Table Of ContentA GENERALIZED DADE’S LEMMA FOR LOCAL RINGS
PETTERANDREASBERGH&DAVIDA.JORGENSEN
6
1
0 Abstract. WeproveageneralizedDade’sLemmaforquotientsoflocalrings
2 by ideals generated by regular sequences. That is, given a pair of finitely
n generated modulesover sucharingwithalgebraicallyclosedresiduefield,we
a proveasufficient(andnecessary)conditionforthevanishingofallhigherExt
J orTorofthemodules. ThisconditioninvolvesthevanishingofallhigherExt
or Tor of the modules over all quotients by a minimal generator of the ideal
6
generated bytheregularsequence.
]
C
A
1. Introduction
.
h
t When is a module over a truncated polynomialring projective? For group alge-
a
bras of elementary abelian p-groups, a criterion was provided by Dade in [Dad]: if
m
the ground field is algebraically closed, then a finitely generated module over such
[
an algebra is projective if and only if its restriction to all the cyclic shifted sub-
1 groupsareprojective. This is now knownasDade’s Lemma. Interms of truncated
v polynomial rings, it can be restated as follows. Let k be an algebraically closed
8
field of positive characteristic p, consider the truncated polynomial ring
3
2 k[X ,...,X ]/(Xp,...,Xp)
1 c 1 c
1
and denote the image of the generatorsX by x . Then Dade’s Lemma saysthat a
0 i i
. finitely generated module over this ring is projective if and only if it is projective
1
over the subalgebra k[α x +···+α x ] for each c-tuple (α ,...,α ) in kc.
0 1 1 c c 1 c
Truncated polynomial rings are local complete intersections, that is, they are
6
1 quotients of regular local rings by ideals generated by regular sequences. For such
: rings,acohomologicalgeneralizationofDade’sLemmaisprovedimplicitly in[Avr]
v
i and [AvB]: instead of providing a criterion for finite projective dimension, a crite-
X
rion for vanishing of cohomology for pairs of modules is given.
r In this paper, we generalize Dade’s Lemma to quotients of arbitrarylocal rings,
a
and provide both a homological and a cohomological version. More precisely, let
(S,n,k) be a local ring with k algebraically closed, I ⊆ S an ideal generated by a
regularsequence,anddenotebyRthe quotientS/I. Inourmainresults(Theorem
3.3 and Theorem 3.4), we provide criteria for the vanishing of TorR(M,N) and
n
Extn(M,N) for n ≫ 0, where M and N are two finitely generated R-modules, in
R
terms of the vanishing of TorS/(f)(M,N) and Extn (M,N) for n ≫ 0 and all
n S/(f)
f ∈I\nI. ThespecialcasewhenN isthe residuefieldk ofR,sothatvanishingof
the (co)homology is equivalent to M having finite projective dimension, is proved
in [HJK, Corollary 2.2].
2010 Mathematics Subject Classification. 13D02,13D07.
Key words and phrases. Regularsequences, localrings,Dade’sLemma.
PartofthisworkwasdonewhilewewerevisitingtheMittag-LefflerInstituteinFebruaryand
March2015. WewouldliketothanktheorganizersoftheRepresentation Theoryprogram.
1
2 PETTERANDREASBERGH&DAVIDA.JORGENSEN
2. Chain maps
Throughout this paper, a local ring is a commutative ring which is both local
and Noetherian, and all modules are assumed to be finitely generated.
