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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 ?

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    $\begingroup$ Please see theorem 3.6 of the book "Syzygies , Evan and Griffith". $\endgroup$ Commented Dec 14, 2019 at 20:26

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Aug 16, 2021 at 23:32
  • $\begingroup$ By the way, when you say "second syzygy," do you mean of any $R$-module? $\endgroup$ Commented Aug 17, 2021 at 0:43
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    $\begingroup$ @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 $n$." $\endgroup$ Commented Aug 17, 2021 at 1:41
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    $\begingroup$ That is precisely what I suspected. I was simply clarifying that the assumption of minimal syzygy was necessary. Thank you very much for clarifying. $\endgroup$ Commented Aug 17, 2021 at 4:49
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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.

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  • $\begingroup$ I'm not sure how to conclude that $\Omega(M^*) \cong \operatorname{coker}(\pi^*)^*.$ Could you elaborate? $\endgroup$ Commented Apr 9, 2022 at 22:48
  • $\begingroup$ @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. $\endgroup$
    – Jian
    Commented Apr 10, 2022 at 13:03
  • $\begingroup$ Why is the map $P^{**} \to M^{***}$ surjective? Ostensibly, it is different from the map $P \to M^*.$ $\endgroup$ Commented Apr 10, 2022 at 18:42
  • $\begingroup$ @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. $\endgroup$
    – Jian
    Commented May 21, 2022 at 5:14

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