multilinear algebra – d g northcott multilinear algebra – werner greub multilinear algebra – marvin marcus multilinear algebra and differential forms for beginners

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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

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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 record for this publication is available from the British Library
Library of Congress Catalogue Card Number: 83-27210

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Preface ix

1 Multilinear mappings 1

General remarks 1
.1 Multilinear mappings 1
.2 The tensor notation 4
.3 Tensor powers of a module 6
1.4 Alternating multilinear mappings 6
.5 Symmetric multilinear mappings 10
i.6 Comments and exercises 13
.7 Solutions to selected exercises 15

2 Some properties of tensor products 19

General remarks 19
2.1 Basic isomorphisms 19
2.2 Tensor products of homomorphisms 22
2.3 Tensor products and direct sums 25
2.4 Additional structure 28
2.5 Covariant extension 29
2.6 Comments and exercises 31
2.7 Solutions to selected exercises 37

3 Associative algebras 42

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vi Contents

3.8 Comments and exercises 58
3.9 Solutions to selected exercises 65

4 The tensor algebra of a module 69

General remarks 69
4.1 The tensor algebra 69
4.2 Functorial properties 72
4.3 The tensor algebra of a free module 74
4.4 Covariant extension of a tensor algebra 75
4.5 Derivations and skew derivations on a tensor algebra 76
4.6 Comments and exercises 78
4.7 Solutions to selected exercises 80

5 The exterior algebra of a module 84

General remarks 84
5.1 The exterior algebra 84
5.2 Functorial properties 87
5.3 The exterior algebra of a free module 89
5.4 The exterior algebra of a direct sum 93
5.5 Covariant extension of an exterior algebra 95
5.6 Skew derivations on an exterior algebra 96
5.7 Pfaffians 100
5.8 Comments and exercises 105
5.9 Solutions to selected exercises 111

6 The symmetric algebra of a module 117

General remarks 117
6.1 The symmetric algebra 118
6.2 Functorial properties 120
6.3 The symmetric algebra of a free module 121
6.4 The symmetric algebra of a direct sum 121
6.5 Covariant extension of a symmetric algebra 122
6.6 Derivations on a symmetric algebra 123
6.7 Differential operators 124
6.8 Comments and exercises 126

7 Coalgebras and Hopf algebras 130

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Contents vii

7.9 Tensor products of Hopf algebras 156

7.10 E(M) as a (modified) Hopf algebra 158
7.11 The Grassmann algebra of a module 160
7.12 S(M) as a Hopf algebra 163
7.13 Comments and exercises 166
7.14 Solutions to selected exercises 169

8 Graded duality 175

General remarks 175
8.1 Modules of linear forms 175
8.2 The graded dual of a graded module 178
8.3 Graded duals of algebras and coalgebras 184
8.4 Graded duals of Hopf algebras 188
8.5 Comments and exercises 191
8.6 Solutions to selected exercises 194

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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 to fill out 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.

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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.

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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, M2, . . . , 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 Mt x M2 x • • • x Mp into M. We use

mi9 m2, . . . , mp to denote typical elements of Mu M2, . . . , Mp respectively

and r to denote a typical element of R. The mapping cf> is called multilinear if
cj)(mu...9m/i + m",... ,mp)

= (t>(mu..., m j , . . . , mp) + </>(m1?..., 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.

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2 Multilinear mappings

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 M2 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

M2 x • • • x Mp 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 Mx x M2 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 
R-homomorphism of this free module into N.

We are now ready to solve Problem 1. Let U(Ml, M2, . . . , Mp) be the

free R-module generated by the cartesian product Mx x M2 x • • • x Mp. Of

course, this has the set of sequences (mls m2, . . . , mp) as a base. The elements

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Multlinear mappings

" , . . . , mp)-(mu ..., m'h..., mp)

- ( m1, . . . , m ;/, . . . ,mp) (1.1.4)

and

(ml9..., rmh . . . , mp)-r(mu . . . , mi9..., mp) (1.1.5)

generate a submodule F(M 1? M2, . . . , Mp) say. Put

M = ( 7 ( M1, M2, . . . , Mp) / K ( M1, M2, . . . , Mp) (1.1.6)

and define

0 : Mx x M2 x • • • x Mp-*M (1.1.7)

so that 0(mx, m2, . . . , mp) is the natural image of (ml5 m2, . . . , mp),

con-sidered as an element of U(Ml9 M2, . . . , Mp), in M. Since the elements of

U(Ml9M2,..., Mp) described in (1.1.4) and (1.1.5) become zero in M, 0

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 M2 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/(M1,M2,... ,MP). Accordingly

there is induced a homomorphism h:M—+N which satisfies
/i(0(m1,m2,... ,mp)) = ^ ( m1, m2, . . . ,mp). 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 Ml 9 M2, . . . , Mp {p > 1) be R-modules. Then the universal

problem for multilinear mappings of Mx x M2 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).

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4 Multilinear mappings

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 <l>(ml9 m2, . . . , mp). Also let hx: M-+M/M' be the natural

homo-morphism 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\, m2, . . . , m'p) + • • *,

where the sum is finite. However,

r(/)(mum2,..., mp) = (f)(rmu m2, . . . , mp)

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

1.2 The tensor notation

Once again Ml9 M2, . . . , Mp, where p> 1, denote R-modules and

we continue to use ml9 m2, . . . , 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

M2 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^ m2, . . . , mp)

of M. When this notation is employed, Mx đR M2 đR ã ã ã ®R Mp is called

the tensor product of Ml9 M2, . . . , 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 đ m2 đ ã ã ã đ 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
ã ã ã đ mp

- - - đmp + m1 đ ã ã ã đ mf( đ ã ã • ®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
Mx ®KM2 ®R"' ®RMp is a finite sum of elements of the form

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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" '

đR Mp into N, such that

h{m^ đ m2 đ ã • • ® mp) = \j/(m1, m2, . . . , mp)

for all ml9 m2, . . . , 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 đR M2 đR * * ã

đRMp is just Mx 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 ® M2 đ ã ã ã đ Mp in place of the typographically

more cumbersome but more explicit Mx ®RM2 ®R" ' ®RMp.

Neverthe-less, 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 f is a free .R-module and that Bt is a base

for Mf. The first point to note is that any mapping ofBl x B2 x • • • x Bp into

an R-module N has precisely one extension to a multilinear mapping of
Mx x M2 x • • • x Mp into N. We use this observation in the proof of our

next theorem.

Theorem 3. Let M, (i = 1, 2 , . . . , p) be a free R-module with a base Bt. Then

Mt đRM2 đR- ã' đRMp is also a free R-module and it has the elements

bt ® b2 ® - • • ® bp, where bt e Bh as a base.

Proof Denote by M the free R-module generated by the set Bx x B2 x • • •

x Bp. Thus the sequences (bl9 b2,..., bp) form a base for M. There is then a

mapping 0, of B1 x B2 x • • • x Bp into M, in which (f>(bu b2,..., bp) is the

base element (bl9 b2,..., bp). Now, as we noted above, the mapping has an

extension (denoted by the same letter) to a multilinear mapping of
Mx x M2 x • • x Mp into M. Clearly all we need to do to complete the proof

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6 Multilinear mappings

Suppose then that we have a multilinear mapping
\//: M1 xM2 x • • • xMp—•N.

In these circumstances there exists an R-homomorphism h: M—•AT which
maps (bl , fc2,..., bp) into \j/{b1 , b2, . . . , bp). Obviously ft ° </> = i/f. Moreover,

if h'\ M^>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/(m1,m2,... ,mp) = 0 whenever the sequence

(m1? m2, . . . , 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 l<i<j<p, then on expanding
n(mu..., mi+nip..., mf + rn^-,..., mp) = 0,

where the element m,- + m} occurs in both the i-th andj-th positions, we find

that

n{mu..., mi9..., mp ..., mp) + >y(m1,..., mp . . . , mi9..., mp) = 0.

Thus if we interchange two terms in the sequence {ml, m2, . . . , mp) the effect

on n(m1, m2, . . . , mp) is to multiply it by — 1. From this observation we at

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Alternating multilinear mappings 7

Lemma 1. Let rj.MxMx'xM—^N be an alternating multilinear

mapping and let (il 5 i2, . . . , ip) be a permutation of ( 1 , 2 , . . . , p). Then

rj(mii9 mi2,..., mip) = ±ri(m1,m2,..., mp\

where the sign is plus if the permutation is even and minus if it is odd.
Another useful observation is recorded in

Lemma 2. Let ri'.MxMx — -xM—>N 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? m2, . . . , 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 
R-modules. 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 đ m2 đ ã ã ã đ mp,

where ( m1, m2, . . . , mp) contains a repetition. (It is understood that

J1(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 đ M đ ã ã ã đ M into K which takes mx đ m2 đ ã ã •

® mp into C(wi1? m2, . . . , 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
and ft' agree on the elements ^/(rnj, m2, . . . , mp) and therefore they agree on a

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8 Multilinear mappings

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 A •

and

ml A • • • Arm; A • • • Amp = r(m1 A • • •

But */ is also alternating. Consequently mx AWI2 A
(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
Am;'

A • * •

A • • •
A • • *

A mp

A m

).

= 0

„ (1.4.3)

(1.4.4)
whenever

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.

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Alternating multilinear mappings 9

The mapping of MxMx- xM 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 ® ' * * ® wp 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 đ ã * * đ 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, m2, . . . , 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 đ m2 đ ã ã ã ® 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 jp( 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,

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10 Multilinear mappings

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 r e p e t i t i o n n{b1, b2,..., bp) is t o

b e z e r o . 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

p e r m u t a t i o n (b[, b'2,..., b'p\ of (bi9 b2,..., bp), s u c h t h a t (b\, b'2,..., b'p)

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 
homo-m o r p h i s homo-m h : N — > K s u c h t h a t h ( b1, b2, . . . , bp) = C ( bl 9b2, . . . , bp)

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 
homo-morphism 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

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Symmetric multlinear mappings 11

co: M x M x • - x M —+K

there exists a unique R-homomorphism h: N-^K such that h°6 = oj.
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

w1đ m2đ *** đmp miiđmi2đ ã ã ã ®m,-, (1.5.2)

where (il9i29..., ip) is a permutation of ( 1 , 2 , . . . , p). Put

N=Tp(M)/Hp(M)

and defined: MxMx- xM—•Nsothat 0(m1,m2,... ,mp) 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)

and

m1m2 . . . mp = 0(ml9 m2, . . . , mp). (1.5.4)

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 i2, . . . , 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 such that

h(mlm2 ... mp) = w(ml,m2,..., mp)

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12 Multilinear mappings

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 đ ãã • ®mi®mi + 1 ® • •• ®mp

-ml®--' ® mx + 1 đ m < đ ããã ® mp,

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

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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
trans-formations. 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

: : : : : (1 6 1)

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14 Multilinear mappings

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
8 l o M ) 2 « i M ) 2 «2( i 4 ) 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 p<q, then a non-trivial solution exists.

Exercise 1 can be used inter alia to establish some fundamental facts
about free modules and their bases. We illustrate this by means of the next
exercises.

Exercise 2. The R-module M can be generated by q elements. Show that any

q + l elements ofM are linearly dependent over R. Show also that if M is a
free module with a base of q elements, then (i) q is the smallest number of

elements which will generate M, and (ii) any q elements that generate M form
a base.

Exercise 3. Let Mbea free R-module and let B, B' be bases of M. Show that

either (i) B and B' are both infinite, or (ii) B and B' are both finite and contain
the same number of elements.

Exercise 3 shows that the number of elements in a base of a free R-module
M is the same for all choices of the base. This number, which is called the
rank of the free module, will be denoted by rank^(M). Thus mnkR(M) is

either a non-negative integer or it is 'plus infinity'.
We now turn our attention to some other matters.

Exercise 4. Let Il9I2,...,Ip be ideals of R such that /x + /2 + • • • +Ip = R,

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x--Solutions to selected exercises 15

is a multilinear mapping of R-modules, then every element ofMl x M2 x • • •

x Mp is mapped into zero.

We recall that if M is an R-module and / is an ideal of R, then M/IM has a
natural structure as an R/7-module as well as an R-module.

Exercise 5*. Let MX,M2,... ,Msbe R-modules and let I be an ideal ofR.

Show that the R/I-modules

(M1 ®RM2®R-- ®R MS)/I(M1 ®RM2®R-- ®R Ms)

and

(MJIMJ ®R/I (M2/IM2) ®R/I • • • ®R/I (MJIMS)

are isomorphic and exhibit an explicit isomorphism.

In Chapter 5 we shall derive the basic facts about free modules, their
bases and their ranks by different means. The central fact in the alternative
approach is embodied in the next exercise.

Exercise 6. Let M^Obea free R-module. Show that rankR(M) is the upper

bound of all integers p such that Ep(M)^0.

Exercise 7. Let M be a free R-module of rank n. Determine the ranks of the

free modules Tp(M), Ep(M) and Sp(M).

The next exercise shows that it is possible for the tensor square of a
non-zero module to be non-zero.

Exercise 8. Show that if Z denotes the ring of integers and Q the field of

rational numbers, then (Q/Z) ®Z((Q/Z) = O.

1.7 Solutions to selected exercises

In this section we shall provide complete solutions to Exercises 1

and 5, and make observations about some of the other exercises.

Exercise 1. Show that the equations (1.6.1) have a non-trivial solution, in the

R-module M, if and only if S&q (M) annihilates a non-zero element of M.

Deduce that if M ^ 0 and p<q, then a non-trivial solution exists.

Solution. First suppose that ml9 m2, . . . , mq satisfy the equations with at

least o n e m, different from zero. We claim that <$lq(A)mi = 0fori=l,2,...,q.

Evidently in establishing this we may suppose that p>q.
From the first q equations, namely

flii»ii +al2rn2 + - - - +alqmq = 0

a21mx +a22m2 + • • • -\-a2qmq = 0

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16 Multilinear mappings

we see that Dmt = Q (i= 1, 2 , . . . , q), where D is the q x q minor

0 2 1

In the same way we can show that any other q x q minor of A annihilates all
the mt. Accordingly ^Lq(A)mi = 0 and the necessity of the stated condition

has been established.

Now suppose that 91^(^4) annihilates a non-zero element of M. If v is the
smallest integer for which ^(A) annihilates a non-zero element, then
0 < v < q and 3^ (A)m = 0 for some m ^ 0 in M. If v = 1 the existence of a 
non-trivial solution is obvious. Otherwise there is a (v— 1) x (v— 1) minor that
does not annihilate m. Without loss of generality we can suppose that the
(v— 1) x (v— 1) minor in question occurs in the top left-hand corner of A.

In the determinant

0 1 01v

0 v - l , 2

let cr be the cofactor of xr. Then cvm^0. Furthermore, for 1 <i<p,
-+aiv(cvm)

0f2 ' ' * 0/v

= m - Ql2 '" a u
0 v - l , 2 ' ' ' 0v-l,v

If 1 
if i>v, then it is zero because Sllv(A)m = 0. Thus by taking

wi1=c1wi, m2 = c

and

we obtain a non-trivial solution of the equations.

Finally suppose that M ^ O and p<q. Then (Hq(A) = 0 and this certainly

annihilates a non-zero element of M. Accordingly a non-trivial solution
exists.

Exercise 1 provides the key to Exercise 2.

Exercise 5. Let M1,M2,... 9Msbe R-modules and let I be an ideal of R.

Show that the R/I-modules

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Solutions to selected exercises 17

and

đR/I (M2/IM2) đR/I ã ã ã đR/I (MJIMS)

are isomorphic and exhibit an explicit isomorphism.

Solution. For m, e Mj let m, denote its natural image in Mj/IMj. Now every
R/7-module has an obvious structure as an R-module. On this
under-standing the mapping

M,xM2X' -xM^iMJIM,) đR/I- ã đR/I(MS/IMS)

which takes (mum2,..., ms) into mx đ m2 đ * * ã đ ms is a multilinear

mapping of R-modules. Accordingly there is induced an R-linear mapping
M, đR ã ã ã đR Ms-+ (MJIM1) đR/Iã • • ®R/I (MJIMS)

and this, as it clearly vanishes on I(M1 đR- ã ã đRMS), gives rise to a

mapping

A: (Af i đ^- ã ã đRMs)/I(Ml đR-" đRMS)

(MJIM,) đR/I - • • ®R/I (MS/IMS).

In fact I is a homomorphism of ^//-modules and if mx đ ã ã ã đ ms denotes

the natural image of mxđ ã ã ã đms in {M1 ®R'" ®RMS)/

1(MX®R-" ®RMS\ then

Mmx ®- — ®ms) = ml®m2®" ' ®ms.

Now suppose that, for each ;, we have elements m, and m}, in Mp such

that rhj^m'j. Then

= ml ® • • • ® (mj — m'j) ® • • • ® ms

and this belongs to I(Ml®R" - ®RMS). Repeated applications of this

observation show that
mlđm2đ" ã đms

= m\đ m'2 đ - - đ mfs (mod I(MX đR ã ã ã ®R Ms)).

It follows that there is a well-defined mapping

( M1/ / M1) x - - - x ( Ms/ / Ms)

-> (M! đn ã ã ã đ R MM)/I(MX đR'"đR Ms)

in which (ml9 m2, . . . , ms) goes into ml®m2®- - ®ms. This is a

multi-linear mapping of ^//-modules and so it gives rise to an 
R/I-homomorphism

®R/I -' ã đR/I (MJIMS)

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18 Multilinear mappings

which satisfies /i(mx đ ã ã ã đ ms) = (mx đ m2 đ ã ã ã đ ms). Finally

( j ) ^ ! s ) j ( l s) = m1 đ ã ã ã đ ms

and

(A0/^)^! đ ã ã • ® ms) = k(m1 ® • • • ®ms) = m! đ ã ã ã đ ms.

It follows that k and ^ are inverse isomorphisms and now the solution is
complete.

Our concluding observation has to do with the answers to the questions
posed by Exercise 7. We record that, for a free /^-module M of rank n,

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2

Some properties of tensor products

General remarks

It was in Section (1.2) that we defined the tensor product of a finite
number of modules over a ring R, but at that time we passed on rapidly to
study the tensor, exterior and symmetric powers of a module. However, the
theory of tensor products is rich in results, some of which will be needed
later. Here we provide an account of the most fundamental properties. As in
Chapter 1 we shall sometimes omit the subscript to the symbol ® if it is
clear which is the ring over which the products are formed.

Throughout Chapter 2 the letters R and S denote commutative rings each
possessing an identity element.

2.1 Basic isomorphisms

Let Mu Af 2, . . . , Mp and Nl9 N2,..., Nq be K-modules. In what

follows ml9m29 - - •, mp denote elements of Ml9 M2, • • •, Mp respectively

and nl9n2,... ,nq denote elements of Nl9N2,... 9Nq.

Theorem 1. There is an R-isomorphism

Mi đR'' - đRMp <g)RN1 đR-" đRNq

ô ( M ! ®R''' ®R Mp) ®R (N1 ®R'"®RNq)

in which mx ® • • • ® mp ® nx ® • • • ® nq is matched with (m1 ® • • • ® mp) ®

{nl®'"®nq).

Proof. By Theorem 2 of Chapter 1, the multilinear mapping
Mx x • • • x Mp x Nl x • • • x Nq

where (mi9..., mp, nl9..., nq) goes to (mx đ ã ã ã đ mp) ®(nl®--® nq),

induces a homomorphism

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20 Some properties of tensor products
f:Mlđ--đMpđN1đ-'đNq

-->(Ml

đ-in which

= (mx đ ã ã ã đ mp) đ (ô! đ ã ã ã đ nq).

p

. Suppose that

f = Z mx ® m2 ® • • • ® mp = ^ ^ ® //2 ® • • • ® \xp

are two such representations of ^ and that

= Y< vi ® V2 ® ' * * ® vq

are two similar representations of rj. The remarks in the last paragraph
show that

Consequently

Y£jm1®'"®mp®nl®'"®nq (2.1.1)

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Basic isomorphisms 21

that takes (£, n) into the element (2.1.1). This mapping, because it is bilinear,

induces a homomorphism

- • Ml ® • • • ® Mp

in which (m! ® • • • g m ^ g ) ^ ! ® • • • ®nq) becomes mx ® • • •

Wi®"'"®wfl. The theorem follows because g r ° / and f°g are clearly

identity mappings.

Corollary. There is an isomorphism

(Ml®RM2)®RM3*M1®R(M2®RM3)

of R-modules in which (mx ® m2) ® m3 corresponds to mx ® (m2 ® w3).

Proo/ The theorem provides isomorphisms

(M! ® M2) ® M3 « MX ® M2 ® M3 ôAf! đ (M2 đ M3)

and the corollary follows by combining them.

Theorem 2. Let (il9 i2,..., ip) be a permutation of ( 1 , 2 , . . . , p). Then there

is an isomorphism

Mx ®RM2®R" ®RMp*Mix ®RMh ®R-" ®RMip

of R-modules which associates m1®m2®'-®mp with mix®mi2®'"

®mip.

Proof The multilinear mapping

M1xM2X" xMp-^Mti ®Mh ®" - ®Mip

taking (ml9 m2,..., mp) into mix đ mi2 đ ã ã ã đ mt induces a

homo-morphism

/ : Mj đ M2 đ ã ã ã ® Mp-+Mti ®Mh®'"® Mip,

where /(mx ® m2 ® • • • ® mp) = mix ® mi2 ® • • • ® mi . For similar reasons

there is a homomorphism

with g(mti đ mi2 đ ã ã ã đ mt ) = m1®m2®- - ® mp. Evidently g ° f and

f°g are identity mappings and therefore / is an isomorphism. The theorem
is therefore proved.

The ring R is itself an K-module. As the next theorem shows, this
particular /^-module plays a very special role in the theory of tensor
products.

Theorem 3. Let M be an R-module. Then there is an isomorphism

R®RM&M (of R-modules) in which r®mis matched with rm. There is a

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22 Some properties of tensor products

Proof Theorem 2 of Chapter 1 shows that there is a homomorphism
/ : R®RM-^M with f(r ® m) = rm. The mapping g: M —> R ® R M given

by g(m)=l®m is certainly a homomorphism. But r ® m = l ® r m and
from this it follows that g ° / and / ° g are identity mappings. Consequently
/ is an isomorphism and the theorem is proved.

2.2 Tensor products of homomorphisms

Let Ml9 M2, . . . , Mp and M i , M2, . . . , M'p be ^-modules, and let

fi'.M^Ml ( i = l , 2 p) (2.2.1)

be given homomorphisms. The mapping

Mi x M2 x • • • xMn—>Mi ® M'2 ® • • • ® M'

in which the image of (ml9 m2, . . . , mp) is fiintx) đ /2(m2) đ ã ã • ® fp(mp) is

multilinear and therefore it gives rise to a homomorphism
MXđ M2 đ ã ã ã đMp-+M'1đM'2 đ ã • • ®M'p

of /^-modules. This homomorphism, which is denoted by fx đ f2 đ ã ã ã

đ fp, satisfies

(/l đ fl ® ' ' ' ® fp) (ml ® m2 ® ' ' ' ® mp)

/ i (mi ) ® fi(mi) ® *' * ®/p(^P). (2.2.2)

Note that if each /j is surjective, then / i ® /2 ® • • • ® fp is surjective as well.

Note too that if MJ = M£ for i = 1 , 2 , . . . , p and /j is the identity mapping of

Mi9 then f1®f2®'"®fp is the identity mapping of M1®M2

®--®MP.

Once again suppose that homomorphisms (2.2.1) are given and that, in
addition to these, we have further homomorphisms

gt: M'l - * Mi (i = 1 , 2 , . . . , p). (2.2.3)

It then follows, from (2.2.2), that

(/i ® fi ® *' * ® fp) ° (#i ® 02 ® * " ® 9P)

= (fi°9i) ® (fi°9i) ®''' ®(fp°9P)' (2.2.4)

We can deduce from this that when each f is an isomorphism, then
fi ® fi ®''' ® fp is als° a n isomorphism. For in this situation we can form
both fi® f2®' " ® fp and f{~x đ f2 1 đ ã ã ã đ fp~1, and when these are

combined, in either order, (2.2.4) shows that we obtain an identity mapping.
Consequently not only is fx ® f2 ® • • • ® fp an isomorphism, but we also

have

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Tensor products of homomorphisms 23

summarized succinctly by the statement that the p-Md tensor product is a
covariant functor of p variables.)

There is an additional item of notation which it is convenient to record
here. Suppose that fi,...<)fi_l9fi + l,...,fp are the identity mappings of

M1 ?. . . ,M,-_1,Ml- + 1, . . . ,Mp respectively, and that f'.M^M'i is a

general homomorphism. Under these conditions fl ® f2 đ''' đ fp is often

written as Mx đ ã • • ® f ® • • • ® Mp. Thus

(Mx đ ã ã ã đ f{ đ ã ã • ® Mp) (mx ® • • • ® mi ® • • • ® mp)

= mx ® • • • ®fi(mi) ® • • • ® mp. (2.2.6)

We recall that if/, # are homomorphisms of an ^-module M into an 
R-module M', and if r e R, then we can form new homomorphisms / + g and rf
of M into M'. Indeed to be explicit ( / + g)(m) = f(m) + g(m) and (rf)(m) =
rf(m). In this connection we note that

P p (2.2.7)

and

/ i ® *'' ® rft đ ' ' ã đ /p = r(/i đ *' ã đ fi đ ' *' ® fP), (2.2.8)

where the notation is self-explanatory.

After these preliminaries we turn our attention to the behaviour of tensor
products in relation to exact sequences of the form

M'l -^-> Mt —• M| —> 0 (2.2.9)

where i = 1 , 2 , . . . , p. Denote by N( the image of Mx đ ã ã ã đ gt đ ã • • ® Mp

in Mx ® M2 ® * * • ® Mp. Thus N{ is generated by the elements of the form

mt đ ã ã ã đ0,-(mj')đ ã ã • ®mp and, in particular, it is contained in the

kernel of /x đ/2 đ ã ã ã đ/p.

Theorem 4. Suppose that for each i (1 <i<p) the sequence (2.2.9) is exact.

Then the homomorphism

->M'1đRM'2đR--đRMp

is surjective and its kernel is N1-\-N2-\- ã • • +iVp, where Nt is the image of

Mi ® • • • ® gt ® - - • ® Mp in Mt đR M2 đR ã ã ã đ* Mp.

