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Tag 030C

Chapter 10: Commutative Algebra > Section 10.36: Normal rings

Lemma 10.36.16. Let $R$ be a ring. Assume $R$ is reduced and has finitely many minimal primes. Then the following are equivalent:

  1. $R$ is a normal ring,
  2. $R$ is integrally closed in its total ring of fractions, and
  3. $R$ is a finite product of normal domains.

Proof. The implications (1) $\Rightarrow$ (2) and (3) $\Rightarrow$ (1) hold in general, see Lemmas 10.36.12 and 10.36.15.

Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the minimal primes of $R$. By Lemmas 10.24.2 and 10.24.4 we have $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_n}$, and by Lemma 10.24.1 each factor is a field. Denote $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ the $i$th idempotent of $Q(R)$.

If $R$ is integrally closed in $Q(R)$, then it contains in particular the idempotents $e_i$, and we see that $R$ is a product of $n$ domains (see Sections 10.21 and 10.22). Each factor is of the form $R/\mathfrak p_i$ with field of fractions $R_{\mathfrak p_i}$. By Lemma 10.35.10 each map $R/\mathfrak p_i \to R_{\mathfrak p_i}$ is integrally closed. Hence $R$ is a finite product of normal domains. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7951–7960 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-reduced-ring-normal}
    Let $R$ be a ring. Assume $R$ is reduced and has finitely many
    minimal primes. Then the following are equivalent:
    \begin{enumerate}
    \item $R$ is a normal ring,
    \item $R$ is integrally closed in its total ring of fractions, and
    \item $R$ is a finite product of normal domains.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    The implications (1) $\Rightarrow$ (2) and
    (3) $\Rightarrow$ (1) hold in general,
    see Lemmas \ref{lemma-normal-ring-integrally-closed} and
    \ref{lemma-finite-product-normal}.
    
    \medskip\noindent
    Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the minimal primes of $R$.
    By Lemmas \ref{lemma-reduced-ring-sub-product-fields} and
    \ref{lemma-total-ring-fractions-no-embedded-points} we have
    $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_n}$, and
    by Lemma \ref{lemma-minimal-prime-reduced-ring} each factor is a field.
    Denote $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ the $i$th idempotent
    of $Q(R)$.
    
    \medskip\noindent
    If $R$ is integrally closed in $Q(R)$, then it contains in particular
    the idempotents $e_i$, and we see that $R$ is a product of $n$
    domains (see Sections \ref{section-connected-components} and
    \ref{section-tilde-module-sheaf}). Each factor is of the form
    $R/\mathfrak p_i$ with field of fractions $R_{\mathfrak p_i}$. 
    By Lemma \ref{lemma-finite-product-integral-closure} each map
    $R/\mathfrak p_i \to R_{\mathfrak p_i}$ is integrally closed. 
    Hence $R$ is a finite product of normal domains.
    \end{proof}

    Comments (4)

    Comment #745 by Wei Xu on June 26, 2014 a 5:19 pm UTC

    In the second sentense of the proof, it uses a lemma tag02xl but that lemma requires an extra condition: $R\setminus (\mathfrak{q}_1\cup\cdots\cup \mathfrak{q}_t)$ is the set of nonzero-divisors. (However this condition can be proved here).

    Comment #757 by Johan (site) on June 28, 2014 a 6:23 pm UTC

    OK, yes, I made an improvement of the treatment of zerodivisors in reduced ring in this commit. Thanks!

    Comment #2180 by David Savitt on August 27, 2016 a 1:50 pm UTC

    In the third paragraph of the proof, it's not clear to me how one is intended to make sense of the claim that $e_i \in R_{\mathfrak p}$ (e.g. since the latter is a domain!). Alternately, how about: each prime $\mathfrak p$ contains a unique minimal prime $\mathfrak q_i$, since $R_{\mathfrak p}$ is a domain. Therefore the irreducible components $V(\mathfrak q_i)$ of $\mathrm{Spec}(R)$ are open and closed, and so $e_i \in R$ for all $i$.

    Comment #2207 by Johan (site) on August 29, 2016 a 7:41 pm UTC

    OK, I agree with this change, thanks! See here.

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