# 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". Commented Dec 14, 2019 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.

• Why do you write "(minimal) syzygy?" Certainly, the claim holds (by definition) if we consider the syzygies of a minimal free resolution. Otherwise, what can be said? Commented Aug 16, 2021 at 23:32
• By the way, when you say "second syzygy," do you mean of any $R$-module? Commented Aug 17, 2021 at 0:43
• @DylanC.Beck For your first question, which claim do you mean? If one takes a nonminimal syzygy of $L$, then this syzygy splits as $R^n \oplus \Omega^R_1(L)$ with $n \ge 1$, where $\Omega^R_1(L)$ denotes the minimal syzygy of $L$, and so one does not have an embedding into $\mathfrak{m} F$; this embedding is equivalent to the minimality. For your second question, yes, "$M$ is a second syzygy" could be rephrased as "There is an $R$-module $L$ and an exact sequence of the form $0 \to M \to R^n \to R^m \to L \to 0$ for some nonnegative integers $m$ and $n$." Commented Aug 17, 2021 at 1:41
• That is precisely what I suspected. I was simply clarifying that the assumption of minimal syzygy was necessary. Thank you very much for clarifying. Commented Aug 17, 2021 at 4:49

The answer is no.

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.

• I'm not sure how to conclude that $\Omega(M^*) \cong \operatorname{coker}(\pi^*)^*.$ Could you elaborate? Commented Apr 9, 2022 at 22:48
• @DylanC.Beck Just because there is a s.e.s. $0\rightarrow Coker(\pi^\ast)^\ast\rightarrow P^{\ast\ast}\cong P\rightarrow M^{\ast\ast\ast}\cong M^\ast$\rightarrow 0.
– Jian
Commented Apr 10, 2022 at 13:03
• Why is the map $P^{**} \to M^{***}$ surjective? Ostensibly, it is different from the map $P \to M^*.$ Commented Apr 10, 2022 at 18:42
• @DylanC.Beck They are isomorphic. I forget to say $P$ is finitely generated.In this case, $P$ is reflexive. $M^\ast$ is reflexive as $M$ is reflexive by the assumption.
– Jian
Commented May 21, 2022 at 5:14