ON A RELATION BETWEEN de RHAM COHOMOLOGY OFH(1f)(R)AND THE KOSZUL COHOMOLOGY OF∂(f)INR/(f)
Tony J. Puthenpurakal and Rakesh B. T. Reddy
Department of Mathematics, IIT Bombay, Powai, Mumbai 400 076, India
e-mails: tputhen@math.iitb.ac.in; rakesh@math.iitb.ac.in
(Received 22 July 2016; accepted 26 May 2017)
LetKbe a field of characteristic zero,R=K[X1, . . . , Xn]. LetAn(K) =K < X1, . . . , Xn, ∂1, . . . , ∂n >be thenthWeyl algebra overK. We consider the case whenRandAn(K)is graded by givingdegXi =ωi anddeg∂i = −ωi fori = 1, . . . , n(hereωiare positive integers). Set ω=Pnk=1ωk. LetIbe a graded ideal inR. By a result due to Lyubeznik the local cohomology modulesHi
I(R)are holonomicAn(K)-modules for eachi≥0. In this article we compute the de Rham cohomology modulesHj(∂;H1
(f)(R))forj≤n−2whenV(f)is a smooth hypersurface inPn(equivalentlyA=R/(f)is an isolated singularity).
Key words : Local cohomology; associated primes; D-modules; Koszul homology.
1. INTRODUCTION
LetK be a field of characteristic zero and let R = K[X1, . . . , Xn]. We considerR graded with degXi = ωi for i = 1, . . . , n; hereωi are positive integers. Set m = (X1, . . . , Xn). Let I be
a graded ideal in R. The local cohomology modules HI∗(R) are clearly graded R-modules. Let An(K) = K < X1, . . . , Xn, ∂1, . . . , ∂n > be the nth Weyl algebra over K. By a result due to Lyubeznik, see [2], the local cohomology modulesHIi(R) are holonomicAn(K)-modules for each
i≥0. We can considerAn(K)graded by givingdeg∂i=−ωifori= 1, . . . , n.
LetN be a graded leftAn(K)module. Now∂ = ∂1, . . . , ∂nare pairwise commutingK-linear
maps. So we can consider the de Rham complexK(∂;N). Notice that the de Rham cohomology modulesH∗(∂;N)are in general only gradedK-vector spaces. They are finite dimensional ifN is holonomic; [1, Chapter 1, Theorem 6.1]. In particularH∗(∂;HI∗(R))are finite dimensional graded K-vector spaces. By [4, Theorem 1] the de Rham cohomology modulesH∗(∂;H∗
I(R))is
Letf be a homogenous polynomial in R and setA = R/(f). By [3, Theorem 2.7] we have H0(∂;H1
(f)(R)) = 0. In this paper we extend a technique from [4]. In that paper the first author relatedHn−1(∂;H(1f)(R)) with Hn−1(∂(f);A). In this paper quite generally we prove that (see, Theorem 3.9):
Theorem 1 — Assume Hi−1(∂f;A) = 0. Then there exists a filtration {Fν}ν≥0 consisting
ofK−subspaces of Hi(∂;Rf) withFν = Hi(∂;Rf) for ν À 0,Fν ⊇ Fν−1 andF0 = 0 and K−linear maps
ηυ : Fυ Fυ−1 →H
i(∂f;A)
(υ+n−i) degf−ω.
such that
(a)ηυis injective for allυ≥2.
(b) Ifi6= 1.Thenη1also injective.
(c) Ifi= 1.Thenker(η1) =K.
IfHn−i(∂f;A) = 0fori≥α+ 1. Then as a corollary we obtain (see, Corollary 3.10)
Hn−i(∂;Rf) =
0 ifα+ 1≤i≤n−2
K ifi=n−1.
WhenAis an isolated singularity, i.e.,AP is regular for all homogeneous prime idealsP 6= m.
Note thatV(f)is a smooth hypersurface inPn. As a corollary we prove that (see, Corollary 3.11)
Hi(∂;H(1f)(R)) =
0 ifi≤n−2 andi6= 1
K ifi= 1.
