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 $n \geq 1$?

1$\begingroup$ 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.) $\endgroup$– David HansenJan 9, 2013 at 19:11

1$\begingroup$ Lemma 18.2 in Matsumura need $x$ is both $R$regular and $M$regular. $\endgroup$– Pham Hung QuyJan 10, 2013 at 2:41

$\begingroup$ So Lemma 18.2 applies at least when $M$ has finite projective dimension. $\endgroup$– Mahdi MajidiZolbaninJan 10, 2013 at 4:22

$\begingroup$ $I^{n1}/I^n$ is a free $R/I$module, so the statement follows from a simple induction. $\endgroup$– AngeloJul 25, 2013 at 5:41

$\begingroup$ I wonder if this is true even in the $n=1$ case? I know it is true if $t=1$, but I wonder if it would be true when $n=1$ and $t>1$? $\endgroup$– user521337Aug 12, 2021 at 9:13
1 Answer
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.

$\begingroup$ 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). $\endgroup$ Jan 12, 2013 at 15:43