C OMPOSITIO M ATHEMATICA
E VA K RISTINA E KSTRÖM
Homological properties of some Weyl algebra extensions
Compositio Mathematica, tome 75, n
o2 (1990), p. 231-246
<http://www.numdam.org/item?id=CM_1990__75_2_231_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
Homological properties of
someWeyl algebra extensions
EVA KRISTINA
EKSTRÖM
Department of Mathematics, University of Stockholm, Box 6701, 11385 Stockholm, Sweden Received 7 March 1989; accepted in revised form 12 February 1990
Introduction
In this work we
study rings
of the formA1(K) Q9K S,
where K is a commutative field of characteristic zero,A1(K)
theWeyl algebra
in one variable and S a K-algebra satisfying
the so called Auslander-Gorenstein condition. TheK-algebra A,(K)
isgenerated by
two elements x and ô such that ôx - x8 = 1. Weget
agraded ring
structure on R =Al (K) Q9 K S
where elements of S arehomog-
eneous of
degree
zero, while x,respectively ô
ishomogeneous
ofdegree - 1, respectively
+ 1.Then,
for afinitely generated
andgraded left(or right)
R-module M = EBMv
we are
going
to establish that thegrade
number of the R-module M is obtained via thegrade
numbers of thehomogeneous components
each of which is afinitely generated
module over thepolynomial ring
in one variable over S.Our main result occurs in Theorem 1.3 and we refer to
§4
for someapplications
to modules overrings
of differentialoperators.
Concerning examples of
Auslander-Gorensteinrings
we mention that if S is a commutative noetherianring
with finiteinjective
dimension then S is Auslander-Gorenstein.Moreover,
if S is apositively
filteredring (or
moregeneral
a zariskian filteredring)
such that the associatedgraded ring
isAuslander-Gorenstein then S is Auslander-Gorenstein. For details we refer to
[Bj:21.
1. Announcements of the main results
Before we announce Theorem 1.3 and 1.4 below we need some
preliminaries:
We consider the
Weyl algebra A 1 (K)
in one variable over a commutative field of characteristic zero. Recall that theK-algebra A1(K)
isgenerated by
twoelements x and 8 such that ôx - xô = 1.
Denote
by vo
theK-subalgebra
ofA1(K) generated by
x and xô.If k
1 weset
It is
easily
seen that{Vk}k~Z
is afiltration,
i.e.vk vm
cVk+m, ~ Vk
={0}
and~Vk = A1(K).
The associated
graded ring ~Vk/Vk-1
is denotedby gr(Al).
Letx,
re-spectively 0
denote theimage
of x inV-1/V-2, respectively a
inVIV,.
It isobvious that the
K-algebra gr(A1)
isgenerated by
x and~.
Moreoverôx - x3
= 1 in thering gr(A1).
It follows that theK-algebras gr(A1)
andA1(K)
are
isomorphic. By
definitiongr(A1)
is agraded ring.
If k is aninteger
we setgrk(A1)
=Vk/Vk-1.
Observe thatgro(A 1 )
=K[x~],
i.e.gro(A 1 )
is thepolynomial ring
in one variable over K.In this paper we
study
tensorproduct rings
of the formgr(A 1 ) 0K S,
where Sis a
K-algebra.
Set R =gr(A 1 ) ~KS.
Thegraded
structure ofgr(A1)
induces agraded
structure of R such thatWe will assume that the
ring
S satisfies certainhomological
conditions and recall first:Auslander-Gorenstein
rings
Let A be a left and a
right
noetherianring.
We say that A has a finiteinjective
dimension if there exists an
integer n
such thatExtvA(M, A)
= 0 for every v > nand any left or
right
A-module M. We refer to the article[Za]
for facts about noncommutative noetherianrings
with finiteinjective
dimension.Following [Bj:2,
Definition1.2]
wegive
1.1. DEFINITION. A
ring
A is called an Auslander-Gorensteinring if A
is aleft
and aright
noetherianring with finite injective
dimension and moreover:for
every
integer
v, everyfinitely generated left
orright
A-module M and every submodule Nof Extv(M, A) it follows
thatExt’(N, A)
= 0for
every i v.Assume A is Auslander-Gorenstein. If M is a nonzero and
finitely generated
left or
right A-module,
then there exists aunique
smallestinteger k
such thatExtkA(M, A)
is nonzero. We set k= jA(M)
and wecall jA(M)
thegrade
number ofM. If M is the zero module we define
jA (M)
= 00.1.2. REMARK. Let A be an Auslander-Gorenstein
ring.
