# Prime ideals and maximal ideals

I know that in a commutative ring with identity element, every maximal ideal is prime. I thought that maybe the converse is true too. Namely, that every prime ideal is a maximal in a commutative ring with identity. But I can't find a way to prove it. Any hints or suggestions? Thanks in advance!

• Proposition from Dummit and Foote. Every nonzero prime ideal in a principal ideal domain is a maximal ideal. I can write the proof if you want. – richitesenpai Aug 13 '15 at 6:11
• Have you tried to think on this: if the converse were true then why people bothered to call them by two names??? – user26857 Aug 13 '15 at 6:57

$\{0\}$ is prime in $\mathbb Z$ but not a maximal ideal.
It is not true in general. For instance take an integral domain which is not a field and you see that $(0)$ is prime, but not maximal.
If you want there is something weaker, in the sense that you can converse your statement, by that I mean: if $A$ is a unital finite commutative ring then $I$ is a prime ideal if and only if $I$ is a maximal ideal. I can sketch a proof, only for the needed part which is this implication $(\Rightarrow)$ then you can make all the details: Suppose $I$ is a prime ideal $\Rightarrow$ $\frac{A}{I}$ is a domain. Now because $\frac{A}{I}$ is a finite domain and every finite domain is a field then $\frac{A}{I}$ is a field $\Rightarrow$ $I$ is maximal.