Proof. For 1

are such that fi(mi) = fi(ni) = m'i. Then there exists m'l eM'l such that

m, — fii = gi(m'l) and so we see that

mx đ ã ã ã đ mt đ • • • ® mp — mx ® • •

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24 Some properties of tensor products

m2 ® * * * ®mp — fi1 ® ^2®"

is in N1 + N2 + * • • + Np and therefore the natural image of m1 đ m2 đ ã * *

đmp in

(Mi đ M2 đ ã ã ã đ MP)/(NX + AT2 + • • • + Np)

depends only on (mi, m'2,..., m'p). It is now easily checked that there is a

multilinear mapping

M\ x M2 x • • • x M'p

which is such that (mi, m2, . . . , m^) is mapped into the image in question. In

this way we arrive at a homomorphism

M\ ® M'2 ® • - • ® M'p

~^(Ml®M2®" -®MP)/(N1 +JV2 + • • • + N , ) (2.2.10)

that takes m\®m'2®- - ® m'p into the natural image of ml ® m2 ® • • •

® mp in the factor module.

Consider the homomorphisms

Mx đ - ã ã đ Mp-ã Mi đ ã • • ® M'p

where the first is fx®f2®' - ®fp and the second is (2.2.10). Their

combined effect is to reproduce the natural mapping of M1 đ M2 đ ã ã •

® Mp onto (Ml®M2®-"® Mp)l{Nl +N2 + • • • + Np) so the kernel of

f\® fi®" ' ® fP must be contained in Nx + N2 + • • • + Np. But we also

have the opposite inclusion because, as was noted earlier, each N( is

contained in the kernel. The proof is now complete.

Corollary. Let M"t —• Mfl —• M'fl —• 0 be an exact sequence of R-modules.

Then the induced sequence

Mi đR' ' ' đRM'1 đR- ã ã ®RMp

- * Mx ®K • • • ®R Mfl ®R • • • ®K Mp

M; ®R-"®R MP->O

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

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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 = Y

J

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 o^A^—»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 
R-module, 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 ậ 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 <rf(Nf). Next from

(B) and (C) it follows that

16/

and now, by using (A), we can deduce that N is the direct sum of its
submodules {0",(Afi)}/e/ in the sense used at the beginning of this section.

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26 Some properties of tensor products

complete representation of N as a direct sum. Furthermore we shall continue
to write

ie/

in the general case, and N = N1®N2(B"'@NS in the case where the

family {Nt}iel reduces to {Nu AT2,..., N

s}-Now suppose that M1, M2, . . . , Mp are R-modules and that each is given

a (generalized) direct sum decomposition, say

M . = L Mi» (d.s.),

where the complete representation is provided by homomorphisms

and

Theorem 5. Suppose that for each \i (1 ^fi^p) we have

M, = I M^ (d.s.),

w/zere the details of the complete representation are as described above. Then
Ml®RM2 ®R-' ®RMp

= Z ^ . ^ M ^ . - ^ M ^ (d.s.),

( » i , . . . , * p )

vv/i^r^ the typical homomorphisms in the complete representation are oix ®

oh ®''' ® oip and nti ® ni2 ® • • • ® 7ilp.

Proo/. Put / = /! x /2 x • • • xlp, N = MX ® M2 ® • • • ®Mp and, for i =
(iui2i...Jp) in / , set

a n

d 7ii = ^il ®ni2 ®' " ®nip- Then a^Ni—^N,

Ui'.N—^Ni and to prove the theorem it will suffice to show that the
conditions which were listed earlier as (A), (B), (C) all hold. That (A) holds is
an immediate consequence of (2.2.4). Indeed (B) and (C) are equally obvious
as soon as it is realized that we need only verify them when n has the form
mx ® m2 ®''' ® mp.

Corollary. Let F be a free R-module with a base {^,}/6/ and let N be an

arbitrary R-module. Then each element of F ®RN has a unique

representa-tion in the form £ (£,. ® n{\ where the nt belong to N and only finitely many of

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Tensor products and direct sums 27

Proof. We have

iel

and N is also a direct sum with N itself as the only summand. Accordingly,
by the theorem,

and so, to establish the corollary, we need only show that for each x in
R£i ® N there is a unique rc e N such that x = £f ® n.

By Theorem 3, there is an isomorphism N&R®N in which n
corresponds to 1 ®n, and of course there is an isomorphism R®N&
R£t ® N which matches 1 ® n with ^ (g) rc. Together these provide an
isomorphism N^R^t ® JV with n corresponding to £f ® rc. Consequently,

as n ranges over N, £t ® n ranges over R£( (x) N giving each element once

and once only.

We recall that a module which is a direct summand of a free module is
called a projective module.

Theorem 6. Let 0 —• M" —• M —> M' —• 0 fee ^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

Zôiđ/ô)) = 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.

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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

an R-homomorphism and therefore Ml®'-'®fs®"'(g)Mp9 where fs

occurs in the /i-th position, is an endomorphism of the K-module Mx đ ã ã •

® M^ ® • • • (g) Mp. For x in Mx đ ã ã ã đ M/4 đ ã • • ® Mp define sx by

5X = (MX ® • • • ® /s ® • • • ® Mp)(x). (2.4.1)

Then, because fs + a = fs+fa and fsa = fs°f(T for s,aeS, it is easily verified

that Mx ® • • • ® 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 đ ã ã ã đ sm^ đ • • • ® 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) ®SC 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 (A<g)RB)®sC and

A®/?(B®SC) are virtually identical bimodules, is known as the associative

law for tensor products.

Theorem 7. Let A be an R-module, B an (R, S)-module, and C an S-module.

Then there is an (R, S)-isomorphism

(A ®RB)®SC&A ®R(B ®SC)

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Covariant extension 29

Proof. We borrow an idea from the proof of Theorem 1. Select an element c
from C and for the moment let it be kept fixed. Consider the mapping

where (a, b) is mapped into a®(b®c). If we regard A,B and
A ®R (B ®s C) simply as K-modules, then the mapping is bilinear and

therefore it induces a homomorphism
A ®R B ã A đR (B đs C)

in which a đ b goes into a ® (b ® c). It follows that if
a ®b + a' ®b' + a" ® b" -\ = 0

in A ®R B, then

a® (b®c) + af ®(b' ®c) + a" ®(b" ®c) + '-=0

inA®R(B®sC).

Let x belong to A ®R B. Then x can be represented in the form

x = ax ®bx +a2 ®b2 + — - +an ®bn

and the remarks of the last paragraph show that

flx ® (h ®c) + a2 ®{b2®c)+--+an® (bn®c) (2.4.3)

depends only on x and c and is independent of the choice of the
representation of x. Consequently there is a well-defined mapping of
(A ®RB)xC into A ®R (B ®s C) in which (x, c) is mapped into the element

(2.4.3). But if we regard A ®RB, C and A ®R (B ®s C) as 5-modules, then

this is a bilinear mapping and therefore it induces an 5-homomorphism

X: {A ®RB) ®SC-^A ®R (B ®SC)

with

k({a ®b)®c) = a®(b®c). (2.4.4)

It is now clear, from (2.4.4), that X is actually an {R, S)-homomorphism.
Entirely similar considerations show that there is an (R, 
S)-homomorphism

fi: A đR (B đs C)-ã {A đR B) ®s C

for which n(a ® (b ®c)) = (a ®b) ®c. Since /i°X and X° \i are identity
mappings, k and \i are inverse isomorphisms. The proof is therefore
complete.

2.5 Covariant extension

Once again R and S denote commutative rings, only now we
assume that a ring-homomorphism

oj.R^S (2.5.1)

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S-30 Some properties of tensor products

module U can be regarded as an K-module by defining rw, where reR and
ue U, to be the same as co(r)u. In fact this turns U into an (R, S)-module. It
follows, as was shown in the last section, that if M is an K-module, then
M ®R U is an (R, S)-module. Note that it is also true of M ®R U that

multiplication of an element by r is the same as multiplication by co(r).
A special case arises if we take U to be S itself. Thus if M is an /^-module,
then M ®R S is an (R, S)-module. Actually it is the S-module structure that

concerns us more because R operates through S in the simple manner just
explained. When M ®R S is considered as an S-module we call it the

covariant extension of M by means of the ring-homomorphism (2.5.1).
Covariant extensions are important because they cover such basic
constructions as localization and the formation of fractions and
polynomials. By combining these constructions under a general heading we
can derive some of their most important properties while still keeping the
details comparatively simple.

Theorem 8. Let Mbea free R-module with {£,-} ieI as a base. If now the ring S

is non-trivial, then the covariant extension M ®R S is a free S-module and it

has {£i ® l }/ 6 / as a base.

Proof. Let x e M ®R S. By Theorem 5 Cor., there is a unique family {st} iel9

of elements of S, with only finitely many st non-zero and x = £ (^ ® s,). The

theorem follows because £t ®^i = si{£>i ® 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 ®RS) 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.

R

S)

Proof. We begin with the case p = 2. By Theorem 7, there is an (R, 
S)-isomorphism

(Mx ®R S) ®s (Af 2 ®R S)«Mx ®* (5 ®s (M2 ®* 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,

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). Finally, the

corollary to Theorem 1 provides an K-isomorphism

Mx ®^ (Af 2 ®R S)ô(Af! đ* M2) đR 5, (2.5.5)

where mlđ(m2đs) 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 
R-isomorphism.

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 
R-module 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' đ N''ã 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.

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32 Some properties of tensor products

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
homo-morphisms. 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 HomR(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) - ã Hom(M đM\NđN') (2.6.2)
of .R-modules.

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Comments and exercises 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') - ã Hom(M đM\N đ 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.

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34 Some properties of tensor products

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 modules is a flat

R-module.

Exercise 9. Show that if M1,M2,...,Mq are flat R-modules, then

Ml®M2®'"®Mqisa flat R-module.

A deeper result is contained in

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
observa-tion 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 
corre-sponding 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

polynomial ring R[X1, X2,..., Xn~\ and covariant extension by means of

the inclusion homomorphism is referred to as adjunction of the 
indeter-minates Xl9 X2,..., Xn. This can be looked at in a more down-to-earth

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Comments and exercises 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
we denote the group by M[Xl 9 X2,. • •, Xn~\, then M\_Xl9 X2,..., Xn~\ can

be regarded as an R[XU X2,..., Xn]-module in which

(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

of R[Xl9 X29...9 Xj-modules in which mX\'X\2... Xvn» is matched with

m ® XyX^2... 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 r' a'r-\-ar'
a a' 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.

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ring-36 Some properties of tensor products

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

V r*

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 <reZ. These new fractions may be

'added', the sum of m/a and m'/cr' being given by
m rri o'm + om!

o o' GO'

In this way we obtain an abelian group which is frequently denoted by
Z ~x (M). In fact we can go further and regard Z "x (M) as a Z "x (K)-module
with

rm

Finally, using Exercise 13, we can readily verify that there is an

iso-morphism

of Z "x (R)-modules in which mJG of Z ~1 (M) corresponds t o r n ® (l/o). We
leave to the reader the task of filling in the details.

Exercise 14*. Let I, be a multiplicatively closed subset of R and letHL~l(R)
be regarded as an R-module by means of the canonical ring-homomorphism
R—•Z"1^). Show that Z- 1( R ) is a flat R-module.

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Solutions to selected exercises 37

Suppose that co:R-^S and X:S—+T are homomorphisms of 
com-mutative rings. By Theorem 7, if M is an ^-module we have an (R, 
S)-isomorphism (M ®RS) ®s T^M ®R (S ®s T) in which (with a

self-explanatory notation) the element (m®s)®t corresponds to m ® (s ® t).
On the other hand, Theorem 3 provides us with an isomorphism
S ®ST^T. Together these yield an isomorphism

(M đR S) đs T ô M đR T (2.6.6)

in which (m ® 5) ® t is matched with m (x) sr. To begin with (2.6.6) is an
isomorphism of K-modules but, because of the way it operates, we can go
further and say that it is actually an isomorphism of T-modules.
Con-sequently if we extend M by means of to and then extend the result by means
of X the total effect is essentially the same as if we had extended M by means
of X ° co. This is what was meant by saying that covariant extension is a
transitive operation.

2.7 Solutions to selected exercises

Exercise 2. Let xl 5x2, . . . , xs be elements of an R-module M and

^i? yi) - - - •> ys belong to an R-module N. Suppose that xx ® yt +x2 ® y^ + ' ' '

+ xs ® ys = 0 in M ® N. Show that there exist finitely generated submodules

M' and Nf, of M and N respectively, such that

(i) the xt are in M' and the yt are in N',

and

(ii) xx (x)^! +x2

Solution. Let U(M, N) be the free K-module generated by M x N and let
V(M9 N) be the submodule of U(M, N) generated by elements having one

or other of the following forms:
{mx +m2, n)- (m1? n)-(m2, n),

(m, nx +n2)- (m, nx) - (m, w2),

(rm, n) — r(m, n),

(m,rn)-r(m,n),

where the notation is self-explanatory. Now in the course of solving
Problem 1 of Chapter 1 it was shown that M ®N=U(M, N)/V(M9 N);

moreover in this context m ® n is the natural image of the basis element
(m, n) in U(M, N)/V(M, N).

Next, because xl đ yx + ã ã • + xs ® ys = 0 in M ®N, we see that

(*i> .Vi)+ ' *' +US» ^s) belongs to V(M, N), that is to say

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38 Some properties of tensor products

where rt e R and ^ is an element of U(M, N) having one or other of the four

forms listed above. Note that (2.7.1) means simply that for each (m, n) in
M x N the sum of the coefficients of all the occurrences of (m, n) in the 
left-hand side equals the sum of the coefficients of its occurrences in the
right-hand side. But only finitely many basis elements (m, n) are actually present
in (2.7.1) and therefore we can choose finitely generated submodules M' and
AT', of M and N respectively, so that if (m, n) appears on either side of (2.7.1),
then (m, n)eM' x N'. This ensures that xl 5 x2, . . . , xs are in M' and that

yi» )^2» • • • j ys a r e *n N'; it also ensures that (xi,yl)+- - + (xs, ys) is in

V(M\ AT) as well as in U(M', AT). But M' ®N'= U(M\ N')/V(M\ AT) and
therefore

+x2

in M' ® AT' as required.

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.

Solution. Let (/>: R —• R/I be the natural homomorphism. Because tensor
products are right exact, the exact sequence

0 —>I —>R —

gives rise to the exact sequence

But M (x) R is isomorphic to M under an isomorphism which matches
m®r with rm (see Theorem 3). It follows that there is a surjective
homomorphism M^>M ® (R/I) which maps m into m ® c/>(l), and 
further-more the kernel of this homomorphism is the image of / ® M in M under
the result of combining the homomorphisms

I ®M —>R®M ^-+M.

Accordingly the kernel of the homomorphism M-+M ® (R/I) is IM and
thus there is induced an isomorphism M/IM ^M ® (R/I).

Exercise 4. Let M be a finitely generated R-module. Show that if M = JM

for every maximal ideal J, then M = 0.

Solution. Let ml9 m2, . . . , mn generate M as an .R-module and let J be a

maximal ideal. Then, since m{ belongs to M = JM, we have

mi = ailml +ai2m2 + • • • +ainmn

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Solutions to selected exercises 39

where the atj are in J. Consequently if D is the determinant

21 0 2 2 -1 • " a2n

then Dmt = 0 for i = l,2, . . . , n . It follows that D belongs to the ideal,

Ann^(M) say, formed by the elements of R that annihilate M. But D =
a + (—1)", where aeJ, and therefore D$J. This shows that AnnR(M)^J,

i.e. it shows that Ann^(M) is not contained in any maximal ideal.
Accordingly AnnR(M) = R and now we see that M = 0.

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.
Solution. Assume (i) and let J be a maximal ideal. Then, since R/J is singly
generated and non-zero, we have M ®(R/J)^0. But, by Exercise 3,
M ® (R/J) is isomorphic to M/JM and so JM # M. Accordingly (i) implies
(ii).

Assume (ii) and let iV be a finitely generated K-module such that

M ®RN = 0. If now J is a maximal ideal, then (Chapter 1, Exercise 5)

(M/JM) ®R/J(N/JN)zz(M ®RN)/J(M ®RN) = 0. But R/J is a field and

M/JM and N/JN are vector spaces over R/J and therefore they are free
K/J-modules. Furthermore, by hypothesis M/JM ^ 0. It follows (Chapter 1,
Theorem 3) that N/JN = 0. We now know that N = JN for every maximal
ideal J and so we may conclude, by virtue of Exercise 4, that N = 0. This
completes the solution.

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.

Solution. Let / : U—> V be an injective homomorphism of K-modules. It
will be enough to show that f ®M: U ®M—>V ®M is an injection as
well. To this end let £ = ul®m1+u2®rn2+- - +us®ms, where uteU

and mteM, belong to the kernel of f®M. Then / ( w1) ® m1+

-+/(ws) ®ms = 0\nV ®M.lt now follows, from Exercise 2, that there exists

a finitely generated submodule V* (of V) containing f(ux), f(u2),..., f(us)

and a finitely generated submodule M* (of M) containing ml9 ra2,..., ms

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40 Some properties of tensor products

/(Mi) ® m1 +f{u2) ® m2 + • • •

in V* ® M * . Choose a flat submodule M' of M so that M*^M'. The
obvious homomorphism K* đ M* ã V đ M' then shows that
f(u1)®ml + -'+f(us)®ma = 0 in V ®M'.

Let ux ® mx + w2 đ m2 + ã ã ã + us ® ms = x in £/ ® M' and consider the

natural commutative diagram

(7 đ M' ã V đ M'

(/ ® M • K ® M.

The image of x in F ® M' is zero, and / ® M' is an injection because / is an
injection and M' is flat. It follows that x = 0. But £ is the image of x under the
homomorphism 1 / ® M ' — • L ^ M so £ = 0 a s well.

Exercise 11. Show that the field Q of rational numbers, when considered as a

Z-module, is flat but not projective.

Solution. Let N be a finitely generated Z-submodule of Q. Then there exists
k e Z, k 7^0 such that N is contained in the Z-submodule of Q generated by
l//c. But the Z-submodule generated by 1/fe is isomorphic to Z and therefore
N is isomorphic to an ideal of Z. It follows that AT is a free Z-module so a
fortiori it is Z-flat. Exercise 10 now shows that Q is a flat Z-module.

We next show that Q is not a submodule of a free Z-module. (This is more
than enough to establish that Q is not projective.) Assume the contrary and
let Q be a submodule of a free Z-module F with a base {£,.} ieI. Then we have

a relation 1 = £ mt£h where mteZ and only finitely many mt's are non-zero.

Choose j e / so that m; ^ 0 and then choose p e Z, /? # 0 so that p does not

divide m^. Now

where the n,- are integers, and it follows from this that pn^ = m, and therefore
p divides my This is the desired contradiction.

Exercise 13. Let m belong to the R-module M and let a belong to the

multiplicatively closed subset Z of R. Show that m®(l/o-) = 0 in
M ®K E ~ * CR) if and only if a'm = 0 for some a' e S.

Solution. First suppose that m ® [I/a) = 0 in M ®R 1 ~x (K). By Exercise 2,

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Solutions to selected exercises 41

respectively, such that m ® (I/a) = 0 in M' ® L. Choose ax e Z so that L c

Ril/aaJ; it then follows that m ® (1/(7) = 0 in M (^Ril/aậ
Consider the exact sequence

4> 1

0 —>I —>R —>R >0

(7(7 x

of /^-modules, where (f)(r) = r/aa1. First we note that / consists of all the

elements r of R such that c'r = 0 for some a ' e l Next there is induced an
exact sequence

M ® / >M ®R >M ®R >0

(7(7 x

and, moreover, m ® al belongs to the kernel of M ® 0, that is to say it

belongs to the image of M ® / in M ® R. The isomorphism M ® R % M in
which m ® r is matched with rm now shows that o^m belongs to IM. We can
therefore find ( j2e l s o that a2Gim = 0. But (72(7! e S so m is annihilated by

an element of Z.

Conversely suppose that a'm = 0 for some a' e Z. Then in M ®^ Z "1 (R)
we have

1 a' 1
m ® — = m ® — = a'm ® — = 0

a aa aa

as required.

Exercise 14. Let Jibe a multiplicatively closed subset of R and letT~l (R) be
regarded as an R-module by means of the canonical ring-homomorphism
K — • Z "1^ ) . Show that Z "1^ ) is a flat R-module.

Solution. Let / : M' —» M be an injective homomorphism of /^-modules. It
is sufficient to show that / ® Z "l (R) is also an injection. To this end let x in
W ® Z " * (R) belong to the kernel of / ® Z ~x (R\ and let x = m' ® (I/a),
where m! e M' and a e Z. (Every element of M' ®^ Z ~x (K) can be written in
this form.) Then f(m') ® (l/(7) = 0 and therefore, by Exercise 13, af(m') = 0
for some a' e Z. Accordingly f(a'm') — 0 and therefore a'm' = 0 because / is
an injection. Finally

x = m'

and the solution is complete.

1

a m' (

a

aa' a'm'(

1
*> —

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3

Associative algebras

General remarks

Our main objective is to discuss the important algebras that can be
derived from a module, and so in this, the last of the introductory chapters,
we shall study those general aspects of the theory of Algebras that are
relevant to our goal. At the outset it needs to be stressed that we shall be
concerned solely with algebras that are associative and possess an identity
element. Accordingly, from now on, the term algebra will only be used in
this restricted sense.

As before R and S denote commutative rings, each with an identity
element, and a ring-homomorphism between them is required to take
identity element into identity element. Later, when we come to consider
homomorphisms of algebras, these too will be required to preserve identity
elements.

Finally when dealing with tensor products of modules we shall
sometimes omit from the symbol (g) the suffix which indicates the
underlying ring. In this chapter whenever the suffix is left out the ring in
question is always R.

3.1 Basic definitions

Let A be a (not necessarily commutative) ring with an identity
element lA, and at the same time let it be a module over the commutative

ring R. We suppose that the sum of two elements of A is the same whether
we use the ring or the module structure. In these circumstances A is called
an R-algebra provided that

r{a1a2) = (ra1)a2=a1(ra2) (3.1.1)

whenever au a2eA and reR. This condition is equivalent to

(r1a1)(r2a2) = (rlr2)(ala2l (3.1.2)

where now r1? r2 denote elements of R.

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Basic definitions 43

Let A be an K-algebra. The mapping

cj):R-+A, (3.1.3)

defined by 0(r) = r\A, is both a ring-homomorphism and a homomorphism

of R-modules. Furthermore, since (j)(r)a = ra = a(j)(r), (j){R) is contained in
the centre of A We shall call (3.1.3) the structural homomorphism of the 
JR-algebra.

This last concept provides an alternative way of looking at K-algebras.
Suppose that A is a ring with identity element and assume that we are given
a ring-homomorphism cp: R—*A which preserves identity elements and
maps R into the centre of A. If we turn A into an ^-module by putting
ra = (t>(r)a, then A becomes an R-algebra with (f> as its structural 
homo-morphism. For example, we see that R itself is an ^-algebra with the
identity mapping as the structural homomorphism.

An important example of an K-algebra arises in the following way. Let M
and N be R-modules. The R-linear mappings of M into N can be added and
they can be multiplied by elements of R. In fact they form the R-module
which is denoted by Hom^ (M, N). When N = M these homomorphisms are

the endomorphisms of M and in this case we use the notation End^(M)
rather than HomR (M,M). Now if /, g belong to EndR(M), then so does

f°g. Indeed it is easy to check that EndR (M) becomes a ring with identity if

the product of / and g is taken to be / ° g. But for r e R we have

so that in fact End^(M) is an K-algebra. We call EndR(M) the

endo-morphism algebra of M. The identity element of EndR{M) is the identity

mapping of M and in the structural homomorphism R—>EndR(M) an

element r, of R, is mapped into the corresponding homothety that is the
mapping M —• M in which meM goes into rm.

We shall now introduce some further terminology. Let A and B be 
#-algebras. A mapping f:A—+B is called an algebra-homomorphism, or a
homomorphism of R-algebras, if it is both a homomorphism of rings
(preserving identity elements) and a homomorphism of R-modules. For
example the structural homomorphism R—+A is a homomorphism of 
R-algebras. Note that if (/>: R —• A and \j/: R—+B are the structural 
homo-morphisms of A and B, then a mapping f:A—+B is an 
algebra-homomorphism if and only if it is a algebra-homomorphism of rings and / ° c/> = \\t.
Naturally a bijective homomorphism is called an 
algebra-isomorphism.

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44 Associative algebras

intersection of any family of subalgebras of A is again a subalgebra. Hence if

U is any subset of A, there will be a smallest subalgebra, B say, containing U.
We call B the subalgebra generated by U. As an K-module B is spanned by
all products ulu1... un, where ute (7 and n > 0 . Here when n = 0we have to

do with the empty product and this is interpreted as being the identity
element of A. Of course the inclusion mapping into A of one of its
subalgebras is an algebra-homomorphism.

Finally suppose that / is a two-sided ideal of the ^R-algebra A. Then / is
necessarily an K-submodule. Accordingly A/I is both a ring and an 
R-module. Indeed A/I is an jR-algebra. Note that the natural mapping of A
onto the factor algebra A/I is a homomorphism of K-algebras.

3.2 Tensor products of algebras

LetAuA2,. • •,;4Pbe R-algebras. Then Ax đRA2 đR- ã ã đRApis

certainly an /^-module. Indeed, as will be established shortly, it has a
natural structure as an K-algebra.

Consider the multilinear mapping

AlxA2X'-xApxA1xA2x-- x Ap—>AX ®A2®- - ®Ap

in which (al9 a2,..., ap, a\,af2,..., a'p) is mapped into a±a\ ® a2a!2 đ ã ã ã

đapa'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 
R-isomorphism

® Ap)

p ® - ' ® A p (3.2.2)

in which, with a self-explanatory notation, (a1đ--- đap)đ(a'lđ ã • •

®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 đ - ã ã ® Ap) ® (Ax ®'' - ® Ap)-+ Ax ® A2 đ ã ã ã đ 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 đ A2 đ ã ã ã đ Ap, to be

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Tensor products of algebras 45

= ala'l ® a2a'2 đ ã ã ã đ apa'p. (3.2.5)

It follows that if x = al®-- ®ap, x' = a\®'' * ®a'p, and xn = a'{®-'

®afp, then

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 ® A2 ® * * • ® Ap.

We are now ready to turn Ax đ A2 đ ã ã ã đ 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 đ A2 đ ã * ã đ 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,.. ã, Ap be R-algebras. Then

AlđRA2 đR'-đRAp

is an R-algebra, where the R-module structure is the usual one and the product
of alđa2đ- - ã đap and a\ đ a!2 đ ã ã ã đ a'p is axa\ đa2a'2 ® • • • ®apdp.

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 and BUB2, ...,Bq be R-algebras. Then

there is an isomorphism

A1đR'"đRApđBlđR"-đRBq

ô ( At đRã ã ã đR Ap) đR (Bx đR'"đRBq)

of R-algebras in which al®---®ap®bl®--'®bq is associated with

(a1®--®ap)®(b1®'-®bq).

