$A$ 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.

  • $\begingroup$ Take $R=Z$, the module $Q$ is torsion-free, but not torsionless. $\endgroup$ – TmobiusX Sep 4 '10 at 7:17

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.

  • $\begingroup$ 1+. Ok then I don't have to finish my answer ;). $\endgroup$ – Martin Brandenburg Sep 2 '10 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.