A primer of algebraic D-modules

a homomorphism of modules. Since the
r" be the subspace of MIN

defined by

A routine calculation shows that r" is a filtration of MIN. Note that

and let 7l'k :

HI H-l ---; r[ I rf-l

be the canonical projection. Putting these together we get a K-linear map,

One easily checks that this map is a surjective homomorphism of Sn-modules. 5.1 LEMMA. Let M be an An-module witl] a filtration

r

compatible with

B. The sequence of Sn-modules

o-+ gr-r' N

4>

-+

.. grr" MIN -+ 0 grr M -+

is exact. PROOF:

Note that the kernel of the map

7l'k

defined above is

62

7. Graded and filtered modules

Thus we have an exact sequence of vector spaces

The sequence of 8 n -modules in the lemma is obtained by adding these vector space sequences for k

~

O. Hence it is exact.

The exact sequence of Lemma 5.1 is very useful. Here is a typical application. Let d be an operator in An of degree r and put M

= AniAnd.

Take l3

to be the filtration of An as a left An-module. The induced filtration in And is l3~

= Bk_rd.

Thus,

Since Bkl B k - 1 is the homogeneous component of degree k of 8 m then

By Lemma 5.1, there is an exact sequence,

G.

= K{x, y}

6.1 Let F

EXERCISES

be the free algebra in two geperators, and I the two-

sided ideal of F generated by the relation xy - )..yt. Show that the quotient ring FI I is a graded ring. 6.2 Let R

= ei2:0 Ri be a graded ring and M

Let M(k);

I = ei2:0 M; a graded R-module.

= Mi-k.

(1) Show that M(k) = EBi2:0 M(k); is a graded R-module. (2) Show that if k > 0 then the identity map M --+ M(k) is an isomorphism of R-modulcs, but. not a graded isomorphism.

7. Graded and filtered modules

63

6.3 In this exercise we define a grading for Al that is not positive. Denote by x and 0 the generators of AI. Define G" = {d E Al : [x8,

dJ = kd}.

Let

K[xo] be the polynomial ring in the operator xo. Show that (1) G"

= K[xo]x" for k ~ 0,

(2) G" = K[x8]0-" for k ~ 0, (3) Al

= IDiEZ G" is a graded ring.

6.4 Let dEAn. Define the principal symbol of d by IT(d)

= 1T,,(d),

where k

is the degree of d and IT" denotes the symbol map of degree k relative to the

Bernstein filtration. Find the principal symbol of the following operators of Aa:

+ x~; x~8i + x~xgol~ + X2X48l + xf; oJ + x~ + xtx~ + 8:2x~ + X2'

(1) otx~

(2) (3)

6.5 Let 'Tk be the symbol map of order k with respect to the order filtration of An. Let

ei =

71 (OJ). Show that g.f An is isomorphic to the polynomial

ring K[X][~1>'" , en]. 6.6 Let R be a filtered K -algebra with a filtration F. Show that if g.,J R is . a domain, then so is R. 6.7 Let J be a left ideal of An. Define the symbol ideal of J to be the ideal gr(J)

= Lk~OIT,,(JnBk) of Sn. Let

M

=

AnjJ.

(1) Let 8' be the filtration of J induced by the Bernstein filtration. Show that g."B/ J

S;:

gr(J).

(2) Show that if 8" is the filtration of M induced by the Bernstein filtration, then g."B"M ~ Snlgr(J).

6.8 The definition of a filtered An-module for the 01'de1' filtration is analogous to that given in §4, except that (4) must be replaced by: M j is a finitely generated K[Xl-module, for i 2: O. Find a filtration of the An-module K[X] with respect to t.he order filt.ration and c:aklllate its associated graded module.

64

7. Gmded and filte1'ed modules

6.9 Let Vbe the K-vector subspace of An with basis {Xl, ... , X n , (h, ... , an}. Let Sp( V) be the symplectic group on V. (1) Show that an element u E Sp{ V) can be extended to an automorphism of An that preserves the Bernstein filtration. (2) Show that this automorphism induces an automorphism of Sn

=

g"J3 An. 6.10 Let (,\b'" ,'\2n) E K2n. Show that the formulae

U(Xi) = Xi - '\; and

U( 0;)

= Oi -

,\n+i

define an automorphism of An(K) that preserves the Bernstein filtration and induces the identity automorphism on Sn. 6.11 Let d E An(C) be an element of degree 2. Show that there exist an automorphism

Ot

of An(C) and constants '\1 •. .. , '\n, TI, . •.

, Tn

such that

n

DI(d)

=L

'\i(X~

+ a;) + Ti·

I

Hint; Use Exercises 6.9 and 6.10, and the fact that quadratic forms are diagonalizable.

CHAPTER 8 NOETHERIAN RINGS AND MODULES There is very little that one can say about a general ring and its modules. In practice an interesting structure theory will result either if the ring has a topology (which is compatible with its operations), or if it has finite dimension, or some generalization thereof. As an example of the former, we have the theory of C" -algebras. The latter class includes many important rings: algebras that are finite dimensional over a field, PI rings, artinian rings and noetherian rings. It is the last ones that we now study. In particular, we prove that the Weyl algebra is a noetherian ring.

1. NOETHERIAN MODULES.

In this book we shall be concerned almost exclusively with finitely generated modules. One easily checks that a homomorphic image of a finitely generated module is finitely generated. However a finitely generated module can have a submodule that is not itself finitely generated. An example is the polynomial ring in infinitely many variables K[xJ , X2,

...

J. Taken as a module

over itself this ring is a cyclic left module: it is generated by 1. However, the ideal generated by all the variables

Xl, X2, ...

every finite set of polynomials in K [Xl, X2,

•••

cannot be finitely generated:

J uses up only finitely many of

the variables. We get around this problem with a definition. A left R-module is called noetherian if all its sub modules are finitely generated. Examples are easy

to come by: vector spaces over K are noetherian K-modules. Every ideal of the polynomial ring in one variable K[xJ is a noetherian K[xJ-module. There are several equivalent ways to define noetherianness. We chose the most natural. Here are two more. l.1 THEOREM . Let M be a left R-module. The following conditions are

equivalent:

(1) M is nueth()j-jaIJ.

8. Noetherian modules

66

(2) For every infinite ascending chain NI there exists k

~

0 such that N;

= N,.

~

N2

~

. . . of SIlbmodules of M

for every i

~

k.

(3) Every set S of submodules of M has a submodule L which is not properly contained in any submodule of S.

Condition (2) is known as the ascending chain condition. This is probably the most common way in which noetherian modules are defined. Condition (3) is the maximal condition: the submodule L is called a maximal element of S. PROOF: Suppose that (1) holds. If NI ~ N2 ~ ... is an infinite ascending

chain of submodules of M, then Q =

Uj~l

N; is a submodule of M. Thus

Q is generated by finitely many elements, say UI,' ..• 'Ut. Hence there exists UI,· . .. • Ut E N k · Thus Q = Nk = N i , for every i ~ k, as

k ~ 0 such that

claimed in (2). Assume now that (2) holds. We prove (3) by reaching a contradiction. Suppose that S does not contain a maximal element. If NI

~

N2

~

... t:::; N,.

is any chain of elements of S, then we can make it longer. Since no element of S is maximal, there must exist Nk+l in S such that Nk

c

NHI . In this

way we construct a proper infinite ascending chain of submodules of M, thus contradicting (2). Finally, assume (3) and let N be a submodule of M . Let S be the set of all finitely generated submodules of N, and let L be a maximal element of S. Suppose LeN, and choose

U

E

N\L. Thus L+ Ru is finitely generated and

contains L properly: a contradiction. Hence L

= N,

is finitely generated.

In the next proposition we collect the basic properties of noetherian modules that are used in later sections. First, a technical lemma. Let M be a left module over a ring R. Let N, PI and P2 be submodules of M such that P2 t:::; Pl. IfN+PI = N+P2 and NnPI = NnP2 ,

1.2

LEMMA.

then PI = P2. PROOF: We need only show that PI t:::; P2 • Suppose that

Ul

E Pl' Since

67

8. Noetherian modules we have that

Ul

= x + 'lIo2, for x

In particular, x E P2 . Hence

Ul

E N and ~ E

P2. Thus

= x + ~ E P2 , as required.

1.3 PROPOSITION. Let M be a left module over a ring R, and let N be a submodule of M. (1) M is noetherian if and only if MIN and N are noetherian. (2) Let N' be another submodule of M and suppose that M

= N + N'.

If N ,N' are noetllerian, then so is M . PROOF:

It is clear that a submodule of a noetherian module is itself noethe-

rian. On the other hand, a submodule of MIN is of the form LIN for some submodule L of M which contains N. If M is noetherian, then L is finitely generated. Thus LIN is finitely generated, and MIN is noetherian. Conversely, suppose that N and MIN are noetherian. Let L1

~

L2

~

...

be an infinite ascending chain of submodules of M . Then

is an ascending chain of submodules of N . Since N is noetherian, this chain must stop. In other words, there exists s such that (L. n N) all i

~

= (L

j

n N) for

s. Similarly, the chain

stops; that is there exists r such that N t = max{s , r}, we have that N +L, Thus by Lemma 1.2, L,

= L, for

To prove (2), note that MIN

+ Lr = N + L j ,

for i

~

r. Taking

= N +Lj and L,nN = Ljn N , for

i ~ t.

i ~ tHence M is a noetherian ring. ~

N'/(N'nN} is noetherian by (1). Since

N is noetherian, we may use (1) again to conclude that M is noetherian.

We have pitifully few examples of noetherian modules, but this situation will be rectified in the next section. 2 . NOETHERIAN RINGS.

We say that a ring R is a left noetherian ring if R is noetherian as a left R-module . Exnlllpl('s are fi elds and any principal ideal domain, like

68

8. Noethe1'ian modules

Z or K[x]. Actually, the most important rings in algebraic geometry are noetherian, because the ring of polynomials in finitely many variables is noetherian. This was proved by D. Hilbert in 1890 and it is the keystone of commutative algebra. It is known as Hilbert's basis theorem. We shall now prove a modernized version of Hilbert's result from which the original theorem follows as an easy consequence. 2.1 THEOREM. Let R be

8

commutative noetherian ring. The polynomial

ring R[xJ is noetherian. PROOF: Suppose that R[xJ is not noetherian. Let I be an ideal of R[x] that is not finitely generated. We shall construct, inductively, an infinite ascending chain of ideals in R; thereby achieving a contradiction. Choose

h

E I of smallest possible degree. Assume, by induction, that

h,··· ,1k

have been chosen. Let fk+1 be the polynomial of smallest possible degree in

I \

ClI, ... ,fk) .

Since I is not finitely generated, this construction produces

an infinite sequence of polynomials aj

h, h, ... E I.

Let nj be the degree and

the leading coefficient of j;. Since R is noetherian, the ascending chain of ideals

must be stationary; say (al , "" ak) = (al, " " ak+l)' Hence ak+l for some bi E R. Put

= :L; bja;,

k

9

= fk+1

-

L

bjx",,+l-n'fi'

I

Since, by construction, the degrees satisfy

nl ~

is indeed a polynomial. Note that gEl , but 9 ¢

'»-2 ~ • •. it follows that 9

(h , ... ,fd . However 9 has

smaller degree than f k+1 , a contradiction. Thus I must be finitely generated . A simple induction using Theorem 2.1 is all that is needed to prove the original basis theorem of Hilbert. 2.2 COROLLARY. Let K be a field. The polynomial ring K[Xl,." ,xnJ is IJoethel"ian.

69

8. Noetherian modules

Although every ideal of K[XI,' .. ,X n ] is finitely generated, it is not true that there is an upper bound on the number of generators of ideals in this ring. For example, it is easy to construct ideals

h of K[x!, ,1;2) which cannot

be generated by less than k elements; see Exercise 4.2. We shall now use Hilbert's basis theorem to prove that An is a left noetherian ring. This will follow from a slightly more general theorem. Recall that

B is the Bernstein filtration of An and Sn

= grI' An.

2.3 THEOREM. Let M be a left An-module with a filtration

r

with respect

to the Bernstein Jiltl'ation B. If grY M is a noethel'ian Sn-module, then M is noetherian. PROOF: Let N be a submodule of M, and r' the filtration of N induced by

rj

see Ch.7, §5. Since g"I" N ~ g"I'M and the latter is noetherian, we

conclude that g"I" N is finitely generated. Since the generators of grY' N are finite in number , they have degree ~ m, for some integer m. We wish to show that N is generated by the elements in

r:n'

Suppose that it is not, and let k be the smallest integer for which there

exists v E k

>

r~

which cannot be generated by the elements of

m. Let J-tk be the symbol map of order k of

g"I" N, there exist

ai

E An and

E

'U;

J.Lk(V) =

r' .

r:,..

Clearly

By the hypothesis on

r;, such that

L• Uk-r.(ai)J.Lr,(Uj) 1

where rj

~

m, for i = 1, 2, ... , S . Hence J

V-

L: aiUj E r k_l , 1

which, by the minimality of k, may be written as an An-linear combination of elements in

r:n . Thus vitself may be written as an An-linear combination

of elements in r:". To finish the proof we must show that a finite number of elements of are enough to generate N. But

r:"

r:,.

is a subspace of the finite dimensional

vector space rln' Hence it has a finite basis, which will be a set of generators for N as An-modnlt-.

70

8. Noetherian modules

2.4 COROLLARY. An is a left noetllel'ian I'ing. PROOF: The graded ring Sn associated to the Bernstein filtration is a polynomial ring in 2n variables by Theorem 7.3.1. Hence, by Hilbert's basis theorem, Sn is noetherian. Thus An is left noetherian by Theorem 2.3. It has already been pointed out in Ch.2 that every left ideal of An can be

generated by two elements. Therefore, there is an upper bound on the number of generators of left ideals in An , which is not the case for a polynomial ring. However, this result is very hard to prove and is beyond our means in this book. There are good reasons why one ought to rejoice that An is a left noetherian ring. For example, it follows from Corollary 2.4 and the next proposition that every finitely generated An-module is noetherian.

2.5 PROPOSITION. Let R be a left noetherian ring. Finitely generated left R-modules are noethel'ian. PROOF: Let M be a finitely generated left R-module. If M is generated by Ie elements then there exists a surjective homomorphism 4J : Rk

---t

M. Since

R is left noetherian, it follows from Proposition 1.3(2) that Rk is noetherian. Hence M is noetherian by 1.3(1). There is really nothing special about left modules in tills context, and all the results that we have stated so far hold for right modules with the obvious changes, Thus the ring An is also right noetherian, and we shall use this in later chapters, without further comment. 3. GOOD FILTRATIONS. Let M be a left An-module and

r

a filtration of M with respect to the

Bernstein filtration B. If g~ M is finitely generated, then it is noetherian by Proposition 2.5. Hence M is finitely generated, by Theorem 2.3. However , it is not always true that if M is finitely generated over An then 9~ M is finitely generated over Sn. When g~ M is finitely generated we say that is a good filtration of M.

r

It is nonetheless true that every finitely generated An-module admits a good filtration . Indeed, if !v! is gencratcd by

'ILl,"

" Us

then the filtration

8. Noetherian modules r of M defined by

n = L~ BkU;

71

is good. The graded module g~ M is

generated over Sn by the symbols of '1.£1> ••• , u•. We need an example ofa filtration that is not good. Before that, however, we shall establish an easy criterion to check whether a filtration is good. 3.1 PROPOSITION. Let M be a left An-module. A filtration

r

of M with

respect to B is good if and only if there exists ko such that ri+k = Bin, for all k ~ ko.

PROOF: Suppose that there exists ko such that The K-vector space

n. has a finite basis.

this basis generate g~ M. Thus

r iH =

BiH, for all k

r is good. '1.£11 • • • ,

be elements of M whose symbols generate g~ M. Assume that Uj E for j

Let k

~

ko .

The symbols of the elements in

Conversely, suppose that g".r M is finitely generated over Sn. Let rlcj-l,

~

u.

r", \

= 1,2, ... ,8, and that ko = max{kb ... , k.}. ko. We prove that ri+k

=

Birk by induction on i.

If i

=

0,

the result is obviously true. Suppose that the equality holds for i-I. Pick v E ri+k. Since g~ M is finitely generated, and denoting by J.Lk the symbol

of order k of r, one has

• J.Li+k(V) E LITk+i-k,(BHi-k)J.LIcj(Uj). j=1

Since B iH - ki

= Bi.Bk-kj, we conclude that

•

VEL B i .Bk - k,'I1:i

+ r iH- I .

j=l

Now Bk-k;'U.j E

n, for every j

and, by the induction hypothesis,

r k +i - 1 =

Bi-lH. Since B i - 1 ~ B i , then v E B in . Thus ri+k ~ Bin . The other inclusion is obvious; hence we have an equality, as desired. It is now easy to check whether a filtration is good or not. For example, let

n be the filtration of the left module An defined by nk = B 2k . Then we have that Bink = B i+2k is properly contained in B 2(i+lo) = niH. By Proposition 3.1, n is not a good filtration of An with respect to B. We end this section with a proposition that will allow us to compare two good filtratiuns.