It follows from the Auslander condition that if 0 ~ M’ ~ M - M" ~ 0 is an exact sequence offinitely generated
A-modules thenFor a
proof
see[Bj:2, Proposition 1.8].
REMARK. Let K be a commutative field and let S be an Auslander-Gorenstein
K-algebra.
Then every skewpolynomial ring
over S is Auslander-Gorenstein.For the
proof
we refer to[Ek;
Theorem4.2].
Inparticular,
thisgives
that if S isan Auslander-Gorenstein
ring,
then so is R =A1(K) 0, S
since R may beregarded
as a skewpolynomial ring
in two variables over S.Moreover, Ro
is apolynomial ring
in one variable over S and hence an Auslander-Gorensteinring.
Let us now consider an Auslander-Gorenstein
K-algebra S
and setR =
gr(A 1 ) 0, S.
Let M = E9M,
be agraded
andfinitely generated
R-module.It follows
easily
that everyhomogeneous component Mv
is afinitely generated Ro-module.
So thegrade numbers jRo(M")
are defined for every v e Z. Now wehave:
1.3. THEOREM. Let S be an Auslander-Gorenstein
K-algebra
and setR =
gr(A1)~KS.
Let M =M, be a graded and finitely generated
R-module.Then
Before Theorem 1.4 below is announced we recall that if A is an Auslander- Gorenstein
ring
then afinitely generated
A-module M is pure if andonly
ifjA(N)
=jA(M)
for every nonzero submodule N of M.1.4. THEOREM. Let S be an Auslander-Gorenstein
K-algebra
and Ma finitely generated
andgraded
R-module. Then the R-module M is pureif
andonly if
eachnonzero
M v
is a pureRo-module
such that1.5. An
application
of Theorem 1.3We are
going
to use Theorem 1.3 in order tocompute projective
dimensions ofgraded
R-modules in the case R =gr(A 1 ) ~KD1, where D1
1 is theK-algebra given by
the skew field extension ofA1(K).
First we recall some
general
facts from[Na-Oy].
Ingeneral,
let R= ~v~ZRv
be a
Z-graded ring.
Assume that R is left andright
noetherian with a finiteglobal homological
dimension. Denoteby Gf(R)
thefamily
ofgraded
andfinitely generated
left R-modules.1.6. DEFINITION. The
graded left global homological
dimensionof
R is thesmallest
integer
w such thatproj.dim.R(M)
wfor
every M inG f(R),
whereproj.dim.R
denotes the usualprojective
dimension in the categoryof
R-modules.1.7. REMARK. Denote the
graded
leftglobal homological
dimensionby l.gr.gl.(R). Replacing
leftby right
we also obtain thegraded right global homological
dimensiongiven by
the smallestinteger
w’ such thatproj.dim.R(N)
w’ for everyfinitely generated
andgraded right
R-module N.By
the results inchapter
1 of[Na-Oy]
we have theequality
Let us
only
remark that(1.8)
is a consequence of the fact that bothl.gr.gl(R)
andr.gr.gl(R)
areequal
to the so-called weakgraded global dimension,
which is the smallestinteger
w" such thatTorR(N, M)
= 0 when v >w",
N is agraded right
and M is a
graded
left R-module.Using (1.8)
wesimply put l.gr.gl(R)
=gr.gl(R)
and refer togr.gl(R)
as thegraded global homological
dimension.By
Theorem Il.8.2 in[Na-Oy]
we haveNow we are
going
to prove1.10. THEOREM. Let S be the skew
field
extensionof A,(K). Then,
withR =
gr(A1) Q9KS
we havegl.dim.(R)
= 2 andgr.gl(R)
= 1.So Theorem 1.10
gives
anexample
when the strictinequality gr.gl(R) gl.dim.(R)
holds.Proof of
Theorem 1.10. Theequality gl.dim.(R)
= 2 isproved by
Hart in[H].
