Top local cohomology and the catenaricity of the unmixed support of a finitely generated module

0.14 MB·English

by  Nguyen Tu Cuong

7
0
0 TOP LOCAL COHOMOLOGY AND THE
2
CATENARICITY OF THE UNMIXED SUPPORT

n
OF A FINITELY GENERATED MODULE
a
J

4
2
NGUYEN TU CUONGa, NGUYEN THI DUNG

]
C Institute of Mathematics
18 Hoang Quoc Viet Road, 10307 Hanoi, Vietnam
A
E-mail adress: [email protected]
.
h
at LE THANH NHANb
m Department of Mathematics
Thai Nguyen PedagogicalUniversity, Thai Nguyen, Vietnam
[
E-mail adress: [email protected]
1
v
5
7 Abstract.1 Let (R,m) be a Noetherian local ring and M a finitely generated
6
R−module with dimM = d. This paper is concerned with the following property
1
for the top local cohomology module Hd(M):
0 m
7
0 Ann(0:Hd(M) p)=p for all prime ideals p⊇AnnHmd(M).
/ m
h
t Inthispaperwewillshowthatthispropertyisequivalenttothecatenaricityofthe
a
unmixed support UsuppM of M which is defined by UsuppM =SuppM/U (0),
m M
whereU (0)isthe largestsubmoduleofM ofdimensionlessthand.Somecharac-
M
:
v terizations of this property in terms ofsystem ofparametersas wellas the relation
i between the unmixed supports of M and of the m-adic completion M are given.
X
c
r
a 1. Introduction
Throughout this paper, let (R,m) be a Noetherian local ring, M a finitely gen-
erated R−module with dimM = d, and A an Artinian R−module. For each ideal
I of R, we denote by V(I) the set of all prime ideals containing I.
An elementary property of finitely generated modules is that Ann(M/pM)= p
for all p∈V(AnnM). The dual question for Artinian modules is to ask whether
(*) Ann(0: p)=p for all p∈V(AnnA).
A
1Keywords Artinianmodule,toplocalcohomology, unmixedsupport,catenaricity.
AMS Classification 13D45, 13E10.
aSeniorAssociateMemberofICTP,Trieste,Italy.
bJuniorAssociateMemberofICTP,Trieste,Italy.
TypesetbyAMS-TEX
1
In case R is complete with respect to the m−adic topology, the property (*) is
satisfied for all Artinian R−modules A because of the Matlis duality between the
category of NoetherianR−modules and the categoryof Artinian R−modules. Un-
fortunately the property (*) is not satisfied in general. For example, let R be the
Noetherian local domain of dimension 2 constructed by D. Ferrand and M. Ray-
naund [FR] (see also [Na, App., Exam. 2]) such that its m−adic completion R has
an associated prime q of dimension 1. Then the Artinian R−module A = Hb1(R)
m
does not satisfy the pbroperty (*), cf [CN1]. However, it seems to us that the prop-
erty (*) is an important property of Artinian modules. For example, the property
(*) is closely related to some questions on dimension for Artinian modules. In
[CN1], it is shown that N-dimA=dimR/AnnA provided A satisfies the property
(*), where N-dimA is the Noetherian dimension of A defined by Roberts [R] (see
also [K2]). Note that this equality does not hold in general. Concretely, with the
Artinian R−module A = H1(R) as above, N-dimA = 1 < 2 = dimR/AnnA al-
m
thoughthisringRiscatenaryandthetoplocalcohomologymoduleH2(R)satisfies
m
the property (*).
The purpose of this paper is to study the property (*) for the top local coho-
mology Artinian module Hd(M), dimM = d, and its applications. We will show
m
that, although N-dimHd(M) and dimR/AnnHd(M) are always equal to d, the
m m
property (*) is not necessarily satisfied for Hd(M). Then we find conditions such
m
thatHd(M)satisfiestheproperty(*). Itseemssurprisingtous,thatthisproperty
m
is equivalent to some important properties of M. In particular, the property (*) is
satisfiedforHd(M)ifandonlyiftheunmixedsupportUsuppM =SuppM/U (0)
m M
is catenary, where U (0) is the largest submodule of M of dimension less than d.
