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ON THE FINITE GENERATION OF A F AMILY OF EXT MODULES TONY J.PUTHENPURAKAL Dedicated to Prof.L.L.Avramov on the occasion of his sixtieth birthday Abstract.Let Q be a Noetherian ring with finite Krull dimension and let f =f 1,...f c be a regular sequence in Q .Set A =Q/(f ).Let I be an ideal in A ,and let M be a finitely generated A -module with projdim Q M finite.Set R (I )=L n ≥0I n ,the Rees-Algebra of I .Let N =L j ≥0N j be a finitely generated graded R (I )-module.We show that M j ≥0M i ≥0Ext i A (M,N j )is a finitely generated bi-graded module over S =R (I )[t 1,...,t c ].We give two applications of this result to local complete intersection rings.1.introduction Let A be a Noetherian ring.Let I be an ideal in A and let M be a finitely gener-ated A -module.M.Brodmann [4]proved that the set Ass A M/I n M is independent of n for all large n .This result is usually deduced by proving that Ass A I n M/I n +1M is independent of n for all large n .Some Generalizations of Brodmann’s result Fix i ≥0.The following sets are independent of n for all large n .1(L.Melkerson and P.Schenzel)[10,Theorem 1](a)Ass A Tor A i (M,I n /I n +1).(b)Ass A Tor A i (M,A/I n ).2(same argument as in 1(a)).Ass A Ext i A (M,I n /I n +1).3(D.Katz &E.West;[8,3.5])Ass A Ext i A (M,A/I n A ).
An example of A.Singh [12]shows that
Ass A lim →Ext i A (A/I n
,M )need not be finite.
So in this example
n ≥1
Ass A Ext i A (A/I n ,M )is not even finite .
2TONY J.PUTHENPURAKAL
We state some questions in this area which motivated me.
(1)(W.Vasconcelos:[13,3.5])Is the set
i≥0Ass A Ext i A(M,A)finite?
(2)(L.Melkerson and P.Schenzel:[10,page936])Is the set
i≥0 n≥0Ass A Tor A i(M,A/I n)finite?
The motivation for the main result of this paper came from a Vasconcelos’s ques-tion.The author now does not believe that Vasconcelos’s question has a positive answer in this generality.However he is unable to give a counter-example.Note that if A is a Gorenstein local ring then Vasconcelos’s question has,trivially,a positive answer.If we change the question a little then we may ask:if M,D are twofinitely generated A-modules then is the set
i≥0Ass A Ext i A(M,D)finite?
This is not known for Gorenstein rings in general.Using Melkerson and Schenzel’s question as a guidepost the questions I was interested to solve was Let(A,m)be a local complete intersection of codimension c.Are the sets
(a) i≥0 j≥0Ass A Ext i A(M,D/I j D)finite?
(b) i≥0 j≥0Ass A Ext i A(M,I j D)finite?
In Theorem5.9we prove that(b)holds.I have been unable to verify whether(a) holds.
The main result in this paper is the following regardingfinite generation of a family of Ext modules.Let R(I)= n≥0I n t n be the Rees algebra of I. Theorem1.Let Q be a Noetherian ring withfinite Krull dimension and let f= f1,...f c be a regular sequence in Q.Set A=Q/(f).Let M be afinitely generated A-module with projdim Q Mfinite.Let I an ideal in A and let N= n≥0N n be a finitely generated R(I)-module.Then
E(N)= i≥0 n≥0Ext i A(M,N n)
is afinitely generated bi-graded S=R(I)[t1,...,t c]-module.
An easy consequence of this result is that(b)holds(by taking N= n≥0I n D); see Theorem5.1.A complete local complete intersection ring is a quotient of a regular local ring mod a regular sequence.So in this case(b)holds from Theorem 5.1.The proof of(b)for local complete intersections in general is a little technical; see Theorem5.9
We next discuss a surprising consequence of Theorem1.Let(A,m)be a local complete intersection of codimension c.Let M,N be twofinitely generated A-modules.Define
cx A(M,N)=inf b∈N|n b−1<∞
ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES3 In this section6we prove,see Theorem6.1,that
(†)cx A(M,I j N)is constant for all j≫0.