Let (S,n,k) be a local ring, f a non-zerodivisor in n2, and denote by R the
quotient S/(f). Consider a (chain) complex
F: ···→Fn+1 −∂−n−+→1 Fn −∂→n Fn−1 →···
of finitely generated free R-modules. By choosing bases for the free modules, we
may view the maps ∂ in F as matrices with coefficients in R. Now lift the whole
i
complex to S, and obtain a sequence
F: ···→Fn+1 −∂−n−+→1 Fn −∂→n Fn−1 →···
e e
offinitely generatedfreeS-modulesandmaps,withF⊗ RisomorphictoF. Thus
e e e e S
for each integer n, the free S-module F is of the same rank as the free R-module
n
F , and the map ∂ can be obtained from ∂ by choeosing preimages in S of all its
n n n
matrix entries. e
In general, theesequence F is not a complex, but the image of the composition
∂ ◦∂ is contained in fF for all n. Therefore, for every element x ∈ F
n+1 n+2 n n+2
there exists an element w ∈e F with ∂ ◦∂ (x) = fw . Since f is a non-
x n n+1 n+2 x
zeerodiveisor in S, this element we is unique, hence the assignment e
x
e e e
t : F →F , x7→w
n+2 n+2 n x
is a well-defined S-homomorphismfor each n. By definition, the equality
e e e
∂ ◦∂ =f ·t
n+1 n+2 n+2
holds for all n.
e e e
Lemma 2.1. For every n, the diagram
F ∂n+3 //F
n+3 n+2
e
e (cid:15)(cid:15)tn+3 e (cid:15)(cid:15)tn+2
F e ∂n+1 //Fe
n+1 n
e
commutes.
e e
Proof. This follows from the proof of [Eis, Proposition 1.1]. For an element x ∈
F , the definition of the maps t gives
n+3 i
e f · ∂n+1◦tn+3(ex) = ∂n+1 f ·tn+3(x)
(cid:16) (cid:17) = ∂ (cid:0)◦∂ ◦∂ (cid:1) (x)
e e en+1 ne+2 n+3
= f ·t ◦∂ (x)
n+2 n+3
e e e
= (cid:0)f · t (cid:1)◦∂ (x)
en+2 en+3
(cid:16) (cid:17)
Since f is a non-zerodivisor on Fn, the result folelows. e (cid:3)
e
A GENERALIZED DADE’S LEMMA FOR LOCAL RINGS 3
Now for each integer n, define an R-homomorphismt : F →F by
n+2 n+2 n
t =t ⊗ R
n+2 n+2 S
Thenby applying −⊗ R to the diagramsinthe lemma, we obtaina commutative
S
e
diagram
··· //F ∂n+3 //F ∂n+2 // F //···
n+3 n+2 n+1
tn+3 tn+2 tn+1
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
··· // Fn+1 ∂n+1 //Fn ∂n // Fn−1 //···
of free R-modules and maps. Consequently, the double sequence
def
t = (...,tn+1,tn,tn−1,...)
constitutes a degree −2 chain map
t: F →F
of complexes of free R-modules; this is [Eis, Proposition 1.1].
Notation. We denote by t(S,f,F) the chain map F → F of degree −2 just con-
structed.
e
This is Eisenbud’s original notation. It emphasizes the fact that the chain map
depends on the overlying local ring S, the non-zerodivisor f ∈ S, and the chosen
lifting F = (F ,∂ ) of the complex F to S. Note that, by [Eis, Corollary 1.4],
n n
if F = (F ,∂ ) is another choice of lifting of F to S, then the two chain maps
n n
t(S,f,Fe) andet(Se,f,F) are homotopic.
In the rest of this section, we shall consider the situation that occurs when the
ringReisobtainedfromS byfactoringoutanidealgeneratedbyaregularsequence
of length two. The above process then produces several degree −2 chain maps
F →F, and we shall link some of these to each other.
Setup. Let (S,n,k) be a local ring, f ,f a regular sequence in n2, and denote
1 2
by R the quotient S/(f ,f ). Furthermore, let α be a unit in S, and define the
1 2
following local rings:
T =S/(f ), T =S/(f ), T =S/(αf +f )
1 2 2 1 α 1 2
TheseeminglyasymmetricindexinginthepresentationsoftheringsT andT is
1 2
made up for by the fact that T /(f )=R=T /(f ). Note also that R=T /(f ).
1 1 2 2 α 1
We havechosennottodistinguishnotationallybetweenthe elementf inS andits
i
image in the various quotient rings.
Now consider a complex
F: ···→Fn+1 −∂−n−+→1 Fn −∂→n Fn−1 →···
of free R-modules again. As before, lift it to S to obtain a sequence
F: ···→Fn+1 −∂−n−+→1 Fn −∂→n Fn−1 →···
b b
of finitely generated free S-modules and maps, with F ⊗ R isomorphic to F.
