Example that M* is not reflexive

Question

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.

Answer 1

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.