Let $R$ be a Artinian commutative ring with $1 \neq 0$. If $I$ is prime, then $I$ is maximal.

Solution 1:

Your clarification in the comments saying that "every descending chain stabilizes" is the Artinian condition, not the Noetherian condition.

Let $M$ be a prime ideal of an Artinian ring $R$. Then $R/M$ is a prime Artinian ring, but such rings are simple (isomorphic to a matrix ring over a division ring!), so by correspondence of ideals in $R/M$ with ideals between $M$ and $R$, $M$ is necessarily a maximal ideal.

Your approach for the commutative case is good one, and it's just a special case of this. Since $R$ is Artinian, so is $R/M$. But $R/M$ is a domain, and an Artinian domain is a field. Thus, $M$ is maximal.

Showing an Artinian domain is a field is also pretty easy: suppose that $aR\neq R$. By examining the chain $aR\supseteq a^2R\supseteq\dots$ you will be able to show $a=0$.

Solution 2:

It's not generally true that every prime ideal is maximal in a commutative ring. Take for example the ring $\mathbb{Z}[x]$ and the ideal $(2)$. It's prime, but it's not maximal, for example $(2)\subset (2,x)\neq\mathbb{Z}[x]$.

Solution 3:

In particular, if you look for a commutative ring where all prime ideals are maximal you have to take an Artinian ring. I think this is the largest case where it holds.