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 $(M,x) = A$.