still torsion-free after nilredution?

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.

-
Note that your translation of the question is not correct: a regular element of $A/\mathrm{nil}(A)$ needs not be the image of a regular element of $A$ (except when the associated primes of $A$ are all minimal primes). An example is given in my answer below. – user18119 Jan 17 '12 at 23:04

This is not true in general.

Consider a noetherian local ring $(A, \mathfrak m)$ of positive Krull dimension and of depth $0$ (e.g. localization of $\mathbb Q[x,y]/(x^2, xy)$ at the origin). Then the regular elements of $A$ are just the units of $A$. So the total quotient ring $\mathrm{Tot}(A)$ of $A$ is $A$ itself, and any $A$-module $M$ is torsion free with your definition because $M\to M\otimes \mathrm{Tot}(A)$ is an isomorphism.

Now take $M=A/\mathfrak m$. Then $M/\mathrm{nil}(A)M=M$. The quotient $B:=A/\mathrm{nil}(A)$ is a reduced noetherian local ring of positive Krull dimension, so there exists a regular and non-invertible element $b$ of $B$. But $bM=0$ (in fact $M\otimes \mathrm{Tot}(B)=0$) and $M\ne 0$.

-
What a beautiful answer! Do you have some intuitive/geometric explanation for this unexpected (by me) behaviour? – Georges Elencwajg Jan 17 '12 at 18:29
Thank you @GeorgesElencwajg. No I don't have any geometric explanations, but only some feeling that the condition ($M$ has the same associated primes than $A$) is not enough because the associated primes behave badly in general with quotients (or geometrically, one can't expect to compare the embedded points of a closed subscheme with that of the ambient scheme). I tried to construct an example with an explicit relation $sm=ux$ with $u$ nilpotent, $x\in M$, $s, m$ as in "Confused"'s question (but $s$ needs not be regular in $A$). I got one, simplified it, and change to $M=A/\mathfrak m$. – user18119 Jan 17 '12 at 23:01
@GeorgesElencwajg This is a test. – Zev Chonoles Jan 18 '12 at 19:39