# Over which Noetherian local rings are any second syzygy module reflexive?

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 ?

• Please see theorem 3.6 of the book "Syzygies , Evan and Griffith". – Mohammad Bagheri Dec 14 '19 at 20:26

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.

Consider $$R=k[X,Y]/(X,Y)^2$$. $$\dim R=0$$ hence $$R$$ is Cohen-Macaulay. The maximal ideal is $$\mathrm{Soc}(R)\cong k\oplus k$$. So there is an exact sequence $$0\rightarrow \mathrm{Soc}(R)\rightarrow R\rightarrow R$$, where the second map is induced by $$k\subset \mathrm{Soc}(R)\subset R$$. We know $$k^\ast\cong k^2$$, so we know $$(\mathrm{Soc}(R))^{\ast\ast}\cong k^8$$. This is a second syzygy module which is not reflexive.

So what is the error in the answer above: if $$0\rightarrow M\rightarrow F\rightarrow G$$ is exact, then the sequence $$0\rightarrow M^{\ast\ast}\rightarrow F^{\ast\ast}\rightarrow G^{\ast\ast}$$ is not exact in general since $$\mathrm{Hom}_R(-,R)$$ is contravariant left exact.

The exercise 1.4.20 in Bruns and Herzog's book is [The converse is not true]:

If $$R$$ is Noetherian ring, $$M$$ is finite generated reflexive $$R$$-module, then $$M$$ is second syzygy.

Proof: Consider $$0\rightarrow \Omega(M^\ast)\rightarrow P\xrightarrow \pi M^\ast\rightarrow 0$$. Then we have the long exact sequence: $$0\rightarrow M\cong M^{\ast\ast}\xrightarrow {\pi^\ast} P^\ast\rightarrow \Omega(M^\ast)^\ast$$. I claim: $$Coker(\pi^\ast)$$ is torsionless. (A finitely generated module $$N$$ is torsionless if $$N\rightarrow N^{\ast\ast}$$ is injective. This is equivalent to $$N$$ can be embedded in a free module.) If the claim is true, then the result follows. Now consider $$0\rightarrow M^{\ast\ast}\rightarrow P^\ast\rightarrow Coker(\pi^\ast)\rightarrow 0$$.Take dual: $$0\rightarrow Coker(\pi^\ast)^\ast\rightarrow P^{\ast\ast}\cong P\rightarrow M^{\ast\ast\ast}\cong M^\ast$$. Compare this with the first s.e.s, we know $$\Omega(M^\ast)\cong Coker(\pi^\ast)^\ast$$. Since $$Coker(\pi^\ast)$$ is submodule of $$\Omega(M^\ast)^\ast$$, the claim is true.

When the module is reflexive, in proposition 3.4 of "Representation theory of Artin algebras":

The following is equivalent for an Artin algebra $$\Lambda$$: 1.$$\Lambda$$ is self-injective;
2. Every finite generated $$\Lambda$$-module is torsionless; 3. Every finite generated $$\Lambda$$-module is reflexive.

Another case is:

Every maximal Cohen-Macaulay module over Gorenstein ring is reflexive.