Algebra Colloquium 21 : 3 (2014) 517–520 DOI: 10.1142/S1005386714000455
Algebra Colloquium c 2014 AMSS CAS ° & SUZHOU UNIV
On the Finiteness Dimension of Local Cohomology Modules Hero Saremi Department of Mathematics, Sanandaj Branch Islamic Azad University, Sanandaj, Iran E-mail:
[email protected]
Amir Mafi† Department of Mathematics, University of Kurdistan Pasdaran St., P.O. Box 416, Sanandaj, Iran E-mail: a
[email protected] Received 9 March 2011 Revised 25 April 2012 Communicated by Zhongming Tang Abstract. Let R be a commutative Noetherian ring, a an ideal of R, and M a non-zero finitely generated R-module. Let t be a non-negative integer. In this paper, it is shown that dim Supp Hai (M ) ≤ 1 for all i < t if and only if there exists an ideal b of R such that dim R/b ≤ 1 and Hai (M ) ∼ = Hbi (M ) for all i < t. Moreover, we prove that dim Supp Hai (M ) ≤ dim M − i for all i. 2010 Mathematics Subject Classification: 13D45, 13E99 Keywords: local cohomology modules, cofinite modules, finiteness dimension
1 Introduction Throughout this paper, we assume that R is a commutative Noetherian ring with non-zero identity, a an ideal of R, and M a non-zero finitely generated R-module. For a non-negative integer i, the i-th local cohomology module of M with respect to a is denoted by Hai (M ). The reader should consult [4] for the definition of local cohomology and its basic properties. An R-module N is called a-cofinite if Supp(N ) ⊆ V (a) and ExtiR (R/a, N ) is finitely generated for all i. This notion was introduced by Hartshorne in [9]. The reader is referred to [3, 5–7, 10–12, 14, 16, 18–20] for more information about cofiniteness with respect to an ideal. We denote by dim Supp Hai (M ) the maximum †
Corresponding author.
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of the numbers dim R/p, where p runs over the support of Hai (M ). Bahmanpour and Naghipour [2] proved that if t is a non-negative integer such that dim Supp Hai (M ) ≤ 1 for all i < t, then Hai (M ) is a-cofinite for all i < t. This extends the main theorem of [7] and [20]. The main aim of this paper is to prove the following result. Theorem 1.1. Let t be a non-negative integer. Then: (i) dim Supp Hai (M ) ≤ 1 for all i < t if and only if there exists an ideal b of R such that dim R/b ≤ 1 and Hai (M ) ∼ = Hbi (M ) for all i < t. i (ii) dim Supp Ha (M ) ≤ dim M − i for all i. 2 The Results We use the notations similar to those used in [13]: for a subset T of Spec(R), we set (T )i := {p ∈ T : dim R/p = i} and (T )>i := {p ∈ T : dim R/p > i}, where i is a non-negative integer. Theorem 2.1. Let t be a non-negative integer. Then the following statements are equivalent: (i) dim Supp Hai (M ) ≤ 1 for all i < t. (ii) There exists an ideal b of R such that dim R/b ≤ 1 and Hai (M ) ∼ = Hbi (M ) for all i < t. Proof. (ii)⇒(i) As Supp Hai (M ) ⊆ V (b) for all i < t and dim R/b ≤ 1, it follows that dim Supp Hai (M ) ≤ 1 for all i < t. © ª St−1 (i)⇒(ii) Let Σ = p ∈ i=0 Supp Hai (M ) : dim R/p = 1 . We show that Σ is a finite set. To do this, suppose that Σ is infinite and look for a contradiction. Then there exists 0 ≤ j ≤ t − 1 such that Σ ∩ Supp Haj (M ) is an infinite set. Since dim Supp Hai (M ) ≤ 1 for all i < t, it follows that the elements of Σ ∩ Supp Haj (M ) j j are minimal in Supp Haj (M ). Hence, Σ ∩ Supp ¡ S∞Ha (M ) ⊆i (Assl Ha (M ¢ ))1 . On the Ass Ext (R/a , M ) is finite for other hand, by [13, Theorem 2.5], the set l=1 1 ¡ S∞ ¢ i l Ass Ext (R/a , M ) , it follows that the set all i < t. Since (Ass Hai (M ))1 ⊆ l=1 1 j i (Ass Ha (M ))1 is finite for all i < t. Therefore, theTset Σ ∩ Supp Ha (M ) is finite, which is a contradiction. Then Σ is finite. Set b := p∈Σ p. Then dim R/b ≤ 1 and we can assume a ( b. Hence, there exists x ∈ b\a. By [4, Proposition 8.1.2], there is a long exact sequence i · · · −→ Hai−1 (Mx ) −→ Ha+Rx (M ) −→ Hai (M ) −→ Hai (Mx ) −→ · · · ,
(∗)
i (Mx ) where Mx is the localization of M at {xi : i ≥ 0}. Note that Hai (Mx ) ∼ = HaR x i i and AssRx (Ha (Mx )) = {pRx : x ∈ / p, p ∈ AssR (Ha (M ))}. In view of [13, Proposition 2.6], (AssR Hai (M ))>1 = ∅ for all i < t and so AssRx (Hai (Mx )) = ∅ for all i < t. i Therefore, Ha+Rx (M ) ∼ = Hai (M ) for all i < t. By assuming b = a + (x1 , . . . , xn ) and applying the argument for finite steps, there is an isomorphism Hai (M ) ∼ = Hbi (M ) for all i < t. This complete the proof. ¤
The following corollary extends [15, Theorem 1.3].