Proof By Theorem 1 of Chapter 2, there is an isomorphism /, of R-modules,
that maps Axđ ã ã ã đApđB1đ ã ã ã đBq onto (Axđ ã ã ã đAp)đ(B1đ

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46 Associative algebras

f(a1đ'-đapđblđ--đbq)

= K đ ã ã ã đ ap) đ (bx đ • •

b'q. Then, by Theorem 1,

Let x = ax ® • • • ® ap ® bx ® • • • ® bq and x' =

xx' = ala'l ® •

and therefore

/(**') = ( a ^ i đ ã ã • ® apa'p) ® ( b ^ ; ® • • • ® bgb;)

But / is R-linear. Hence if y and / a r e any two elements of Ax đ ã ã ã đ 4pđ

# 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.

Theorem 3. Let Al9A2,..., Ap be R-algebras and let (il5 i2, . . . , ip) be a

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 đ a2 đ ã ã ã đ ap corresponds to aix đ aiz đ ã ã ã

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 ^A 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 ® ' " ' đ fP- Ax đR A2 đR ã ã ã ®R Ap

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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 đ A2 đ ã ã ã đ 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 đ a2 đ ã * * đ ap and a\ đ a'2 đ ã • • ® 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=^An (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<0. A

grading with this additional property will be called a non-negative grading.
Suppose that A is graded by {An} neI. Then an element a e A has a unique

representation in the form

where an e An and only finitely many terms on the right-hand side are

non-zero. We call the summands in the representation the homogeneous
components of a.

Theorem 5. Let {An}neI be a grading on the R-algebra A. Then the identity

element \A belongs to Ao.

Proof. Let e be the homogeneous component of lA of degree zero and let

aneAn. By comparing the components of degree n in the equations \Aan =

an = anlA we see that san = an = ane. But every element of A is a sum of

homogeneous elements and therefore ex = x = xe for all xeA. Accordingly
lA=s and the theorem is proved.

Corollary. Ao is an R-subalgebra of A.

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R-48 Associative algebras

algebra, by its homogeneous elements of degree one. The relevant facts are
recorded in the next lemma.

Lemma 1. Let {An}nel be a grading on the R-algebra A, and suppose that A

is generated, as an R-algebra, by Av Then the grading is non-negative, and,

for p>\, 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 
R-algebra A. Put

Hn = HnAn. (3.3.1)

Then

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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 <^H = Hn. Thus we have isomorphisms

AJHH*(Am + H)/H (3.3.5)

of R-modules and these can be combined to give an isomorphism

A/H*ZAJHn (d.s.). (3.3.6)

nsZ

In the first instance (3.3.6) is just an isomorphism of /^-modules, but of
course there is a unique way to turn the right-hand side into a graded
JR-algebra so that (3.3.6) becomes an isomorphism of graded JR-algebras.

We next investigate the tensor product of A(1\ A{2\..., Aip\ where each
A{i) is a graded R-algebra. By Theorem 5 of Chapter 2,

Ail) ® Ai2) ® • • • ® Aip) = £ A^ đ A\2) đ ã ã ã đ A\p) (d.s.),
where the direct sum is taken over all sequences (ils i2,..., ip) of p integers.

Put

A = A{1) đ A(2) đ ã • • ® Aip) (3.3.7)
and for each sequence / = (il9 i2,..., ip) let

A^A^QAW® - - - ®A\P\ (3.3.8)

Then (to recapitulate) A is an R-algebra and

A = Z*i (d-s.) (3.3.9)
/

this being a direct sum of/^-modules. Furthermore if J= (JiJ2> • • • JP) is a

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50 Associative algebras

It will be convenient to put

\I\ = ii+i2 + ~' + ip (3.3.10)

and

?

R

"- ®

R

A\

P

K (3.3.11)

\l\ = n \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
graded K-algebras, then A{1) đRAi2) đR- ã ã đRA{p) is also a graded

R-algebra, 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) is a permutation of ( 1 , 2 , . . . , p). Then the isomorphism

A(1)đRAi2)đR'"đRAip)

ôi40<l) đRAUl2) đR'" đRAUlo) (3.3.14)

that we get from Theorem 3 preserves degrees and so it too is an
isomorphism of graded algebras.

In the case of Theorem 4 a little more explanation is needed. If B is an 
R-algebra we can obtain a grading on B by putting B0 = B and Bn = 0

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A modified graded tensor product 51

graded K-algebra. Then Theorem 4 provides algebra-isomorphisms

R®RA*A (3.3.15)

and

A®RR*A. (3.3.16)

It is readily checked that these are isomorphisms of graded algebras
provided that we endow R with the trivial grading.

Our final observation in this section concerns homomorphisms
ft: A{i)-+B{i\ where i = 1 , 2 , . . . , p, of graded algebras. It was observed in

Section (3.2) that / i ® /2 ® * * * ® /p is a homomorphism

A{1) đRAi2) đR> ã ã đRA(p) B{1) ®RBi2) ®R-" ®RB(P)

of K-algebras. However, in the present situation it has the additional
property of preserving degrees and therefore, this time, / i đ f2 đ ã ã * ® fp is

a homomorphism of graded algebras.

3.4 A modified graded tensor product

Let A{1\ A{2\ . . . , Aip) be graded ^-algebras. We have shown that
A{1) ® A(2) ® • • • ® Aip) = A say has a natural structure as a graded 
K-algebra. We propose to modify this structure. The outcome will be that A
will remain a graded ^-algebra with the same R-module structure, the same
grading and the same identity element. Only the definition of the product of
two elements of A will be changed.

Let / = (I'I, i2, - -, ip) be a sequence of p integers and put

as we did before. Then A is the direct sum of the A{ so that each as A has a

unique representation in the form

fl = Ia,, (3-4.1)

/

where aleAl and only finitely many of the summands are non-zero.

Next let J = {ji,j2i • - • JP) be a second sequence of p integers. The

multiplication already introduced on A provides a bilinear mapping

M/J: ^./ x Aj >A
I+J,

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

(3.4.2)

JV(J,J)=£y.

r>s
and

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52 Associative algebras

Suppose that

I J

are elements of A. Then the modified product of a and a' is defined to be

5 > ( / , J W a , , a ; ) . (3.4.4)
IJ

Thus if ait đ ai2 đ ã ã ã đ aip belongs to Aj and a^ đ fl}2 đ ã ã • ® a}p belongs

to Xj, then, whereas their ordinary product is aixa'h đ ôl2ô}2 đ * ã * đ 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) đ e(2) đ ã ã ã đ 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.

L e t 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 đ ã ã ã ® at , y = a'jx ® • • • ® a) and z =

0*, ® ''' ®a'k • ^u t n o w5 in yie 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)

in order to distinguish it from A{1) ® Ai2) ® • • • ® Aip). Note that the total
grading on A{1) đ A{2) đ ã ã ã đ Aip) also serves as a grading on
A{1) đ A(2) đ ã ã • ® 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 
R-algebras. Then

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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)đ ã ã ã đaip)đb{1)đ • • • ®biq) with (a{1)® • • • ®aip))®

Proof. We already know that there is an isomorphism / , of graded
algebras, which maps A{1)đ ã ã ã đAip)đB{1)đ • • • ®Biq) onto
(Ail) ® • • • ® A{p)) đ (B{1) đ" ã đ 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 =

O'i»7*2» • • • Jq) a n (i W = (wls w2, . . . , 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 = aiiđ--đaipđbjiđ---đbjq and y = ocViđ ã • • ®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 đ ã ã ã đ 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
p r o d u c t o f (ati ®-- ®aip)®(bji ®-- ®bjq) a n d (ay.(x) ã ã ã đ<xVp)đ

(PWl đ- " đPWq) in (3.4.8). However, this latter product is

where { is the product of ati đ- ã ã đ a{ and ocVi đ • • • ® ocVp in

A{1) ® A{2) ® • - - ® A{p\ and n is the product of bh đ - ã ã đ bjq and

PWlđ" ' đ 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).

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54 Associative algebras

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 đ f2 đ ã ã * đ 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 đ * ã ã đ fp is a

homomorphism

đRB{2) đR'" ®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 đ at2 đ ã ã ã đ at and y = aji đ ocJ2 đ ã ã •

® (Xj belong to A^ ® • • • ® A\p) and Af^ đ ã ã ã đ 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 amaB (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 a2 = 0 for every

a^eA^ then A is anticommutative.

Proof From Lemma 1 we see that An = 0 for all n<0 and also that Ao is

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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 ! X2 . . . xmy ^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,

ujui = (-l)m2uiUj=-uiuj

and now it follows that a^ = 0. Thus the lemma is proved.

Theorem 7. The modified tensor product of a finite number of

anticommuta-tive R-algebras is itself an anticommutaanticommuta-tive 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+s2a2r®b2s

is zero.

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56 Associative algebras

3.6 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 
ring-homomorphism

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 
R-algebra A is the S-R-algebra A ®R 5. Note that the structural homomorphism

5ã A đRS maps s into lA đs.

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--' ®s (Ap ®R 5)

*{Al®RA2®R'-®R Ap) ®R 5 (3.6.3)

of 5-modules in which (ax ® st) ® (a2 ® s2) ® • • * ® (ap ® sp) corresponds

to (al ® a2 đ ã ã * đ ap) đ s1s2 . . . sp. This isomorphism is compatible with

multiplication and therefore (3.6.3) is actually an isomorphism of
5-algebras. 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

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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(axa2 . . . a j = £ ax. . . ai_1(Z)ai)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 algebra A is generated (as an

R-algebra) 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? a2, . . . , 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(aj)D'(al.))a1... af. . . a , . . . aw,

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

... aJ

so that Dr°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

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58 Associative algebras

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 al5 a2, . . . , 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 
homo-morphism 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.

We begin by noting that if Xl9 X2,..., Xn are indeterminates, then the

polynomial ring K[Xl 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.

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Comments and exercises 59

This said, let Sx and E2 be two multiplicatively closed subsets of R. Then

S f1^ ) and Z^"1^) are ^-algebras so we can form their tensor product

^HR)®RZ;HR) (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 <r2eZ2, is again a multiplicatively

closed subset of R. Let this new multiplicatively closed subset be denoted by
I1Z2 a nd let us form the K-algebra

(L^r^n (3.8.2)

The next exercise shows that (3.8.1) and (3.8.2) are virtually identical.

Exercise 2*. Let Zx and Z2 be multiplicatively closed subsets of R and put
Z = Z!Z2. Establish an isomorphism

sr'w®*^

1

^)^"

1

^)

of R-algebras.

The existence of a grading on an R-algebra has some interesting
consequences. Let A be a graded K-algebra and let {An}neZ be its grading.

First we make the rather trivial observation that if a belongs to the centre of
A, then all its homogeneous components belong to the centre as well. To see
this let

a = - - - + a _1+ a0 + a1+ a2 +

---be the decomposition of a into its homogeneous components, where the
suffix indicates the degree of an individual component. If now am e Am, then

from otam = am<x we obtain <xnam = am(xn by comparing components of degree

m + n. But now it follows that (xnx = x(xn for all xeA, i.e. aw belongs to the

centre of A.

Exercise 3*. Let A be a non-negatively graded R-algebra and let a belong to

the centre of A. If OL2 = a show that a is homogeneous of degree zero.
Of great importance is the notion of the free algebra generated by a given
set. To explain what this is let X be an arbitrary non-empty set. We consider
formal products x^x2 . . . xn (n >0) of elements of A' or, to use an alternative

terminology, words formed by using the elements of X as letters. (Note that,
because the possibility n = 0 is allowed, we have included the empty word.)
Next we form the free .R-module which has the set of words as a base, and
afterwards turn the free module into an .R-algebra by defining the product
of two of its elements by means of the self-explanatory formula

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60 Associative algebras

Let A be this algebra. For n > 0 let An be the K-submodule generated by

all words with n letters, and put Aq = 0 when q<0. It is easy to check that

{As}seI is a non-negative grading on A. Note that A0 = R, that Ax is the free

R-module generated by X, and that this module generates the algebra. In
the next chapter where with each module we shall associate an algebra
called its tensor algebra, we shall see that A is the tensor algebra of the free
module generated by X.

Another interesting example arises in the following way. Let M be an 
R-module and this time consider the set A formed by all 2 x 2 'matrices'

where reR and meM. If addition and multiplication on A are now

defined by

and

[

r m~\ Yr' m'~\ Yr + r' m + m'~|
0 r \ 0 r' 0 r + r'

r m l f r ' rri~\ Yrr' rm' + r'm~\

0 r l o r'J

=

L0

rr' J

respectively, then it is easily checked that A is a commutative ring with

as its identity element. Furthermore, if for peR we put

[

r m l Ypr pm~\

0 rj

=

[o pr\

then A is turned into a commutative K-algebra.
Now let Ao consist of all matrices of the form

let Al consist of all matrices

0 ml

0 Oj'

and for n ^ 0 , 1 put An = 0. Then {Aq}qeZ is a family of R-submodules of A.

Exercise 4. With the above notation, show that {Aq} q€l is an algebra-grading

for the R-algebra formed by all matrices of the form

[;:]•

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Comments and exercises 61

Let A(l\ A{2\..., Aip) be graded K-algebras. We have discussed two
graded algebras that can be obtained from them by using tensor products.
These algebras are the ordinary tensor product A{1) đ A{2) đ ã ã ã ® Aip)
and the modified tensor product A{1) ® A{2) ® • • • ® Aip\ The latter will
also be known as the twisted tensor product.

In establishing (3.3.13) we showed that the operation of combining
graded algebras by means of the ordinary tensor product is associative and
the discussion of (3.3.14) shows that it is commutative as well. Furthermore,
from (3.3.15) and (3.3.16), we see that when R is endowed with the trivial
grading it acts like an identity in this context.

For modified, that is twisted, tensor products the facts are not so
transparent. We have indeed shown (Theorem 6) that the associative law
continues to be satisfied, but so far we have not examined whether modified
tensor multiplication is commutative. This will be our next concern.

Let A and B be graded ^-algebras and {An}neI and {Bn}n€l their

respective gradings. There is an isomorphism Am ® Bn % Bn ® Am in which

am ® bn is matched with bn ® am. (Here, of course, am e Am and bn e Bn.) It

follows that there is an isomorphism

Tmn:Am®Bn^-+Bn®Am (3.8.3)

in which

Tmn(am®bn) = (-\rnbn®am. (3.8.4)

But A®B = Y,Am®Bn (d.s.) and B®A = Y,Bn®Am (d.s.) so we can

combine the various Tmn to obtain an isomorphism

T:A®B-^-+B®A (3.8.5)

of K-modules which is such that

T(am®bn) = (-irnbn®am (3.8.6)

whenever am e Am and bneBn.T is known as the twisting isomorphism. Now

A ® B and A ® B are identical as ^-modules and the same is true of B ® A
and B ® A. Consequently T provides a bijection

T:A®B-+B®A. (3.8.7)

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 - ^

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62 Associative algebras

The next exercise extends the result embodied in Exercise 5 to twisted
tensor products with an arbitrary number of factors.

Exercise 6*. Let A{1\ A(2\..., A{p) be graded R-algebras and let
(s1? s2, . . . , sp) be a permutation of ( 1 , 2 , . . . , p). Show that

A(l) ® A{2) ® •• • ® A{p)
and

A(Sl) Q A(s2) ® . . . g) A(sP)

are isomorphic graded algebras.

Let A be an R-algebra. The mapping A x A —> A in which (a, a) is taken
into aoc is certainly bilinear and so it induces an ^-linear mapping
fi: A đ A ã A in which n(a ® a) = aa. If A is a commutative /^-algebra, then
fi: A ®A—+Aisa homomorphism of algebras, and if A is a commutative
graded algebra, then \i is a homomorphism of graded algebras. The next
exercise deals with the corresponding result for anticommutative algebras.
Note that \i can also be regarded as an K-linear mapping of A ® A into A.

Exercise 7*. Let A be an anticommutative R-algebra. Show that the mapping

fi\ A ® A-^Ain which fi(a ® a) = ace is a homomorphism of graded algebras.
Ordinary and modified tensor products of graded algebras have so many
features in common that the idea of finding an approach that will deal with
both types simultaneously has obvious attractions, and in fact something
can be done on these lines as will now be explained.

Let A and B be graded /^-algebras with {An}neI and {Bn}neI their

respective gradings. Suppose that An = 0 whenever n is an odd integer. Then

A ® B and A ® B not only coincide as modules, but ordinary and modified
multiplication are the same. Consequently A ® B and A ® B are identical
algebras and we may write

A®B = A®B. (3.8.8)

Likewise (with the same assumption on A) we have

B®A=B®A. (3.8.9)
For example, (3.8.8) and (3.8.9) hold if the grading on A is trivial. In
particular, if R is endowed with the trivial grading, then, for any graded 
K-algebra £, R®B = R®B and B ® R = B ® R and therefore we have
isomorphisms

R®B*B (3.8.10)
and

B®R*B (3.8.11)

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Comments and exercises 63

Suppose now that C is a graded K-algebra and let Cn denote its

submodule of homogeneous elements of degree n. Put
fO if n is odd,

c:=

Znj2 if n is even.

Then C = J ] Q (d.s.) and indeed {C'n}neI is an algebra-grading on C.

Accordingly we have a new graded R-algebra, C say. The connection
between C and C is described by saying that C is obtained from C by
doubling degrees.

Exercise 8. Let C and D be graded R-algebras. Show that

C ®D' = (C® D)\

where C", D' and (C ® D)' are obtained from C, D and C ®D respectively by
doubling degrees.

Let us retain, for the moment, the notation of Exercise 8. Then, by (3.8.8),
C ® D' = C ® Df and therefore

C ®D' = {C®D)f. (3.8.12)

Now the operation of doubling degrees is easily reversed. Consequently
(3.8.12) provides a means whereby an ordinary tensor product can be
expressed in terms of a modified tensor product and this, on occasions,
offers some advantages.

We next observe that the notions of derivation and skew derivation, as
defined in Section (3.7), can be considerably extended. For instance, let A be
a graded .R-algebra. Then by a generalized derivation of degree i we
understand an .R-linear mapping D: A—>A of degree i such that

for all a, oc in A. Thus the derivations considered in the main text are
precisely the generalized derivations of degree — 1.

Exercise 9. Let D and D' be generalized derivations, on the graded R-algebra

A, of degrees i and j respectively. Show that D' ° D — D° D' is a generalized
derivation of degree i +j.

As is to be expected, generalizing the notion of a skew derivation is a
more complicated matter. It is helpful to make some preparatory
observations.

Once again let {An} nsI be a grading on an K-algebra A. For each n e Z we

can define an automorphism Jn, of An, by putting Jn(x)= (— l)nx for all

xeAn; and then these automorphisms can be combined to give an

automorphism

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64 Associative algebras

of the J?-module A which satisfies

J(an) = (-l)nan (3.8.14)

for all an in An. Actually J is more than a module-automorphism as the next

exercise shows.

Exercise 10. Show that J, as defined in (3.8.13), is an automorphism of the

R-algebra A.

Indeed this is not the end of the matter. Not only is J an automorphism of
the algebra, but J2 is the identity automorphism, so that, in other words, J is
an involution of the graded algebra. It is known as the main involution of A.
Note that if i e Z, then J1 is the identity automorphism if i is even whereas it
is J itself if i is odd. Note too that if ameAm, then

Ji(am) = (-l)i*nam. (3.8.15)

After these preliminaries let A: A—-• A be an R-linear mapping of degree
— 1. Then A is a skew derivation (as defined in Section (3.7)) if and only if

A(aa) = (Aa)a + J(a)A(a) (3.8.16)
for all a, a in A.

We are now ready to generalize the notion of a skew derivation.
Specifically a mapping A: A —• A will be called a generalized skew derivation
of degree i if it is R-linear of degree i and

A(aa) = (Aa)a + Jl(a)A(a) (3.8.17)
for a, a in A. Consequently the skew derivations which occur in the main
text are none other than the generalized skew derivations of degree — 1.
Exercise 11*. Let A and A' be generalized skew derivations, of degrees i and]

respectively, on the graded R-algebra A. Show that

A'°A-(-l)"A°A'

is a generalized skew derivation of degree i +j. Show also that if the integer i is
odd, then A ° A is a generalized skew derivation of degree 2i.

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Solutions to selected exercises 65

3.9 Solutions to selected exercises

Exercise 2. Let Ex and E2 be multiplicatively closed subsets of R and put
Z = Z1Z2. Establish an isomorphism

of R-algebras.

Solution. In what follows au a\ denote elements of Ex; <r2, G'i elements of

E2; and r, r', p, p' elements of R.

If r/a1=r'/(T'1 in Z f1^ ) and p/(J2 = p'laf2 in Z ^1^ ) , then it is easily

verified that
rp r'p'

in E "1^ ) . Hence there is a well-defined mapping

in which (r/a1,p/a2) is mapped into rpJGxo2, and a straightforward check

shows that the mapping is bilinear. Accordingly there is an K-linear
mapping

where ^{r/a1 ® p/a2) =

We claim that <\> is an isomorphism of R-algebras. Indeed it is obvious that
<j) is surjective and that it preserves identity elements. Next, if M is any 
R-module, then we know that each element of M ®R Z ^l (R) can be written in

the form m (g) (l/a2), where meM. Consequently every element of

E j "1^ ) (x^E^T1^) can be expressed in the form (r/a^ (x) (p/a2). Let x =

(rfci) ® (p/(T2) and x' = (r'la\) ® (p'l&2). Then

(j){xx') = (t){{rr'lG1G\) ® {pp'l<J2o'2)}

= rr'pp'/(T1 G\O2O'2
= (rp/<J1(T2)(r'p'/<T'1(T'2)

= 4>(x)ct)(x')

so that 0 is a homomorphism of .R-algebras.

Finally let x = (r/(T1) ® (p/(T2) and suppose that </>(x) = 0. Then

rp/G1(T2 = 0 in E "1 (R) and therefore there exist G\ e Ex and a'2 e E2 such that

(j'1G2rp = Q. But now

x = (G'lr/(Ti(j'1) ® (o'2plo2of2)

= (v\<r'2rplG1o\) ® (l/a2(J2)

= 0.

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66 Associative algebras

Exercise 3. Let A be a non-negatively graded R-algebra and let a belong to

the centre of A. If a2 = a show that a is homogeneous of degree zero.

Solution. Let a = a0 + ax + a2 + * • *, where am is homogeneous of degree m,

and put jS = ax -f- a2 + * • • so that a = a0 + /?. Then a2) = a0 and both a0 and P

are in the centre of A. Now

and from this we see that (1 — ao)j3 is equal to its own square. Accordingly

((1 — (xo)P)m = (1 — (xo)p for all m > 1. But (1 — ao)j8 is a sum of homogeneous

elements of strictly positive degrees and therefore we may conclude that
(l-ao)j3 = 0, i.e. that p = (xop. Next

whence p2 = - p. It follows that pm = ( - l)m + 1 p for all m > 1. However, this
implies that jS = 0 because p is a sum of homogeneous elements of strictly
positive degrees. Thus a = a0 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.

Con-sequently

= T(x * y)

which is what we were seeking to prove.

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Solutions to selected exercises 67

(3.9.1)
and

A{Sl) đ A^ đ ãã ã đ A{SJ (3.9.2)
are isomorphic graded algebras.

Solution. Suppose that \<i<p. By repeated applications of Theorem 6
together with the remarks made concerning (3.4.10), we obtain the
following isomorphisms of graded algebras:

A® ® A{i + 1) ® Aii + 2) đ ã ã ã đ Aip)

)} đ (A(i + 2) đ ã ã ã đ Aip))
đ (X(l + 2) đ ã ã ã

In a similar manner we obtain an isomorphism

2)

đ ã ã ã đ Aip)

But we know from Exercise 5 that A{i) ® Aii + 1) and v4(l + 1 )® A( 0 are
isomorphic graded algebras. It follows that the algebra (3.9.1) remains
unchanged up to an isomorphism of graded algebras if we interchange the
adjacent factors A{i) and A(i + 1). However, the factors may be brought into
whatever order we please by a succession of adjacent interchanges and with
this observation the solution is complete.

The reader may like to refine this result by constructing an explicit
isomorphism between the algebras (3.9.1) and (3.9.2).

Exercise 7. Let A be an anticommutative R-algebra. Show that the mapping

fi: A đ A ã A in which \i{a ® a) = act is a homomorphism of graded algebras.

Solution. It is clear that fi is K-linear and that it preserves identities and
degrees. Let x and xf belong to A ® A and let x * xr denote their product in
A ® A. We need only prove that ii{x *x') = fi(x)fi(xf) and in doing this we
may suppose that x = ar ® as, x' = a'p ® a^, where ar, a'p, as, ct!a are

homo-geneous elements of A whose degrees are indicated by their suffixes. Now
x * x' = (— \)spara'p ® (xsa'a and therefore

^(x * x') = ( - \)spara'pas(x'a = arasa'pcda

because A is anticommutative. Accordingly fx{x * x') = fi(x)jn(x') as required.

Exercise 11. Let A and A' be skew derivations, of degrees i andj respectively,

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68 Associative algebras

is a generalized skew derivation of degree i +7. Show also that if the integer i is
odd, then A° A is a generalized skew derivation of degree 2i.

Solution. It is clear that A'°A and A ° A' are endomorphisms of degree i +7
and therefore the same is true of A' ° A — (— 1)°A ° A'. Now let a, a belong to
A. Then A(aa)=(Aa)a + JJ(a)(Aa) and therefore

A'(A(aa)) = (A'(Afl))a + Jj(Aa) (A'a)

Similarly

A(A'(aa)) = (A(A'fl))a + f {A'a) (Aa)

+ A(J'"(a)) (A'a) + J* (Jj(a)) (A(A'a)).

But Jj(Ji(a)) = Ji(Jj(a)) = Ji+j(a) and therefore it will follow that A ' ° A 
-(— 1)°A ° A' is a skew derivation of degree i +7 provided we show that

Jj(Aa) = (-l)ijA(Jj(a)) (3.9.3)
and

Ji(A'a)=(-l)ijA'(Ji(a)). (3.9.4))

Indeed it is enough to prove the first of these relations for then the second
will follow by symmetry; and furthermore it will suffice to prove (3.9.3)
when a is a homogeneous element.

Suppose therefore that aeAm. Then

which is what we were aiming to prove.

From here on we shall suppose that i is an odd integer and that a, a are
general elements of A. Then

and so we shall have proved that A ° A is a skew derivation of degree 2i if we
show that

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4

The tensor algebra of a module

General remarks

In Section (1.3) we defined the tensor powers of an /^-module M. In
this chapter we shall show how the different tensor powers can be fitted
together to produce an K-algebra. This algebra, which is known as the
tensor algebra of M, contains M as a submodule. In fact the tensor algebra
solves a certain universal problem concerned with algebras that contain a
homomorphic image of M. The tensor algebra is particularly important in
our context because it has, as homomorphic images, the exterior and
symmetric algebras which we shall study later.

