Is a domain all of whose localizations are noetherian itself noetherian ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:19:32Z http://mathoverflow.net/feeds/question/114715 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian Is a domain all of whose localizations are noetherian itself noetherian ? KBuck 2012-11-27T23:27:14Z 2012-11-29T19:23:36Z <p>Is a domain $D$, all of whose localizations $D_P$ for $P \in Spec(D)$ are noetherian, itself noetherian ? </p> <p>The question is motivated by proposition 11.5 of Neukirch's Algebraic Number Theory: </p> <blockquote> <p>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. </p> </blockquote> <p>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. </p> <p>By googling I found a counterexample for a ring with zero-divisors: </p> <p><a href="http://math.stackexchange.com/questions/73421/a-non-noetherian-ring-with-all-localizations-noetherian" rel="nofollow">http://math.stackexchange.com/questions/73421/a-non-noetherian-ring-with-all-localizations-noetherian</a></p> <p>But I couldn't find a counterexample for a domain. </p> http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian/114719#114719 Answer by Faisal for Is a domain all of whose localizations are noetherian itself noetherian ? Faisal 2012-11-28T00:46:06Z 2012-11-28T00:46:06Z <p>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 (<em>J. Sci. Hiroshima Univ. Ser. A. <strong>16</strong></em>, 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 $p$.</p> http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian/114721#114721 Answer by Ralph for Is a domain all of whose localizations are noetherian itself noetherian ? Ralph 2012-11-28T01:11:51Z 2012-11-28T01:11:51Z <p>No, this isn't true. In the paper <a href="http://www.ams.org/journals/tran/1971-158-02/S0002-9947-1971-0280472-2/S0002-9947-1971-0280472-2.pdf" rel="nofollow">[Heinzer, Ohm: Locally Noetherian Commutative Rings]</a> 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. </p> <p>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. </p> http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian/114750#114750 Answer by François Brunault for Is a domain all of whose localizations are noetherian itself noetherian ? François Brunault 2012-11-28T10:02:48Z 2012-11-28T11:41:46Z <p>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.</p> <p><strong>EDIT</strong> 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.</p> <p>The nonzero ideals of $\mathcal{O}$ are of the form $I_{\geq \alpha} = \{x \in \mathcal{O} : v(x) \geq \alpha\}$ with <code>$\alpha \in \mathbf{Q}_{&gt;0}$</code>, and $I_{> \alpha} = \{x \in \mathcal{O} : v(x) > \alpha\}$ with <code>$\alpha \in {\bf R}_{\geq 0}$</code>. Here $v$ is the $p$-adic valuation on <code>$\mathbf{C}_p$</code>. The ring $\mathcal{O}$ is one-dimensional : its only prime ideals are $(0)$ and the maximal ideal $I_{>0}$.</p> http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian/114903#114903 Answer by YACP for Is a domain all of whose localizations are noetherian itself noetherian ? YACP 2012-11-29T17:30:44Z 2012-11-29T19:23:36Z <p>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</p> <blockquote> <p>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. </p> </blockquote>