A Noetherian domain $R$ with fraction field $K$ is said to satisfy $N$-1 if the integral closure of $R$ in $K$ is module finite over $R$.
My question is: Let $R$ be a Noetherian domain of finite Krull dimension such that $R$ has prime characteristic. If for every prime ideal $P$ of $R$, the domain $R_P$ satisfies $N$-1, then does $R$ itself necessarily satisfy $N$-1 ?
I know the answer is yes if we moreover assumed $R_f$ is normal for some $0\ne f \in R$. (https://stacks.math.columbia.edu/tag/0333). Also, the reason I required prime characteristic is because in characteristic zero, $N$-1 is equivalent to $N$-2 (https://stacks.math.columbia.edu/tag/032M) and it is known that $N$-2 is not a local property , counterexample exists in any equicharacteristic (https://mathoverflow.net/questions/40935/on-noetherian-and-japanese-rings) .
Please help.
No, being N-1 is not a local property. In fact, the same counter-example as in my MathOverflow answer works, which again, appears in a note that Rankeya Datta and I wrote, and is available on his website.
Example [Hochster, Ex. 1]. 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 one-dimensional domain such that all of its local rings are excellent (hence N-1), and such that the normal locus is not open in $\operatorname{Spec}(R)$ (in fact, the normal locus consists of only the generic point) [Hochster, Prop. 1].
We claim that $R$ is not N-1. Consider the short exact sequence $$0 \longrightarrow R \longrightarrow R^N \longrightarrow Q \longrightarrow 0,$$ where $R \to R^N$ is the normalization of $R$ in its fraction field. Since normalization commutes with localization, we know that $\operatorname{Supp}(Q)$ is the non-normal locus of $\operatorname{Spec}(R)$, which is not closed by the previous paragraph. Thus, $Q$ cannot be finitely generated as an $R$-module. Using the surjection $R^N \twoheadrightarrow Q$, this implies that $R^N$ cannot be finitely generated as an $R$-module, either.