Table Of ContentGRADED COMPONENTS OF LOCAL COHOMOLOGY
MODULES
7
1
0 TONY.J.PUTHENPURAKAL
2
Dedicated to Prof. Gennady Lyubeznik
b
e
F Abstract. Let A be a regular ring containing a field of characteristic zero
5 and let R=A[X1,...,Xm]. Consider R as standard graded with degA=0
1 and degXi = 1 for all i. In this paper we present a comprehensive study of
gradedcomponentsoflocalcohomologymodulesHi(R)whereIisanarbitrary
I
homogeneous idealinR. Ourstudyseemstobethefirstinthisregard.
]
C
A
.
h
Contents
t
a
m 1. Introduction 2
2. Preliminaries 6
[
3. Generalized Eulerian modules 9
2 4. Graded holonomic modules 13
v
5. Vanishing 20
0
6. Tameness and Rigidity 21
7
2 7. Examples and Proof of Theorem 1.5 25
1 8. Infinite generation 27
0 9. Bass numbers 28
.
1 10. Proof of Theorem 1.9 30
0 11. Growth of Bass numbers 31
7 12. Associate Primes 34
1
13. Dimension of Support and injective dimension 35
:
v References 36
i
X
r
a
Date:February16,2017.
1991MathematicsSubjectClassification. Primary13D45,14B15,Secondary13N10,32C36.
Key words and phrases. local cohomology, graded local cohomology, ringof differential oper-
ators, Weyl Algebra, De Rham (and Koszul) cohomology of holonomic modules over the Weyl
algebra.
1
2 TONY.J.PUTHENPURAKAL
1. Introduction
Let S = S be a standard graded Noetherian ring and let S be it’s
n≥0 n +
irrelevant ideal. The theory of local cohomology with respect to S is particularly
+
L
satisfactory. Itiswell-known(cf. [2,15.1.5])thatifM isafinitelygeneratedgraded
S-module then for all i 0
≥
(1) Hi (M) is a finitely generated S -module for all n Z.
S+ n 0 ∈
(2) Hi (M) =0 for all n 0.
S+ n ≫
Itisnaturaltoexpectwhetherlocalcohomologywithrespecttootherhomogeneous
ideals exhibit similar results (or predictable results). It has been well-known for
many years that the answer to the latter question is clearly in the negative (even
in the case when S =B[X ,...,X ] is a polynomial ring). Even in the case of S
1 n +
the local cohomology module Hi (M) need not be tame, i.e., Hi (M) = 0 for
S+ S+ n 6
infinitely many n<0 does not imply that Hi (M) =0 for all n 0, see [4, 2.2].
S+ n 6 ≪
The purposeofthis paperisto showthatifAisaregularringcontainingafield
ofcharacteristiczeroandifR=A[X ,...,X ]isstandardgraded(withdegA=0)
1 n
then the theory of local cohomology of R with respect to arbitrary homogeneous
ideals of R exhibit striking good behavior. We should note that local cohomology
modules over regular rings does indeed show good behavior. For instance see the
remarkable papers [5], [7] and [8]. However there has been no previous study of
graded components of graded local cohomology modules of polynomial rings over
regular rings.
1.1. StandardAssumption: FromhenceforthAwilldenotearegularringcontaining
a field of characteristic zero. Let R = A[X ,...,X ] be standard graded with
1 m
degA=0anddegX =1foralli. Wealsoassumem 1. LetI beahomogeneous
i
≥
ideal in R. Set M = Hi(R). It is well-known that M is a graded R-module. Set
I
M = M .
n∈Z n
We first give a summary of the results proved in this paper.