By (1.9)
weget
1gr.gl(R)
2. So there remains to prove thatgr.gl(R) 1,
i.e.we must show that
proj.dim.(M)
1 for everyfinitely generated
andgraded
R-module. To prove this we argue
by
acontradiction,
i.e. assume that there exists afinitely generated
andgraded
R-module M such thatproj.dim.(M)
> 1. Notice that we then obtainproj.dim.(M)
= 2 sincegl.dim.(R)
= 2.Then,
standardhomological algebra implies
thatExtR(M, R)
is non-zero. Set N =ExtR(M, R)
and notice that N is a
graded right
R-module. Auslander’s conditionimplies
thatjR(N)
2 and sincegl.dim.(R)
= 2 we must havejR(N)
= 2. Now N = E9Nv
and Theorem 1.3
gives
At this
stage
weget
the contradiction sinceRo
is thepolynomial ring
in onevariable over the skew field
D1. Thus, gl.dim.(Ro)
= 1 sojR0(Nv) 1
whenNv ~
0 and(i)
cannot hold.2.
Preliminary
resultsIn this section we collect various results about Auslander-Gorenstein
rings.
In
general,
let S be aring
and T one element of S such that T is neither a leftnor a
right
zero-divisor.Moreover,
we assume that ST= TS and the twosided idealgenerated by
T is denotedby (T).
If M is a left S-module such that TM =
0,
then weget
a leftS/(T)-module
structure on M.
Conversely,
if N is a leftS/(T)-module
we see that N has a left S-module structure such that TN = 0. Let J1 be the functor from the
category
offinitely generated
leftS/(T)-modules
to thecategory
offinitely generated
left S-modules.
Similarly
we have a functor from thecategory
offinitely generated right S/(T)-modules
to thecategory
offinitely generated right
S-modules.2.1. PROPOSITION. Let M be a
finitely generated S/(T)-module.
ThenExtv+1S(03BC(M), S) = 03BC(ExtvS/(T)(M, S/(T))) for ever y v
0.Proof.
For the case T is a central element ofS,
this is Theorem 9.37 of[Rot].
When T is not central the same methods can still be
applied
to prove theproposition.
REMARK.
Concerning
theinequality
we mention that
equality
holds if S is commutative. But in the noncommutativecase we can have a strict
inequality.
Forexample
consider theWeyl algebra An(K)
where K is a commutative field of characteristic zero.Using
thepositive Bernstein filtration
we construct the associated Reesring
denotedby
S. If T is thecentral element in S which
corresponds
to theidentity
inAn(K) put
indegree
oneof the
graded ring S,
thenS/(1 - T)
isisomorphic
withA,,(K). Moreover, An(K)
is a
ring
with a finiteglobal homological
dimension which isequal
to itsinjective
dimension. Hence we have that
On the other hand since
gr(An(K))
hasglobal homological
dimension2n,
itfollows that
Finally,
theinequality
is
proved
in[Ek]
and hence weget inj.dims(S)
= 2n + 1. So weget
the strictinequality
n =inj.dim T))
«inj.dims (S)
= 2n + 1 for everypositive integer
n.2.2. PROPOSITION. Let M be a
finitely generated S/(T)-module.
Thenis (/À(m» = jS/(T)(M)
+ 1.Proof.
This followsimmediately
fromProposition
2.1.2.3. THEOREM. Let S be an Auslander-Gorenstein
ring containing
an element Tas above. Then
S/(T)
is an Auslander-Gorensteinring.
Proof.
