Let $A$ be a ring. Let $\mathfrak a$ be an ideal contained int he Jacobson Radical of $A$. Let $u:M \rightarrow N$ be a morphism. Let $N$ be finitely generated. Show that if $w:M/\mathfrak a M \rightarrow N/ \mathfrak a N$ is surjective then so is $u$.
Firstly this screams Nakayamma's lemma.
We have the following commutative diagram which we can apply the snake lemma to.
$\DeclareMathOperator{\coker}{coker}$
If we suppose that $w$ is surjective then the $\coker(w)=0$. Then we have $\ker(w)\rightarrow \coker(v)\rightarrow \coker(u)\rightarrow 0$.
I'm not to sure what to do from here. The map $v$ is just the restriction of $u$ so there should be some nice between their cokernels but I cannot see what.
Best Answer
You don't even need exact sequences here. $N$ is finitely generated, $u(M)$ is its submodule, and it is easy to show that $N=\mathfrak{a}N+u(M)$. Now just apply corollary $2.7$ from Atiyah-Macdonald.