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Corollary 2.2. Let t be a positive integer such that Supp Hai (M ) is finite for all i < t. Then there exists an ideal b of R such that dim R/b ≤ 1 and Hai (M ) ∼ = Hbi (M ) for all i < t. Proof. By [13, Corollary 2.7], (Ass Hai (M ))>1 = ∅ for all i < t, so dim Supp Hai (M ) ≤ 1 for all i < t. Hence, by Theorem 2.1, the result follows. ¤ The following theorem improves [1, Proposition 3.2(ii)]. Theorem 2.3. For any non-negative integer i, dim Supp Hai (M ) ≤ dim M − i. Proof. Set d := dim M and we can assume d < ∞. We proceed by induction on d. The case d = 0 is obvious. Thus, suppose, inductively, that d ≥ 1 and we have established the result for non-zero finitely generated R-modules of dimension smaller than d. Set M = M/Γa (M ) and so Hai (M ) ∼ = Hai (M ) for all i > 0. If dim M < d, then by the inductive hypothesis dim Supp Hai (M ) ≤ dim M − i for all i > 0 and since Γa (M ) ⊆ M , it follows that dim Supp Hai (M ) ≤ d − i for all i ≥ 0. Now suppose that dim M = d. Then there exists x ∈ a which is non-zero divisor x on M . The exact sequence 0 −→ M −→ M −→ M /xM −→ 0 induces an exact sequence Hai−1 (M /xM ) −→ (0 :H i (M ) x) −→ 0. a
Since x is not in any minimal member of Supp(M ), we have dim M /xM ≤ d − 1, so that by the inductive hypothesis, dim Supp Hai (M /xM ) ≤ d − i. Therefore, in view of the above exact sequence, dim Supp(0 :H i (M ) x) ≤ d − i. Since Supp(0 :H i (M ) x) a a ¤ = Supp Hai (M ), it follows that dim Supp Hai (M ) ≤ d − i, as required. Corollary 2.4. (cf. [17, Corollary 2.5]) Let M be a non-zero finitely generated R-module with dim M = d. Then Supp Had−1 (M ) ⊆ (Ass Had−1 (M ))1 ∪ Max(R). Proof. Let p ∈ Supp Had−1 (M ). Then by Theorem 2.3, dim Supp Had−1 (M ) ≤ 1. Therefore, p is either a maximal ideal or a minimal element of Supp Had−1 (M ). Thus, p ∈ (Ass Had−1 (M ))1 ∪ Max(R) and so the result follows. ¤ Lemma 2.5. Let R be a semi-local ring and M a non-zero finitely generated R-module with dim M = d. Then Supp Had−1 (M ) is finite. Proof. By induction on d and using the same argument as that used in the proof of Theorem 2.3, the result immediately follows. ¤ The following consequence extends [8, Corollary 2.2]. Corollary 2.6. Let R be a semi-local ring and M a non-zero finitely generated R-module with dim M = d. If Had (M ) 6= 0, then there exists an ideal b of R such that dim R/b ≤ 1 and Had (M ) ∼ = Hbd (M ). T Proof. Set Σ := Supp Had−1 (M ) ∪ Supp Had (M ) and b := p∈Σ p. In view of Lemma 2.5 and Corollary 2.4, we have dim R/b ≤ 1. Now we can assume a ( b. Hence, there exists x ∈ b\a. By using the exact sequence (∗), we conclude that
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d Had (M ) ∼ (M ). Now the assertion follows by assuming b = a + (x1 , . . . , xn ) = Ha+Rx and applying the argument for finite steps. ¤
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