First we observe that thering S/(T)
is left andright
noetherian since S is.Next it follows from
Proposition
2.1 thatinj.dims/T>(S/(T)) 5 inj.dims(S) -
1.Now,
let M be afinitely generated S/{T)-module
and N a submodule ofExtvS/(T)(M, S/(T))
for some v > 0. Thenby Proposition
2.103BC(N)
is an S-submodule
of Extv+1S(03BC(M), S)
so the Auslander condition of thering
Sgives
thatjs(03BC(N))
v + 1.Finally Proposition
2.2gives
thatjS/(T)(N)=jS(03BC(N))
- 1 v + 1 - 1 = v
and the theorem isproved.
Let us consider a
ring A
and let p: A - A be aring automorphism.
2.4. DEFINITION. Let M be a
left
A-module and v c- Z. Then~v(M)
is theleft
A-module whose additive group is M and
if a
E A and m E~v(M)
then a * m isequal
top’(a)m
in theleft
A-module~v(M).
It is obvious that M -
~(M)
is an exact functor on thecategory
of left A- modules and we leave out the detailed verification of thefollowing:
2.5. LEMMA. Let M be
a finitely generated
A-module.Then jA(M) = jA(~(M)).
Next,
theautomorphism
p enables us to construct the skewpolynomial ring
B =
A[x, p]
where xa =
p(a)x
for every a in A.The
ring
B isgraded
withBv
= Ax" forevery v
0 whileBv
= 0 if v 0.From now on we assume that A is Auslander-Gorenstein.
By [Ek,
Theorem4.2]
B is also an Auslander-Gorensteinring.
Let us now consider a
finitely generated
andpositively graded
B-moduleM =
~v0Mv.
We note that every
M"
is afinitely generated
A-module.2.6. LEMMA. There exists an
integer
w such that x:M, -+ Mv+
1 isbijective for
every v w.
Proof.
Set K ={m~M:xm
=01
and N =M/xM.
We note that both K andN are
graded
andfinitely generated
B-modules such that xK = xN = 0. So wefind an
integer
v w suchthat v
wyields K,
=N"
= 0 and Lemma 2.6 follows.2.7. COROLLARY. Let M be a
graded
andfinitely generated
B-module. Thenthere exists an
integer
w such that v wyields jA (Mv)
=jA(Mw)’
Proof.
Lemma 2.6 and the construction of the skewpolynomial ring
B showthat
if v w
then the A-modulesM,
and~(Mv+1)
areisomorphic.
Then Lemma2.5
gives Corollary
2.7.Using Corollary
2.7 weget
2.8. DEFINITION. Let M = (9
M" be a graded
andfinitely generated
B-module. Then
limv~~jA(Mv)
exists and is denotedby j~(M).
2.9. PROPOSITION. Let M be
a finitely generated
andgraded
B-module. ThenjB(M)
=infv~Z{j~(M),
1+ jA(Mv)}.
Proof.
Choose w so thatx: Mv~Mv+1
isbijective
for all v w. Set N= ~vw Mv.
It follows that thegraded
B-submodule N of M isisomorphic
with the
graded
B-module B~A Mv- Moreover jA(Mw)=j~(M).
We have the exact
sequence 0 -
N - M -M/N ~
0 and the Auslander conditiongives:
The
equality jB(N)
=jA(Mw)
followsby
standardhomological algebra.
Finally, M/N
is agraded
B-module such thatxw+k(MjN)
= 0 for someinteger
k. Then standard
homological algebra gives:
The A-module
M/N
isisomorphic
with~vwMv
and hencejA(MjN) = infvw{jA(Mv)}.
Now weget Proposition
2.9.3. Proof of the main results
In this section we set R =
gr(A1) ~KS.
Here S is an Auslander-Gorenstein K-algebra.
Let M = ~M,
be afinitely generated
andgraded
R-module.3.1. LEMMA.
Every homogeneous
componentMv
is afinitely generated Ro-
module.Proof.
We notice thatif k 1
thenRk
=Rock
andR-k
=Roxk.
Now Lemma3.1 follows since M is a
graded quotient
of a free R-module of finite rank.Next, Ro
=gr0(A1) Q9KS
andgr0(A1)
is thepolynomial ring KV]
withV =
x~.
HenceRo
=S[V],
i.e. thepolynomial ring
in one variable over S.The next result is a crucial
step
towards theproof
of Theorem 1.3.3.2. LEMMA. For every v E 7L and n 1 we have
Proof.