8. Noetherian modules

72

3.2 PROPOSITION. Let M be a left An-module. Suppose that rand [} are two filtrations of M with respect to B.

(1) If r is good then there exists kJ such that rj ~ [}j+k•. (2) If rand [} are good then tllere exists k2 such that [}j-k, ~ rj ~ [}j+ko·

PROOF: By Proposition 3.1 there exists ko such that Birj

=

r i+j , for all

j ~ ko and i ~ O. Since rko is a finite dimensional vector space over K,

there exists kJ such that rko

~ [}k •.

Thus rj+ko

=

Bjrko is contained in

Bj.[}k. ~ [}j+k•. Hence rj ~ Ij+ko ~ [}j+kl

for all j

~

0, which proves (1). To prove (2) apply (1) twice, swapping

r

and [}. The full power of good filtrations will only be felt in the next chapter, where they will be used to define a dimension for finitely generated An-modules. 4. EXERCISES.

4.1 A K-algebra R is affine if there exist elements rJ, ... , r, E R so that the monomials

r;nl ... r;n.

form a K-basis for R.

(1) Show that if R is an affine commutative K-algebra then it is a homomorphic image of a polynomial ring in finitely many variables over

K. (2) Use (1) and Hilbert's basis theorem to conclude that a commutative affine K-algebra is noetherian. (3) Show that (2) is false if the commutative hypothesis is dropped. Hint: Construct an infinite ascending chain of left ideals for the free algebra

K{x, y}. 4.2 Let h be the ideal of the polynomial ring K[Xl,X2) generated by the monomials of degree k. In other words, the generators of h are of the form xi~ with i+j = k. Let M =

h.

(1) Show that M is a maximal ideal of K[Xl' X2], and that K[xJ, X2]/ M

K. (2) Show that h/Mh is a K-vector space of dimension k + l. (3) Show that h canllot be generated by less than k + 1 elements.

8:'

8. Noetherian modules

73

4.3 Show that the results of §3 remain true if we replace the Bernstein filtration B by the filtration by order. Hint: In some of the proofs it will be necessary to use Hilbert's basis theorem and Proposition 2.5. 4.4 Let M be a left An-module with a filtration that i

Ui E

r k., for

= 1, ... s,

r with respect to B.

Suppose

i = 1, ... ,s. Show that if grf M is generated by JJ k, (Ui), for

then M is generated by

UJ, . .. , U •.

4.5 The purpose of this exercise is to show that the converse of Exercise 4.4 is false. Let J be the left ideal of A3 generated by 01 and fh.

+ XlOa.

Recall

that by Exercise 7.6.7, if 8' is the filtration of J induced by the Bernstein filtration of A3 , then g".B'J ~ grJ. (1) Using Exercise 2.4.2, show that grJ is generated by Ul(Oi) for i

=

1,2,3. (2) Show that although 01 and fh.+XI0a generate J, their symbols do not generate grJ. 4.6 Let M be a left An-module with a good filtration annihilator of grf M is a homogeneous ideal of Sn.

r.

Show that the

CHAPTER 9 DIMENSION AND MULTIPLICITY Using the filtered and graded methods of the previous chapters we shall define a dimension for An-modules. This is a very useful invariant and it comes naturally associated with another invariant: the multiplicity. The first section contains a result in commutative algebra that is the key to the definition of dimension in §2.

1. THE HILBERT POLYNOMIAL.

Without further ado we state the main theorem of this section.

It is

expected that the reader will feel somewhat surprised by its content.

1.1

THEOREM.

Let M

=

ffii~O

M; be a finitely generated graded module

over the polynomial ring K[Xl, ... , xnJ. There exist a polynomial x( t) E Q[ ~ and a positive integer N such tllat

•

L dimK(M

i)

= Xes)

o

for every s ;::: N. The polynomial X(t) is known as the Hilbert polynomial of M. We need a technical lemma before we come to the proof of the theorem. The following definitions and notations make the proofs more bearable. A numerical polynomial is a polynomial p( t) of Q[ 4 such that p( n) E Z for all integers n »0. The difference function of a function f : C

LlJ(z)

= fez + 1) -

---t

C is defined by

J(z). Finally, set

G)

=t(t-l) ... (t-r+1)/r!

where 'r iR a positive illtogOl' and t is a variable. If r

= 0 we put

(~) = 1.

9. Dimension

75

1.2 LEMMA. The following are true:

(1) L1(~) = C~l)' (2) Let p( t) E Q[ t] be a numerical polynomial.

There exist integers

Co, . .. , Ck such that

In particular p( n) E Z, for all n E Z. (3) Let f : Z

--t

Z be a function. Suppose that there exists a numerical

polynomial q(t) E Q[t] such that L1f(n) = q(n), for all n» O. Then there exists a numerical polynomial p( t) E Q[ ~ such that f (n) = p( n),

for all n PROOF:

» O.

(1) follows from a straightforward calculation that we omit. Let

us prove (2) by induction on the degree of p( t). If p( t) has degree 0, then it equals an integer. Suppose that the result holds for all numerical polynomials of degree

~

k - 1. Let p( t) be a numerical polynomial of degree k. Since the

leading term of (~) is t r / r!, there exist rational numbers Co, •.. , Ck such that

pet)

= 2:~ Ck-iG)·

To complete the proof we must show that

Since L1p( t) has degree k - 1, it follows by induction that Thus p( t) -

Ck

Ci E

Z. By (1),

Ck-l,'" ,Co

E Z.

is integer valued. Since p( t) is a numerical polynomial, it

follows that for r» 0 both per) and per) -

Ck

are integers. Hence c/o E Z,

and we have proved (2). We now prove (3). By (2), the numerical polynomial q(t) may be written in the form

q(t) =

~ Ck-.(:) ,

with Co, ... , Ck E Z. Let k

pet)

= ~ Ck-i(i:

1)'

76

9. Dimension

By (1),

we

= q(t).

have that ..1p(t)

Hence..1(J - p)(n)

=

°for all n» O.

But this is possible only if (J - p)(n) equals a constant integer, to be called Ck+l,

for all n» 0. Thus f(n)

= p(n) + Ck+b for n» 0, as required.

PROOF OF THE THEOREM:

The proof is by induction on the number n of variables. If n

=

0, we

are reduced to the base field K. In this case the finitely generated module

M is a vector space of finite dimension. Hence it has only finitely many homogeneous components. Thus

L• dimJ( Mi = dimK M, a

for s

» 0, and we may choose X(t) = dimJ( M, a constant.

Suppose, by induction, that the theorem holds for any finitely generated graded module over K[xJ, .. . , Xn-l]' Let M

=

E9i~O

Mi be a finitely gen-

erated graded module over K[xI, ... , xn]. Define a function

f :Z

--t

Z

by 8

1(s)

=L

dimJ( Mi

-00

where Mi

= 0 if i < O.

Let ¢Ji : Mi

--t

M,+l be the linear map of vector

spaces defined by multiplication by x n' Set Qi = ker (
Now Q

= E9i~O Qi and

over K[Xl"" ,Xn].

L

= E9,~o Li

are finitely generated graded modules

They are respectively the kernel and cokernel of the

endomorphism of M defined by multiplication by X n • Moreover, the elements of Q and L are annihilated by

X n,

hence these are in fact finitely generated

graded K[Xl' ... ,xn_l]-modules. Thus, by induction, there exist polynomials Xl (t),

X2( t) E Q[ t] such that s

XI(S) =

L ()

•

dimJ( Qi and X2(S) = Ldiml( L j , 0

9. Dimension for s

77

» o.

On the other hand, the dimensions of the vector spaces in (1.3) are related by the formula dimK

Qi-l -

Adding these up for 0

:s;

dimK M i - 1 + dimK M; - dim]( Li i:S; sand s

Xl(S -1)

»

= O.

0, one obtains

+ <1f(s) -

X2(8)

= o.

Hence <1f (8) is a numerical polynomial for s »0. We may now conclude from Lemma l.2(3) that f(s) itself is a numerical polynomial for all s

»

0,

as required.

2. DIMENSION AND MULTIPLICITY. We are ready to define the dimension of a module over An. Note that the Bernstein filtration B is essential for the construction that follows. Let M be a finitely generated left An-module. Suppose that

r

is a good fil-

tration of M with respect to the Bernstein filtration B. Denote by X( t, r, M) the Hilbert polynomial of the graded module grI'M over the polynomial ring Sn. By Theorem 1.1, we have, for

t» 0,

t

X(t,r,M) = EdimK(r;jr;-l) o

(2.1)

= dim](rt ).

The last equality in (2.1) follows from the fact that dimK is additive over exact sequences of vector spaces. The dimension d(M) of M is the degree of X( t, r, M). Let ad(M) be the leading coefficient of X(t,

r, M).

The multiplicity of Mis m(M)

Both numbers are non-negative integers.

= d!ad(M)'

This is obvious for d(M), and

follows from Lemma 1.2(2) for m(M). The definitions of dimension and multiplicity apparently depend on the good filtration

r from which the Hilbert polynomial is calculated. Let us

show that any choice of good filtration will give the same result. Suppose that rand

n

are two good filtrations of M. By Proposition 8.:3.2, there exists

78

9. Dimension

Ie, such that

dimK

[}j+k.

{}j-k ~

rj

~ {}j+k.

In particular, dimK

We conclude from Theorem 1.1 that for :f ~

x(;i - k, [},M)

xC,;", r,M)

Since the behaviour of a polynomial at

00

~

[}j-k

»

< dimK Fj <

0

xC;' + k, [}, M).

is determined by its leading term,

it follows that X( t, [}, M) and X( t, r, M) have the same degrees and leading coefficients. Hence d(M) and m(M) are independent of the choice of good filtration for M. Let us consider a few examples. First let M be the left An-module An. The Bernstein filtration B is a good filtration of M and it is possible to calculate X(t,B,M) explicitly in this case. By (2.1) we must determine the dimension of B t . But the monomials x a Ef1 with

lal + 1.81

~ t form a basis

of B t as a K -vector space. So it is enough to count the elements of this basis. To do this we must count the non-negative solutions of the equation

+ ... + an + f31 + ... + f3n

t. It is an easy exercise in combinatorics to show that there are c~~n) such solutions. Hence X( t, B, M) = (~~n). As a polynomial in t it has degree 2n and leading coefficient 1/(2n)!. Thus d(An) = 2n and meAn) = 1. al

~

Another An-module that we know very well is K[X] eh. 7 we defined a filtration polynomials of degree

~ t.

r

of K[X] such that

= K[Xl,'"

rt

is the space of all

Since Bi contains the polynomials in

of degree i, we have that Birt

,x n]. In

Xl, ... ,

Xn

= r t+j. Hence r is a good filtration. It is easy

to show that dimK r t = (n~t). Thus

xC t, r, M) =

(n~t) is a polynomial of

degree nand leading coefficient lin!. Hence d(K[X])

= nand m(K[X]) = 1.

A large class of examples is obtained by twisting a module by an automorphism, as in Ch. 5, §2. Curiously, this will lead us to a different definition for the dimension. Recall that if u is an automorphism of An and M is a left

An-module, then the action of a E An 011

U

EM" is defined by a. u = u(a)u.

We begin with the Fourier transform of a module. In this case the automorphism :F is defined by :F(x;)

for 1

~

:F(B,)

= OJ and

F(o;)

=

-Xj,

i ~ n. This automorphism :F preserves the Bernstein filtration; thus

= B i . This is very convenicut.

9. Dimension

79

2.2 PROPOSITION. Let M be a finitely generated left An-module. Then M

and M:r have the same dimension and multiplicity. PROOF: Write

= BkrO

M. Then H by

ro for a K-vector space whose basis is a set of generators of is a good filtration for M. Now M:r is also generated

Fa, so we may construct a good filtration for M:r by setting [h = Bk • roo

Since :F preserves Bk, we have that ilk

= rio.

Hence M:r and M have the

same Hilbert polynomial, therefore also the same dimension and multiplicity. Things are a lot more complicated if the automorphism does not preserve the filtration. To get around the problem we must give a different definition for the dimension.

Start by choosing a finite number of elements which

generate An as a K-algebra and let Vbe the K-vector space generated by these elements and by 1. Put

Ua = K and Uk = V'. This is a filtration of An which satisfies dimK Uk < then Uk

= Bk

00.

Note that if V

= BJ,

is the Bernstein filtration of An.

Now let M be a finitely generated left An-module, with a good filtration

r

with respect to the Bernstein filtration. Without loss of generality we may assume that r k = BkrO, for k

,sCM, V)

~

O. Put ilk

= inf{v : dimK [l.

= UkrO ~

and

s" for s

»

O}.

2.3 PROPOSITION. Let V be a vector SIJace whose basis is a finite set of

generators fol' An. Then ,s(M, V)

= d(M).

PROOF: Since V is finite dimensional, there exists r E N such that V <; B r . Thus Uk <; Brio. Furthermore,

Hence dimK ilk ~ dim]( rrk.

In particular, 6(M, V) ~ 6(M, B J ).

The

= 6(M, BJ)'

But

opposite inequality is proved similarly. Thus 6(M, V)

dimJ( r. = Xes, M, r) for s » O. Since d(M) is the degree of the polynomial Xes, M, r), then SCM, B 1 ) = d(M), and the proof is complete. The result

w(~

want is a simple corollary of this proposition.

9. Dimension

80

2.4 COROLLARY. Let M be a finitely generated left An-module and u an

automorphism of An, then d(M,,) PROOF: Let V

= u(BJ ).

= d(M).

Since u is an automorphism, we have that V has

generators of An for its basis. Define Uk as above. Let

r be a good filtration

of M with respect to the Bernstein filtration, and assume that r k

= Bkra.

Now let

This a good filtration for M", hence by Proposition 2.3, d(M,,) = c(M", BJ)' But ilk = U(Bk)ra. Since U(Bk) = Uk, we have that ilk = Ukra and so c(M", Bd = c(M, V). By Proposition 2.3, the latter equals d(M). In ring theory 6(M, V) is called the Gelfand-Kirillov dimension of a module. It is discussed in detail, including many applications, in [Krause and Lenagan 85J and [McConnell and Robson 87, Ch. 8]. 3.

BASIC PROPERTIES.

Let M be a finitely generated left An-module and

ra good filtration of

M with respect to B. Let N be a submodule of M. Denote by r' and the filtrations induced by

r in N

r"

and MIN, respectively; see Ch. 7, §5. It

follows from Lemma 7.5.1 that we have an exact sequence of Sn-modules, namely

o - t grr'N - t grrM - t grr"MIN - t O. Since

r

is good,

gr M

is finitely generated. But Sn is a noetherian ring.

Hence gr' Nand grl! (MIN) are also finitely generated. Therefore both r' and

r"

are good filtrations.

On the other hand, since the sequence of vector spaces

is exact, we have that

Summing these terms for k = 0, 1, ... , s and assuming that s (3.1)

xCs, r', N) + xC·~,r", MIN) = xes, r, M).

»

0 one obtains

81

9. Dimension

3.2 THEOREM. Let M be a finitely generated left An-module and N a submodule of M. (1) d(M)

= max{d(N),d(MjN)}.

(2) If d(N)

= d(MjN)

then m(M)

= meN) + m(MjN).

PROOF: Since M is finitely generated it admits a good filtration

r.

Let r'

and r" be the good filtrations of Nand MjN induced by r. Thus (3.1) holds for an infinite number of values of s. Hence we have an equality of polynomials:

x(t, r', N)

+ X(t,r", MjN) = X(t, r, M).

Since d(M) corresponds to the degree of X( t, r, M), we have that d(M)

~

max{d(N), d(MjN)}.

But the leading coefficients of these polynomials are positive. Thus we must have equality in the formula above, and (1) is proved. Now, if d(MjN)

=

deN) then all these polynomials have the same de-

gree. Thus the leading term of x( t, r, M) is the sum of the leading terms of X(t,r',N) and X(t,r",MjN), and (2) follows immediately.

This theorem is useful in calculating the dimension of some modules. We saw in §2 that deAn)

= 2n.

We may use the theorem to calculate the dimen- .

sian and multiplicity of a free module of finite rank r: it has dimension 2n and multiplicity r. This follows from the following result. 3.3 COROLLARY. Let MJ. .. . , Mk be finitely generated left An-modules, and M = Ml EEl ..• EEl M k . (1) d(M) = max{d(M\) , ... , d(Mk)}.

(2) If d(M) = d(M;) for 1 ~ i ~ k, then m(M) = I:~ m(Mi ). PROOF: The prooffollows by induction if we apply Theorem 3.2 to the exact sequence

We may also usc the theorem to get an upper bound on the dimension of a finitely generated An-module.

9. Dimension

82

3.4 COROLLARY. Let M be a finitely generated An-module. Then d(M)

~

2n. PROOF:

Suppose that M is generated by r elements.

a surjective homomorphism
A~ ~

= max{ d(M), d(ker
that d(A~)

Then there exists

M. It follows from the theorem

Since d(A~)

= 2n by

Corollary 3.3, we

conclude that d(M) ~ 2n. This upper bound may be sharpened if the module is a quotient of An by a left ideal. 3.5 COROLLARY. Let I be a non-zero left ideal of An. Then d(Anl I)

~

2n-1. PROOF:

I

First consider the case of a cyclic left ideal. Let dEAn, and put

= And.

