**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.
$\square$

## Comments (4)

Comment #745 by Wei Xu on

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