Let v~Zand n
1. SetKv,n
={m~Mv: ~nm
=01
andTv,n
={m~Mv:
x"m =01.
We notice that
Kv,n
andTv,n
both areS[V]-submodules
ofMv.
Now we have:SUBLEMMA.
Kv,n
nTv,n
= 0.Proof of
Sublemma. Assume thecontrary
and take0 ~
m EK","
nTv,n’
Sincegr(A1)
is aK-subalgebra
of R weget
thecyclic gr(A 1 )-module generated by
m.By assumption
x"m= ~nm
= 0 and thisgives dimK(gr(A1)m)
oo. Thisgives
acontradiction since
gr(A1)
isisomorphic
toA 1 (K)
andby
the result from[G-R]
there does not exist any non-zero
A 1 (K)-module
whoseunderlying K-space
isfinite dimensional.
Proof
continued. We have the exact sequence:Next,
consider theinjective
map~n: Mv/Kv,n ~ Mv+n.
In thering
R we have~n~
=(V
+n)ô".
Let p be theautomorphism
on thering S[V]
definedby p(s)
= sfor every SES and
03C1(~) = ~
+ 1.Using
the notation from 2.4 we havethat
~-n(Mv/Kv,n)
isisomorphic
to aS[V]-submodule
ofMv+n.
Sincejs[v] (MvjKv,n)
it follows from remark 1.2 thatJS[VI (M,,
+n)-
Next the sublemma
implies
thatK",n
isisomorphic
with anS[V]-submodule
of
Mv/Tv,n. By
similar methods as above we haveNow we
get
and then Remark 1.2
gives
Lemma 3.2.3.3. The 03C4-filtration on R
If v 0 we
set 03C4v(R) = ~|k|v Rk.
We see that{03C4v(R)}v0
is apositive
filtrationon the
ring
R. Letgrt (R)
= ~03C4v(R)/03C4-1(R)
be the associatedgraded ring.
Theimage
of x inil(R)/io(R)
is denotedby
X.Similarly,
Y is theimage of 3
in03C41(R)/-03C40(R).
In order to describe the
ring gr,(R)
we also notice that if we set03C4(v) = 03C4v(R)/03C4v-1(R)
thenT(O)
=To(R)
=Ro
and hence theRo-element V
isidentified with an element in
T(O),
i.e. V ishomogeneous
ofdegree
zero in thegraded ring gr03C4(R). Moreover,
sincex·x·~-x·~·x=-x
holds inR,
it followsthat
holds in
gr03C4(R). Similarly,
we obtainLet us now consider the
ring S[V].
Denoteby
pi theautomorphism
onS[V]
defined
by 03C11(s)
= s for every s E S and03C11(~)
= V +1,
then Xa =p,(a)X
holdsin
gr03C4(R)
for everya~S[~]
=ï(0). Similarly,
if p. is theS[V]-automorphism
defined
by 03C12(s)
= s and03C12(~) = ~ - 1,
then Ya =p2(a)Y
holds ingr,(R)
forevery a E
S[V].
Now we notice that theautomorphisms
pi and 03C12 commute and hence we can construct the skewpolynomial ring S[~][z1,
z2; pl,P2].
Denoteby (z 1 z2)
the two-sided idealgenerated by
z1z2. With these notations we have:3.6. LEMMA. The
ring gr03C4(R)
isisomorphic
withProof.
We notice that(3.4)
and(3.5) yield
asurjective K-algebra morphism:
9:
S[VI[Zl,
Z2; Pl,03C12] ~ gr03C4(R)
which is theidentity
onS[V]
while z, ismapped
into X and Z2 into Y
Next,
in thering
R we havex3
inRo
and then X Ybelongs
to
io(R)
so theimage
of X Y in03C42(R)/03C41(R)
is zero. Thus XY= 0 ingr,(R).
Similarly
YX = 0 ingr03C4(R).
So the map 9 induces asurjective K-algebra morphism 03C8
from thequotient ring S[~][z1,
z,; 03C11,03C12]/(z1z2)
intogr,(R).