Then we have an exact sequence

where the map () is defined by 8(a) contradiction, that d(AnIAnd)

Since m(An)

= ad, for every a

E

An. Suppose, by

= 2n. Then by Theorem 3.2(2), we have that

= 1 and the multiplicity is a positive number, this equation is

impossible. Hence d(Anl And)

~

2n - l.

Now fur the general case. Let I be a non-zero left ideal of An and choose

o :f dEl.

Since And £; I, we have that Ani I is a quotient of Ani And. Since

the latter has dimension

~

2n - 1, so does Ani I, by Theorem 3.2(1).

4. BERNSTEIN'S INEQUALITY.

At the end of the previous section we obtained an upper bound on the dimension of a finitely generated An-module. In this section we establish a lower bound. We begin with a lemma from linear algebra. 4.1 LEMMA. Let M be a finitely generated left An-module wit}l filtration

with respect to B. Suppose that

ra :f O.

The K -linear transformation

r

9. Dimension

83

which maps a E B j to the linear transformation <Pa( u) = au is injective. PROOF:

The statement of the lemma is equivalent to ari

o =f a E Bj.

We will prove this by induction on i. If i

the statement above is equivalent to

ra =f O.

Suppose that if 0 =f bE Bi-l then br;_l of B j • If arj = 0, then a

~

monomial. Then

OjCX

then Bo

=K

and

This is true by hypothesis.

=f O. Let a be a non-zero element

K. Hence the canonical form of a has a term

cx o Ef3 with c E K \ {O} and O

= 0,

=f 0 whenever

101 + 1.81 >

O. Suppose that

OJ

=f 0 for this

-£iEf3 is a summand in the canonical form of [a, a;].

Thus [a, a;] is a non-zero element of B j -

[a, 8i ]r.- 1

~

1•

Since ari = 0, we conclude that

aair.- 1 .

= 0, which contradicts the induction hypothesis. We may reach a similar contradiction by assuming that .8; =f o.

But O;ri - 1

~

r i , hence [a, a;jr;_l

Thus the lemma is proved. 4.2 THEOREM.

If M is a finitely generated non-zero left An-module, then

d(M)? n. PROOF:

Choose a set of generators for M and let

r

be the good filtration

obtained by giving each of these generators degree zero, see Ch. 8, §3. Then

ra =f O.

Let x( t)

= x( t, r, M) be the corresponding Hilbert polynomial.

By the lemma, B j can be embedded in HomK(rj , r2i). In particular,

But HomK (r;, r 2i ) has dimension dimK F; . dimK

Hi.

Thus, assuming that

i» 0, we get that dimK Bi ~ X(i)x(2i).

On the other hand, dimJ( B; = c~~n) is a polynomial in i of degree 2n. Hence, as a polynomial in i, X( i)X(2i) must have degree? 2n. But the degree of X(i)X(2t) is 2d(M). Thus d(M) ? n, as required. This inequality was first proved by I. N.Bernstein in [Bernstein 72j; and is often called the Bernstein inequality. The proof above is due to A. Joseph. We have already seen that both bounds for the dimension are attained. For example, d(An)

=

211, and d(K[X]) = n. In fact. there exist An-modules of

9. Dimension

84

dimension k for every k in the interval n to 2n; see Exercise 5.3. The modules of minimal dimension are so important that Ch. 10 will be devoted to them.

5.

EXERCISES

5.1 Show that if M is a finitely generated torsion An-module, then d(M)

:s;

2n-1.

5.2 Show that if d is a non-zero operator of An then d(AnIAnd)

= 2n-1.

Hint: Let k be the degree of d. There is an exact sequence,

where 8" is the filtration of Ani And induced by B. Using this sequence show that the Hilbert polynomial of Ani And is (2~~t)

-

C~~~k), and that it has

degree exactly 2n - 1. 5.3 Let k be a positive integer such that 1 :s; k subalgebra of An generated by

Xl, ... ,X n

:s;

n. Denote by Rk the

and 01, ... , Ok. Let J

= AnOk+l +

... + AnOn. (1) Show that the complex vector space Bi+JIJ is isomorphic to BinRk. (2) Show that the Hilbert polynomial of Ani J with respect to the filtration induced by B is

(n:tt).

(3) Show that Ani J has dimension n + k. What is its multiplicity? 5.4 Let D be a (noncommutative) domain. We say that D satisfies the Ore condition if Dan Db =I 0, for any two non-zero elements a, bED. A division ring Q is a left quotient 1'ing of D if (1) D is a subring of Q;

(2) every non-zero element of D is invertible in Q; (3) every element of Q is of the form b-ia where a, bED and b =I O. Show that if a domain has a left quotient ring then it satisfies the Ore

COll-

dition. The converse is also true, but it is harder to prove. For details see [Cohn 79, Theorem 12.1.2].

9. Dimension

85

5.5 Show that An satisfies the Ore condition. Hint: Let a, b be non-zero elements of An. If Ana n Anb

= 0,

then An maps

injectively to An/Ana ffi An/ Anb. This contradicts Corollary 3.5. 5.6 Let dE An(C) be an operator of degree 2. Set M

= An/And.

(1) Show that there exists a submodule N of M such that d(M/N) (2) Show that M cannot be an irreducible module if n

~

= n.

2.

Hint: By Exercise 7.6.11, there exists an automorphism a of An(C) such that a(d)

=

2:=~ d; where di

=

~i(X~

+ af) + Ti

for ~i'

Ti E

C. Let J be

the left ideal generated by d1 , ... ,dn. Then An/ J is a homomorphic image of Ma-l

~

An/And and d(An/ J)

of An/J see Exercise 13.5.8.

= n.

For an easy method to find the dimension

CHAPTER 10 HOLONOMIC MODULES

The most important An-modules are the holonomic modules, also known among PDE theorists as maximally overdetermined systems. An An-module is holonomic if it has dimension n.

Ordinary differential equations with

polynomial coefficients correspond to holonomic modules. In this chapter we begin the study ofholonomic modules, which will be one of the central topics of the second half of the book.

1.

DEFINITION AND EXAMPLES.

A finitely generated left An-module is holonomic if it is zero, or if it has dimension n. Recall that by Bernstein's inequality this is the minimal possible dimension for a non-zero An-module. We already know an example of a holonomic An-module, viz. K[X]

= K[XI, .. . , x n ]. We also know that

An itself is not a holonomic module: it has dimension 2n. It is easy to construct holonomic modules if n ideal of AI' By Corollary 9.3.5, d(AdI) inequality, deAl I I)

= 1.

~ 1.

If I

= 1.

Let I

=I 0 be a left

=I AI, then, by Bernstein's

Hence AdI is a holonomic AI-module.

This is wonderful source of examples, which will pour forth with the help of the next proposition. 1.1 PROPOSITION. Let n be a positive integer.

(1) Submodules and quotients of llOlonomic An-modules are holonomic.

(2) Finite sums of lwlonomic An-modules al'e holonomic. PROOF:

(1) These follow from Bernstein's inequality. Let M he a left An-

module, and N a suhmodule of M. From Theorem 9.3.2, deN) d(MjN)

~

d(M). Since d(M)

~

d(M) and

= n, and using Bernstein's inequality, we

deduce that deN) = d(MIN) are also equal to n. Thus N and MIN are holonomic. Now (2) follows from Corollary 9.3.3 anu (1).

10. Holonomic modules

87

1.2 COROLLARY. Finitely gene1·ated torsion AI-modules are holonomic.

PROOF: Let M be a finitely generated torsion AI-module. Suppose that it is generated by UI, ... , U r • Since M is torsion, for each i exists 0

=I bi

E Al such that

biu;

= O. Hence

= 1,2, ... , r there

Al U; is a quotient of At! Al bi

which is a holonomic module. Thus each Al U; is holonomic. Since M is the sum of AIuj, for i= 1,2, ... , r, it is holonomic by Proposition 1.1(2).

=I

~

J are left ideals of AI, then the quotient I I J is a torsion AI-module. A good It is easy to construct torsion AI-modules. For example, if 0

I

exercise, which the reader may wish to try, is to prove that I I J is holonomic using only Proposition 1.1(1). Still on the subject of torsion modules, we have the following proposition.

1.3 PROPOSITION. Holonomic An-modules are torsion modules. PROOF: Let M be a holonomic left An-module. Suppose that 0 element of M. Consider the map
~

M, it follows that d(im
= n.

=I 0,

u is an

M defined by
Thus by Theorem 9.3.2,

2n = deAn) In particular ker ¢

~

=I

= d(ker ¢).

and u is a torsion element of M.

Putting 1.2 and 1.3 together, we conclude that for finitely generated A I modules, torsion and holollOmic are equivalent concepts. However, it is not true that all torsion An-modules are holonomic when n consider the module M

= Ani AnOn.

~

2. For example,

The Bernstein filtration induces in M

a filtration defined by

r i is isomorphic to the vector space generated by the monomials in XI, . .. , Xn and Bt, ... , On-I of degree::; i. We may use this to calculate the Note that

Hilbert polynomial of M explicitly, using combinatorics. It equals (2;!~~I). As a polynomial in i, this binomial number has degree 2n- 1; see Exercises

9.5.2 and 9.5.3. Hence d(M)

= 2n-1 > n, if n ~

2. On the other hand, it is

very easy to check that M is a torsion module: its elements are annihilated by powers of Dn·

10. Holonomic modules

88

2. BASIC PROPERTIES. Many interesting properties ofholonomic modules follow from the fact that they are artinian. A module M over a ring R is said to be arlinian if given a descending sequence Nl k such that Nk

=

Nj

,

~

N2

~

for every ;"

~

... of submodules of M, there exists k. In other words, every descending

chain of submodules of M is stationary. This is like the noetherian property. The following proposition collects some of the basic properties of artinian modules.

2.1 THEOREM. Let M be a left module over a ring R and let N be a submodule of M.

(1) M is artinian if and only if evel"J set S of submodules of M lJas an element (a minimal element) which does not contain any other element of S. (2) M is artinian if and only if N and MIN are artinian. (3) Let Nt be anotllel" submodule of M, and suppose that M = N + NI. If N, Nt are al·tinian, tllen so is M. The proof of (1) is like that of Theorem 8.1.1 and the proofs of (2) and (3) are like those in Proposition 8.1.3. The details are left to the reader. Let us now see why we are interested in artinian modules. 2.2 THEOREM. Holonomic modules are artinian. PROOF: Let M be a holonomic left An-module.

Suppose that M has a

descending chain of submodules

By Theorem 9.3.2 and Proposition 1.1(1), it follows that meN;)

= m(Ni+l)+

m(Ni/Ni+l). Putting these together, we get that r-1

m(M) =

L: m(N;jN +

i 1)

+ m(Nr )

~ r.

a Hence M cannot have a descending chain of more than r sub modules. In particular M cannot have an infinite descending chain.

10. Holonomic modules

89

Let us dispel any hopes that the reader may have formed, by saying that not all An-modules are artinian. For example, An is not artinian as a module over itself. It is easy to construct an infinite descending chain; take for instance

More examples of non-artinian An-modules are found in Exercises 4.2 and 4.3. A ring R that is artinian as a left R-module is called a left artinian

ring. The argument above shows that An is not a left artinian ring. Artinian rings have a very nice representation theory; see [Reiten 85]. We return, for a moment, to the general situation. Let M be a left nonzero module over a ring R. Assume that M is both artinian and noetherian. Suppose we have constructed a chain No <;;; Nl <;;; ..• <;;; Nk of submodules of M such that N;/Ni -

1

is irreducible. It follows from Theorem 2.1(1) that

there exists a submodule Nk+l of M such that Nk+llNk is a minimal nonzero submodule of MINk.

Since Nk+llNk is minimal in the set of non-

zero submodules of MINk, then it must be irreducible. Thus in the chain

No <;;; ... <;;; Nk <;;; Nk+l, the quotient of two adjacent terms is irreducible. This process cannot be continued forever, since M is also noetherian. Thus we get a chain

o = No <;;; ..• <;;; N r = M of submodules of M such that N;/ N i -

1

is irreducible for i

= 1, ... , r.

Such a

chain is called a composition series of M. It follows from the Artin-Schreier theorem [Cohn 84, 9.2 Theorem 2) that any two composition series of M have the same number of submodules. The number r is called the length of M. Since a holonomic module is noetherian and artinian, it must have a composition series. Hence its length is well-defined. It may be calculated using the argument in the proof of 2.2. 2.3

SCHOLIUM.

The lengtll of a holonomic module cannot exceed its multi-

plicity. 2.4 COROLLARY. A llOlonomic An-modules of multiplicity 1 is b-reducible. PROOF:

Let M be a left holonomic module of multiplicity 1 and suppose

that it has a lIoll-zero submodul0. N. Then Nand A1jN are also holollomic

10. Holonomic modules

90

= m(N) +

by Proposition 1.1. Hence by Theorem 9.3.2, we have that m(M)

m(MIN}. Since m(M) = 1 and N of 0, we have that MIN = O. Thus

= N and M must be irreducible.

M

It is surprisingly easy to calculate the number of generators of a module

that is artinian and noetherian. 2.5 THEOREM. Let R be a simple left noetherian ring and M a finitely

generated left R-module. If M is aJ:tinian but R is not artinian (as a left R-module), then M is a cyclic module. PROOF: The statement that we shall prove is even sharper: if M is generated by

Ub""

Ur

then there exist

a2, ... , a r

M. It is enough to prove the case r

E R such that

Ul

+ L:; aiU; generates

= 2. The general case follows by

induction. Suppose that M is generated by two elements, say u and v. Since M is artinian and noetherian, it must have finite length. We shall prove, by induction on the length of M, that there exists a E R such that M

=

A(u+ av). If M has length 1, then it is irreducible and there is nothing to prove. Suppose that the result holds for all R-modules of length < r, and that M has length r. In particular Rv has finite length, so it must have a non-zero irreducible submodule. This must be cyclic, hence generated by an element of the form

cv, for some e E R. Since M I Rev has smaller length than M, we may use the induction hypothesis to construct an element A E R such that

M

= R(u+ AV) + Rev.

It is better to rephrase this by saying that one may assume that M

= Ru+ Rv,

with Rv an irreducible module. Let : R

-+

M be the homomorphism of modules defined by (x) = xu.

If is injective, then R ~ im is a non-artillian submodule of M, which

contradicts Proposition 2.1. Hence ker of O. Let d be a non-zero element of ker . Since R is a simple ring, one has that RdR = R, and so RdR . v of O. In particular, there exists b E R, such that dbv of 0 . We claim that M is

10. Holonomic modules generated by u + bv. First note that d( u + bv)

= dbv is a

91 non-zero element

in Rv. Since Rv is irreducible, v is a multiple of dbv. Hence v, and therefore

u, are elements of R( u + bv), which must then equal M. 2.6

COROLLARY.

PROOF:

Holonomic modules are cyclic.

We know from Theorem 2.2 that a holonomic An-module must be

artinian. We have also seen that An is not artinian. The result follows from Theorem 2.5.

3.

FURTHER EXAMPLES.

In this section we construct a family of holonomic An-modules which is very important in applications. We begin with a technical lemma. Note that in this lemma we do not assume that the module is finitely generated: this will be proved as part of the lemma.

3.1

LEMMA.

Let M be a left An-module with a filtmtion

r

with respect to

the Bernstein filtration of An. Suppose t1lat tIl ere exist constants cI, c2 such

»0

that for j

dimK rj :::;

cdn/nJ + C2(;j + l)n-l.

Then M is a holonomic An-module whose multiplicity cannot exceed particular M is finitely generated. PROOF:

CI'

In

The proof breaks up naturally into two parts.

Part 1: Every finitely generated submodule of Mis holollomic of multiplicity :::; CI·

Let N be a finitely generated sub module of M. Thus N admits a good filtration, say [} . By Proposition 8.3.2, there exists a positive integer r such that

[}j ~

IJ+r n N. III particular, dimK

[}j :::;

dimK rj+r' Let X(t) be the

Hilbert polynomial of N for the filtration [} . Using the polynomial bound of the hypothesis, we conclude that, for very large J',

In particular the degree of X(t) cannot exceed n. Hence d(N) :::; n. From the Bernstein inequality we conclude that d( N)

= n equals the degree of X( t).

Going back to the polynomial inequality above, we obt.ain that. meN) $

Cl.

10. Holonomic modules

92

Part 2: M is finitely generated. Consider an ascending chain Nl

~

~

N2

'"

~

N r of finitely generated

sub modules of M. Each of these is holonomic. By Theorem 9.3.2, m(Ni )

m(Ni - 1 )

+ m(Ni/Ni-l), for

=

i = 1,2, ... , r. Adding them up one has

r

L m(Ni/Ni-l) + m(N

l)

= m(NT ) ~

Cl·

2

In particular, all ascending chains of finitely generated submodules of M have less than

Cl

steps. Thus M must be finitely generated. We may now apply

Part 1 to M itself, and conclude that it is holonomic of multiplicity

Let K(X)

~ Cj.