L
I: (Vanishing:) The first result we prove is that vanishing of almost all graded
components of M implies vanishing of M. More precisely we show
Theorem 1.2. (with hypotheses as in 1.1). If M =0 for all n 0 then M =0.
n
| |≫
II (Tameness:) In view of Theorem 1.2, it follows that if M = Hi(R) =
I
M is non-zero then either M =0 for infinitely many n 0, OR, M =0
n∈Z n n 6 ≪ n 6
for infinitely many n 0. We show that M is tame. More precisely
L ≫
Theorem 1.3. (with hypotheses as in 1.1). Then we have
(a) The following assertions are equivalent:
(i) M =0 for infinitely many n 0.
n
6 ≪
(ii) M =0 for all n m.
n
6 ≤−
(b) The following assertions are equivalent:
(i) M =0 for infinitely many n 0.
n
6 ≫
(ii) M =0 for all n 0.
n
6 ≥
We complement Theorem 1.3 by showing the following
Example 1.4. There exists a regular ring A and homogeneous ideals I,J,K,L in
R=A[X ,...,X ] such that
1 m
(i) Hi(R) =0 for all n m and Hi(R) =0 for all n> m.
I n 6 ≤− I n −
GRADED COMPONENTS OF LOCAL COHOMOLOGY MODULES 3
(ii) Hi(R) =0 for all n 0 and Hi(R) =0 for all n<0.
J n 6 ≥ I n
(iii) Hi (R) =0 for all n Z.
K n 6 ∈
(iv) Hi(R) =0 for all n m, Hi(R) =0 for all n 0 and Hi(R)=0 for all
L n 6 ≤− L n 6 ≥ L
n with m<n<0.
−
An easy way to construct examples of type (ii) above is as follows: Choose an
ideal Q in A with Hi (A)=0. Then Hi (R)=Hi (A) R will satisfy (ii). The
Q 6 QR Q ⊗A
authorwasalsoableto constructexampleofahomogeneousidealJ inRandideal
Q in A with J QR such that (ii) is satisfied and √J = √qR for any ideal q in
6
A. Surprisingly the following general result holds:
Theorem 1.5. (with hypotheses as in 1.1). Further assume A is a domain. Sup-
pose J is a proper homogeneous ideal in R such that Hi(R) = 0 for all n 0
J n 6 ≥
and Hi(R) = 0 for all n < 0. Then there exists a proper ideal Q of A such that
J n
J QR.
⊆
III(Rigidity:) Surprisinglynon-vanishingofa singlegradedcomponentofM =
Hi(R) is very strong. We prove the following rigidity result:
I
Theorem 1.6. (with hypotheses as in 1.1). The we have
(a) The following assertions are equivalent:
(i) M =0 for some r m.
r
6 ≤−
(ii) M =0 for all n m.
n
6 ≤−
(b) The following assertions are equivalent:
(i) M =0 for some s 0.
s
6 ≥
(ii) M =0 for all n 0.
n
6 ≥
(c) (When m 2.) The following assertions are equivalent:
≥
(i) M =0 for some t with m<t<0.
t
6 −
(ii) M =0 for all n Z.
n
6 ∈
IV (Infinite generation:) Recall that each component of Hm (R) is a finitely
R+
generated R-module, cf., [2, 15.1.5]. We give a sufficient condition for infinite
generation of a component of graded local cohomology module over R.
Theorem1.7. (withhypothesesasin1.1). FurtherassumeAisadomain. Assume
I A=0. If Hi(R) =0 then Hi(R) is NOT finitely generated as an A-module.
∩ 6 I c 6 I c
V (Bass numbers:) The jth Bass number of an A-module E with respect to a
prime ideal P is defined as µ (P,E)=dim Extj (k(P),E ) where k(P) is the
j k(P) AP P
residue field of A . We note that if E is finitely generated as an A-module then
P
µ (P,E) is a finite number (possibly zero) for all j 0. In view of Theorem 1.7
j
≥
it is not clear whether µ (P,Hi(R) ) is a finite number. Surprisingly we have the
j I n
following dichotomy:
Theorem1.8. (withhypotheses as in1.1). LetP beaprimeideal inA. Fix j 0.
