# The Stacks Project

## Tag 030C

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}

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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