TITLE: Torsion submodule
QUESTION [8 upvotes]: $A$ 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.

REPLY [12 votes]: 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.