We now describe in brief the contents of the paper. In Section 2 we discuss a few preliminaries that we need. In Section 3 we construct certain functions which we need to defineην. In Section 4
we construct our filtration ofHi(∂;H(1f)(R))and prove our result.
2. PRELIMINARIES
In this section we discuss a few preliminary results that we need.
Remark 2.1 : Although all the results are stated for de Rham cohomology of aAn(K)-module
An(K)-moduleM. LetS = K[∂1, . . . , ∂n]. Consider it as a subring of An(K). Then note that
Hi(∂, M)is theithKoszul homology module ofMwith respect to∂.
2.2 LetA be commutative ring anda = a1,· · · , an ∈ A.Let I ⊆ {1,· · · , n}, |I| = m.Say
I ={i1 < i2 <· · ·< im}.Let
eI:=ei1 ∧ei2 ∧ · · · ∧eim.
Then the Koszul complex ofAwith respect toais
K(a;A) := 0→Kn−→φn Kn−1 φn−→ · · · →−1 K1−→φ1 K0 →0.
HereKm= L
|I|=mAeI.
Letξ =
X
|I|=m
ξIeI ∈Km we write ξ = (ξI| |I|=m). For the mapKp −→φp Kp−1,
sayφp(ξ) =U.WriteU = (UJ | |J|=p−1). Then
UJ =X i6∈J
(−1)σ(J∪{i})¡aiξJ∪{i} ¢
.
HereJ ={j1 < j2 <· · ·< jp−1}and
σ(J∪ {i}) =
0, ifi < j1;
p, ifi > jp−1
r, ifjr< i < jr+1.
2.3 Letf ∈Rbe a homogeneous polynomial. We consider elements ofRmf as column-vectors. For x∈Rm
f we write it asx= (x1, . . . , xm)0; here0indicates transpose.
2.4 Let f ∈ R be a homogeneous polynomial. Set∂ = ∂1,· · · , ∂n. Consider the commutative
subringS =K[∂1,· · · , ∂n]ofAn(K). The de Rham complex on a holonomic moduleN is just the
Koszul complexK(∂;N)ofN with respect toS. In particular whenN =Rf we have,
K(∂;Rf) = 0→Kn−→φn Kn−1 → · · ·Kp φp
−→Kp−1 → · · · →K1 −→φ1 K0 →0.
Here K0 =Rf and Kp= M
|I|=p
The mapsKp φp
−→Kp−1, sayξ= (ξI | |I|=p)0.Then
φp(ξ) =
X
i6∈J
(−1)σ(J∪{i}) µ
∂ ∂xiξJ∪{i}
¶
: |J|=p−1
.
2.5 Letf ∈ Rbe a homogeneous polynomial. SetA =R/(f), and∂f =∂f /∂x1,· · · , ∂f /∂xn.
Consider the Koszul complexK0(∂f;A)onAwith respect to∂f.
K0(∂f;A) = 0→K0n−→ψn K0n−1· · · →K0p −→ψp Kp0−1→ · · · →K01 ψ1
−→K00→0.
Here K0p =
M
|I|=p
A(−pdegf+ωi1 +· · ·+ωip).
2.6 By [4, Theorem 1],Hi(∂;Rf)j = 0forj6=ω,whereω=ω1+· · ·+ωn.
2.7 Letξ ∈Rm
f \Rm. The element(a1/fi, a2/fi, . . . , am/fi)0, withaj ∈Rfor allj, is said to be
a normal form ofξif
(1) ξ = (a1/fi, a2/fi, . . . , am/fi)0.
(2) f does not divideaj for somej.
(3) i≥1.
It can be easily shown that normal form ofξexists and is unique (see [4, Proposition 5.1]).
2.8 Letξ ∈Rmf . We defineL(f)as follows.
Case 1 :ξ∈Rmf \Rm.
