equality of modules I'm reading a proof of Nakayama's theorem; it says at a certain step that:
For $M$, a finitely generated module on a ring $R, N$ a submodule, and $I$ an ideal of the ring $R$:
If $M = N + IM$, then $M/N = I(M/N)$. 
I am definitely sure that this question might seem stupid, but I still can't convince myself about that implication.
The proof hints at the fact that for $x$ say representing a certain class in $M/N$ there is $x'$ in $M$ such as $x+N = I (x'+N)$, but again, isn't that equivalent to saying that $M=IM$?
Is there an isomorphism between $IM/N$ and $I(M/N)$?
I know that I am confused so I thank you in advance for clarifying this for me.
 A: In general $N$ might not be a submodule of $IM$, so "$IM/N$" doesn't make sense.
Also, the statement "$M=IM$" means that for any $m\in M$, there is some $m'\in M$ and $r\in I$ such that $m=rm'$. The statement "$M/N=I(M/N)$" means that for any coset $m+N$, there is some $m'+N \in M/N$ and $r\in I$ such that $m+N=r(m'+N)=rm'+N$, but this does not imply that $m=rm'$ unless $N=0$. Thus, $M/N=I(M/N)$ is not equivalent to $M=IM$.
A: You need $I$ to be contained in the Jacobson radical of R so you can apply Nakayama's lemma. 
First observe that $M/N$ is a finitely generated $A$-module (because M is), now once showing that $M=N + IM$ implies $M/N = I(M/N)$ (just do it by inclusions) then we can conclude by Nakayama's lemma that $M/N$ is the zero-module, i.e $M=N$. 
Now to show $M/N = I(M/N)$ well pick $x \in I(M/N)$ then by _definition, we have:
$x = \sum_{i} a_{i} (m_{i}+ N)$ where the sum is finite and $a_{i} \in $I, $m_{i} \in M$.
By definition of the operations in the quotient module $M/N$ we have:
$x = \sum_{i} a_{i}m_{i} + N = (\sum_{i} a_{i}m_{i}) + N$.
Now $M$ is an $R$-module and $a_{i} \in I \subseteq R$ so that the product $a_{i}m_{i}$ lies in $M$ because $M$ is an $R$-module. Since modules are abelian groups then $(\sum_{i} a_{i}m_{i}) \in M$ so that $x \in M/N$ and we are done. Can you see the other inclusion? (hint: it is immediate)
