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

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    I know that if $P$ is a prime then $R/P$ is a domain and any finite domain is a field so $P$ is a maximal ideal. If $P$ is maximal ideal how can we prove that $P$ is a prime ideal? Is that correct? – 2017-02-27

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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.

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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.

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    Can you give me counterexample for non commutative ring? – 2017-02-27
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    First, since any field is an integral domain, so, any maximal is prime in commutative algebra, it's done. Second, counterexample for what? I don't understand what you mean. – 2017-02-27
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    Why any maximal is not prime in a non-commutative ring? – 2017-02-27
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    Maximals and primes behave very different in non-commutative algebra. I prefer to remember you the definition of these in non-commutative algebra: an ideal is maximal if it is not contained in a proper ideal while we say an ideal P is prime when if P contains the product of two ideals ,then it contain one of them entirely. – 2017-02-27
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    @user420255 Yes, maximal and prime ideals are defined slightly differently, but not so differently that the same argument (basically) does not work for both. – 2017-02-27