= K(Xl,"" xn) be the field of rational functions. We may

extend the left action of An on the polynomial ring K[X] to the field of rational functions. The

Xi

continue to act by multiplication. The action

of Oi on a rational function

f / 9 is defined by the rule for differentiation of

quotients, namely 8i(J /g) = (8;(J)g - f8;(g))/i.

A routine calculation shows that this action satisfies the required properties. Note that this module is not finitely generated. Suppose that a polynomial p is chosen in K[X]. Let K[X, p-l] denote the set of rational functions of the form

f /pr, where f

is a polynomial. Note

that a partial derivative of f / pr has denominator rfT. Hence these rational functions are preserved by partial differentiation and by multiplication by a polynomial. In other words, K[X,p-l] is a left An-submodule of K(X). 3.2 THEOREM. The An-module K[X, p-l] is holonomic and its multiplicity

is

~

(deg(p)

+ 1)n.

PROOF: To save on notation, put M

= K[X, p-l].

Suppose that p has degree

m. Set

r k = {f/l : deg(f) We first check, in detail, that

r

~ (m+ l)k}.

is a filtration for M.

10. Holonomic modules

93

Let IIrY' be an element of M and assume that I has degree s. Then IIrY' = I· P'1P'+k· But Ip' has degree s(m+ 1), which is ~ (m+ 1)(s + k). Hence

f IV<

E r.+k' It follows that

Next suppose that f tiplication by

Xi

II E n.

M is the union of all r k, for k ~ O.

I

Equivalently,

has degree ~ (m+l)k. Mul-

increases the degree of f by 1, thus Xj(J IIi)

rk+l' Differentiating

I frY'

with respect to

The numerator has degree ~ (m + l)k

Xi

= x;jplIi+ 1 E

we get

+ (m -

1), so that OJ(J IIi) E n+l.

This may be summed up as

Since B; = B1 we also have that Birk ~ ri+/:o Finally, the dimension of

rk

space of polynomials of degree

cannot exceed the dimension of the vector ~

rk . d111lK

~

rk is finite dimenr is a filtration of M, and shows that

(m + 1)k. Hence each

sional. This concludes the proof that

(m+ l)k n

+

n)

•

Since the two terms of highest degree in k of this binomial number are

it follows that

for very large values of k. By Lemma 3.1, M must be a holonomic module of multiplicity ~ (m + 1), as required. We may use the theory we have developed so far to construct the Bernstein polynomial of a differential operator with polynomial coefficients. As before, we start with a polynomial p E K[xl> . .. ,xnJ. Let s be a new variable. We are going to construct a holollomic module over the ring A,,(K(s», wherE'.

10. H olonomic modules

94

K (8) is the field of rational functions on s. The generator of this module will be denoted by p'. This is a formal symbol on which 8j acts by

It follows from this formula that An(K{s»P' is an An(K{s))-submodule of

K{s)[X,p-l]p'. Now define an automorphism t of K(s)[X,p-ljp' by the formula t(SiP')

= (s + l)ip. p'.

Note that this is An(K)-linear, but not

An(K{s))-linear, see Exercise 5.8.

3.3

THEOREM.

Let p E K[X]. There exist a polynomial B(s) E K[s] and a

differential operator D(s) in the polynomiall'ing An(K)[sj such that B(s)p' PROOF:

= D(s)pp'.

The An{K(s))-module K(s)[X,p-ljp' is holonomic. The proof is

like that of Theorem 3.2 and is left to the reader; see Exercise 4.6. Since An(K(s»P' is a submodule of K(S)[X,p-l]P', it must be holonomic. In particular An(K(s))P' has finite length by Scholium 2.3. Thus the descending sequence

An(K(s))p' ;2 An(K(s»p· p' ;2 An{K(s»p'l. p' ;2 '" must terminate. This means that there exists k > 0 such that

Applying t- k to both sides of this equation, we get p' E An(K(s»)p· p'. Clearing denominators, we conclude that there exists a polynomial B(s) E

K[s] such that B(s )p'

E

An(K)[s]p. p'

and the theorem is proved. The polynomial B(s) and the operator D(s) are not uniquely determined by p. However, the set of all the possible polynomials B(s) satisfying Theorem 3.3 is an ideal of K[s], as one can easily verify. The monic generator of this ideal is thus unique; it is called the Bernstein polynomial of p and denoted by bp(s).

10. Holonomic modules

95

The calculation of the Bernstein polynomial for a given p can be a very complicated affair. Here is a famous example that can actually be done by hand. Let p = x~

+ '" + x!. If we denote by

D the differential operator

8? + ... + o~ then D.]I+!

=

4(8

+ 1)(8 + n/2)]I.

Thus bp (8) = 4(8 + 1)(8 + n/2). Note that in this case D is independent of 8. The polynomial bp(s) was introduced by Bernstein in [Bernstein 71] as part of his solution of Gelfand's problem on the meromorphic extension of a certain analytic function. Bernstein's work is discussed in detail in [Krause and Lenagan 85, Ch. 8] and requires a basic knowledge of distributions. The Bernstein polynomial is also very important in singularity theory; see [Malgrange 76]. Building on work of Malgrange, Kashiwara showed that the roots of the Bernstein polynomial are strictly negative rational numbers; see [Kashiwara 76]. Kashiwara's approach works over any analytic manifold. For details see [Granger and Maisonobe 93, Ch. VI] and [Bjork 79, Ch. 6]. The explicit calculation of the Bernstein polynomial and its roots has been the subject of many papers; see for example [Yano 78] and [Cassou-Nogues 86].

4.

EXERCISES.

4.1 Show that An/Anan is a torsion An-module.

Hint: Every class in An/AnOn contains a representative of the form 2:=~ djx~, where d j E An-I' Show that this element is annihilated by O~+I E An/Anon. 4.2 Let I be a non-zero left ideal of An. Show that I is not an artinian

An-module. 4.3 Show that An/Anon is not an artinian An-module. Hint: The modules

form an infinite descending chain in An/ Anan. 4.4 Let p E K[X] be a non-constant polynomial. Is the module K[X, p-I) irreducible?

10. Holonomic modules

96 4.5 For r ~ n, let P AnP

~

= 2:; Eft -

2:~+1

at.

Find a left ideal J of An such that

J and Anj J is holonomic.

4.6 Let p E K[X].

Show that K(s)[X,p-l]V is a holonomic An(K(s»-

module. Hint: Let m be the degree of p. The K-vector spaces,

rk =

{q. p-k . V

:

deg(q) :::; (m+ l)k} give rise to a filtration of K(s)[X,p-l]. What are their dimensions? 4.7 Let p be a polynomial in K[Xl, ... , xn]. Denote by An [s] . V the submodule of K[s, X, p-l] generated by

V over

the polynomial ring An[s]. This is

also an An-module, but in this case it is not clear whethe1' it is finitely

generated! Show that if p belongs to the ideal of K[X] generated by its partial derivatives, then An[s] . V is finitely generated as an An-module. Hint: Let D

= 2:~ -I;; a;.

Then D· f

= sf·

= An[s]V be as in Exercise 4.7. Let map defined by t(D(s). V) = D(s + l)p· V.

4.8 Let p and M

t:

M

-t

M be the

(1) Show that t is an endomorphism of M as an An-module but not as an An[s]-module. (2) Show that [t, s]

= t.

(3) Use (2) to show that MjtM is an An[s]-module, even though tis not

An[s]-linear. It is true that MjtM is a holonomic module, but this is ha1'de1' to prove; see [Granger and Maisonobe 93, eh. VI].

= exp(Alxl+" +Anx n). = {P E An(lR) : P . f = a}.

4.9 Let AI,'" ,An be real numbers. Setf(xl, ... ,Xn) (1) Find generators for the left ideal J

(2) Show that An(R)j J is a holonomic module over An(R).

CHAPTER 11 CHARACTERISTIC VARIETIES In this chapter we give a geometrical interpretation of the dimension of an An-module. In order to do this we will have to use a number of results of algebraic geometry and linear symplectic geometry. All the algebraic geometry that we need can be found in [Hartshorne 77, Ch. 1]. Throughout the chapter we assume that the base field is C. 1. THE CHARACTERISTIC VARIETY. Let An be the 'Ilrth complex Weyl algebra and let M be a finitely generated left An-module with a good filtration

r.

Then grf M is a finitely generated

module over the polynomial ring Sn. Let ann(M, r) stand for the annihilator of grf M in Sn. Then ann(M, r) is an ideal of Sn. It depends not only on

M, but also on the choice of the good filtration

r;

see Exercise 4.1. The

radical of ann(M, r) however is independent of the filtration. 1.1 LEMMA. Let [} be another good filtration of M. Then

rad(ann(M, r)) PROOF:

= rad(ann(M, [})).

The ideals ann(M, r) and ann(M, [}) are homogeneous ( see Exer-

cise 8.4.6), hence so are their radicals (Exercise 4.2). Choose a homogeneous element

f

of degree s in rad(ann(M,r)). There exists dEB., the compo-

nent of degree ~ s of the Bernstein filtration, such that

f = used).

Since f E rad(ann(M, r)), there exists mEN such that f'n E ann(M, r). Hence,

dTnr ~ r m.+i-l, for every i ~ O. j

Iterating q times we get that

(1.2) On the other h.and, by Proposition 8.3.2 there exists k ~ 0 such that

for all i ~ D. Together with (1.2) for q

= 2k + 1, this leads to

r e re 'i+",-,(2k+l)-k-l _ .'i+ln1(2k+l)-1'

t"l C d rn (2k+l) d m (2k+l) Hi _ i+k _

t"l

11. Characteristic varieties

98 Thus d,n(2k+l)[}i ~

[}i+m8(2k+l)-b

and so rad(ann(M, r))

~

for all i ~ O. Hence f'n(2k+l) E ann(M, [}),

rad(ann(M, .a)). The opposite inclusion follows

by swapping rand [}. The ideal I(M)

=

rad(ann(M, r)) is called the characteristic ideal of

M. It follows from Lemma 1.1 that it is independent of the good filtration

r used to calculate it.

In other words, I(M) is an invariant of M. Thus, so

is the affine variety Ch(M)

=

Z(I(M)) ~ C 2n

that it determines. This variety is called the characteristic variety of M. Since I(M) is a homogeneous ideal ( see Exercise 4.2), the variety Ch(M) is homogeneous. This means that if p E Ch(M) and

~ E

C, then

~p E

Ch(M).

Note that Ch(M) is a subvariety of C2n, since Sn is a polynomial ring in 2n variables! Here is a simple example. Let dEAn be an element of degree r and put

M

= An/And.

If B" is the filtration of M induced by the Bernstein filtration

of An, then

as we saw in Ch. 7, §5. Therefore, ann(M, B")

= SnUT( d)

and so Ch(M)

=

Z(uT(d)) is a hypersurface.

1.3 PROPOSITION. Let M be a finitely genel'ated left An-module and N a

submodule of M. Then Ch(M) PROOF: Let

and

r"

r

= Ch(N) U Ch(M/N).

be a good filtration of M. It induces good filtrations

r

in N and MIN respectively. By Lemma 7.5.1, there exists an exact

sequence of finitely generated graded Sn-modules,

Clearly

ann(M, r) ~ anneN, r) n ann(M/N, r")

11. Characteristic varieties

99

and so Ch(N)UCh(M/N) ~ Ch(M), by [Hartshorne 77, Ch. 1, Proposition

1.2J. The opposite inclusion follows from ann(N, r)ann(M / N, r") £; ann(M, r). Before we proceed with the study of the characteristic variety, let us review some basic facts about the dimension of an affine variety. Let J be an ideal of Sn

=

C[YI, ... , YlnJ and put V

=

Z(J). If p is a point of V then the

Zariski tangent space of Vat p is the linear subspace of C2n defined by the equations 2n

of

L-(p)Yi=O lOY;

for all F E J. This space is denoted by 7;,( V)j it is a complex vector subspace of C2n. The dimension of V equals inf{ dime 7;,( V) : p E V}. Actually one need not look at every point of V. dim( V)

= dime 1;,( V).

If p is a non-singula1' point of V, then

This is equivalent to the definition in terms of heights

of prime ideals, see [Hartshorne 77, Ch. 1, Exercise 5.lOJ. We may also define 1;,( V) in terms of linear forms on linear form defined on the vector Y

= (YI, ... , Y2n) 2n

dpJ(Y) Consider the map d p : Sn

=

- t «((!2n)*

((!2n.

Let dpf be the

= dpf.

Let f, 9 E Sn

by

8f

L "UYi (P)Yi. I

defined by dp(f)

and), E C. Then dp satisfies dp(f

+ ),g) = dpf + )'dpg, dp(fg) = J(p)dpg + g(p)dP/.

We may rephrase the definition of Tz,( V) using d p , as follows: u E Tz,( V) if and only if dpF( u)

= 0, for every F E I( V).

This may be improved, by restricting

the elements of I( V) to a finite set. Suppose that FI, ... , Fm generate I( V). It follows from the properties of d p stated above that u E

if

Tz,( V) if and only

11. Characteristic varieties

100

The characterization of 7;,( V) in terms of dp will be used in the next section. The following theorem is an immediate consequence of the fact that if N is a graded module over Sn then the degree of its Hilbert polynomial is

dimZ(anns.N)j see [Hartshorne 77, Ch. 1 Theorem 7.5].

1.4

THEOREM.

dim Ch(M)

Let M be a finitely generated left module over An. Then

= d(M).

It is now very easy to show that if d

=I

0 is an operator in An, then

Ani And has dimension 2n - 1. Suppose that d has degree r

> O.

seen, the characteristic variety of Ani And has equation a"r{ d)

As we have

= O.

This is a

hypersurface of (:2nj hence its dimension is 2n-1. Thus d(AnIAnd) = 2n-1, by Theorem 1.4. Compare this example with Exercise 9.5.2. We may also use the characteristic variety to achieve a geometrical interpretation of Bernstein's inequality. This depends on results of symplectic geometry that we summarize in the next section. 2. SYMPLECTIC GEOMETRY.

A symplectic structure on (:2n is determined by a non-degenerate skew-

symmetric form. The standa1'd symplectic structure on (:2n is expressed by means of the matrix

where In is the nx n identity matrix. Given vectors u, v E (:2n, the symplectic form is

The matrix of any non-degenerate skew-symmetric form can be written in the form above for a suitable choice of coordinates [Cohn 84, §8.6, Theorem

1]. If U is a subspace of (:2n, then its skew-orthogonal complement is

UJ.

= {v E (:2n : w( 1.1, v) = 0 for

all

1.1

E

U}.

Note that it is not true that (:2n is the direct sum of a subspace and its skeworthogonal complement. For example, since w( 1.1,

1.1)

= 0 for

every u E (:2n,

11. Characteristic varieties

101

then everyone-dimensional space is contained in its complement. A subspace that is contained in its skew-symmetric complement is called isotmpic. However, since w is non-degenerate, it is always true that U and UJ. have complementary dimensions. The proof is short enough to be included here. First, some definitions. For wE c2n, define a linear form
--t

C by

cPw( v) = w( w, v). Let !l> : c 2n --t (c 2n)* be the linear map which associates cPw to w E C2n. 2.1

PROPOSITION.

Let w be a non-degenerate skew-symmetric form in

and U a subspace of c

2

n.

c2

n

Then:

(1) !l> is an isomorphism. (2) If a linear form

(3) dim U + dim UJ. PROOF:

cPw restricts to zero on U, then = 271.

w E UJ..

Note that the kernel of !l> is zero, because w is non-degenerate.

Hence, dim !l>(C 2 n) Since dim(C2n)*

= 271 -

dim ker(!l»

= 2n.

= 2n, it follows that !l> is surjective. Thus !l> is an isomor-

phism of vector spaces, and (1) is proved. Now let !l>lu :

c2

n

--t

U" be the map defined by !l>1U(w)

= rPwlv, the

restriction of cPw to U. Since every linear form on U extends to a linear form . on c2n, we have that !l>lv is surjective by (1). Hence the image of !l>lu is U". Now (2) is clear, and it implies that the kernel of !l>1 u is UJ.. Hence, 2n = dimker!l>1 v + dim!l>1 u(c 2 n) = dim UJ.

+ dim U*,

which implies (3). The subspaces that are important, from our point of view, are the ones that contain their skew-complement. They are called co-isotropic or involutive. A hyperplane is always an involutive subspace. Indeed, if H is a hyperplane of C2n that is not involutive, then there exists v ¢ H such that w(v, u) for all u E H. Since C2n

= H + Cv,

= 0,

this contradicts the non-degeneracy of

w. If V is an affine variety of C2n, then we will say that it is i71volutive if the tangent space 7;,( V) <; c2n is invollltive at every Iloll-singular point p E v:

11. Characteristic varieties

102

In particular, a hypersurface will always be involutive. The dimension of an involutive variety satisfies an inequality similar to Bernstein's. 2.2 PROPOSITION. The dimension of an involutive variety ofC 2n is greater

than or equal to n. PROOF: Let V be an involutive variety of

c2

n

and p a non-singular point

of V. By definition, we have that 7;.( V) 1. ~ 7;.( V). Hence dim 7;.( V)1.

