Normal points and normal varieties

Assume the Axiom of Choice for the affine-coordinate and localization interfaces. Let $k$ be an algebraically closed field and let $X$ be a classical variety over $k$, in the reduced separated finite-type register of Classical algebraic prevarieties, regular maps, and varieties; the irreducible case is the one of Integral classical varieties in the compatible affine-atlas register. Let $x\in X$ and let ${\mathcal{O}}_{X,x}$ be the local ring of germs of regular functions at $x$ (Germs of regular functions and the local ring at a point of a classical affine variety); when $X$ is affine with maximal ideal ${\mathfrak{m}}_x$ of $x$, this ring is the localisation $k[X{]}_{{\mathfrak{m}}_x}$ (The local ring at a point of an affine variety is the localization at its maximal ideal). For a reducible affine chart, the same identification follows by representing germs on principal neighbourhoods and using Regular functions on a principal open are the principal localization.

The point $x$ is normal when ${\mathcal{O}}_{X,x}$ is an integrally closed domain (Integral closure in an extension ring and integrally closed domains). The variety $X$ is normal when every one of its points is normal; equivalently, when every local ring ${\mathcal{O}}_{X,x}$ is an integrally closed domain.

This is the pointwise form of the ring-theoretic notion of normal noetherian ring: a Noetherian ring is normal when all of its prime localisations are integrally closed domains, and the local rings of the points of a classical variety are the localisations of the coordinate rings of its affine charts. Prime localizations and classical point localizations give the same normality condition, as proved in the affine-chart criterion ↗; reducibility is allowed.

Recorded consequence. A point that lies on two distinct irreducible components of $X$ cannot be normal. On an affine chart $U$ with reduced coordinate ring $R=k[U]$, the components through $x$ correspond to two distinct minimal primes $\mathfrak{p}\mathrm{\ne }\mathfrak{q}$ of $R$ contained in the maximal ideal ${\mathfrak{m}}_x$. Pick $a$ in the intersection of the minimal primes other than $\mathfrak{q}$ with $a\mathrm{\notin }\mathfrak{q}$, and $b$ in the intersection of the minimal primes other than $\mathfrak{p}$ with $b\mathrm{\notin }\mathfrak{p}$; such elements exist because if that intersection were contained in $\mathfrak{q}$, then some minimal prime other than $\mathfrak{q}$ would be contained in $\mathfrak{q}$, hence equal to it. Then $ab$ lies in every minimal prime of the reduced ring, hence $ab=0$. Moreover $a\mathrm{/}1\mathrm{\ne }0$ in $R_{{\mathfrak{m}}_x}$: if $sa=0$ with $s\mathrm{\notin }{\mathfrak{m}}_x$, then the class of $s$ is a nonzero element of the domain $R\mathrm{/}\mathfrak{q}$ annihilating the nonzero class of $a$, a contradiction, so $s\in \mathfrak{q}\subseteq {\mathfrak{m}}_x$; the same argument applies to $b$. Thus ${\mathcal{O}}_{X,x}$ has zero divisors and is not a domain, so $x$ is not normal. In particular, distinct irreducible components of a normal classical variety are disjoint.