Under AC, proper integral finite-type schemes over fields with multiple points are not affine

Let $f:X\to \mathrm{Spec}k$ be the structure morphism of the $k$-scheme $X$ [F4]. Suppose first that $f$ is affine. Since $\mathrm{Spec}k$ is an affine open subscheme of itself, [F3] gives that $X=f^{-1}(\mathrm{Spec}k)$ is an affine scheme.

Now let $k$ be any field and consider the projective line ${\mathbb{P}}_k^1$ of [F9]. Its charts $U_0=\mathrm{Spec}k[t]$ and $U_{\mathrm{\infty }}=\mathrm{Spec}k[u]$ are open subschemes covering ${\mathbb{P}}_k^1$ and $W=U_0\cap U_{\mathrm{\infty }}$ is the affine open $\mathrm{Spec}k[t,t^{-1}]$, identified with $\mathrm{Spec}k[u,u^{-1}]$ by $t\mapsto u^{-1}$. By [F11] the global sections are $\mathrm{\Gamma }(U_0,\mathcal{O})=k[t]$ and $\mathrm{\Gamma }(U_{\mathrm{\infty }},\mathcal{O})=k[u]$, and by [F12] the restrictions to the overlap are the localisations $k[t]\to k[t,t^{-1}]$ and $k[u]\to k[u,u^{-1}]$, the second followed by the identification $u\mapsto t^{-1}$.

Two distinct points of ${\mathbb{P}}_k^1$ lie in the chart $U_0$: the evaluation map $\epsilon :k[t]\to k$, $t\mapsto 0$, is a unital ring homomorphism by [F15], and its kernel is $(t)$, because $t\in \ker \epsilon$ and every $h$ with $h(0)=0$ is divisible by $t$ by [F18]; hence by [F17] $k[t]\mathrm{/}(t)\cong \mathrm{im}\epsilon =k$, which is a field and so a domain [F7], and [F16] makes $(t)$ a prime ideal, i.e. a point of $U_0=\mathrm{Spec}k[t]$, while [F31] makes it maximal, since the quotient $k[t]\mathrm{/}(t)$ is a field. The zero ideal $(0)$ is also prime, since $k[t]$ is a domain by [F19] and $ab=0$ then forces $a=0$ or $b=0$ [F6, F7]. The ideals differ because $t\in (t)$ but $t\mathrm{\notin }(0)$. Since the chart inclusion $U_0\hookrightarrow {\mathbb{P}}_k^1$ is an open immersion and therefore injective on points [F20], these are two distinct points of ${\mathbb{P}}_k^1$.

By the quasi-inverse property in [F2], an affine scheme is canonically isomorphic to the spectrum of its global sections, so $X\cong \mathrm{Spec}\mathrm{\Gamma }(X,{\mathcal{O}}_X)$. Let $\mathrm{\Gamma }:=\mathrm{\Gamma }(X,{\mathcal{O}}_X)$.

By the sheaf axioms [F10], restriction to the open cover $\{U_0,U_{\mathrm{\infty }}\}$ identifies the global sections of the structure sheaf with the matching pairs $\mathrm{\Gamma }({\mathbb{P}}_k^1,\mathcal{O})=\{(a,b)\in k[t]\times k[u]:a(t)=b(t^{-1})\text{ in }k[t,t^{-1}]\},$ where the second entry is read through the identification $u=t^{-1}$ of the overlap.

The projective line also satisfies the hypotheses of the preceding positive-dimension assertion. It is nonempty by step 1.3 and proper over $k$ by [F22], hence of finite type by [F29]. By [F23] the chart rings $k[t]$ and $k[u]$ are Noetherian, so their spectra $U_0,U_{\mathrm{\infty }}$ are Noetherian topological spaces by [F24]. A descending chain of closed subsets of ${\mathbb{P}}_k^1$ stabilizes after restriction to each of these two opens and therefore stabilizes globally, since they cover the space; thus ${\mathbb{P}}_k^1$ is Noetherian by [F25].

The prime ideals $(0)⊊(t)$ of $k[t]$ from step 1.3 give a strict chain $V((t))=\{(t)\}⊊V((0))=U_0$ of nonempty closed subsets of $U_0$: the inclusion is strict because $(t)\mathrm{\ne }(0)$; $(t)$ is maximal by [F31], so by [F27] the only prime containing it is $(t)$ itself and $V((t))=\{(t)\}$; and every prime contains $(0)$, so $V((0))=U_0$ [F27]. Both subsets are irreducible in the sense of [F30]: a singleton is irreducible, since a cover $\{p\}=F_1\cup F_2$ by closed subsets has $p\in F_i$ for some $i$ and then $F_i=\{p\}$; and $U_0$ has the point $(0)$ generic for $V((0))=U_0$ by [F35], so the closure of $\{(0)\}$ is $U_0$ [F36], every closed subset of $U_0$ containing $(0)$ equals $U_0$, and a closed cover $U_0=F_1\cup F_2$ has some $F_i$ containing $(0)$ and hence equal to $U_0$. Hence $\dim U_0\ge 1$ by [F8], and the open-cover formula [F26] gives $\dim {\mathbb{P}}_k^1\ge 1$.

