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 $(M,x) = A$.

Hint: $\ \exists\, y\!:\ xy-1\in M\! \iff\! (x) + M = (1),\: $ which is true since $M$ is maximal and $x\not\in  M$
