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.



Characterisation of reflexive modules
This is true even with weaker assumptions (finitely generated modules for Noetherian rings, or for non-Noetherian semiperfect rings).


If $M\cong M^{\*\*}$ then $M$ is a dual, and for any dual the natural map $M\to M^{\*\*}$ is a split monomorphism, so if $M$ is not reflexive then $M\cong M\oplus N$ for some non-zero $N$.


This is not possible if $M$ is a Noetherian module, since $M$ would have an increasing chain of submodules $N< N\oplus N<N\oplus N\oplus N<\dots$.


It's also not possible if $M$ is finitely generated and $A$ is semiperfect, since then $M/MJ(A)\cong M/MJ(A)\oplus N/NJ(A)$ are finite direct sums of simple modules, so Krull-Schmidt can be applied.


