Show that a proper ideal M of a commutative ring R is maximal if and only if R/M is a field.
What I know:
Because M is a proper ideal $M \neq R$. The ideal M is maximal if it is a maximal element among the proper ideals. R is a field if it is commutative and $R^{x} = R^{*} $ where $R^{x}$ is the group of units of R and $R^{*}$=$R$ \ {$0$}.
I need help working through and writing the actual proof.
Some ideas:
I know that a ring is a field iff the zero ideal is maximal.
I assume I will need to reference the Correspondence Theorem for rings.
Any advice or help would be greatly appreciated!
Best Answer
If $M$ is a maximal ideal and $a\notin M$ then the ideal $(a)+M$ must equalize $R$ so there are elements $r\in R$ and $m\in M$ with $ra+m=1$ or equivalently $(r+M)(a+M) =1+M$ in ring $R/M$.
Conversely if $R/M$ is a field and $a\notin M$ then $(r+M)(a+M) =1+M$ for some $r\in R$. Then $ra-1\in M$ so that $1\in (a)+M$ and consequently $R=(a)+M$. This for every $a\notin M$ so apparently $M$ is maximal.