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

I got stuck on this. I understand that every ideal is generated by finitely many elements. Here is my approach:
It's enough to prove the following implication:
$$ R/I \ \text{is a domain} \quad \Rightarrow \quad R/I \ \text{is a field}$$
Then only finding an inverse is left. So let $x \in R$ in arbitrary. We have to find an inverse element of $x+I \in R/I$, that as an element y so that $x \cdot y \in I+1$.
We denote $I = (x_1, x_2, \cdots, x_n)$. 
Now I have to find y so that $xy -1 = r_1 x_1+ r_2 x_2 + \cdots + r_n x_n$. This is where I no more knew what to do. 
Your advise will be appreciated. 
 A: 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 $a=0$.
A: 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 $(2)\subset (2,x)\neq\mathbb{Z}[x]$.
A: 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.