Thereremains
only
to seethat 03C8
isinjective.
To prove theinjectivity,
we notice that if v 1 then we have a direct sumdecomposition
Moreover, Rv
=ôWRo
andR-v
=xVRo
for v > 0. Then iteasily
followsthat 03C8
isinjective.
3.7. COROLLARY. The
ring gr03C4(R)
is Auslander-Gorenstein.Proof. By
the remark after Remark 1.2 the skewpolynomial ring S[~][z1,
z2;p i,
P2]
is Auslander-Gorenstein. With T = z1z2 weapply
Theorem 2.3 and then Lemma 3.6gives Corollary
3.7.Let us now consider a
finitely generated
andgraded
R-module M = (BMv.
Set
We notice that
03C4j(R)0393v
c0393j+v
andhence {0393v}
is a filtration on M when R isequipped
with thepositive
03C4-filtration. Set0393(v)
=0393v/0393v-1
and then we noticethat
3.10. LEMMA. 0
0393v/0393v-1
is afinitely generated grt(R)-module.
Proof.
From Lemma 3.1 it follows thatr(v)
is afinitely generated Ro-module
for every v.
Moreover,
since M is agraded
andfinitely generated
R-module it follows that there exists anon-negative integer
w such that v wyields Mv+1
=~Mv
andM-v-1
=-XM -,.
Then Lemma 3.10 followseasily.
At this
stage
we aregoing
to use some results concerned with filteredrings.
First,
the i-filtration on the Auslander-Gorensteinring R
ispositive,
i.e. Tv = 0 ifv 0. Then the material in
[Bj:2] yields
where
grr(M) =
~rvjrv-l
is thefinitely generated
andgraded gr03C4(R)-module
from Lemma 3.10.
Hence Theorem 1.3 will follow if we have
proved:
3.12. PROPOSITION. We have that
Proof. By (3.9)
we havewhere
r(v)
=Mv
~M-v
if v i 1 andr(O)
=Mo.
Put
N + = ~v 1 MU -
We have that thegr03C4(R)-elements
X and Y from(3.4)
and
(3.5)
which areimages
of xrespectively ~.
SincexM"
cMv-1
hold for every v1,
it follows thatXN + =
0.Also,
since~Mv
cMv +
1 we obtainYN +
cN +
and conclude that
N +
is agraded gr03C4(R)-submodule of gr0393(M)
annihilatedby
X.Next, put
B =S[V][zi,
z,; 03C11,P2]
and recall from Lemma 3.6 thatgrt(R)
=B/(z1, Z2).
So we can considerN +
as a left B-module and sinceXN + =
0 we obtainz1N+ =
0.Now
Proposition
2.2gives
Next, BI(z,)
isisomorphic
withS[~][z2; P2]
whichimplies
thatN +
also is agraded S[~][z2; p2]-module.
NowProposition
2.9gives
where joo(N +)
=limv -
+ 00js[~](Mv).
Let us also consider the element
z 1 z 2 in
thering
B. Anotherapplication
ofProposition
2.9gives
Let us now consider the
quotient
modulegr0393(M)/N+
and denote itby
N_.We notice that N - is a
graded griR)-module
with N_ =~v0N_(v)
where theS[~]-module N-(v)
isequal
toM-v
for every v 1 0. SinceôM -v
cM - v , 1 for
every u 0 we
easily get
YN_ = 0.Repeating
thesteps (a), (b)
and(c) reversing
the roles between X and T we obtain.
where j oo(N -)
=limv- + ~js[~](M-v).
At this
stage
we aregoing
to use Lemma 3.2 to finish theproof
ofProposition
3.12.
Namely,
Lemma 3.2yields
Then
(d), (e)
and(f) yield
Finally,
Remark 1.2 and the exact sequenceyield
So the
equality
in(g) gives Proposition
3.12.We have now finished the
proof of Theorem
1.3. Let us now consider afinitely generated
leftRo-module
N. We obtain agraded
left R-module X = ROR. N,
where we remark that if k is some
integer
then thehomogeneous component Rk
in the
ring
R is an R-module and thusRk 0 Ro
N is a leftRo-module
which isequal
to the k-thhomogeneous component
of thegraded
R-module X. Now Theorem 1.3gives
Concerning
theright
hand side in(3.13)
we have thefollowing:
3.14. LEMMA.
infv~Z{jR0(Nv)} =jR0(N).
