Is a domain all of whose localizations are noetherian itself noetherian ? - MathOverflow

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

2012-11-27T23:27:14Z

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:

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

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

Answer by Faisal for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-28T00:46:06Z

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 $p$.

Answer by Ralph for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-28T01:11:51Z

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.

Answer by François Brunault for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-28T10:02:48Z

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 $I_{>0}$.

Answer by YACP for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-29T17:30:44Z

Actually I find it more interesting to know when a commutative ring whose all localizations are Noetherian is itself Noetherian. The paper of Heinzer and Ohm invoked by Ralph gives such conditions, but there is another one, much more famous, found by Nagata in order to help him to build an example of Noetherian ring with infinite Krull dimension. Nagata's result is the following

Let $R$ be a commutative ring such that $R_m$ is Noetherian for all $m\in\operatorname{Max}(R)$. If every non-zero element of $R$ belong to finitely many maximal ideals, then $R$ is Noetherian.