Let $M$ be an $R$module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism.
I'd like to know if there exists a module isomorphic to its bidual but not reflexive, do you know an example?
Let $M$ be an $R$module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism. I'd like to know if there exists a module isomorphic to its bidual but not reflexive, do you know an example? 

Yes, there are such examples. In the case of Abelian groups for example, one can have a group A which is not reflexive, but which is isomorphic to its double dual. The book "Almost Free Modules" by Eklof and Mekler (North Holland) contains much of what is known. As a specific example, take E to be a stationary and costationary subset of $\omega_1$, and let $X=\omega_1 + 1\backslash E$, given the subspace topology (of the space $\omega_1 +1$ with the interval topology). Then $C(X, Z)$ (continuous functions to the integers with the discrete topology) is such a group. In fact for this group $C(X,Z)^{**} = \sigma[C(X,Z)] \oplus Z$ where $\sigma$ is the natural map, so that $C(X,Z)$ is not reflexive, but $C(X,Z)\oplus Z \cong C(X,Z)$ because for example $C(X,Z)$ has $Z^\omega$ as a summand. 


$J$
(the James space) that is isometrically isomorphic to$J^{\ast\ast}$
but for which the image of$J \to J^{\ast\ast}$
has codimension 1. See mathoverflow.net/questions/43986/… – Faisal Sep 20 '11 at 21:51