On the Annihilation of local homology modules - SSMR

66 Amir MafiProof: First we show, for k ≤ n, that (c) (k i ) H i (x t 1, . . . , x t k; A) = 0, for 0 ≤ i < kand t ≥ 1. To this end we make an induction on k. If k = 1 and i = 0, thenH 0 (x t 1; A) ∼ = A/x t 1A is annihilated by c. Assume k ≥ 2. We show the statementby induction on i. For i = 0 we have the exact sequenceA/x t 1A −→ H 0 (x t 1, . . . , x t k; A) −→ 0and the assertion is true. For i ≥ 1 there is a short exact sequence0 −→ H i (x t 1, . . . , x t k−1; A)/x t kH i (x t 1, . . . , x t k−1; A) −→ H i (x 1 , . . . , x t k; A)−→ (0 : Hi−1(x t 1 ,...,xt k−1 ;A) x t k) −→ 0,t ≥ 1. If i < k − 1, the induction hypothesis yields the statement. In the casei = k − 1 we getH i (x t 1, . . . , x t k−1; A)/x t kH i (x t 1, . . . , x t k−1; A) ∼ = R/(x t 1, . . . , x t k)⊗(0 : A (x t 1, . . . , x t k−1)).Hence the short exact sequence proves the statement on the annihilation.In particular, we have (c) (k i ) H i (x t 1, . . . , x t k; A) = 0 (0 ≤ i < n, t ≥ 1).By [2, Theorem 3.6], we have Hi a (A) = limH ←− i (x t 1, . . . , x t n; A). Note thatt(c) (n i ) limH i (x t ←−t 1, . . . , x t n; A) ⊆ lim(c) (n i ) H i (x t ←−t 1, . . . , x t n; A) = 0. Therefore it follows(c) (n i ) Hi a (A) = 0, which proves the statement.Acknowledgement . The author is deeply grateful to the referee for carefullyreading of the original manuscript and the valuable suggestions. The author alsowould like to thank Dr. Hero Saremi for many helpful discussions during thepreparation of this paper.References[1] J. Bartijn, Flatness, completions, regular sequences, un ménage à trois.Thesis, Utrecht, (1985).[2] N.T. Cuong and T. T. Nam, The I-adic completion and local homologyfor Artinian modules, Math. Proc. Camb. Phil. Soc., 131(2001), 61-72.[3] N.T. Cuong and T.T. Nam, A local homology theory for lineary compactmodules, J. Algebra, 319 (2008), 4712-4737.