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If $R$ is a one-dimensional Noetherian domains, not Dedekind. It is true that each localization with a maximal ideal is a DVR, since the localization is a one dimensional Noetherian local domain. My question is: is this reasoning correct, even if $R$ is not integrally closed?

Rick88
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    Of course is not! If all localizations are DVRs, then the ring is Dedekind. – user26857 Dec 09 '20 at 14:05
  • Yes, I understand. I have another doubt. If I have a one-dimensional Noetherian domain, not ntegrally closed, from the algebraic point of view, why are some localizations not DVRs? I thnk that can increase the dimenson of the localization, but I can't view an exemple. – Rick88 Dec 13 '20 at 09:26

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