≥
EXACTLY one of the following hold:
(i) µ (P,M ) is infinite for all n Z.
j n
∈
(ii) µ (P,M ) is finite for all n Z. In this case EXACTLY one of the following
j n
∈
holds:
(a) µ (P,M )=0 for all n Z.
j n
∈
(b) µ (P,M )=0 for all n Z.
j n
6 ∈
(c) µ (P,M )=0 for all n 0 and µ (P,M )=0 for all n<0.
j n j n
6 ≥
4 TONY.J.PUTHENPURAKAL
(d) µ (P,M )=0 for all n m and µ (P,M )=0 for all n> m.
j n j n
6 ≤− −
(e) µ (P,M ) = 0 for all n m, µ (P,M ) = 0 for all n 0 and
j n j n
6 ≤ − 6 ≥
µ (P,M )=0 for all n with m<n<0.
j n
−
We also give easy examples where (i) and (ii) hold. The only examples where
the author was able to show (i) hold had m 2. Surprisingly the following result
≥
holds.
Theorem 1.9. (with hypotheses as in 1.1). Assume m = 1. Let P be a prime
ideal in A. Then µ (P,M ) is finite for all n Z.
j n
∈
Remark1.10. AnintriguingconsequenceofTheorem1.8isthefollowing: Suppose
M =Hi(M)=0 but M =0 for some c. Then for any prime ideal P and j 0 we
I 6 c ≥
have µ (P,M )=0 is finite. So µ (P,M )< for all n Z.
j c j n
∞ ∈
VI ( Growth of Bass numbers). Fix j 0. Let P be a prime ideal in A such
≥
that µ (P,Hi(R) ) is finite for all n Z. We may ask about the growth of the
j I n ∈
function n µ (P,Hi(R) ) as n and when n + . We prove
7→ j I n →−∞ → ∞
Theorem 1.11. (with hypotheses as in 1.1). Let P be a prime ideal in A. Let
j 0. Suppose µ (P,M ) is finite for all n Z. Then there exists polynomials
j n
≥ ∈
fj,P(Z),gj,P(Z) Q[Z] of degree m 1 such that
M M ∈ ≤ −
fj,P(n)=µ (P,M ) for all n 0 AND gj,P(n)=µ (P,M ) for all n 0.
M j n ≪ M j n ≫
By 1.10 if M = 0 for some c then µ (P,M ) is finite for all n Z, j 0 and
c j n
∈ ≥
prime P of A. In this case we prove:
Theorem 1.12. (with hypotheses as in 1.1). Let P be a prime ideal in A. Fix
j 0. Suppose µ (P,M )=0 for some c (this holds if for instance M =0). Then
j c c
≥
fj,P(Z)=0 or degfj,P(Z)=m 1,
M M −
gj,P(Z)=0 or deggj,P(Z)=m 1.
M M −
VII (Associate primes:) If E = E is a graded R-module then there are
n∈Z n
two questions regarding asymptotic primes:
L
Question 1:(Finiteness:) Is the set Ass E finite?
n∈Z A n
Question 2: (Stability:) Does there exists integers r,s such that Ass E =
A n
S
Ass E for all n r and Ass E =Ass E for all n s.
A r A n A s
≤ ≥
For graded local cohomology modules we show that both Questions above have
affirmative answer for a large class of regular rings A.
Theorem 1.13. (with hypotheses as in 1.1). Further assume that either A is local
or a smooth affine algebra over a field K of characteristic zero. Let M =Hi(R)=
I
M . Then
n∈Z n
(1) Ass M is a finite set.
L n∈Z A n
(2) Ass M =Ass M for all n m.
A n A m
S ≤−
(3) Ass M =Ass M for all n 0.
A n A 0
≥
VIII (Dimension of Supports and injective dimension:) Let E be anA-module.