M
Fromnowon,wedenotebyR(resp. M)the m−adiccompletionofR(resp. M)
and UsuppbM the unmixed subpport ofcM as an R−module. The main result of
R
this paper is cthe following theorem. c b
Theorem. The following statements are equivalent:
(i) Ann(0:Hd(M) p)=p for all p∈V(AnnHmd(M)).
m
(ii) UsuppM is catenary.
(iii) UsuppM ={p∩R: p∈UsuppbM}.
R
(iv) For every sequbence x ,b... ,x of elcements in m, (x ,... ,x ) is a system of
1 d 1 d
parameters of Hd(M) if and only if it is a system of parameters of M/U (0).
m M
Here,thenotionofsystemofparametersforArtinianmodulesisdefinedaccord-
ing to Section 2.
Asanimmediateconsequenceoftheabovemaintheorem,wehavethefollowing
characterizationfor the catenaricity of a Noetherian local domain.
Corollary. Supposethat(R,m)isaNoetherianlocaldomain ofdimensiond.Then
R is catenary if and only if Hd(R) satisfies the property (*).
m
This paper is divided into 4 sections. In Section 2 we introduce the property
(*) for Artinian modules and recall some basic facts that we need in the sequel.
In the last two sections we present the proof of the above main theorem. The
characterizations of the property (*) for Hd(M) in terms of system of parameters
m
andtherelationbetweentwosetsUsuppM andUsuppbM (thepart(i)⇔(iii)⇔(iv)
R
2 c
ofthemaintheorem)areprovedinSection3. InSection4,weprovetheequivalence
between the property (*) for Hd(M) and the catenaricity of UsuppM (the part
m
(i)⇔(ii) of the main theorem). We also examine some non-catenary domains to
clarify the results.
2. When is Ann(0: p)=p for all prime ideals p⊇AnnA?
A
For each Noetherian R−module M, it is clear that Ann(M/pM) = p for each
prime idealp∈V(AnnM).Therefore,foreachArtinianR−moduleA, weconsider
the following property:
(*) Ann(0: p)=p for all prime ideals p∈V(AnnA).
A
As we mentioned in the introduction, the property (*) is not satisfied for all
Artinian modules. In this section, we give some conditions such that this property
is satisfied.
First we have the following result which is proved in [CN1].
Proposition 2.1. A satisfies the property (*) if one of the following conditions
holds.
(i) R is complete with respect to the m−adic topology.
(ii) A contains a submodule which is isomorphic to the injective hull of R/m.
Itshouldbe mentionedthatSuppM ={p∩R : p∈SuppbM}foreachfinitely
R
generatedR−moduleM.ThismeansthatVb(AnnM)b={p∩R :cp∈V(Annb(M)}.
R
We also note that, for each Artinian R−module A, A hbas a nabtural structucre as
an Artinian R−module (cf. [Sh]), and with this structure, a subset of A is an
R−submodulebofAifandonlyifitisanR−submoduleofA.Thereforeitisnatural
to ask whether b
V(AnnA)={p∩R : p∈V(AnnbA}.
R
b b
Below we show that this equality holds if and only if A satisfies the property (*).
Recall that A has a minimal secondary representation A = A +...+A , where
1 n
A is p −secondary, and the set {p ,... ,p } does not depend on the choice of the
i i 1 n
minimal secondaryrepresentationofA.This setis denotedbyAttAandcalledthe
set of attached prime ideals of A, cf. [Mac]. It follows by [Sh] that
AttA={p∩R : p∈AttbA}.
R
b b
Proposition 2.2. The following conditions are equivalent:
(i) A satisfies the property (*).
(ii) V(AnnA)={p∩R: p∈V(AnnbA)}.
R
b b
Proof. (i)⇒(ii). Let p ∈ V(AnnbA). Then p ⊇ q for some minimal prime ideal
R
q containing AnnbA.bNote that any minimalbprimbe ideal containing AnnbA is a
R R
bminimal element of AttbA, cf. [Mac]. Therefore q ∈ AttbA. So, q∩R ∈ AttA.
R R
Henceq∩R∈V(AnnA)andhencep∩R∈V(AnnAb).Conversely,letbp∈V(AnnA).