We now describe in brief the contents of this paper.In section one we give a module structure to E(N)over S(as in Theorem1).We also discuss a few preliminaries.The local case of Theorem1is proved in section2while the global case is proved in section3.In section4we prove our results on asymptotic primes. In section5we prove(†).
generatedAcknowledgements:The author thanks Prof.L.L.Avramov and Prof.J.Herzog for many discussions regarding this paper
Let Q be a Noetherian ring and let f=f1,...f c be a regular sequence in Q. Set A=Q/(f).Let M be afinitely generated A-module with projdim Q Mfinite. We will not change M throughout our discussion.Let I an ideal in A and let N= n≥0N n be afinitely generated R(I)= n≥0I n t n-module.Set
E(N)= i≥0 n≥0Ext i A(M,N n).
In this section we show E(N)is a bi-graded S=R(I)[t1,...,t c]-module.We also discuss two preliminary results that we will need later in this paper.
2.1.Let F:···F n→···F1→F0→0be a free resolution of M as a A-module.
Let t1,...t c:F(+2)→F be the Eisenbud-operators[6,section1.]Then
(1)t i are uniquely determined up to homotopy.
(2)t i,t j commute up to homotopy.
2.2.Set T=A[t1,...,t c]with deg t i=2.
In[7]Gulliksen shows that if projdim Q M isfinite then i≥0Ext i A(M,L)is a finitely generated T-module.
2.3.Let N= n≥0N n be a f.g module over R(I).Let u=xt s.The map
N n u−→N n+1yields
Hom(F,N n)
u
Hom(F,N n)(+2)
u
Hom(F,N n+s)(+2)
Taking homology gives that
E(N)= i≥0 n≥0Ext i A(M,N n)is a bi-graded S=R(I)[t1,...,t c]-module. Remark2.4.1.For each i,we have n≥0Ext i A(M,N n)is afinitely generated R(I)-module.
2.For each n,we have i≥0Ext i A(M,N n)is afinitely generated A[t1,...,t c]-module.
We state two Lemma’s which will help us in proving Theorem1.
4TONY J.PUTHENPURAKAL
2.5.Notation
(1)Let N= n≥0N n be a graded R(I)-module.Fix j≥0.Set
N≥j= n≥j N n.
E(N≥j)is naturally isomorphic to the submodule
E(N)≥j= i≥0 n≥j E(N)ij
of E(N).
(2)If A→A′is a ring extension and if D is an A-module then set D′=D⊗A A′. Notice that if D isfinitely generated A-module then D′is afinitely generated A′-module.
(3)Set S′=S⊗A A′.Notice S′is afinitely generated bi-graded A′-algebra.Let U= i≥0 n≥0U i,n be a graded S-module.Then
U′=U⊗A A′= i≥0 n≥0U′i,n
is a graded S′-module.
Lemma2.6.If E(N≥j)is afinitely generated S-module then E(N)is afinitely generated S-module.
Proof.Set D=E(N)/E(N≥j)We have the following exact sequence of S-modules
0−→E(N≥j)−→E(N)−→D−→0.
Using Gulliksen’s result it follows that D is afinitely generated T=A[t1,...,t c]-module.Since T is a subring of S,we get that D is afinitely generated S-module. Thus if E(N≥j)is afinitely generated S-module then E(N)is afinitely generated S-module Lemma2.7.[with notation as in2.5(3)]Let A→A′be a faithfullyflat extension of rings and let U= i≥0 n≥0U i,n be a graded S-module such that U i,n is afinitely generated A-module for each i,n≥0.If U′is afinitely generated S′-module.then U is afinitely generated S-module.
Proof.Choose afinite generating set L of U′.Suppose deg i u≤a and deg n u≤b for each u∈L.For0≤i≤a and0≤n≤b choose afinite generating set C i,n of U i,n.Set
C=
a
i=0b n=0C i,n.
Set C′={u⊗1|u∈C}.Let V be the submodule of U generated by C.By construction C′generates U′.So U′=V′.Thus(U/V)⊗A A′=0.Since A′is a faithfullyflat A-algebra we get U=V.So U is afinitely generated S-module.
3.The local case
In this section we prove Theorem1when(Q,n)is local.Let m be the maximal ideal of A.Set k=A/m.Let I be an ideal in A.Set F(I)=R(I)⊗A k= n≥0I n/m I n thefiber cone of I.
ON THE FINITE GENERATION OF A FAMILY OF EXT MODULES5 3.1.Assume N= n≥0N n is afinitely generated R(I)-module.Notice
F(N)=N⊗A k= n≥0N n/m N n
is afinitely generated F(I)-module.Set
spread(N):=dim F(I)N/m N the analytic spread of N.
Proof of Theorem1in the local case.
Case1.The residuefield k=A/m is infinite.
We induct on spread(N).
