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Suppose $R$ is a noetherian ring and $M$ a finitely generated $R$-module. I was trying to prove that if $M$ is isomorphic to the double dual then it is reflexive. I reduced the problem to proving that the double dual of $M$ is reflexive. Is that true? and if it is true how can I prove it?

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Is'nt it just the definition of reflexive module ? – user18119 Oct 26 '12 at 21:36
A module is reflexive if the canonical map $M\rightarrow M^{**}$ is an iso. There are modules that are not reflexive but isomorphic to their double dual. – Chris Oct 26 '12 at 22:10
thanks for clarification. – user18119 Oct 27 '12 at 10:36
up vote 2 down vote accepted

The double dual of a finitely generated module over a noetherian ring is not necessary reflexive. T.Y. Lam gives the following example: $R=K[X,Y]/(X,Y)^2$ and $M=R/\mathfrak{m}$, where $\mathfrak{m}=(X,Y)/(X,Y)^2$. Actually $M\cong K$ (as $R$-modules) and $M$ is the only simple $R$-module. Then $M^*\cong\mathrm{socle}(R)$, and we can see easily that $\mathrm{socle}(R)\cong M\oplus M$. Now we get that the $r$th dual of $M$ is isomorphic to the direct sum of $2^r$ copies of $M$. This shows that none of these is reflexive.

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just for curiosity, which book of Lam are you refering to? – Chris Oct 28 '12 at 21:21
@Chris, it looks like Lectures on Modules and Rings, pages 517-518. – Manny Reyes Nov 1 '12 at 15:48
Sorry Chris, but I didn't noticed your comment before. Manny Reyes is right (thanks Manny for the answer!). This also appears in the Lam's book Exercises in Modules and Rings as exercise 16.13. This example shows also that there exists a finitely generated $R$-module which is $2$-syzygy, but not reflexive (namely $\mathfrak{m}$), despite the wrong claim made by Huneke and Leuschke on page 783 that these two notions are equivalent for f.g. modules over noetherian rings. – user26857 Nov 1 '12 at 17:55

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