Then Abnn(0 :A p) = p by the hybpothesis (i). Since p is a minimal prime ideal
containing Ann(0 : p), it follows that p ∈ Att(0 : p). Therefore there exists a
A A
3
prime ideal p∈AttRb(0:A p) such that p∩R=p. Since p∈AttRb(0 :A p), we have
p⊇AnnRb(0b:A p), and hence p∈V(AnnbRbA) with p∩Rb=p.
b(ii)⇒(i). Let p ∈ V(AnnA).bBy the hypothesis (iib), there exists p ∈ V(AnnbA)
R
such that p∩R=p. It follows by Matlis duality that AnnRb(0:A p)b=p. Therefore
b b b
p⊆Ann(0:A p)⊆AnnRb(0:A p)∩R=p∩R=p.
b b
Thus, Ann(0: p)=p. (cid:3)
A
Roberts [R] introduced the concept of Krull dimension for Artinian modules.
Kirby [K2] changed the terminology of Roberts and used the terminology of Noe-
therian dimensionto avoidconfusion with Krulldimension defined for finitely gen-
erated modules. In this paper we use the terminology of Kirby [K2]. The Noe-
therian dimension of A, denoted by N-dim A, is defined inductively as follows:
R
when A = 0, put N-dim A = −1. Then by induction, for an integer d ≥ 0,
R
we put N-dim A = d if N-dim A < d is false and for every ascending sequence
R R
A ⊆A ⊆... ofsubmodulesofA,thereexistsn suchthatN-dim (A /A )<d
0 1 0 R n+1 n
for all n>n .
0
The following result gives some good properties of Noetherian dimension for
Artinian modules which are in some sense dual to that of Krull dimension for
Noetherian modules, cf [R], [K1].
Lemma 2.3. ℓ(0: mn) is a polynomial for n≫0, and
A
N-dim A=degℓ(0: mn)=inf{t:∃x ,... ,x ∈m:ℓ(0: (x ,... ,x )R)<∞}.
R A 1 t A 1 t
ItfollowsbyLemma2.3thatthereexistsasequence(x ,... ,x ),d=N-dim A,
1 d R
such that ℓ(0: (x ,... ,x )R)<∞. A such sequence is called a system of param-
A 1 d
eters of A.
Note that, with the natural structure as an Artinian R−module, N-dim A =
R
N-dimRbA.Thereforethereis no confusioninwriting N-dimbAinsteadofN-dimRA
or N-dimbA. Moreover,it follows by Matlis duality that
R
N-dimA=dimR/AnnbA=max{dimR/p : p∈AttbA}.
R R
b b b b
Note that the set of all minimal elements of AttA is exactly the set of all minimal
prime ideals containing AnnA. So, we have
dimR/AnnA=max{dimR/p : p∈AttA}.
The following result shows the relation between N-dimA and dimR/AnnA.
4
Proposition 2.4. [CN1]. The following statements are true.
(i) N-dimA6dimR/AnnA.
(ii) If A satisfies the property (*) then N-dimA=dimR/AnnA.
Remark 2.5.
(i) As we mentioned in the introduction, there exist Artinian modules A such
that N-dimA<dimR/AnnA.
(ii) The converseof Proposition2.4,(ii) is not true. In the next sections, we will
show that the top local cohomology module A = Hd(M) of a finitely generated
m
R−moduleM ofdimensionddoesnotsatisfythe property(*)ingeneral,although
it always satisfies the condition N-dimA=dimR/AnnA=d.
3. The property (*) for the top local cohomology modules
From now on, let M be a finitely generatedR−module with dimM =d. In this
section,weexaminethe property(*)forthetoplocalcohomologymoduleHd(M).
m
We first present a nice property of Hd(M), cf. [CN1, Corollary 3.6].
m
Lemma 3.1. N-dimHd(M)=dimR/AnnHd(M)=d.
m m
Let U (0)be the largestsubmodule of M of dimensionless thand. Note that if
M
0= N(p) is a reduced primary decomposition of the zero submodule of M
\
p∈AssM
then U (0)= N(p), cf. [CN2]. Therefore we have
M \
dimR/p=d
AssM/U (0)={p∈AssM : dimR/p=d}.
M
Hence
SuppM/U (0)= V(p).
M [
p∈AssM, dimR/p=d
The set SuppM/U (0) is called the unmixed support of M and denoted by
M
UsuppM.
