How to directly prove that $M$ is maximal ideal of $A$ iff $A/M$ is a field? An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field.
This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove it directly?  Take $x + M \in A/M$.  How can you construct $y + M \in A/M$ such that $xy - 1 \in M$?  All I can deduce from the maximality of $M$ is that $(M,x) = A$.

Since $(M,x)=A$, you have that $1\in (M,x)$. The elements of $(M,x)$ are expressions of the form $$am+bx$$
where $a,b\in A$ and $m\in M$ (of course, since $M$ is an ideal, we in fact have $am\in M$ as well.)
Thus, if $1\in (M,x)$, there exists a $m\in M$ and an $y\in A$ such that $$1=m+xy$$
and modding out by $M$, we get that
$$1+M = (x+M)(y+M).$$
