem is proved. be a homomorphism of R-modules. Then the Theorem 12. Let f:E—*N mapping E(f): E(M)—+E(N) defined in Section (5.2) is a homomorphism of modified Hopf algebras. Proof. We know, from Chapter 5, that E(f) is a homomorphism of graded algebras. We also know that and e£(M): E(M)~>R are homomorphisms of algebras and that a similar observation applies to AE[N) and eE(N). This makes it very easy to verify that £ ( / ) : E(M)^>E(N) is a homomorphism of graded coalgebras. But this is all that is needed to complete the proof of the theorem. The details are left to the reader. 7.11

The Grassmann algebra of a module Let M be an R-module. Then, by the results of the last section, E(M) is a modified Hopf algebra, and so inter alia it is a graded coalgebra. We recall that its comultiplication mapping A satisfies A(m) = m ® 1 + 1® m for all meM, and that the counit mapping e is the projection of E(M) onto

The Grassmann algebra of a module

161

E0(M). Furthermore, A: £(M)->£(M) ® E(M) and e:E(M)-^R are homomorphisms of /^-algebras. By Theorem 8 we can now obtain a graded R-algebra by considering the linear forms on E(M). It is this algebra that we are about to investigate. Suppose that p> 1 is an integer. In what follows I = (il9 i2, • • •, is) will denote a sequence of s (0 < 5 0 and consider the K-endomorphisms of E(M) of degree — h. Such an endomorphism induces a linear form on Eh(M). In what follows p denotes an integer satisfying p > h, I = (ix, i2,..., ih) is a sequence of integers with 1 A* is a homomorphism of R-algebras. Solution. With the usual notation we have (j)*(eB) = sB°(j) = sA because <j> preserves counits. Consequently 0* preserves identity elements. Now suppose that hx and h2 belong to B*. By (7.7.4), their product in this algebra is fiR ° (h1 ® h2) ° AB and, moreover,

because $ is compatible with comultiplication. But = ^ ° ^ i ) ® <j)*(h2)°AA and this is just the product of (^(/i^ and 0*(/i2) i n 4*- Thus 0* is compatible with multiplication and the solution is complete. Exercise 7. Let (j) and \j/ be differential forms on E(M) of degrees h>0 and k>0 respectively. Show that i// ° (j) and (po\j/ are differential forms of degree h + k and that \jj ° = ( - 1)^0 ° i//. Show also that if h is odd, then (j) ° (j) = 0. Solution. Evidently \jj ° and 0 ° i/^ are endomorphisms of £(M) and each has degree — (ft + /c). Now suppose that p>h + k and that ml9 m2,... ,mp belong to M. In what follows I = (il9 i2,..., ih) and J=UiJ2,... ,7fc) will denote a pair of sequences of integers, where 1 < ix < • • • < ih < p, 1 associated with \\i ° (j) satisfies i// ° = (j> A \j/9 where
and \jj be h > 0 and k > 0 respectively, and let ml9 m 2 , . . . , mh+k belong to M. By (7.13.2), Am 2

A

and therefore Am2 A • • • Amh+k) = Y, sgn(/,

174


Here / = (ix, i2,..., ih) is an increasing sequence of integers between 1 and h + k and I denotes the residual sequence. We now consider ( A\j/)(ml Am2 A • • • Amh+k). This can be obtained from Lemma 10; and, if we remember that (Mx ® M 2 • • • ® M p )* which takes (fl9 / 2 , . . . , / p ) into ^jf} ° (fi f2®' Accordingly there is an K-linear mapping

" ® fp) is multilinear.

c»: Mf ® Mf ® • • • ® MJ - • (Mx M 2 ® • • • ® M p )*

(8.1.11)

which is such that (A®A)n

A®A

is commutative. If we make the identifications (A ® A)n = (Af ® Af)f = An ® An we find that A ® A —• (A ® A)n, considered as a homomorphism of A ® A into An ® An, is the tensor product of the homomorphism A—>An with itself. This shows that A-+Aff is compatible with comultiplication. Next we consider the commutative diagram

R

>Rn

and observe that when we put the mapping R —• Rn becomes the identity mapping of R. Accordingly A —• An preserves counits and with this the proof is complete.