S
b b b b
The image of ∂ ◦∂ is contained in f F +f F, hence there exist two graded S-
1 2
homomorphisms b
b b t : F →F,b t :bF →F
1 2
b b b b b b
4 PETTERANDREASBERGH&DAVIDA.JORGENSEN
both of degree −2, with
∂◦∂ =f ·t +f ·t
1 1 2 2
For each i∈{1,2}, apply −⊗ T to the sequence F to get a sequence
S i
b b b b
F ⊗S Ti: ···→Fn+1⊗S Ti −∂−n−+1−⊗−S−T→i Fn⊗S Tib−∂−n−⊗−S−T→i Fn−1⊗S Ti →···
b b
offreeT -modulesandmaps. Thisisalifting ofthe originalcomplexF toT ,since
i i
b b b b
(F ⊗ T )⊗ R is isomorphic to F. Moreover,the map
S i Ti
t ⊗ T : F ⊗ T →F ⊗ T
i S i S i S i
b
is a graded T -homomorphism of degree −2, with
i
b b b
(∂⊗ T )◦(∂⊗ T ) = (∂◦∂)⊗ T
S i S i S i
= f ·t +f ·t ⊗ T
1 1 2 2 S i
b b b b
= (cid:0)fi·ti⊗S Ti (cid:1)
b b
= f · t ⊗ T
i i S i
b
We then know from the first part of this section(cid:0) that th(cid:1)e map
b
t(T ,f ,F ⊗ T )=(t ⊗ T )⊗ R
i i S i i S i Ti
becomes a chain map F →F of degree −2.
Next, we repeat all this, butbnow for thebring T =S/(αf +f ). The sequence
α 1 2
F ⊗S Tα: ···→Fn+1⊗S Tα −∂−n−+−1⊗−−S−T→α Fn⊗S Tα −∂−n−⊗−S−T→α Fn−1⊗S Tα →···
b b
offreeT -modulesandmapsisaliftingofF toT . Here,thegradedT -homomorphism
α α α
b b b b
F ⊗ T →F ⊗ T of degree −2 of interest is (t −αt )⊗ T , since
S α S α 1 2 S α
(∂⊗ T )◦(∂⊗ T ) = (∂◦∂)⊗ T
S α S α S α
b b b b
= f ·t +f ·t ⊗ T
1 1 2 2 S α
b b b b
= (cid:0)f1·t1−f1·αt(cid:1)2+(αf1+f2)·t2 ⊗S Ti
b b
= (cid:0)f1·t1−f1·αt2 ⊗S Ti (cid:1)
b b b
= f(cid:0)1· (t1−αt2)⊗(cid:1)S Ti
b b
The map (cid:0) (cid:1)
b b
t(T ,f ,F ⊗ T )= (t −αt )⊗ T ⊗ R
α 1 S α 1 2 S α Tα
then becomes another degree −2 chain map F → F. The following result shows
(cid:0) (cid:1)
that this chain map equalsbthe obvious lbinear cbombination of the two chain maps
constructed above.
Proposition 2.2. Let (S,n,k) be a local ring, f ,f a regular sequence in n2, and
1 2
denote by R the quotient S/(f ,f ). Furthermore, let α be a unit in S, and define
1 2
the following local rings:
T =S/(f ), T =S/(f ), T =S/(αf +f )
1 2 2 1 α 1 2
Finally, take a complex (F,∂) of finitely generated free R-modules, and lift it to a
sequence (F,∂) of free S-modules and maps. Then for every unit α ∈ S, there is
an equality
b b
t(T ,f ,F ⊗ T )=t(T ,f ,F ⊗ T )−αt(T ,f ,F ⊗ T )
α 1 S α 1 1 S 1 2 2 S 2
of chain maps F →F of degree −2.
