# Example that M* is not reflexive

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.

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.
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 )