As in previous chapters R and S always denote commutative rings. All
algebras are understood to be associative, rings and algebras are assumed to
have identity elements, and homomorphisms of rings and algebras are
required to preserve identity elements. Finally, we shall often omit the suffix
from the tensor symbol (x) when the underlying ring can easily be inferred
from the context.

4.1 The tensor algebra

Let M be an ^-module. We shall define its tensor algebra in terms
of a certain universal problem. To this end suppose that A is an K-algebra
and that </>: M —• A is a homomorphism of K-modules. If now h: A —• B is a
homomorphism of K-algebras, then, of course, /z°c/>:M—•£ is an 
R-module homomorphism of M into B. This observation leads us to pose the
following universal problem.

Problem. To choose A and (f> so that given any R-module homomorphism

\j/:M—+B {B is an R-algebra) there shall exist a unique homomorphism
h: A^»B, of R-algebras, such that h°(f) = il/.

It is clear that if (A, (/>) and (A\ </>') both solve this problem, then there will

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70 The tensor algebra of a module

be inverse algebra-isomorphisms k. A —>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) <f>\M—*A is an injection;

(ii) A is generated, as an R-algebra, by (j){M);
(iii) there is a grading {An}neI on A with Ax =4>(M);

(iv) for p>l, Ap is the p-th tensor power of M with ml đ m2 đ ã ã ã

(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

(m1,m2,... ,mp) into <t>(m1)<l)(m2).. • (j>(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 đM đ ã • • ® 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. (4.1.2)

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The tensor algebra 71

yq) = ryq (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) 
R-algebra, 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 (m1,m2,... ,mp) into

^ ( m i ) ^ ^ ) . . . i//(mp), and this will induce a homomorphism hp: Ap—+B,

where

hp{m1 ®m2 ®''' ®mp) = ^ ( m1) ^ ( m2) . . . 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

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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 
algebra-homomorphism (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, m2, . . . , mp belong to M, then

TifUnti đ m2 đ ã ã ã đ mp) = f(m1)đf(m2) đ ã ã ã đf(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

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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 To(/): 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 
two-sided 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 đ ã * ã ® mp, where mi, m2, . . . , mp belong to M and at least one m{ is in K.

Since Tp(f) = f đ / đ ã ã • ® / , 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.

Let h: T(M)^T(M)/I(K) be the natural homomorphism of graded
algebras. Since T1(M) = M and 71(K) = X, the submodule of T(M)/I{K)

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74 The tensor algebra of a module

homomorphism N-+T(M)/I(K) and this, by Theorem 2, extends to an
homomorphism T(N)-+ T(M)/I(K). Let us combine the 
algebra-homomorphism 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
R-module 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 ®

b2®---®bp, where bt e B. Accordingly T(M) is itself a free R-module and

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

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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(<T) is the identity

mapping of T(P). The following lemma allows us to conclude that T(P) is
isomorphic to a direct summand of T(F); and, since T(F) is free, this means
that T(P) is projective.

Lemma 1. Let u: N —>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

5-algebra.

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

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76 The tensor algebra of a module

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 Sã TS(M đRS) 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 
S-algebras 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

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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 m2, . . . ,rap) becomes

p

X / K M ^ i đ ã ã' đ fht đ ••' ® 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)ã

Tp_l(M\ where Dp(m1 đ m2 ® * * • ® mp) is the element (4.5.1). Note that
Dx=f.

Next the homomorphisms Dl 9 Z)2, D3, . . . give rise to an

R-endomorphism 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 m2, . . . , 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 + lf(^i)(^i ®"-®rhi®'"® mp).

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78 The tensor algebra of a module

which extends /. Moreover, in place of (4.5.2) we obtain
A(mx đ ã ã ã ® mp ® mi ® • • • ® m'q)

= A(m1 đ ã ã ã đ mp) đ (mi đ • • • ® 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

[

r m

o

where reR and meM. Let us denote this algebra by T(M) and let Tn{M)

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Comments and exercises 79

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

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

R-algebras 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 
R-module 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/I-algebra. 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)-^A1 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

<t>: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

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80 The tensor algebra of a module

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? m2, . . . ,mp belong to M, then it is

easily seen (using induction on p) that
p

D(m1 ® m2 ® • • • ® mp) = £ mi ® ' * * ® ^mv ® ' ' ' ® "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 đ ã ã ã đ mp)

= £ ( - l )( v + 1 ) lm1® - - - ® A mv® - - - ® 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 Solutions to selected exercises

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 đ m2 đ ã ã * đ mp, Ê' = m\ đ m2 đ ã ã ã đ 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

algebra-homomorphism 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

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Solutions to selected exercises 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

x2 + • • • +xh (

where xt e T{{M) and y^e 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 w2, . . . , 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 and fc^ # 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

R-algebras 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? m2, . . . , mp belong to M. Then X{m1 đ m2 đ ã ã ã (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), T4(M),... into

zero.

Let x belong to T(M) and let
x = xo + x1 + x2 + • • •

be its decomposition into homogeneous components. Then xoeR, x1eM

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82 The tensor algebra of a module

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
<t>: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

homo-morphism 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/I-algebra 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

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Solutions to selected exercises 83

Suppose that p > 1. The mapping of the p-fold product MxMx - - xM
into Ti+p(M) which takes (ml9 m2, . . . , mp) into

is multilinear and therefore it induces an R-linear mapping
Ap:Tp(M)-^Tp+i(M)

which is such that

ll

) ® • • • ® m

.

A* = i

Next we define Ao: T0(M)-+ T((M) to be the null homomorphism. Then

Ao, A1, A2, . . . 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\ đ m'2 đ ã ã ã ® 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)ipx ® (Ay)

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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

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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' -^A,oiK-algebras, such that A ° </> = (/>' and A'°<£' = 0.
The next theorem shows inter alia that there is always a solution.

Theorem 1. The above universal problem possesses a solution. If {A, (/>) is

any solution, then

(i) (/>: M —+ A is an injection;

(ii) A is generated, as an R-algebra, by (j)(M);

(iii) there is a grading {An}nel on A such that Ax = (/>(M);

(iv) for each p>l,Apisthe p-th exterior power of M where, for m1,m2,

. . . ,mpin M, we have m1 r\m2 A • • • Amp = (j)(ml)(t)(m2) • • • 0(WP)»*

(v) the structural homomorphism R^>A maps R isomorphically onto
Ao.

Remark. As in the case of the tensor algebra, condition (ii) ensures that
there can only be one grading with the property described in (iii).

Proof. It is sufficient to construct one solution of the universal problem that
has all the five properties. To this end let T(M) be the tensor algebra of M
and let m belong to M = Tx (M). Then m®m belongs to T2(M) and therefore

such elements generate a homogeneous two-sided ideal, J(M) say, in T(M).
Let Js(M) = J(M)nTs(M). Then Jo(M) = 0 and J1(M) = O. Furthermore

Jn(M) = O f o r a l l n < 0 .

Since J(M) is homogeneous, the factor algebra T(M)/J(M) = A (say) has
a grading {,4S}S6Z in which As is the image of TS(M). Bearing in mind that

Tx (M) = M, we let </>: M —• A be the homomorphism induced by the natural

mapping T(M)^> T(M)/J(M). Then A and 0 satisfy (i) because J1(M) = 0.

Since Tx (M) generates T(M) as an algebra, (/>(M) generates A as an algebra

so that (ii) holds as well. Evidently (iii) is satisfied. Again T0(M) is mapped

isomorphically onto Ao and therefore the requirement (v) is also met.

Now suppose that p > 1. The natural mapping of T(M) onto A induces a
surjective homomorphism of Tp(M) onto Ap and the kernel of this

homomorphism is Jp(M). But, by the definition of J(M), Jp(M) is the

submodule of Tp(M) generated by all products ml®m2®-'-®mp with

mi = mi + 1 for some i. That condition (iv) holds therefore follows from

Theorem 5 of Chapter 1.

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86 The exterior algebra of a module

Finally suppose that \j/: M —• B is an K-linear mapping (of M into an 
R-algebra 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 
R-algebras 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.

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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, m2, . . . , 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

anticommuta-tive.

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

£(/)(»»! Am2 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)-^E(N)
is a homomorphism of graded algebras and therefore, for each peZ, it
induces a homomorphism

Ep(f):Ep(M)^Ep(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 / ( m2) 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)

and also

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88 The exterior algebra of a module

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 /

=

£ ( A (5.2.9)

and

/\pf = Ep(f). (5.2.10)

Accordingly (5.2.5) and (5.2.6) become (/\f)°(f\g) = f\(f°g) and
(AP / ) ° (AP 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,M2,...,Mr

of M satisfy relations

U

s = elCls + 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

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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 . . . cl>?

(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

r<n.

Suppose that 1 <j1 <j2 < • • • <jr<n and let (il9 i2,..., ir) be a

permuta-tion of (JiJ2,... Jr). Then, because ep A eq = — eq A ep, we have

eh A ^ A " " A ei p= ± ^ A ^ A • • • A 6j r,

where the plus (respectively minus) sign is to be taken if (iu i2,..., ir) is an

even (respectively odd) permutation of (jlj2, - • • Jr)- Accordingly the

permutations of {j1j2,... ,jr) contribute to (5.2.13) an amount equal to

(eh A ^-2 A • • • A eJr)CJiJ2 jr. The lemma follows.

As an application of the lemma we have

Theorem 5. Let the R-module M be generated by el9 e2,..., en. Then

Er(M) = 0 if r>n. If l<r<n, then Er(M) is generated, as an R-module, by

the elements eji A eJ2 A • • • A ejr, where (JiJ2,... Jr) is a typical sequence of r

integers satisfying 1 <j1 <j2 < • • • <jr < n.

5.3 The exterior algebra of a free module

Throughout Section (5.3) we shall assume that the ring R is 
non-trivial.

Let M be a free .R-module. If p > 1, then Ep(M) is free by Theorem 6 of

Chapter 1. Moreover E0(M) is free because it is isomorphic to R. But

£ ( M ) = £ En(M) (d.s.)

n>0

so that E(M) too is a free .R-module.

Theorem 6. Let P be a projective module. Then E(P), considered as an

R-module, is projective. Consequently, for all s, ES(P) (since it is a direct

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90 The exterior algebra of a module

Proof. The corresponding result for tensor algebras is Theorem 5 of
Chapter 4. The argument used there, with only trivial modifications, also
yields the desired result on this occasion. The details are left to the reader.
Let M be a free R-module and let B be a base for M. If B is infinite, then, by
Theorem 6 of Chapter 1, all the exterior powers E0(M\ EX{M\ £2( M ) , . . .

are non-zero free /^-modules.

Now consider what happens when B consists of a finite number of
elements, say w1? w2, . . . , un, where n>0. 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<n.

Thus Er(M) has a base consisting of (") elements and, in particular,

0. Of course Eo(M)^0 because it is isomorphic to R.

Theorem 7. Let M be a free R-module and B be a base of M. Then the

number of elements in B is the upper bound of all integers k such that

This is an immediate consequence of our previous remarks.

Corollary. Let B and B' be bases of the free module M. Then either B and B'

are both infinite, or they are both finite and contain the same number of
elements.

The number of elements in a base of a free K-module M is called the rank
of M and it will be denoted by rank/?(M). Thus rankK(M) is either a

non-negative integer or else it is 'plus infinity'. If rankR(M)= oo, then

rankfl (£0(M)) = 1 and rank^ (Ep(M)) = oo for all p > 1. On the other hand if

rankR(M) = n, then

mnkR(Es(M)) = (n) (5.3.1)

for all s > 0 .

Let F be a free K-module of rank n and let f:F—+F be an 
JR-endomorphism of F. Then En(f) is an endomorphism of En(F) and, by

(5.3.1), En(F) has a base consisting of a single element. Consequently there

exists aeR such that En(F): En(F)—>En(F) is the homothety a and,

moreover, a is uniquely determined. This scalar a is called the determinant of
/ and it is denoted by Det(/). Hence

/ ( * i ) A / ( X2) A • • • A / ( x J = Det(/)(x1 AX2 A • • • Ax„), (5.3.2)

where xx, x2, . . . , xn are any n elements of F.

Suppose now that e1, e2,... ,en is a base of F. Then

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The exterior algebra of a free module 91

where

C =

C12

C22 -In (5.3.4)

is the matrix of / with respect to the given base. By Lemma 1,

f(ex) Af(e2) A • • • A / ( e J = Det(C)(^ Ae2 A • • • Aen). (5.3.5)

Since ex A e2 A • • • A en is a base for En (F) we see, from (5.3.2), that Det(/) =

Det(C). Thus the determinant of / equals the determinant of any of its
representative matrices. Again if g: F—• F is also an .R-endomorphism of F,
then from En(g° f) = En(g)°En(f) it follows that Det(#°/) =

Det(gf) Det(/). When this is expressed in terms of representative matrices
for / and g, we obtain, of course, the usual formula for the determinant of
the product of two square matrices.

Once again letel9e29... ,enbea. base of F, let / be an #-endomorphism

of F, and let f(er) be given by (5.3.3). Suppose that 1

we shall use K = (kl9 k2,..., kp) (respectively J= (j 1J2,... Jp)) to denote a

sequence of p integers satisfying 1 </cx <k2 < • • • < kp< n (respectively 1 <

h <A<' ' ' <jp<n), and K' (respectively J') will stand for the sequence

obtained from (1, 2, 3 , . . . , n) by deleting the terms that belong to K
(respectively J).

Putxi = f(ei),xK = xk i Axki A • • • Axk ,andej = eji A ^ A - -1 A ^ - . T h e n ,

by Lemma 1,

where

CJK =

L

J2kl ^J2k2

Similarly

where T= (tl912,..., tp) and T are to be understood in the same way as K

and K'. We now have

A er )CJK CrK'

J,T

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92 The exterior algebra of a module

because e3 A er = 0 unless J=T. Next, from (5.3.5) we obtain

= £KDet(C)(el A £ ?2A - " Aen),

where eK is the sign of (fel5...,/cp,/c'l9/c2,...) considered as a permutation of

(1,2, 3 , . . . ,n); that is to say e^ is —1 raised to the power (/cx — 1) +

(k2 — 2) + - • • +(/cp — p). We also have e, AeJ' = 8J(e1 Ae2 A • • • A 4 If we

now substitute in (5.3.6) and remember that ex A e2 A • • • A en is actually a

of En(F), we find that

Thus

Det(C) = X £ ^ C ^ Cr r (5.3.7)

where eJK = SjSK is — 1 raised to the power

O'I +h + - ' +Jp) + (ki + k2 + -- + fcp).

This expression for Det(C) is known as Laplace's expansion of the
determinant of the matrix C using the columns of the matrix that are
indexed by fcl9 fc2,..., kp.

We give one more illustration of the connection between the theory of
exterior algebras and the theory of determinants. To this end consider free
/^-modules U, V and W of finite rank. We select bases for these, namely
ul9 t*2,..., um for (7, vuv2,..., vn for V and wl5 w2, . . . , wq for W Let

X\ I/—• F and JU: K—> VF be R-linear mappings and let their matrices, with
respect to the chosen bases, be A = ||an\\ and B=\\bkj\\ respectively. Thus A

is an n x m matrix, B is a q x M matrix,

and

k = i

Furthermore the matrix C of /i° A is given by C = BA.

After these preliminaries let / = (il5 i2, . . . , ip) and J=(JiJ2,... ,7P) be

sequences of p integers, where 1 < i\ < z2 < • • • < ip < m and 1 <j1 <j2 < • • •

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The exterior algebra of a direct sum 93

where

In a similar manner we can show that

where K = (kl9 k2,..., kp) is a sequence of integers satisfying l<fex<

k2< • • • <kp<q,wK = wki A wfe2 A • • • Awk , and the definition of B^) mimics

that of AW. On the other hand

because C = BA is the matrix of (j, ° L But the wK form a base for Ep(W) and

£p(^ o ^) = £p(^) o j E p ( / l)< Consequently

(5.3.8)

We mention briefly that if we fix p and regard the A$ as entries in a

matrix Aip\ then A(p) is known as the p-th exterior power of the matrix A.
The relation (5.3.8) then yields (BA)ip) = Bip)Aip\ which is the usual formula
for the p-th exterior power of the product of two matrices.

5.4 The exterior algebra of a direct sum

Let Ml,M2,.. • ,Mq be ^-modules and let £(Mt-) denote the

exterior algebra of Mt. Then, using the construction set out in Section (3.4),

we can form the modified tensor product £(MX) ® E(M2) ® ''' ® E(Mq).

This, of course, is an K-algebra and we know (Chapter 3, Theorem 7) that it
is anticommutative.

Next there is an R-linear mapping

given by

E(M2) E(Mq)

2, . . . ,mq)= m

t

(5.4.1)

(5.4.2)

(Naturally in 1 đ ã ã ã đ mt ® • • • ® 1 it is understood that mi occurs in the

i-th position.) Note that (t>(M1 đ ã • • ® Mq) generates E{MX) ® • • • ® E(Mq)

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94 The exterior algebra of a module

Theorem 8. The modified tensor product

£(MX) ®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 Mx â M2 đ '' * © 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 Mx © M2 © * * * © Mq.

Proof For the moment let us leave on one side the assertions about
gradings. For the rest, the isomorphisms

(Mx â ã ã ã â Mq_l) â 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 
homo-morphisms h: E(U)^E(M) and k: E(V)^E(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).

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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

<l>{Ml © • • • © Mq) is the module of elements of degree one. But

(j)(M1 â ã ã ã â 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 algebra graded by the

S-modules {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) <g)R S is

anticommutative.

By Theorem 2, the inclusion mapping M đR Sã ER (M) ®R S extends to

a homomorphism

X:ES(M®RS)^ER(M)®RS

of 5-algebras. (Here ES(M ®RS) denotes the exterior algebra of the

S-module M ®R S.) Note that X is degree-preserving.

Theorem 9. The mapping

X:ES(M®RS)^ER(M)®RS,

described above, is an isomorphism of graded S-algebras.

Proof There is a close similarity between this result and Theorem 6 of
Chapter 4. It will be found that it is a simple matter to adapt the arguments
used to prove the earlier result, so the details have been omitted.

In view of Theorem 9 we may regard ER (M) ®R S as the exterior algebra

of the S-module M ®R S. We shall therefore write

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96 The exterior algebra of a module

and, because k is an isomorphism of graded algebras, we can add to this
En(M®RS) = En(M)®RS. (5.5.2)

The latter relation holds for every integer n.

5.6 Skew derivations on an exterior algebra

The linear mappings of a module M into the ring R form the 
R-module UomR (M, R). We put

M* = HomR(M,R). (5.6.1)

The members of M* are, of course, the linear forms on M. M* itself is known
as the dual of M.

We recall that the structural homomorphism of the exterior algebra
E(M) maps R isomorphically onto E0(M). This fact will be used throughout

Section (5.6) to identify the two so that in what follows we have

E0(M) = R. (5.6.2)

Now let A be a skew derivation on E(M). Then A induces an 
R-homomorphism El{M)^>E0(M). But El(M) = M and E0(M) = R.

Con-sequently 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, m2, . . . , 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 m2 A • • • A mp)

t (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

R-endomorphism A: £(M)—>£(M) that has degree — 1 ; and now it only
remains for us to verify that A is a skew derivation.

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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 + 1f(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 , /2e 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 = AU i+AU 2, (5.6.9)

Aru = rAu, (5.6.10)

Au°Au = 0, (5.6.11)

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98 The exterior algebra of a module

U->EndR(E(M)) (5.6.12)

in which u has image AM. Note that if q > 1 and m1,m2,..., mq are in M,

then, by (3.7.4) and (5.6.8),

Au(m! A m2 A • • • A mq)
q

= £ (—l)l + iy(u,mi)m1 A - • - Arhi A - • - Amq (5.6.13)
i= 1

for all u in U. But AU°AU = O. Accordingly (5.6.12) extends to a

homo-morphism

F: E(U)-+ End^ (E(M)) (5.6.14)
of K-algebras. Note that if ul9 u2,..., u belong to U, then

and therefore

F ^ AU2 A - • • AW/,) = AM I° AM 2° - •
< O

AV (5.6.15)

This shows that when F(w1 A U2 A • • • A up) operates on a homogeneous

element of E(M) it lowers its degree by p. We also see that for p>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.

Suppose that \<p<q. We shall use T=(tl,t2,... ,tp) to denote a

sequence of p integers satisfying I<t1<t2<- - • <tp<q; and if

m1, m2,..., mq belong to M, then {mx Am2 A • • • Amq)T will denote the

result of striking out from m1 A m2 A • • • A mq the terms mh, mh,..., mt .

Thus

(mt Am2 A • • • Amq)T

= mx A • - - Arhti A - - - Amt A • • • Amq. (5.6.17)

(If q = p, then (mx A m2 A • • • A mq) T is understood to be the identity element

of E(M).) Finally we put

y(ul9mti) y(ul9mt2) ••• y(ul9mt

and

DT =

l ^ l = ^i

(5.6.18)

' • + f n

Lemma 2. Suppose that \<p<q and let ut,u2,... ,up belong to U and

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Skew derivations on an exterior algebra 99

A u2 A • • • A Mp))(mj A m2 A • • • A m9)

= (-1)"X ( - D ' ^ r K Am2 A • • • Am,)r.

T

Proof. We argue by induction on p. If u e U, then

(5.6.19)

Am2 A • • • Am2 A • • •

A • • • A m{ A • • • A mn

by (5.6.13). This shows that (5.6.19) holds when p= 1.

We now assume that p > 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 |^ ( m1A m2A - - - A m , )K. (5.6.20)

K

Here the sequence K = (/c2, k3,..., kp) is required to satisfy 1 < k2 < k3 < • • •

<kp<q and

K~ :

p 2 y(up,mk3) ••• y(up,mkp)

Next, if we operate with AMi on (5.6.20) and use (5.6.16) we find that

(Ht/j AU2 A" ' A M p ) ) ^ Am2 A ' ' ' Atnq)

)i,). (5.6.21)

We must now identify the coefficient of (m1 A m2 A • • • A mq)T when we

expand the right-hand side of (5.6.21). To this end, for 1 < v

( * ! , . . . , £ , , . . . , tp). Then when AMi((m! A m2 A • • • A mq)K) is developed it

provides the term

y(t/1,mfv)(-l)fv+v(m1 A m2 A - - - A mq)T.

We also have \KV \ + tv = \T\. Hence the coefficient of (m1 A m2 A • • • A mq)T

in the expansion of the right-hand side of (5.6.21) is

(-I)'"1 t (-Dm+vy(«i^rv)D5,v.

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100 The exterior algebra of a module

However, this follows on expanding the determinant (5.6.18) by means of its
first row.

Corollary. Suppose that p > 1 . Let wl5 u2,..., upbelongto U andmi,m2,...,

mp to M. Then

y(ul9m2) ••• y(ul9mp)

y(up9 mx) 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)

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->EndR(£(M)) (5.7.6)

in which m of M becomes Am is R-linear. Furthermore Am° Am = 0 for all

meM.

Proof The first assertion is clear. Now let xeE(M). Then (Lm°Lm)(x) =

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Pfaffians 101

But y(m, m) = 0 because y is an alternating bilinear form. Consequently

and therefore Am°Lm + Lm° 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

Am i.m 2...m p=(Lm,+Am i)o(Lm 2 + Am2)o---o(Lmp + Amp) (5.7.8)

whenever ml5 m2, . . . , mp belong to M.

We next define

Q:E(M)->E(M) (5.7.9)

Q(x) = Ax(l£(M)). (5.7.10)

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

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102 The exterior algebra of a module

= 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)-Am(Q(u)). (5.7.15)

We also have, this time from (5.7.8),

A.l A.2 A...A m p=(Lm i-Am i)o(Lm 2-Am 2)o---o(Lm p-Am p) (5.7.16)

for ml5 m2, . . . , 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

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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, m2, . . . , mp (p > 1) belong to M. Then

y(mum2) ••• y(mump)

= co\ (5.7.17)
y(m2, mx) y(m2, m2) -m y(m2, mp)

y(mp, mx) y(mp, m2) • • • y(mp, mp)

where w is the component of degree zero of the result of applying

(Lmi +Am,)° (Lm2 + Am2)°- • - (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))
= AX A X-(1£ ( M )) = Q ( X A X ) .

But x = Q(x) is the result of applying

(Lm i-Am i)°(Lm 2-Am 2)o---°(Lm p-Am 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)). (5.7.18)

Next, by (5.7.8),

Ax = (Lmi +AMi)o (L + Am, ) o . . .o ( L + A
p

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104 The exterior algebra of a module

where Ap_2s,x is an endomorphism of E(M) of degree p — 2s, and now we

may conclude, from (5.7.16), that
_ p

Ax= X (-1)SAP
-2s,x-s=0

Accordingly

) = A

X( 1£ ( M )) — YJ Ap_2s,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

12

C?2i ^22

fli

aPI

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 
R-module M with a base el9e2,-..,ep of p elements and afterwards an

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

that

(5.7.19)
and when p is odd P((A) is zero.

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Comments and exercises 105

A. =

12

— a

12

"13 "14

023 024

0 fl34

" 0 3 4 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

- 0 i 2 0 a23

- 0 1 3 - 0 2 3 0
- 0 1 4 - 0 2 4 - 0 3 4

as may be verified directly.

024
034
0

= (012034-013024+014023)2

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 
homo-morphism 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)

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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
deter-minant 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 
In what follows J = (JiJ2, • • • Jp) respectively K = (kl9 / c2, . . . , kp) will

denote a sequence of p integers with 1 <j\ < • • • <jp < n respectively 1 <

kx<- • <kp<n, and as before we put

C

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Comments and exercises 107

Again J' respectively K' will be used to denote the increasing sequence that

is obtained by striking out from (1, 2 , . . . , « ) the terms that belong to J
respectively K. If now we put

\J\=h+J2 + '-+Jp> (5.8.6)
then (5.3.7) can be rewritten as

Det(C) = £ ( - iyJl + lKlCJKCfK>. (5.8.7)

j

This, as we remarked at the time, is Laplace's expansion of the determinant
of C using columns fcl5 k2,..., kp. Now there is a companion formula,

namely

Det(C) = Y (—1)'J| + 'K'C C> ' (5 8 8)
K

which is Laplace's expansion using rows jl9j2,... Jp. This can be derived

by applying (5.8.7) to the transpose of C.

