Let $A$ be a ring and $\mathfrak a \subseteq A$ an ideal. Let $N \to M$ be a homomorphism of $A$-modules such that the induced homomorphism $N/\mathfrak a N \to M/\mathfrak a M$ is surjective. If $M$ is a finite $A$-module, then there exists an element of the form $b = 1+a$ with $a \in \mathfrak a$, such that the induced homomorphism $N_b \to M_b$ of $A_b$-modules is surjective.
Unfortunately, I don't understand how to use Nakayama's lemma to solve this exercise. Can someone give me an hint/solution, please?
Let's call $\phi$ the map $N \to M$. Since $\bar\phi:N/\mathfrak aN\to M/\mathfrak aM$ is surjective we have $\phi(N)+\mathfrak aM=M$, that is, $\mathfrak a(M/\phi(N))=M/\phi(N)$. Nakayama Lemma (Statement 1) tells you that there is $a\in\mathfrak a$ such that $(1+a)(M/\phi(N))=0$, that is, $(1+a)M\subseteq\phi(N)$. Can you conclude from here?