The Stacks project

Lemma 10.37.16. 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 10.37.12 and 10.37.15.

Let $\mathfrak p_1, \ldots , \mathfrak p_ n$ be the minimal primes of $R$. By Lemmas 10.25.2 and 10.25.4 we have $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_ n}$, and by Lemma 10.25.1 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 10.22 and 10.24). Each factor is of the form $R/\mathfrak p_ i$ with field of fractions $R_{\mathfrak p_ i}$. By Lemma 10.36.10 each map $R/\mathfrak p_ i \to R_{\mathfrak p_ i}$ is integrally closed. Hence $R$ is a finite product of normal domains. $\square$

Comments (4)

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: 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.g. since the latter is a domain!). Alternately, how about: each prime contains a unique minimal prime , since is a domain. Therefore the irreducible components of are open and closed, and so for all .

There are also:

  • 3 comment(s) on Section 10.37: Normal rings

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