5

dim 7;.( V). Thus, by Proposition 2.1,

2n = dim 7;.( V)1. and so dim 7;.( V) ~

1t.

+ dim 7;.( V) :5 2 dim 7;.( V),

Therefore, dim( V) ~ n.

The involutive varieties of dimension n are called lagrangian. Note that

if V is lagrangian in

c2n,

then 7;.( V)1. C 7;.( V) and both subspaces have

dimension n. Hence 7;.( V)

=

7;.( V)1.. In particular, V is also isotropic.

Thus lagrangian varieties can also be characterized as varieties which are involutive (co-isotropic) and isotropic. Talk of an affine variety, and one immediately thinks of its defining ideal. How can one decide whether a variety is involutive by looking at its ideal? The answer lies with the Poisson bracket of Sn. Let I denote the inverse ofthe map !l> defined above. The Poisson bracket of I, 9 E Sn is

An explicit calculation using coordinates shows that

{/,g}(p)

~ 81 8g = L..,{-(P)· -(p) I

8Yn+i

8Yi

8g

81

-(p). -(p)}. 8Yn+i 8y,

From this formula it is easy to see that {f, g} is a polynomial in Sn and that the map

{I, .} : Sn -+ Sn is a derivation of Sn.

An ideal J of Sn is closed for

the Poisson bracket if {f,g} E J whenever I,g E J. We want to show that a variety is involutive if its ideal is closed for the Poisson bracket. The proof will make use of a technical fact about the tangent space that we isolate ill a lemma.

11. Chamcteristic varieties

103

2.3 LEMMA. Let V be an affine variety ofC2n and p a point of V. If 8 is a form on C2n whose restriction to 1;.( V) is zero, then there exists

J E I( V)

such that 8 = dpJ. PROOF: Let F I , ... ,Fm be the generators of I( V). Set Y

= (Y1. ... .1l2n).

Then 1;.( V) is the solution set of the equations

in C2n . But 8( y)

=

0 is satisfied by the vectors of 1;.( V), by hypothesis.

Thus

where ai E C. Let I

= L:~ aiFi E I( V).

A straightforward calculation shows

that 8 = dp/. 2.4 PROPOSITION. An affine variety V of c2n is involutive if and only if its

ideal I( V) is closed for the Poisson bl·acket. PROOF: Suppose that V is involutive. Let p be a non-singular point of V. If I E I(V), then dpl restricted to 1;.(V) is zero. By Proposition 2.1(2),

I dp/ E 1;.( V).l.. Since V is involutive, then Idpl E 1;.( V). Now if 9 E I( V), then dpg is zero on 1;.( V), hence

{I, g}(p) =

-dpg(I dP/) =

o.

Since this identity holds for all non-singular points p of V, we conclude that the polynomial

{I, g} is in I( V), which is then closed for the Poisson bracket.

Conversely, assume that I( V) is closed for the Poisson bracket. Choose WE 1;.( V).l.. By Lemma 2.3, there exists 9 E I( V) such that

cPw = dpg. But

dpg( u) = w(I dpg, u). Since w is non-degenerate, w = I dpg. If I E I( V) then

Hence wE Tp( V). Thus Tp( V).l.

~

1;.( V).

We have seen that hypersurfaces are involutive, since their tangent spaces are hyperplanes. This is very

ca~y

t.o prove using Proposition 2.4, since the

11. Chamcteristic varieties

104

ideal of a hypersurface is principal, and the Poisson bracket of a polynomial with itself is always zero. The relation between characteristic varieties and symplectic geometry is the subject ofthe next theorem. The first proofs ofthis result were analytical. There is now a purely algebraic proof due to O. Gabber [Gabber 81]. We shall not include the proof here as it would take us too far from our intended course. 2.5 THEOREM. Let M be a finitely genel'ated left An-module. Then Ch(M) is involutive with respect to the standard symplectic stl'uctw'e ofC 2n . Equiv-

alently, I(M) is closed f01' the Poisson bracket. It is not obvious at first sight why this theorem should be difficult to prove.

In fact, it is easy to prove that if r is a good filtration of M, then ann(M, r) is closed for the Poisson bracket, The theorem does not follow easily from this fact because it is not true that the radical of a closed ideal is closed! Here is an example. Let J be the ideal of 8 1 generated by y'f, wand YI Y2. It is closed for the Poisson bracket, but its radical contains YI and Y2, and {YI, Y2}

= l.

Thus rad(J) is not closed for the Poisson bracket. Theorem 2.5

states the very subtle fact that the radical of the annihilator of the graded module of an An-module is closed, even though radicals of closed ideals arc)

not in general closed. Putting together Theorem 2.5 and Proposition 2.2 we get Bernstein's inequality. 2.6 COROLLARY. Let M be a finitely genel'ated left An-module. Then

d(M)

= dim Ch(M)

~

n.

It also follows from Theorem 2.5 that a holonomic module has a homoge-

neous lagrangian characteristic variety. A very spectacular use of the involutivity of the characteristic variety is discussed in the next section. 3. NON-HOLONOMIC IRREDUCIBLE MODULES

It is a very curious fact that, until 1985, it was widely believed that every irreducible An-module had to be holonomic. There was no good reason for

11. Characteristic varieties

105

this, except a lack of examples. The truth carne to light in [Stafford 85]. Stafford showed that if n

> 2 and A2 •...• An E C are algebraically indepen-

dent over Q, then the operator n S

n

= 01 + (~: A;XIX;O; + X;) + I)Xi 2

generates a maximal left ideal of An. Let M ducible and d(M)

= 2n -

Oi)

2

= Ani Ans.

Then M is irre-

1 > n, since n > 2. The fact that Ans is maximal

is proved by a long calculation, which can also be found in [Krause and Lenagan 85, Theorem 8.7].

In 1988, J. Bernstein and V. Lunts found a different and more geometric way of constructing irreducible modules of dimension 2n-1 over An. As we have seen in §§1 and 2, the characteristic variety of an An-module is always homogeneous and involutive. We will say that a homogeneous involutive variety of

c 2n is

minimal if it does not contain a proper involutive homo-

geneous subvariety. For example, an irreducible lagrangian variety must be minimal, since varieties of dimension less than n cannot be involutive. The work of Bernstein and Lunts depends on the following result. 3.1 PROPOSITION. Let d be an opel'ator of degree k in An and suppose that:

(1) the symbol uk(d) is ineducible in Sni (2) the hypersurface Z(uk(d» is minimal. Then the left ideal And is maximal. In particular, the quotient Ani And is an irreducible module of dimension 2n - lover An. PROOF: Suppose that

J is a left ideal of An such that And C J. Since these

are submodules of An we may consider their graded modules with respect to the filtrations induced by B; we get

The corresponding varieties are

o~ Z(gr(J»

C Z(uk(d).

11. Characteristic varieties

106

Note that the last inclusion is proper, because ITj,(ll) is irreducible. But

gr(An/J) ~ 8 n/ gr(J), and so ann(An/J, B) = gr(J). Therefore, Z(gr(J») = Ch(An/ J) is involutive by Theorem 2.5. Since Z(Uk(d) is minimal, then Z(gr(J)) = 0. Hence gr(J) = 8 n and so J = An. Thus And is a maximal left ideal of An. To put this proposition to good use we have only to construct minimal hypersurfaces in

[:2n.

That these hypersurfaces exist is the heart of the work

of Bernstein and Lunts. In fact they show that most hypersurfaces in

[:2n

are minimal. To make this into a precise statement we need a definition. Let 8 n (k) be the homogeneous component of degree k of 8 n . This is a finite dimensional complex vector space; so it makes sense to talk about hypersurfaces in 8 n (k). We say that a property P holds generically in 8 n (k) if the set

{f

8 n (k) : P does not hold for J}

E

is contained in the union of countably many hypersurfaces of 8 n (k). 3.2 THEOREM. The property 'Z(f) is minimal' holds generically in 8 n (k), ~

whenever k

4 and n

~

2.

TIns result was proved for n = 2 in [Bernstein and Lunts 88] and later generalized to n

~

2 in [Lunts 89]. The proof of Theorem 3.2 uses a lot

of algebraic geometry and also some results on differential equations, one of which goes back to Poincare's thesis! The proficient reader will find more details in the original papers. The gist of the work of Bernstein and Lunts is that 'most' irreducible Anmodules are not holonomic: almost the exact opposite of what was believed before 1985. 4. EXERCISES 4.1 Let M

= Ad A1x.

filtration defined by

ann(M, [})

Let

[}k

r

be the filtration of M induced by Band [} the

= Bk' (8 + A1x).

= 8 1 y'f + 8 1 Yl Yl.

Show that ann(M, r)

= 81 Yl

and

Compute their radicals.

4.2 Show that if J is a homogeneous ideal of a graded algebra R then Tad(J) is also homogeneous.

11. Chamcte7'istic varieties 4.3 Let J be a left ideal of An and put M

= An/J.

107 Show that Ch(M)

=

Z(gr(J».

4.4 Is the union of two involutive varieties involutive? What about their intersection? 4.5 Show that if V is an involutive homogeneous variety of C2 n then its irreducible components are also homogeneous and involutive. 4.6 Let s be the operator of An defined in §3. Let M = Ani Ans. (1) Compute Ch(M). (2) Is it an irreducible variety of c2n? (3) Is it a minimal hypersurface? 4.7 Let J be a left ideal of An. Show that if gr(J) is a prime ideal of Sn and Z(gr(J» is minimal, then J is a maximal ideal of An.

4.8 When is a hypersurface of C 2 n lagrangian? Give an example of a lagrangian variety of c2n, when n ~ 2. 4.9 A holonomic left An-module M is regular if there exists a filtration

r

for

M such that anns.grf M is a radical ideal of Sn. Let N be a submodule of . a regular holonomic module M. Show that N and MIN are also regular. 4.10 Show that if M is a regular holonomic module whose characteristic variety is irreducible, then M is an irreducible module. Why is the regular hypothesis required? 4.11 Let J be an ideal of Sn. Show that J2 is always closed for the Poisson bracket.

CHAPTER 12 TENSOR PRODUCTS All the operations on An-modules to be defined in the next chapters make use of the tensor product, which we are about to study. The construction of the tensor product presented in §2 usually seems artificial on a first encounter. Fear not; it is the neat universal property of §3 that is most often used in the applications. The final two sections before the exercises contain a number of results that will be required later. 1. BIMODULES.

To discuss tensor products in sufficient generality it is necessary to introduce bimodules. Let Rand S be rings and let M be an abelian group. To qualify as an R-S-bimodule, the group M must be a left R-module and a right S-module, and the R-action and the S-action must be compatible in the sense that if r E R, s E Sand u E M, then

r(us) If S

=

=

(ru)s.

R then we simply say that M is an R-bimodule.

A few examples and counter-examples will make the definition clear. If R is a ring then Rk is an R-bimodule. More generally, if R is a subring of S, then Sk is an S-R bimodule. In particular this applies to Weyl algebras; for if m < n, then Am is a subring of An. Thus A~ is an An-Am-bimodule. On the other hand, K[x] is a left AI-module and a right K[x]-module, but it is not an A1-K[x]-bimodule because the two actions are not compatible. Indeed, suppose that

f

E

K[x] is acted on the left by 8 and on the right by

x. Then 8(fx) = x8(f)

but (8. f)x

+f

= x8(f). More generally, K[X] is not an An-K[X]-bimodule.

Let M be an R-S-bimodule. A subgroup N of M is a sub-bimodule of M if it is stable under both the R-action and the S-action on M. In this

case, the quotient group !vII N has a natural ::;tructure of R-S-bimodule.

12. Tensor products

109

The following example will often come up in applications. Consider subring of

An+!

in the usual way. The left ideal

of the An+l-An-bimodule

An+!'

An+1Xn+l

An

as a

is a sub-bimodule

Thus the quotient

is an An+l-An-bimodule. We may similarly define a homomorphism of R-S-bimodules as a homomorphism of the underlying abelian groups which preserves both the left and right module structures. 2. TENSOR PRODUCTS.

Let R, Sand T be rings. Let M be an R-S-bimodule and let N be an ST-bimodule. We will define the tensor product of M and N over S, denoted by M®sN. First consider the set M x N of all pairs (u, v), with u EM, v EN. Let A be the free abelian group whose basis is formed by the elements of M x N. The elements of A are formal (finite) sums of the form (2.1)

with aj E Z,

La;(u"v;)

Ui

EM,

Vi

E N. Note that in this sum the pairs are mere

symbols: the sum is not an element of the direct sum M ffi N. In fact, if we assume that different indices correspond to different pairs in (2.1), then the sum is zero if and only if each

aj

= O.

r(u, v)

If r E Rand t E T, put

= (ru, v),

(u, v) t = (u, vt), where u EM and

V

EN. These are well-defined actions that make an R-T-

bimodule of A. Now consider the subgroup B of A generated by elements of the following types:

(U+ oJ, v) - (u, v) - (oJ, v), (u,

1)

+ V') -

(u., v) - (u., d),

12. Tensor prod1.lcts

110

(U8, v) - (u,sv), where r E R, s E S, t E T, u, 'Ii EM, v, 'If E N. One immediately checks that B is a sub-bimodule of A. The tensor product M @s N is the quotient AlB. The image of (u, v) in M @s N is denoted by 1.1 @ v. Since A has the pairs (1.1, v) as basis, the elements of the form u@ v generate M @s N. However, it is not true that every element of the tensor product is of the form u@ v. The elements 1.I@vsatisfyrelations which are determined by the generators of B, as follows (1.1 + J) @ v = 1.1 @ V

u@(v+v')

+ 1.1' @ v,

= u@v+u@v',

which are a sort of distributivitYi and also

U8@V= 1.I@SV, which means that the tensor product is balanced; finally

r(u@ v)

= ru@ v,

(1.1@ v)t= 1.I@vt, are, respectively, the left R-action and right T-action on M @s N. Note that if R is a K-algebra then a left R-module may be seen as an R-Kbimodule. This trivial observation must be kept in mind in the applications. For example, if M is a left R-module and N a right R-module, then N @R M is a K-vector space. We may also calculate M@l(N, which is an R-bimodule.

On the other hand, if R is commutative then an R-module is automatically an R-bimodule. In particular if M and N are R-modules and R is commutativc, then M

@R

N is an R-module.

Let us end this section by calculating a tensor product using the definition above. This is a simple example which, at the same time, carries a warning: the tensor product can behave in very unexpected ways. Consider the Zbimodules Zp and Zq, and assume that p and q are co-prime. We wish to calculate Zp 0z Zq. We know tllflt it is generated by elemcnts of the form

12. Tensor products

111

m 0 n, where m E Zp, n E Zq. But since p and q are co-prime, there exist integers a, b such that ap + bq = 1. Thus m0 n= (ap+ bq)m0 n= bq· m0 n

= O. Since the product is = bm0 qn. But qn = 0, Zp 0z Zq = o.

because pm

balanced, we deduce from the above

that m0 n

and so we end up with m® n = O.

Therefore,

3. THE UNIVERSAL PROPERTY The first property of the tensor product that we shall study is also the key to proving all the other ones. To make its statement more succinct it is convenient to introduce some terminology. For the rest of this section let R, Sand T be rings; let M be an R-S-bimodule and N an S- T-bimodule. Suppose that L is an R- T-bimodule and that ¢ : M x N

--t

L is a map. We

say that ¢ is bilinear if

¢(ru + u, v)

= r¢(u, v) + ¢( u, v), ¢( u, vt+ d) = ¢( u, v)t + ¢( u, d),

for all u, 11 EM, v, 'If EN, r E R, t E T. The map ¢ is said to be balanced, if

¢(us, v)

= ¢(u,sv)

for all s E S, u E M, v E N. The canonical example of a bilinear and balanced map is the projection 71' :

defined by 71'(u, v)

M x N

--t

M 0s N

= u® v.

Let U be an R- T-bimodule, and suppose that there is a bilinear and balanced map

'T} :

M x N

--t

U. We say that

'T}

is a universal bilinear

balanced map if given al1Y R- T-bimodule L and a bilinear and balanced map ¢ : M x N - - t L, there exists a unique R- T-bimodule homomorphism

;p: U --> L such that I/J = 1>. '1/.

12. Tensor products

112

3.1 THEOREM. The map

71' ;

M x N

M ®s N is a universal bilinear

--+

balanced map.

PROOF: First of all, we may extend ¢ to get a map 'IjJ : A 'IjJ(L: ai( U;, Vi»

=L

--+

L. In detail,

ai¢( Uj, Vi).

i

Since ¢ is bilinear and balanced, it follows that 'IjJ is zero on every element of the submodule 13. Thus 'IjJ induces an R-T-bimodule map

M ®s N

= AI13 --+ L.

