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Let $R$ be a noetherian ring. Set $(-)^\ast={\rm Hom}_R(-,R)$. For each $R$-module $N$, let $\pi_N:N\rightarrow N^{\ast\ast}$ be the map which maps $n\in N$ to $(f\mapsto f(n))$. $N$ is called reflexive if $\pi_N$ is isomorphism.

Question: Does there exist a finitely generated $R$-module $M$ such that $M^\ast$ is not reflexive?

I guess the answer is yes. But I can’t find any example.

For each $R$-module $N$, we can check directly that the composition $N^\ast\xrightarrow{\pi_{N^\ast}}N^{\ast\ast\ast}\xrightarrow{(\pi_N)^\ast}N^\ast$ is identity. In particular, $\pi_{N^\ast}$ is always invective. I searched the internet. It is proved in Yoshino’s paper that if $R$ is Gorenstein in depth one, then $M^\ast$ is reflexive for each finitely generated $R$-module $M$; see Lemma 4.4 of HOMOTOPY CATEGORIES OF UNBOUNDED COMPLEXES OF PROJECTIVE MODULES.

Thank you in advance.

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2 Answers 2

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Let $k$ be a field, and let $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a three-dimensional local commutative $k$-algebra.

Let $S$ be the (one-dimensional) simple $R$-module $S=R/(x,y)$. Then $S^\ast\cong S\oplus S$, and so $S^{\ast\ast\ast}$ is the direct sum of eight copies of $S$, and so $S^\ast$ is not reflexive.

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Let $\text{Tr}(-)$ denote Auslander-Bridger transpose. Recall that $\text{Tr} { \space}\text{Tr } M$ is stably isomorphic to $M$, and $\left(\text{Tr }(M)\right)^*$ is stably isomorphic to $\Omega^2 M$. This gives the following equivalent statements:

$M^*$ is reflexive for every finitely generated $R$-module $M$ if and only if $\left(\text{Tr }(M)\right)^*$ is reflexive for every finitely generated $R$-module $M$, if and only if $\Omega^2 M$ is reflexive for every finitely generated $R$-module $M$, if and only if $R$ satisfies Serre's condition $(S_1)$ and is generically Gorenstein. (The last equivalence follows from Theorem 2.3 of https://link.springer.com/article/10.1007/s00013-017-1020-9 )

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