Depth of tensor product of finitely generated reflexive modules over regular local rings

Question

Let $M,N$ be finitely generated reflexive modules over a regular local ring $(R, \mathfrak m)$ of dimension at least $3$ (Hence, $M,N$ both have depth at least $2$) .

Then, is it necessarily true that $\text {depth}_R(M \otimes_R N)\ge 1$ ? i.e. is it true that $\mathfrak m \notin \text {Ass}_R (M \otimes_R N)$ ?

Answer 1

This is true in dimension $3$ and false in dimension greater than $3$. For the dimension $3$ case consider the following:

Theorem: Suppose $(R,\mathfrak{m})$ is a local ring. Let $M$ be a finitely generated $R$-module with projective dimension $\operatorname{pd}(M)=1$ and $N$ a finitely generated $R$-module with $\operatorname{depth}(N) \ge 2$. Suppose one of $M$ or $N$ is locally free on the punctured spectrum of $R$ (or more generally that their nonfree loci intersect only at $\mathfrak{m}$). Then $\operatorname{depth}(M \otimes_R N) \ge 1$.

Proof: Since $M$ has projective dimension $1$, there is an exact sequence of the form $0 \to R^m \to R^n \to M \to 0$. Tensoring with $N$ and considering the long exact sequence in $\operatorname{Tor}$ gives an exact sequence of the form

$$0 \to \operatorname{Tor}^R_1(M,N) \to N^m \to N^n \to M \otimes_R N \to 0.$$

As one of $M$ or $N$ is locally free on the punctured spectrum, $\operatorname{Tor}_1^R(M,N)$ has finite length. But it embeds in $N^m$ which has positive depth, and so $\operatorname{Tor}_1^R(M,N)$ must be $0$. Since $\operatorname{depth} N \ge 2$, the depth lemma implies $\operatorname{depth}(M \otimes_R N) \ge 1$.

Remark: In fact, the argument here concludes that $\operatorname{Tor}^R_i(M,N)=0$ for all $i>0$. It's easy to see this $\operatorname{Tor}$-vanishing implies $\operatorname{pd}_R(M \otimes_R N)=\operatorname{pd}_R(M)+\operatorname{pd}_R(N)$ in case both terms on the right hand side are finite, and in fact, a deeper result of Auslander shows the formula

$$\operatorname{depth}(R)+\operatorname{depth}(M \otimes_R N)=\operatorname{depth}(M)+\operatorname{depth}(N)$$ holds when $\operatorname{Tor}^R_i(M,N)=0$ for $i>0$ and only one of $M$ or $N$ has finite projective dimension.

For your question one needs only to note that, in your dimension $3$ setting, reflexive modules are locally free on the punctured spectrum, and, are either free, or have depth $2$ and projective dimension $1$. In fact, the remark shows that the depth of $M \otimes_R N$ is $3$ if both $M$ and $N$ are free, $2$ if exactly one of $M$ or $N$ is free, and $1$ if both are not free.

In your dimension $d=4$ or higher setting, the syzygies $\Omega^2_R(k)$ and $\Omega^{d-2}_R(k)$ are reflexive and we have $\operatorname{Tor}^R_i(\Omega^2_R(k),\Omega^{d-2}_R(k)) \cong \operatorname{Tor}^R_{i+d}(k,k)=0$ when $i>0$. Either Auslander's result or a projective dimension/Auslander-Buchsbaum formula argument implies $\operatorname{depth}(\Omega^2_R(k) \otimes_R \Omega^{d-2}_R(k))=0$.