Let(a1/fi, a2/fi, . . . , am/fi)0 be the normal form ofξ. SetL(ξ) = i. NoticeL(ξ) ≥1in this
case.
Case 2 :ξ∈Rm\ {0}.
SetL(ξ) = 0.
Case 3 :ξ= 0.
SetL(ξ) =−∞.
The following properties of the functionLcan be easily verified.
(1) IfL(ξ1)< L(ξ2)thenL(ξ1+ξ2) =L(ξ2).
(2) IfL(ξ1) =L(ξ2)thenL(ξ1+ξ2)≤L(ξ2).
(3)L(ξ1+ξ2)≤max{L(ξ1), L(ξ2)}.
(4) Ifα∈K∗thenL(αξ) =L(ξ).
(5)L(αξ)≤L(ξ)for allα∈K.
(6)L(α1ξ1+α2ξ2)≤max{L(ξ1), L(ξ2)}.
(7) Letξ1, . . . , ξr∈Rmf and letα1, . . . , αr∈K. Then
L
Xr j=1
αjξj
≤max{L(ξ1), L(ξ2), . . . , L(ξr)}.
3. CONSTRUCTION OFCERTAINFUNCTIONS
In this section we construct few functions.
We define a function,θ:Zp(∂;Rf)\R( n
p) −→Hp(∂f;A), as follows.
Letξ ∈Zp(∂;Rf)\R( n
p) and let(ξ
I/fc | |I|=p)0 be the normal form ofξ.Asφp(ξ) = 0.We
have for everyJ such that|J|=p−1,
X
i6∈J
(−1)σ(J∪{i}) ∂
∂xi µ
ξJ∪{i} fc
¶ =0.
This implies
X
i6∈J
(−1)σ(J∪{i}) Ã ∂
∂xi(ξJ∪{i})
fc −cξJ∪{i} ∂f ∂xi
fc+1
! =0.
So f.X
i6∈J
(−1)σ(J∪{i}) µ
∂
∂xi(ξJ∪{i}) ¶
=c.X
i6∈J
(−1)σ(J∪{i}) µ
ξJ∪{i}∂f ∂xi
¶
.
Thusf dividesPi6∈J(−1)σ(J∪{i})
³
ξJ∪{i}∂xi∂f
´
.Therefore
( ¯ξI | |I|=p)0 ∈Zp(∂f;A).
Setθ(ξ) = [( ¯ξI | |I|=p)0]∈Hp(∂f;A).
The following Lemma identifies the degree ofθ(ξ).
(a)ξ∈R(
n p) f \R(
n p).
(b) IfL(ξ) =c.Thenθ(ξ)∈Hp(∂f;A)(c+p) degf−ω.
PROOF:(a)Letξ= (ξI)0 be non-zero inZp(∂;Rf)−ω. Note that
ξ ∈ M
|I|=p
(Rf(ωi1+· · ·+ωip))−ω.
It follows that
ξI∈(Rf)−ω+Pp s=1ωis.
It follows thatξ ∈R(
n p) f \R(
n p).
(b) Let(aI/fc | |I| = p)0 be the normal form ofξ.As aI/fc ∈ Rf(ωi1 +· · ·+ωip)−ω.It follows that
deg(aI) =cdegf−ω+ p X
s=1 ωis.
Asθ(ξ) = [(¯aI | |I|=p)0].Let
¯
aI ∈A(−pdegf + p X
s=1 ωis)t.
Then
¯
aI ∈A(−pdegf+Pps=1ωis)+t.
It follows that
t= (c+p) degf−ω.
Thusθ(ξ)∈Hp(∂f;A)(c+p) degf−ω. 2
A natural condition we want inθis that it vanishes on boundaries. The following result gives a sufficient condition when this happens.
Proposition 3.2 — Letp < n. AssumeHp+1(∂f;A) = 0. Thenθ(Bp(∂;Rf)−ω\{0}) = 0.
