ebook img

Graded components of local cohomology modules PDF

0.42 MB·
Save to my drive
Quick download
Download
Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.

Preview Graded components of local cohomology modules

GRADED 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)

See more

The list of books you might like

Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.