# Can finitely generated reflexive module have strictly larger depth than the depth of the ring?

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 )

• I would at least try the following example. Let $R=k[[x,y,z,t]]/(x^2,xy)$ and $M=R/xR$. Then depth of $R=2$, depth of $M=3$. I have not checked reflexivity of $M$. May 28, 2020 at 23:40
• @Mohan: I'm not sure if $M$ is reflexive either ... May 30, 2020 at 0:55

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.