b b b
A GENERALIZED DADE’S LEMMA FOR LOCAL RINGS 5
Proof. This follows from the construction of the three chain maps. Let t and t
1 2
be the degree −2 homomorphisms on F with the property that
b b
∂◦∂ =f ·t +f ·t
b1 1 2 2
Then
b b b b
t(T ,f ,F ⊗ T ) = (t −αt )⊗ T ⊗ R
α 1 S α 1 2 S α Tα
= (cid:0)t1⊗S R −α t2(cid:1)⊗S R
b b b
= (cid:0)b(t1⊗S T(cid:1)1)⊗T(cid:0)1bR −α(cid:1)(t2⊗S T2)⊗T2 R
= (cid:0)t(T1,f1,F ⊗S T1)(cid:1)−αt(T(cid:0)2,f2,F ⊗S T2) (cid:1)
b b
(cid:3)
b b
3. Dade’s Lemma
Having gone through the necessary machinery on chain maps, we now start
establishing the main result. As before, let (S,n,k) be a local ring, f a non-
zerodivisor in n2, and denote by R the quotient S/(f). In order to prove our first
result, we need the following fact. It implies that if a complex of free S-modules
becomes exact after it is reduced modulo f, then the complex itself is exact.
Lemma 3.1. Suppose that
p q
X −→Y −→Z
is a complex of S-modules, with Y finitely generated and f a non-zerodivisor on Z.
If the reduced complex
p q
X/fX −→Y/fY −→Z/fZ
is exact, then so is the original complex.
Proof. Ify ∈Kerq,theny+fY ∈kerq,andsobyexactnessofthereducedcomplex,
thereexistsanelementx+fX ∈X/fX suchthaty+fY =p(x+fX)=p(x)+fY.
Then p(x)−y ∈ fY, so there exists an element y′ ∈ Y such that p(x)−y = fy′.
Now notice that
fq(y′)=q(fy′)=q(p(x)−y)=q(p(x))−q(y)=0
Since f is anon-zerodivisoronZ,it followsthatq(y′)=0,i.e. y′ ∈Kerq. We have
shown that
Kerq =Imp+fKerq
ThemoduleKerq isfinitelygenerated,beingasubmoduleofY,henceKerq =Imp
by Nakayama’s Lemma. (cid:3)
Now let M be an R-module, and choose a free resolution
F: ···→F −∂→3 F −∂→2 F −∂→1 F
3 2 1 0
of this module, not necessarily minimal. Thus F is a complex of finitely generated
free R-modules, with nonzero homology only in degree zero, where H (F) ∼= M.
0
As in the previous section, lift the complex to a sequence
F: ···→F −∂→3 F −∂→2 F −∂→1 F
3 2 1 0
e e e
e e e e e
6 PETTERANDREASBERGH&DAVIDA.JORGENSEN
of finitely generated free S-modules and maps, with F ⊗ R isomorphic to F. Let
S
t: F → F be the graded S-homomorphism of degree −2 satisfying ∂ ◦∂ = f ·t.
Thus for every n≥0 there is an S-homomorphism t e : F →F with
n+2 n+2 n
e e e e e e
∂ ◦∂ =f ·t
n+1 n+2 n+2e e e
The corresponding degree −2 chain map t⊗ R: F →F, denoted by t(S,f,F), is
e e S e
pictured in the commutative diagram
e e
··· // F ∂4 //F ∂3 //F ∂2 // F ∂1 //F
4 3 2 1 0
t4 t3 t2
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
··· //F ∂2 //F ∂1 //F
2 1 0
with t =t(S,f,F) . The mapping cone C of this chain map is the complex
i i t(S,f,F)
e
e −∂3 0 −∂2 0
··· //F ⊕F h t3 ∂2i // F ⊕F h t2 ∂1i //F ⊕F [−∂1 0] // F
3 2 2 1 1 0 0
with (Ct(S,f,F))n =Fn−1⊕Fn−2.
Thereisacanonicalwayofliftingthismappingconecomplextoasequenceover
e
S, in terms of the lifting F: replace everything by the appropriate map or module
over S, and insert the element f in the upper right corner of all the matrices.
The following result showes that this canonical lifting is an S-free resolution of the
module M. This result will appear in [Ste]; we include a proof for completeness.
