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Let $(R,\mathfrak m)$ be a regular local ring. Let $I,J$ be proper ideals of $R$ such that $R/(I+J)$ has finite length i.e. $\sqrt{I+J}=\mathfrak m.$ Since $I+J$ annihilates $\text{Tor}_n^R(R/I, R/J)$ for each $n$, so each $\text{Tor}_n^R(R/I, R/J)$ has finite length. Consider Serre's intersection multiplicity $$\chi(R/I, R/J):=\sum_{n=0}^\infty (-1)^n \operatorname{length}_R\text{Tor}_n^R(R/I, R/J)$$

Now, also assume $\dim(R/I)+\dim(R/J)=\dim R$. (i.e. $V(I)$ and $V(J)$ intersect properly).

My question is: If $\chi(R/I, R/J)=\operatorname{length}_R (R/(I+J))$, then is it true that $\text{Tor}_i^R(R/I, R/J)=0$ for all $i\ge 1$ ? (or equivalently, are both $R/I$ and $R/J$ Cohen-Macaulay?)

If needed, I am willing to assume $I,J$ are prime ideals.

Note: Here is an argument that why the vanishing of all positive Tor is equivalent to saying $R/I, R/J$ are Cohen-Macaulay: Indeed, since $R$ is regular and each $\text{Tor}_n^R(R/I, R/J)$ has depth $0$ (since finite length), so putting $q:=\sup\{n: \text{Tor}_n^R(R/I, R/J)\ne 0 \}$, we see by Theorem 2.2 of https://doi.org/10.1080/00927879808826375 that $$q=\text{depth }R- \text{depth }(R/I) - \text{depth }(R/J)=\dim(R/I)+\dim(R/J)- \text{depth }(R/I) - \text{depth }(R/J),$$ where we used the assumption that $V(I)$ and $V(J)$ intersect properly i.e. $\dim(R/I)+\dim(R/J)=\dim R$. Hence, $q=0$ if and only if $\dim(R/I)- \text{depth }(R/I) +\dim(R/J) - \text{depth }(R/J)=0$ if and only if $\dim(R/I)- \text{depth }(R/I) =0=\dim(R/J) - \text{depth }(R/J)$ i.e. both $R/I$ and $R/J$ are Cohen-Macaulay.

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2 Answers 2

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Essentially, you are asking if $\chi_1(R/I,R/J)=0$ implies $\text{Tor}^R_{>0}(R/I, R/J)=0.$ If $R$ is an unramified regular local ring, then this is true and is the main Theorem of Hochster's paper https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-28/issue-2/Euler-characteristics-over-unramified-regular-local-rings/10.1215/ijm/1256065276.full . Note that Hochster even proves this for any modules $M,N$ whose tensor product has finite length (so you can just take $M=R/I, N=R/J$). Here, recall that a regular local ring $(R, \mathfrak m,k)$ is said to be unramified if either $R$ contains a field, or else $\text{char } R=0$ and $\text{char } k=p\notin \mathfrak m^2$ (this ensures that the completion of $R$ is a power series ring over a field or a complete discrete valuation ring).

I do not know if in general it is still unknown or not. Perhaps someone who knows more intersection theory will stumble upon this question and can illuminate on the current status.

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This is indeed true. Given any finitely generated modules M, N over a regular local ring A such that their tensor product has finite length, $\text{Tor}_i(M,N)=0$ for $i\geq 1$ if and only if M, N are Cohen Macaulay A-modules of complementary dimension. Of course, this is what you essentially have.

(For a proof, see Serre’s book “Local Algebra” Chapter V)

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    $\begingroup$ This is not at all what I asked ... I asked if $\chi(R/I,R/J)=\mathcal l_R(R/(I+J))$ implies $\text{Tor}^R_{>0}(R/I, R/J)=0$ or not ... $\endgroup$
    – Alex
    Commented Oct 6, 2022 at 2:18
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    $\begingroup$ My bad. But again, this is true in equal characteristic according to Serre. Given your hypothesis, $\chi_1(R/I, R/J) = 0$ and this is if and only if $\text{Tor}_{i+1}(R/I, R/J) =0$. Serre attributes this result, which holds more generally for $\chi_r$, to Auslander & Buchsbaum. $\endgroup$
    – Nawaj
    Commented Oct 6, 2022 at 15:22
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    $\begingroup$ Instead of Auslander and Buchsbaum, it should be Auslander and Lichtenbaum I think ... respectively projecteuclid.org/journals/illinois-journal-of-mathematics/… and projecteuclid.org/journals/illinois-journal-of-mathematics/… ... and later in slightly more generality by Hochster as I have mentioned in my answer $\endgroup$
    – Snake Eyes
    Commented Oct 14, 2022 at 5:12

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