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?
Is the double dual reflexive?
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$\begingroup$
commutative-algebra
ring-theory
modules
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0A 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. – 2012-10-26
1 Answers
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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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1Sorry 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](http://www.sciencedirect.com/science/article/pii/S002186930300557X) on page 783 that these two notions are equivalent for f.g. modules over noetherian rings. – 2012-11-01