Let $M$ be a finitely generated module over a local Cohen-Macaulay ring $(R, \mathfrak m)$ such that there is an exact sequence $0\to M \to F \to G$ for some finitely generated free modules $F$ and $G$ . Then is it true that $M$ is reflexive i.e. the natural map $M\to M^{**}$ is an isomorphism ?
Over which Noetherian local rings are any second syzygy module reflexive
cohen-macaulaycommutative-algebrahomological-algebra
Best Answer
The answer is no. For a nice class of examples, any local ring $(R,\mathfrak{m},k)$ with $\mathfrak{m}$ not principal and $\mathfrak{m}^2=0$ cannot have any syzygies be reflexive, for a sort of trivial reason. Any (minimal) syzygy embeds into $\mathfrak{m}F$ where $F$ is a finitely-generated free module. As $\mathfrak{m}^2=0$, this forces any syzygy to be a $k$-vector space. But no k-vector space can be reflexive over such a ring, as $\operatorname{Hom}_R(k^n,R) \cong \operatorname{Hom}_R(k,R)^n$ has dimension equal to $r(R)n$ where $r(R):=\dim \operatorname{Soc} R$ is the type of $R$. Since $\mathfrak{m}^2=0$, the socle of $R$ is $\mathfrak{m}$ which, by assumption, has dimension greater than $1$.
However, the claim is true with the additional (relatively mild) condition that $R$ be Gorenstein in codimension $1$. Recall $M$ satisfies Serre's condition $(S_n)$ if $\operatorname{depth}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}}) \ge \min\{n,\dim R_{\mathfrak{p}}\}$ for all $\mathfrak{p} \in \operatorname{Spec}(R)$. The point is, if $R$ is Cohen-Macaulay and $M$ satisfies Serre's condition $(S_2)$, then one can check reflexivity of $M$ by checking it in codimension $1$; it's a nice exercise. Note that any second syzygy in a Cohen-Macaulay ring must satisfy $(S_2)$, by the depth lemma. From there one appeals to the fact that maximal Cohen-Macaulay modules over a Gorenstein ring are reflexive. There are several good references for this discussion; I'm partial to A.15 Corollary in Cohen-Macaulay Representations by Leuschke and Wiegand.