First assume spread(N)=0.This implies that N n/m N n=0for all n≫0. By Nakayama Lemma,N n=0for
all n≫0;say N n=0for all n≥j.Then E(N≥j)=0and it is obviously afinitely generated S-module.By2.6we get that E(N)is afinitely generated S-module.
When spread(N)>0then there exists u=xt∈R(I)1which is N⊕F(N)-filter ,there exists j such that
(0:N u)n=0and(0:F(N)u)=0for all n≥j.
Set N≥j= n≥j N n and U=N≥j/uN≥j.Notice we have an exact sequence of R(I)-modules
0−→N≥j(−1)u−→N≥j−→U−→0.
For each n≥j the functor Hom A(M,−)induces the following long exact sequence of A-modules
0−→Hom A(M,N n)u−→Hom A(M,N n+1)−→Hom A(M,U n+1)
−→Ext1A(M,N n)u−→Ext1A(M,N n+1)−→Ext1A(M,U n+1)
···························
−→Ext i A(M,N n)u−→Ext i A(M,N n+1)−→Ext i A(M,U n+1)
···························
Using the naturality of Eisenbud operators we have the following exact sequence of S-modules
−−−→E(N≥j)−→E(U)
E(N≥j)(−1,0)(u,0)
By construction
spread(U)=spread(N≥j)−1=spread(N)−1.
By induction hypothesis E(U)is afinitely generated S-module.Therefore by Lemma3.2we get E(N≥j)is afinitely generated S-module.Using2.6we get that E(N)isfinitely generated S-module.
Case2.The residuefield k isfinite.
In this case we do the standard trick.Let Q′=Q[X]n Q[X].Set A′=A⊗Q Q′. Notice A′=Q[X]m Q[X]is aflat A-algebra with residuefield k(X)which is infinite. Set I′=IA′and M′=M⊗Q Q′=M⊗A A′.Notice projdim Q′M′isfinite.Set R(I)′=R(I′)the Rees algebra of I′.Notice that N′=N⊗A A′is afinitely generated R(I)′-module.Also note that E(N′)=E(N)⊗A A′.
By Case1we have that E(N′)is afinitely generated S′-module.So by Lemma 2.7we get that E(N)is afinitely generated S-module.
The next Lemma is a bi-graded version of Lemma2.8(1)from[11].
6TONY J.PUTHENPURAKAL
Lemma  3.2.Let R be a Noetherian ring(not necessarily local)and let B= i,j≥0B i,j be afinitely generated bi-graded R-algebra with B0,0=R.Set
B x= i≥0B(i,0)and B y= j≥0B(0,j).
Let V= i,j≥0V i,j be a bi-graded B-module such that
(1)V i,j is afinitely generated R-module for each i,j≥0.
(2)For each i≥0,V i= j≥0V i,j isfinitely generated as a B y-module
(3)For each j≥0,V j= i≥0V i,j isfinitely generated as a B x-module
(4)There exists z∈B(r,0)(with r≥1)such that we have the following exact
sequence of B-modules
V(−r,0)z−→Vψ−→D
where D is afinitely generated bi-graded B-module.
Then V is afinitely generated B-module
Proof.Step1.We begin by reducing to the case whenψis surjective.
Notice D′=imageψis afinitely generated bi-graded B-module.Ifψ′:V→D′is the map induced byψthen we have an exact sequence
V(−r,0)z−→Vψ′−→D′−→0.
Thus we may assumeψis surjective.
Step2.Choosing generators:
2.1:Choose afinite set W in V of homogeneous elements such that
ψ(W)={ψ(w)|w∈W}
is a generating set for D.
2.2:Assume all the elements in W have x-co-ordinate≤c.
2.3:For each i≥0,by hypothesis V i is afinitely generated B y-module.So we may choose afinite set P i of homogeneous elements in V i which generates V i as a
B y-module.
2.4:Set
G=W  c i=0P i .
Clearly G is afinite set.
Claim:G is a generating set for V.
Let U be the B-submodule of V generated by G.It suffices to prove that U i,j=V i,j for all i,j≥0.By construction we have that for0≤i≤c
(*)U i,j=V i,j for each j≥0
Let be the lex-order on X=Z≥0×Z≥0.It is well-known that is a total order on X.So we can prove our result by induction on X with respect to the total order .
The base case is(0,0).
In this case U0,0=V0,0by(*).
Let(i,j)∈X\{(0,0)}and assume that for all(r,s)≺(i,j);we have U i,j=V i,j.
Subcase1.i≤c.
By(*)we have U i,j=V i,j.

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