This is the map that we call ¢. Since the canonical projection of A onto

AI13 is induced by

¢

now show that

M ®s N

--+

the equation ¢

71',

= ¢ . 71' follows

immediately. We must

is unique. Note that any bilinear and balanced map 8 :

L for which ¢ = 8· 71' satisfies 8(u® v) = ¢(u, v), for any u EM,

v EN. But this equation is enough to characterize ¢; hence 8 = ¢.

It follows from this theorem that the universal property uniquely characterizes the tensor product. 3.2 COROLLARY. Let

1] :

M x N --+ U be a universal bilinear balanced

map. Then U ~ M ®s N as R-T-bimodules. PROOF: Since N

--+

71'

is a universal bilinear balanced map, there exists 1i ; M ®s

= 17' 71'. On the other hand, since "I is also universal, U --+ M ®s N such that 71' = 7i' • 1]. Thus 1] = (1i' 7i') . 7)

U such that

there exists 7i' :

1]

. Using again the universality of 1], this time with respect to itself, and its uniqueness, we get that 1i . 7i' =

~,

where

~

; U

--+

U is the identity map.

Similarly, 7i"1i is the identity on M ®s N. We must now consider what happens with maps when we take tensor products. Let M' be an R-S-bimodule and N' be an S- T-bimodule. Let

M

--+

M' and 'IjJ : N

the map 9; M x N

--+

--+

N' be two bimodule homomorphisms. Consider

M' ®s N' defined by 9(u,v) = ¢(u) ® 'IjJ(v). It

follows from the properties of the tensor product that this map is bilinear and balanced. Hence, by the universal property of bilinear balanced maps, there exists a bimodule homomorphism: (j: M ®s N --+ M' 0s N'.

12. Tensor products

113

This map is usually denoted by cP@ 'I/J . Note that

(cP@ 'IjJ)(u@ v)

= cP(u) @ 'I/J(V).

4. BASIC PROPERTIES. We now use the universal property to prove some of the basic properties of the tensor product of modules. We illustrate the method with some examples, and leave the others as exercises for the reader. 4.1 PROPOSITION. Let R, S be rings, and M be an R-S-bimodule. Then

R@RM

~ M~

M@sS.

PROOF: Let 8 be the map from R x M to M defined by 8(r, u)

=

ru. It

follows from the definition of an R-module that B is bilinear and balanced. By the universal property of the tensor product there exists a bimodule homomorphism 9 from R@RM to M. This map satisfies the equation 9(r@

u)

= ru.

It is clear that

9 is onto, we show that it

is injective. Since we are

tensoring up over R, we have that r @ u = 1 @ ru. Thus every element of

R@RM can be written in the form l@u, for some u E M. But 9(1 @u)

= u,

hence '8 must be injective. The proof of the other isomorphism is analogous and is left to the reader. The property that we now prove is the distributivity of the tensor product with respect to the direct sum. It will be used often in later chapters. 4.2 THEOREM. Let R, S, T be l'ings, and I be an index set. Suppose that

M is an R-S-bimodule and that Ni are S-T-bimodules, for every i E I. Then

PROOF: The elements of EBi Ni will be denoted by every i E I. Consider the map

8: M x ffiN i

--+

EBi Vi,

ffi(M@sN i )

where

Vi

E N; for

12. Tensor products

114 defined by 9(U,EBiVi)

= EBi(u0 Vi).

It is a bilinear and balanced map. By

the universal property of the tensor product, there exists a bimodule map

8:M which satisfies 8(UI8>

0s

(EB; Vi»

EB N;

-+

EB(M 0s N

i)

EB;(U0 Vi).

=

On the other hand, let /-Ij : Nj

->

EBi Ni be the natural embedding.

Thus

we have maps

By the universal property of direct sums [Cohn 64, p.311j, there exists a unique map

such that /-I (EBi(U 0 Vi»

= u0 (EBiVi). One now checks that

/-I and Bare

inverse to each other by a simple calculation on generators; hence both maps are isomorphisms. A similar result holds when the direct sum is the first entry of the tensor product. Thus the following corollary is an immediate consequence of Theorem 4.2. 4.3 COROLLARY. Let R be a commutative ring and II, I2 be index sets. Let

Fi be free R-modules with bases {v~ : a E Ii}, for i is free with basis {v~ 18> ~ :

Q'

E

= 1,2.

Then H 0R F2

Ibf3 E I 2 }. Furthermore, if FI and F2 have

finite bases, then

Let us now state a few more properties of the tensor product: the proofs follow the general pattern of 4.1 and 4.2. 4.4 PROPOSITION. Let Ri be rings and Mi be R;-R;+l-bimodules, for i 1,2,3. Then

=

12. Tensor products

l15

4.5 PROPOSITION. Let R be a commutative ring and let M and N be R-

modules. Then

We must also consider the effect of the tensor product on exact sequences. The first result holds in great generality.

4.6

THEOREM.

Let R, Sand T be rings and let rjJ

tI>

M' --; M --; Mil --; 0

be an exact sequence of S-T-bimodules. If B is an R-S-bimodule, then the

sequence B 0s M'

1®1/> ---+

B 0s M

l®tI> ---t

B 0s Mil --; 0

is exact. PROOF:

Since r/J is surjective, the elements of B 0s Mil can be written in the

form

where bi E Band

11; E

M. Since this is equal to

we conclude that 1 0 r/J is surjective. On the other hand, (1 0 r/J) . (1 01jJ) equals zero, since ker(r/J)

= im(1/I).

= (10 r/J. 1jJ)

Hence

im(10 1/1) <; ker(1 0 r/J) and the proof will be complete if we show that the opposite inclusion holds. Let N

= im(l 0

1/1). Then r/J induces a map, (): (B ('0s M)jN --; B 08 Mil

12. Tensor products

116 defined by B(b 0 u + N)

=

b 0 cP(u). Note that () is well-defined because

N ~ ker(l 0 cP). Now let 7T be the projection 7T: B 0s M One easily checks that (). 7T

(B0s M)/N.

-+

= 1 0 cPo

Suppose we have shown that () is an

isomorphism, then ker(l0
= ker«(}. 7T) =

= N = im(l 0 1/J), by definition.

ker(7T).

Thus the proof will be complete

if we show that () is an isomorphism. We do this by explicitly constructing an inverse. Define a map, {3: B x M"

by (3(b, v)

(B0s M)/N,

-+

= b0u+N, where ¢(u) = v.

Assume that ¢(u)

= ¢(u) = v, where

We must check that it is well-defined. u, 'Ii E M. Then
u- U E ker(¢)

= im(1/J).

b0u=.b0U

(modN)

= 0 and so

But we have that

and {3 is well-defined. On the other hand, {3 is clearly bilinear and S-balanced. By the universal property ofthe tensor product there exists a homomorphism of R-T-bimodules,

73 : B 08 M"

-+

(B 0s M)/N.

A straightforward calculation shows that () and {3 are inverse to each other.

Of course there is a version of Theorem 4.6 with B tensoring on the right, and we shall use it whenever necessary without any further comment. The same applies to the next result.

12. Tensor products 4.7 COROLLARY.

117

Let S be a K-algebra and R a subalgebra of S. Suppose

that I <;;; S is an S-R-bimodule and that M is a right S-module. Then

M 0s S/I

~

M/MI

S

->

as right R-modules. PROOF: Tensoring the sequence

I

/I

->

S/I

->

0 with Mover S, and

using Theorem 4.6, we obtain the exact sequence

M0sI But M 0s S

~

I®/I

--+

M0s S

->

M0S/I

->

O.

M by Proposition 4.1. The image of the composition of 10 B

with this isomorphism is MI. Note that this is a right R-submodule of M. Thus, M0sS/I~M/MI

as right R-modules. An important warning: injective maps need not be preserved by tensor products. Consequently, tensor products do not always preserve short exact sequences. An example is given in Exercise 6.6. See also Exercise 6.5.

5.

LOCALIZATION.

It is time for our first application of the tensor product to the Weyl algebras. In it we generalize a result proved in Ch.10, §3. Let p be a non-zero polynomial in K[X] and consider the subset K[X, p-I] of all rational functions whose denominator is a power of p. We have proved that this is a holonomic left An-module in Theorem 10.3.2. Let M be a K[X]-module and put

Since K[X] is commutative, this is a K[XJ-module. Note that every element I

of M[p-I] can be written in the form p-k 0 u, for some u E M and k 2: 0, but this form is not

uIliq\l(~.

This posps the question: when is

p-k

0

U

= O?

12. Tensor products

118

5.1 PROPOSITION. Let M be a K[X]-module. Suppose that u EM satisfies

10 u = 0 in M[p-l]. Then there exists an integer k ~ 0 such that '1u = o. PROOF:

There exist an index set I and a surjective map IJ

K[XjI -tM, where K[XJZ denotes the direct sum of copies of K[X] indexed by I: a free

K[X]-module. Let F be the kernel of this map. Tensoring by K[X, p-l] over K[X], we get an exact sequence, (5.2)

By Theorem 4.2,

It is easy to check that the image of the first map of (5.2) in K[X, p-l]I is

K[X, p-l]F, the submodule generated by F. Therefore

as K[X]-modules. Choose v E K[XjI such that fJ(v) (1 0 fJ)(l 0 v)

= u.

Thus

= 1 0 u = O.

Identifying 10 v with v in K[X,p-IV, we have that v E K[X,p-l]F. Hence there exists k ~ 0 such that

'Iv E F;

from which we conclude that

'1u = 0

inM. Now assume that M is a left An-module. Then M[p-l] is a K[X]-modulc, and we want to turn it into a left An-module. We do that by defining the action of 8; as follows

fJ; . (p-k 0 u)

= p-k-I ® (pB; . u -

kfJ;(p)u).

It is not immediately clear that this formula gives a well-defined action,

because an element of l\IJ[p-l) does not have a unique representation in the

12. Tensor products

119

form p-k @ u. However, applying Proposition 5.1 one can show that it is indeed well-defined; the details are left to the reader. It is also easy to check that the relations of An are preserved by this action. Hence M[p-l] is a left

An-module by Appendix 1. It is called the localization of M at p. The name comes from the way this construction is used in algebraic geometry.

If we also assume that M has a good filtration

r, then we can construct a

filtration {} for M[p-l]. If m is the degree of p, put

(5.3) To check that this is a filtration of M[p-l], proceed as in the proof of Theorem 10.3.2.

5.4 THEOREM. If M is a holonomic An-module, then so is M[p-lj. PROOF: We have that

Let x( t, M, r) be the Hilbert polynomial of r. Then for k

»

0,

dimK (}k ~ x(k(m+ 1), M, r) a polynomial of degree n. Hence M[p-l] is holonomic by Lemma 10.3.1.

6. EXERCISES

6.1 Let p, q E Z, and let m be their greatest common divisor. Show that

Zp@z Zq

~

Zm.

6.2 Prove the following isomorphisms of Z-modules:

(1)

Q@zQ~Q.

(2) Q @z (Q/Z)

~ (Q/Z) @z (Q/Z).

6.3 Let R be a ring. ShaWl that Rn @R R1n

~

6.4 Prove Propositions 4.4 and 4.5 in detail.

Rmn.

120

12. Tensor products

6.5 Let R be a K-algebra and let k

~

0 be an integer. Show that if N

--t

M

is an injective map of left R-modules, then so is

6.6 Let cP : Z2 by cP(l

--t

Z4 be the injective homomorphism of Z-modules defined

+ 2Z) = 2 + 4Z.

Show that:

(1) Z2 0z Z2 ~ Z2. (2) Z2 0z Z4 ~ Z2. (3) The map Z2

--t

Z2 induced by tensoring cP by Z2 over Z is identically

zero. Conclude that the tensor product by Z2 does not preserve injectivity. 6.7 Show that for any non-zero p E K[X], there exist An-module isomorphisms:

(1) K[X,p-l]0J\'[xj K[X] ~ K[X, P-l]i

(2) K[X, p-l]0I<[Xj K[BJ, . .. , on]

= O.

CHAPTER 13 EXTERNAL PRODUCTS In this chapter we discuss the simplest operation to be performed on modules over the Weyl algebra: the external product. We also discuss the computation of the dimensions of the external product from the dimension of its factors. In particular, we will show that the external product of holonomic modules is itself holonomic.

1. EXTERNAL PRODUCTS OF ALGEBRAS. We begin with a general definition. Let A,B be K -algebras. The tensor product A 0]( B is a K-vector space, on which we define a multiplication. For a, a' E A and b, 1J E B, let

(a 0 b)(a' 01J) = aa' 0 blJ. It is routine to check that A 0]( B with this product is a K-algebra. This is called the external product of A and B. Since we use this construction very often, it is convenient to have a special notation for it: A®B. If we apply the construction to polynomial rings or to the Weyl algebra' we do not get anything new: that is the secret of its power. The key result is the following theorem. 1.1 THEOREM. Let R be a K-algebl·a and let A and B be subalgebras of R.

Suppose that (1) R = AB,

(2) [A, B] = 0, (3) thel·e exist K-bases {a; : i E N} and {bJ, : i E N} of A and B, respectively, such that {aibj : i, i E N} is a K -basis of R. Then R

8;:

A®B.

PROOF: Define a map r/J ; ~ x B

->

R by r/J( a, b) == abo It is clearly K -bilinear

and K-balanced. Therefore, there exists a map rp; A 01( B sli(a @ b) = abo This is called the 'I1'w.1Uplic.ation map.

->

R such that

13. External products

122

Let a, a! E A and b, If E B. Then

!Ii«a 18) b)(a'

@

IJ» = !Ii(aa'

@

bY) = aa'bli.

On the other hand,

!Ii(a@ b)!li(a' Thus (2) implies that

iJj

@

Ii)

= aba'Y.

is a K-algebra homomorphism.

Since, from (1), every element of R is a linear combination of monomials

ab, with a E A and bE B, we conclude that !Ii is surjective. The injectivity is a consequence of (3) and Corollary 12.4.3. A carefully chosen notation makes the application of Theorem 1.1 to polynomial rings and the Weyl algebra almost tautological. Our choice is the following. Let K[X]

= K[Xb ... ,x n] and

K[Y]

= K[Yl, ... , Ym]

be polyno-

mial rings. Write K[X, Y] for the polynomial ring on the x's and y's. Then

An will be the Weyl algebra generated by the x's and ax's, and Am the Weyl algebra generated by the y's and ay's. Both are subalgebras of Am+n> the Weyl algebra generated by the x's, y's and their derivatives. We shall retain this notation for the rest of the chapter. 1.2 COROLLARY. The following isomorphisms are induced by the multiplication map:

(1) K[X]®K[Y] ~ K[X, Y]. (2) Am®An ~ Am+n' The isomorphisms defined in the corollary are so natural that we shall take them to be equalities. Thus, from now on, we shall use without further comment that Am®An = Am+n' In other words if a E An and bEAm, we will identify the monomial ab with a @ b. Similarly for polynomial rings. 2.

EXTERNAL PRODUCTS OF MODULES.

Once again we turn to the general situation. Let A,B be K -algebras. Suppose that M is a left A-module and that N is a left B-module. Then we may turn the K-vector space M

18)K

N into a left (A®B)-module. The

action of a 18) b E Ai§B on u@ v EM 18)K N is given by the formula (a@b)(u0 'u) = atL@bv.

13. External products

It is routine to check that M

(9K

123

N is a module for this action. We shall

write M®N for this (A®B)-module. 2.1 PROPOSITION. If M and N are finitely generated modules, then so is M®N. PROOF:

Suppose that M is generated by Ut, ... , Us over A and that N is

generated over B by 'Ut, •.• , Vt. The elements of M®N are K-linear combinations of elements of the form u (9 v, for u EM, v E N. But u and v =

Ei bj Vj.

= E~ aiUj

Thus, u(9 v

= ~)a; (9 bj)(Uj (9 Vj). iJ

Hence M®N is generated over A®B by Uj (9 Vj, for 1 SiS sand 1 S j S t. Now let M be a left Am-module and N a left An-module. The external product M0N is a left Am+ n-module, under the convention that Am+n

=

Am0An. This will be used so often, that even at the risk of being pedantic,

we write the action explicitly. Let u(9 v E M®N. A monomial of Am+n can be written, in only one way, in the form ab with a E Am and bEAn. Thus, (ab)(u(9 v)

= au (9 bv.

The situation for polynomial rings is entirely similar. The next lemma will be very useful in calculations. We make use of these conventions in its statement and proof. 2.2 LEMMA. Let I be a left ideal of Am and J a left ideal of An. Denote by Am+nI + Am+nJ the left ideal of Am+n genel'ated by the elements of I and J. Then

as Am+n-modules. PROOF:

Since every monomial in Am+n may be written, uniquely, in the

form ab, for a E Am, bEAn, there exists a K-linear map,

13. External products

124 given by 'IjJ(ab)

=

(a + I) (/9 (b+ J). An easy calculation shows that this is an

Am+n-module homomorphism. It is clearly surjective. On the other hand, a monomial ab with a E lor bE J, gives 'IjJ(ab)

=

O. Hence Am+nI + Am+nJ ~

ker'IjJ . By the universal property of the tensor product, there is a K -linear map,

defined by c/J((a+ 1) (/9 (b+ J»

= ab+ Am+nI + Am+nJ.

