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As posted by navigetor23 in this question the dual of a finitely generated module over a noetherian integral domain is reflexive. Could you tell me how to prove it?

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I think I have a proof.

I will use this theorem:

Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module. Then $M^*$ is reflexive if and only if $M^*_P$ is reflexive for every $P$ such that $\mathrm{depth}\;A_P=0$; in short, a dual is reflexive if and only if it is reflexive in depth $0$.

If the ring is a domain then the only prime of depth $0$ is the zero ideal. If $M$ is finitely generated, then $M^*_{(0)}$ is a finite dimensional vector space over the ring of fractions and so it is reflexive.

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he edited it after you – Chris Nov 2 '12 at 0:31

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