Theorem 3.2. Let (S,n,k) be a local ring, f a non-zerodivisor in n2, and denote
by R the quotient S/(f). Furthermore, let F = (F ,∂ ) be an R-free resolution
n n
of an R-module M, and lift it to a sequence F = (F ,∂ ) of free S-modules and
n n
maps. Finally, for each n ≥0, let t : F →F be the S-homomorphism with
n+2 n+2 n
∂ ◦∂ =f ·t . Then the sequence e e e
n+1 n+2 n+2
e e e
e e e −∂3 −f −∂2 −f
··· //F ⊕F (cid:20) t3e ∂2 (cid:21) // F ⊕F (cid:20) t2e ∂1 (cid:21) //F ⊕F [−∂1 −f] // F
3 2 2 1 1 0 0
e e e e e
is an S-free resolution of M.
e e e e e e e
Proof. Denote the sequence by C(F,t), and note that it follows directly from the
definition of the maps t that it is a complex. We must show that the homology of
i
this complex is given by e e
e
M for n=0
H C(F,t) ∼=
n 0 for n6=0
(cid:16) (cid:17) (cid:26)
In what follows, denote by ˚ΣCe(Fe,t) the complex C(F,t) shifted one degree to
the left, that is, ˚ΣC(F,t) = C(F,t)n−1, and without changing the sign of the
n e e e e
differential. (cid:16) (cid:17)
Foreachn≥0,lette e=t ⊗eRe,andconsiderthe correspondingdegree−2
n+2 n+2 S
chain map
e t(S,f,F): F →F
e
A GENERALIZED DADE’S LEMMA FOR LOCAL RINGS 7
with t(S,f,F) =t . We see directly that its mapping cone C is isomorphic
i i t(S,f,F)
tothe complex ˚ΣC(F,t) ⊗SR. Nowconsiderthecanonicalshorteexactsequence
e
(cid:16) (cid:17)
(1) 0→F →C →F →0
e e t(S,f,F)
of complexes of free R-modules, where eache(Ct(S,f,F))n =Fn−1⊕Fn−2 sits in the
middle of the canonical split short exact sequence of free R-modules
e
0→Fn−2 →Fn−1⊕Fn−2 →Fn−1 →0
This short exact sequence of complexes gives rise to a long exact sequence of ho-
mology groups, and since H (F) = 0 for n ≥ 1, it follows that H (C ) = 0
n n t(S,f,F)
for n ≥ 3. This implies, by Lemma 3.1, that Hn(˚ΣC(F,t)) = 0 for n ≥ 3,egiving
H (C(F,t))=0 for n≥2. The isomorphisms
n
e e
M ∼=F /Im∂ ≃ F /fF / (Im∂ +fF )/fF ∼=F /(Im∂ +fF )
e e 0 1 0 0 1 0 0 0 1 0
show that H0(C(F,t))∼=(cid:16)eM, aend(cid:17)so(cid:16)it remeains oenly toes(cid:17)howethat H1e(C(F,et))=0.
Let [ab]T be an element in Ker[−∂1 −f]. Then ∂1(a)=−fb, and so a+fF1 ∈
Ker∂1 = Im∂2. Ceheoose an element ue∈ F2 such that a+fF1 = ∂2(−ue+efF2) =
−∂ (u)+fF . Then−∂ (u)−a belongsto fF , andetherefore−∂ (u)−fv=aefor
2 1 2 1 2
some v ∈F . Finally, the equalities e e e
1
e e e e e
f · t (u)+∂ (v) =∂ ◦∂ (u)+∂ (fv)=∂ ∂ (u)+fv =∂ (−a)=fb
2e 1 1 2 1 1 2 1
(cid:16) (cid:17) (cid:16) (cid:17)
combineed with tehe fact theat feis a noen-zerodiveisoreon F0 gives t2(eu)+∂1(v) = b.
We have shown that
−∂2 −f u = a e e e
" t2 ∂1 #(cid:20) v (cid:21) (cid:20) b (cid:21)
e
that is, that H (C(F,t))=0. (cid:3)
1 e e
Wecannowprovethemainresult,thegeneralizedDade’sLemmaforlocalrings.
