Is a domain $D$, all of whose localizations $D_P$ for $P \in Spec(D)$ are noetherian, itself noetherian ?

The question is motivated by proposition 11.5 of Neukirch's Algebraic Number Theory:

Let $\mathfrak{o}$ be a noetherian integral domain. $\mathfrak{o}$ is a Dedekind domain if and only if, for all prime ideals $\mathfrak{p}\neq 0$, the localizations $\mathfrak{o}_\mathfrak{p}$ are discrete valuation rings.

If the question above has a positive answer, this proposition would give an unconditioned (i.e. without precondition "noetherian") characterization of Dedekind domains by a local property.

By googling I found a counterexample for a ring with zero-divisors:


But I couldn't find a counterexample for a domain.

  • $\begingroup$ I am quite sure the answer to your first question is yes: If A_P is noetherian for all prime P, then A must be noetherian. This is proposition 3.2 in chapter 2 of hartshorne. $\endgroup$ Nov 27, 2012 at 23:37
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    $\begingroup$ @Robert: No that proposition says that noetherian is local with respect to the Zariski topology. And boolean rings show that this property cannot be checked stalkwise. $\endgroup$ Nov 28, 2012 at 0:03
  • $\begingroup$ @Martin: But isn't $\mathbf{F}_2$ the only boolean domain? $\endgroup$ Nov 28, 2012 at 0:14
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    $\begingroup$ @Filippo: Yes, I just wanted to repeat that "If A_P is noetherian for all prime P, then A must be noetherian." is wrong. $\endgroup$ Nov 28, 2012 at 0:30
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    $\begingroup$ @Robert: In the proof of Hartshorne's Prop. 3.2, the statement of the proposition is translated into: $A$ a ring, $f_1,...,f_r$ elements of $A$, which generate the unit ideal and each localization $A_{f_i}$ is noetherian. Then $A$ is noetherian. So this doesn't cover the OP's question. $\endgroup$
    – Ralph
    Nov 28, 2012 at 1:26

4 Answers 4


I had the exact same question not too long ago. Apparently if you drop the noetherian precondition in Neukirch's definition of "Dedekind domain" then you get what some people call an "almost Dedekind domain". There are indeed examples of almost Dedekind domains that aren't Dedekind (i.e. aren't noetherian). The first of these was given by Nakano (J. Sci. Hiroshima Univ. Ser. A. 16, 425–439 (1953)): take the integral closure of $\mathbb Z$ in the field obtained by adjoining to $\mathbb Q$ the $p$th roots of unity for all primes $p$.

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    $\begingroup$ By the way, almost Dedekind domains appear in the wikipedia article on Dedekind domains. (I put them there.) $\endgroup$ Dec 26, 2013 at 16:20

No, this isn't true. In the paper [Heinzer, Ohm: Locally Noetherian Commutative Rings] the authors (who also mention Nakano's example from Faisal's answer which they call "quite involved" ) construct two counter-examples: see Examples 2.2, 2.3.

Their example 2.3 shows moreover that $D$ doesn't have to be noetherian, even if the space $Spec(D)$ and all $D_P$ are noetherian.

  • $\begingroup$ 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. $\endgroup$ Nov 28, 2012 at 1:31
  • $\begingroup$ @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. $\endgroup$
    – Faisal
    Nov 28, 2012 at 2:02
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    $\begingroup$ 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. $\endgroup$
    – Ralph
    Nov 28, 2012 at 2:21
  • $\begingroup$ @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. $\endgroup$
    – user28172
    Nov 28, 2012 at 2:32
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    $\begingroup$ @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! $\endgroup$ Nov 28, 2012 at 3:59

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

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    $\begingroup$ wow!! this is a very nice example, I find this behavior very pathological $\endgroup$
    – user147820
    May 30, 2021 at 18:33

The ring of integers $\mathcal{O}_{\mathbf{C}_p}$ of $\mathbf{C}_p$ is not noetherian, but its only nontrivial localization is $\mathbf{C}_p$, which is noetherian.

EDIT This doesn't answer the question : the ring $\mathcal{O}$ is local, so its localization at the maximal ideal is $\mathcal{O}$ itself, which isn't noetherian.

The nonzero ideals of $\mathcal{O}$ are of the form $I_{\geq \alpha} = \{x \in \mathcal{O} : v(x) \geq \alpha\}$ with $\alpha \in \mathbf{Q}_{>0}$, and $I_{> \alpha} = \{x \in \mathcal{O} : v(x) > \alpha\}$ with $\alpha \in {\bf R}_{\geq 0}$. Here $v$ is the $p$-adic valuation on $\mathbf{C}_p$. The ring $\mathcal{O}$ is one-dimensional : its only prime ideals are $(0)$ and the maximal ideal $I_{>0}$.

  • $\begingroup$ Could you explain this a little bit more? What are the prime ideals of $\mathcal{O}$ and why are their localizations $\mathbb{C}_p$? $\endgroup$ Nov 28, 2012 at 10:37
  • $\begingroup$ Sorry this doesn't answer the original question. See my edit for a more detailed explanation. $\endgroup$ Nov 28, 2012 at 11:43
  • $\begingroup$ Yes local domains don't qualify. $\endgroup$ Nov 28, 2012 at 14:31
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    $\begingroup$ Is the fontaine-fargues curve an example? $\endgroup$
    – Asvin
    Apr 14, 2020 at 17:57

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