Lemma 3.2. Let p∈SuppM. Then p∈UsuppM if and only if p⊇AnnHd(M).
m
In particular, UsuppM =V(AnnHd(M)).
m
Proof. We have by [BS] that
AttHd(M)={q∈AssM : dimR/q=d}.
m
Moreover, the set of all minimal prime ideals containing AnnHd(M) and the set
m
of all minimal elements of AttHd(M) are the same. Therefore
m
V(AnnHd(M))= V(p)=UsuppM.
m [
p∈AssM, dimR/p=d
(cid:3)
There are some nice relations between associatedprimes and the supports of M
and of its m−adic completion M. For example, AssM = {p∩R : p ∈ AssbM}
R
and SuppM ={p∩R : p∈SucppbM}. Moreover, b b c
R
b b c
{p∈AssM : dimR/p=d}={p∩R :p∈AssbM, dimR/p=d}.
R
b b b b
So, it is natural to ask about the relation between UsuppM and UsuppbM. First
R
we have the following lemma. c
5
Lemma 3.3. UsuppM ⊇{p∩R: p∈UsuppbM}.
R
b b c
Proof. Letp∈UsuppM.Thenp⊇qforsomeq∈AssbM satisfyingdimR/q=d.
R
It follows thbat q∩R∈cAssM anbd dbimR/(q∩Rb)=d. Sicnce p∩R⊇q∩Rb, wbe get
p∩R∈UsuppbM. b b b (cid:3)
b
In general, the two sets UsuppM and {p∩R : p ∈ UsuppbM} are different
R
(cf. Proposition 4.6). The following theoremb shows tbhat they arecthe same if and
only ifHd(M)satisfiesthe property(*). Acharacterizationofthe property(*)for
m
Hd(M) in term of systems of parameters is also given.
m
Theorem 3.4. The following statements are equivalent:
(i) Hd(M) satisfies the property (*).
m
(ii) UsuppM ={p∩R: p∈UsuppbM}.
R
(iii) For every seqbuence xb,... ,x of eclements in m, (x ,... ,x ) is a system of
1 d 1 d
parameters of Hd(M) if and only if it is a system of parameters of M/U (0).
m M
Proof. (i) ⇔ (ii). We get by Lemma 3.2 that V(AnnHd(M)) = UsuppM and
m
V(AnnRbHmd(M)) = UsuppRbM. Therefore the condition (ii) is equivalent to the
condition c
V(AnnHmd(M))={p∩R : p∈V(AnnRbHmd(M))}.
b b
So, our claim follows by Proposition 2.2.
(i)⇒(iii). Let(x ,... ,x )beasystemofparametersofHd(M).LetI betheideal
1 d m
generated by x ,... ,x . For each prime ideal p of R containing I +AnnHd(M),
1 d m
we have by (i) that
p=Ann(0:Hd(M) p)⊇Ann(0:Hd(M) I).
m m
Therefore
rad(cid:0)I+AnnHmd(M)(cid:1)= \ p⊇rad(cid:0)Ann(0:Hmd(M) I)(cid:1).
p⊇I+AnnHd(M)
m
Hence rad(cid:0)I+AnnHmd(M)(cid:1)=rad(cid:0)Ann(0:Hmd(M) I)(cid:1). Since (x1,... ,xd) is a sys-
temofparametersofHmd(M),the lengthof(0:Hd(M) I)isfinite. So,wegetbythe
m
lastequalitythatI+AnnHd(M)isanm−primaryideal. Sincerad AnnHd(M) =
m (cid:0) m (cid:1)
rad Ann(M/U (0)) ,(cf. Lemma3.2),theidealI+Ann(M/U (0))ism−primary.
M M
(cid:0) (cid:1)
Hence(x ,... ,x )isasystemofparametersofM/U (0).Conversely,assumethat
1 d M
(x ,... ,x ) is a system of parameters of M/U (0). Then I +Ann(M/U (0)) is
1 d M M
m−primary, and hence so is I +AnnHmd(M). Therefore ℓ(0 :Hd(M) I) < ∞, i.e.
m
(x ,... ,x ) is a system of parameters of Hd(M).
