Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.
My attempt: Let $I$ be an ideal of $R$. Then we have $I$ is maximal $\Leftrightarrow$ $R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain $\Leftrightarrow$ $I$ is a prime ideal.
Is my proof valid ?
Yes, your proof is valid, but note that the second implication relies on $R$ being finite. It’d be clearer if written as
$R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain
The whole thing would be even cleaner if written as
Since $R$ is finite, we have the following equivalences:
$I$ is maximal $\Leftrightarrow$ $R/I$ is a field $\Leftrightarrow$ $R/I$ is an integral domain $\Leftrightarrow$ $I$ is a prime ideal
though only the second relies on $R$ being finite.
The heart of the proof is good, and I wanted to comment that you could easily prove a “cousin” of that theorem for noncommutative Artinian rings!
I am, of course, using the noncommutative definition of prime ideals which generalizes the commutative one.
Proposition: A prime ideal in a right Artinian ring $R$ is maximal iff it is prime.
Proof ($\Rightarrow$) If $M$ is a maximal ideal, $R/M$ is a simple ring, which is certainly a prime ring. It follows that $M$ is a prime ideal.
Proof ($\Leftarrow$) (This is where the theme of your proof can be applied again!) Suppose $P$ is a prime ideal. Then $R/P$ is a prime ring. Since $R$ is right Artinian, so is $R/P$. But the Artin-Wedderburn theorem says that such a ring is simple, hence $P$ is maximal. $\Box$
The connection is that Wedderburn’s little theorem is like the Artin-Wedderburn theorem: one says that a finite domain is a field, one says that a right Artinian prime ring is a simple ring. (“Right Artinian” is a weaker form of “finite” and “prime” is a weaker form of “domain”.)
R/p is semiprime and right artinian, so it is semisimple. Since R/p is in fact prime, it can have only one simple component.
Therefore, R/p is simple, so p is a maximal ideal.