On Noetherian and Japanese rings

Question

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

Answer 1

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.