1 d m
(iii)⇒ (i). Let p ∈ V(AnnHmd(M)). Assume that N-dim(0 :Hd(M) p) = d−r. By
m
[TZ, Proposition 2.10], there exist x ,... ,x ∈p which form a part of a system of
1 r
parameters of Hd(M) in p, and it is clear that this part of a system of parameters
m
is maximal. Let
0:Hd(M) (x1,... ,xr)R=A1+...+An
m
be a minimal secondary representation of 0 :Hd(M) (x1,... ,xr)R, where Ai is
m
q −secondary. For each element y ∈ m, note that y is a parameter element of
i
6
0:Hd(M) (x1,... ,xr)Rifandonlyify 6∈qi forallisatisfyingN-dimAi =d−r (cf.
m
[TZ,Lemma2.14]). Since(x ,... ,x )isamaximalpartofasystemofparameters
1 r
of Hd(M) in p, we have
m
p⊆ q
[ i
N-dimAi=d−r
andhencep⊆q forsomeisatisfyingN-dimA =d−r.Fromthehypothesis(iii),we
i i
cancheckthat(x ,... ,x )isamaximalpartofasystemofparametersofM/U (0)
1 r M
inp.So,thereexistsaprimeidealq∈Ass M/U (0)/(x ,... ,x )M/U (0) such
M 1 r M
(cid:0) (cid:1)
thatdimR/q=d−randp⊆q. Sincep∈Supp M/U (0)/(x ,... ,x )M/U (0) ,
M 1 r M
(cid:0) (cid:1)
it follows that p=q. Hence dimR/p=d−r. Since A is q −secondary,we have by
i i
Lemma 2.4,(i) that N-dimA 6dimR/q . Because p⊆q , we have
i i i
d−r =N-dimA 6dimR/q 6dimR/p=d−r.
i i
Hence p = qi and hence p ∈Att(0 :Hd(M) (x1,... ,xr)R). Therefore there exists a
m
prime ideal p ∈ AttRb(0 :Hmd(M) (x1,... ,xr)R) such that p∩R = p. This implies
that b b
p⊆Ann(0:Hmd(M) p)⊆AnnRb(0:Hmd(M) p)∩R=p∩R=p.
b b
Thus Ann(0:Hd(M) p)=p as required. (cid:3)
m
4. The catenaricity of UsuppM
WesaythatSuppM iscatenaryifforanyprimeidealsp,q∈SuppM withp⊂q,
all saturated chains of prime ideals starting from p and ending at q have the same
length. It is clear that SuppM is catenary if and only if the ring R/AnnM is
catenary. Therefore, SuppM is catenary and dimR/p = d for all minimal prime
idealsp∈AssM ifandonlyifdimR/p+dimM =dforallp∈SuppM,cf. Remark
p
18.5of[HIO].Inparticular,sincedimR/p=dforallp∈AssM/U (0),theumixed
M
supportUsuppM =SuppM/U (0)iscatenaryifandonlyifdimR/p+dimM =d
M p
for all p∈UsuppM.
The following theorem is the main result of this section, which shows that the
property (*) for Hd(M) is equivalent to the catenaricity of UsuppM.
m
Theorem 4.1. The following statements are equivalent:
(i) UsuppM is catenary.
(ii) Hd(M) satisfies the property (*).
m
Before proving Theorem 4.1, we need the following lemmas.
Lemma 4.2. Assume that R is complete with respect to the m−adic topology and
M a finitely generated R−module such that dimR/p = d for all p ∈ AssM. Then
dimR/p = d−r for any part of system of parameters (x ,... ,x ) of M and any
1 r
minimal associated prime ideal p of M/(x ,... ,x )M.
1 r
Proof. As (x ,... ,x ) is a part of a system of parameters of M, we have
1 r
dim(R/AnnM +(x ,... ,x )R)=dim(M/(x ,... ,x )M)=d−r.
1 r 1 r
Moreover p is a minimal prime divisor of AnnM +(x ,... ,x )R, so that dimR/p
1 r
is at most d−r. There is a minimal prime divisor q of AnnM which is contained
7
in p. As q belongs to Ass (M) it follows from our assumptions that dimR/q=d.
