The Stacks project

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 $I = R$. $\square$


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