Let injdim E denotes the injective dimension of E. Also Supp E = P E =
A A { | P 6
0 and P is a prime in A is the support of an A-module E. By dim E we mean
A
}
the dimension of Supp E as a subspace of Spec(A). We prove the following:
A
GRADED COMPONENTS OF LOCAL COHOMOLOGY MODULES 5
Theorem 1.14. (with hypotheses as in 1.1). Let M =Hi(R)= M . Then
I n∈Z n
we have
L
(1) injdimM dimM for all c Z.
c c
≤ ∈
(2) injdimM =injdimM for all n m.
n −m
≤−
(3) dimM =dimM for all n m.
n −m
≤−
(4) injdimM =injdimM for all n 0.
n 0
≥
(5) dimM =dimM for all n 0.
n 0
≥
(6) If m 2 and m<r,s<0 then
≥ −
(a) injdimM =injdimM and dimM =dimM .
r s r s
(b) injdimM min injdimM ,injdimM .
r −m 0
≤ { }
(c) dimM min dimM ,dimM .
r −m 0
≤ { }
Techniques used to prove our results: Weusethreemaintechniquestoproveour
results:
1.15. (a) For the first technique A is arbitrary regular ring containing a field of
characteristiczero. LetR=A[X ,...,X ],gradedwithdegA=0anddegX =1
1 m i
for all j. Let be the mth Weyl-algebra on A. We consider it a graded ring with
S
degA=0,degX =1anddeg∂ = 1. WenotethatRisagradedsubringof . If
i i
− S
E is a graded -module and e is a homogeneous element of E then set e =dege.
Consider thSe Eulerian operator = m X ∂ . If f R is homoge|ne|ous then
E i=1 i i ∈
it is easy to check that f = f f. We say a graded -module W is Eulerain if
E | | P S
w = ww for each homogeneous element w of W. Notice R is an Eulerian -
E | | S
module. We say W is generalized Eulerian if for each homogeneous w of W there
exists a depending on w such that ( w)aw =0.
E −| |
The notionofEulerianmodules wasintroducedinthe caseAis a fieldK by Ma
andZhang[9](they alsodefinedthenotionofEulerianD-modulesincharacteristic
p>0,whereDistheringofK-lineardifferentialoperatorsonR=K[X ,...,X ]).
1 m
Unfortunately however the class of Eulerian D-modules is not closed under exten-
sions(see3.5(1)in[9]). Torectifythis,theauthorintroducedthenotiongeneralized
Eulerian D-modules (in characteristic zero), see [12].
One can define the notion of graded Lyubeznik functors on ∗Mod(R) the cat-
egory of all graded R-modules, see 2.5. Our techniques in [15, Theorem 1.7]
generalize to prove the following:
Theorem 1.16. [with hypotheses as in 1.15.] Let be a graded Lyubeznik functor
G
on ∗Mod(R). Then (R) is a generalized Eulerian -module.
G S
1.17. (b) For the second technique we look at A = K[[Y ,...,Y ]] where K is a
1 d
field of characteristic zero. Let R = A[X ,...,X ], graded with degA = 0 and
1 m
degX = 1 for all j. Let D (A) = A < δ ,...,δ > (where δ = ∂/∂Y ) be
i k 1 d j j
the ring of k-linear differential operators on A. Furthermore set = A (D (A))
m k
D
be the mth-Weyl algebra over D (A). We can consider a graded ring with
k
D
degD (A) = 0, degX = 1 and deg∂ = 1. We note that R is a graded subring
k i i
−
of .
D
It is well-known that the global dimension of is d+m, see [1, 3.1.9]. Fur-
D
thermorethere is a filtration of suchthat the associatedgradedgr ( ) is the
T D T D
polynomial ring over A with d+2m-variables, [1, 3.1.9]. Thus a -module M is
D
holonomic if either it is zero or there a compatible filtration of M such that
T F
gr M is a finitely generated gr -module of dimension d+m.
F T D
The main technical result we show is
6 TONY.J.PUTHENPURAKAL
Theorem 1.18. [with hypotheses as in 1.17.] Let be a graded Lyubeznik functor
G
on ∗Mod(R). Then (R) is a graded holonomic generalized Eulerian -module.
