$A$ 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.



Torsion submodule
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.


