Let $R$ be a Noetherian ring and $M$ a finite $R$-module. Let $x=x_1,\dots,x_n$ be an $R$-sequence such that it is also an $M$-sequence and let $I=(x_1,\dots,x_n)$.

Question: Is it true that $\operatorname{Tor}_i^R(R/I,M)=0$ for every $i>0$? If yes, how can we see that? If no, how does the situation change if $R$ is $\mathbb{Z}$-graded and $x_i$ homogeneous of positive degree?

Motivation: Let $k$ be a field and $S=k[x_0,\dots,x_n]$ and define two ideals $I=(x_0,x_1,\dots,x_j), J=(x_k,x_{k+1}\dots,x_m)$ such that $j < k$. Then $I$ is $J$-regular but it is also $S$-regular. J. Sidman in "Resolutions and Subspace Arrangements, 2007", proof of proposition 9.2.5 says that "it follows easily that $\operatorname{Tor}_i(I,J)=0, \forall i >0$."

  • $\begingroup$ Why do you suspect that to be true? Have you considered special cases, like that of $n=1$? What happens there? $\endgroup$ – Mariano Suárez-Álvarez Jan 14 '14 at 23:10
  • $\begingroup$ @MarianoSuárez-Alvarez: I added what makes me think that this might be true. $\endgroup$ – Manos Jan 14 '14 at 23:24
  • $\begingroup$ What about the case where n is 1? $\endgroup$ – Mariano Suárez-Álvarez Jan 14 '14 at 23:27
  • $\begingroup$ @MarianoSuárez-Alvarez: Let me think about $n=1$ :) $\endgroup$ – Manos Jan 15 '14 at 0:53
  • $\begingroup$ Sigh. Did you do the case n=1 at least before you got this learning opportinuty ruined? $\endgroup$ – Mariano Suárez-Álvarez Jan 16 '14 at 17:24

Since $x_1, ..., x_n$ is an $R$-regular sequence, we can resolve $R/I$ by the Koszul complex. We can then use this resolution to compute $\text{Tor}_i(R/I, M)$: this will be the homology of the complex obtained by tensoring the Koszul complex of $I$ with $M$. But $x_1, ..., x_n$ is also $M$-regular, so the tensored complex will remain acyclic (e.g. by the following proposition), so $\text{Tor}_i(R/I, M) = 0$ for all $i > 0$.

Proposition: $R$ Noetherian ring, $M$ finite $R$-module, $\underline{x} = x_1,...,x_n \subseteq R$, $I = (\underline{x})$, $IM \neq M$. TFAE:

1) $\underline{x}$ is $M$-quasiregular
2) $K_*(\underline{x}; M)$ ($ = K_*(\underline{x}) \otimes_R M$) is acyclic
3) $H_1(\underline{x}; M) = 0$
4) $\text{grade}(I, M) = n$


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