G D
1.19. (c)The finaltechniquethatweuse isthetechnique ofdeRhamcohomology,
Koszul homology of generalized Eulerian modules (when A=K, a field). We also
prove
Theorem1.20. LetK beafieldofcharacteristiczero. LetA=K[[Y ,...,Y ]]and
1 d
R = A[X ,...,X ] and is as in 1.17. Let E be a graded holonomic generalized
1 n
D
Eulerian -module. Then for all i 0 the Koszul homology H (Y ,...,Y ;E) is
i 1 d
D ≥
a graded holonomic generalized Eulerian A (K)-module (here A (K) is the mth-
m m
Weyl algebra over K).
Our technique to prove all our results is as follows: Localize A at an appropri-
ate prime ideal and complete it. Then we use Theorem 1.16, Corollary 1.18 and
Theorem 1.20 to reduce to the case A is a field.
Remark 1.21. A natural question is what happens when A contains a field of
characteristic p > 0. We note that technique (a) and (b) have analogues in this
case. Howeverwedo notknowanyanalogueofourtechnique (c). So ourproofsdo
not work in this case.
Acknowledgment: In 2008, I started learning about applications of D-modules
in local cohomology theory. At that time Prof. G. Lyubeznik visited IIT-Bombay.
I asked him whether de Rham cohomology of local cohomology modules will be
interesting. He told me that it will be of interest. I (and co-authors) developed
techniques to study de Rham cohomology and Koszul cohomology of local coho-
mologymodulesinaseriesofpapers[11],[12],[13], [14]and[15]. Thesetechniques
have proved to be fantastically useful in this paper. I thank Prof. G. Lyubeznik
for his advice and to him this paper is dedicated.
2. Preliminaries
In this section, we discuss a few preliminary result that we need.
2.1. Lyubeznik functors:
Let B be a commutative Noetherian ring and let X =Spec(B). Let Y be a locally
closed subset of X. If M is a B-module and Y be a locally closed subscheme of
Spec(R),wedenotebyHi (M)theith-localcohomologymoduleofM withsupport
Y
inY. SupposeY =Y Y whereY Y aretwoclosedsubsetsofX thenwehave
1 2 2 1
\ ⊆
an exact sequence of functors
Hi ( ) Hi ( ) Hi ( ) Hi+1( ) .
···→ Y1 − → Y2 − → Y − → Y1 − →
ALyubeznikfunctor isanyfunctoroftheform = whereevery
1 2 m
T T T ◦T ◦···◦T
functor is either Hi ( ) for some locally closedsubset ofX or the kernel,image
Tj Y −
or cokernel of some arrow in the previous long exact sequence for closed subsets
Y ,Y of X such that Y Y .
1 2 2 1
⊆
2.2. Lyubeznik functor under flat maps:
We need the following result from [7, 3.1].
Proposition2.3. Let φ: B C be a flat homomorphism of Noetherian rings. Let
→
be a Lyubeznik functor on Mod(B). Then there exists a Lyubeznik functor on
T T
Mod(C) and isomorphisms T(M ⊗BC)∼=T(M)⊗B C which is functorial in M.
b
b
GRADED COMPONENTS OF LOCAL COHOMOLOGY MODULES 7
2.4. Graded Lyubeznik functors:
Let B be a commutative Noetherian ring and let R= B[X ,...,X ] be standard
1 m
graded. We say Y is homogeneous closed subset of Spec(R) if Y = V(f ,...,f ),
1 s
where f′s are homogeneous polynomials in R.
i
WesayY isahomogeneouslocallyclosedsubsetofSpec(R)ifY =Y′′ Y′,where
−
Y′,Y′′ are homogeneous closed subset of Spec(R). Let ∗Mod(R) be the category
of graded R-modules. We have an exact sequence of functors on ∗Mod(R),
(2.4.1) Hi ( ) Hi ( ) Hi ( ) Hi+1( ).