The following observation will serve to introduce the next exercise.
Consider the 2 x 4 matrix

X^ X2 X-$

y\ yi ^3

>

x4

r

\yx

X2 X3

=0.

X2 X4

y2 y4

There is an identity that connects its 2 x 2 subdeterminants which takes the
form

(5.8.9)

Using Laplace's expansion, (5.8.9) can be generalized as follows.

Exercise 3*. Let A = | | al J| | be an nx2n matrix with entries in R, let
/ = (i'i, i2,... ,in) be a typical sequence of n integers satisfying 1

<in<2n, and let V be the sequence obtained from ( 1 , 2 , . . . , 2n) by deleting

the terms in I. If

D,= "

show that

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108 The exterior algebra of a module

Show also that if k (where 1 <k<2n) is fixed and n is even, then

' ) 0 / 0r = O. (5.8.11)

Here it is to be understood that the summation is over the different
sequences /, and that if/' = (i'i, i"2,..., i'n), then by sgn(7, /') is meant the sign

of (il, i2,..., i'i, i2, . . . ) considered as a permutation of ( 1 , 2 , . . . , 2n). Of

course (5.8.9) is what (5.8.11) becomes when n = 2.

We return to (5.8.8). Let S = (sl5 s2, . . . , sp), where 1 <Sj < • • • <sp<n,

and let S' (the corresponding residual sequence) be (si, s2, . . . , s'n_p).

Further, let a n n x n matrix Q be formed as follows: the first p rows of Q are
toberows;1 ?;2>-.. Jp of the matrix C and the last n — p rows of Q are to be

rows si, s2, . . . ,sj,_pofC. Evidently, if J = 0 ' i , ;2, . . . Jp) and S are different,

then the determinant of Q is zero whereas if J = S, then
Det(6) = sgn(J,J')Det(C)

and therefore

( - 1 )1 + 2 +-+ p Det(Q) = (~ l )| s | Det(C).
On the other hand, for arbitrary J and S

as may be seen by employing Laplace's expansion on the first p rows of Q. It
follows that

Z

( nlsi + |Kir r
\—l

) ^JK^S'K'
K

is zero if J ^S and it has the value Det(C) if J = S. Hence if we put
rKs = ( - l ) ' *l + | 5 |Q ' r (5.8.12)
then we obtain

t '

K A S

"lDet(C) if J = S.

(5>8

-

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

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Comments and exercises 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

l<p<n let C{p) be the p-th exterior power of C. Show that

where by (r, s) is meant the binomial coefficient r \ / s \ (r — s)l.

To be fair to the reader it should be remarked that the solution provided
for this exercise uses the fact that polynomials with integer coefficients form
a unique factorization domain and also the result that the general
determinant, considered as a polynomial in its entries, is irreducible.

We shall now interpret the result contained in Exercise 5 in terms of
exterior algebras. To this end let F be a free K-module of rank n, and let
/ : F —• F be an endomorphism of F. Further, suppose that 1 
el9 e2, . . . , en is a base for F,f(er) = exclr + e2c2r+ • • • +encnr, where c^eR,

and C=||co.||, then, of course, Det(/) = Det(C). Let J = 0'iJ2> • • • JP\

where 1 <j^ < • • • <jp < n, and put e3 = eji A eJ2 A • • • A ejp. Then the e3 form

a base for EP(F) and, as we remarked earlier,

J

But this means that, with respect to the base {ed} of Ep(F), the

endomorphism Ep{f) has matrix Cip). Consequently Det(£p(/)) =

Det(C(p)). Exercise 5 now shows that

Det(£p(/)) = (Det(/))("~1'p"1 ) (5.8.16)

which is the relation we were seeking.

We end this section with some comments on Pfaffians. The simplest
alternating matrix of even order has the form

0 a"

ro

L-and the Pfaffian of this is a. At the end of Section (5.7) we calculated the
Pfaffian of the general 4 x 4 alternating matrix. The reader may like to verify
by the same method that the Pfaffian of

0

-al2

- ^ 1 4

-015

-016

012

0

- 0 2 3

- 0 2 4

- 0 2 5

- 0 2 6

013

023

0

- 0 3 4

- 0 3 5

- 0 3 6

014

024

034

0

- 0 4 5

- 0 4 6

015

025

035

045

0

- 0 5 6

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110 The exterior algebra of a module

is

+ a14(a23a56-a25a36+a26a35)

-a15(a23a46-a24a36+a26a34)

+ a16(a23a45-a24a35+a25a34).

(5.8.17)

The general formula, of which this is a special case, will be derived later.
It will be noticed that the Pfaffian of an alternating matrix is a polynomial
in its entries with integer coefficients. Let us make this more precise.
Consider the generic alternating matrix

0 xl2 x13 • - • xln

— x12 0 x23 - • • x2n

— x13 —x23 0 • • • x3n

— xln —x2n —x3n • • • 0

where the x(j (i<j) are n(n— l)/2 different indeterminants. We regard this

matrix as having its entries in the polynomial ring Z[x1 2, x13,..., xn_l M]

and then its Pfaffian is a polynomial, say Q(x12, x1 3, . . . , xn_x J , with

integer coefficients in the indeterminates in question. Now let R be an
arbitrary commutative ring and A = ||ao|| annxn alternating matrix with

entries in R. Then

with the same polynomial Q as before. This follows from the method given
in Section (5.7) for calculating Pfaffians.

For the next exercise we observe that if A and B are alternating matrices,
then

is an alternating matrix as well.

Exercise 6*. Let A and B be alternating matrices with entries in R. Show that

the Pfaffian of

~AA 0 1

o

B\

is the product of the Pfaffian of A and the Pfaffian of B.

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Solutions to selected exercises 111

matrix, so that we can compare the Pfaffian of ATAA with that of A. The
next exercise records the facts.

E x e r c i s e 7*. Let A be an nxn alternating matrix and A an arbitrary nxn

matrix. Show that ATy4A is an alternating matrix and that
Pf(ArylA)=(Det(A))Pf(A).

Let Abeannxn(n>2)alternating matrix and suppose that 1 <i<j<n.
If we strike out the i-th andj-th rows and also the i-th andj-th columns, then
the resulting matrix will also be alternating.

Exercise 8*. Let A = || apq \\ beannxn(n>2) alternating matrix and, for 1 <

i <j < n, let %j be the matrix obtained from A by striking out the i-th andj-th
rows and columns. Show that

7=2

This result gives a convenient way of evaluating Pfaffians. We
encoun-tered a special case of it in (5.8.17).

Exercise 9. Let In be the identity matrix of order n. Show that the Pfaffian of

is equal to

(-5.9 Solutions to selected exercises

Exercise 3. Let A = \\aij\\ be an nx2n matrix with entries in R, let 1 =

(i1,i2,...,in) be a typical sequence of n integers satisfying l ^ i ^ " *

<in<2n, and let I be the sequence obtained from ( 1 , 2 , . . . , 2n) by deleting

the terms of I. If

a2i
n

show that

Xsgn(/,/')D7/)r = 0.

Show also that if k (where 1 <k<2n) is fixed and n is even, then

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112 The exterior algebra of a module

Solution. Let C be the In x 2n matrix

Then Det(C) = 0 because there are repeated rows. Hence if we use Laplace's
expansion to evaluate the determinant (applying it to the first n rows) we
find that

This is the first assertion. (Here we have used the fact that

which was noted previously.) Observe that
sgn(/, /') sgn(/\ /) = (—l)1/|-|-|/'l =
Consequently

when n is even.

From here on we suppose that k (1 <k<2n) is a given integer which we
keep fixed, and we suppose also that n is even. Let /l 5 J2, . . . , /s be the

sequences / that contain k. Then those that remain are /'l5 / '2, . . . , I's and we

have

= t sgn(J,,/;)D

D

+ t sgn(r

v

r;)D

r

p

r;

= 2 £ sgn(/

, /;)/),,D

;;

f = 1

because /'/ = /f and sgn(/,, rt) = sgn(I't, It). Thus

(5.9.1)

and now we have the problem of getting rid of the 2.
We proceed as follows. Let

X =

*11 X

\2 ' ' ' X

\,2n

X

2\ * 2 2 ' ' ' X

2,2n

X

nl X

n2 X

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Solutions to selected exercises 113

where the xtj are indeterminates, and consider X as a matrix with entries in

Z [ xn, x1 2, • • •, xn,in\- If n o w A7 denotes the subdeterminant of order n that

comes from columns iu i2,..., in, then (5.9.1) shows that

kel

but now, because 2 is not a zerodivisor in the polynomial ring, we may
conclude that

Xsgn(/,/')A7Ar = 0. (5.9.2)

Finally we observe that there is a ring-homomorphism of Z [ xn, x1 2, - •.,

xn2n\ into R in which m (in Z) is mapped into m\R and x0- is mapped into atj.

If we apply this homomorphism to (5.9.2) we obtain the required relation. A
convenient way to describe the last step is to say that

kel

is obtained from (5.9.2) by specialization.

Exercise 5. Let 0=11^11 be an nxn matrix with entries in R and for

\<p<n let C(p)

be the p-th exterior power of C. Show that

where by (r, s) is meant the binomial coefficient r\/s\ (s — r)\.
Solution. Let

X =

X

21

Xnl Xn2

2n

where the xtj are indeterminates and X is regarded as a matrix with entries

in Z[x11? x1 2, . . . , xnn~\. We shall show that

Once this has been proved the desired result will follow by applying the
ring-homomorphism

in which xtj is mapped into ctJ.

By (5.8.15), we can find a square matrix U (say), of the same size as Xip\
such that (i) the entries in U belong to Z [ xn, x1 2, . . . , xn n], and (ii) X{P)U is

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114 The exterior algebra of a module

and therefore Det(Ar(p)) is a factor, in the polynomial ring, of (Det(X))in>p\

But Det(X) is irreducible in Z\_xl 1? x1 2, . . . , xnn~\ and + 1 are the only units

in this ring. We now see that

where k is an integer and s is either + 1 or — 1. By comparing degrees we
find at once that

k

(p-l)\(n-p)\'

To determine £ we give xl l9 x2 2, • • • > xnn ^e value 1 and all the other xtj the

value zero. Then X specializes to the identity matrix /„ and we obtain e =
Det(/ip))- B u t I(np) i s i t s e l f a n identity matrix so that e = + 1 .

Exercise 6. Let A and B be alternating matrices with entries in R. Show that

the Pfaffian of

B_, (5-9.3)

is the product of the Pfaffian of A and the Pfaffian of B.

Solution. Let A have p rows and p columns and let B have q rows and q
columns. Put n = p-\-q. Next, let M be a free K-module with a base el9 e2,

..., en and let y: M xM—+R be the alternating bilinear form in which

\\y{eh £y)|| is the matrix (5.9.3). Then, with the notation of Section (5.7), the

desired Pfaffian is the component of degree zero belonging to

Now

for suitable elements xl 9 x2, . . •, xq in E(M). It is therefore enough to show

that

has 0 as its component of degree zero. But this is clear because if 1 
then Ae(ep+j) = 0fovj= 1, 2 , . . . , q, whereas operating with Le replaces an

element of the form ep + 1 AXX + • • • +ep+q Axq by another element of the

same form.

Exercise 7. Let A be an nxn alternating matrix and let A be an arbitrary

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Solutions to selected exercises 115

Solution. Let X be the alternating matrix

0 x1 2 x1 3
 n

v-12 ^ 2 3
— X13 ~ ^ 2 3 0

— xln —x2n —x3n

xln
X2n

and let

Y =

ynn

Here the xtj {i<j) and the y,a, are distinct indeterminates and X and Y are

regarded as matrices with entries in the ring formed by adjoining all these
indeterminates to the ring of integers. Now

(YTXY)T= YTXTY= - (YTXY)

and so (because we are in characteristic zero) YTXY is an alternating
matrix. Note that it will suffice to show that Pf(YTXY) = (Det(Y)) Pf(X)
because the general result will then follow by specialization. But

= (Det(Y))2Det(X)
= ((Det(Y))Pf(X))2

so either Pf(YTXY) = (Det(Y)) Pf(X) or else Pf(YTXY) =

— (Det(7)) Pf (X). However, we have only to specialize Y so that it becomes
the identity matrix to see that it is the former of these which is correct.

Exercise 8. Let A = \\apq \\ beannxn(n> 2) alternating matrix and, for 1 <

i <j < n, let %j be the matrix obtained from A by striking out the i-th andj-th
rows and columns. Show that

Solution. It is convenient to deal with the problem in its generic form, that is
we consider the matrix

X =

0 x1 2

~xl

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116 The exterior algebra of a module

where the xtj (i<j) are indeterminates and X is regarded as having its

entries in Z[xl 2, xx 3, . . . , xn _ x „]. Let Etj be the matrix obtained by striking

out the f-th and j-ih rows and columns. Then it will suffice to prove that

j=2

since all other cases will follow by specialization. Note that we can suppose
that n is even for otherwise Pf(X) and the Pf(Sl7) will all be zero thus

making the assertion trivial.

Suppose that 1 < r < n. The indeterminates xA/1 (X < /i) with r e (A, fi) occur

only in the r-th row and the r-th column. Each term of Det(X) is of degree 2
in these particular indeterminates, and therefore, since Det(X)= (Pf(X))2,
each term of Pf (X) will contain exactly one x^t with k < \i and r e (A, JX). It

follows that

where Uj is a polynomial not involving any indeterminate xpq (p < q) with

either 1 or j in (p, q).

Suppose that 2 <j<n. Replace xpq (p < q) by zero if exactly one of 1 and;

is in (p, q) and replace xXj by 1. (All the other indeterminates are to remain

unaltered.) Then X becomes a new alternating matrix, Xj say, and Uj is
unchanged. We therefore have Uj = Pf(Xj).

Let A be the matrix obtained from the n x n identity matrix by taking its
;-th column and moving it step by step to the left until it becomes the second

column. Then Det(A) = ( - l)j and
0 1 0 ••• 0
- 1 0 0 ••• 0

0 0

Consequently (see Exercises 6 and 7)
Pf(S1<;.) = Pf(ATXjA) = (Det(A))

whence Uj=(—l)jPf(Elj). Accordingly

3=2

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6

The symmetric algebra of a module

General remarks

We have examined the origins and properties of the tensor and
exterior algebras of a module. The symmetric algebra, which concerns us
now, fits into the same general pattern and, as in the other cases, we shall
motivate its investigation by means of an appropriate universal problem.
However, from a different standpoint, one can say that the exterior algebra
can be obtained from the tensor algebra by making it anticommutative.
Viewed from this position, the symmetric algebra is the result of making the
tensor algebra commutative.

Because much of the theory of the symmetric algebra runs closely parallel
to that of the exterior algebra, we can frequently borrow proofs and adapt
them without difficulty, and when this is the case we shall restrict the
amount of detail that is given.

There is, however, one area in this account of symmetric algebras which is
not foreshadowed in Chapter 5. The symmetric algebra of a module can be
regarded as a generalization of a polynomial ring; and for a polynomial
ring the partial differentiations with respect to the indeterminates generate
a commutative algebra. When this observation is analysed it leads naturally
to the algebra of differential operators and this is one of the topics we shall
discuss. In the next chapter this particular algebra will be examined from a

more general standpoint and, as a result, we shall be able to complete the
analogy between exterior and symmetric algebras.

The general conventions that apply to this chapter are the same as those
set out in the General Remarks at the commencement of Chapter 5 with one
variation. Where we have to consider another commutative ring besides R
we shall denote the second ring by R' rather than by S. This is because the
letter S is widely used as part of the standard notation for symmetric
algebras and symmetric powers.

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118 The symmetric algebra of a module

6.1 The symmetric algebra

Let M be an .R-module, A an K-algebra, and suppose that we are
given an K-linear mapping (ft: M-^A which is such that

(ft(m)(ft(m

f

) = (ft(m')(ft(m) (6.1.1)

for all m and m' in M. A natural universal problem connected with this
situation can be posed as follows.

Problem. To choose A and (ft so that given any R-linear mapping \ft: M^>B

(B is an R-algebra) such that ^/(m)\ft{m') = \jj(m')\j/(m) for all m, m! in M, there
shall exist a unique homomorphism h:A-+B, of R-algebras, such that
h°(ft = \jj.

This type of problem will, by now, have become familiar. Clearly the
problem has at most one solution in the sense that if (A, (ft) and (A', (ft') both

meet the requirements, then there exist inverse isomorphisms k: A —> A' and
k': A —>A,of ^-algebras, such that k° (ft = (ftf and k' ° (ft' = (ft. As for the rest
the essential facts are contained in the following theorem.

Theorem 1. The universal problem just enunciated possesses a solution.

Furthermore if (A, (ft) is any solution, then

(i) (ft: M—+A is an injection;

(ii) (ft(M) generates A as an R-algebra;

(iii) there is a, necessarily unique, algebra-grading {An}n£Z on A with

A,=(ft{M);

(iv) for each p>\, Ap is the p-th symmetric power of M, with

mlm2 . . . mp = (j)(ml)(ft(m2)... (ft(mp) for elements m1? m2, . . . , mp

in M;

(v) the structural homomorphism R^>A maps R isomorphically onto
Ao.

Remark. This result should be compared with Theorem 1 of Chapter 5. As
will be seen the proofs are very similar.

Proof Because any two solutions of the universal problem are copies of
each other, it is sufficient to find one solution that satisfies all of (i)-(v).

Suppose that m, m! e M. Then m (g) m' — m! ® m is a homogeneous
element of degree two in the tensor algebra T(M), and therefore such
elements generate a homogeneous two-sided ideal, H(M) say, in T(M). Put
Hp(M) = H{M) n Tp{M) and A = T(M)/H(M). Of course A has a natural

structure as a graded .R-algebra; let {An}nel be the grading which it inherits

from T(M). Then, in the natural homomorphism T(M)—+A of graded
algebras, Tn(M) is mapped onto An. In particular, since M = TX(M), the

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The symmetric algebra 119

that, because m®m' — m' ®m belongs to H(M), its image in A is zero and
therefore $(m)0(m') = 0(m')0(m).

It is a simple matter to check that A and 0 satisfy conditions (i), (ii), (iii)
and (v) in the statement of the theorem. To see that (iv) also holds suppose
that p>\. Then the surjective homomorphism Tp(M)-^Ap induced by

T(M)-+A has kernel Tp(M) nH(M) = Hp(M). Furthermore Hp(M) is the

submodule of Tp(M) generated by all elements of the form

mx (x) • • • (x) mt ® mi + x (x) • • • (x) mp

- m1 ® • • • <g) mi +! (x) mt- <g) • •

where the second term is the same as the first except that mt and mt + x have

been interchanged. That Ap is the p-th symmetric power of M now follows

from Theorem 8 of Chapter 1. Thus (iv) has been verified as well.

All that remains of the proof is for us to verify that (A, <f>) solves the
universal problem. Suppose then that \j/: M —> B is an K-linear mapping, of
M into an K-algebra B, and that \jj(m)\f/(m') = \jj(m')\lj(m) for all m, m' in M.
The extension of \\i to an algebra-homomorphism T{M)^>B vanishes on
H(M) and so induces a homomorphism h: A-^B of K-algebras. Clearly h
satisfies h ° (f) = \\J ; and, since </>(M) generates A as an algebra, it is the only
homomorphism of A into B to do so. This ends the proof.

We next introduce the standard notation and terminology. Let (A, 0) be
a solution of our universal problem. We put

S(M) = A (6.1.2)

and we employ {Sn(M)}nel to denote the grading referred to in Theorem 1.

Also we use the fact that <f> maps M isomorphically onto Al =Sl(M) to

make the identification

M = SX{M\ (6.1.3)

and then M generates S(M) as an R-algebra. Of course the grading on S(M)
is non-negative.

We shall use ordinary juxtaposition to denote multiplication in S(M) so
that, because <}){m)(j)(m') = (f){m')(j)(m), we have from (6.1.3)

mm'= m'm (6.1.4)
for m, m' in M. Next, when p > 1, Sp(M) is the p-th symmetric power of M so

naturally we call S(M) the symmetric algebra of M. Finally Theorem 1 tells
us that the structural homomorphism R^>S(M) maps R isomorphically
onto S0(M). It follows that we may make the identification S0{M) = R

whenever it is convenient to do so.

Theorem 2. Let S(M) be the symmetric algebra of M and let \jj: M —> B be an

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120 The symmetric algebra of a module

for all m, m' in M, then i// has a unique extension to a homomorphism

S{M)-^B of R-algebras.

Of course Theorem 2 holds because of the connection between S{M) and
the universal problem with which we started.

The proof of Theorem 1 reveals the relation of the symmetric algebra to
the tensor algebra. However, we shall now describe this relation in a less
encumbered context. To this end observe that the inclusion mapping
M-+S(M) extends to a surjective homomorphism

T(M)->S(M) (6.1.5)

of g r a d e d R - a l g e b r a s , a n d s o for e a c h peZ t h e r e is i n d u c e d a surjective
h o m o m o r p h i s m

Tp(M)^Sp(M) (6.1.6)

of R-modules. If /?>1, then the mapping (6.1.6) takes ml®m2®''' ®mp

into m1m2... mp and therefore it is none other than the canonical

homomorphism which we first met in Section (1.5). Thus (6.1.5) is effectively
a combination of canonical homomorphisms.

The next result is the counterpart of Theorem 3 of Chapter 5.
Theorem 3. The symmetric algebra S(M) is commutative.

This follows from (6.1.4) and the fact that M generates S(M) as an 
R-algebra.

6.2 Functorial properties

The functorial properties of the symmetric algebra now follow in a
straightforward manner. Thus if f:M—»iV is a homomorphism of 
R-modules, then N = SX (N) by (6.1.3) and therefore (see Theorem 2) / can be

extended to a homomorphism

S(f):S(M)-+S(N) (6.2.1)
of R-algebras. But S(f) preserves the degrees of homogeneous elements.
Consequently for each integer p there is induced an K-linear mapping

Sp(f):Sp(M)-+Sp(N). (6.2.2)

Note that Sl(f) = f and that if we make the identifications S0(M) = R =

S0{N), then S0(f) is the identity mapping of R. Note too that if p > 1 and

ml, ra2,..., mp belong to M, then

Sp(f)(m1m2 . . . mp) = f(ml)f(m2)... f(mp). (6.2.3)

Of course, if / is the identity mapping of M, then S{f) is the identity
mapping of S(M). Also if / : M —> N and g: N —• K are homomorphisms of
K-modules, then

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The symmetric algebra of a direct sum 121

and hence, for all p e Z,

Sp(g°f) = Sp(g)oSp(f). (6.2.5)

Thus S(M) can be regarded as a covariant functor from ^-modules to
commutative graded ^-algebras. Naturally if / is an isomorphism of 
R-modules, then S(f) is an isomorphism of graded ^-algebras and

1

). (6.2.6)

6.3 The symmetric algebra of a free module

Throughout Section (6.3) we shall assume that the ring R is 
non-trivial.

Let M be a free K-module with a base B. If p>l, then (Chapter 1,
Theorem 9) Sp(M) is a free module and it has a base consisting of the

different products blb2... bp, where bfeB. We also know that two such

products, say b1b2...bp and b\b'2...bp, are equal if and only if

(b'l9 b2,..., b'p) is a rearrangement of (fo1? b2,..., bp). Of course 50(M) is a

free K-module having the identity element of S(M) as a base.

The above remarks show that S(M) itself, considered as an K-module, is
free and has as a base the set of all distinct products blb2 . . . bn9 where bt eB

and n > 0 ; moreover two products blb2... bn and frifr'2 . . . b'q are the same if

and only if n = q and (b\, b'2,..., b'q) is a rearrangement of (bl9 b2,..., bn).

Since the multiplicative structure of S(M) is determined by the way the basis
elements are to be multiplied, it follows that

S(M) = R[Bl (6.3.1)

Here # [ £ ] denotes the ordinary ring of polynomials, with coefficients in R,
in which the elements of B are treated as distinct commuting
indeterminates.

Theorem 4. Let P be a projective module. Then S(P), considered as an

R-module, is also projective. Hence, for each neZ, Sn(P) is a projective

R-module.

Proof This result resembles Theorem 5 of Chapter 4, and there should be
no difficulty in adapting the proof of the earlier theorem to meet the needs of
the present one.

6.4 The symmetric algebra of a direct sum

If Al9 A2,..., Aq are commutative .R-algebras, then A1 ® A2 ® • • •

® Aq is also commutative. Hence if Ml9 M2, . . . , Mq are ^-modules, then

S(MX) đ S(M2) đ * ã' đ S(Mq) (6.4.1)

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122 The symmetric algebra of a module

the ordinary and not the modified tensor product.) The elements of degree
one in this algebra form the module

M1®R®--®R-\-R®M2®--®R

+ '-+R<g>R®-'®Mq, (6.4.2)

where each term in the sum contains q factors; moreover this module
generates S(Af J ® S{M2) ®''' ® S{Mq) as an K-algebra.

Consider the K-linear mapping

đS(M2) đ ã ã ã đ S(Mq), (6.4.3)

where

q

(/>(m1,m2,... , mp) = Ê 1 đ ã ã ã ® mt- ® • • • ® 1 (6.4.4)
;= I

it being understood that in 1 đ ã ã ã đ mt đ ã ã ã đ 1 the factor mt occurs in the

i-th position.

Theorem 5. The tensor product

S(Af 2) ®RS(M2) ®R--- ®RS(Mq) (6.4.5)

together with the R-linear mapping

<t>: Mx ®M2 ® • • • 0 Mq-^S(M1)®R" • ®RS(Mq)

constitute the symmetric algebra of M1đ M2 đ ã ã ã â Mq. Furthermore the

grading which (6.4.5) possesses by virtue of being a tensor product of graded
algebras is the same as its grading when considered as the symmetric algebra
of M1®M2®---®Mq.

Proof This result should be compared with Theorem 8 of Chapter 5 and
once again it is a simple matter to adapt the proof of the corresponding
result for exterior algebras. For instance it is possible to reduce the general

result to the case where q = 2. The only point which possibly deserves
mention is that whereas in Chapter 5 we used the isomorphism

đ E(M2) đ ã ã • ® E(Mq)

this time all we need is the less sophisticated

*S(Ml)®S(M2)®--®S(Mq)

which, of course, is a special case of (3.3.13).

6.5 Covariant extension of a symmetric algebra

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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

R-algebra.

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)

6.6 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.

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 
p-fold product M x M x • • • x M into Sp-l(M), in which (ml9 m2, . . . , 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

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124 The symmetric algebra of a module

Next, by combining DX,D2,D3,... we obtain an K-endomorphism

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 m2, . . . , mp belong to M. Let / = (j1? i2,

. . . , ih) be a sequence of h integers with 1 <ix < i2 < • • • < ih <p, and let /' be

the increasing sequence obtained from ( 1 , 2 , . . . , p) by deleting the terms in
/. Finally put

mI = mimii...mih (6.7.1)

and define mr similarly.