8.4

Graded duals of Hopf algebras We can extend the results of the last section to Hopf algebras, but first of all it is necessary to see how tensor products of graded algebras and coalgebras are effected by graded duality. We begin with algebras. Let A{1\ A(2\... ,Aip) be algebras with nonnegative gradings whose components are finite free modules. Then A{1) ® A{2) ® • • • ® A{p) is a graded algebra of the same kind. We propose to compare (A(1) ® Ai2) ® • • • ® Aip))f and A(1)+ ® 4 ( 2 ) + ® • • • ® Aip)f as graded coalgebras. Let A(v4(1), Ai2\..., Aip)) be the isomorphism

Graded duals of Hopf algebras

189

® Aip)) ® U ( 1 ) ® • • • (x) Aip)) - ^ (A(1) ® A{1)) ®-"®

(A{p) ® Aip))

(8.4.1)

of graded modules which operates as in (8.2.15) and let the inverse isomorphism "-®

(A{p) ® A{p))

-^-> (A(1) ® • • • ® A(p)) ® (A(1) ® • • • ® Aip)) (1

i2

(8.4.2)

(p)

be denoted by V{A \ A \ . . . , A ) (see (8.2.16)). If now fit and rjt are the multiplication and unit mappings of A{i\ then the multiplication mapping \ . . . , Aip))

P

and its unit mapping is (f7 1 ®f/ 2 ®---®f/ p )°Ajf ) . At this point let us make the identifications ip)

Y = A(1)+ ® A(2)+ ® • • • ® Aip)\

(A{p) ® Aip))Y ® • • • ® (A(p)f ® Aip)f) and (04 ( 1 ) ® • • • ® Aip)) ® {A{1) ®"-® = {Ail)f

®-"®

Aip)f)

® {A(1)f

Aip)))f ® • • • ® Aip)f).

Then, using (8.2.18), we see that the comultiplication mapping of (,4(1) Ai2) ® "-® A(p))f is

= V(A{1)\ Ai2)\ . . . , Aip)f) o (fi\ ®fif2®'"®

n\)

(8.4.3)

and that its counit mapping is

P

(8.4.4)

(see (8.2.24)). Now the right-hand sides of (8.4.3) and (8.4.4) are none other than the comultiplication and counit mappings of Ail)f ® • • • ® A{p)f. Consequently we have proved that the natural isomorphism

-^-* (Ail) ® Ai2) ® • • • ® Aip))f (8.4.5) of graded modules (see Lemma 4) is actually an isomorphism of graded coalgebras.

190

Graded duality

Of course (8.4.5) may also be regarded as a module-isomorphism

-^-+ (A{1) ® A{2) ® • • • ® Aip)Y

(8.4.6)

and this too can be shown, in very much the same way, to be an isomorphism of graded coalgebras. We record both these results in Theorem 6. Let A{1\ A{2\ . . . , A{p) be non-negatively graded algebras whose components are finite free modules. Then the graded coalgebras ,4 (1)+ ® Ai2)f ® • • • ® Aip)iand (Ail) ® A{2) ® • • • ® A{p)Y can be identified by means of (8.4.5); and the graded coalgebras Ail)f ® A(2)f ® • • • ® A{p)f and (A(l) ® A{2) ® • • • ® Aip)Y can also be identified by means of the same mapping (see (8.4.6)). Naturally this result has an analogue in which the roles of algebras and coalgebras are interchanged. This is recorded as Theorem 7. The adjustments that need to be made to the proof of Theorem 6 are perfectly straightforward so no details will be given. Theorem 7. Let A{1\ A{2\... ,A{p) be non-negatively graded coalgebras with finite free components. Then the isomorphism Ad)f

g) A(2)f g) . . . g) A(P)^

(^(1) (g) A(2) g) • • . 0

A(P))\

(of graded modules) provided by Lemma 4 enables us to make the following identifications of graded algebras: (i) A(1)i ® Ai2)f ® • • • ® Aip)f

with (Ail) ® A(2) ® • • • ® Aip))f

and (ii) A{l)f ® A{2)i ® • • • ® A(p)f with (A{1) ® A{2) ® • • • ®

A{p))\

We are now in a position to discuss the graded duals of Hopf algebras. By Theorems 1 and 3, if (rj, /a, A, A, s) is a non-negatively graded Hopf algebra with finite free components, then (A\ A+, e+) is a graded algebra and (A\ fi\ rf) is a graded coalgebra. Next, because A is a Hopf algebra, ^: A ® A —•• A and rj: R —> A are homomorphisms of graded coalgebras and so (Theorem 2) jaf: Af—• (A ® A)f and tf\ Af-+Rf are homomorphisms of graded algebras. But Theorem 6 and Lemma 5 show that the familiar module-isomorphisms (A ® A)f&Af ® Af and Rf^R are, in fact, isomorphisms of graded algebras. Consequently / / : Af—+Af ® Af and rjf: Af-+R are algebra-homomorphisms. Theorem 8. Let (rj, \i, A, A,s)be a non-negatively graded Hopf algebra with finite free components. Then (e+, Af, Af, //, rf) is a graded Hopf algebra of the same kind. This follows immediately from our previous remarks as soon as we