Y′ − −→ Y′′ − −→ Y − −→ Y′ −
Definition 2.5. A graded Lyubeznik functor is a composite functor of the form
T
= ... , where each is either Hi ( ), where Y is a homogeneous
T T1◦T2◦ ◦Tk Tj Yj − j
locally closed subset of Spec(R) or the kernel of any arrow appearing in (2.4.1)
with Y′ = Y′ and Y′′ = Y′′, where Y′ Y′′ are two homogeneous closed subsets
j j j ⊂ j
of Spec(R).
2.6. Graded Lyubeznik functors with respect to some standard operations on B.
(a) Let S be multiplicatively closed set in B. Set
R =R B =B [X ,...,X ].
W B W W 1 m
⊗
If Y is a homogeneous closed subset of Spec(R), say Y = V(f ,...,f ), set Y =
1 s W
V(f /1,...,f /1) where f /1 is the image of f in R . We note that Y is a
1 s i i W W
homogeneous closed subset of Spec(R ). Furthermore it is clear that we have
W
a homogeneous isomorphism HYi (−)⊗B BW ∼= HYiW(−). If Y is a homogeneous
locally closed subset of Spec(R), say Y = Y′′ Y′. set Y = Y′′ .Y′ , a
− W W − W
homogeneous locally closed subset of Spec(R ). Furthermore localizing (2.4.1) at
W
W yields a homogeneous isomorphism HYiW(−)∼=HYi (−)⊗BBW. More generally
if is a graded Lyubeznik functor on R then B is a graded Lyubeznik
B W
T T ⊗
functor on ∗Mod(R ).
W
(b)AssumeB islocalwithmaximalidealm. LetB bethecompletionofB with
respect to m. Set
R=R B =B[X ,...,Xb ].
B 1 m
⊗
If Y is a homogeneous closed subset of Spec(R), say Y = V(f ,...,f ), set Y =
b b b 1 s
V(f ,...,f ) where f is the image of f in R. We note that Y is a homogeneous
1 s i i
closed subset of Spec(R ). Furthermore it is clear that we have a homogenebous
W
isomborphisbm HYi (−)b⊗B B ∼= HYi (−). If Y isba homogeneous lbocally closed subset
b
of Spec(R), say Y = Y′′ Y′. set Y = Y′′ Y′, a homogeneous locally closed
b− −
subset of Spec(R). Furthermore applying the functor ( ) B to (2.4.1) yields a
B
− ⊗
homogeneousisomorphismHYi (−)∼=bHYi (−c)⊗BBc. MoregenerallyifT isagraded
Lyubeznik functbor onR thenb B is agradedLyubeznik fubnctor on ∗Mod(R).
B
T ⊗ b
2.7. Weyl Algebra’s:
b b
Let Γ be a ring (not necessarily commutative). The first Weyl algebra over Γ is
denoted by A (Γ) and it is the ring Γ < x,y > /(xy yx 1). Alternatively we
1
− −
can consider the polynomial ring Γ[X] and let δ be the derivation on Γ[X] defined
by formal differentiation with respect to X (treating elements of Γ as constants):
δ γ Xi = iγ Xi−1.
i i
(cid:16)X (cid:17) X
8 TONY.J.PUTHENPURAKAL
ItcanbeeasilyshownthatthedifferentialpolynomialringΓ[X][Y;δ]isisomorphic
to the first Weyl algebra over Γ, see [6, p. 12]. We note that YX =XY +δ(X)=
XY+1. ThusY canbethoughtas∂/∂X. TheidentificationofA (Γ)asΓ[X][Y;δ]
1
also yields that a Γ basis of A (Γ) is given by
1
XiYj: i,j 0 as well as by YjXi: i,j 0 .
{ ≥ } { ≥ }
Itfollows that if Z(Γ), the center of Γ, containsa field k then A (Γ)=Γ A (k).
