Global functions on proper integral schemes form a finite extension of the base field

The scheme $X$ is nonempty, reduced and irreducible by [F2], with generic point $\eta$ satisfying $\overline{\{\eta \}}=X$ by [F3]; the structure morphism $X\to \mathrm{Spec}k$ is proper, hence separated and of finite type, by [F6]. By [F1], restriction of global sections to the generic point embeds $\mathrm{\Gamma }:=\mathrm{\Gamma }(X,{\mathcal{O}}_X)$ into $K$ and $K\mathrm{/}k$ is finitely generated.

Fix $g\in \mathrm{\Gamma }$. Since $\mathrm{\Gamma }$ is a $k$-algebra, there is a unique $k$-algebra homomorphism ${\theta }_g:k[u]\to \mathrm{\Gamma }$ with ${\theta }_g(u)=g$, and by [F4] it corresponds to a $k$-morphism ${\phi }_g:X\to \mathrm{Spec}k[u]=U_0$; composing with the open immersion $U_0\hookrightarrow {\mathbb{P}}_k^1$ of the standard chart [F5] gives a $k$-morphism ${\bar{\phi }}_g:X\to {\mathbb{P}}_k^1$ with $\bar{\phi }{}_g=\iota \circ {\phi }_g$, so $\bar{\phi }{}_g(X)={\phi }_g(X)\subseteq U_0$.

The morphism ${\bar{\phi }}_g$ is proper: $X\to \mathrm{Spec}k$ is proper, ${\mathbb{P}}_k^1\to \mathrm{Spec}k$ is separated by [F7], and ${\bar{\phi }}_g$ is a $k$-morphism, so [F8] applies. By [F9] ${\bar{\phi }}_g$ is a closed map, so $\bar{\phi }{}_g(X)$ is closed in ${\mathbb{P}}_k^1$; since $\bar{\phi }{}_g(X)={\phi }_g(X)$ is contained in the open chart $U_0$, it is closed in $U_0=\mathrm{Spec}k[u]$.

The closure of ${\phi }_g(X)$ in $\mathrm{Spec}k[u]$ is exactly $V(\ker {\theta }_g)$. The map ${\phi }_g$ factors as $X\stackrel{c}{\to }\mathrm{Spec}\mathrm{\Gamma }\stackrel{\mathrm{Spec}{\theta }_g}{\to }\mathrm{Spec}k[u]$ by [F4]. Since the restriction $\mathrm{\Gamma }\hookrightarrow K$ is injective by [F1], the point $c(\eta )$ is the zero prime of the domain $\mathrm{\Gamma }$, and ${\phi }_g(\eta )=\ker {\theta }_g$ under contraction. Every point of ${\phi }_g(X)$ contains $\ker {\theta }_g$, so its closure is contained in $V(\ker {\theta }_g)$ by [F10]. Conversely the closure of the point $\ker {\theta }_g$ is $V(\ker {\theta }_g)$ by the Zariski closed-set description [F10], and this point lies in ${\phi }_g(X)$; hence the reverse containment holds.

The image ${\phi }_g(X)$ is not all of $U_0$: otherwise $U_0$, being the image of the closed map ${\bar{\phi }}_g$ of step 2.1, would be closed in ${\mathbb{P}}_k^1$. But $\mathrm{\infty }:=V(v)\in U_1$ lies in the closure of $U_0$: given an open neighbourhood $N$ of $\mathrm{\infty }$ in ${\mathbb{P}}_k^1$, the open set $N\cap U_1$ contains a distinguished open $D(f)\ni \mathrm{\infty }$ with $f\in k[v]$ by [F11]; here $f\mathrm{\notin }(v)$, so $f\mathrm{\ne }0$, and with $vf\mathrm{\ne }0$ in the domain $k[v]$ the point $(0)$ lies in $D(vf)=D(v)\cap D(f)\subseteq (U_0\cap U_1)\cap N$ by [F5] and [F10]. Hence every neighbourhood of $\mathrm{\infty }$ meets $U_0$, so $\mathrm{\infty }\in \overline{U_0}$ and $U_0$ is not closed — a contradiction.

Consequently $\ker {\theta }_g\mathrm{\ne }0$: step 2.1 makes ${\phi }_g(X)$ closed in $U_0=\mathrm{Spec}k[u]$, while step 2.2 identifies its closure with $V(\ker {\theta }_g)$, so ${\phi }_g(X)=V(\ker {\theta }_g)$. By step 3.1 this is a proper subset of $\mathrm{Spec}k[u]$, whereas $V(0)=\mathrm{Spec}k[u]$; hence $\ker {\theta }_g\mathrm{\ne }(0)$. Choosing $0\mathrm{\ne }h\in \ker {\theta }_g$ gives $h(g)={\theta }_g(h)=0$ in $\mathrm{\Gamma }$, so $g$ is algebraic over $k$. As $g\in \mathrm{\Gamma }$ was arbitrary, $\mathrm{\Gamma }\subseteq k_{\mathrm{a}\mathrm{l}\mathrm{g}}:=\{a\in K:a\text{ algebraic over }k\}$ inside $K$.

