# Is a domain all of whose localizations are noetherian itself noetherian ?

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:

https://math.stackexchange.com/questions/73421/a-non-noetherian-ring-with-all-localizations-noetherian

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

• 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. – Robert Garbary Nov 27 '12 at 23:37
• @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. – Martin Brandenburg Nov 28 '12 at 0:03
• @Martin: But isn't $\mathbf{F}_2$ the only boolean domain? – Filippo Alberto Edoardo Nov 28 '12 at 0:14
• @Filippo: Yes, I just wanted to repeat that "If A_P is noetherian for all prime P, then A must be noetherian." is wrong. – Martin Brandenburg Nov 28 '12 at 0:30
• @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. – Ralph Nov 28 '12 at 1:26

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 $p$.

• By the way, almost Dedekind domains appear in the wikipedia article on Dedekind domains. (I put them there.) – Pete L. Clark Dec 26 '13 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.

• 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. – Filippo Alberto Edoardo Nov 28 '12 at 1:31
• @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. – Faisal Nov 28 '12 at 2:02
• 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. – Ralph Nov 28 '12 at 2:21
• @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. – user28172 Nov 28 '12 at 2:32
• @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! – Filippo Alberto Edoardo Nov 28 '12 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 $$\dim D=2$$.

• I fixed what I believe is a typo in the first bullet point: $S$ should be $\mathbb{Z} \setminus \{ 0 \}$, not $D \setminus \{ 0 \}$. Please change back if I'm confused. And nice example! – David E Speyer Nov 5 '19 at 16:43
• Yes, it was a typo. Thank you. – user26857 Nov 5 '19 at 19:37

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 $I_{>0}$.

• Could you explain this a little bit more? What are the prime ideals of $\mathcal{O}$ and why are their localizations $\mathbb{C}_p$? – Martin Brandenburg Nov 28 '12 at 10:37
• Sorry this doesn't answer the original question. See my edit for a more detailed explanation. – François Brunault Nov 28 '12 at 11:43
• Yes local domains don't qualify. – Martin Brandenburg Nov 28 '12 at 14:31
• Is the fontaine-fargues curve an example? – Asvin Apr 14 at 17:57