This is a surjective

map, and

Hence Am+nI + Am+nJ

= ker 'IjJ and

'IjJ is an isomorphism of Am+n-modules.

Although we have not used in the proof of 2.2 that c/J is a homomorphism of Am+n-modules, this is easily shown to be true.

2.3

COROLLARY.

Let I be a left Am-module. Then

is an isomOl'phism of Am+n-An-bimodules. PROOF:

It follows from Lemma 2.2 that there exists an isomorphism of left

Am+n-modules,

given by 'IjJ(ab + Am+nI) = (a since I

~

+ I) (/9 b,

where a E Am, bEAn. Note that

Am, the elements of I commute with the elements of An. Hence

Am+nl Am+nI is a right An-module. So is (Ami I)0An, with An acting on the right component of the tensor product. Let c E An. Then

tf;((ab)c + A m+n1)

= (a + 1) (/9 bc = tf;(ab + A'n+n1)c.

Therefore, 'IjJ is also a right An-module isomorphism.

3.

GRADUATIONS AND FILTRATIONS.

We will now construct a good filtration for the external product of modules over the Weyl algebra and calculate its associated graded module. Throughout this section, let M be a finitely generated left Am-module with good

13. External products

filtration reM)

= {r;(M) :

125

i E N} and let N be a finitely generated left

An-module with good filtration rCN). We will proceed through a series of lemmas. 3.1 LEMMA. The Bernstein filtration of Am+n satisfies

L

BtCA m+n) =

Bp(Am)Bq(An).

p+q=t PROOF:

First note that

Now every monomial in Bt(Am+n) may be written as a product ab, with a E

Am and bEAn. Assume that a, b have degrees p, q, respectively. Since ab has degree p+q, it follows that ab E Bp(Am)Bq(An). But an element of Bt(Am+n) is a sum of such monomials. Therefore, Bt(Am+n) 3.2 LEMMA.

= L.:p+q=tBp(Arn)Bq(An).

The K -vector spaces n(M0N)

=

L r;(M) \Of{ rj(N)

;+J=k

form a good filtration of M0N as an Am+n-module. Note that we are identifying ri(M) \OK Tj(N) with a subspace of M0N, in the obvious way. PROOF:

Since the summation in the definition of n(M0N) is finite, we have

by Corollary 12.4.3 that nCM0N) is a finite dimensional K-vector space. Now every element of M0N is a finite linear combination of elements of the form u\O v, with u EM, v E N. Assume that u E T;CM) and v E Tj(N). Then

u\O v E Ti(M) \0 Tj(N) ~ ri+j(M0N). Hence, M0N = Uk~O Tk(M0N). Finally, since

13. External products

126

we conclude, using Lemma 3.1, that B t(A",+n)rk(M'0N)

= n+t for t» o.

Thus n(M'0N) is a good filtration of M'0N. We will now calculate the graded module with respect to this good filtration. The ring Sn

= g".8 An is a polynomial ring by Theorem 7.3.1.

Using the

conventions of §1 we may identify Sm'0Sn with Sm+n' Denote the homogeneous component of degree t of Sn by Sn(t). By an argument similar to that used in Lemma 3.1, we have that Sn+",(t) = EBp-tq=tSp(Am)Sq(An). Finally, write gr,.(M)

= rk(M)/ n-l (M).

3.3 LEMMA. Thel'e is an isomOl'phism of K -vector spaces

grk(M'0N)

9!

EB gri(M) 0]( gTj(N). (+j=k

PROOF: Consider first the K-linear map,

defined by 1)ij(u0v)

= J1.i(U) 0J1.j (v), where we are denoting the symbol maps

of both M and N by J1.. It is easy to see that 1);j is surjective and that its kernel is the subspace

Putting these maps together, we get a linear map

i+j=k

i+j=k

Note that the canonical projection gives a map

E9 (r;(M) 0 rj(N» i+j=k But ker 1)ij ~ ri+j-l (M'0N), thus

--) n(M'0N).

127

13. External products factors through

n(M0N)/Fk_l(M0N) --;

EB (grj(M) ®grj(N». i+j=k

Since the former map is bijective, so is the latter, which gives the required isomorphism. Taking direct sums for k

~

0, and using Lemma 3.3, we conclude that

there exists an isomorphism of K-vector spaces,

B : gr(M0N) --; gr(M)09r(N). Since M0N is an Am+n-module, gr(M0N) is an Sm+n-module. On the other hand, since Sm+n

= Sm0Sn, it follows that gr(M)0gr(N) is an Sm+n-module.

3.4 THEOREM. The lineal' map B is an isomorphism of Sm+n-modules. PROOF: We have only to check that B is compatible with the action of Sm+n. Choose a monomial of Srn+n, and write it ill the form

f g, with f E Sm(P),

9 E Sn(q). There exist operators a, b E Am+no such that

f

= O"p(a) and 9 = O"q(b).

In particular, a E Am and bEAn. Now let i

+ ;j = k.

U

® v E rj(M) ® Tj(N), with

Then

(jg)J.'k(u0 v) = J.'k+p+q«ab)(u0 v». But (ab)(u0 v)

= au0 bv; therefore

The latter is mapped onto J.'i+p(au) 0 J.'j+q(bv) by 8. Since this may be rewritten as (fg)(p,j(u) ®/-lj(v», we have

which completes the proof. 4. DIMENSIONS

AND MULTIPLICITIES

(

We are now ready to calculate the dimension and multiplicity of an external product.

128

13. External products

4.1 THEOREM. Let M be a finitely generated left Am-module and N a finitely generated left An-module. Then

(1) d(M@N)

= d(M) + d(N),

(2) m(M@N) ~ m(M)m(N). PROOF: We retain the notation of §3. It follows from Lemma 3.3 that k

dimKr,.(M@N)

=L

L dimKgri(M)dimKgrj(N).

r=O i+j=r

Therefore, dim](n(M@N) ~

k

k

;=0

;"=0

L dim](gri(M) L dimKgrj(N).

From this inequality and Lemma 3.2 we have that

Assuming that k

»

0, and using the Hilbert polynomials of the corresponding

filtrations we get that X(k, M@N) ~ X(k, M)X(k, N) ~ X(2k, M@N).

Since this holds for all large values of k, both (1) and (2) immediately follow. The corollary is an easy consequence of the theorem.

4.2 COROLLARY. Let M be a llOlonomic Am-module and N a holonomic An-module. Then M@N is a holonomic Am+n-module.

5. EXERCISES 5.1 Let Bn be the ring of differential operators of the rational function field

K(X)

= K(xJ,""

xn). Is it true that Bn@B m ~ Bn+m as K-algebras?

5.2 Show that the multiplication map induces an isomorphism of Am+nmodules, K[X,

11

~ K[X)&K[11.

13. External p1'oducts

129

5.3 Let p E K[X] and q E K[ Y]. Show that there exists an isomorphism of Am+n-modules,

5.4 Let

(7, (7'

be automorphisms of Am and An, respectively. Let M be a left

Am-module and N be a left An-module. (1) Show that there exists a unique automorphism 8 of Am+n whose restriction to Am is

0'

and whose restriction to An is

0".

(2) Show that

5.5 Let M be a left Am-module and N a left An-module. Show that the Fourier Thansform satisfies

5.6 Let M be a holonomic Am-module and N a holonomic An-module. Show that if M and N have multiplicity 1, then M0N is a simple holonomic Am+n-module. 5.7 In this exercise we use the results of eh. 11. Let M be a left Am-module and N a left An-module. Show that Ch(M0N) 5.8 Let 1

sis n.

= Ch(M) x Ch(N).

Suppose that d; is an operator of An of degree

is a linear combination of monomials in

Xi

~

1 which

and 8;. Let J be the left ideal

generated by d), . .. ,dn. Show that Ani J is a holonomic module over An. Hint: An/J ~ (AdA 1d1 )0 ... 0(AJ/A 1 dn).

CHAPTER 14 INVERSE IMAGES In this chapter we show how to change the ring of scalars of a modulll,

over a Weyl algebra. This is very important and leads to the construction of many new modules. As we shall see later, holonomic modules are preserved under this construction. 1. CHANGE OF RINGS

We start with a general construction for rings and modules. Let R, S be rings, and let <jJ : R

-+

S be a ring homomorphism. If M is a left R-module,

then we may use ¢ to turn it into a left S-module. This is called changing the base ring. The key to the construction is that S may be considered as

a right R-module in the following way. Let r E R, s E S; the right action of R on S is defined by S*

r

= s<jJ(r).

The juxtaposition on the right hand side of the equation denotes multiplication in the ring S. With this in mind, we may consider S as an S-R-bimodule. We are now allowed to t!lke the tensor product S

@R

M which is a left S-module. Thus

starting with the left R-module M we have constructed the left S-module S

@R

M. If it is necessary to call attention to the homomorphism ¢, we will

write S@¢M. As an application, let us rephrase the twisting construction of Oh. 5, §2 in terms of change of rings. 1.1 LEMMA. Let R be a ring and u an automorphism of R. Let M be a left

R-module. Then

PROOF:

Let a E R. The left action on M" is given by a. u

= u(a)u, where

u EM,,; and the right action on R by s*a = su- (a), where s E R. Consider 1

the map rjJ:RxM-M"

14. Inverse images defined by
= a _ u.

Since


131

= a_ bu = (u(a)b)u

=
is also bilin-

ear. Thus, by the universal property of the tensor product, there exists a homomorphism

Since M"

=M

as abelian groups, the map

¢ is surjective.

Let v E R@,,-l M. There exists u EM such that v = 1 @ u. Hence

O=¢(v)=u, which implies that u

= O. Thus v = 0 and ¢ is injective. Therefore, ¢ is an

isomorphism of R- modules. We shall now apply the change of rings construction to polynomial maps. We often use the results of Ch. 4, §1 in the coming chapters, so you may wish to read that section again before you proceed. A good notation will be of great help: we shall abide by the conventions we are about to make right to the end of the book.

1.2

NOTATION.

Put X

=

Kn. The polynomial ring K[XI, ... ,xnl will be.

denoted by K [X]; and the Weyl algebra genel'ated by tile x's and

a" 's

by

An. The n-tuple (Xl, ... ,x n) will be denoted by X. Similar conventions will hold for Y

= Km

and Z

= K r,

with polynomial rings K[ 11 and K[Z] and

Weyl algebras Am and A r . The space Km+n equals the cartesian product of X and Y and will be denoted by X x Y. FOl' the polynomial l'ing in the x's and y's we shall write K[X,

11.

The (n+ m)-th Weyl algebra Am+n will

always denote the algebra genel'ated by K[X,

11, a", 's and Oy's.

Note that

with these conventions Am+n = Am0An. Although these conventions may cause some initial confusion, they payoff later on by making the formulae easier to digest than they would otherwise be. I

Let F: X - t Ybe a polynomial map. Its comorphism FI : K[11-> K[X] is an algebra homomorphism. If M is a K[ YJ-module, we may use FI to

14. Inve1'se images

132

construct the K[X]-module K[X]0[([y) M. It is called the inverse image of

M by the polynomial map F, and denoted by F* M. Let us consider some examples. Let F : X

--l

X be a polynomial isomorphism, with inverse C. The

comorphism Fd is an automorphism of K[X] with inverse Cd. Let M be a

K[X]-mod ule. By Lemma 1.1, its inverse image F* M is isomorphic to MG'. Consequently, if u E F* M and hE K[X], then hu = C~(h)u. The second example is the projection 7r : X x Y coordinate: 7r(X, y)

--l

Yonto the second

= Y The comorphism 7rd maps a polynomial p E K[ Y]

to itself, but considered as an element of K[X, Y]. Let M be a module over

K[n then 7r* M

= K(X, Y] 0[([11 M.

A monomial in K[X, Y] may be written in the form pq, with p E K[X] and q E K[Y]. If u EM then

pq0u= p@qu as elements of 7r* M. Using this identity we may construct an isomorphism, 7r* M ~ K[X]0M, of K[X, Y]-modules. 2. INVERSE IMAGES. Let F : X

Y

= Km,

--+

Y be a polynomial map and M a left Am-module. Since

it follows that M is in particular a K[ Y]-module. Thus we may

compute the inverse image of M by F defined in §l:

F*(M)

=

K[X]0[([y) M

is a module over K[X]. We want to make this module into an An-module. At first sight this construction may seem very puzzling. Why do change of rings as if we had only a module over a polynomial ring and then turn it back into a module over a Weyl algebra? Why not do the change of rings at the Weyl algebra level? First of all, one is allowed to do change of rings at the level ofWcyl

alg{~bras,

since t.he construction works for any ring. However, in

14. Inverse images

133

doing that one is limited by the fact that Weyl algebras are simple rings; and homomorphisms of simple rings are necessarily injective. A more satisfactory answer is that this is really a construction in algebraic geometry, that we are lifting to the realm of noncommutat ive rings; see [Hartshorne 77, Ch. 2, §5]. Anyhow, the efficacy of the construction is its best justification, as we shall see later on. We know how a polynomial of K[X] acts on F*(M). We now give the recipe for the action of 0"". Let

q@uE F*(M), where q E K[X] and u EM. Let F l , ... , Fm be the coordinate functions of F. The action of ox; is defined by m

(2.1)

ox;(q@ u)

= ox.(q) @ u+ L qox, (Fk) @ oYk u k=l

for i

= I, ... , n.

The x's and ox's generate An, and we know how they act on F*(M). According to Appendix I, these formulae will make an An-module of F*(M) if they are compatible with the relations satisfied by the generators. The relations are

[O"'i,Xi] = 8ij l, [Xi,Xj]

= [OX"OXi] = 0,

for 1 S i, j S n. Let us carefully check that the first of these relations is compatible with the actiolls of x's and ox's defined above. Let wE F*(M). We want to show that

Clearly it is enough to check this when w is of the form q@ u, for q E K[X] and u E M. Let us begin by calculating ox) (Xi(q applying (2.1) to Xjq @ u, which gives

@

u)). We do this by

m

(2.2)

8xj (Xjq)

@

u+

L xiq8

xi

(Fk) ® Oy, u.

14. Inverse images

134

Using Leibniz's rule and the fact that OXj(Xi) = 6i:J" we may rewrite (2.2) in the form

Note that the expression in brackets equals OXj(q 0 u). Thus we have shown that

But this is equivalent to

which is what we wanted to prove. The proof that the other relations are compatible with these actions is an exercise for the reader. We will calculate some concrete examples in the next section. 3. PROJECTIONS.

Let us return to the projection 7r : X x Y --; Yof §l. Let M be a left Am-module. We have already seen that, as a K[X, l1-module, the inverse image 7r* M is isomorphic to K[X]0M. However, K[X] is an An-module and

M is an Am-module. So K[X]0M has a natural Am+n-module structure. We want to show that this structure coincides with the one determined by (2.1).

It is enough to check how

ox; and oy,

act on K[X]0M. For this we return

to the definitions. The K[X, l1-module isomorphism between 7r*(M} and K[X]0M maps qaxO 0 U to x a 0 qou, where a E Nn and qa E K[n We must show that this is compatible with the action of the derivatives. Consider first the action of ow' as defined by (2.1), m

Ow (qaxO0 u) = Ow (qa x O) 0 u+ L(OW(Yk)qa x O0 o!J>u}. I

14. Inverse images

135

Using Leibniz's rule, this may be rewritten as

ay; (q"x" 18> u) = x" 18> ay; (q"u). Summing up: if we identify n*(M) with K[X]0M, then

a!h

acts only on M,

in the usual way. Consider now the action of ax. . From the definition, we have that m

ax;(q"x'" 18> u)

= ax;(q"x"') 18> u + E ax;(Yk)q"x'" 18> a

Zk

u.

1

Since az;(Yk)

= ox,(q,,) = 0,

for all k and all ex, we end up with

ax;(q"x'" 18> u) = q",ox;(x") 18> u. Thus, under the previous identification, Ox, acts only on K[X]. We may sum up our calculations as follows. If the module 7r*(M) is identified with K[X]0M then the x's and o,,'s act only on the first factor K[XJ, and the y's and ay's act only on the second factor M. Besides, the actions are the natural ones. We may use this description to calculate the dimension of an inverse image. 3.1 THEOREM. Let M be a finitely generated left Am-module and 7r the

projection defined above. Then:

(1) 7r* M is a finitely generated A'n+n-module. (2) d(7r*M)=n+d(M). (3) m(7r*M)Sm(M). PROOF: (1) follows from Proposition 13.2.1; whilst (2) and (3) are consequences of Theorem 13.4.1 since, as an An-module, K[X] has dimension n and multiplicity 1; see Ch. 9, §2. 3.2 COROLLARY. Let M be a holonomic Am-module. Then 7r* M is a holo-

nomic Am+n-module. 4. EXERCISES 4.1 Let R, S, T be rings and cp: R

-+

Sand 'IjJ: S

phisms. If M is a left R-n\'odu]e. show that

--+

T be ring homomor-

136

14. Inverse images

4.2 Let F : K

---t

K be the polynomial map defined by F(x)

= x2•

Let M

be the algebra Al considered as a left module over itself. Show that F* M is not finitely generated over AI. 4.3 Let u be an automorphism of An. Suppose that the restriction B of (T to K[X] is an automorphism of this ring. Put F

=

B~

for the polynomial map

determined by B. Let M be a left An-module. Is it true that

as An-modules?