By [F12] the set $k_{\mathrm{a}\mathrm{l}\mathrm{g}}$ is a finite extension of $k$, and $\mathrm{\Gamma }\subseteq k_{\mathrm{a}\mathrm{l}\mathrm{g}}$ is a $k$-linear subspace, hence finite-dimensional with ${\dim }_k\mathrm{\Gamma }\le [k_{\mathrm{a}\mathrm{l}\mathrm{g}}:k]$ by [F13]. Moreover $\mathrm{\Gamma }\subseteq K$ is a subring of a field, hence a domain, and multiplication by $0\mathrm{\ne }x\in \mathrm{\Gamma }$ is an injective $k$-linear self-map of $\mathrm{\Gamma }$; by [F14] its image has dimension ${\dim }_k\mathrm{\Gamma }$, so by [F13] the image is all of $\mathrm{\Gamma }$, $x$ is invertible, and $\mathrm{\Gamma }$ is a field. Thus $\mathrm{\Gamma }$ is a finite field extension of $k$ contained in $K=k(X)$, which is the first assertion.

Now assume that $X$ is geometrically integral over $k$ for the chosen algebraic closure $\bar{k}$, so that $K{\otimes }_k\bar{k}$ is a domain by [F1]. Suppose, for contradiction, that the field $\mathrm{\Gamma }$ of step 5.1 is not $k$, and let $L:=\mathrm{\Gamma }$ and $n:=[L:k]\ge 2$. Then $L{\otimes }_{k}L$ is a commutative $k$-algebra by [F16], free with the product basis of [F15]; hence $L{\otimes }_{k}L\mathrm{\ne }0$ and ${\dim }_k(L{\otimes }_{k}L)=n\cdot n=n^2$ by [F15]. The multiplication map $\mu :L{\otimes }_{k}L\to L$, $a\otimes b\mapsto ab$, is a surjective $k$-algebra homomorphism by [F16], and $\mu (1\otimes 1)=1\mathrm{\ne }0$.

If $L{\otimes }_{k}L$ were a domain, then it would be a field: for $0\mathrm{\ne }x\in L{\otimes }_{k}L$ multiplication by $x$ is an injective $k$-linear self-map of the finite-dimensional space $L{\otimes }_{k}L$, its image has dimension ${\dim }_k(L{\otimes }_{k}L)$ by [F14] and hence equals $L{\otimes }_{k}L$ by [F13], so $x$ is invertible. A $k$-algebra homomorphism from a field $L{\otimes }_{k}L$ to the nonzero ring $L$ is then injective, because its kernel is an ideal of a field and does not contain $1$; so $n^2={\dim }_k(L{\otimes }_{k}L)\le {\dim }_{k}L=n$, contradicting $n\ge 2$. Hence $L{\otimes }_{k}L$ is not a domain.

But $L{\otimes }_{k}L$ is a domain. Since $L\subseteq K$ is a linear subspace, [F17] gives a $k$-linear complement, hence a $k$-linear retraction $r:K\to L$ of the inclusion $i:L\to K$; then $r\otimes {\mathrm{id}}_L$ retracts $i\otimes {\mathrm{id}}_L$ as a $k$-linear map on $L{\otimes }_{k}L$, so $i\otimes {\mathrm{id}}_L:L{\otimes }_{k}L\to K{\otimes }_{k}L$ is injective. By [F18] the embedding $k\hookrightarrow \bar{k}$ extends to a $k$-algebra embedding $\varphi :L\to \bar{k}$ (the extension $L\mathrm{/}k$ is finite, hence algebraic); its underlying $k$-linear injection is retracted by a $k$-linear map $\bar{k}\to L$ supplied by [F17], so ${\mathrm{id}}_K\otimes \varphi :K{\otimes }_{k}L\to K{\otimes }_k\bar{k}$ is injective. The composite $L{\otimes }_{k}L\to K{\otimes }_{k}L\to K{\otimes }_k\bar{k}$ is thus an injective $k$-algebra map [F15, F16] into the domain $K{\otimes }_k\bar{k}$ of [F1]; the image of an injective ring map is a subring of a domain, hence a domain, and $L{\otimes }_{k}L$ is isomorphic to that image, so $L{\otimes }_{k}L$ is a domain, contradicting step 7.1.

Therefore $n=[\mathrm{\Gamma }:k]=1$, i.e. $\mathrm{\Gamma }(X,{\mathcal{O}}_X)=k$, whenever $X$ is geometrically integral over $k$. If $k$ is algebraically closed and $X$ is a nonempty proper integral finite-type $k$-scheme, the chosen algebraic closure satisfies $\bar{k}=k$: an algebraic closure is algebraic over $k$ by [F19], and an element algebraic over an algebraically closed field lies in it because its minimal polynomial has a root there, so the geometric fibre $X_{\bar{k}}=X{\times }_k\mathrm{Spec}\bar{k}$ is $X$, which is integral; the previous conclusion applies and $\mathrm{\Gamma }(X,{\mathcal{O}}_X)=k$. The empty scheme is excluded by hypothesis, and the zero ring does not occur since $X$ is nonempty. The Axiom of Choice [F20] is assumed and is used exactly through the four AC-carrying suppliers [F1], [F8], [F17] and [F18], as recorded in the AC-use line; every other ingredient and computation is choice-free.