Maximal and prime ideal in a finite rings

Question

Suppose that $R$ is a finite ring with $1$.Could we prove that the numbers of maximal ideal and prime ideal is equal?

Answer 1

Yes: every maximal ideal is prime, and in a finite ring every prime ideal is maximal.

Proof of the first claim: let $M$ be maximal and $A,B$ be ideals such that $AB\subseteq M$. If $A$ is not contained in $M$, then $A+M=R$. Multiplying by $B$ on the right, $AB+MB=B$. The left hand side is a subset of $M$ , so $B \subseteq M$.

Prof of the second claim: $R/P$ is a prime ring. If it is finite, it is Artinian, hence simple by the Artin-Wedderburn theorem. Thus $P$ is maximal as well.

Answer 2

Let $R$ be commutative. We know that any maximal ideal is prime. Conversely, for any prime ideal $P$ of $R$, the quotient ring $R/P$ is a finite integral domain, hence a field. Thus in commutative finite rings, prime ideals coincide with maximal ideals.

For non-commutative ring, the argument above does not work.