# The Stacks Project

## Tag 030B

Lemma 10.36.10. Let $R$ be a domain. The following are equivalent:

1. The domain $R$ is a normal domain,
2. for every prime $\mathfrak p \subset R$ the local ring $R_{\mathfrak p}$ is a normal domain, and
3. for every maximal ideal $\mathfrak m$ the ring $R_{\mathfrak m}$ is a normal domain.

Proof. This follows easily from the fact that for any domain $R$ we have $$R = \bigcap\nolimits_{\mathfrak m} R_{\mathfrak m}$$ inside the fraction field of $R$. Namely, if $g$ is an element of the right hand side then the ideal $I = \{x \in R \mid xg \in R\}$ is not contained in any maximal ideal $\mathfrak m$, whence $I = R$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7822–7832 (see updates for more information).

\begin{lemma}
\label{lemma-normality-is-local}
Let $R$ be a domain. The following are equivalent:
\begin{enumerate}
\item The domain $R$ is a normal domain,
\item for every prime $\mathfrak p \subset R$ the local ring
$R_{\mathfrak p}$ is a normal domain, and
\item for every maximal ideal $\mathfrak m$ the ring $R_{\mathfrak m}$
is a normal domain.
\end{enumerate}
\end{lemma}

\begin{proof}
This follows easily from the fact that for any domain $R$ we have
$$R = \bigcap\nolimits_{\mathfrak m} R_{\mathfrak m}$$
inside the fraction field of $R$. Namely, if $g$ is an element of
the right hand side then the ideal $I = \{x \in R \mid xg \in R\}$
is not contained in any maximal ideal $\mathfrak m$, whence $I = R$.
\end{proof}

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