PROOF : Let U ∈ Bp(∂;Rf)−ω be non zero. As Bp(∂;Rf)−ω ⊂ Zp(∂;Rf)−ω. We get by
Lemma 3.1(a) thatU ∈R(
n p) f \R(
n p).Set
Noticec≥1. Letξ ∈(Kp+1)−ωbe such thatL(ξ) =candφp+1(ξ) = U.Let(bG/fc | |G|=
p+ 1)0be the normal form ofξ. LetU = (U
I | |I|=p)0. Then
UI=X i6∈I
(−1)σ(I∪{i}) ∂
∂xi µ
bI∪{i} fc
¶
=X
i6∈I
(−1)σ(I∪{i}) Ã ∂
∂xibI∪{i}
fc −c
bI∪{i}∂xi∂f fc+1
!
= f
fc+1
X
i6∈I
(−1)σ(I∪{i}) µ
∂
∂xi(bI∪{i}) ¶
− c
fc+1
X
i6∈I
(−1)σ(I∪{i}) µ
bI∪{i} ∂f ∂xi
¶
.
Set
VI =−c X
i6∈I
(−1)σ(I∪{i}) µ
bI∪{i}∂f ∂xi
¶
.
Then
UI = ff∗c++1VI.
Claim :f does not dividesVI for someI with|I|=p.
First assume the claim. Then U = (UI | |I| = p)0 = ( (f ∗+VI)/fc+1 | |I| = p)0 is the
normal form ofU. Therefore
θ(U) = [( ¯VI)0] = [ψp+1(−c¯bG | |G|=p+ 1)0] = 0.
We now prove our claim. Suppose if possiblef|VI for allI with|I|=p. Letb= (bG | |G|=
p+ 1)0.Then
ψp+1(−c¯b) = ( ¯VI)0= (0,0,· · ·,0)0.
So−c¯b∈Zp+1(∂f;A).AsHp+1(∂f;A) = 0, we get−c¯b∈Bp+1(∂f;A).Thus
Forp=n−1we have
−c¯b= 0
⇒ cbG= ˜αGf for some α˜G∈R.
Thus f /bG for all G. This contradicts that (bG/fc| |G| = p+ 1)0 is the normal form of ξ.
Thereforep < n−1. So
−cbG= X
k6∈G
(−1)σ(G∪{k}) µ
rG∪{k} ∂f ∂xk
¶
+ ˜αGf. (3.2.1)
Now we compute the degrees ofrL. Note thatξ ∈(Kp+1)−ω.So
bG
fc ∈Rf(ωi1+· · ·+ωip+1)−ω.
It follows that
degbG=cdegf−ω+ p+1
X
s=1
ωis. (3.2.2)
It can be easily checked that
¯bI∪{i} ∈A(−(p+ 1) degf+ωi
1 +· · ·+ωip+1)(c+p+1) degf−ω.
So
¯
rL∈A(−(p+ 2) degf+ωi1+· · ·+ωip+2)(c+p+1) degf−ω.
It follows that
degrL= (c−1) degf−ω+ p+2
X
j=1
ωij. (3.2.3)
Case (1) : Letc= 1. Then by equation (3.2.1) we getα˜G= 0. Also notice
degrL=−ω+ p+2
X
j=1
ωij <0 if n > p+ 2.
Now considern =p+ 2.ThenrL = r12···n =constant. SayrL = r. So by equation (3.2.1) it
follows
b= (bG | |G|=n−1)0 = (−r∂f /∂x1,· · ·,(−1)nr∂f /∂xn)0. (3.2.4)
As UI = X
i6∈I
(−1)σ(I∪{i}) µ
∂ ∂xi
(bI∪{i}/f) ¶
.
Using equation (3.2.4) we getUI = 0for allIwith|I|=p.ThereforeU = 0a contradiction.
Case (2) : Letc≥2.By equation (3.2.1) we have
−cbG fc =
1
fc X
k6∈G
(−1)σ(G∪{k}) µ
rG∪{k} ∂f ∂xk
¶ + α˜G
fc−1.
