Exercise with Nakayama's lemma

Solution 1:

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.