Lemma 10.37.10 (030B)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.37: Normal rings
Lemma 10.37.10 (cite)

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Lemma 10.37.10. Let $R$ be a domain. The following are equivalent:

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

Proof. We deduce (1) $\Rightarrow$ (2) from Lemma 10.37.5. The implication (2) $\Rightarrow$ (3) is immediate. The implication (3) $\Rightarrow$ (1) follows 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 $I = R$ . $\square$

Comments (2)

Comment #8508 by Elías Guisado on June 18, 2023 at 07:11

For the implication (1) $\Rightarrow$ (2), shouldn't one perhaps mention 10.37.5?

Comment #9113 by Stacks project on May 10, 2024 at 13:20

Thanks and fixed here.

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