Let $(R, \mathfrak m)$ be a Valuation ring of finite Krull dimension such that every non-maximal ideal i e. every ideal which is not $\mathfrak m$, is principal. Then is $R$ Noetherian i.e. a discrete valuation ring ?
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If $\mathfrak{m}=0$, then $R$ is a field. Otherwise, let $r\in\mathfrak{m}$ be nonzero and let $I=r\mathfrak{m}$. Then $r\not\in I$ so $I$ is a non-maximal ideal, and hence is principal. But multiplication by $r$ is an isomorphism of $R$-modules from $\mathfrak{m}$ to $I$, so this means $\mathfrak{m}$ is principal as well. Thus every ideal in $R$ is principal and $R$ is Noetherian.
answered Jun 9, 2018 at 4:41
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$\begingroup$ Ah ... that was easy ... what do you think would happen if I only required that all non-maximal radical ideals are principal ? $\endgroup$– user495643Commented Jun 9, 2018 at 5:37
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$\begingroup$ If the value group is $\mathbb{Q}$, there are no nonzero non-maximal primes. $\endgroup$ Commented Jun 9, 2018 at 5:39
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$\begingroup$ Yes yes ... I realized we can get counterexample with Value group a subgroup of real line but not the integers ... what if I want all non-maximal radical ideals to be principal ? $\endgroup$– user495643Commented Jun 9, 2018 at 5:40
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$\begingroup$ Radical ideals are the same as prime ideals in a valuation ring. (An ideal is radical iff it is the intersection of all prime ideals that contain it, but an intersection of a chain of prime ideals is prime.) $\endgroup$ Commented Jun 9, 2018 at 5:43
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$\begingroup$ Ah right ... thanks ... in your answer you only use that $R$ is local and integral domain ... how much can any of these two properties be relaxed you think ? $\endgroup$– user495643Commented Jun 9, 2018 at 5:48