Maps from proper integral schemes to the affine line have closed-point image

The affine line is ${\mathbf{A}}_k^1=\mathrm{Spec}k[x]$ with structure morphism $\pi$ induced by the coefficient inclusion $k\hookrightarrow k[x]$, and by [F4] its global sections are $\mathrm{\Gamma }({\mathbf{A}}_k^1,\mathcal{O})=k[x]$ while $\mathrm{\Gamma }(\mathrm{Spec}k,\mathcal{O})=k$. By [F2] with $A=k[x]$ the $k$-morphism $f$ corresponds to the ring map $\theta :=f^{\mathrm{♯}}:k[x]\to \mathrm{\Gamma }$ on global sections, where $\mathrm{\Gamma }:=\mathrm{\Gamma }(X,{\mathcal{O}}_X)$; the canonical morphism $c:X\to \mathrm{Spec}\mathrm{\Gamma }$ corresponds to ${\mathrm{id}}_{\mathrm{\Gamma }}$, so naturality of the bijection gives $f=\mathrm{Spec}(\theta )\circ c$.

By [F1], $\mathrm{\Gamma }$ is a finite field extension of $k$ contained in $K$: in particular $\mathrm{\Gamma }$ is a field and ${\dim }_k\mathrm{\Gamma }<\mathrm{\infty }$; if $X$ is geometrically integral over $k$ for the chosen algebraic closure, then $\mathrm{\Gamma }=k$.

Since $f$ is a $k$-morphism, $\pi \circ f=s$; applying the global-sections functor, which reverses composition, gives $\theta \circ (k\hookrightarrow k[x])=s^{\mathrm{♯}}:k\to \mathrm{\Gamma }$ by [F13] and [F4], where $s^{\mathrm{♯}}$ is the $k$-algebra structure of $\mathrm{\Gamma }$. Hence $\theta$ is a $k$-algebra homomorphism: it is a unital ring map and $\theta (\lambda )=s^{\mathrm{♯}}(\lambda )=\lambda$ for $\lambda \in k$.

Put $I:=\ker \theta$. Then $\theta$ factors as $k[x]\twoheadrightarrow k[x]\mathrm{/}I\stackrel{\bar{\theta }}{\to }\mathrm{\Gamma }$ with $\bar{\theta }$ injective, and $\bar{\theta }$ is $k$-linear by step 2.1, so $k[x]\mathrm{/}I$ is a $k$-subspace of $\mathrm{\Gamma }$ and by [F14] is finite-dimensional over $k$ with ${\dim }_{k}k[x]\mathrm{/}I\le {\dim }_k\mathrm{\Gamma }<\mathrm{\infty }$. Since $\mathrm{\Gamma }$ is a field by step 1.2, it has no zero divisors, so its subring $k[x]\mathrm{/}I$ has none and, having $1\mathrm{\ne }0$, is a domain by [F12].

For the second assertion assume that $X$ is geometrically integral over $k$ for the chosen algebraic closure; then $\mathrm{\Gamma }=k$ by [F1]. Put $a:=\theta (x)\in k$. With $\theta$ a $k$-algebra homomorphism by step 2.1, the universal property [F16] applied to $\phi ={\mathrm{id}}_k$ and $s=a$ gives $\theta ={\mathrm{ev}}_{{\mathrm{id}}_k,a}$, that is, $\theta (g)=g(a)$ for every $g\in k[x]$.

The ring $k[x]\mathrm{/}I$ is a field: for $\alpha \mathrm{\ne }0$ in $k[x]\mathrm{/}I$ the multiplication map $m_{\alpha }:k[x]\mathrm{/}I\to k[x]\mathrm{/}I$, $\beta \mapsto \alpha \beta$, is $k$-linear and injective because $k[x]\mathrm{/}I$ is a domain by step 3.1, so [F15] gives ${\dim }_k\ker m_{\alpha }+{\dim }_k\mathrm{im}{m}_{\alpha }={\dim }_{k}k[x]\mathrm{/}I$ with $\ker m_{\alpha }=0$; hence ${\dim }_k\mathrm{im}{m}_{\alpha }={\dim }_{k}k[x]\mathrm{/}I$, and [F14] applied to the subspace $\mathrm{im}{m}_{\alpha }$ gives $\mathrm{im}{m}_{\alpha }=k[x]\mathrm{/}I$. Thus $1=m_{\alpha }(\beta )=\alpha \beta$ for some $\beta$, so every nonzero element of $k[x]\mathrm{/}I$ is invertible.