Notice
rG∪{k}∂f /∂xk
fc =
∂ ∂xk
µ
rG∪{k}/(1−c)
fc−1
¶
− ∗
fc−1.
Put r˜G∪{k} =
rG∪{k}
c(c−1). We obtain
bG fc =
X
k6∈G
(−1)σ(G∪{k}) µ
∂ ∂xk
µ ˜
rG∪{k} fc−1
¶¶
+ ∗
fc−1.
Set
δ= µ
˜
rG∪{k} fc−1
¶
and ξ˜= µ
∗
fc−1
¶
.
Then
ξ =φp+2(δ) + ˜ξ.
So we haveU =φp+1(ξ) =φp+1(˜ξ)andL(ξ)≤c−1. This contradicts our choice ofc. 2
4. CONSTRUCTION OF AFILTRATION ONHp(∂;Rf)
In this section we construct a filtration of Hp(∂;Rf). Throughout this section 1 ≤ p < n and
Hp+1(∂f;A) = 0.
4.1 By 2.6 we have
Hp(∂;Rf) =Hp(∂;Rf)−ω =
Zp(∂;Rf)−ω
Letx∈Hp(∂;Rf)be non-zero. Define
L(x) =min{L(ξ)| x= [ξ], where ξ ∈Zp(∂;Rf)−ω}.
Let ξ = (ξI/fc | |I| = p)0 ∈ Zp(∂;Rf)−ω be such that x = [ξ]. So ξ ∈ (Kp)−ω. Thus
ξ ∈ R(
n p)
f (ωi1 +· · ·+ωip)−ω. So if ξ 6= 0thenξ ∈ R
(n p) f \R(
n
p).It follows thatL(ξ) ≥ 1.Thus
L(x)≥1.Ifx= 0setL(x) =−∞.
We now define a function
˜
θ:Hp(∂;Rf)→Hp(∂f;A)
x→
0 ifx= 0
θ(ξ) ifx6= 0, x= [ξ], and L(x) =L(ξ).
Proposition 4.2 — (with hypothesis as above).θ˜(x)is independent ofξ.
PROOF: Supposex = [ξ1] = [ξ2]is non zero andL(x) =L(ξ1) = L(ξ2) =c.Let(aI/fc)0 be
the normal form ofξ1 and(bI/fc)0 be the normal of ξ2.As[ξ1] = [ξ2]it follows thatξ1 = ξ2 +δ for someδ ∈Bp(∂;Rf)−w.We getj =L(δ)≤cby 2.9. Let(cI/fj)0 be the normal form ofδ.We
consider two cases.
Case (1) :j < c.Then note thataI =bI+fc−jcI, for|I|=p.It follows that
θ(ξ1) = [( ¯aI)0] = [( ¯bI)0] =θ(ξ2).
Case (2) :j=c.Note thataI =bI+cI for|I|=p.It follows that
θ(ξ1) =θ(ξ2) +θ(δ).
However by Proposition 3.2θ(δ) = 0.Soθ(ξ1) =θ(ξ2).Henceθ˜(x)is independent of choice of
ξ. 2
4.3 We now construct a filtrationF={Fυ}υ≥0ofHp(∂;Rf).Set
Fυ={x∈Hp(∂;Rf) | L(x)≤υ}.
Proposition 4.4 —(1)Fυ is aK−subspace ofHp(∂;Rf).
(3)Fυ =Hp(∂;Rf)for allυÀ0.
(4)F0 = 0.
PROOF: (1)Letx ∈ Fυand letα ∈ K.Letx = [ξ]withL(x) =L(ξ) ≤υ.Thenαx= [αξ].
So
L(αx)≤L(αξ)≤υ.
Soαx∈ Fυ.
Letx, x0 ∈ Fυbe non-zero. Letξ, ξ0 ∈ Zp(∂;Rf)be such thatx = [ξ], x0 = [ξ0]andL(x) =
L(ξ), L(x0) =L(ξ0).Thenx+x0 = [ξ+ξ0].It follows that
L(x+x0)≤L(ξ+ξ0)≤max{L(ξ), L(ξ0)} ≤υ.