Proof.
Recall thatRo
=S[V]
and let p be thering-automorphism
onS[V]
defined
by p(s)
= s for every SES andp(V) = V
+ 1.Next,
if v 1 thenRv
=ô"Ro
andusing
theq-functor
from Definition 2.4 we find that the R-module
N,
isisomorphic to ri - "(N)
andhence jR0(Nv)=jR0(N). Next,
we haveR-v
=x"Ro
and find that theRo-modules N-v
andil’(N)
areisomorphic.
Soagain jRo(N -v)
=jRo(N)
and Lemma 3.14 follows.3.15.
Proof of
Theorem 1.4. Let M = Ef)Mv
be afinitely generated
andgraded
R-module. Assume first that every non-zero mv is a pure
Ro-module
withiR0(Mv)
=jR(M).
Then we show that the R-module M is pure. To prove this weapply
thetheory
in[Ek:§3]
whichimplies
that M is a pure R-module ifjR (N)
=jR (M)
for every non-zero andgraded
submodule N of M. If N =~ Nv
isa
graded
R-submodule then thehypothesis yields jR0(Nv)
=jR (M)
whenNv i=-
0.Now we see that Theorem 1.3
gives jR (N) = jR (M).
Conversely,
assume that M is a pure R-module.Then,
if some non-zerohomogeneous component M,
fails to be a pureRo-module with jR0(Mv)
=jR (M),
it follows that there exists an
integer
v and a non-zeroRo-submodule
N ofMv
such that
jRo(N)
>jR(M). Using N,
we construct thegraded
left R-module%=R(8)RoN
where
%k
=Rk-v ~R0Nv for every k.
Lemma 3.14givesjR(%)
=jRo(N). Next,
we notice that there exists an R-linear map ~: N ~ M such that
qJ(%)
is thegraded
R-submodule of Mgenerated by
N. Thepurity
of Mgives jR(~(N)) = jR (M).
On the other hand wehave jR(N)) jR(N) >jR(M)
whichis a contradiction.
4.
Applications
Let R =
gr(A1) (8) K S
where S is an Auslander-GorensteinK-algebra
and letM = (B
Mv
be afinitely generated
andgraded
R-module. InProposition
3.1 wenoticed that every
homogeneous component Mv
is afinitely generated gro(R)-
module where
gro(R)
is thepolynomial ring S[V].
In the
special
case when everyM"
is afinitely generated
S-module we say thatthe
graded
R-module M isstrongly finitely generated.
We shall
give examples
ofstrongly finitely generated
R-modules in 4.5. But first we establish thefollowing:
4.1. THEOREM. Let M = (B
Mv be a graded
andstrongly finitely generated
R-module. Then
jR(M)
= 1 +infv~Z{jS(Mv)}.
Proof.
Ingeneral,
if N is anS[V]-module
such that itsunderlying
S-module isfinitely generated
thenJs(N) = js[VI (N) -
1by Proposition
2.2. Now we see that Theorem 1.3gives
Theorem 4.1.4.2. V-filtrations on
rings
of differentialoperators
Following example
1.5.5 inChapter
II of[Sch]
we consider anon-singular complex analytic hypersurface
Y of acomplex
manifold X. Let9-y
be the ideal of(9x
whose sections vanishon Y,
thatis,
thedefining
ideal of Y in X. If k E Z we setThen
{FkDX}
is a filtration onDX,
called theV-filtration of DX along
Y Ifdim(X)
= n + 1 and(x1, ... ,
xn,t)
are local coordinates such that Y ={(x, t):
t =01
thenFoéox
is thesubring of DX generated by
the zero-order differentialoperators
andDx1,..., Dxn
andtDt . If k
1 we haveand
Moreover,
letgDX = ~FkDX/Fk-1DX
be the associatedgraded
sheaf ofrings.
