# The Stacks Project

## Tag 0332

Lemma 10.155.14. Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_f$ is normal then $$U = \{\mathfrak p \in \mathop{\rm Spec}(R) \mid R_{\mathfrak p} \text{ is normal}\}$$ is open in $\mathop{\rm Spec}(R)$.

Proof. It is clear that the standard open $D(f)$ is contained in $U$. By Serre's criterion Lemma 10.151.4 we see that $\mathfrak p \not \in U$ implies that for some $\mathfrak q \subset \mathfrak p$ we have either

1. Case I: $\text{depth}(R_{\mathfrak q}) < 2$ and $\dim(R_{\mathfrak q}) \geq 2$, and
2. 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 10.62.5) and $$\mathop{\rm Spec}(R) \setminus U = \bigcup\nolimits_{\mathfrak q \in E} V(\mathfrak q)$$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42839–42848 (see updates for more information).

\begin{lemma}
\label{lemma-openness-normal-locus}
Let $R$ be a Noetherian domain.
If there exists an $f \in R$ such that $R_f$ is normal
then
$$U = \{\mathfrak p \in \Spec(R) \mid R_{\mathfrak p} \text{ is normal}\}$$
is open in $\Spec(R)$.
\end{lemma}

\begin{proof}
It is clear that the standard open $D(f)$ is contained in $U$.
By Serre's criterion Lemma \ref{lemma-criterion-normal} we see that
$\mathfrak p \not \in U$ implies that for some
$\mathfrak q \subset \mathfrak p$ we have
either
\begin{enumerate}
\item Case I: $\text{depth}(R_{\mathfrak q}) < 2$
and $\dim(R_{\mathfrak q}) \geq 2$, and
\item Case II: $R_{\mathfrak q}$ is not regular
and $\dim(R_{\mathfrak q}) = 1$.
\end{enumerate}
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 \ref{lemma-finite-ass}) and
$$\Spec(R) \setminus U = \bigcup\nolimits_{\mathfrak q \in E} V(\mathfrak q)$$
as desired.
\end{proof}

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