# The Stacks Project

## Tag 0333

Lemma 10.155.15. Let $R$ be a Noetherian domain. Assume

1. there exists a nonzero $f \in R$ such that $R_f$ is normal, and
2. for every maximal ideal $\mathfrak m \subset R$ the local ring $R_{\mathfrak m}$ is N-1.

Then $R$ is N-1.

Proof. Set $K = f.f.(R)$. 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 10.155.14 the set of primes $\mathfrak p' \in \mathop{\rm Spec}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\rm Spec}(R')$. Since $\mathop{\rm Spec}(R') \to \mathop{\rm Spec}(R)$ is closed the image of this set is closed in $\mathop{\rm Spec}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R)$ 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 $K$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42878–42888 (see updates for more information).

\begin{lemma}
\label{lemma-characterize-N-1}
Let $R$ be a Noetherian domain.
Assume
\begin{enumerate}
\item there exists a nonzero $f \in R$ such that $R_f$ is normal, and
\item for every maximal ideal $\mathfrak m \subset R$
the local ring $R_{\mathfrak m}$ is N-1.
\end{enumerate}
Then $R$ is N-1.
\end{lemma}

\begin{proof}
Set $K = f.f.(R)$. 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 \ref{lemma-openness-normal-locus}
the set of primes
$\mathfrak p' \in \Spec(R')$ with $R'_{\mathfrak p'}$ non-normal
is closed in $\Spec(R')$. Since $\Spec(R') \to \Spec(R)$
is closed the image of this set is closed in $\Spec(R)$.
For such a ring $R'$ denote $Z_{R'} \subset \Spec(R)$ this image.

\medskip\noindent
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 \ref{lemma-integral-closure-localize}
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 $\mathfrak m \not \in Z_{R'}$.

\medskip\noindent
As $\Spec(R)$ 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 $K$.
\end{proof}

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