Thusx+x0 ∈ Fυ.
(2)This is clear from the definition.
(3)LetB={x1,· · · , xl}be aK−basis ofHp(∂;Rf) =Hp(∂;Rf)−ω. Let
c=max{L(xi) | i= 1,· · · , l}.
We claim that
Fυ =Hp(∂;Rf) for all υ≥c.
Fixυ≥c.Letξi ∈Zp(∂;Rf)−ωbe such thatxi = [ξi]andL(xi) =L(ξi)fori= 1,· · ·, l.
Letu ∈ Hp(∂;Rf).Sayu = Pl
i=1αixi for someα1,· · · , αl ∈ K. Thenu = [ Pl
i=1αiξi]. It
follows that
L(u)≤L( l X
i=1
αiξi)≤max{L(ξ) | i= 1,· · · , l}=c≤υ.
Sou∈ Fυ. HenceFυ=Hp(∂;Rf).
(4)Ifx∈Hp(∂;Rf)is non-zero thenL(x)≥1.ThereforeF0 = 0. 2
4.5 LetG =Lυ≥1Fυ/Fυ−1. Forυ≥1we define
ηυ : Fυ Fυ−1
→Hp(∂f;A)(υ+p) degf−ω
u→
0 ifu= 0 ˜
Proposition 4.6 — (with hypothesis as above).ηυ(u)is independent of choice ofx.
PROOF: Supposeu=x+Fυ−1 =x0+Fυ−1 be non-zero. Thenx =x0+ywherey ∈ Fυ−1. Asu 6= 0we havex, x0 ∈ Fυ\Fυ−1. SoL(x) = L(x0) = υ. Sayx = [ξ],x0 = [ξ0], andy = [δ]
whereξ, ξ0, δ ∈Zp(∂;Rf)−ωwithL(ξ) = L(ξ0) =υandL(δ) = L(y) = k≤υ−1. So we have
ξ=ξ0+δ+αwhereα∈B
p(∂;Rf)−ω. LetL(α) =r.Not thatr ≤υ.
Let(aI/fυ)0,(a0I/fυ)0,(bI/fk)0and(cI/fr)0be normal forms ofξ, ξ0, δandαrespectively, here |I|=p.So we have
aI=a0I+fυ−kbI+fυ−rcI with |I|=p.
Case(1):r < υ. In this case we have that¯aI = ¯a0IinA. Soθ(ξ) =θ(ξ0). Thusθ˜(x) = ˜θ(x0).
Case(2): r =υ. In this case we have¯aI = ¯a0I + ¯cIinA. Soθ(ξ) = θ(ξ0) +θ(α). However
θ(α) = 0asα∈Bp(∂;Rf)−ω. Thusθ˜(x) = ˜θ(x0). 2
Proposition 4.7 — (with notation as above). For allυ≥1,ηυisK−linear.
PROOF: Letu, u0 ∈ Fυ/Fυ−1. We first show thatηυ(αu) =αηυ(u). Ifα= 0oru= 0we have
nothing to show. So assumeα6= 0andu6= 0. Sayu=x+Fυ−1.Thenαu=αx+Fυ−1.It can be easily shown thatθ˜(αx) =αθ˜(x). So we get the result.
Next we show thatηυ(u+u0) =ηυ(u) +ηυ(u0). We have nothing to show ifuoru0is zero. Now
consider the case whenu+u0= 0. Thenu=−u0. Soηυ(u) =−ηυ(u0). Thus in this case
ηυ(u+u0) = 0 =ηυ(u) +ηυ(u0).
