Suppose $R$ is a ring possessing a unique maximal ideal $m$. The claim is that if $M$ is an $R$-module such that $M=mM$, them $M=0$.
The proof begins as follows: Suppose first that $M$ is generated by a single element $x$. Then the map $\phi:R\rightarrow M$, $a\mapsto ax$ induces an isomorphism $M\cong R/I$ where $I=\ker\phi$. If $I\neq R$, then $I$ must be contained in $m$ (as it must be contained in some maximal ideal, and hence the unique one).
Now, the next part of the proof confuses me: It follows that $M/mM\cong R/m\neq 0$, a contradiction.
Why do we have $M/mM\cong R/m$? It looks like we are modding out both sides of the previous isomorphism by $m$, but I'm not sure if this is valid or even if it's the case. What am I missing?
Another way of saying the same thing: suppose $M$ is generated as an $R$-module by a single element $x \in M$. The hypothesis $M = \mathfrak m M$ tells you that $x = a x$ for some $a \in \mathfrak m$. Pulling this equation back via the isomorphism $R/I \xrightarrow{\cong} M$ gives you the equality of cosets $$1 + I = a + I$$ in $R/I$. This tells you that $1 - a \in I \subset \mathfrak m$, and hence $1 \in \mathfrak m$.