This seems like it ought to be true and easy, but somehow I'm stymied.
Let $A$ be a commutative ring (Noetherian if you like) and let $M$ be a finitely generated $A$-module. Suppose that $M$ is torsion-free in the sense that $M$ is without torsion by zero-divisors in $A$ (equivalently, $M$ injects into its scalar extension to the total quotient ring of $A$).
My question is whether this torsion-free property persists after nilredution. That is, is $M/\mathrm{nil}(A)M$ torsion-free over $A/\mathrm{nil}(A)$ in the same sense?
It seems strange that nilreduction should introduce torsion, but I can't seem to put it together this morning.
It seems to come down to the following: if $m\in M$ and $s$ is a non zero-divisor in $A$, and $sm\in \mathrm{nil}(A)M$, then is $m\in \mathrm{nil}(A)M$ as well? If $M=A$ this is certainly true since then $s^km^k=0$ for some $k$, so $m^k=0$ because $s$ is a non zero-divisor. My geometric intuition suggests that this should be true in the general case, but my algebra is failing me. Then again, perhaps by intuition is failing me and it's just false...
Any thoughts would be appreciated.
I've edited to replace "$rad(A)$" by "$\mathrm{nil}(A)$" so as not to confuse with the Jacobson radical.