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).
Finitely generated modules over noetherian rings isomorphic to their double duals
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0Since I would like to know how to prove what you stated I posted a question here http://math.stackexchange.com/questions/225790/the-dual-of-a-finitely-generated-module-over-a-noetherian-integral-domain-is-ref – 2012-10-31
1 Answers
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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0This proof shows that there is no reason to consider the ring $R$ noetherian. – 2012-10-31