Isomorphism of modules + tensor product Is it true that:
$$M{\otimes}_{A}(A/I) \cong M/IM$$
and
$$IM \cong I {\otimes}_AM$$
where $A$ is a commutative ring, $M$ an $A$-module, and $I \subset A$ an ideal.
 A: The 1st claim ok. The 2nd not: $A=\mathbf{Z}/4\mathbf{Z}$, $I=2A$, $M=A/2A$. In that case $IM=0$, but because $I$ is isomorphic to $A/2A$ as an $A$-module, we have $I\otimes_AM$ isomorphic to $M$ as well. Basically you are tensoring the short exact sequence
$$
0\rightarrow I\rightarrow A\rightarrow A/I\rightarrow 0
$$
with $M$. The resulting sequence will be exact except possibly at the first spot. Look up flat module for more information. The proof of the first result is also there, because we can identify the image of $I\otimes M\rightarrow A\otimes M\cong M$ with $IM$ (as pointed out by Dylan).
A: There is a proof of your first equation given in page 17 of Osborne's Basic Homological Algebra book (presuming you have defined tensor product in terms of the usual universal property). I'll give an outline here
Define $\phi':M \times A/I \to M/IM$ by $\phi'(m,a+I) = am + IM.$ Check that this is well defined. Suppose there is a bilinear $\psi:  M \times A/I \to G$. Using the fact that $A \otimes M \simeq M$ we have the following diagram
[image lost]
$\theta$ and $\phi$ come from the above tensor product ($A \otimes M \simeq M$). Convince yourself that $\theta'$ is induced from $\theta$, that the diagram commutes and that $\theta'$ is unique. 