Thenwhere i: Y- X is the closed
embedding
andgr(A1(C))
is thegraded
one-dimensional
Weyl algebra
wheredeg(t) =- -1
anddeg(Dt)
= 1. If(xo, 0)
is apoint
in Y then the stalkE0y(xo)
is thering E0n
of differentialoperators
with coefficients in the localring On.
We see that the stalkand recall that
Dn
is an Auslander-GorensteinK-algebra
whoseglobal
hom-ological
dimension isequal
to n. So with S= Dn
we canapply
the main results to stalks of coherent andgraded G-9,-modules.
To
get specific examples
we first consider a coherent left-9x-module
M.Working locally if necessary
we assume that uIt has agood
V-filtration whichby
definition consists of an
increasing
sequence{FkM}
such that there existslocally
a finite set of
generators
m1, ..., ms of theDX-module M
andintegers k1 ··· ks
with
for every
integer
k. Then GM = ~FkM/Fk-1M
is a coherent andgraded GDX-
module. So for every
(xo, 0)
in Y weget
thegrade
numberTheorem 1.3 shows that this
grade
numberequals
Next, following
Definition 1.6.1 inChapter
III of[Sch]
we say that the coherentDX-module M
iselliptic along
Y if thefollowing
holds: the-9,-module
M is
generated by
a coherent(9,-submodule M0
such thatlocally
there existspolynomials b
= V" +a1(x, t)~m-1
+ ··· +am (x, t)
where V =tDt
anda, (x, t)~OX
andbM0
cF-1(DX)M0.
Following [Sch,
p.134]
we have the result below:4.3. PROPOSITION. Let M be a coherent
!!fix-module
which iselliptic along
thehypersurface
Y. Thenfor
everygood
V-filtration{FkM}
and(xo, 0)
EY,
itfollows
that
~FkM(x0,0)/Fk-1M(x0, 0) is
astrongly finitely generated G-9x(xo, 0)-
module.
Now Theorem 4.1 can be
applied
andidentifying
the stalk-qy(xo)
with thering !!fin
we obtain4.4. COROLLARY. Let Jt be a coherent
-9x-module
which iselliptic along
Y and{FkM}
agood V-filtration.
ThenjGDX(x0,0)(GM(x0, 0))
= inffor every
(xo, 0)
E Y.4.5. REMARK. In the
special
case when A is a coherent and pure-9,-module,
i.e. when every stalk
A(xo)
is a pureDX(x0)-module
where xobelongs
toSupp M,
it followsby
thegeneral
result in[Bj: 2,
Theorem3.8]
that therelocally
exists a
good
V-filtration{FkM}
such that if GA = ~FkM/Fk-1 M
and(xo, 0)
ESupp
GAI thenGM(x0, 0)
is a pureGDX(x0, 0)-module
withSo in the
special
case when M is purealong
Y we can use such aspecial good
V-filtration and
applying
Theorem 1.4 it follows first that if(xo, 0)
ESupp GM,
thenis
equal
tojDX(x0)(M(x0)
forevery k
such that(xo, 0) belongs
toSupp(FkM/Fk-1 1
Finally,
if M is both pure andelliptic
then we first notice that Theorem 1.4 and the actualproof of
Theorem V.1implies
thatif (xo, 0)
ESupp(FkM/Fk-1 M),
then the
DY(x0)-module FkM(x0, 0)/Fk-1M(x0, 0)
is pure withgrade
numberLet us remark that if M is a
holonomic DX-module
then vit iselliptic along
everynon-singular hypersurface
Y For a detailedproof
of this fact we refer to the article[L-S].
Also,
any holonomicDX-module
is pure and theequality
in(4.6) yields
where
(xo, 0)~Supp(FkM/Fk-1M).
This means thatFkM/Fk-1M
is aholonomic
DY-module
for everyinteger
k. We remark that this conclusion isalready
contained in the work[L-S]
while our main resultsyield
new factsconcerned with non-holonomic
DX-modules
and their associatedGDX-modules
obtained
by good
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