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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 )

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    $\begingroup$ 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 $M$. $\endgroup$
    – Mohan
    May 28, 2020 at 23:40
  • $\begingroup$ @Mohan: I'm not sure if $M$ is reflexive either ... $\endgroup$
    – user102248
    May 30, 2020 at 0:55

1 Answer 1

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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.

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