1 k 1
⊗
ThehigherWeylalgebra’saredefinedinductivelyasA (Γ)=A (A (Γ). For
m 1 m−1
us the ring Γ will always contain a field K of characteristic zero in its center. So
A (Γ)=Γ A (K). The description of A (K) can be found in [1, Chapter 1].
m K m m
⊗
2.8. Koszul homology: Let Γ be a not-necessarilycommutative, Z-gradedring.
We assume that Γ has a commutative field K in its center with degK = 0 Let
u ,...,u be homogeneouscommuting elements in Γ. Consider the (commutative)
1 c
subring S =K[u ,...,u ] of Γ.
1 l
LetM beagradedΓ-module. LetH (u ,...,u ;M)betheith Koszulhomology
i 1 c
module of M with respect to u ,...,u . It is clearly a graded S-module. (with its
1 c
natural grading). The following result is well-known.
Lemma 2.9. Let u = u ,u ,...,u and let u′ = u ,...,u . For each i 0
r r+1 c r+1 c
≥
there exists an exact sequence of graded S-modules.
0 H (u ;H (u′;M)) H (u;M) H (u ;H (u′;M)) 0.
0 r i i 1 r i−1
→ → → →
2.10. Examples
(1) Γ = A (K), the mth-Weyl algebra over K. The operators ∂ , ,∂
m 1 m
···
commutewitheachother. Inthis caseH (∂;M)isusuallycalledtheith de
i
Rham homology of M.
(2) Γ = A (K), the mth-Weyl algebra over K. The elements X ,...,X
m 1 m
commute with each other. We can consider the usual Koszul homology
modules H (X,M).
i
2.11. We willrecallthefollowingcomputationofKoszulhomologywhichweneed.
Let A=K[[Y ,...,Y ]]. Let E be the injective hull of K as an A-module. Then
1 d
K if ν =d
H (Y,E)=
ν
(0 otherwise.
Althoughwe believe thatthis resultis alreadyknown,we givesketchof aprooffor
the benefit of the reader. Set A = K[[Y ,...,Y ]] and E to be the injective hull
i 1 i i
of K as an A -module. We have an exact sequence
i
0 A Yi A A 0
i i i−1
→ −→ → →
Taking Matlis dual’s with respect to E yields
i
0 E E Yi E 0.
i−1 i i
→ → −→ →
Now an easy induction of Koszul homology ( compute H (Y ,Y , ,Y ;E ))
j i i+1 d d
···
yields the result.
2.12. Wewillusethefollowingwell-knownresultoften. LetB beaNoetherianring
and let M be an A-module not necessarily finitely generated. Let P be a minimal
prime of M. Then the B -module M has a natural structure of an B -module
P P P
c
GRADED COMPONENTS OF LOCAL COHOMOLOGY MODULES 9
(here B is the completion of B with respect to it’s maximal ideal PB . In fact
P P P
MP ∼=MP ⊗BP BP.
c
c 3. Generalized Eulerian modules
The hypotheses in this section is a bit involved. So we will carefully state it.
3.1. Setup: In this section
(1) B is a commutative Noetherian ring containing a field K of characteristic
zero.
(2) We also assume that there is a not necessarily commutative ring Λ con-
taining B such that K Z(Λ), the center of Λ. Furthermore we assume
⊆
that
(a) Λ is free both as a left B-module and as a right B-module.
(b) N = B is a left Λ-module such that if we restrict the Λ action on N
to B we get the usual action of B on N.
(3) Set R = B[X ,...,X ]. We consider R graded with degB = 0 and
1 m
degX =1 for all i.
i
(4) SetΓ=Λ[X ,...,X ]. WeconsiderΓgradedwithdegΛ=0anddegX =
1 m i
1 for all i. Notice
(a) R is a graded subring of Γ.
(b) Γ is free both as a left and right R-module.
(c) N′ =R is a gradedleft Γ-module such that if we restrict the Γ action
on N′ to R we get the usual action of R on N′.
(5) We also assume that for each homogeneous ideal I of R and a graded Γ-
module E the set H0(E) is a graded Γ-submodule of E. Here
I
H0(E)= e E Ise=0 for some s 1 .
