Isomorphism $M/IM\simeq A/I\otimes_A M$ I'm trying to prove that $M/IM\simeq A/I\otimes_A M$. I defined $\varphi:M\to A/I\otimes_A M$ with $\varphi(m)=[1]\otimes m$. I've already proven that $\varphi$ is $A$-linear and also surjective. 
But I'm having trouble to prove that $\ker(\varphi)=IM$. The inclusion $IM\subset \ker(\varphi)$ is clear. On the other hand, if $m\in \ker(\varphi)$, then $[1]\otimes m=0$. If $m=0$, then $m\in IM$, and we are done. On the other hand, if $m\neq 0$, then I must prove that there are $i\in I$, $n\in M$ such that $m=in$. I really don't know how to prove that.
What is the trick?
 A: Here is another (more advanced) way to prove it:
Consider the exact sequence of $A$-modules
$I \to A \to A/I \to 0$
Tensoring with $M$, we get an exact sequence:
$I \otimes M \to A \otimes M \to A/I \otimes M \to 0$
Now $A \otimes M$ is naturally isomorphic to $M$ via $a \otimes m \to am$ and it is easy to see that if we take this isomorphism as an identification, the image of $I \otimes M$ in $M$ is $IM$. Thus by exactness, we have $M/IM \cong A/I \otimes M$.
A: I think it is more convenient to find an inverse map, rather than showing injectivity and surjectivity. As you showed, $IM\subset\text{Ker}(\varphi)$, which implies that $\varphi$ induces a map
$$\overline{\varphi}:M/IM\rightarrow A/I\otimes_{A}M:[m]\mapsto[1]\otimes m.$$
To make an inverse map, we can consider
$$\psi:A/I\times M\rightarrow M/IM:([a],m)\mapsto[am],$$
which induces
$$\overline{\psi}:A/I\otimes_{A}M\rightarrow M/IM:[a]\otimes m\mapsto[am].$$
You can check that $\overline{\varphi}$ and $\overline{\psi}$ are inverses of each other.
