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Tag 0332

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

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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