Weil divisor normal noetherian scheme

The finite affine cover in the Noetherian-scheme condition [F1] gives, for every point $x$, a chart $U=\mathrm{Spec}A$ and a prime $\mathfrak{p}\subseteq A$ with $x\leftrightarrow \mathfrak{p}$. By [F3], ${\mathcal{O}}_{X,x}\cong A_{\mathfrak{p}}$; by normality [F2], this stalk is a domain.

Let $C$ be a global irreducible component and $U$ an affine open meeting it. By [F7, F13], $C\cap U$ is irreducible and dense in $C$; it is closed in $U$ because $C$ is closed [F8]. If an irreducible closed subset $E$ of $U$ contains $C\cap U$, its closure in $X$ is irreducible by [F8] and contains the dense subset $C\cap U$. It therefore contains $C$, and equals $C$ by maximality [F6]. Since $E$ is closed in $U$, it equals its closure intersected with $U$ by [F14], so $E=C\cap U$. Thus $C\cap U$ is an irreducible component of $U$.

Let $D,E$ be locally finite formal sums. The support of $D+E$ is contained in the union of their supports. Around any point, intersect a neighbourhood witnessing local finiteness for $D$ with one witnessing it for $E$; this neighbourhood meets only finitely many terms in either support. Hence $D+E$ is locally finite. Negation preserves support, and coefficientwise addition has zero, inverses, associativity, and commutativity, so the Weil divisors form an abelian group.

If $X$ is quasi-compact and a family of closed subsets is locally finite, take a witnessing neighbourhood at each point and then a finite subcover. The union of the finite sets met by those neighbourhoods contains the whole support. Hence the support is finite, as stated in the definition.

Suppose distinct global components $C,D$ meet at $x$, and choose an affine chart $U=\mathrm{Spec}A$ containing $x\leftrightarrow \mathfrak{p}$. Their intersections with $U$ are components by [step 1.2]. They are distinct: each is dense in its global component by [F13], so equality would imply $C=D$. Thus they correspond to distinct minimal primes ${\mathfrak{q}}_C,{\mathfrak{q}}_D$ of $A$ by [F9]. Since $x$ belongs to both, ${\mathfrak{q}}_C,{\mathfrak{q}}_D\subseteq \mathfrak{p}$. By [F10], extension to $A_{\mathfrak{p}}$ preserves their distinction and minimality. But [step 1.1] identifies $A_{\mathfrak{p}}$ with a domain, which has the unique minimal prime $(0)$. This contradiction shows that distinct irreducible components of $X$ are disjoint.

Fix an enumeration of the finite affine cover from [F1]. Each chart has finitely many irreducible components by [F11]. Every global component meets at least one chart by [F8]; assign it the first such chart. Its intersection with that chart is a component there by [step 1.2]. Two distinct global components assigned to the same chart have distinct intersections, since each is dense in its global component [F13]. Thus the finite cover and its finite chartwise component sets give only finitely many global components.

Every stalk is reduced by [step 1.1]. By [F4], the nilpotent ideal sheaf has zero stalk at every point. Its map to the zero sheaf is an isomorphism on stalks, hence an isomorphism by [F5]; thus $X$ is reduced.

The global components are closed and cover $X$ by [F8]. By [step 2.1] they are disjoint, and by [step 2.2] they are finite in number. The complement of each is a finite union of closed components, so each component is also open. An irreducible closed subscheme meets at most one member of this open disjoint cover; since the cover is exhaustive, it lies in exactly one. This proves the componentwise interpretation.