Take the 2-minute tour ×

Let $R$ be a commutative ring with 1. If every maximal ideal of $R$ is principal, is $R$ a principal ideal ring?

share|improve this question

1 Answer 1

up vote 10 down vote accepted

The link gap between principal maximal ideals and principal ideal rings is bridged by two theorems (and their combination):

(Kaplansky): For a commutative Noetherian ring $R$, $R$ is a principal ideal ring iff every maximal ideal is principal.

(Cohen): For a commutative ring $R$, $R$ is Noetherian iff every prime ideal is finitely generated.

(Cohen-Kaplansky): A commutative ring $R$ is a principal ideal ring iff every prime ideal is principal.

Since there can be more prime ideals than just the maximal ideals, you can see that $R$ has a chance to fail if only its maximal ideals are principal.

Two references to examples appear at this MO post: http://mathoverflow.net/questions/30715/a-remark-on-cohens-theorem . I wouldn't be able to fit in a full example any better than they could, so I hope the references are sufficient. (Beyond the reference in the accepted solution, there is a solution in the comments that might suit you better, depending on your background. To be clear, they have principal maximal ideals, but they aren't Noetherian and hence not principal ideal rings. )

For a proof of Kaplansky's theorem, you may as well look at the original paper which is free to view online. (Theorem 12.3)

share|improve this answer
Very interesting. Do you know an elementary proof of Kaplansky theorem? –  Joseph Curwen Jun 20 '13 at 19:02
Sorry, but i'm looking for a proof of the theorem with only the maximal ideals principal... –  Joseph Curwen Jun 21 '13 at 12:49
@JosephCurwen Ugh, sorry, I didn't read closely enough to realize that he calls what I call the C-K theorem just "Kaplansky's theorem." I replaced it with a link to the original paper, where the proof is also still good. –  rschwieb Jun 21 '13 at 13:15

Your Answer


