Let $M$ be a finitely generated module over a Noetherian ring $R$ such that $M$ is isomorphic with its double dual $M^{\*\*}=Hom(Hom(M,R),R)$. 


Then is the natural map $j:M \to M^{\*\*}$ defined as $j(m)(f)=f(m),\forall m\in M, \forall f\in M^\*$, an isomorphism ? 



Noetherian module, over Noetherian ring, which is isomorphic to its double dual
I asked this question (or at least a similar one) here:
[Characterisation of reflexive modules](https://mathoverflow.net/questions/288878/characterisation-of-reflexive-modules)
By an answer of Jeremy Rickard it is even true for noetherian rings that are not necessarily commutative.
It seems to be an open question for finitely generated modules over general (not necessarily noetherian) rings.