Recall first that if (eS,en,k) is a local ring and I an ideal of S minimally generated
by c elements, then I/nI is a c-dimensional k-vector space. The image in I/nI
of any element f ∈ I \nI can be completed to a basis, and lifting these elements
back to S gives a minimal generating set (containing f) for the ideal I. By [BrH,
page 52], if I is generated by a regular sequence f ,...,f contained in n2, then
1 c
any other minimal generating set for I is also a regular sequence.
Recall also that a residual algebraic closure of S (the terminology is taken from
[AvB]) is a faithfully flat extension S ⊆ S♯ of local rings with the following prop-
erties: the maximal ideal of S♯ equals nS♯, and its residue field S♯/nS♯ is alge-
braicallyclosed. By [Bou, App., Th´eor`eme1, Corollarie],suchanextensionalways
exists. Now suppose that f ,...,f is a regular sequence in S, generating an ideal
1 c
I, and let R = S/I and R♯ = S♯/IS♯. By [AvB, Lemma 2.2], the sequence is also
regular in S♯, and R♯ is a residual algebraic closure of R. We denote by n♯ the
maximal ideal nS♯ of S♯, and by I♯ the ideal IS♯.
Theorem 3.3. Let (S,n,k) be a local ring, I ⊆ S an ideal generated by a regular
sequencef ,...,f containedinn2,anddenotebyRthequotientS/I. Furthermore,
1 c
letM andN betwofinitelygeneratedR-modules. Then thefollowing areequivalent
(1) TorR(M,N)=0 for all n≫0,
n
8 PETTERANDREASBERGH&DAVIDA.JORGENSEN
(2) there exists a residual algebraic closure S♯ of S, for which
TorS♯/(f)(S♯⊗ M,S♯⊗ N)=0
n S S
for all n≫0 and all f ∈I♯\n♯I♯,
(3) for every residual algebraic closure S♯ of S, the vanishing condition in (2)
holds.
In particular, if k itself is algebraically closed, then TorR(M,N)=0 for all n≫0
n
if and only if TorS/(f)(M,N)=0 for all n≫0 and all f ∈I\nI.
n
Proof. LetS♯ bearesidualalgebraicclosureofS,andR♯ thecorrespondingclosure
of R. Then R♯ is faithfully flat over R, and
TorR♯(R♯⊗ M,R♯⊗ N)∼=R♯⊗ TorR(M,N)
n R R R n
for all n. Therefore TorR(M,N)=0 if and only if TorR♯(R♯⊗ M,R♯⊗ N)=0.
n n R R
Consequently,wemayassumethatk isalgebraicallyclosed,andproveonlythelast
statement.
If c = 1, then I = (f ), and if f is an element in I \nI then f = αf for some
1 1
unit α∈S. Thus S/(f )∼=S/(f), and the result follows.
1
Next, suppose that c ≥ 2. We argue by induction on c, starting with the case
c=2, i.e. I =(f ,f ). Take an element f ∈I\nI, complete to a regularsequence
1 2
f,g generating the ideal I, and note that R∼=T/(g), where T =S/(f). Let (F,∂)
be anR-freeresolutionofM, andlift itto a sequence (F,∂)offree S-modules and
maps, with F ∼=F ⊗ R. The machinery in Section 2 produces a degree −2 chain
S
map b b
b t(T,g,F ⊗ T): F →F
S
which,forsimplicity,wedenotebyjustt. Thischainmapgivesrisetoashortexact
sequence b
0→F →C →F →0
t
ofcomplexesoffreeR-modules,asin(1),whereC isthemappingconeoft. Ineach
t
degree the short exact sequence of modules splits, and so when we apply −⊗ N
R
we obtain another short exact sequence
0→F ⊗ N →C ⊗ N →F ⊗ N →0
R t R R
of complexes of R-modules. The corresponding long exact sequence of homology
groups is
···→H (C ⊗ N)→TorR(M,N)−→s TorR (M,N)→H (C ⊗ N)→···
n+1 t R n n−2 n t R
where the connecting homomorphism s is induced by the chain map t, that is, s=
TorR(t,N). Moreover,byTheorem3.2anditsproof,thereexistsaT-freeresolution
F of M with the property that the mapping cone complex C is isomorphic to
T t
˚ΣF ⊗ R. Here ˚ΣF denotes the complex F shifted one degree to the left, but
T T T T
with no sign change on the differential. The homology of the complex C ⊗ N is
t R
therefore given by
H (C ⊗ N) ∼= H (˚ΣF ⊗ R)⊗ N
n t R n T T R
(cid:16) (cid:17)
∼= H ˚ΣF ⊗ N
n T R
∼= Tor(cid:16)T (M,N) (cid:17)
n−1
A GENERALIZED DADE’S LEMMA FOR LOCAL RINGS 9
for all n, and so the long exact homology sequence takes the form
···→TorT (M,N)→TorR (M,N)−→s TorR (M,N)→TorT(M,N)→···
n+1 n+1 n−1 n
where we have denoted by s the map TorR(t,N).
