Closed-point local rings and function fields of integral curves

Choose a nonempty finite-type affine open $U=\mathrm{Spec}A$, which exists by F1. Its ring is a nonzero domain. The point $(0)$ is dense in $U$ by F2. Every nonempty open of an irreducible space is dense, because two disjoint nonempty opens would express that space as the union of two proper closed subsets. Thus $U$ is dense in $T$, and the closure in $T$ of its zero prime is all of $T$. It is therefore a generic point $\eta$, meaning $\overline{\{\eta \}}=T$. Apply F2 to obtain uniqueness, Noetherianity of the space, and closedness of all other points. Every nonempty open contains $\eta$, since otherwise its proper closed complement would contain the closure of $\eta$. On each nonempty affine open, its zero prime is likewise generic, hence represents $\eta$. The generic point is not closed: otherwise $T$ would be a singleton of chain dimension zero. Now fix any closed point $x$ and choose a finite-type affine neighbourhood $U=\mathrm{Spec}A$ by F1. Since $x\mathrm{\ne }\eta$, its prime $\mathfrak{p}={\mathfrak{p}}_x$ is nonzero. It is maximal because $x$ is closed in $U$. Every nonzero prime of $A$ similarly represents a point different from $\eta$, so it is maximal.

The only ideals of $k$ are $(0)$ and $k$, so F4 makes $k$ Noetherian. Hence the finite-type algebra $A$ and its localization $R=A_{\mathfrak{p}}$ are Noetherian, and every ideal of $R$ is finitely generated. By F3 this localization is the stalk at $x$. Its primes correspond to primes of $A$ contained in $\mathfrak{p}$. By step 1.1 those primes are exactly $(0)$ and $\mathfrak{p}$: any nonzero prime below $\mathfrak{p}$ is maximal and equals $\mathfrak{p}$.

Identify $K={\mathcal{O}}_{T,\eta }$ with $\mathrm{Frac}A$ by F5. Sending $a\mathrm{/}s\in A_{\mathfrak{p}}$ to the same fraction in $K$ is well defined and injective: equality in the localization means $v(at-bs)=0$ for some $v\mathrm{\notin }\mathfrak{p}$, and the domain property gives $at=bs$; the converse is immediate. Thus $R$ is a domain. If $a\mathrm{\notin }\mathfrak{p}$, then $s\mathrm{/}a$ is its inverse; if $a\in \mathfrak{p}$, then $a\mathrm{/}s$ belongs to the proper prime $\mathfrak{p}R$ and cannot be a unit. In particular every element outside $\mathfrak{p}R$ is a unit, so every proper ideal lies in $\mathfrak{p}R$ and this is the unique maximal ideal. Its quotient is the residue field by F3. Since $\mathfrak{p}\mathrm{\ne }0$, the injection makes $\mathfrak{p}R\mathrm{\ne }0$; the prime chain $(0)⊊\mathfrak{p}R$ therefore has length one and no longer prime chain exists. Hence $\dim R=1$ by F7. The fraction equivalence and arithmetic used here are those of F6.

Every element of $K=\mathrm{Frac}A$ is a quotient of elements of $A\subset R$, and every quotient of elements of $R$ with nonzero denominator defines an element of $K$. These inverse fraction maps identify $\mathrm{Frac}R$ with $K$. A germ at $x$ is represented on a principal affine neighbourhood by a fraction with denominator outside $\mathfrak{p}$; F3 and F5 send its restriction to $\eta$ to that same fraction. The embedding is therefore canonical and independent of the affine neighbourhood. The function-field supplier F5 applies on every nonempty affine chart, and the preceding argument applies at every closed point using a finite-type affine neighbourhood from F1. AC is used only through the suppliers in F2 and F4.