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Let $(R, \mathfrak{m})$ be a commutative Noetherian local ring and $M$ a finitely generated $R$-module. Let $x_1,...,x_t$ be an $M$-regular sequence and $I = (x_1,...,x_t)$. Is it true that $$\mathrm{Tor}_1^R(R/I^n, M) = 0$$ for all $n \geq 1$?

flag	asked Jan 9 at 18:56 Pham Hung Quy 1,163●1●11

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It's true for $n=1$ if the sequence is also $R$-regular, since then $\mathrm{Tor}_{1}^{R}(R/I,M) \simeq \mathrm{Tor}_{1}^{R/I}(R/I,M/I)$. (see Lemma 18.2.iii in Matsumura's CRT.) – David Hansen Jan 9 at 19:11

Lemma 18.2 in Matsumura need $x$ is both $R$-regular and $M$-regular. – Pham Hung Quy Jan 10 at 2:41

So Lemma 18.2 applies at least when $M$ has finite projective dimension. – Mahdi Majidi-Zolbanin Jan 10 at 4:22

YACP · Accepted Answer · 2013-01-11 20:26:59Z UTC

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I've posted a proof here for the special case when $M$ is cyclic. Furthermore, I've mentioned that the result holds for finitely generated modules when the sequence is $R$-regular and $M$-regular.

answered Jan 11 at 20:26

YACP
365●5

Very nice proof. Thanks you very much YACP. The case $M$ is cyclic is an exercise in the book about tight closure of C. Huneke. Your proof even better than its solution (in my opinion). – Pham Hung Quy Jan 12 at 15:43

Vanishing of Tor

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