**Proof.**
First assume $R$ is N-1. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. By assumption we can find $x_1, \ldots , x_ n$ in $R'$ which generate $R'$ as an $R$-module. Since $R' \subset K$ we can find $f_ i \in R$ nonzero such that $f_ i x_ i \in R$. Then $R_ f \cong R'_ f$ where $f = f_1 \ldots f_ n$. Hence $R_ f$ is normal and we have (1). Part (2) follows from Lemma 10.159.3.

Assume (1) and (2). Let $K$ be the fraction field of $R$. Suppose that $R \subset R' \subset K$ is a finite extension of $R$ contained in $K$. Note that $R_ f = R'_ f$ since $R_ f$ is already normal. Hence by Lemma 10.159.14 the set of primes $\mathfrak p' \in \mathop{\mathrm{Spec}}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\mathrm{Spec}}(R')$. Since $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is closed the image of this set is closed in $\mathop{\mathrm{Spec}}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\mathrm{Spec}}(R)$ this image.

Pick a maximal ideal $\mathfrak m \subset R$. Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral closure of the local ring in $K$. By assumption this is a finite ring extension. By Lemma 10.35.11 we can find finitely many elements $r_1, \ldots , r_ n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $r_1, \ldots , r_ n$ over $R_{\mathfrak m}$. Let $R' = R[x_1, \ldots , x_ n] \subset K$. With this choice it is clear that $\mathfrak m \not\in Z_{R'}$.

As $\mathop{\mathrm{Spec}}(R)$ is quasi-compact, the above shows that we can find a finite collection $R \subset R'_ i \subset K$ such that $\bigcap Z_{R'_ i} = \emptyset $. Let $R'$ be the subring of $K$ generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset $. Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_ i$ such that $(R'_ i)_{\mathfrak p'_ i}$ is normal. This implies that $R'_{\mathfrak p'} = (R'_ i)_{\mathfrak p'_ i}$ is normal too. Hence $R'$ is normal, in other words $R'$ is the integral closure of $R$ in $K$.
$\square$

## Comments (2)

Comment #5006 by Rankeya on

Comment #5245 by Johan on

There are also: