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Tag 030B

Chapter 10: Commutative Algebra > Section 10.36: Normal rings

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 7823–7833 (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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