# If Serre's intersection multiplicity $\chi(R/I, R/J)$ equals $\operatorname{length}_R (R/(I+J))$, then are $R/I, R/J$ Cohen-Macaulay?

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.

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.

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)

• 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 ...
– Alex
Commented Oct 6, 2022 at 2:18
• 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. Commented Oct 6, 2022 at 15:22
• 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 Commented Oct 14, 2022 at 5:12