Suppose that TorR(M,N) = 0 for n ≫ 0. Then we see directly from the long
n
exact sequence that TorT(M,N) also vanishes for n ≫ 0. Since T = S/(f), and
n
f is an arbitrary element in I \nI, this proves one direction of the statement.
Conversely,suppose that TorS/(f)(M,N)=0 for all n≫0 and all f ∈I \nI. Let
n
α be a unit in S, and consider the three local rings
T =S/(f ), T =S/(f ), T =S/(αf +f )
1 2 2 1 α 1 2
Fori∈{1,2},denotebyt thedegree−2chainmapt(T ,f ,F⊗ T ): F →F,and
i i i S i
denote the degree −2 chain map t(T ,f ,F ⊗ T ): F →F by t . The vanishing
α 1 S α α
assumption implies in particular that TorT(M,N) = 0 forbn ≫ 0, if we replace
n
T by each of the rings T , T and T . Cobnsequently, in the long exact homology
1 2 α
sequence, the map
TorR (M,N)−→s TorR (M,N)
n+1 n−1
is an isomorphism for high n, if we replace the map s by each of the maps s =
1
TorR(t ,N), s = TorR(t ,N) and s = TorR(t ,N). However, it follows from
1 2 2 α α
Proposition 2.2 that s = s −αs . This implies that for every unit α ∈ S, the
α 1 2
maps s s−1 and
2 1
s s−1 =1−αs s−1
α 1 2 1
are isomorphisms TorR(M,N) → TorR(M,N) of R-modules for all n ≫ 0. If
n n
TorR(M,N) is nonzero, then this is impossible: reduce the R-module TorR(M,N)
n n
andthe isomorphismsmodulo the maximalidealin R. Thens s−1 and1−αs s−1
2 1 2 1
are k-vector space automorphisms for all α ∈ k. However, since k is algebraically
closed, the automorphism s s−1 has a nonzero eigenvalue λ ∈ k, and then 1 −
2 1
λ−1s s−1cannotbeanisomorphism. ThereforeTorR(M,N)mustvanishforn≫0.
2 1 n
We havenow establishedthe casec=2,andproceedbyinduction withthe case
c ≥ 3. First, suppose that TorR(M,N) = 0 for all n ≫ 0, and take an element
n
f ∈ I \ nI. Complete to a regular sequence f,f ,...,f generating I, and let
2 c
T = S(f,f2,...,fc−2). Then R ∼= T/(fc−1,fc), and the established case c = 2
gives TornT/(fc−1)(M,N) = 0 for n ≫ 0. Since T/(fc−1) ∼= S/(f,f2,...,fc−1), it
follows by induction that TorS/(f)(M,N)=0 for n≫0.
