Let $M$ be a non-zero finitely generated module over a Noetherian local ring $(R, \mathfrak m)$. Then $\operatorname {depth}(M)\le \dim M\le \dim R$. So if $R$ is Cohen-Macaulay, then
$\operatorname {depth}(M)\le \operatorname{depth}(R)$.
My question is: If $M$ is finitely generated and reflexive and
$\operatorname {depth}(R)\ge 2$ , then
can $\operatorname {depth}(M)$ be strictly larger than $\operatorname {depth}(R)$ ?
(Note that since $R$ has depth at least $2$ and $M$ is reflexive, so $\operatorname {depth}(M)\ge 2$ by https://stacks.math.columbia.edu/tag/0AV5 )
Best Answer
Here is an example. See [Hochster] for details. We state the simplest version of Hochster's example.
Example (see [Hochster, Example 5.9]). Consider the Segre product $$R := \frac{\mathbf{C}[X_1,X_2,X_3]}{(X_1^3+X_2^3+X_3^3)} \mathbin{\#} \mathbf{C}[Y_1,Y_2] \subseteq \frac{\mathbf{C}[X_1,X_2,X_3,Y_1,Y_2]}{(X_1^3+X_2^3+X_3^3)} =: S.$$ This is an integrally closed domain of dimension 3 that is not Cohen–Macaulay. Now consider the prime ideal $$Q = Y_1S \cap R.$$ Then, $$\operatorname{depth}_{\mathfrak{m}}(Q^{(i)}) = 3$$ for $i$ sufficiently large, where $\mathfrak{m}$ is the irrelevant ideal and $Q^{(i)}$ denotes the $i$-th symbolic power of $Q$.
Now to get an example for your question, we note that $\operatorname{depth}_{\mathfrak{m}}(R_{\mathfrak{m}}) = 2$, and hence $\operatorname{depth}_{\mathfrak{m}}(R_{\mathfrak{m}}) < \operatorname{depth}_{\mathfrak{m}}(Q^{(i)}_{\mathfrak{m}})$ for $i$ sufficiently large. Finally, $Q^{(i)}_{\mathfrak{m}}$ is reflexive by [Leuschke–Wiegand, Corollary A.14], for example.