191

invoke Lemma 5 Cor. of Chapter 7. Naturally we refer to (e+, A+, A\ n\ rf) as the graded dual of the Hopf algebra (Y\, fi, A, A, e). Finally, by using Lemma 6 Cor. of Chapter 7, we can adapt the above argument so that it applies to modified Hopf algebras. We thus obtain Theorem 9. Let (rj, n,A,A,e) be a modified Hopf algebra with a nonnegative grading and finite free components. Then (ef,A\A\fif,rjf) is a modified Hopf algebra of the same kind. Finally let us see how R itself fits into this theory. It is, of course, both a graded Hopf algebra and a modified Hopf algebra, and therefore the same is true of R\ Furthermore, by Lemma 5, the usual module-isomorphism Rf&R is an isomorphism of Hopf algebras. Hence, under graded duality, the Hopf algebra R remains essentially unchanged. 8.5

Comments and exercises Let M!, M 2 , . . . , Mp be ^-modules. Then, by (8.1.11), we have a homomorphism Mf ® Ml ® • • • ® MJ -> (M1 ® M 2 ® • • • ® Mp)*; and if we take M 1? M 2 , . . . , Mp to be the same module M, the homomorphism takes the form Tp(M*)-+Tp(M)*,

(8.5.1)

where Tp(M) denotes the p-th tensor power of M. In this section we shall consider certain natural homomorphisms Ep(M*)—>EP(M)* and By (7.11.8), if G(M) is the Grassmann algebra of M, then M* is the submodule of G(M) formed by the homogeneous elements of degree one; and, by Theorem 13 of Chapter 7, G(M) is an anticommutative algebra. It follows that the inclusion mapping M*^>G(M) extends to a homomorphism E(M*)-+G(M)

(8.5.2)

of /^-algebras which preserves degrees. In degree zero (8.5.2) induces a homomorphism R —• Hom(R, R) which maps the identity element of R into the identity mapping of R; and in degree p > 1 it induces an K-linear mapping Ep(M*)Êp(M)*.

(8.5.3)

This is one of the homomorphisms we are seeking. Let us see how it operates. Suppose that / i , / 2 , . . . , fp belong to M* and let the image of fx A f2 A • • • A fp under (8.5.3)be 0. Since (8.5.2) is a homomorphism of algebras, is the

192

Graded duality

product o f / i , / 2 , . . . , /pin G(M). Consequently,if ml,m2,... M, then (Chapter 7, Lemma 11) /iK) Am 2 A • • • A m p ) =

fi(m2) '"

M m , ) f2{m2)

,mpbelong to

Mm

••• Mm

(8.5.4)

> i ) / P (^ 2 ) ••• / P K ) Thus the image of/x A f2 A • • • A fp under (8.5.3) is the linear form on Ep(M) that maps ml A m2 A • • • A mp into the determinant of the matrix H/Km^H. Now suppose that M is a finite free R-module with bl9 b2,... ,bs as a base, and let the base of M* that is dual to this be fl9 / 2 , . . . , fs. Suppose that 1 G(M) (see (8.5.2)) is an isomorphism of graded algebras. In the case where M is a finite free module, E(M) is not only a modified Hopf algebra with a non-negative grading, but also it has finite free components. Accordingly (Theorem 9) its graded dual E(M)f is a modified Hopf algebra as well. Furthermore, the proof of Theorem 1 shows that the algebras E(M)f and G(M) coincide, so that by the last paragraph we have an isomorphism E(M*)-Ê(M)f of graded algebras. However, more than this is true as the next exercise shows. Exercise 2*. / / M is a finite free R-module, show that the mapping E{M*)—>£(M)+, defined above, is an isomorphism of modified Hopf algebras. As is to be expected there are some analogous results connected with symmetric algebras though, as we shall see later, there is one important difference. We examine the details in the following paragraphs. If M is an arbitrary R-module, then S(M) is a graded Hopf algebra and so a fortiori it is a graded coalgebra; and we have already used Theorem 8 of Chapter 7 to derive from it a graded algebra. We have not given a name to this algebra or introduced a special notation for it though we have shown


