Timeline for Is a domain all of whose localizations are noetherian itself noetherian ?
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| Nov 28, 2012 at 3:59 | comment | added | Filippo Alberto Edoardo | @Faisal&Ralph: Thanks, I agree that dvr are of dimension 1, so I was puzzled in proving it a DVR. But you are right in saying its Spec won't be noetherian. I also agree with nosr, I was just puzzled in proving that the closure of $\mathbb{Z}$ is really that one (clear if you add only one $\zeta_p$, but I never tried with all of them...); I will think about thay, thanks a lot everybody! | |
| Nov 28, 2012 at 2:32 | comment | added | user28172 | @Filipps: Intuition for the normal domain $O = \mathbf{Z}[\zeta_2,\zeta_3,\zeta_5,\dots]$ can be gained by noting that its strict henselization at any maximal ideal of residue characteristic $p$ coincides with a strict henselization of the dvr $\mathbf{Z}[zeta_p]_{(p)}$ (as this latter strict henselization contains all prime-to-$p$ roots of unity, and strict henselization of a normal local domain is again a normal domain). Since the strict henselization $A'$ of a local ring $A$ is faithfully flat over $A$, if $A'$ is noetherian then so is $A$. Hence, every local ring of $O$ is noetherian. | |
| Nov 28, 2012 at 2:21 | comment | added | Ralph | The Krull dimension of a ring and its integral closure coincide by Cohen-Seidenberg, hence Nakano's example (call it $\mathcal{O}$) has dim 1. But I think you can deduce from Heinzer-Ohm that $Spec(\mathcal{O})$ isn't noetherian. | |
| Nov 28, 2012 at 2:02 | comment | added | Faisal | @Filippo: That can't be right. The localization of that ring at any any nonzero prime $\mathfrak p$ is a dvr, and this forces the ring to be one-dimensional. | |
| Nov 28, 2012 at 1:31 | comment | added | Filippo Alberto Edoardo | Combining the two answers, it follows that the closure of $\mathbb{Z}$ inside $\mathbb{Q}[\zeta_2,\zeta_3,\zeta_5,\zeta_7,\dots]$ is not of dimension 1 (according to Corollary 1.4, as observed at the beginning of Example 2.3, of Heinzer and Ohm). But althohgh I can imagine it is not noetherian, I confess this surprises me a lot. | |
| Nov 28, 2012 at 1:11 | history | answered | Ralph | CC BY-SA 3.0 |