# Finitely generated modules over noetherian rings isomorphic to their double duals

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module. Suppose that $M$ is isomorphic to the double dual, how can I prove that $M$ is reflexive? (i.e. it is isomorphic to the double dual through the canonical map).

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By the canonical map you mean the map $f\mapsto \mbox{ev}_f$, where $\mbox{ev}_f(\lambda) = \lambda(f)$? – Sh4pe Oct 30 '12 at 14:01
yes, the map $m\mapsto(\varphi\mapsto\varphi(m))$ – Chris Oct 30 '12 at 16:14
For noetherian integral domains it seems to be okay, because the dual of any finitely generated module is reflexive. BTW, the posted question can be found here as exercise 11, page 38. This makes a connection with a remark from Mathoverflow where someone said that this result was proved (but not published) by a student of Huneke. – user26857 Oct 31 '12 at 1:31
Since I would like to know how to prove what you stated I posted a question here math.stackexchange.com/questions/225790/… – Chris Oct 31 '12 at 3:31

For any module $X$, I will let $\alpha_X:X\rightarrow X^{**}$ be the canonical map you described; for $a\in X, b\in X^*$, $\alpha_X(a)(b)=b(a)$. I claim that for any $X$, $\alpha_{X^*}:X^*\rightarrow X^{***}$ is a split injection; in fact it is split by the map $\alpha_X^*$ which is dual to $\alpha_X$. The splitting map is explicitly given by $\alpha_X^*(c)(a)=c(\alpha_X(a))$ for $c\in X^{***}$ and $a\in X$.

To verify this claim, let $a\in X, b\in X^*$. Then $\alpha_X^*(\alpha_{X^*}(b))(a)=\alpha_{X^*}(b)(\alpha_X(a))=\alpha_X(a)(b)=b(a)$. Since this holds for all $a$ and $b$, $\alpha_X^*\circ\alpha_{X^*}=\mathrm{id}$ as desired.

Suppose now that $M$ is a noetherian module, and let $f:M\rightarrow M^{**}$ be an isomorphism. Applying the above to $X=M^*$ shows $\alpha_{M^{**}}$ is a split injection. Using the naturality of $\alpha$ and the fact that $f$ is an isomorphism, it is clear that $\alpha_M$ is a split injection. We wish to show it is an isomorphism. Let $\beta:M^{**}\rightarrow M$ be the splitting map. It suffices to show $\beta$, or equivalently $\beta\circ f$, is injective.

That $\beta \circ f:M\rightarrow M$ is an isomorphism is a special case of the fact that an epimorphism from a noetherian module $M$ to itself is always an isomorphism. To prove this fact, let $\phi$ be such an epimorphism, e.g. $\phi=\beta \circ f$. Consider the chain of submodules $\ker \phi^n\subseteq M$. Because $M$ is noetherian there exists n such that $\ker \phi^n=\ker \phi^{n+1}$. If $x\in \ker \phi$, then $x=\phi^n(y)$ for some $y$. Since $\phi(x)=0$, $y\in \ker\phi^{n+1}=\ker\phi^n$, so $x=0$. Hence $\phi$ is injective.

Since $\beta\circ f$ is injective, $\alpha_M$ is surjective, which completes the proof.

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Cool. I have to wait 10 hours before giving you the 50 points of reputation. – Chris Oct 31 '12 at 3:28
wait....I have a question, since every surjective endomorphism of a finitely generated module is an isomorphism (see math.us.edu.pl/zatl/szymiczek/referaty/NAKsurj.pdf theorem 1 on page 3), it seems to me that we don't have to require the noetherianity of the ring, am I wrong? – Chris Oct 31 '12 at 3:52
So every finitely generated module isomorphic to the double dual is reflexive? We don't need noetherianity? – Chris Oct 31 '12 at 16:44
This proof shows that there is no reason to consider the ring $R$ noetherian. – user26857 Oct 31 '12 at 19:25