
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 10.23). 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. $\square$

Comment #745 by Wei Xu on

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 on

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

Comment #2180 by David Savitt on

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 $i$.

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