R
Moreover p is a minimal prime divisor of q+(x ,... ,x )R, so that ht(p/q) does
1 r
notexceedr (cf. [Mat,Theorem18]). AsR/qis catenaryofdimentiond, itfollows
that dimR/p=d−ht(p/q) is at least d−r. Therefore dimR/p=d−r. (cid:3)
Lemma 4.3. Let p ∈ V(AnnHd(M)) such that dimM +dimR/p = d. Then
m p
Ann(0:Hd(M) p)=p.
m
Proof. Let p⊇AnnHd(M) be a prime ideal suchthat dimM +dimR/p=d. Set
m p
dimR/p = d−r. It follows by the hypothesis that dimM = r. Therefore there
p
exists a prime ideal q∈AssM such that q⊆p and ht(p/q)=r. Since
dimR/q≥dimR/p+ht(p/q)=d,
it follows that dimR/q=d. It should be noted that dimR/pR=dimR/p=d−r.
So, there exists a prime ideal p ∈ AssbR/pR such thatbdimbR/p = d−r. Since
R
p∈AssbR/pR, we get p∩R∈bAssR/p,ib.e. pb∩R=p. Note thbatbthe naturalmap
R
bR −→ R ibs fabithfully flabt, and therefore the bgoing down theorem holds (see [Mat,
Theoremb 4]). So, there exists a prime ideal q∈SpecR such that q∩R=q, q⊆p
and ht(p/q)≥r. These facts imply that b b b b b
b b
d=dimR/q=dimR/qR≥dimR/q=dimR/p+ht(p/q)≥d−r+r =d.
b b b b b b b b
Hence dimR/q = d. Moreover, since the natural homomorphism Rq −→ Rbq is
faithfully flabt abnd Mq 6=0, we have b
Mq⊗Rq Rbq ∼=Mbq 6=0.
b c
Hence q ∈ SuppRbM. Since dimR/q = d and p ⊇ q, we have p ⊇ AnnRbHmd(M).
Therefobre we get bcy the Matlis dubalbity that AnbnRb(0b:Hmd(M) p)b=p. Now we have
b b
p⊆Ann(0:Hmd(M) p)⊆AnnRb(0:Hmd(M) p)∩R=p∩R=p.
b b
Thus Ann(0:Hd(M) p)=p. (cid:3)
m
Proof of Theorem 4.1. (i) ⇒ (ii) follows by Lemma 4.3 and Lemma 3.2.
(ii) ⇒ (i). Let p ∈ UsuppM. We need to show that dimR/p+dimM = d. The
p
case p=m is trivial. Assume that p6=m. Let dimR/p=d−r. Then it is enough
to prove dimM =r. Since p⊇AnnM/U (0), we have
p M
dim M/U (0) p(M/U (0)) =dimR/p=d−r.
(cid:16) M (cid:14) M (cid:17)
So,thereexistsamaximalpartofasystemofparameters(x ,... ,x )ofM/U (0)
1 r M
in p. Since p ∈ UsuppM, there exists by Theorem 3.4, (i)⇔(ii) a prime ideal
p ∈ UsuppRbM such that p∩R = p. Set M1 = M/UMc(0). Since (x1,... ,xr) is a
bpart of a systcem of paramebters of M/UM(0c), it isca part of a system of parameters
\
of the m−adic completion M/U (0) of M/U (0). Because M is a quotient of
M M 1
\ \
M/U (0) and dimM = dimM/U (0), it follows that (x ,..c. ,x ) is a part of a
M 1 M 1 r
systemofparametercsofM1.Notethatp∈SuppRbM1/(x1,... ,xr−1)M1.Therefore
c b8 c c
p⊇p1 forsomeminimal primeidealp1 ∈SuppRbM1/(x1,... ,xr−1)M1.Since xr is
ba pabrameter element of M /(x ,... ,bx )M , wecget by Lemma 4.2cthat x ∈/ p .
1 1 r−1 1 r 1
Set p1 = p1 ∩ R. Thencxr ∈/ p1. Therefocre p ⊃ p1 and p 6= p1. By the sambe
arguments,bthere exists a minimal prime ideal p2 ∈ SuppRbM1/(x1,... ,xr−2)M1
such that p ⊇p . Set p =p ∩R. Then p ⊃pband p 6=pcsince x ∈p \cp .
