Let $A$ be a semiperfect noetherian ring. A module $M$ is called reflexive in case the canonical map $f_M: M^{**} \cong M$ is an isomorphism, when $(-)^{*}:=Hom_A(-,A)$. This is equivalent to say that $Ext_A^i(Tr(M),A)=0$ for $i=1,2$ when $Tr$ denotes the Auslander-Bridger duality.

Question: Assume $M$ is finitely presented. Is $M$ reflexive iff $M^{**} \cong M$? (You are welcome to give example for any kind of ring, I know none)

This should be true for Artin algebras: Assume $M^{**} \cong M$. We have $M^{**} \cong \Omega^2 Tr \Omega^2 Tr(M)$ and thus $M \in \Omega^1(mod-A)$ and thus $M$ is torsionfree, which is equivalent to $f_M$ being injective. But $M^{**} \cong M$ gives us that the modules have the same length and thus $f_M$ is even an isomorphism and $M$ is relfexive. In the book of Auslander and Bridger I found that this should also be true in case the ring is additionally commutative Gorenstein (we dont need semiperfect here). Remark with regards to the previous (deleted) thread: I decided to split up the bigger confusing thread into smaller questions to make things less confusing.