A normal noetherian ring is a finite product of normal domains

If $a\mathrm{\ne }0$ in $A$, its annihilator is proper, so by [A1] it is contained in a maximal ideal $\mathfrak{m}$. Then $a\mathrm{/}1\mathrm{\ne }0$ in $A_{\mathfrak{m}}$: otherwise some $s\mathrm{\notin }\mathfrak{m}$ annihilates $a$. A nilpotent element maps to zero in every prime localization because these localizations are domains. Hence $A$ has no nonzero nilpotents, so its zero ideal is radical.

By [F2], $0={\bigcap }_{i=1}^r{\mathfrak{p}}_i$ for the finitely many minimal primes. Distinct ${\mathfrak{p}}_i,{\mathfrak{p}}_j$ are comaximal. Otherwise their sum lies in a maximal ideal $\mathfrak{m}$ by [A1]. Their extensions to $A_{\mathfrak{m}}$ are distinct minimal primes: contraction of a prime below either extension is a prime below the corresponding ${\mathfrak{p}}_i$, and distinctness follows by choosing $a\in {\mathfrak{p}}_i\setminus {\mathfrak{p}}_j$ and using primality of ${\mathfrak{p}}_j$. This contradicts $A_{\mathfrak{m}}$ being a domain, whose only minimal prime is zero.

For each $i$, choose $e_i$ congruent to $1$ modulo ${\mathfrak{p}}_i$ and to $0$ modulo every other ${\mathfrak{p}}_j$: multiply, over $j\mathrm{\ne }i$, elements of ${\mathfrak{p}}_j$ congruent to $1$ modulo ${\mathfrak{p}}_i$. Therefore $A\to {\prod }_{i}A\mathrm{/}{\mathfrak{p}}_i$ is surjective by the sum ${\sum }_i{e}_i{a}_i$, and it is injective because its kernel is the intersection in step 2.1. Each quotient is a noetherian domain. If $r=0$, the same kernel equality gives $A=0$ and the product is empty.

Put $B=A\mathrm{/}{\mathfrak{p}}_i$. For every prime $\mathfrak{q}$ of $B$, the corresponding localization of $B$ is a localization of the product in step 3.1 at a prime and hence is an integrally closed domain by [F1]. If $x$ in the fraction field of $B$ is integral over $B$, then its image is integral over every $B_{\mathfrak{q}}$ and therefore belongs to each. If $x\mathrm{\notin }B$, the ideal $I=\{b\in B:bx\in B\}$ is proper; choose a maximal ideal $\mathfrak{m}\supseteq I$ using [A1]. The membership $x\in B_{\mathfrak{m}}$ would give an $s\mathrm{\notin }\mathfrak{m}$ with $sx\in B$, a contradiction. Thus $B$ is integrally closed, completing the proof.