Lemma 10.37.10. 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 $I = R$. $\square$

Comment #8508 by on

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

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