1
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ Please see theorem 3.6 of the book "Syzygies , Evan and Griffith". $\endgroup$ – Mohammad Bagheri Dec 14 '19 at 20:26
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.