# On Noetherian and Japanese rings

Let $R$ be a Noetherian ring all of whose local rings are Japanese. Is $R$ necessary Japanese?

• "In commutative algebra, an integral domain A is called an N-1 ring if its integral closure in its quotient field is a finite A module. It is called a Japanese ring (or an N-2 ring) if for every finite extension L of its quotient field K, the integral closure of A in L is a finite A module." - wikipedia – Guntram Oct 3 '10 at 15:32
• EGA also only defines "Japanese" for noetherian domains. Basic result of Nagata (see EGA IV$_2$, 6.13.6) is that when localizations of a noetherian domain $R$ at all primes are Japanese then integral closure $R'$ in a finite extension $K'/K$ of fraction field is $R$-finite if and only if $R'_r$ is $R_r$-finite for some nonzero $r \in R$. Using EGA IV$_2$, 5.10.17 and 6.13.2--6.13.4, it then follows that your $R$ is Japanese iff each $R$-finite ring extension that is a domain has non-empty open normal locus in Spec. Not sure if it's of any use... – BCnrd Oct 3 '10 at 17:26
• In fact, since your hypothesis is inherited by any domain that is a module-finite ring extension of $R$, to prove an affirmative answer in general it is equivalent to prove in general that your hypothesis implies that the normal locus in Spec($R$) contains a non-empty open set (which actually implies it is open, and conversely such openness is always true in the Japanese case; see 6.13.2--6.13.4 of EGA IV$_2$). Still seems like it may be a useless viewpoint on the question... – BCnrd Oct 3 '10 at 19:33
• Wow. For 18 months this question had 0 votes and today it got 5 upvotes after a minor syntax edit. I guess syntax is pretty important 'round here. – David White Apr 26 '12 at 19:25

No, $$R$$ is not necessarily Japanese. What follows is a explicit example from a note that Rankeya Datta and I wrote, now available on his website.
Example. We use Hochster's example [Hochster, Ex. 1]; see one of my other answers for the relevant results therein. Let $$I$$ be the set of positive integers, and set $$R_i := k[x_i^2,x_i^3] \qquad\text{and}\qquad P_i := (x_i^2,x_i^3) \subseteq R_i$$ for every $$i$$, where $$k$$ is a fixed algebraically closed field. Then, setting $$R' := \bigotimes_{i \in I} R_i$$, the ring $$R := \Bigl(R' \smallsetminus \bigcup_{i \in I} P_iR'\Bigr)^{-1}R'$$ is a domain such that all of its local rings are excellent (hence Japanese), and such that the regular locus in $$\operatorname{Spec}(R)$$ is not open [Hochster, Prop. 1].
We claim that $$R$$ is not Japanese. By Hochster's construction (see [Datta–M, Rem. 4]), the ring $$R$$ is one-dimensional. But the regular locus is open for any one-dimensional Japanese domain [Datta–M, Lem. 5], and hence $$R$$ cannot be Japanese.