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$A$ a commutative Noetherian domain, $M$ a finitely generated $A$-module. How can I show that the kernel of the natural map $M\rightarrow M^{**}$, where $ M^{ * *}$ is the double dual (with respect to $A$), is the torsion submodule of $M$?

I do know that in this situation torsionlessness coincides with torsion-freeness. According to Auslander this result is ``well-know'' but I can't seem to prove it or find any reference on this.

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Take $R=Z$, the module $Q$ is torsion-free, but not torsionless. – TmobiusX Sep 4 '10 at 7:17
up vote 12 down vote accepted

Let $K$ be the fraction field of $A$. Then there is a natural isomorphism $M^*\otimes_A K \cong (M\otimes_A K)^*$ (where the dual on the left is the $A$-dual, and on the right is the $K$-dual). Thus the double dual map $M \to M^{* *}$ becomes an isomorphism after tensoring with $K$ over $A$, and hence its kernel is contained in the kernel of the natural map $M \to K\otimes_A M,$ which shows that its kernel is torsion. On the other hand, clearly the torsion submodule of $M$ is contained in this kernel, since $M^{* *}$ is torsion free. This proves the result.

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1+. Ok then I don't have to finish my answer ;). – Martin Brandenburg Sep 2 '10 at 14:13
Thank you very much! That was an awesome proof. – ashpool Sep 2 '10 at 14:25

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