By [F1] the ring $\mathrm{\Gamma }$ is a finite field extension of $k$, in particular a field; so by [F5] its only ideals are $(0)$ and $\mathrm{\Gamma }$, and $(0)$ is a prime ideal: for $ab\in (0)$, that is $ab=0$, the domain property in [F7] gives $a=0$ or $b=0$, i.e. $a\in (0)$ or $b\in (0)$ [F6]. Since $(0)\mathrm{\ne }\mathrm{\Gamma }$, the spectrum $\mathrm{Spec}\mathrm{\Gamma }$ consists of the single point $(0)$.

Such a pair is constant: write $b={\sum }_{j=0}^e{b}_j{u}^j$ with $b_j\in k$, taking $e=0$ when $b=0$ [F13, F14]. Multiplying the identity $a(t)=b(t^{-1})$ by $t^e$ and using $u=t^{-1}$ gives $t^{e}a(t)={\sum }_{j=0}^e{b}_j{t}^{e-j}\in k[t]$, a polynomial whose displayed exponents are at most $e$, so its degree is at most $e$ [F13]. If $a\mathrm{\ne }0$, then $t^{e}a$ has degree $e+\mathrm{deg}a$, because the coefficient of $t^{e+\mathrm{deg}a}$ is the nonzero leading coefficient of $a$ and there are no terms above [F13]; hence $e+\mathrm{deg}a\le e$, so $\mathrm{deg}a=0$ and $a$ is constant. Then $b=a$ is the same constant, and if $a=0$ also $a,b\in k$. Thus $\mathrm{\Gamma }({\mathbb{P}}_k^1,\mathcal{O})=k$.

Hence $X$ has exactly one point, because the isomorphism of step 2.1 is a bijection on underlying sets. This contradicts the hypothesis that $X$ has more than one point. Therefore the structure morphism $f$ is not affine, and $X$ is not an affine scheme over $k$.

If ${\mathbb{P}}_k^1$ were affine over $k$, then by steps 1.1-1.2 applied to its structure morphism ${\mathbb{P}}_k^1\to \mathrm{Spec}k$ it would be ${\mathbb{P}}_k^1\cong \mathrm{Spec}\mathrm{\Gamma }({\mathbb{P}}_k^1,\mathcal{O})=\mathrm{Spec}k$ by step 3.2, and $\mathrm{Spec}k$ has exactly one point by [F5, F6] as in step 3.1, contradicting the two distinct points of step 1.3. Hence the projective line is not affine over $k$.

For the dimension refinement, suppose the underlying Noetherian space of $X$ has positive dimension. By [F8] there are nonempty irreducible closed subsets $Z_0⊊Z_1$ of the underlying space; picking a point of $Z_0$ and a point of $Z_1\setminus Z_0$ exhibits two distinct points of $X$, a selection from two nonempty sets and so not a use of AC. Hence the hypothesis of step 4.1 is satisfied and such an $X$ is not affine over $k$.

Finally, ${\mathbb{P}}_k^1$ is integral by [F28]: it is nonempty by step 1.3; it is reduced because the ideal sheaf $\mathcal{N}$ of nilpotent germs [F32] restricts over the open chart $U_0$ to the nilpotent-germ sheaf of $\mathrm{Spec}k[t]$, which vanishes since $k[t]$ is a domain [F19], hence a reduced ring [F33], hence a reduced affine scheme [F34], with the same holding over $U_{\mathrm{\infty }}$, so that $\mathcal{N}=0$ by the sheaf property over the cover $\{U_0,U_{\mathrm{\infty }}\}$ [F10]; and it is irreducible because $U_0$ is open, irreducible, and dense: the overlap $U_0\cap U_{\mathrm{\infty }}=D(u)$ contains the generic point $(0)$ of $U_{\mathrm{\infty }}$ [F35], so it is dense in $U_{\mathrm{\infty }}$, and $U_0$ is dense in ${\mathbb{P}}_k^1$: every nonempty open subset either meets $U_0$ directly or lies in $U_{\mathrm{\infty }}$ and, being open there, meets its dense subset $D(u)\subseteq U_0$; then any closed cover ${\mathbb{P}}_k^1=F_1\cup F_2$ restricts to the closed cover $U_0=(F_1\cap U_0)\cup (F_2\cap U_0)$ of the irreducible $U_0$, so $U_0\subseteq F_i$ for some $i$, and $F_i$, being closed, contains the closure ${\mathbb{P}}_k^1$ of the dense subspace $U_0$ [F36], so $F_i={\mathbb{P}}_k^1$. The Axiom of Choice [F21] is used through the AC-carrying suppliers [F1], [F22] and [F24]; all other selections are finite.