Let $R$ be a ring with $1$. Prove that a nonzero proper ideal $I$ of $R$ is a maximal ideal if and only if the quotient ring $R/I$ is a simple ring.
My attempt:-
$I$ is maximal $\iff$ $R/I$ is a field. $\iff$ $R/I$ has no non-trivial ideals $\iff$ $R/I$ is simple.
Is it correct?
You won't necessarily get a field in the quotient without commutativity, but you have a decent notion, nonetheless.
The rightmost equivalence is just the definition of simple ring.
If $I$ isn't maximal, then there is a proper ideal $J$ of $R$ with $I\subsetneq J$. Show that $J/I$ is a non-trivial ideal of $R/I$. Thus, simpleness of $R/I$ implies maximality of $I$.
On the other hand, suppose that $R/I$ isn't simple, so that there is a non-trivial ideal $\overline J$ of $R/I$. Let $J$ be the preimage of $\overline J$ under the quotient map $R\to R/I$, and show that $J$ is a proper ideal of $R$ containing $I$, so that $I$ is not maximal.
Cameron's answer will give you all you need, but I'll sketch a proof of $I$ maximal $\implies R/I$ simple that doesn't involve contradiction, just in case you're interested in such things.
Suppose $I$ is maximal, let $\pi : R\twoheadrightarrow R/I$ be the natural projection, and let $J$ be a nonzero ideal of $R/I$. Then $\pi^{-1}(J)\subseteq R$ is an ideal of $R$ (strictly) containing $I$, so that $\pi^{-1}(J) = R$. However, this implies that $J = R/I$, so that $R/I$ is simple.