193

(Chapter 7, Theorem 16) that it is isomorphic to the algebra of differential operators on S(M). However, the proof of Theorem 1 shows that when M is a finite free module the algebra in question is the graded dual of the coalgebra 5(M); hence there is no harm in using 5(M) + to denote the algebra in question in all cases. Thus, to sum up, S(M) + is defined for an arbitrary module M, and when S(M) has finite free components it coincides with the graded dual of the coalgebra S(M). Note that S(M)f is always a commutative algebra and that the module formed by its homogeneous elements of degree one is none other than M*. Because S(M)+ is commutative, the inclusion mapping M*—>S(M)f extends to a homomorphism (of algebras) A:S(M*) — S(M) +

(8.5.5)

which in degree p (p> 1) induces an ^-linear mapping Xp:Sp(M*)^Sp(M)*.

(8.5.6)

The way in which (8.5.6) operates is described in the next exercise. fp) Exercise 3. L e f / l 9 / 2 , . . . , fp(p> 1) belong to M*. Show that Xp(flf2... is the linear form on Sp(M) that maps m1m1.. .mp into the permanent of the matrix \\ft(mj)\\. The permanent of a square matrix was defined in Section (6.8). Now suppose that M is a finite free module. Let bx, b2,... ,bs be a base of M and fl9 f 2 , . . . , fs the base of M* that is dual to it. Then the elements b^lb^2 ...b^'9 where jUf>0 and jUx + /x2 + ' " + Ais = P> f ° r m a base of SP(M); and likewise the elements f\1f22.. .fvsM, where v,>0 and vl +v 2 + a

Exercise 4. Show that ^(/ivi/2V2.../

has the value (vx !)(v2 ! ) . . . (vs 1)1^ if (vl, v 2 , . . . , vs) = (^, / i 2 , . . . , fis) and is

zero otherwise. Exercise 4 shows that it is possible for kp{flxfl2... // s ) to be zero. Consequently, when M is a finite free module the homomorphism kp\ SP(M*)-+SP(M)* need not be an isomorphism and therefore the algebra-homomorphism X: S(M*)—>S(Af ) f need not be an isomorphism either. However, in this situation S(M*) and S(M)f are certainly graded Hopf algebras. Exercise 5. / / M is a finite free module show that the mapping X\ S(M*)—>5(M)+, defined in (8.5.5), is a homomorphism of graded Hopf algebras.

194

Graded duality

This can be solved by arguments very similar to those used in connection with Exercise 2. 8.6


Exercise 1. Let the notation be as explained. Show that the base {fj} of Ep(M*) is mapped by (8.5.3) into the base of Ep(M)* that is dual to the base {&,} of Ep(M). Solution. Suppose that (8.5.3) takes f, into the linear form cj)j. Then, by (8.5.4),

A) fhK) A> fhK)

'" -

W fhK)

Now if (il9 i 2 ,... ,ip)*t UiJi* • • • JP)> ^ e n we can find t so that), is not in / and hence (f>J(bI) = O because the determinant has a row of zeros. On the other hand, if / = J, then (t>J(bI)= 1. Exercise 2. / / M is a finite free module, show that the mapping E(M*)—+ E(MY defined above is an isomorphism of modified Hopf algebras. Solution. We already know that E(M*)—>E(M)f is an isomorphism of graded algebras so it will suffice to show that it is also a homomorphism of coalgebras. Let us denote the homomorphism under consideration by 6 and for g in E(M*) use g° to denote its image in E(M)\ To show that 6 is compatible with comultiplication we need only show that E(M*)

• £(M*) ® £(M*)

is a commutative diagram, where Ax and A2 are comultiplications. But all the mappings are homomorphisms of algebras and M* generates £(M*) as an algebra. Thus we need only show that A2(f°) = (0 ® 6) (Ax (/)) for all / in M*. Now (6 ® 0)(Ai(/)) = (0 ® 0 ) ( / ® 1 + 1 ® / ) = / ' ' ® e +e ® / ° , where e is the identity of £(M) + , that is to say it is the projection of E(M) onto E0(M) = R. Let pi be the multiplication mapping of the algebra E{M). Then A2 is


195

obtained by combining the homomorphisms — • (E(M) ® E(M))f - ^ E(MY ® E{M)\ ° /i. Consequently it suffices to show that the isomorphism Also ^(f°)=f° (£(M) ® E{M)Y*£(M)+ ® £(M) + matches f° ° ^ with / * ® £ + e ® f° and for this it is enough to establish that f°(xAy) = 8(y)fe(x) + 8(x)f°(y) (8.6.1) for all homogeneous elements x and y of E(M). Now, except in the case where one of the elements has degree zero and the other has degree one, both sides of (8.6.1) are zero; and if we do have this special situation, the two sides are obviously equal. Thus 6 is compatible with comultiplication. It remains to be shown that 6 also preserves counits, that is to say we must show that the diagram ,E(M*)

is commutative, where the unlabelled mappings are counits. But, once again, the mappings are homomorphisms of algebras, so that we need only examine what happens to an element of M*. However, for such an element its image in R is zero by either route. Accordingly the solution is complete.