Under the additional geometric-integrality hypothesis of step 3.2, $a\in k$ is defined and $I=\ker \theta =\{g\in k[x]:g(a)=0\}=(x-a)$ by the factor theorem [F17]: $x-a\in I$ and every element of $I$ is divisible by $x-a$. The k-point $a:\mathrm{Spec}k\to {\mathbf{A}}_k^1$ corresponding by [F5] to the ring map $\alpha :k[x]\to k$, $x\mapsto a$, has source $\mathrm{Spec}(k[x]\mathrm{/}(x-a))\cong \mathrm{Spec}k$ and residue field $k$ by [F11], so it is a $k$-rational point of ${\mathbf{A}}_k^1$.

Consequently $I$ is a maximal ideal of $k[x]$: since $k[x]\mathrm{/}I$ is a field by step 4.1, $I$ is proper, and if $J$ were an ideal with $I⊊J\subseteq k[x]$, then $J\mathrm{/}I$ would be a nonzero proper ideal of the field $k[x]\mathrm{/}I$, which is impossible; this is exactly the maximality of [F8].

Therefore $V(I)=\{I\}$ in $\mathrm{Spec}k[x]$: every $\mathfrak{p}\in V(I)$ is a prime with $I\subseteq \mathfrak{p}$, hence a proper ideal containing the maximal ideal $I$, so $\mathfrak{p}=I$ by [F8], while $I\in V(I)$ because $I$ is prime.

Moreover $I$ is a closed point of $\mathrm{Spec}k[x]={\mathbf{A}}_k^1$ by [F9] (equivalently [F10], since $I$ is maximal), and [F11] gives $\kappa (I)\cong \mathrm{Frac}(k[x]\mathrm{/}I)=k[x]\mathrm{/}I$, which is finite-dimensional over $k$ by step 3.1; so the residue field of this closed point is finite over $k$.

The image of $f$ is this point: for $\mathfrak{q}\in \mathrm{Spec}\mathrm{\Gamma }$, [F6] identifies $\mathrm{Spec}(\theta )(\mathfrak{q})={\theta }^{-1}(\mathfrak{q})$, and $I={\theta }^{-1}(0)\subseteq {\theta }^{-1}(\mathfrak{q})$ is a prime of $\mathrm{Spec}k[x]$ as in [F7], so $\mathrm{Spec}(\theta )(\mathrm{Spec}\mathrm{\Gamma })\subseteq V(I)=\{I\}$ by step 6.1. Since $f=\mathrm{Spec}(\theta )\circ c$ by step 1.1, also $f(X)\subseteq \{I\}$, and $X\mathrm{\ne }\mathrm{\varnothing }$ makes $f(X)$ nonempty, so $f(X)=\{I\}$. Together with step 6.2 this is the first assertion of the statement.

Under the additional geometric-integrality hypothesis, finally $f=a\circ s$: the structure morphism $s$ corresponds under the bijection [F2] to $s^{\mathrm{♯}}:k\to \mathrm{\Gamma }$ by [F4], the $k$-point $a$ corresponds to $\alpha$, and the composite $a\circ s$ therefore corresponds to the composite $s^{\mathrm{♯}}\circ \alpha :k[x]\to k\to \mathrm{\Gamma }$; that composite is a $k$-algebra map sending $x$ to $s^{\mathrm{♯}}(a)=a\in \mathrm{\Gamma }$, so it equals $\theta$ by the uniqueness in [F16], and $\theta$ corresponds to $f$ by step 1.1. Injectivity of the bijection [F2] gives $f=a\circ s$, so $f$ factors through the $k$-rational point $a$ of step 4.2. The Axiom of Choice [F18] is used exactly through the AC-carrying suppliers [F1], [F9] and [F10] cited in steps 1.2 and 6.2; all other steps use no choice principle.