Now consider the case whenu, u0 are non-zero andu+u0 non-zero. Sayu = x+Fυ−1 and u0 =x0+F
υ−1. Note that asu+u0 is non-zerox+x0 ∈ Fυ\Fυ−1. Letx= [ξ]andx0 = [ξ0]where ξ, ξ0 ∈Z
p(∂;Rf)−ωandL(ξ) = L(ξ0) =υ. Thenx+x0 = [ξ+ξ0]. Note thatL(ξ+ξ0)≤υ. But
L(x+x0) =υ.SoL(ξ+ξ0) =υ. Let(a
I/fυ)0,(a0I/fυ)0be the normal forms ofξ,andξ0respectively.
Note that((aI+a0I)/fυ)0is the normal form ofξ+ξ0. It follows thatθ(ξ+ξ0) =θ(ξ) +θ(ξ0). Thus ˜
θ(x+x0) = ˜θ(x) + ˜θ(x0). Therefore
ηυ(u+u0) =ηυ(u) +ηυ(u0).2
Surprisingly the following result holds.
(a)ηυis injective for allυ≥2.
(b) Ifp6=n−1.Thenη1also injective.
(c) Ifp=n−1.Thenker(η1) =K.
PROOF: Suppose if possibleην is not injective. Then there exists non-zerou ∈ Fν/Fν−1 with ην(u) = 0. Sayu=x+Fν−1. Also letx= [ξ]whereξ∈Zp(∂;Rf)−ωandL(ξ) =L(x) =ν. Let (aI/fυ | |I|=p)0be the normal form ofξ. So we have
0 =ην(u) =θe(x) =θ(ξ) = [(aI)0].
It follows that(aI)0 =ψp+1(b), whereb= (bG | |G|=p+ 1)0. It follows that
aI= X
i6∈I
(−1)σ(I∪{i}) µ
¯bI∪{i}∂f¯
∂xi
¶
.
It follows that for|I|=pwe have the following equation inR:
aI=X i6∈I
(−1)σ(I∪{i}) µ
bI∪{i}∂f ∂xi
¶
+dIf, (4.8.1)
for somedI ∈ R. Note that the above equation is of homogeneous elements inR. So we have the
following
aI fυ =
P
i6∈I(−1)σ(I∪{i})bI∪{i}∂xi∂f
fυ +
dI
fυ−1. (4.8.2)
We consider two cases:
(a): Letν ≥2. SetebI∪{i} =−bI∪{i}/(υ−1). Then note that
bI∪{i}∂xi∂f
fν =
∂ ∂xi
à eb
I∪{i}
fν−1
!
− ∗
fν−1.
By equation (4.8.2) we have
aI
fν = X
i6∈I
(−1)σ(I∪{i}) ∂
∂xi
à eb
I∪{i}
fυ−1
!
− ∗
fν−1.
Putξ0 =¡∗/fν−1:|I|=p¢0 andδ =
³ eb
I∪{i}/fν−1 |i6∈I,|I|=p ´0
. Then we have
So we havex= [ξ] = [ξ0]. This yieldsL(x)≤L(ξ0)≤ν−1. This is a contradiction.
(b): Letν = 1andp6=n−1.Note thatn≥p+ 2.Also note thatξ ∈(Kp)−ω. Thus for|I|=p
we have
aI
f ∈(Rf(ωi1+· · ·+ωip)−ω.
It follows that
degaI = degf −ω+ (ωi1+· · ·+ωip).
Also note thatdeg∂f /∂xi= degf−ωi. By comparing degrees in equation (4.8.1) we getaI = 0
for allI with|I|=p. Thusξ= 0. Sox= 0.Thereforeu= 0a contradiction.
(c): Letp=n−1.By comparing degrees in equation (4.8.1) we getdI = 0andbI∪{i}=constant.
But
ξ= µ
∂f ∂xn/f,−
∂f
∂xn−1/f,· · · ,(−1)
n−1 ∂f ∂x1/f
¶0
.
It is easily verified thatξ∈Zn−1(∂;Rf)and that ifx= [ξ]thenη1(x) = 0.
We prove thatξ6∈Bn−1(∂;Rf). Suppose if possible letg∈R, g.c.d(g, f) = 1and µ
∂ ∂xn,−
∂
∂xn−1,· · ·,(−1)
n−1 ∂ ∂x1
¶0
(g/fc) =ξ.