Index

adjunction of indeterminates, 34 algebra, 42 algebra generated by a set, 44 algebra-homomorphism, 43 algebra-isomorphism, 43 algebra of differential forms, 168 algebra of differential operators, 126 alternating bilinear form, 100 alternating matrix, 104 alternating multilinear mapping, 6 anticommutative algebra, 54 associative law for tensor products, 28

base of a free module, 2 bilinear form, 97 coalgebra, 133 coassociative mapping, 133 commutative coalgegra, 150 commutative Hopf algebra, 154 complete representation of a module as a direct sum, 26 comultiplication mapping of a coalgebra, 133 contravariant functor, 176 counit of a coalgebra, 133 covariant extension of a module, 30 covariant extension of an algebra, 56 covariant extension of an exterior algebra, 95 covariant extension of a symmetric algebra, 122 covariant extension of a tensor algebra, 75 covariant functor, 23 derivation on an algebra, 57 determinant of an endomorphism, 90

diagonalization mapping of a coalgebra, 133 diagonalization mapping of £(M), 159 diagonalization mapping of S(M), 164 differential form, 168 differential operator, 124 doubling of degrees, 63 dual of a graded module, 179 dual of a homomorphism, 176 dual of a module, 96

endomorphism algebra of a module, 43 exterior algebra of a direct sum, 93 exterior algebra of a free module, 89 exterior algebra of a module, 86 exterior powers of a matrix, 93 exterior powers of a module, 8

finite free module, 177 flat module, 33 free algebra generated by a set, 59 free module, 2 free module generated by a set, 2

generalized derivation, 63 generalized skew derivation, 64 generic alternating matrix, 110 graded algebra, 47 graded coalgebra, 134 graded dual of a graded algebra, 187 graded dual of a graded coalgebra, 185 graded dual of a graded Hopf algebra, 191 graded dual of a graded module, 179 graded Hopf algebra, 155 grading on a module, 132 Grassmann algebra of a module, 163

197

198

Index

homogeneous element, 47 homogeneous ideal, 48 homomorphism of algebras, 43 homomorphism of coalgebras, 134 homomorphism of graded algebras, 48 homomorphism of graded coalgebras, 134 homomorphism of graded modules, 179 homomorphism of Hopf algebras, 154 homothety, 43 Hopf algebra, 153

isomorphism isomorphism isomorphism isomorphism

of of of of

coalgebras, 134 graded algebras, 48 graded coalgebras, 134 Hopf algebras, 154

Laplace's expansion of a determinant, 92 linear form on a module, 46

main involution of a graded algebra, 64 McCoy's theorem, 14 modified Hopf algebra, 155 modified tensor product of graded algebras, 52 modified tensor product of graded coalgebras, 146 multilinear mapping, 1 multiplication mapping of an algebra, 131 multiplicatively closed subset of a ring, 35

Nakayama's lemma, 32 non-negative grading on an algebra, 47

Pfaffian of an alternating matrix, 104 projective module, 27 rank of a free module, 14, 90 right exactness of tensor products, 24 ring of fractions, 35

skew-commutative algebra, 150 skew-commutative coalgebra, 150 skew-commutative modified Hopf algebra, 156 skew derivation on an algebra, 57 structural homomorphism of an algebra, 43 subalgebra, 43 symmetric algebra of a direct sum, 121 symmetric algebra of a module, 119 symmetric multilinear mapping, 10 symmetric powers of a module, 11

tensor algebra of a free module, 74 tensor algebra of a module, 72 tensor powers of a module, 6 tensor product of algebras, 44 tensor product of coalgebras, 138 tensor product of homomorphisms, 22 tensor product of Hopf algebras, 157 tensor product of modules, 4 total grading on a tensor product, 50, 132 trivial grading, 50 trivial ring, 1 twisting isomorphism, 61 twisted tensor product of algebras, 61

opposite of a ring or algebra, 169

unit mapping of an algebra, 131 universal problem for multilinear mappings, 2

permanent of a square matrix, 129

wedge notation for exterior algebras,

Multilinear Algebra

Multilinear Algebra

Multilinear algebra