TITLE: Module isomorphic to second dual
QUESTION [11 upvotes]: Is there a simple condition on a module $M$ over a ring $R$ which will ensure that $M$ is isomorphic to its double dual, $M^{**} = \operatorname{Hom} (\operatorname{Hom}(M,R),R)$? What about a condition on $R$ which guarantees this will hold for all $R$-modues $M$? Can we find conditions which are necessary, as well as sufficient?

REPLY [14 votes]: A module which is isomorphic to its double dual under the natural map is called reflexive. These are objects that have been rather carefully studied in commutative algebra.
As noted in the other answers/comments, a finitely generated projective module over a ring $R$ is necessarily reflexive.  Conversely if $R$ is a regular local commutative ring of dimension $\leq 2$, then any f.g. reflexive module is projective (equivalently, free, since we are over a local ring).  This is not true if the dimension is $> 2$; in that case there are f.g. reflexive modules that are not free.  
This MO answer provides more information, including a characterization of f.g. reflexive modules over an integrally closed domain in terms of other standard module theoretic properties.