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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 bi-dual but not reflexive, do you know an example?

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    $\begingroup$ Yes. In fact there is an example of a Banach space $J$ (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/… $\endgroup$
    – Faisal
    Commented Sep 20, 2011 at 21:51
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    $\begingroup$ For finitely generated modules over a Noetherian ring, no such examples exist. A student of Huneke proved this around 2004, but I don't think he ever published it. (It's possible it was already known at that time, but I never found a reference.) $\endgroup$
    – Graham Leuschke
    Commented Sep 21, 2011 at 0:23

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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.

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