I { ∈ | ≥ }
(6) Let be the mth-Weyl algebra over Λ. Note =Λ A (K) where
m m K m
D D ⊗
A (K) is the mth-Weyl algebra over K. We can consider graded by
m m
D
giving degΛ=0, degX =1 and deg∂ = 1. Notice
i i
−
(a) Γ is a graded subring of .
m
D
(b) is free both as a left and right module over Γ. Thus is free
m m
D D
both as a left and right R-module.
(c) Theoperators∂ actasderivationsonR. ItfollowsthatRisagraded
i
-module.
m
D
3.2. Examples where our hypotheses 3.1 hold:
(1) K =B =Λ.
(2) B =K but Λ=B.
6
(3) B = K[[Y ,...,Y ]] and Λ = D (B) is the ring of K-linear differential
1 d K
operators on B, i.e., Λ = B < δ , ,δ > where δ = ∂/∂Y . We have
1 d j j
···
to verify hypotheses (5) of 3.1. We note that in the action of Γ on R the
elementsδ actasderivationsonR. LetI beahomogeneousidealofRand
j
letE beagradedΓ-module. WenotethatH0(E)isagradedR-submodule
I
of E. Let e H0(E). Say Ise=0.
∈ I
(a) Wefirstshowthatδ e H0(E). We claimIs+1δ e=0. Letu Is+1.
j ∈ I j ∈
Noticeuδ =δ u+δ (u). We notethatδ (u) Is. Thus uδ e=0. So
j j j j j
∈
our claim is true.
10 TONY.J.PUTHENPURAKAL
(b) We now note that Γ is a free left R-module with generators
δi1δi2 δid i ,i , ,i 0 .
{ 1 2 ··· d | 1 2 ··· d ≥ }
It follows that H0(E) is a Γ-submodule of E.
I
3.3. Generalized Eulerian -modules:
m
D
The Euler operator on , denoted by , is defined as
m m
D E
m
:= X ∂ .
m i i
E
i=1
X
Note that deg = 0. Let E be a graded -module. If e E is homogeneous
m m
E D ∈
element, set e =dege.
| |
Definition 3.4. Let E be a graded -module Then E is said to be Eulerian if
m
D
for each homogeneous element e of E.
e= e e.
m
E | |·
We note that R is an Eulerian -module.
m
D
Definition 3.5. A graded -module M is said to be generalized Eulerian if for
m
D
each homogeneous element e of E there exists a positive integer a (depending on
e) such that
( e)a e=0.
m
E −| | ·
The main result of this section is
Theorem 3.6. Let be a graded Lyubeznik functor on ∗Mod(R). Then (R) is
T T
a generalized Eulerian -module.
m
D
The first step in proving Theorem 3.6 is that (R) is a graded -module.
m
T D
Notice that (R) is a graded R-module. So we have to show both that (R) is
T T
a -module and that this action is compatiable with the grading on (R). We
m
D T
isolate this fact as a seperate
Lemma 3.7. [with hypotheses as in Theorem 3.6] (R) is a graded -module.
m
T D
Remark 3.8. If = is a graded but not-necessarily commutative ring
S n∈ZSn
then the category ∗Mod( ) of graded left -modules has enough injectives (the
L S S
proof given in Theorem 3.6.3 of [3] in the case is commutative extends to the
S
non-commutative case).
The following result is an essential ingredient in proving Lemma 3.7.
Proposition 3.9. Let V be a ∗-injective left -module. Then V is a ∗-injective
m
D
R-module.
Proof. We note that
∗Hom ( ,V)= ∗Hom ( , ∗Hom ( ,V)),
R − R − Dm Dm
= ∗Hom ( ,V).
Dm Dm⊗R−
As is free as a graded right R-module we have that is an exact
m m R
D D ⊗ −
functor from ∗Mod(R) to ∗Mod( ). Also by hypothesis V is a ∗-injective
m
D
-module. Itfollows that ∗Hom ( ,V)is anexactfunctor. So V is ∗-injective
m R
Das a R-module. − (cid:3)