On this understanding an #-endomorphism <p: S(M)—>5(M) of degree
— h is called a differential operator of degree h provided that

^ ( m ^ . . . mp) = YJ (t>(mi)^i' (6.7.2)

for all p >h and arbitrary elements m1? m2, . . . , 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 sl5 s2, . . . ,

sq are non-negative integers with s1+s2 + m" +sq>h. If we replace

mlm2.. .mp by xjxx22... x j , then (6.7.2) becomes

= 1 (l

1

) • • • M*!

1

""

1

*?""

2

• • • K

q

~

aq

> (

6

-

7

-

3

)

where the summation is over all sequences (a1? a2, . . . , aq) of integers with

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Differential operators 125

and

- " - + a , = /i. (6.7.5)
Note that we have allowed the possibility that some of sx, s2, . . . , 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 <f> 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 that rax, m2, . . . , mp belong to M. In what

follows / = (il9 i2,..., ih),J = (JiJ2> • • • Jk) and L = ( /l 9l2, . . . , 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, . . . , jl 9 j2, . . . ) 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

^. (6.7.6)

= L J

This shows that \j/ ° <j> = <f> ° 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 m2, . . . , mp) becomes

(Here / = (il9 i2,..., ih) satisfies 1 <i1 < • • • <ih have the

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126 The symmetric algebra of a module

with the property that

Thus far we have assumed that p > h. For 0 p to be the null

homomorphism of Sp(M) into Sp-h(M).

We can now combine 0O, (/>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 = S1(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)

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Comments and exercises 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 a2, . . . , an)andj5 = (jS1, /?2, •. •,/?„) be two sequences of n

non-negative integers and let us put

and

(P

(6.8.5)

We shall use a</? to mean that OL{ </?,- for f = 1 , 2 , . . . , n so that, on this

understanding, (?) = 0 whenever a $ /?. Lastly, a + /? and a — ft will denote the
sequences (a1 + / ?l 9. . . , an+/?n) and (oq — j 8l 9. . . , an — /?„) respectively.

Now let M be a free K-module having Xx, Z2, . . . ,Xn as a base. Then, by

(6.3.1), S(Af) is the polynomial ring R[Xl9 X2,..., Xn~\ and, if p > 0 , 5p(M)

consists of all the homogeneous polynomials, i.e. forms, of degree p. It will
be convenient to denote the power-product XI1 X22... X^n by Xa. We can

then say that Sp{M) is a free K-module having the X* with |a| =p as a base.

Suppose next that </>: 5(M) —• S{M) is an /^-linear mapping of degree — h
(h>0). By (6.7.3) this will be a differential operator of degree h provided
that, whenever \/3\>h, we have

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128 The symmetric algebra of a module

g\v\
dXyS dXy

22... 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^} ^|= pasabase. 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

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

R-linear 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)^EndR{S(M)) (6.8.9)

of K-algebras which has the property that

T { ulu2. . . up) = DU x° DU 2° - - - ° DU p (6.8.10)

for uu u2,. • •, up in U. But, because a derivation is a differential operator of

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Comments and exercises 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

C =

12

Cn2

• • * 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 ( u2, m2) ••• y ( u2, mp)

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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
corre-sponding 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 <g> A —> A which is such that /J.(X ® y) = xy and this, because 
multiplica-tion on A is associative, makes

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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 < Al®R

also commutes. (Here R ® A ã A and A đ R ã A are the isomorphisms
provided by Theorem 3 of Chapter 2.)

At this point we make a fresh start. Suppose that A is an R-module and
that we are given K-linear mappings /i: 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 
algebra-homomorphism, in the sense of Section (3.1), if and only if the conditions
(i) / is /Minear, (ii) fo^iA = (f®f)o^iB, and (iii) f°nA = nB are all satisfied.

Now assume that, for 1 <i<p, {At, nt, rjt) is an .R-algebra. Let

ã (zlj đ Ax) đ - - - đ (Ap ® Ap) (7.1.3)

be the isomorphism (of /^-modules) which matches (a đ ã ã ã đap) đ

(a\ đ '' - ® a'p) with [ai ® a\) ® • • • ® (ap ® a'p). Further, let

A(RP):R-+R ®R ®--- ®R, (7.1.4)

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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> đ'' ã đ Aff) (7.1.11)
onto

(Al? đ i4jj>) đ (A\? ®Af

2)

)®-"® (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)®--® (aip ® a'jp).

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Coalgebras 133

(7.1.13)

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

AV>®- ® ^ = 0*i ® /*2 ® ã ' ' đ /*P) AP (7.1.14)

and

^( 1> ® - ® ^ = (>7i ® ^2 ® • •' ® r\p)° Ajf> (7.1.15)

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 module and this time suppose that we are given 
R-linear mappings A: A ã A đ A and e: A —• R. The triplet (A, A, e) is called
an R-coalgebra provided that the diagrams

A ã A đA

AđA (7.2.1)

AđA

A > A đAđA

and

_ eđA

RđA < Ađ A ã A"đ R

(7.2.2)

are commutative. Here A ã A đ R maps a into a ® 1 and correspondingly
for A-+R® A.

If (A, A, e) is a coalgebra, then A: A ã A đ A is called the comultiplication
or diagonalization mapping of the coalgebra, and the fact that (7.2.1) is
commutative is described by saying that comultiplication is coassociative.
The mapping e: A—+R is known as the counit.

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134 Coalgebras and Hopf algebras

take e: R —• R to be the identity mapping. Whenever we speak of R as being
a coalgebra it is always this structure that we have in mind.

Now suppose that (A, AA, sA) and (B, AB, eB) are coalgebras. A mapping

f:A—*B is called a homomorphism of coalgebras or a 
coalgebra-homomorphism provided that (i) / is ^-linear, (ii) AB° / = ( / ® / ) ° A ^ and

(iii) eB ° / = eA. (We can describe conditions (ii) and (iii) by saying that / has

to be compatible with comultiplication and it must preserve counits.) If a
homomorphism of coalgebras is also a bijection, then it is called an
isomorphism of coalgebras and when this is the case the inverse mapping is
also a coalgebra-isomorphism. In the last paragraph we observed that R
itself is a coalgebra. Now if AR is the comultiplication of R and eR is its

counit, then sR ° BA = eA and AR°eA = (eA ®eA)° AA. (The latter follows from

the commutative property of (7.2.2).) Accordingly the counit mapping
sA: A—+R is a homomorphism of R-coalgebras.

Finally, homomorphisms of coalgebras can be combined, that is to say
that if / : A —> B and g: B —• C are coalgebra-homomorphisms, then g ° f is
a homomorphism of the coalgebra A into the coalgebra C.

7.3 Graded coalgebras

Let (A, A, e) be an /^-coalgebra and let {An}neZ be a grading on A

considered, for the moment, simply as an R-module. Then there is induced
on the module A ® A the usual total grading in which (A ® A)n is given by

(7.1.8). Let R be given the trivial grading.

Definition. We say that 'A is graded as a coalgebra by {An}ne^ or that
i

{An}neI is a coalgebra-grading on A' if A and e preserve the degrees of

homogeneous elements.

Thus, in the case of a graded coalgebra, the counit e maps all those
homogeneous elements of A whose degree is different from zero into the
zero element of R.

Any coalgebra can be turned into a graded coalgebra by giving it the
trivial grading (i.e. all non-zero elements are treated as being homogeneous
and of degree zero). When R itself is considered as a graded coalgebra, it is
always to be understood that the grading in question is the trivial one.

Now let A and B be graded coalgebras. By a homomorphism A-^B of
graded coalgebras is meant a coalgebra-homomorphism which preserves
degrees. Naturally if such a mapping happens to be a bijection, then it is
called an isomorphism of graded coalgebras in which case its inverse will also
be an isomorphism of graded coalgebras.

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Tensor products of coalgebras 135

7.4 Tensor products of coalgebras

Let Al,A2,..., Ap be K-modules. There is an isomorphism

Vp:(Ai®A1)®---®(Ap®Ap)

-^-+(Al®--®Ap)®(Al®---®Ap) (7.4.1)

in which, with a self-explanatory notation, the element (ax đa\) đ ã ã ã

đ (ap đ a'p) of the first module is matched with the element

(ax ® - " đ ap) đ (a\ đ - - ã ® a'p) of the second. Thus Vp is an isomorphism

which accomplishes a certain type of rearrangement. In like manner we
shall use

Wp: {Ax đ A! đ A!) đ ã ã ã ® (Ap ® Ap ® Ap)

- ^ (v4t (x) • • • <g> Ap) ® (Ax ® - •• ® Ap) đ (Ax đ ã ã ã đ Ap)

(7.4.2)
for the rearranging isomorphism which makes

(ax ® a\ ® a'{) ® (a2 ® a'2 ®a"2)®'"® (ap ®a'p® a"p)

and

ap)® {a\ ®af2®-- ®a'p)

®'-® a"p)

correspond.

Now suppose that, for \<i<p, (A^A^Ei) is an /^-coalgebra (not

necessarily graded) and put

A=A1®A2®-"®Ap. (7.4.3)

At this stage A is just an R-module. We note that the mapping Aj đ
A2đm ã' đAP has domain Ax đ A2 đ- ã ã đAp and codomain

{Ax ®AX)® (A2 ®A2) ® • • - ® (Ap® Ap). Consequently, if we put

A = Vp o (Aj ® A2 ® - - • ® Ap\ (7.4.4)

then A is an R-linear mapping of A into A® A.

Consider the product R ® R ® • • • ® R where the number of factors is p.
There is a homomorphism

$>: R ® R đ ã ã ã đ R -> R (7.4.5)

which satisfies ^p )(ri ® ri ®''' ® rP) — riri • • • rP> a n (l using this we may

define

e:A-+R (7.4.6)
by

e = ixtf ° (et ® e2 ® • • • ® ep). (7.4.7)

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136 Coalgebras and Hopf algebras

Theorem 1. For l<ih Ahet) be an R-coalgebra. Then with the

above notation {A, A, e) is also an R-coalgebra.
Proof. The mapping

04! ® Ax) o Ax ® (A2 ® A2) ° A2 ® • • • ® (Ap ® Ap) ° Ap

takes X1® ^2 <$••• ® Ap into ( ^ ® ^ ® Ax) ® • • • ® (Ap ® Ap ® Ap).

Consequently

Wp°((Al®A1)°A1®--®(Ap®Ap)°Ap) (7.4.8)

maps A into /I ® A ® A. We claim that (7.4.8) is the same as (A ® A) ° A.
In order to verify this claim we shall introduce some notation which is
rather cumbersome but which is fully explicit. Later, when dealing with
similar but sometimes more complicated situations, we shall simplify the
notation in order to prevent it becoming unwieldy. The following argument
will provide a model for expanding the simplified expressions if the reader
wishes to have more details.

Suppose that al9 a2,..., ap belong to Al9A2,..., Ap respectively and let

us write

and so on until finally we have

(The simplification that we shall use later consists in replacing these
relations by AA.(ai) = Yl^i ®a'l for 1 = 1 , 2 , . . . , p. Thus there will be no

explicit references to the parameters k, ft,..., x used to index the terms in

the various sums. The reader is to take their presence as understood.) It then
follows that

(Ax đ A2 đ ã ã • ® Ap)(fl! ® a2 ® • • • ® ap)

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Tensor products of coalgebras 137

and therefore

(A ® A) (A(fl! đ a2 đ ã ã ã

= Z (fli

On the other hand,

((A! đ Aj) Ax đ ã ã • ® (Ap ® Ap) ° Ap) (fl!

from which it follows that

>---®fZ

fl

p(

T

)®Vp(

T

))

applied to ax ® a2 ® ' *' ® aP yields

• • • ®a'p(T))

This establishes our claim.
It has now been shown that

and similarly it can be proved that

(A ® A) ° A = Wp° ((Ax ® Ai) o Aj đ ã ã ã đ (Ap đ Ap) Ap).

But (Xf ® At) ° Af = (Af ® At) ° Af because Af is coassociative. Accordingly

(A ® A) ° A = A ° (A ® A) and therefore A is coassociative as well.
It remains for us to demonstrate that

R®A < A ® A

is a commutative diagram, and, as the two triangles can be treated similarly,
we shall confine our attention to the one on the left-hand side. In this case it
will suffice to show that the effect, on ax ® a2 ® * * * ® ap, of applying in

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138 Coalgebras and Hopf algebras

A r.đA

A ã A ®A >R®A -^—> A

is to leave the element unchanged. Now, when these homomorphisms are
applied the outcome is

/

]ep(a'p(x))a"p(r)\

But ^]/le1(ai(A))a'][(A) = a1 because the diagram

-A,

commutes, and similarly in the other cases. Accordingly the image of ax ®

di ®''' ® dp is itself and with this observation the proof is complete.

Definition. The coalgebra (A,A,e)of Theorem 1 is called the tensor product

of the coalgebras AX,A2,. • • ,Ap and it is denoted by Ax ® A2 ® ' ' ' ® Ap.

Since R is both an algebra and a coalgebra, the p-fold product R đ
R đ ã ã ã đ R is also an algebra and a coalgebra. Now by (7.1.4) and (7.4.5)
we have jR-linear mappings

Atf: R-+R ® R ®- - ®R
and

in fact Ajf} is the unit mapping of the algebra R ® R ® • • • ® R whereas ^
is the counit mapping of the coalgebra R ® R ® • • • ® R. It follows that A{f}
is an algebra-homomorphism and /xjf* is a coalgebra-homomorphism.
However, more is true as is shown by

Lemma 1. Ajf* is a homomorphism of both R-algebras and R-coalgebras;

and the same is true of fi^\

Proof It is clear that ///> is an algebra-homomorphism so we need only
show that Ajf) is a homomorphism of coalgebras. Evidently Ajf) preserves
counits so we are left with showing that the diagram

R -> R đ R

Rđ'- đR > (R đ - ã - đ R) đ (R đ ã ã ã đ R)

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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,
Af l |(/^)

and therefore

= Z (/iai ® ' *' ® fp<*p) ® (/ifl'i ® ' ' ' ® fPa"P

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140 Coalgebras and Hopf algebras

In the case of the counits we have

(efl/)(fli đ ' ã' đap) = eBi(f1a1)eB2(f2a2)... eBp(fpap).

But &B. (ftai) ~ ÊAt (ai)- Consequently

(sBf)(a1 đ ã ã ã đ ap) = eAi(a1)eA2(a2)... e^fo,)

= *U(0i®"-®0p).

Thus sB° f = eA and the proof is complete.

We turn next to the basic isomorphism theorems for coalgebras and
prove results which are the counterparts of Theorems 2, 3 and 4 of
Chapter 3.

Theorem 2. Let Al9 A2,..., Ap arcd B1,B29...,Bq be R-coalgebras. Then

the canonical isomorphism
A1®--®Ap®B1®'-®Bq

of modules (see Chapter 2, Theorem 1) is actually an isomorphism of 
R-coalgebras.

Proof Put A=AXđ ããã đAp,B = Bx đ '-đBq and C =

Axđ- ã - đApđB1đ" - đBq. In what follows a( will denote an element

of At and bj an element of By, furthermore we shall suppose that AAi(at) =

X a'i ® af' a n (^ ABj(bj) = Y, b'j ® fc}. Lastly we shall use

to denote the canonical module-isomorphism. Of course, it will suffice to
show that (j) is a homomorphism of coalgebras.

To this end we first observe that the result of applying Ac to

a1đ-" đapđb1đ"

' * ã đ a'p ® b\ ® • • • ® bq) ® {a'[ ® • • • ® d'p ® b\ ® • • • ® bq)

and therefore when {(j) ®(j>)° Ac is applied to the same element what we

obtain is

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Tensor products of coalgebras 141

say, and we also have

Ay4(a) = S (fli đ ' ã ã đa'p) đ (a'[ ®--'®a"p)

and

Afi(fr)=£ (b\ ® • • • ® *>;) ® (//; ® • • • ® / 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 p r o d u c e s

the element (7.4.10) and so we have established that ((/> ® 4>) ° Ac = A^ ^ ° </>.

Finally when £A®B°(I) and ec are made to act on ax ®- • • ® ap®

^i ® * * * ® bq the result in both cases is

e^di)... ^p(0p)£B l(M . . . eBq(bq).

It follows that £A0BO(t) = ec a nd 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®---®Ap^Aii ®Ai2®--- ®Aip

of modules (see Chapter 2, Theorem 2) is in fact an isomorphism of 
R-coalgebras.

Proof Let 0 : Ax đ A2 đ ã ã ã đ Ap-^Aix đ Ai2 đ ã ã ã đ Ai be the

module-isomorphism in question. For 1 and suppose that
A

^(ôi) = Zf lJđf li'- P u t ^ = ^ 1 ® ^ 2 ®'"®Ap and B = Aix ®Ai2

®---® A{ . If now ((/) ® (j>) ° A^ and AB °<j) are applied to a1 đ a2 đ ã ã ã ® ap, then

the result in both cases is

£ K ® • •' ® a'ip) ® (a'lx ® • • • ® a'(p),

whereas if sB ° </> and sA are made to operate on the same element, then the

result, again in both cases, is eAi(a1 )eAi(a2)... eA (ap). The theorem follows.

We recall that R itself is a coalgebra.

Theorem 4. Let A be an R-coalgebra. Then the canonical

module-isomorphisms A ® R &A and R đ A ô A are module-isomorphisms of R-coalgebras.

Proof Let (/>: A đ R ã A be the module-isomorphism in which (f)(a ®r) =
ra, let a belong to A, and let A^ (a) = £ a' ® a''. Then

so that (0 ® 0) ° A^ 0R = AA ° 0. Again

and therefore &AO4>=&A®R' This shows that ,4 ® # ^ y 4 is a

coalgebra-isomorphism and the other coalgebra-isomorphism is dealt with similarly.

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142 Coalgebras and Hopf algebras

A =Ail) đ A(2) đ ã ã ã ® Aip) (7.4.11)
has a natural structure as a coalgebra. In what follows A^ and eA will denote

the comultiplication and the counit of this coalgebra.

Let / = (il5 i2, • • •, ip) be a sequence of p integers, and (as in the case of

graded algebras) put

AI = A\1i)®A\22)®- -®A\P) (7.4.12)

so that

A=YtAI (7.4.13)

this being a direct sum of /^-modules. If we now set

|/| = i1+ i2 + - " + ip (7.4.14)

and

An= Z A!9 (7.4.15)

where the sum is taken in A, then {An} «6Z is a grading on the module A. It is

clear that if R is graded trivially, then eA: A —• JR preserves the degrees of

homogeneous elements; and it follows from (7.4.4) that A^: A—+A ® A is
also degree-preserving provided that A ® A is given the usual total grading.

We can sum up these observations as follows. Let A{1\ A{2\ . . . , Aip) be
graded coalgebras. Then A{1) đ A{2) đ ã ã ã đ A{p) has the structure of a
coalgebra by virtue of Theorem 1. Furthermore the usual total grading on the
module A{i) ® Ai2) đ ã ã * đ A{p) is a coalgebra-grading.

In the next section we shall discuss a way of modifying this structure. For
the present, however, we shall leave this refinement on one side and begin by
observing that if A{1\ A{2\ . . . , A{p) and B{1\ B{2\ . . . , B{q) are graded
coalgebras, then the isomorphism

Aa) ® • • • ® A{p) ® B(1) ® • • • ® B{q)

% (Ail) ® • • • ® Aip)) đ (B(1) đ ã ã ã đ Biq)\ (7.4.16)
provided by Theorem 2, is an isomorphism of graded coalgebras; and that if
(il9 i2, • • •, ip) is a permutation of ( 1 , 2 , . . . , /?), then the isomorphism

Ail) đ A{2) đ ã ã • ® A{p) % A{il) ® A{il) ® • • • ® A{i^ (7.4.17)
of Theorem 3 is also an isomorphism of graded coalgebras. Yet again, if A is
a graded coalgebra, then the coalgebra-isomorphisms

A®R*A (7.4.18)
and

R®A^A (7.4.19)

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Modified tensor products of coalgebras 143

Finally, suppose that
ft:A®^>B®

is a homomorphism of graded coalgebras for i = 1 , 2 , . . . , p. Then, because
ã ã ã đ Aip)->Bil) đ • • • ® 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) ® (4

2

2)

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

and these various isomorphisms can be combined to give an isomorphism
Vp: (A(1) đ A{1)) đ ã ã ã đ (Aip) đ Aip))

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144 Coalgebras and Hopf algebras

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) đ ã • •

® A{p)) ® (A(1) ® • • • ® Aip)) onto ,4, ® i4y, 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) đ Ai4(2> đ ã ã ã đ A^(P»). (7.5.8)

Now, by Theorem 1, A = A{1) ® y4(2) ® • • • ® A{p) 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) đ ã ã ã đ 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

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Modified tensor products of coalgebras 145

and

K , , K ° 0 4 ® A ) ° A ) ( a )

= % e({*'}9 {*'})*,(*') ®nJJC(Aắ).

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 = nu 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 đ^R < Ađ A ã /Tđ A

commutes. Now the combined effect, on a, of the mappings

A A®e

A >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 A®e

A > A ®A >A ®R -^—^ A

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146 Coalgebras and Hopf algebras

The graded coalgebra (A{1) đ ã ã ã đ 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) ®--® Ai 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) đ A(2) đ ã ã ã đ A{p) into the
coalgebra B{1) đ B{2) đ ã ã ã đ Bip\ But changing to modified tensor
products does not affect the underlying module structures. Consequently
/ i ® - " ® /p: ^( 1 ) ®'" ® A{ P )- > B{ 1 ) ® - - - ® Bi p ). (7.5.12)

Note that (7.5.12) is JR-linear and preserves degrees.

Lemma 4. / / fr đ f2 đ ã ã • ® fp is considered as a mapping of A{1) ®

A{2) ® • • • ® Aip) into B{1) ® B{2) đ ã ã ã đ 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

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Modified tensor products of coalgebras 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 
R-coalgebras. Then, by Theorem 1 of Chapter 2, there is an ^-linear bijection

®A{p) ®B(1) < Đ ã ã ã đBiq)

ã ã ã đ A(p)) đ (B(1) đ ã ã ã đ Biq)) (7.5.13)

in which

4>{ah đ--đaipđbhđ--đ bjq) = (ati đ ã ã ã đ aip) đ (bji đ ã ã ã đ bjq).

Theorem 5. The bijection (7.5.13) is an isomorphism

ã ã ã đ Aip) đ 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) đB{2) đ"- đB(q\
and

C = Aa) đ ã ã • ® 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

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148 Coalgebras and Hopf algebras

it follows that

= 1 *(Ki đ ã ã ã đ *>;}, {a\ ® • • • ® fc;})(ai ® • • • ® vq) ® (*; ® • • • ® &;),

where by {ai đ ã ã ã đ 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" đ &"),

where a' = ai đ ã ã ã ® ap, a" = a? ® • • • ® a,, &' = & i ® - - - ® ^ and ft" =

Put / ' = {a\ ® • • • ® ^p}, J' = {b[ ® • • • ® fo^} and define J" and J"
similarly. Also, if / ' = (i'i, r2, . . . , 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 đ (j>)o Ac)(ati đ ã ã ã đ 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 đ ã ã ã đ afp) đ (bh đ • • • ® ^ ) )

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 + ã ã ã + fp.

Accordingly

đ ( ^ ® ' *' ® 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 ) ° AC and

&A®BO(}> produce the same result when applied to a^®- - ®a

{ ®

bjt ® * * * ® bj. It follows that (0 ® (/>) ° Ac = A^^^ ° 0 and with this the

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Modified tensor products of coalgebras 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){*'}{flV ® V) ® (a" ® b")

and therefore

(T®T)(AA®B(am®bn))

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150 Coalgebras and Hopf algebras

= AB(^A(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 
skew-commutative algebra provided that the diagram

T

A đA ã A đA

(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

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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.

7.7 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

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 đ R ã 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

fđgđh

A > Ađ Ađ A ã RđRđR > R, (7.7.5)

where A^A ®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 HR

A >A đA ã RđA >RđR >R

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152 Coalgebras and Hopf algebras

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)). 
Con-sequently, 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

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

commutative algebra.

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

u f®g nR

A® A > A ®A >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 (

-

)

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 -ã A đ A preserves degrees

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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-^A ®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) (A, \i, n) is an R-algebra;
(ii) (A, A, e) is an R-coalgebra;

(iii) A: A-^A ® A and e: A^>R are homomorphisms of algebras;
(iv) [i\ A ® A-^A 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) is a

coalgebra. Then

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154 Coalgebras and Hopf algebras

(b) fi: A ® A—>A preserves counits if and only if e: A^>R is 
com-patible 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 đ A ã 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 
comulti-plication 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 ã A đ A and e: A ằ R are homomorphisms of
algebras.

Now suppose that (rjA, fiA,A,AA,sA) and (rjB, fiB,B, AB,eB) are Hopf

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.

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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-^A 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-^A 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-^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.

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156 Coalgebras and Hopf algebras

where T: A ®A^*A ®A is the twisting isomorphism, so that what we
require is

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 = fiA^A° (A ® A) and this is

equivalent to (7.8.6) because fiA^A = (/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-^A ®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 Af, ef) be a Hopf algebra (over the ring R)

and put

A=A1®A2®"-®Ap. (7.9.1)

Then A has a natural structure both as an algebra and as a coalgebra.
Indeed, by (7.1.5) and (7.1.6),

° Ap 9 (7.9.2)

pA^\ (7.9.3)

and, by (7.4.4) and (7.4.7),

A^ = Vp o (Ax ® A2 ® • • • ® Ap), (7.9.4)

*A = HRP) o(e1®e2®'"® ep). (7.9.5)

(The notation is explained in Sections (7.1) and (7.4).) Now, because At is a

Hopf algebra, Ax đ A2 đ ã • * ® Ap is an algebra-homomorphism of Ax ®

A2®'"®Ap into (A1®A1)®(A2®A2)®'"®(Ap®Ap), and it

follows, from Theorems 2 and 3 of Chapter 3, that Vp is an

algebra-isomorphism of

(Ax ®A1)®(A2®A2)®-"®(Ap® Ap)

onto - (A1đA2đ-"đ Ap) đ(A1đA2đ-"đ Ap). Accordingly

A^: A ã A đ A is a homomorphism of i^-algebras. Next p^ (see (7.4.5)) is a
homomorphism of R-algebras and, of course, £1®e2®- - ®ep is an

algebra-homomorphism of Ax đ A2 đ ã ã ã ® Ap into R ® R ® • • • ® JR.

