Tensor product of a module with an ideal is isomorphic to their standard product Let $A$ be a commutative ring and $M$ an $A$-module. Let $I$ be any ideal of $A$. 
We have an epimorphism $M \otimes_A I \rightarrow IM$. It seems to me that this is not in general an isomorphism. 
Q1: Any counterexample? 
If $M$ is flat, then $M \otimes_A I \cong IM$. However, flatness seems to be a too strong condition for this equality to be true for any ideal $I$. I am interested in finding a characterization of $M$ such that $M \otimes_A I \cong IM$ for any ideal $I$ of $A$. 
Q2: Any suggestions?
 A: It is easy to come up with a bunch of counterexampless, without knowing the notion of flatness, just by looking at simple special cases.
For $M=A/J$ we have $M \otimes_A I = I/IJ$, which injects to $M$ if and only if $IJ= I \cap J$. For $I=J$ this says $I^2=I$. But of course there are ideals with $I^2 \neq I$ (for example maximal ideals in a PID, or ideals with $I^2=0 \neq I$ as in Jim's example).
A: An $A$-module $M$ is flat if and only if for every finitely generated ideal $I$ of $A$, $M\otimes_AI\rightarrow IM$ is an isomorphism. This is proved, e.g., as Theorem 1.2.4 of Liu's algebraic geometry textbook. He actually states it with all ideals $I$, but a direct limit argument reduces to the case of finitely generated ideals.
A: Here is your counterexample: Let $A = k[x]/x^2$ and let $I = (x)$.  Then $I$ is itself an $A$-module, $I^2 = 0$ but $I \otimes_A I \neq 0$.
Edit: To show that $I \otimes_A I \neq 0$.  Intuitively the idea is that we cannot do $x \otimes x = 1 \otimes x^2$ because $1 \notin I$.  To make this formal show that $I \times I \to I$ defined by $(ax, bx) \mapsto abx$ is a well-defined and nonzero bilinear map.  It must then factor through $I \otimes_A I$.
