Function field of an integral finite-type scheme

Fix a nonempty affine open $U=\mathrm{Spec}A$. Since $\eta$ is generic, a nonempty open cannot omit $\eta$: its closed complement would contain $\overline{\{\eta \}}=X$. Thus $\eta \in U$, and let $\mathfrak{p}$ be its corresponding prime. By F1, $A$ is a nonzero domain, so $(0)$ is a prime. If $\mathfrak{p}\mathrm{\ne }(0)$, choose $a\in \mathfrak{p}$ with $a\mathrm{\ne }0$. Then $D(a)$ is a nonempty open by F16--F17, since it contains $(0)$, but it does not contain $\eta$. This contradicts genericity as above, so $\mathfrak{p}=(0)$.

Restricting the structure sheaf from $X$ to its open subscheme $U$ does not change the stalk at $\eta$. By F5 it is $A_{(0)}$, and F6--F7 identify this localization canonically with $\mathrm{Frac}A$. By F3, $\mathrm{\Gamma }(U,{\mathcal{O}}_X)\cong A$. These identifications all pass through the same stalk $K={\mathcal{O}}_{X,\eta }$, so for every such $U$ they give the canonical isomorphism $K\cong \mathrm{Frac}\mathrm{\Gamma }(U,{\mathcal{O}}_X)$.

Apply F2 at $\eta$ to obtain an affine neighbourhood $W=\mathrm{Spec}B$ for which $k\to B$ is of finite type. By F1, $B$ is a domain, and step 2.1 identifies $K$ with $\mathrm{Frac}B$. Choose finite algebra generators $b_1,\dots ,b_r$ for $B$ over $k$. Then $\mathrm{Frac}B=k(b_1,\dots ,b_r)$, so F9 gives that $K\mathrm{/}k$ is finitely generated. The list may be empty, in which case $B=k$ and $K=k$.

Restriction to the generic stalk is a ring map $\mathrm{\Gamma }(X,{\mathcal{O}}_X)\to K$. If a global section maps to zero, then on every nonempty affine open $U=\mathrm{Spec}A$ its restriction maps to zero in $\mathrm{Frac}A$ under step 2.1. F8 makes $A\to \mathrm{Frac}A$ injective, so the section vanishes on each such $U$. Affine opens cover $X$; the sheaf uniqueness axiom therefore makes the global section zero. Hence the restriction map is injective.

Assume the geometric-integrality clause and fix a nonempty affine open $U=\mathrm{Spec}A$. By F12, the $k$-module $A$ is flat: the only finitely generated ideals of the field $k$ are $(0)$ and $k$, and the corresponding multiplication maps are injective. Tensoring the injection $k\hookrightarrow \bar{k}$ with $A$ gives an injection $A\cong A{\otimes }_{k}k\hookrightarrow A{\otimes }_k\bar{k}$. Thus $R=A{\otimes }_k\bar{k}$ is nonzero. By AC and F14, $R$ has a maximal ideal, so $\mathrm{Spec}R=U_{\bar{k}}$ is nonempty. By F11 it is an affine open in the chosen geometric fibre $X_{\bar{k}}$; F10 makes that fibre integral, so F1 implies $R$ is a domain. Let $S=A\setminus \{0\}$. The injection just proved shows that its image $S^{\mathrm{\prime }}=\{a\otimes 1:a\in S\}$ avoids zero in $R$. There is a canonical ring isomorphism $K{\otimes }_k\bar{k}\cong (S^{-1}A){\otimes }_k\bar{k}\cong (S^{\mathrm{\prime }}{)}^{-1}R:$ the forward map sends $(a\mathrm{/}s)\otimes \lambda$ to $(a\otimes \lambda )\mathrm{/}(s\otimes 1)$, and its inverse sends $(a\otimes \lambda )\mathrm{/}(s\otimes 1)$ to $(a\mathrm{/}s)\otimes \lambda$; F15 verifies these maps extend through the indicated localizations and are inverse on the generators. Since a localization of a domain at a multiplicative set avoiding zero is a domain, $K{\otimes }_k\bar{k}$ is a domain.

Steps 2.1, 3.1 and 3.2 prove the canonical function-field identification, finite generation and injectivity for every nonempty affine chart; step 3.3 proves the geometric-integrality implication under its stated AC assumption. The zero-generator case is included in step 3.1, and if $\bar{k}=k$ the tensor claim reduces to the field property of $K$. ∎