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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)

eAđeB HR

AđB ã R đR ã 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

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158 Coalgebras and Hopf algebras

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) đ A{2) đ ã ã ã

đ A{p~1} is a modified Hopf algebra and therefore, by Lemma 7, the same is
true of

{A{1) đ A{2) đ ã ã ã đ Ato-V) ®A{p).

Next, Theorem 1 of Chapter 2 provides a degree-preserving 
module-isomorphism

(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
left-hand 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

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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 
anti-commutative R-algebra (see Chapter 3, Theorem 7). Consequently the
mapping <t>: M ã E(M) đ 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 (7.10.4)

A®£(M) ^

E(M) ® E(M) > E(M) ® E(M) ® E(M)

and

^E(M).

(7.10.5)

£®£(M) y

E(M)®e

R ® E(M) < E(M) ® 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.

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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 
skew-commutative.

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) đ E(M) and
e: E(M) ã 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.

Theorem 12. Let f:E—*N be a homomorphism of R-modules. Then the

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

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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 <p) integers satisfying 1 < i1 < i2 < • • * < is <p,

and by /' we shall understand the sequence obtained from ( 1 , 2 , . . . , p) by
deleting the terms that belong to /. Furthermore, if /' = (*i, i'2,...)> then

sgn(7,/') will denote the sign of ( i1 ?. . . , is, ii,i"2,...) considered as a

permutation of ( 1 , 2 , . . . , p).

Now let ml9 m2, . . . , mp belong to M. Put mI = mii A mi2 A • • • Amfj and

define mr similarly, it being understood that the natural conventions apply

when s = 0 and s=p.

Lemma 9. With the above notation

Am2 A • • • Amp) = YJ sgn(J, 7')m7

Proof. Since A: E(M)—>E(M) ®E(M) is a homomorphism of algebras,
A(ml Am2 A • • • A mp) is the product, in E(M) ® E(M), of A(ml)9A(m2),...,

A(mp), that is to say it is the product of (mx ® 1 -f 1 (g) mx), (m2 ® 1 + 1 ® m2)

and so on. Accordingly A(mx A m2 A • • • A mp) is the sum, taken over all

sequences / = (il9 i2, . . . , is), of the product in £(M) ® £(M) of the elements

1 ® mx, . . . , mfi ® 1 , . . . , mis ® 1 , . . . , 1 ® mp.

But £(M) ® E(M) is an anticommutative algebra and here we are

concerned with homogeneous elements of degree one. Consequently the
product in question is equal to sgn(/, /') times the product of

mfi ® 1, mi2 ® 1 , . . . , 1 ® mi;, 1 ® mi2,...

and therefore it has the value sgn(7,1')mI ® m7'. The lemma follows.

We now put

E(M)* = Homi?(£(M), R) (7.11.1)

so that E(M)* consists of all the linear forms on E(M); we also put
En(M)* = HomR(En(M),R) (7.11.2)

and identify En(M)* with the module formed by the linear forms on E(M)

that vanish on Ek(M) for all k^n.

By Theorem 7, E{M)* is an JR-algebra whose identity element is the linear
form e: E(M) —> R which projects E{M) onto E0(M). For / and g in E(M)*

we shall use / A g to denote their product in this algebra. Naturally / A g is
also a linear form on E(M) and, by (7.7.4),

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162 Coalgebras and Hopf algebras

(We recall that A is the comultiplication mapping of E(M) and that fiR is the

multiplication mapping of R.)

Lemma 10. Let ml9 m2, . . . , mp (p>0) belong to M and let f g belong to

£(M)*. Then (with the same notation as in Lemma 9)

Am2 A • • • Amp) = £ sgn(/, /')

Remark. When p = 0 this is to be interpreted as asserting that ( / A g)(i) =

Proof. The desired result follows by combining (7.11.3) with Lemma 9.

Lemma 11. Let /l 9/2, . •. , /p ( p > l ) belong to M* = E1(M)* and let fx A

f2A"'Afpbe their product in the algebra £(M)*. Furthermore, let m1? m2,

. . . , mp belong to M. Then

( / i A /2 A - - - A /p) ( m ! A m2 A - - - )

/p(m2)

(7.11.4)

Proof. We use induction on /?. The assertion is obviously true when p = 1 so
we shall assume that p > 1 and that the lemma has been established for all
smaller values of the inductive variable. Put g = f2A"'AfP' Since

fxim^ AmV2 A ••• Amv) = 0 except when h = l, it follows, by Lemma 10,

that

A m2 A - - - A mp)

/\mp)
( / i A /2A

-p

/2K)

because, by the inductive hypothesis, g(mt A • • • A mt A • • • A mp) is equal to

the determinant obtained by striking out the first row and the i-th column
in (7.11.4).

Let us now put

(7.11.5)

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S(M) as a Hopf algebra 163

(7.11.6)
Then, by Theorem 8, G(M) is an K-subalgebra of E(M)* and it has
{Gn(M)}neZ as an algebra-grading. Since

G0(M) = Hom^(£0(M),JR) = Homi?(^,JR), (7.11.7)

G0(M) is a free R-module of rank one. Note that e is the identity element of

G(M) and that it forms a base for G0(M). We also have

Gl(M) = HomR(El(M)iR) = M*. (7.11.8)

Finally, we note that if M is a finitely generated R-module, then G(M) =
E(M)*. This is because finite generation ensures that En (M) = 0 for all large

values of n.

Definition. The graded R-algebra G{M) is called the 'Grassmann algebra' of

M.

Theorem 13. The Grassmann algebra G(M) is an anticommutative algebra.

Proof. By Lemma 8, E(M) is a skew-commutative coalgebra and therefore,
by Theorem 8 Cor., G(M) is a skew-commutative algebra. Now suppose
that feGp(M), where p is an odd integer. Then, to complete the proof, we

need only show that / A / = 0. Suppose therefore that mi,m2,... , m2p

belong to M. Since / A / belongs to G2p(M), the theorem will follow if we

establish that

( / A / M m ! Am2 A - - - Am2p) = 0.

But, by Lemma 10,

( / A / ) ( mx Am2 A • • • Am2p) = YJ sgn(/, //) / ( w/) / ( mr) ,

where (because feGp(M)) we have only to sum over all sequences / =

(il5 i2, • • •, ip) of length p with ! < / ! < • • • <ip<2p. (Here / ' is obtained by

deleting the terms in / from ( 1 , 2 , . . . , 2p).) But, for such an /, sgn(7, /') =
— sgn(/', /) because p is odd, and therefore

( / A / ) ( * * ! ! A m2 A ' " Am2 p) = 0

as required.

7.12 S(M) as a Hopf algebra

Let M be an K-module and let S(M) be its symmetric algebra. We
propose to show how the structure of S(M) as an algebra can be
complemented so that the final result is a graded Hopf algebra. In what
follows we identify S0(M) with R.

Let s: S(M)-^R be the projection of S(M) onto S0(M). This will be the

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164 Coalgebras and Hopf algebras

To define the comultiplication or diagonalization mapping on S(M), we
first observe that there is an R-linear mapping M —> S(M) ® S{M) in which
m, in M, becomes m ® 1 + 1 ® m. However, S(M) ® 5(M) is a commutative
algebra and therefore the /Minear mapping can be extended to a
homomorphism

A:S(M)->S(M)®S(M) (7.12.1)
of algebras. We record that, by construction,

A(m) = m ® l + l ® m (7.12.2)
for all m in M.

Lemma 12. With the above notation, (S(M), A, e) is a coalgebra. It is graded,

as a coalgebra, by the symmetric powers {Sn(M)}neI of the module M. The

coalgebra (S(M), A, e) is commutative.

Proof. This lemma is the counterpart of Lemma 8 and it can be proved in
very much the same way. The details will therefore be omitted.

In the next theorem, \i and rj denote the multiplication and unit mappings
of the symmetric algebra.

Theorem 14. Let the notation be as explained above. Then (rj, JJL, S(M), A, s)

is a graded Hopf algebra, the grading being given by the symmetric powers
{Sn(M)}neI of the module M. As a Hopf algebra, S(M) is commutative.

Proof That S(M) is a Hopf algebra follows from the corollary to Lemma 5.
It is obviously commutative.

Theorem 15. Let f: M —• N be a homomorphism of R-modules. Then the

mapping S(f): S(M)-^S(N), which was defined in Section (6.2), is a
homomorphism of graded Hopf algebras.

Proof Theorem 15 corresponds to Theorem 12. In the case of Theorem 12
we described the essential features of the demonstration. Similar

considera-tions apply here.

In Section (7.11) we used the fact that E(M) is a graded coalgebra in order
to derive from it the Grassmann algebra of M. In like manner we can use the
graded coalgebra S(M) to derive what we might expect to be a new algebra.
Let us see what the construction produces.

To this end we consider

Un = Sn(M)* = HomR(Sn(MlR) (7.12.3)

as a submodule of S(M)* = HornR(S(M),R) and put

H=IH, (7.12.4)

nel

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S(M) as a Hopf algebra 165

by the family {9lw}neZ. Furthermore, the corollary to Theorem 7 shows that

the algebra is commutative. For / , g in 91 we shall use fg to denote their
product. Then, by (7.7.4),

fg = HR°(f®g)°A, (7.12.5)
where A is the diagonalization mapping of S(M).

Next, we recall that in Section (6.7) we defined an algebra V5(M) which
we named the algebra of differential operators. By construction it is a
commutative subalgebra of End^(5(M)) and, moreover, it is graded. The

module of differential operators of degree h was denoted by Vh5(M).

Suppose that </>eVhS(M), where h>0. Then </> is an K-endomorphism of

S(M) of degree —h and it induces a linear form, ^ say, on Sh(M). Also, by

Theorem 8 of Chapter 6, the mapping 0h-x/> is an isomorphism of VhS(M)

onto Sh(M)*. Accordingly we have an isomorphism

^ > 91, (7.12.6)

for each value of h and, by combining these isomorphisms, we arrive at a
degree-preserving isomorphism

(7.12.7)

of ^-modules.

Theorem 16. The module-isomorphism (7.12.7) is actually a

degree-preserving algebra-isomorphism of the algebra VS(M) of differential
operators onto the algebra 91.

Proof Suppose that 0eVfcS(M) and i^eVkS(M), where h>0 and /c>0.

Then </> ° \jj is a differential operator of degree h + k and its image in 91,,+k is

its restriction </>°^ to Sh+k(M). The theorem will follow if we show that

(j) ° \\t is the product of cj) and ^ in 91.

To see this let ml5 m2, . . . , rnh+k belong to M. It will now suffice to prove

that {(f)°\jj)(m1m2 . . . rnh+k) and (^{j/)(m1m2 .. .mh+k) are the same. In this

connection we note that
...mh+k)

imi 2. . . mih)il/(mjmJ2... mjk),

where (i1,i2,--4h) anc* (/iJ2>""Jk) constitute a variable pair of strictly

increasing sequences of integers which, between them, include all the
positive whole numbers from 1 to h + k (see (6.7.6)). Next, from (7.7.4), we see
that (j)ij/ = tiR ° (0 (g) ^ ) ° A, where in this context 0 and ij/ are regarded as

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166 Coalgebras and Hopf algebras

S{M) ® S(M), of the elements

mf ® 1 + 1 ® mt (i = 1 , 2 , . . . , h

Of course, these elements are of degree one and S(M) ® 5(M) is a
commutative algebra; also </> is null on Sn(M) if n ^h, whereas \j/ is null on

Sn(M) if n=£k. From these observations we conclude that

((f ®\j/)o A)(mim2 ... mh+k)

where the sum is taken over the same pairs of sequences as before. But
extends </> and \j/ extends \j/. Consequently

= £ ^(m^m^ . . . mih)\l>{mhmh . . . mJk)

and with this the proof is complete.

Thus, to sum up, when we consider linear forms on the coalgebra S(M)
we do not obtain an essentially new graded algebra; instead we recover up
to isomorphism the algebra of differential operators, on S{M), that we
encountered in Chapter 6.

7.13 Comments and exercises

It will be clear to the reader, and indeed it has been mentioned
already, that the analogy between the theory of algebras and the theory of
coalgebras can be developed further than was done in the main text. Thus
our first exercise corresponds to the fact that the tensor product of a number
of commutative algebras is itself commutative, and the second exercise is
suggested by Theorem 7 of Chapter 3. The two exercises can be solved in
very much the same way, so a solution will be provided only for the second
one. As on earlier occasions, those exercises for which a solution is given are
marked with an asterisk.

Exercise 1. Show that if At, A2, • . . , Ap are commutative R-coalgebras, then

Al đ A2 đ ã * • ® Ap is also a commutative R-coalgebra.

Exercise 2*. Show that if A{1\ A{2\..., Aip) are skew-commutative 
R-coalgebras, then A{1) ® A{2) ® • • • ® A{p) is also a skew-commutative R~

coalgebra.

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Comments and exercises 167

modified tensor product remains unchanged to within an isomorphism of
graded coalgebras.

The formal difference between the definition of an anticommutative
algebra and that of a skew-commutative algebra is clear enough. The next
exercise invites the reader to construct an example to show that the
distinction is a real one.

Exercise 4*. Give an example of a skew-commutative algebra which is not

anticommutative.

Next we return to the original definition of a graded algebra. Suppose
then that A is an K-algebra and {An}neI is a family of R-submodules of A

that grades A as a module. By the definition given in Section (3.3), {An}neZ

grades A as an algebra if from ar e Ar and as e As it follows that aras belongs

to Ar+S. Thus we have an algebra-grading provided only that the

multiplication mapping ix: A® A-+A preserves degrees. However, by
Theorem 5 of Chapter 3, in these circumstances the identity element of A is
homogeneous of degree zero and so the unit mapping rj.R^A 
auto-matically preserves degrees as well. The following exercise shows that a
similar situation exists in the case of coalgebras.

Exercise 5*. Let A be an R-coalgebra and let {An}neI be a family of

R-submodules of A such that

A=YJAn (d.s.).
neZ

Further, let the comultiplication mapping A: A —• A (x) A preserve the degrees
of homogeneous elements. Show that if R is given the trivial grading, then the
counit mapping a: A-+R also preserves degrees.

We now add some comments concerning the results of Section (7.7).
There we saw that if A is an R-coalgebra and A* = HomR (A, R), then A* has

a natural structure as an R-algebra. Let us examine the functorial aspects of
the way the algebra A* depends on the coalgebra A.

First we note that each ^-linear mapping </>: A—+B, of A into an 
R-module B, gives rise to a R-module-homomorphism

</>*:B*-+A* (7A3A)
in which <£*(/) = / °(/>.

Exercise 6*. / / (j>: A—>B is a homomorphism of R-coalgebras, show that

cj)*: B*^>A* is a homomorphism of R-algebras.

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168 Coalgebras and Hopf algebras

Let M be an ^-module. In Theorem 16 we saw that when S(M) is
regarded as a graded coalgebra and we use Theorem 8 to derive from it a
graded algebra, what we obtain is isomorphic to the algebra of differential
operators on S(M); in particular the application of Theorem 8 produces an
algebra that is isomorphic to a certain graded subalgebra of EndR(S(M)).

This suggests that the Grassmann algebra G(M) may be related to some
graded subalgebra of EndR(E(M)). As will now be shown this is indeed the

case.

Suppose that h > 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 <*! <i2<'' • <ih<p, and / ' denotes the increasing

sequence obtained from ( 1 , 2 , . . . , p) by deleting the terms that belong to /.
Finally, if m1,m2,..., mp belong to M, then m7 will be used for the element

mti A mi2 A • • • A/Mj and mr is to be defined similarly.

By a differential form on E(M) of degree h, we shall understand an 
R-endomorphism cj) of degree — h that has the following property: for all p > h
and all mx, m2, . . . , mp in M

^(m! Am2 A - - - Amp) = £sgn(/,/')</>(w/)"V- (7.13.2)

As usual, if I = (il,i2,...) and r = (i'l9i2,.. .)> then sgn(/,/') stands for the

sign of (il5 i2,..., i\, i'2,...) considered as a permutation of ( 1 , 2 , . . . , p).

Evidently the differential forms of degree h constitute an K-submodule of
EndK (£(M)); and if two differential forms of degree h induce the same linear

form on Eh(M), then they coincide.

Exercise 7*. Let (j) and if/ be differential forms on E(M) of degrees h>0and

k>0 respectively. Show that ij/ ° (/> and (f)o{j/ are differential forms of degree
h + k and that xj/ ° 0 = (— \)hk(j) ° \j/. Show also that if h is odd, then 0 ° 0 = 0.

The next exercise should be compared with Theorem 8 of Chapter 6.

Exercise 8*. Let M be an R-module and f a linear form on Eh(M), where

h>0. Show that, on E(M), there is exactly one differential form of degree h
that extends f

From Exercises 7 and 8, we see that there is a subalgebra of End# (E(M))
that is non-negatively graded by the modules of differential forms. Let us
call this the algebra of differential forms on E(M). By Exercise 7, the algebra
is anticommutative.

Let (/> be a differential form on E(M) and let the degree of <f> be h. Then <j>
restricts to a linear form, </> say, on Eh(M). Note that 0 belongs to the

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Solutions to selected exercises 169

Exercise 9*. Let <f> and i// be differential forms on E(M). Show that the linear

form \f/o(f> associated with i// ° cj) satisfies \\i ° </> = 0 A \j/, where (j> A \j/ is the
product of (j> and \j/ in G(M).

Let A be a ring with identity and let A be the ring that is opposite to A.
(Thus A and A coincide as sets and addition is the same for both of them;
however, the product of two elements of A is their product in A but with the
order of the factors reversed.) If A happens to be an fl-algebra, then A will
also be an K-algebra with the same R-module structure as A. Naturally we

refer to A as the opposite algebra to A. Finally, if {An}neZ grades A as an

R-algebra, then the same family provides an algebra-grading on the opposite
algebra.

This terminology enables us to sum up our main conclusions as follows:

the Grassmann algebra G(M) and the opposite of the algebra of differential
forms on E(M) are isomorphic graded algebras.

7.14 Solutions to selected exercises

Exercise 2. Show that if A(1\ A{2\... ,Aip) are skew-commutative 
R-coalgebras, then A(1) ® Ai2) ® * • • ®A(p) is also a skew-commutative 
R-coalgebra.

Solution. The property of being skew-commutative is preserved under an

isomorphism of graded coalgebras, and because of this it is enough to prove

the result in question in the case of two coalgebras. We shall therefore
assume that A and B are skew-commutative coalgebras and prove that

A ® B is skew-commutative as well.

We use our usual notation. Let am belong to Am and bn to Bn. Further let

&A(flm) = If l' ® a" a r id AB(bn) = £ b' ® ft", where a', a" are homogeneous
elements of A and b\ b" homogeneous elements of B. Then

*A9B(am ® K) = l (~ l){a"m(a' ® b') ® (an ® b"),

where it is understood that {ar}, {a") denote the degrees of a! and a", and

similarly in the case of {b'} and {b"}.

Let TA:A®A^>A ® A, TB\B®B^+B®B and
T: (A ®B)®(A ® B) -^-> (A ® B) ® {A ® B)

be twisting isomorphisms. Then

(T°AA®B)(am®bn)

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170 Coalgebras and Hopf algebras

and now it follows that

b") ® (a' ® V)

Accordingly T° AA0B = AA0B and with this the solution is complete.

Exercise 4. Give an example of a skew-commutative algebra which is not

anticommutative.

Solution. Let F be a field of characteristic two and let A be the free 
F-algebra generated by a set {X,Y} of two elements. Then XY+YX
generates a homogeneous two-sided ideal, / say, in A. Let B denote the
graded F-algebra A/1 and denote by x and y the natural images of X and Y
in B.

The homogeneous elements of B that have degree one form the vector
space B^ = Fx + Fy, and this generates B as an algebra. By construction
xy + yx = 0 and now, because F has characteristic two, we have bb' + b'b = O
for all b, b' in Bv That £ is skew-commutative is a simple consequence of

this.

Finally X2$I and therefore x2=^0. This shows that B is not
anticommutative.

Exercise 5. Let A be an R-coalgebra and let {An}neI be a family of

R-submodules of A such that

Further, let the comultiplication mapping A: A —• A ® A preserve the degrees
of homogeneous elements. Show that if R is given the trivial grading, then the
counit mapping e: A-+R also preserves degrees.

Solution. Suppose that ameAm, where m / 0 . It is sufficient to show that

s(am) = 0.

We can write A(flm) = ]T af ® fii9 where a,-, pt are homogeneous elements

of A the sum of whose degrees is m. Now, because £ is the counit mapping,
we have

flm = Zfi(«i)A = Ze(i8l)a< (7.14.1)

and therefore e(am) = ^e(ai)£(j?l).

Let us use {a,}, respectively {Pt}, to denote the degree of af, respectively

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Solutions to selected exercises 111

and therefore

£ e(af)£(ft) = 0. (7.14.2)

/:{a,-}*0

But, again from (7.14.1),

and so

/:{«,} = 0

If now we combine this with (7.14.2) we find that £ e(ai)e()8f) = 0, that is to

say a(am) = 0. This completes the solution.

Exercise 6. / / </>: A-^B is a homomorphism of R-coalgebras, show that
(/>*: B*—>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) in 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 <7\ < • • •

<jk A • • •

A m^ and define my similarly. Further we use /', respectively J\ to denote the

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172 Coalgebras and Hopf algebras

By (7.13.2),

^(m! Am2 A - - - Amp) = ^ s g n ( / , / ' ) 0 ( m/)

and now if we operate on this with \j/ we obtain

--- Amp)

= £ sgn(/, /') sgn(J,

u

Note that the sequence J followed by the sequence (IJ)' is a permutation of
/'. It is the sign of this permutation that is denoted by sgn(J, (IJ)').

Now /, J and (IJ)', taken in order, provide a permutation of ( 1 , 2 , . . . , p).
If sgn(/, J, (U)f) denotes the sign of this permutation, then

sgn(/, /') sgn(J, (/J)') = sgn(J, J, (IJ)')

and therefore

(if/ ° ^)(m! Am2 A - - - Amp)

= £ sgn(/, J, (/jy^K^^K,), (7.14.3)

Similarly we find that

X K ^ (7.14.4)
and from (7.14.3) and (7.14.4) it is clear that \\i ° 0 = ( - l)*k0 °

<A-For the last step we assume that h is odd and we take \j/ = </> so that ft = /c.
By (7.14.3),

(0 °0)(m1 Am2 A • • • Amp) =

u
= 0
because, under the present assumptions,

sgn(/, J, (JJ)') + sgn(J, /, (J/)') = 0.
Accordingly 0 ° (/> = 0 as required.

Exercise 8. Let M be an R-module and f a linear form on Eh(M\ where

h>0. Show that, on E(M), there is exactly one differential form of degree h
that extends f

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Solutions to selected exercises 173

where the notation is the same as in (7.13.2). This mapping is clearly
multilinear.

We claim that it is also an alternating mapping. For suppose that \<j<p
and that m}f = m7- + v If both j and j+1 are in 7,thenm7 = O;andifneitherisin

/ then m7' = 0. Thus in both these cases f(mI)mI=0.

The remaining /'s are those which contain just one of j and j +1. Such
sequences can be arranged in pairs, Ix and I2 say, where each member of a

pair becomes the other member when j and j + 1 are interchanged. Now
/(w/^m^ is equal to /(m/2)my2, because mj = mj+l; and

s g n ( /1, / i ) = - s g n ( /2, //2) .

Consequently

sgn(/1? ry)f(mh)mVx + sgn(/2, I2)f(ml2)ml2 = 0

and therefore we have

Xsgn(/,/')/(m7)mr = 0.

This establishes our claim (see Chapter 1, Lemma 2).

It now follows that there is an R-linear mapping of Ep(M) into Ep_h(M)

in which

mx A m2 A •••

When /? = /* this is just / Accordingly if we combine these various
homomorphisms into an endomorphism of E(M) it will extend / ;
moreover, by construction, the endomorphism so obtained will be a
differential form of degree h. This provides all that is needed because the
part of the exercise that is concerned with uniqueness has been dealt with
already.

Exercise 9. Let (j) and \\/ be differential forms on E(M). Show that the linear

form \\i ° (/> associated with \\i ° (j) satisfies i// °</> = (j> A \j/9 where

product of (j) and ij/ in G(M).

Solution. Let the degrees of (/> and \jj be h > 0 and k > 0 respectively, and let
ml9 m2, . . . , mh+k belong to M. By (7.13.2),

Am2 A

and therefore

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174 Coalgebras and Hopf algebras

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 <^ and \jj (considered as elements
of £(M)*) belong to Eh(M)* and Ek(M)* respectively, we find that the

relation provided by the lemma is

! Am2 A • • • Amh+k) = YJ s g n ( / , / ' ^ ( m ; ) ^ ^ ) ,

where / ranges over the same set of sequences as before. However, 0 (mz) =

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8

Graded duality

General remarks

From a graded coalgebra it is possible, by considering linear forms
on the submodules which make up the grading, to derive a graded algebra.
In fact, this is the conclusion of Theorem 8 of Chapter 7. An attempt to
perform a similar construction starting instead with a general graded
algebra soon runs into difficulties; but when the algebra is suitably
restricted the process can be carried through successfully to produce a
graded coalgebra as the end-product.

In the present chapter these ideas are developed in detail and it is shown
that, for non-negatively graded algebras and coalgebras whose grading
modules are free modules of finite rank, one has a full duality theory. This
not only interchanges algebras and coalgebras, but also preserves ordinary
and modified Hopf algebras.

The first two sections of the chapter are used to establish the results on

modules that provide the basis of this theory. As always, R denotes a
commutative ring with an identity element, and the symbol ®, when used
without a subscript, indicates a tensor product formed over R.

8.1 Modules of linear forms

It was remarked in the introduction to this chapter that the duality
we shall be discussing arises from the study of linear forms on free modules
with finite bases. However, before we restrict our attention to such modules
it is convenient to make some observations concerning linear forms on
arbitrary modules.

Suppose then that M and N are R-modules and put

M* = HomR(M,R), (8.1.1)

N* = HomR(N, R) so that M* and N* are the so-called duals of M and N

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176 Graded duality

respectively. If now k: M—>N is K-linear, then the dual homomorphism
k*:N*->M* (8.1.2)

is defined by

k*((t)) = <j)°k (8.1.3)

for all <j> in N*. Note that if idM denotes the identity mapping of M, then

(idM)* = idM*; (8.1.4)

and that if we have a second homomorphism, \i\ N-^K say, then

(^°/l)* = /l*0/**. (8.1.5)

(In the language of Category Theory, the taking of duals is a contravariant
functor from ^-modules to R-modules.)

If we consider R as a module and form its dual R*, then this is isomorphic
to R itself; in fact we have an isomorphism

R**R (8.1.6)

in which / in R* is matched with / ( I ) in R. This will be called the canonical
isomorphism of R* onto R and, in the sequel, it will often be used to identify
R* with R.

