A note on the associated primes of local cohomology modules

A note on the associated primes of local cohomology
modules
Keivan Borna Lorestaniab , Parviz Sahandiab , and Tirdad Sharif
a
a
Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran
b
Department of Mathematics, University of Tehran, Tehran, Iran
Abstract
In this note we give a simple proof of the folowing result: Let R be a commutative noetherian
ring, a an ideal of R and M a finite R-module, if Hia (M ) has finite support for all i < n then
Ass (Hna (M )) is finite.
1. Introduction
Throughout this note all rings are commutative noetherian with nonzero identity and all
modules are finite (i.e. finitely generated). Let R be a commutative noetherian ring, a an
ideal, and M a finite R-module. A crucial problem in local cohomology is determining when
the set of associated primes of the i-th local cohomology module, Hia (M ), with support in a is
finite. The question of finiteness of associated primes of the local cohomology module Hia (R)
when R is a regular local ring was first raised by C. Huneke [H]. In this direction, when R
contains a field of positive characteristic, C. Huneke and R. Sharp [HS] gave an affirmative
answer to this question. Further, when R is regular contains a field of zero characteristic or
is mixed characteristic unramified, it was proved by G. Lyubeznik [L]. On the other hand, M.
Katzman [K] constructed for any field k, a local cohomology module of a local finite k-algebra
(not regular) with an infinite set of associated prime ideals. When R is not required to be local,
A. K. Singh [S] made an example showing that the set of associated primes of local cohomology
modules may be infinite. Furthermore, Singh and Swanson [SS] in a deeper investigation made
examples of a unique factorization domain R which is a standard graded hypersurface over a
field k such that a local cohomology module Hja (R) has infinitely many associated primes. In
2000 Mathematics subject classification. 13D45, 13E10
Key words and phrases. associated primes, support of modules , local cohomology modules.
The third author is supported by a grant from IPM (No. 83130311)
Email addresses:[email protected], [email protected], [email protected]
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[KH] Khashayarmanesh and Salarian proved that the local cohomology module Hna (M ) has
finite associated primes, when n is an integer for which all the local cohomology modules Hia (M )
for i < n have one of the following properties: (a) are finitely generated, (b) have finite support.
Their approach is mainly by using filter regular sequences. In [BL] M. Brodmann and A.
Lashgari proved the first case of the above theorem by using induction. In this note we treat
the second case of the above theorem by induction and no use of such sequences at all. Our
terminology and notation of local cohomology come from [BS].
2. Main results
Lemma 2.1. Let M be an a-torsion R-module i.e., M =
Then
Ass (M ) = Ass (0 :M a).
S∞
n=1 (0 :M
an ) for an ideal a of R.
Proof. It is trivial. ¤
The following lemma is a substantial tool in the inductive step in the theorem below.
Lemma 2.2. Let R be a ring and M be an R-module. If N is a submodule of M then,
Ass (M/N ) ⊆ Ass (M ) ∪ Supp (N ).
In particular, if the set Supp (N )is finite, then Ass (M/N ) is finite if and only if Ass (M ) is finite.
Proof. Let p ∈ Ass (M/N ) \ Supp (N ). p
So there is a T
non zero element x of M such that
p = (N :R x), so we have px ⊆ N . Set Ann (px) = ni=1 qi . Then there exists a positive
integer t such that (q1 . . . qn )t px = 0. Set q = (q1 . . . qn )t . So that qpx = 0 and therefore
p ⊆ Ann (qx) ⊆ (N :R qx). Now let a ∈ (N :R qx), then aqx ⊆ N and therefore aq ⊆ p. If a ∈
/p
this means that q ⊆ p and thus qi ⊆ p for some 1 ≤ i ≤ n. Since qi ∈ Supp (px) and px ⊆ N
we obtain p ∈ Supp (N ) and this is a contradiction. It follows that a ∈ p and p = Ann (qx),
therefore p ∈ Ass (qx) and hence p ∈ Ass (M ), this ends the proof. ¤
Theorem 2.3. Let M be a finite R-module and a be an ideal of R. Suppose that there is a
positive integer n such that for all i < n the set Supp (Hia (M )) is finite. Then Ass (Hna (M )) is
finite.
Proof. We will induct on n. If n = 0, there is nothing to prove. If n = 1, the statement
follows easily by [BL, Theorem (2.2)]. Assume inductively that n > 1 and the result settled for
all i < n. It is harmless to assume that M is a-torsion free R-module, note that there is an
isomorphism Hia (M ) ∼
= Hia (M/Γa (M )) for all i ≥ 1. Thus there is an M -regular element x ∈ a.
2
x
Now the exact sequence 0 → M → M → M/xM = M → 0 induces the long exact sequence
x
g
f
· · · → Hn−1
(M ) → Hn−1
(M ) → Han−1 (M ) → Hna (M ) → · · · .
a
a
It can be seen that Supp (Hia (M )) is a finite set for all i < n − 1. Using induction hypothesis we obtain Ass (Hn−1
(M )) is finite. Furthermore note that Supp (Im g) is a subset
a
of Supp (Hn−1
(M
))
which
is
finite
by the hypothesis. By applying lemma (2.2) to the exact
a
n−1
sequence 0 → Im g → Ha (M ) → Im f → 0 we deduce that Ass (Im f ) is finite. The result now
follows by noting that Im f = (0 :Hn (M ) x), and using lemma (2.1). ¤
a
Remark 2.4. From Katzman’s example [K] it follows that the condition of finiteness of support
can not be weakened by finiteness of associated primes.
Corollary 2.5. Suppose that Supp (Hia (M )) is finite for all i < n and N is a submodule of
Hna (M ) such that Ass (Ext 1R (R/a, N )) is finite then, Ass (Hna (M )/N ) is finite.
Proof. The exact sequence
0 → N → Hna (M ) → Hna (M )/N → 0
induces the long exact sequence
· · · → Hom R (R/a, Hna (M )) → Hom R (R/a, Hna (M )/N ) → Ext 1R (R/a, N ) → · · · .
Note that Ass (Hom R (R/a, Hna (M ))) is finite by lemma (2.1) and Ass (Ext 1R (R/a, N )) is finite
by hypothesis, hence Ass (Hna (M )/N ) is finite. ¤
In [N] L. T. Nhan proved the following result when the ring is local. Since Artinian modules
have finite support the following corollary is an immediate consequence of theorem (2.3).
Corollary 2.6. Let M be a finite R-module and a be an ideal of the ring R. Suppose that
there is a positive integer n such that for all i < n, Hia (M ) is Artinian. Then Ass (Hna (M )) is
finite. ¤
ACKNOWLEDGMENT
We are grateful to the referee for his/her suggestions. We also would like to thank to the Institute
for Studies in Theoretical Physics and Mathematics (IPM) for the facilities.
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References
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