# Proving a ring is Noetherian when all maximal ideals are principal generated by idempotents

Let $R$ be a commutative ring with unity such that all maximal ideals are of the form $(r)$ where $r\in R$ and $r^2=r$. I wish to show that $R$ is Noetherian.

I know that if all prime (or primary) ideals in $R$ are finitely generated, then $R$ is Noetherian, so my plan was to show that all prime or primary ideals in $R$ are maximal and therefore of the above form (finitely generated), but I seem to be missing what exactly I should do to show that.
Any help is appreciated!

• In fact, $R$ is artinian. Mar 25, 2015 at 11:56
• Artinian and semisimple, also. Mar 28, 2015 at 11:21

Let $\mathfrak m = (r)$ be maximal. Note $r^2 = r$ can be written as $r(1 - r) = 0$ so every prime $\mathfrak p$ either contains $r$ or $1 - r$. As $\mathfrak m$ does not contain $1 - r$ this means every prime $\mathfrak p$ either equals $\mathfrak m$ or is not contained in $\mathfrak m$. In other words, all primes are maximal because they must equal the maximal ideal that contains them.
If $R$ is a commutative unitary ring such that all its maximal ideals are generated by idempotents, then $R$ is isomorphic to a finite direct product of fields.
The other answer proved that every prime ideal is maximal. In particular, $R$ is artinian, so it is isomorphic to a finite direct product of artinian local rings. But a local ring whose maximal ideal is generated by an idempotent is a field, and we are done.