An ^-module M is connected with its double dual M** by means of an 
R-linear mapping

M^M** (8.1.7)
in which m in M becomes the homomorphism M*-^>R given by/i—>f(m).
The homomorphism M^>M** is natural in the sense that if k: M—>N is # 
-linear, then

M >M**

(8.1.8)

N > N**

is a commutative diagram.

We must now consider the duals of finite direct sums and of tensor
products. The former are easily described for if Ml9 M2, . . . , Mp are

R-modules, then we have an isomorphism

here / in (Mi â M2 â ã ã * â Mp)* corresponds to (/l9 /2, . . . , /p), where /•

is the restriction of / to Mj(

For tensor products the situation is more complicated. Let us again
assume that Mt, M2, . . . , Mp are /^-modules and let ft belong to Mf for i =

1,2,... ,p. If now

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Modules of linear forms 177

denotes p-fold multiplication (so that /ijf

r

iri" - rp)> then /4JP) °{fi® f2®" ' ® fp

M2 ®' • • ® Mp and the mapping

Mx x M2 x • • • x Mp- > (Mx ® M2

which takes (fl9 /2, . . . , /p) into ^jf} ° (fi

Accordingly there is an K-linear mapping
cằ: Mf đ Mf đ ã ã ã ® MJ - • (Mx

which is such that

<o(fi ®fi®" ' ® /P) = 4P) o(fi®f2

and therefore, when mf e M, for i = 1 , 2 , . . . , p,

i ® ri ® " ' ® rP) is equal to

is a linear form on Mi đ

ã • • ® Mp)*

f2®' " ® fp) is multilinear.

M2 ® • • • ® Mp)* (8.1.11)

"®fP) (8-1-12)

^fi(m1)f2(m2)...fp(mp). (8.1.13)

It will be noticed that there is an ambiguity surrounding the use of fx ®

f2®" ' ® fp- On the one hand it can denote an element of Mf đ MÊ đ ã • •

® M* and on the other it can stand for a mapping of Mx đ M2 đ ã ã • ® Mp

into R ® R ® • • • ® R; indeed in (8.1.12) both interpretations occur in the
same formula. The reader will therefore need to examine the context of each
occurrence to discover the appropriate meaning.

It will be seen shortly that when we restrict our attention to free modules
with finite bases the homomorphisms (8.1.7) and (8.1.11) become
isomorphisms. To simplify the language used to describe these matters, we
shall employ the term finite free module to describe a free module of finite
rank.

Let M be such an K-module and let bx, fc2,..., bs be one of its bases. Then

for each integer i satisfying 1 <f < s there is a unique ft in M* such that

It is easily seen that fx, /2, . . . , fs is a base of M* and therefore M* is a finite

free module of the same rank as M itself. The particular base (/1? /2, . . . , fs)

is said to be dual to the base (bl9 fe2,..., bs) of M.

Still supposing that M is a finite free module, let 6: M-+M** be the
homomorphism (8.1.7). Then, using the same notation as in the last
paragraph, we find that O(bi)(fj) = fj(bi) has the value 1 if i=j and is zero

otherwise. Thus d{bx), 6(b2),..., 6{bs) is the base of M** that is dual to the

base fl9f29...,fa of M*.

Lemma 1. Let Mbea finite free R-module. Then the natural homomorphism

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178 Graded duality

This is clear because we have just seen that M^>M** takes a base of M
into a base of M**. Of course, when M is a finite free module, Lemma 1 may
be used to identify M** with M.

Now suppose that Ml 5 M2, . . . , Mp are all of them finite free modules, let

Bt be a base of Mi9 and let Bf be the base of Mf that is dual to Bt. In what

follows bt denotes an element of B{ and fj an element of Bf.

By Theorem 3 of Chapter 1, the products bx đ fo2 đ ã ã • ® fopform a base

of Ml ® M 2 ® • • • ® Mp whereas the products / / ® /2 ® * *' ® /p

constitute a base of Mf đ MJ đ ã ã • ® M*. If now

co: Mf ® Mf ® • • • ® M* -> (M1 ® M2 ® • • • ® Mp)*

is the homomorphism (8.1.11), then, by (8.1.13),
*' * đ fp) (bi đ b2 đ ã ã ã ® bp)

But this is zero unless

in which case it has the value 1. It follows that {co(f{ ® f2 ® • • • ® fp)} is the

base of (Mx ® M2 đ ã ã ã đ Mp)* that is dual to the base {bx đ fo2 đ ã ã ã

đ fop} of Mx đ M2 đ * ã ã đ Mp. Accordingly co maps a base of Mf ®

Mf ® • • • ® M* into a base of (Mx ® M2 ® • • • ® Mp)* and we have

proved

Lemma 2. Let Ml 5 M2, . . . , Mp foe y?m£e /ree R-modules. Then the natural

homomorphism

Mf đ Mf đ ã ã ã đ M* -> (Mi đ M2 đ ã ã ã đ Mp)*

{see (8.1.11)) is an isomorphism.

This lemma shows how, when M1, M2, . . . ,MP are finite free

mod-ules, it is possible to identify M f ® M f ® ã ã ã đ M * with
( M1đ M2đ - - - ® Mp) * .

8.2 The graded dual of a graded module

Finite free modules are, of course, very special. More general than
these are graded modules with finite free components (the components of a
graded module are the submodules which make up its grading), and the aim
of this section is to extend our results to this wider class. However, we shall
begin by considering arbitrary graded modules.

Let M be an ^-module and {Mn}neZ a grading on M. Then M* =

HomR (Mn, R) can be regarded as a submodule of M* = HomR (M, R); more

precisely, if we fix n, then the linear forms on M that vanish on Y*k*n Mk

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The graded dual of a graded module 179

will be embedded in M* by means of this isomorphism. The reader will
recall that we have already met this kind of situation in Section (7.7).

This said, we can form

ZM: = Mf (8.2.1)

n

say, where the summation is carried out in M*. It is easy to check that the
sum in (8.2.1) is direct. Accordingly M+ is graded by {M*}neZ.

Definition. The graded module M+ is called the 'graded dual' of the graded
module M.

Now suppose that M and N are graded /^-modules with {Mn}neZ and

{Nn}neZ as their respective gradings; further, let

X:M-^N (8.2.2)

be a degree-preserving ^-linear mapping, that is to say X is a 
homo-morphism of graded modules. By (8.1.2), A* maps N* into M* and it is easy
to see that X*(Nk¥)^Mk¥ for all k. Accordingly there is induced a

homomorphism

^ A T —M+ (8.2.3)

of graded modules; moreover if Xk: Mk^>Nk is the homomorphism in

degree k induced by X, then the homomorphism AT£ —• Mjf induced by Xf is
X£. Evidently

(idM)+ = idMt (8.2.4)

and if \i\ N-+K is another homomorphism of graded modules, then

(fi°Xy = kf°fi\ (8.2.5)

Thus the formation of graded duals is a contravariant functor from the
category of graded fl-modules (and their degree-preserving
homo-mo rphisms) to itself. Naturally if X: M—>N is an isohomo-morphism of graded
modules, then so too is Xf: AT+^M+ and

(^)-l=(X-l)\ (8.2.6)

We next consider the homomorphism M—>M** of (8.1.7). This
associates with each element of M a linear form on M* and hence, by
restriction, a linear form on M+. Thus we arrive at a homomorphism
M—>(Mf)* and now an easy verification shows that the image of M
belongs to (M+)+. Accordingly we have a homomorphism

Af-»M+ t (8.2.7)

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180 Graded duality

homomorphism of graded modules, then

M >M++

t (8.2.8)

N > Nn

is a commutative diagram.

Lemma 3. Let M be a graded R-module with finite free components. Then

the homomorphism M —• M++ of (8.2.7) is an isomorphism of graded modules.

This follows from Lemma 1 as soon as we consider components.

Now suppose that M(1), M( 2 ), . . . , M(p) are graded K-modules. We can

turn M( 1 )® M( 2 )® - - - ® M( p ) and M(1)+ đ M( 2 ) + đã ã ã ®M( p ) + into

graded modules by giving them the usual total gradings. Let /1e M( 1 ) t,

f2 e M(2)+ and so on. Then /4f} ° (/i ® fi ® * * * ® fP\ where /^p) denotes

p-fold multiplication, is a linear form on M( 1 ) ® M(2) ® • • • ® M(p) and

among the submodules Af jf* đ Mj2) đ ã ã ã đ M\p\ of M( 1 ) đ ã ã ã đ M(p),

there are only finitely many on which jijf} ° (/x ® /2 ® • • • ® fp) does not

vanish. Consequently the linear form belongs to (M(1) ® M(2) ® • • •

® M(p))+. It follows that there is an K-linear mapping

ã ã ã đ M(p))+ (8.2.9)

in which /x đ f2 đ ã ã * đ fp (considered as an element of the left-hand side)

is mapped into /#> (fx đ f2 đ ã ã ã đ /p). But M( 1 ) t đ M(2)+ đ ã ã • ® M(p)+

is the direct sum of its submodules M^* ® M[2)* ® • • • ® M[P)*, and

(Mj^ ® M|2) đ ã ã ã đ M|p))* is a submodule of the component of degree

*'i+1''2 + - ' -2+ ;p of ( A /( 1 )® M( 2 )® - - - ® M( P ))+. Furthermore (8.2.9)

induces the homomorphism

ã ã ã đ M\P))* (8.2.10)

which operates exactly as in (8.1.11). In particular we see that (8.2.9) is a
homomorphism of graded modules.

Lemma 4. Let M( 1 ), M( 2 ), . . . ,M( P ) be graded modules with finite free
components, and let all the gradings be non-negative. Then the homomorphism

Md ) t (g M(2)t g ) . . . g) M( p ) +- > (M(1) đ M(2) đ ã ã ã đ Mip)Y

(see (8.2.9)) is an isomorphism of graded modules.

Proof. Since the components are finite free modules all the

homo-morphisms (8.2.10) are isohomo-morphisms; and, because the gradings are

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The graded dual of a graded module 181

direct sum of its submodules

(M£> ® M<2) ® • • • ® M\P))* (i! + i2 + • • • + ip = fc)

(see (8.1.9)). The lemma follows.

We next consider some important situations involving homomorphisms
like (8.2.9). First suppose that we have homomorphisms

^:M('> — JV<'> (i = l , 2 p)

of graded modules. It is then easily verified that

đ JV(2)+ đ ã ãã đ N(p)f > (N(1) đ iV(2) đ ã ãã đ Nip)Y

Md ) t (g M(2)t (g . . . 0 M(P)t > (M(l) g) M(2) 0 .

(8.2.11)

is a commutative diagram. Of course, it is understood that the horizontal
mappings are derived from (8.2.9).

By Lemma 4, if all the modules have non-negative gradings and finite free
components, then the horizontal mappings in (8.2.11) are isomorphisms.
Hence in these circumstances we may make the identifications

AT(1)+ ® Ni2)f đ ã ã ã đ Nip)f = (N{1) đ JV(2) đ ã ã ã đ Nip)Y
and

Md ) t ^ M(2)t 0 . . . 0 Mip)f = (M(1) ® M(2) ® • • • ® M(p))\

The commutative property of the diagram then ensures that

4+i đ A2 đ ã • • ® A^ = (Ai ® A2 ® • • • ® /lp)f. (8.2.12)

Another situation which will concern us arises in the following way. Let
M(1), M(2), . . . , M{p) and K(1), X(2), . . . , K{q) be graded /^-modules. The
homomorphisms

Md ) t ^ M(2)t 0 . . . 0 M( p ) +-> (M(1) đ M(2) đ ã ã ã đ M(p))+

and

Ê(Dt ^ X(2)t 0 . . . 0 X ^t- > (X^> 0 X(2) ® ' ' ' ® K(9))t

induce a homomorphism

(Md)t 0 . . . 0 M( p ) + ) đ (X(1)t đ ã ã ã đ X(^)+)

- ã ( M( 1 ) đ ã ã • ® Mip)y ® ( x( i ) ® • • • ® Kiq)y

of graded modules. But, again from (8.2.9), we have a homomorphism

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182 Graded duality

® • • • ® Mip)f)

(8.2.13)
This too is a homomorphism of graded modules.

Now consider the diagram

® Kiq))f

(8.2.14)
(Here the upper horizontal mapping comes from (8.2.9) and the lower one
from (8.2.13). The left vertical isomorphism comes from Theorem 1 of
Chapter 2 whereas the one on the right-hand side is the graded dual of the
isomorphism

ã ã ã đ Mip)) đ (X(1) đ ã ã • ® Kiq))

• • • ® Mip) ® X(1) ® • • • ® Kiq)

provided by the same result.) It is a straightforward matter to check that
(8.2.14) is commutative. We shall describe this fact by saying that the taking
of graded duals is compatible with the associative property of tensor products
as embodied in Theorem 1 of Chapter 2. Note that all the mappings
preserve degrees and that when the gradings are non-negative and all
components are finite free modules, the horizontal homomorphisms are
actually isomorphisms.

There is one more result connected with (8.2.9) that it is useful to place on
record. Let M(1), M( 2 ), . . . , Mip) be graded modules and let A(M(1), M(2),
. . . , M(p)) denote the isomorphism

(M(1) ® • • • ® Mip)) ® (M(1) ® • • • ® Mip))

(Mip) ® Mip)) (8.2.15)

of graded modules in which

(mti đ mh đ ã ã ã đ mip) đ (m'h đ m'h đ ã ã ã đ m'jp)

is mapped into (mti đ m'h) đ (ml2 đ m'h) đ ã • • ® (m; ® m)). (The notation

is the normal one, i.e. mix and m'h belong to Mj^ and MJ*} respectively and

so on.) The isomorphism

(Mip) ® Mip))

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The graded dual of a graded module 183

that is inverse to (8.2.15) will be denoted by V(M{1\ M( 2 ), . . . ,
It is now a simple matter to verify that the diagram

) đ ã ã ã đ (Mip)f

ã • • ® Mip))

(8.2.17)
is commutative. (Of course, the horizontal mappings are obtained by using
homomorphisms of the kind typified by (8.2.9).) Once again, all the
mappings preserve degrees.

Should it happen that M(1), M( 2 ), . . . , M{p) have non-negative gradings
and finite free components, tfren the horizontal mappings in (8.2.17) will be
isomorphisms so that in each case we can identify the domain with the
codomain. If this is done we find that

K(M(1)t, M( 2 ) +, . . . , Mip)f) = A(M(1), M( 2 ), . . . , Mip))f (8.2.18)
and now, by taking inverses, we can add the relation

A(M(1)+, M( 2 ) +, . . . , Mip)f)= F(M(1), M( 2 ), . . . , Mip))\ (8.2.19)
The isomorphisms (8.2.15) and (8.2.16) may be modified by introducing
signs just as we did in (7.1.3) and (7.5.6). Let A(M( 1 ),... ,M(P)) and
F ( M( 1 ), . . . , M(p)) be the results of making these changes. We then find that
when M(1), M(2), . . . , Mip) have non-negative gradings and finite free
components we can replace (8.2.18) and (8.2.19) by

V(M{1)\ M( 2 ) +, . . . , Mip)f) = A(M(1), M( 2 ), . . . , Mip))f (8.2.20)
and

A(M(1)+, M( 2 ) +, . . . , M(p)i)= F(M(1), M( 2 ), . . . , M(p))+ (8.2.21)
respectively.

Our final comments in this section involve much simpler situations.
Evidently if the /^-module K is graded trivially, then Kf is just K* with the

trivial grading. In particular

Rf = R**R, (8.2.22)

where the isomorphism R*zzR comes from (8.1.6). We shall use (8.2.22) to
identify Rf and R as graded modules.

Now consider the homomorphisms

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184 Graded duality

trivially graded module. Then

/ijf)+: £+- > (R đ R đ ã ã ã đ R)f = Rf đ Rf đ ã ã
and

Ajf> +: Rf đ K+ đ ã ã ã đ Rf = {R đ R đ ã ã ã đ R)f - ã R\

so that the identification Rf — R turns /4f)+ into a mapping R—>R®
K ® - - - ® R and it turns Ajf)+ into a similar mapping in the reverse
direction. On this understanding we find that

A«jr = Aj? (8.2.23)

and

^ . (8.2.24)

8.3 Graded duals of algebras and coalgebras

In Section (7.7) we considered an algebra that was formed by linear

forms on a coalgebra, and now that we are about to discuss the graded duals
of algebras and coalgebras, it is convenient to point out the connections
between the present discussion and the ideas that were developed earlier. It
is for this reason that we begin with coalgebras.

Suppose then that (A, A, e) is a non-negatively graded coalgebra whose
components are finite free modules. Then A+: (A ® A)f—>Af and
ef: JR+—>y4f. But, by Lemma 4, (A ® A)f can be identified with Af ® A^ and,
by (8.2.22), Rf can be identified with R. In this way we arrive at 
degree-preserving K-linear mappings

A1: ^ ® ^1" - ^1" (8.3.1)
and

e+:R —X+. (8.3.2)

Theorem 1. Let (A, A, e) be a graded R-coalgebra with a non-negative

grading and finite free components. Then (A\ A+, a+) is a graded R-algebra.
Proof. Because (A\ A\ ef) will turn out to be a special case of a graded
algebra that was encountered in the last chapter, we shall use an argument
that exhibits this relationship rather than one which is based directly on the
definition.

Let f,geA\ By (8.2.9), when A^®Af is identified with (A®AY the
element / ® g of the former becomes fiR°(f ® g), where fiR denotes the

multiplication mapping of R. Thus for Af: Af ® Af—>Af we have
A+( / ® 0 ) = / i *o( / ® 0 ) ° A . (8.3.3)
On the other hand

e+(l) = id^oe = e. (8.3.4)

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Graded duals of algebras and coalgebras 185

natural structure as a graded /^-algebra and now (8.3.3) and (8.3.4) show
that Af and e+ are the multiplication and unit mappings of the previously
encountered algebra. This proves the theorem. It should be observed,
however, that Theorem 8 of Chapter 7 was proved under much more
general conditions.

Definition. With the above notation the graded algebra (A\ A+, ef) is called
the 'graded duaV of the graded coalgebra (A, A, e).

Note that the graded dual also has a non-negative grading and finite free
components.

Theorem 2. Let A and B be graded coalgebras with non-negative gradings

and finite free components. Further, let X: A-+B be a homomorphism of
graded coalgebras. Then 1+: Bf—> Af is a homomorphism of graded algebras.

This follows at once from Exercise 6 of Chapter 7. However, a direct
proof can be obtained by adapting the arguments, given below, to establish
the dual result (Theorem 4).

We turn now to algebras. Suppose that (A, \i, rj) is a non-negatively
graded algebra with finite free components. Then /x+ maps A+ into (A ® A)f
and *7+ maps Af into R\ Consequently if we make the identifications
(A ®A)f = Af®Af and Rf = R, then we obtain K-linear mappings

fS:Af->A+®Af (8.3.5)
and

nf:A'-^R (8.3.6)

which preserve the degrees of homogeneous elements.

Theorem 3. Let (A, jj,,n) be a graded R-algebra, where the grading is

non-negative and has finite free components. Then (A\ / / , nf) is a graded 
R-coalgebra.

Proof. Since \i ° {\i ® A) = pt ° {A ® fi) we have

(n°(n®A)y = (ti°(A®ii))f. (8.3.7)

Now, to be quite precise, \i ° {\i ® A) is the result of combining the
homomorphisms

A® A® A - ^ - > {A® A)® A > Ađ A ã A

so that (fi (/i đ A))f is the total mapping

A* > (A đ AY ã ((A đA)đ AY

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186 Graded duality

Let us make the identifications (A ®AY = Af ® Af and
((A ® A) ® A)+= (A ® AY ®Af = (Af®Af)® A\

Then the first mapping in (8.3.8) becomes the mapping (8.3.5) and, by
(8.2.12), the second one is turned into / / ® A\ Accordingly (8.3.8) becomes

((A

and if next we identify {A ® A ® A)f with Af ®Af ® A\ then the final
mapping is the module-isomorphism (Af ® Af) ® Af-^Af ® Af ® Af (see
(8.2.14)). Thus (fi°(fi®A))\ considered as a mapping of Af into Af ®
Af ® A\ is simply (/jf ® Af) ° /z+; likewise

{li-{A ® /x))f: A*^>A* ®A^®Af
is none other than (Af ® /x+) ° //. Accordingly

so that fif: Af—>A^ ® Af is coassociative.
Next, the total mapping

A -^-+ ^ ^

is just the identity mapping of A and therefore

A' — (A ® AY ~ (R ® AY - ^ A' (8.3.9)

is the identity mapping of A\ If now we put {A ® AY = A^ ® Af and
(R®AY = R*® Af = R®A*

we find that (8.3.9) becomes

{R®AY -^A\

Consider the final mapping in this sequence. If / e A\ then 1 ® /
(regarded as a member of (R ® AY) maps r®a into rf(a), hence the image
of 1 ® / in Af is just / itself. Accordingly the total mapping

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Graded duals of algebras and coalgebras 187

Definition. The graded coalgebra (A\fi\rjf) of Theorem 3 is called the
'graded duaV of the graded algebra (A, fi, n).

Naturally the grading on the graded dual is non-negative and the
components are finite free modules.

Theorem 4. Let A and B be graded R-algebras with non-negative gradings

and finite free components, and let X: A-+B be a homomorphism of graded
algebras. Then A+: J5+—+Af is a homomorphism of graded coalgebras.
Proof If we apply graded duality to the commutative diagram

PA

A®A >A

B®B >B

the result is a new commutative diagram; and if we then make the
identifications (A ® A)f = Af ® Af and (B ® B)+ = B+ ® B\ the new 
diag-ram becomes

Af

B+ ® Bf < £+

where the horizontal mappings are comultiplications. Thus Xf is compatible
with comultiplication. A similar exercise carried out on the commutative
diagram

"B

shows that 2+ also preserves counits. The proof is therefore complete.
We know that R is both a graded K-algebra and a graded R-coalgebra;
indeed the results of this section are applicable to it. Consequently R* is also
a graded algebra and a graded coalgebra.

Lemma 5. The isomorphism Rf&Rof (8.2.22) is an isomorphism of graded
algebras and graded coalgebras.

The verification is trivial if we use (8.2.23) and (8.2.24).

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188 Graded duality

Theorem 5. Let A be a graded algebra, respectively coalgebra, with a

non-negative grading and finite free components. Then the canonical isomorphism
A-^An of graded modules (see Lemma 3) is actually an isomorphism of
graded algebras, respectively coalgebras.

Proof We shall deal only with the case where A is a coalgebra; the other
case can be dealt with similarly.

Let A and e be the comultiplication and counit mappings of A. By (8.2.8),

the diagram

A®A >(A®A)n

is commutative. If we make the identifications (A ® A)n = (Af ® Af)f =
An ® An we find that A đ A ã (A đ 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
non-negative gradings whose components are finite free modules. Then A{1) đ
A{2) đ ã ã ã ® A{p) is a graded algebra of the same kind. We propose to
compare (A(1) đ Ai2) đ ã ã ã ® Aip))f and A(1)+ ® 4( 2 ) + ® • • • ® Aip)f as
graded coalgebras.

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Graded duals of Hopf algebras 189

® Aip)) ® U( 1 ) đ ã ã ã (x) Aip))

- ^ (A(1) đ 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) đ ã ã ã đ A(p)) đ (A(1) đ • • • ® Aip)) (8.4.2)
be denoted by V{A(1\ Ai2\ . . . , A(p)) (see (8.2.16)). If now fit and rjt are the

multiplication and unit mappings of A{i\ then the multiplication mapping

P \ . . . , Aip))

and its unit mapping is

(f71®f/2®---®f/p)°Ajf).

At this point let us make the identifications

ip)

Y = A(1)+ đ A(2)+ đ ã ã ã ® Aip)\
(A{p) ® Aip))Y

® • • • ® (A(p)f ® Aip)f)

and

(04( 1 ) đ ã ã ã đ Aip)) đ {A{1) đ"-đ Aip)))f

= {Ail)f đ-"đ Aip)f) đ {A(1)f đ ã • • ® Ai p ) 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

</div>

multilinear algebra – d g northcott multilinear algebra – werner greub multilinear algebra – marvin marcus multilinear algebra and differential forms for beginners

<i><b>Multilinear algebra</b></i>

D. G. NORTHCOTT F.R.S.

CAMBRIDGE UNIVERSITY PRESS

Contents

Preface

Multilinear mappings

<b>2</b>

Some properties of tensor products

<i>in which m</i>

<i>®m</i>

<i>®' — ®m</i>

<i> is mapped into m</i>

<i>®--®m</i>

<i>®n</i>

<i></i>

<i>Let £ eM</i>

<i> ® M</i>

<i> ® • • • ® M</i>

<i> and rjeN</i>

<i>®N</i>

<i>®--®N</i>

<i>. We know</i>

<i>m</i>

<i>®m</i>

<i>®'"®m</i>

<i>. Suppose that</i>

<i><b>n = Y</b></i>

<i><b>n</b></i>

<b> (2.3.1)</b>

<i><b>= Z ^ . ^ M ^ . - ^ M ^ (d.s.),</b></i>

<i><b>(M, ®</b></i>

<i><b> S) ®</b></i>

<b> (M</b>

<i><b> đ</b></i>

<i><b> S) đ</b></i>

<i><b> - - ã đ</b></i>

<i><b> (M</b></i>

<i><b> đ</b></i>

<i><b> S)</b></i>

<i><b>*(M! ®</b></i>

<i><b>M</b></i>

<i><b> ®</b></i>

<i><b>-- ®</b></i>

<i><b>M</b></i>

<i><b>) ®</b></i>

<i><b>S</b></i>

<b>3</b>

Associative algebras

<b>v</b>

<i><b>?</b></i>

<i><b>"- ®</b></i>

<i><b>A\</b></i>

<i><b>K (3.3.11)</b></i>

JV(J,J)=£y.

<b>... aJ</b>

<i>(L^r^n (3.8.2)</i>

<b>sr'w®*^</b>

<b>^)^"</b>

<b>^)</b>

<i>[</i>

<b>0 r l o r'J</b>

<b> L0</b>

<i><b><sub>rr' J</sub></b></i>

<i>[</i>

<b>0 rj</b>

<i><b> [o pr\</b></i>

<b>0 Oj'</b>

<b>[;:]•</b>

<b>c:=</b>

<i><b>= 4>(x)ct)(x')</b></i>

<b>4</b>

The tensor algebra of a module

<i>[</i>

<i><b>o</b></i>

<b>o oj</b>

<b>o oj</b>

<b>Lo o j</b>

<i>A</i>

<i>(m</i>