1 2 2 2 1 2 1 2 r−1 1 2
Continue tbhe abbove process,baftebr r steps, we get a chain p ⊃ p1 ⊃ p2... ⊃ pr of
primeidealscontainingAnnM suchthatp 6=p foralli=1,...r−1.Therefore
i i+1
dimM =r. (cid:3)
p
Proof of the main theorem. The equivalences between (i), (iii) and (iv) are
provedbyTheorem3.4. Theequivalencebetween(i)and(ii)isprovedbyTheorem
4.1. (cid:3)
Remark 4.4. The catenaricity of UsuppM is equivalent to the property (*) for
the top local cohomology module Hd(M) (see Theorem 4.1), but it is not related
m
to the property (*) of other local cohomology modules of M. In fact, let R be
the Noetherian local domain constructed by Ferrand and M. Raynaund [FR] of
dimension2suchthatthem−adiccompletionhasanassociatedprimeofdimension
1. It is clear that SuppR = UsuppR is catenary, but H1(R) does not satisfy the
m
property (*).
Let 0=M ⊂M ⊂...⊂M =M be the filtration of submodules of M, where
0 1 t
M isthelargestsubmoduleofM ofdimensionlessthandimM fori=1,... ,t.
i−1 i i
Such a filtration always exists, and it is unique. We call this filtration to be the
dimension filtration of M (cf. [CN2]). Let dimM =d for i=1,... ,t. Then it is
i i
easy to check that
SuppM = SuppM /M .
[ i i−1
i=1,...,t
Foreachi=1,... ,t,itshouldbenotedthatdimR/p=d forallp∈AssM /M .
i i i−1
Therefore we obtain by Theorem 4.1 the following result.
Corollary 4.5. SuppM is catenary if and only if Hdi(M /M ) satisfies the
m i i−1
property (*) for all i=1,... ,t.
Now we examine some non-catenary Noetherian local domains. Note that any
domain of dimension 2 is catenary, but there exist non-catenary Noetherian local
domains of dimension d for any d≥3 (cf. [B, (8)]).
Proposition 4.6. Let R be a non-catenary Noetherian local domain of dimension
3. Set
U ={p∈SpecR:dimR/p+htp=2};
V ={p∈SpecR:dimR/p+htp=3}.
Then the following statements are true
(i) UsuppR=SpecR=U ∪V and U,V 6=∅.
(ii) Ann(0:H3(R) p)=p for all p∈V. But Ann(0:H3(R) p)6=p for all p∈U.
m m
(iii) For each p∈V, there always exists p∈SuppR/Ub(0) such that p∩R=p.
R
But for each p∈U, there does not exist p∈bSuppR/Ubb(0) such that p∩bR=p.
R
b9 b b
(iv) N-dimH2(R)=2 and dimR/AnnH2(R)=3.
m m
Proof. (i). This is clear since R is a non-catenary domain.
(ii). It follows by the proof of Theorem 4.1 that Ann(0:H3(R) p)6=p for all p∈U
m
and Ann(0:H3(R) p)=p for all p∈V.
m
(iii). This follows by (ii) and by the proof of Theorem 3.4.
(iv). Let p ∈ U. Then dimR/p = 1. Let p ∈ SpecR such that p∩R = p. Then
dimR/p = 1. It follows by (iii) that p 6⊇ AbnnRbHm3(Rb). Moreoverb, htp ≥ htp = 1
by thbebgoing down theorem [Mat, Thbeorem 4]. Therefore there existbs q ∈ AssR
such that q⊂ p and q 6=p. Hence dimR/q≥ 2. Since p6⊇ AnnRbHm3(R),bit followbs
that dimRb/q b= 2. Sbo wbe have by [BSb, bCorollary 11b.3.3] that q ∈ AttRbHm2(R)
and hencebqb⊇AnnRbHm2(R). Therefore N-dimHm2(R)=dimR/AbnnRbHm2(R)≥ 2.
Note that Nb-dimHm2(R) 6 2 by [CN1, Theorem 3.1]. So N-dibmHm2(R) = 2. Since
q∈AttRbHm2(R)∩AssR, we have
b b
q∩R∈AttH2(R)∩AssR.
m
b
Since R is a domain, we have q∩R = 0. It follows that 0 = AnnH2(R). Thus
m
dimR/AnnHm2(R)=3. b (cid:3)
Acknowledgment. We wish to express our gratitude to the refree for his/her
useful suggestions and, especially, the shorter proof of Lemma 4.2.
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