Thus
∂ ∂xi
(g/fc) = ∂f
∂xi
/f for i= 1,· · ·, n.
By computing left hand side we see thatf dividesg∂f /∂xifori= 1,· · · , n.
Let g∂f
∂xi =f hi.
Let
f =fa1
1 f2a2· · ·fsas, fjirreducible andaj ≥1.
Then
∂f ∂xi =
s X
j=1 fa1
1 · · ·f
aj−1
j−1 (ajfjaj−1
∂fj
∂xi)f aj+1
j+1 · · ·fsas.
Ifaj ≥2⇒fjaj−1divides∂f /∂xiandfjaj does not divides∂f /∂xi.So we can write
∂f ∂xi =f
a1−1
1 f2a2−1· · ·fsas−1Vi, wherefj does not dividesVi∀j.
LetU = f1f2· · ·fs.Then we havegVi = U hi.As g.c.d(g, f) = 1so g.c.d(g, U) = 1.Sofj
dividesVia contradiction. Thereforeξ6∈Bn−1(∂;Rf).Henceker(η1) =K. 2
By summarizing the above results, we obtain our main Theorem:
Theorem 4.9 — Assume Hp+1(∂f;A) = 0. Then there exists a filtration{Fν}ν≥0 consisting
ofK−subspaces of Hp(∂;Rf) with Fν = Hp(∂;Rf) forν À 0,Fν ⊇ Fν−1 andF0 = 0 and K−linear maps
ηυ: FFυ
υ−1 →Hp(∂f;A)(υ+p) degf−ω.
such that
(a)ηυis injective for allυ≥2.
(b) Ifp6=n−1.Thenη1also injective.
(c) Ifp=n−1.Thenker(η1) =K.
Corollary 4.10 — IfHi(∂f;A) = 0fori≥α+ 1. Then
Hi(∂;Rf) =
0 ifα+ 1≤i≤n−2
K ifi=n−1.
PROOF: Letα+ 1≤i≤n−2..By Theorem 4.9 there exist a filtration{Fυ}υ≥0ofHi(∂;Rf)
and injective maps
ηυ: Fυ Fυ−1
−→Hi(∂f;A).
Note thatF0= 0. AsHi(∂f;A) = 0we getF1= 0.Continuing this way we getFυ= 0for all
υ.AsFυ=Hi(∂;Rf)forυÀ0.HenceHi(∂;Rf) = 0.
Leti=n−1.Thenker(η1) =K.SoF1/K= 0.ThusF1 =K.
As dimKHn−1(∂;Rf) = X
dimK(Fυ/Fυ−1)
= dimKF1
WhenAis smooth we get: Ä
Corollary 4.11 — Letf ∈Rbe quasi homogeneous. LetA=R/(f)be smooth. Then
Hi(∂;H(1f)(R)) =
0 if 2≤i≤n−2ori=n
K if i=n−1.
PROOF: AsAis smooth, soHi(∂f;A) = 0fori≥2. Therefore by Corollary 4.10.
Hi(∂;Rf) =
0 ifα+ 1≤i≤n−2
K ifi=n−1.
By [3, Theorem 2.7]
Hn(∂;H(1f)(R)) = 0 and Hi(∂;Rf)≡Hi(∂;H(1f)(R))for i < n.
Hence the result. 2
REFERENCES
1. J.-E. Bj¨ork, Rings of differential operators, North-Holland Mathematical Library, 21, North-Holland Publishing Co., Amsterdam-New York, 1979.
2. G. Lyubeznik, Finiteness properties of local cohomology modules (an application of D-modules to com-mutative algebra), Invent. Math., 113(1) (1993), 41-55.
3. T. J. Puthenpurakal, de Rham cohomology of local cohomology modules, Algebra and its applications, 159-181, Springer Proc. Math. Stat., 174, Springer, Singapore, 2016.