n
Conversely, suppose that TorS/(f)(M,N) = 0 for n ≫ 0 and for all f ∈ I \nI,
n
andconsidertheringT =S/(fc). Then(theimageinT of)f1,...,fc−1isaregular
sequenceinT,generatinganidealJ ⊆T withT/J ∼=R. Denotethemaximalideal
of T by m, and let g be an arbitrary element in J \mJ. Lift this element back
to S and obtain an element g′ ∈(f1,...,fc−1)\n(f1,...,fc−1), and note that the
sequence f ,g′ is regular. Now take any element f in (f ,g′)\n(f ,g′). Then f
c c c
also belongs to I \ nI, hence TorS/(f)(M,N) = 0 for n ≫ 0. The case c = 2
n
then gives TorS/(fc,g′)(M,N)=0 for n≫0. Since S/(f ,g′)∼=T/(g), this implies
n c
that TorT/(g)(M,N) = 0 for n ≫ 0. As g was an arbitrary element in J \mJ, it
n
now follows from the induction hypothesis that TorR(M,N) = 0 for n ≫ 0. This
n
completes the proof. (cid:3)
10 PETTERANDREASBERGH&DAVIDA.JORGENSEN
WhatistheconnectionbetweenthisresultandDade’sLemma,i.e.[Dad,Lemma
11.8]? Let E be a finite elementary abelian p-group and k an algebraically closed
field of characteristic p. Then Dade’s Lemma says that a finitely generated kE-
module is projective if and only if its restriction to each of the cyclic shifted sub-
groups of E is projective. Now recall that the group algebra kE is isomorphic to
the truncated polynomial ring
k[X ,...,X ]/(Xp,...,Xp)
1 c 1 c
where c is the rank of E. For simplicity, we identify kE with this ring. Let x
i
denote the coset X +(Xp,...,Xp) in kE. Given a c-tuple α=(α ,...,α ) in kc,
i 1 c 1 c
denotetheelementα x +···+α x inkE byx . Thenxp =0,andsotheelement
1 1 c c α α
x corresponds to an element in E generating a cyclic shifted subgroup of E. The
α
subalgebrak[x ] ofkE generatedby x is isomorphic to the truncated polynomial
α α
ring k[y]/(yp).
WhenweidentifykEandthetruncatedpolynomialringasabove,Dade’sLemma
takes the following form: a finitely generated kE-module is projective if and only
if it is projective over the subalgebra k[x ] for every α∈kc. However, a module is
α
projectiveoverk[x ]ifandonlyifitisprojectiveoverk[X ,...,X ]/(αpXp+···+
α 1 c 1 1
αpXp); this is implicit in the proof of [Avr, Theorem 7.5], and proved explicitly
c c
in [BeJ, Lemma 4.3]. Thus with S = k[X ,...,X ], the regular sequence being
1 c
Xp,...,Xp, and N =k, Theorem 3.3 above is equivalent to Dade’s Lemma.
1 c
Remark. The change of rings long exact homology sequence
···→TorT (M,N)→TorR (M,N)−→s TorR (M,N)→TorT(M,N)→···
n+1 n+1 n−1 n
in the proof of Theorem 3.3 is standard and appears many places in the literature
(hereR=T/(g)andg isanon-zerodivisorinT). Itisnothingbutthedegenerated
change of rings spectral sequence
TorR M,TorT(N,R) =⇒TorT (M,N)
p q p+q
p
(cid:0) (cid:1)
butcanalsobederivedinacompletelyelementaryway,asin[Mur,proofofLemma
1.5]. However, in order to obtain an explicit description of the connecting homo-
morphism
TorR (M,N)−→s TorR (M,N)
n+1 n−1
we have to involve techniques as those in Section 2. The cohomology version is
extensively treated in [Avr, Sections 1 and 2].
We endthis paperwiththe cohomologyversionofTheorem3.3. Itsproofisjust
aslightvariationofthatofthehomologyversion. Inthe specialcasewhenthering
R is a complete intersection, this result can be deduced from the results in [AvB].
Theorem 3.4. Let (S,n,k) be a local ring, I ⊆ S an ideal generated by a regular
sequencef ,...,f containedinn2,anddenotebyRthequotientS/I. Furthermore,
1 c
letM andN betwofinitelygeneratedR-modules. Then thefollowing areequivalent
(1) Extn(M,N)=0 for all n≫0,
R
(2) there exists a residual algebraic closure S♯ of S, for which
Extn (S♯⊗ M,S♯⊗ N)=0
S♯/(f) S S
for all n≫0 and all f ∈I♯\n♯I♯,