4.4 Let M and N be left An-modules. In particular these are modules over K[X). Hence the tensor product M 0K[X] N is a well-defined K[X]-module.

Define the action of 8x ; on u0 v E M 0J([x] N by the formula

8x ;(u0 v)

= 8x ;u0 v+ u0 8x ;v.

(1) Proceeding as in §2, show that this action makes M

01qX]

N into an

An-module.

(2) Give an example of two finitely generated An-modules M and N such that M 0K[X] N is not finitely generated over An.

4.5 Let F : C

---t

C be the map defined by y = F(x) = x 7n , where m

~

2 is

an integer. Let 6 be the Dirac microfunction; see Ch. 6, §3. The purpose of this exercise is to show that the inverse image F*(AI «(;)6) is isomorphic to the Al (C)-module generated by

(1) Show that x 7n and x8"

6'n, the

+m

m-th derivative of 6.

annihilate 1 06.

(2) Show by induction that there exist non-zero complex numbers cpq such that

for q = 0, 1, ... , m. (3) Using (2) show that

1 ® 8;;+16

= 2-8~rH)m(1 ® 6). c1nn

14. Inverse images

137

(4) Conclude from (3) that 1 08 generates F"(A l (C)8). (5) By (1), (3) and Exercise 6.4.10, F"(Al(C)8) is a homomorphic image of Al (C)8 m . Since the latter is irreducible, we have the desired isomorphism.

In the language of microfunctions we have proved that the inverse image of 8 under F is 8m .

CHAPTER 15

EMBED DINGS In the previous chapter we defined the inverse image of a module and calculated a formula for its dimension and multiplicity under projections. In this chapter we turn to embeddings. As we shall see, the behaviour of the inverse image under embeddings is a lot less regular than under projections. 1. THE STANDARD EMBEDDING

In this chapter we shall retain the notation of 14.1.2. We begin by considering the embedding

£ :

X

~

X x Y defined by LeX)

= (X, 0), where 0

denotes the origin of Y. We will call £ the standard embedding. The comorphism £1 : K[X, J1

---t

K[X) is defined by LI(g(X, Y))

= g(X, 0).

It may be

used to make K[X] into a K[X, J1-modulej as such, K[X] is generated by 1, since

g(X, Y) ·1

= £~(g(X,

Y)) . 1 = g(X, 0).

The annihilator of 1 in K[X, J1 is the ideal generated by Yt, ... , Yrn' Denoting this ideal by (Y) we have that K[X] ~ K[X, J1/(Y) as K[X, J1-modules. ~

Now consider a left Am+n-module M. Since An

A m+n, the Am+n-module

M is also an An-module. The elements of An commute with the variables Yt, ... , Yrn, thus (Y)M is an An-submodule of M. Hence M / ( Y)M is an An-

module. Note however that it is not an Am+n-submodule.We want to show that

£7M

= K[X] @K[X,y) M

~

M/(Y)M

as An-modules.

If u EM, let

u denote its image in M / ( Y)M.

Define the map

¢: £*M---tM/(Y)M' by ¢(q@u) = qu. It is a homomorphism of K[X]-modules. We want to show that the action of the derivations OXI' ... , that

oz. is compatible with ¢.

8q ~ OXk. 8z ,(q@u)=-8@u+L..,[email protected]. Xi

k=l

Xi

We have

139

15. Embeddings and so

8q 8x ,.(q® u) = 8x; ® u+ q® 8Xi u. The right hand side is mapped by r/J onto

8q_ -8 u+q8zi u, Xi

which is equal to

8Xi (qu)

= 8x ,.(r/J(q ® u)).

Thus r/J is an isomorphism of An-modules. Summing up: ~*M ~ Mj(Y)M as An-modules. Since M/(y)M is a quotient of An-modules, the action of x's and 8's is the natural one. Let us specialize this construction to the standard embedding ~ : K

--l

K2,

which takes x to (x,O). Let M be the ring A2 itself, considered as a left A2module. Applying the above results, we have that

~* M

taken as a left AI-module. Note that although M

= A2 is a finitely generated

left A2-module, the module

L* M

is the quotient A2/ yA2

is not finitely generated. In fact,

~* M

is a

free AI-module whose basis is the image of the powers of 8y in A2/ yA 2. Since this basis is infinite,

~* M

is not a finitely generated left AI-module.

We conclude from this example that the inverse image under embed dings does not preserve the noetherian property. This is very important, since we have defined dimensions only for noetherian modules. In particular there can be no general formula for the dimension of an inverse image under an embedding. Despite this, holonomic modules are preserved by inverse images under embeddings. This mystifying fact will have to wait until Oh. 18 for an explanation. It will be necessary, for future purposes, to consider a more general kind

of embedding, defined as a composition of the standard embedding and a polynomial isomorphism. Before we do this we must study the behaviour of the inverse image under composition of polynomial maps. 2.

COMPOSITION.

Let us first consider what happens to the change of rings construction of Cll. 14, §1 under composition. Let R, Sand T be rings and
->

S

15. Embeddings

140

and 1/J : S

---+ T

be ring homomorphisms. Let M be a left R-module. We

may turn M into a T-module in two different ways. On the one hand, we may apply the change of rings construction twice to get T®", S ®¢ M. On the other hand, the ring T is a right R-module via the homomorphism 'l/JcP , therefore T0",¢ M is a T-module, where

t® ru = t'I/JcP(r) ® u for t E T, r E Rand u E M. Using Propositions 4.1 and 4.4 of

eh. 12, we

conclude that these T-modules are isomorphic; thus,

It is not difficult to give a direct proof of this isomorphism. The T-module

T 0", S 0¢ M is generated by elements of the form t 0 s ®

U,

where t E T,

s E Sand u E M. But

to s 0 u = t'I/J(s) ® 10 u. Define a T-module homomorphism

by O(t0 s ® u)

= t'IjJ(s) 0

u. Note that if r E R then the element

t ® 1 0 ru

= t1/JcfJ( r) ® 1 0

u

is mapped to t'IjJcP( r) 0 u. It is easy to construct an explicit inverse for 0; hence it is an isomorphism. We will now apply this to An-modules. Recall that by Theorem 4.1.1, if

F: X

---+

Yand G: Y ---+ Z are polynomial maps, then (GF)~

= FaQII.

Note

the change in the order of the maps. 2.1

THEOREM.

Let F : X

---+

Yand G: Y

---+

M be an Ar-module. TlJeIl F'G*M

~

(GF)*M

Z be polynomial maps and

15. Embeddings

141

as An-modules.

By definition,

PROOF:

(CF)* M = K[X]0K[Z] M.

In this formula K[X] is a right K[Z]-module. Recall that the action of

f

E K[Z] on hE K[X] is given by h(J . C· F). On the other hand, F*G*M

= K[X]0ICfY) K[t]0K[Z] M.

We have already seen that these two modules are isomorphic as K[X]modules. The isomorphism B is defined by B(h 0 g 0 u)

= h(g . F) 0

u.

We must check that B is compatible with the action of the derivatives

ax.

By

definition,

az,(h 0 g0

u)

=

ah 0 (g 0 u) + hL (aF a-: ~ 0 a!/j(g 0 u) ) In

j=1

X,

X,

a

where F I , ... , Fm are the coordinate fUllctions of F. Replacing w(g 0 u) by its formula, the term

becomes

Applying B to this expression:

(a9 L (aF 8x, 8YJ In

_J.

-.'

F ) 0 u)

j=1

m L r +L j=1 k=1

(

aF (aCk) F

(9' F) _ J ax,

-.'

aYJ

)

0 OZk U ,

which, by interchanging the summations and applying the chain rule, becomes

~(aFj ~ Ox . j=1

(a

I

g

~. F ) ® u )

,.1J

+~ ~ (g. F) k=1

(O(Ck'F) {). 0 8 Zk U) . X,

15, Embeddings

142

Substituting the last expression in a(8",;(h ® 9 ®

L:

8h,® u + h m-8 (8 F (g , F) -8 J 89 ( ,F8®)u) x, j=1 x, YJ

u», one obtains

+ h Lr (g , F) 8(G8" " F) x,

k=1

®

8'k U.

Another application of the chain rule and Leibniz's rule to the first two terms shows that this is equal to

and hence to 8:z;;{h(g· F) ® u). We have thus proved that

Thus () is an isomorphism of An-modules. Applying this theorem to polynomial isomorphisms we get the following corollary.

2.2 COROLLARY. Let F ; X --; X be a polynomial isomorphism and G its

inverse. If M is a left An-module, then F* G* M

~

M.

3. EMBED DINGS REVISITED.

In this section we study a more general type of embedding. Let F : X --; Y be a polynomial map. Define a new polynomial map j ; X j(X)

= (X, F(X».

-->

X x Y by

This is an injective map, as one easily checks.

We shall write j as a composition of two maps, as follows. Let

~

:X

--l

Y be the standard embedding of §1. Let G : X x Y --; X X Y be the polynomial map defined by G(X, Y) = (X, Y + F(X». Then G is bijective X

X

andi=G'L Now let M be a left Am+n-module. We will calculate j* M. By Theorem 2.1, it equals

~*G* M.

Let us compute G* M. First of all, by Corollary 1.3.2,

there is an automorphism 0' of Am+n which maps

Xi

and satisfies

O'(Yj) = Yj - Fj(X), n

0'(8,,><>

= 8",; + E (8Fj/8xi) 8'Jj' j-1

and 8'Jj to themselves

143

15. Embeddings

Note that u restricts to an automorphism of K[X,

11 which we will also call

u. This automorphism is the inverse of the comorphism

G. By Lemma

14.1.1, G*M~M(7

as a K[X, l1-module. Let 'IjJ stand for this isomorphism; it satisfies 'if;(h@u) u(h)u, for hE K[X,

11 and

=

u E M.

3.1 THEOREM. Let M be a left A'n+n-module. Then G* M

~

M(7 as A m + n -

modules.

PROOF: We know that the isomorphism holds for K[X, l1-modules. Let us investigate the behaviour of the action of 8",. under 'if;. By definition, 8",(h @ u) equals

Applying 'if; to this formula, we get n

(3.2)

u(8x ;(h)u+ u(h)8,,;u+

L u(h8F /8x.j)81J) u. j

Since u leaves the elements of K[X] unchanged, we have that u(Fj)

= Fj .

Thus (3.2) is equal to u(8",(h»u+ u(h)u(8,,,)u.

But [8"" h]

= 8x ,(h).

Hence

from which we deduce that 'if;(8z ;(h@ u» = u(8"Ju(h)u.

Thus

as required. One may similarly check that the action of 8!h' is compatible I with 'if;. The next corollary is a combination of Theorem 3.1 and Corollary 9.2.4.

15. Embeddings

144

3.3 COROLLARY. Let M be a E.nitely generated left Am+n-module. Then

G* M and M have the same dimension. We will put all this together in a theorem. 3.4. Let M be a left Am+n-module. Then

THEOREM

i*M~M"/(Y)M"

as An-modules. As we have seen, j*M

PROOF:

~

£*G*M. By Theorem 3.1, G*M

~

M" .

By §1,

as claimed. 4. EXERCISES

K

K2 be the standard embedding. Compute the inverse image

4.1 Let

£ :

under

of the following A2-rnodules.

£

---+

(1) A2/A2fh

(2) K[XbX2J

(3) A2/A2x2 (4) A2/A2X2~ (5) A21A2~ 4.2 Which of the inverse images of Exercise 4.1 are finitely generated over AI?

4.3 Let

£ :

X --; X x K be the standard embedding, and denote by y the

coordinate of K. Let An+! be the Weyl algebra generated by x's and ax's and by y and Oy. Suppose that the term of highest order of dEAn+! is 8~. Show that L*(An+I/An+Id) is a free module of rank k over An4.4 Keep the notations of Exercise 4.3. Suppose that I is a left ideal of An+1

which properly contains An+! d. Show that £*(An+!/1) is finitely generated over An.

145

15. Ernbeddings

4.5 Let F : X -+ X x X be the polynomial map defined by F(X)

= (X, X).

This is a generalized embedding in the sense of §3. Let M and N be left modules over An. Then M0N is a left module over A 2n = An0An. Show that

F"(M0M) S!! M

@K[X)

N.

The latter was defined in Exercise 14.4.4. 4.6 Show that if £ : X -+ X x Y is the standard embedding, then

where K[oy] is the left Am-module K[oy" ... ,oYm].

CHAPTER 16 DIRECT IMAGES In the previous chapter we saw that starting with an Am-module and a

polynomial map F : Kn

-t

Km we can construct an An-module, its inverse

image. We shall now study a similar construction, which uses F to associate an Am-module to an An-module, its direct image. Curiously, the direct image is easier to define for right modules; and it is with them that we start. Throughout this chapter, the conventions of 14.1.2 remain in force. 1. RIGHT MODULES Let us briefly recall the definition of the inverse image. Let F : X

-t

Y

be a polynomial map. Let M be a left Am-module. The inverse image of M under F is F" M = K[X]

@]([Y]

M. This is a K[X]-module. It becomes an

An-module with the OX; acting according to the formula

oZi(h@u)

oh

m

Xi

1

op.

= {) @u+ Eh{/- @o'JjU' X;

Let us rewrite this definition in a slightly different way. Since

Am@A .. M 9!

M, we have that

Writing DX-tY for F* Am

= K[X] @K[y] Am,

one has that

Note that D x -+ Y is a left An-module and a right Am-module. One easily checks, that these two structures are compatible and that it is in fact an An-Am-bimodule. The bimodule D

X-+Y

is the key to the direct image construction. Let N

be a right An-module. The tensor product

16. Direct images

147

is a right Am-module, which is called the direct image of N under the polynomial map F. Consider the projection

11' :

X

X

Y -> Y defined by 1I'(X, Y)

= Y.

From

Ch.14, §3, it follows that

As a left An-module, K[X] is isomorphic to Ani L~ An8"". Hence, by Corollary 13.2.3, n

11'* Am

~ Am+nl

L A m+n8"" 1

is an isomorphism of Am+n-Am-bimodules. Now if N is a right Arn+n-module, then by Corollary 12.4.7,

as right Am-modules.

2. TRANSPOSITION We will now see how one can turn right modules into left modules, and vice versa. That will allow us to do direct images for left modules. We present the construction for general algebras. Let R be a K -algebra. A tmnsposition of R is an isomorphism r : R

->

R

of the underlying K -vector space that satisfies

(1) r(ab) = r(b)r(a), (2) -r2 = r. Examples oftranspositions abound. The most familiar occurs in the matrix ring over a field. An example in the Weyl algebra is provided by the map r : An -> An defined by

where h E K[xj, ... , xnJ and Q E !"in. We will refer to it as the standard tran.sposition of A". If dEAn, we also use the notation d for T(d). T

148

16. Direct images

Let us go back to the general construction. Let R be a K -algebra and

7'

a

transposition of R. If N is a right R-module then we define a left R-module

Nt as follows. As an abelian group, Nt = N. If a E Rand

U

E Nt then the

left action of a on U is defined by aou = U7'( a). The left action 0 is called the

transposed action. Instead of checking in detail that Nt is a left R-module, let us prove just one typical property, namely

(ab)

0

u = a 0 (bo u)

where a, b E Rand u EN. The left hand side is, by definition, u7'(ab). Since 7'

is a transposition, it equals u7'(b)7'(a). But this is ao (bo u). The same construction can be used to turn left modules into right modules.

Indeed, let M be a left R-module. The right R-module Mt equals M as an abelian group. The right action of a E R on u E M is defined by uoa Note that since

-? =

7' ,

= 7'( a) u.

the two constructions above are inverse to each

other. If N is a right R-module, then (Nt)t = Nj and the same holds for left modules. It is very easy to transpose the action when the module is cyclic. Let

R be a K-algebra with a transposition

7'.

If J is a left ideal of R, define

Jt = {7'(a) : a E J}. This is a right ideal of R. 2.1 PROPOSITION. Let R be a K -algebra with a transposition

7'

and J a left

ideal of R. The right R-module (R/ J) t is isomorphic to R/ P. PROOF:

If a E R, denote by

defined by ¢(a)

= 7'(a).

a its image in (R/ J)t.

Consider the map

It is clearly a homomorphism of additive groups. If

a, b E R, then

¢(ab) which equals 7'(b)7'(a)

= ¢(a) 0 b.

= 7'(ab),

Thus ¢ is a homomorphism of right R-

modules. One can easily check that it is surjective; let us calculate its kernel. Suppose that a E R satisfies
16. Direct images

149

conclude that ker