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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?

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If $M$ is a finitely-generated free $R$-module, then there is such an isomorphism. I suspect the only ring for which double-dualisation is an isomorphism is the trivial ring. –  Zhen Lin Jun 20 '12 at 15:16

2 Answers 2

up vote 2 down vote accepted

This is true if $M$ is finitely-generated projective (see for example these notes). I'm not aware of a more general condition that's reasonably easy to state.

I agree with Zhen that probably no nontrivial ring has this property for all $M$. This is not even true for fields (if $M$ is an infinite-dimensional vector space then the natural map to the double dual is never an isomorphism).

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I should mention that that last fact is contingent on the axiom of choice. It may actually be consistent with ZF that this is not true... –  Qiaochu Yuan Jul 1 '12 at 5:36

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.

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