Lemma 10.161.14. Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_ f$ is normal then

is open in $\mathop{\mathrm{Spec}}(R)$.

Lemma 10.161.14. Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_ f$ is normal then

\[ U = \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid R_{\mathfrak p} \text{ is normal}\} \]

is open in $\mathop{\mathrm{Spec}}(R)$.

**Proof.**
It is clear that the standard open $D(f)$ is contained in $U$. By Serre's criterion Lemma 10.157.4 we see that $\mathfrak p \not\in U$ implies that for some $\mathfrak q \subset \mathfrak p$ we have either

Case I: $\text{depth}(R_{\mathfrak q}) < 2$ and $\dim (R_{\mathfrak q}) \geq 2$, and

Case II: $R_{\mathfrak q}$ is not regular and $\dim (R_{\mathfrak q}) = 1$.

This in particular also means that $R_{\mathfrak q}$ is not normal, and hence $f \in \mathfrak q$. In case I we see that $\text{depth}(R_{\mathfrak q}) = \text{depth}(R_{\mathfrak q}/fR_{\mathfrak q}) + 1$. Hence such a prime $\mathfrak q$ is the same thing as an embedded associated prime of $R/fR$. In case II $\mathfrak q$ is an associated prime of $R/fR$ of height 1. Thus there is a finite set $E$ of such primes $\mathfrak q$ (see Lemma 10.63.5) and

\[ \mathop{\mathrm{Spec}}(R) \setminus U = \bigcup \nolimits _{\mathfrak q \in E} V(\mathfrak q) \]

as desired. $\square$

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