This exercise is supposed to be resolved using Nakayama's lemma:
suppose that $A$ is a local ring with $\mathfrak m $ as maximal ideal, that $M,N$ are $A$-modules and $N$ is finitely generated, and that $\phi:M\to N$ is $A$-linear. If the induced map $\bar\phi:M/\mathfrak m M\to {N/\mathfrak m N}$ is surjective, also $\phi$ is surjective.
I should prove that the cokernel of $\phi $ is zero, and in order to use Nakayama I would write try to prove that $$\frac{N/\phi M}{\mathfrak mN/\phi M}=0.$$ So $$\frac{N/\phi M}{\mathfrak mN/\phi M}=\frac{N/\mathfrak mN}{\phi M}=\frac{N/\mathfrak mN}{\bar\phi M}=0$$ by hypothesis. Are the passages correct? (They should follow from the isomorphism theorems). I ask because my teacher used another proof that I don't understand, so maybe I'm missing something here too.
Best Answer
The notation maybe obscures what is going on. Let $f : M \to N$ be your map, and note that what you want to show is that $C=\operatorname{Coker}f=0$. Thus, by Nakayama, it suffices that you prove $\mathfrak m C=C$ or, what is the same, that $A/\mathfrak m\otimes_A C=0$.
Since tensor products are right exact they preserve cokernels, so that the cockerel of $M/\mathfrak mM\to N\mathfrak mN$ is precisely $A/\mathfrak m\otimes_A C$, which is zero because this map is onto.
Thus, it follows that $C=0$ by Nakayama, since $C$ is a quotient of $N$ and thus also finitely generated.