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!

  • $\begingroup$ In fact, $R$ is artinian. $\endgroup$
    – user26857
    Mar 25, 2015 at 11:56
  • $\begingroup$ Artinian and semisimple, also. $\endgroup$
    – rschwieb
    Mar 28, 2015 at 11:21

2 Answers 2


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.

The maximal ideals are finitely generated by hypothesis.


One can prove even more:

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.


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