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Is it true that if $R$ is a commutative ring such that every maximal ideal of $R$ is generated by an idempotent, then $R$ is a PIR?

I thought this solution: If $P$ is a prime ideal such that isn't maximal, then by Krull's theorem $P$ is contained properly in a maximal ideal $M$. By hypothesis, $M=(e)$, where $e^2=e$. So $e(1-e)=0\in P$, so since $P$ is prime, either $e\in P$ or $1-e\in P$. If $e\in P$, then $(e)\subseteq P$, i.e., $M\subseteq P$, contradiction. So $1-e\in P\subset M$, and thus $e+(1-e)=1\in M$, hence $M=R$, contradiction again. Therefore, every prime ideal is maximal, and then every prime ideal of $R$ is principal (by hypothesis), so by a well-known theorem (I think this result is a theorem of Kaplansky analogous to Cohen's theorem for noetherian rings), $R$ is a PIR.

Is my proof right? Thanks in advance.

user26857
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Xam
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Any ideal of $R$ generated by an idempotent is a direct summand of $R$, and therefore they aren't essential ideals of $R$.

Saying that all maximal ideals are direct summands implies that all ideals are direct summands. This is because given any ideal $I$, you can find another ideal $J$ such that $I\oplus J$ is an essential ideal of $R$, and then any maximal ideal containing that would be essential. Because this is impossible, $I\oplus J=R$.

Therefore you are looking at a commutative semisimple Artinian ring: a finite product of fields. This is, in fact, a principal ideal ring.

Of course, idempotents in domains can only be $0$ or $1$, so I'll let you work out why you won't get many maximal ideals generated by those... To summarize, a ring with the above property is a PID iff it is a field.

And yes, your idea to show that every prime ideal is principal is a valid approach, albeit fairly high powered. That is also discussed at the link I gave above.

rschwieb
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