Proper normal curve rational function map

The scheme $C$ is integral, proper, finite type, and has chain dimension one; its nonempty affine coordinate rings are Noetherian domains, and $K\mathrm{/}k$ is finitely generated.

The function field has transcendence degree one: ${\mathrm{trdeg}}_{k}K=1$. By the chain-dimension definition, there are nonempty irreducible closed subsets $Z_0⊊Z_1$ of $C$. The proper closed subset $Z_0$ contains a point $x$, which is not the generic point and therefore is closed by [F10]. Choose an affine neighbourhood $\mathrm{Spec}A$ of $x$, and let $\mathfrak{p}$ be its prime. Since $x$ is not generic, $\mathfrak{p}\mathrm{\ne }(0)$; because $A$ is a domain, $(0)⊊\mathfrak{p}$, so $\dim A\ge 1$. A prime chain of length at least two in $A$ would give a strict chain of the same length of irreducible closed subsets in the open chart and, by taking closures in $C$, contradict $\dim C=1$; strictness is preserved because each closure meets the open chart in its original closed subset. Hence $\dim A=1$, and [F4] gives ${\mathrm{trdeg}}_{k}K=1$.

Algebraic case. Suppose $f$ is algebraic over $k$. Then both $f$ and $f^{-1}$ are algebraic over $k$ and have monic polynomial equations over $k$.

Every closed point $x$ has a discrete valuation ring ${\mathcal{O}}_{C,x}$. In an affine neighbourhood $\mathrm{Spec}A$ of $x$, the corresponding prime $\mathfrak{p}$ is nonzero and has height one: it has height at least one, and a longer prime chain would contradict the chain dimension of $C$ as in step 1.2. The ring $A$ is Noetherian by [F2]. Its prime localizations are the local rings of $C$, all integrally closed by normality, so [F5] makes $A$ integrally closed. Now [F7] applies to $A_{\mathfrak{p}}={\mathcal{O}}_{C,x}$, whose fraction field is $K$ by [F3].

Transcendental case. Suppose $f$ is transcendental over $k$. Then $k(f)$ has transcendence degree one over $k$. Since $K\mathrm{/}k$ is finitely generated, the same finite list of field generators also generates $K$ over $k(f)$, so $K\mathrm{/}k(f)$ has finite transcendence degree. By step 1.2 and additivity of transcendence degree, that relative transcendence degree is zero, and $K\mathrm{/}k(f)$ is algebraic. It is a finitely generated algebraic extension, hence finite by [F17]. Write $d=[K:k(f)]$.

In the algebraic case, for every closed point $x$ one has $v_x(f)\ge 0$ and $v_x(f^{-1})\ge 0$. Consider a monic equation $p(f)=f^n+{\sum }_{i<n}{a}_i{f}^i=0$. If $v_x(f)<0$, omit the zero coefficients: each remaining $a_i\in k^{\times }$ is a unit in ${\mathcal{O}}_{C,x}$ because $k$ is a field, so it has valuation zero, and $v_x(f^n)=n{v}_x(f)<i{v}_x(f)=v_x(a_i{f}^i)$ for every remaining term. Thus the leading term is the unique term of least valuation. By [F8] the sum has finite valuation $n{v}_x(f)$, contradicting $v_x(0)=\mathrm{\infty }$. The same argument applied to a monic equation for $f^{-1}$ proves the second inequality. Therefore both valuations are nonnegative and, since $v_x(f^{-1})=-v_x(f)$, both are zero.

For each closed point $x$, step 2.1 gives a DVR, so either $f\in {\mathcal{O}}_{C,x}$ or $f^{-1}\in {\mathcal{O}}_{C,x}$. Both belong to ${\mathcal{O}}_{C,\eta }=K$. Let $U_0$ be the set where $f$ is regular and $U_1$ the set where $f^{-1}$ is regular. Membership in a stalk is represented by a section on a neighbourhood; therefore each $U_i$ is open. They contain the generic point and, by the DVR alternative at every closed point, cover $C$. This argument uses no algebraicity of $f$.

Hence in the algebraic case $f$ and $f^{-1}$ belong to every affine coordinate ring $A$. Indeed, by [F5] each such $A$ is integrally closed; it is Noetherian by [F2], so [F9] expresses $A$ as the intersection of its height-one localizations. Each height-one prime corresponds to a closed point by [F10], and [F7] identifies its localization with the DVR at that point; step 3.1 puts both rational functions in each such localization. These functions on the affine cover agree in $K$ and glue by [F6] to global sections whose product is $1$. Thus $f\in \mathrm{\Gamma }(C,{\mathcal{O}}_C{)}^{\times }$.

The regular section $f{\mathrm{\mid }}_{U_0}$ gives a morphism $U_0\to \mathrm{Spec}k[t]$, with $t\mapsto f$, by [F12]; compose it with the standard chart inclusion into ${\mathbb{P}}_k^1$. Similarly $f^{-1}{\mathrm{\mid }}_{U_1}$ gives a morphism $U_1\to \mathrm{Spec}k[s]\subseteq {\mathbb{P}}_k^1$, with $s\mapsto f^{-1}$. These constructions use sections on the actual opens $U_0,U_1$; they do not require $f$ to belong to an arbitrary affine coordinate ring or use a localization such as $A_f$.

On $U_0\cap U_1$, the sections $f$ and $f^{-1}$ are reciprocal, so $f$ is a unit there and the chart transition is $s=t^{-1}$. The two morphisms of step 4.2 therefore agree on the overlap. By [F13] they glue to a morphism ${\phi }_f:C\to {\mathbb{P}}_k^1$ with ${\phi }_f^{\mathrm{\#}}(t)=f$ and ${\phi }_f^{\mathrm{\#}}(s)=f^{-1}$.

The morphism ${\phi }_f$ is dominant. On function fields, its pullback sends the indeterminate $t$ to the transcendental element $f$, so $k(t)\to K$ is injective and the generic point of $C$ maps to the generic point of ${\mathbb{P}}_k^1$.

The morphism ${\phi }_f$ is proper: $C$ is proper over $k$ by [F1], and ${\mathbb{P}}_k^1$ is separated over $k$ by [F11], so [F14] applies.

The morphism is quasi-finite. First let $y$ be a closed point of ${\mathbb{P}}_k^1$. Its fibre is closed and is not all of $C$, since ${\phi }_f$ is dominant. By [F10] it is a finite set of closed points, so each point is isolated in the fibre. The residue extension $\kappa (x)\mathrm{/}\kappa (y)$ is finite for each such point: [F16] makes $\kappa (x)\mathrm{/}k$ finite, and the point map embeds $\kappa (y)$ into $\kappa (x)$. Now let $y$ be the generic point of ${\mathbb{P}}_k^1$. A closed point $x$ mapping to $y$ would induce an embedding $k(t)=\kappa (y)\hookrightarrow \kappa (x)$, impossible because $\kappa (x)\mathrm{/}k$ is finite and $t$ is transcendental. The only point of $C$ left is its generic point $\eta$, which maps to $y$; it is isolated in this one-point fibre and its residue extension is $K\mathrm{/}k(f)$, finite of degree $d$ by step 2.2. The fibre criterion [F15] now gives quasi-finiteness.

Since ${\phi }_f$ is proper by step 6.2 and quasi-finite by step 7.1, it is finite by [F15].

Let $V_0=\mathrm{Spec}k[t]$ and $V_1=\mathrm{Spec}k[s]$ be the target charts. Finiteness gives affine preimages ${\phi }_f^{-1}(V_i)=\mathrm{Spec}{B}_i$, where each $B_i$ is a finite module over the corresponding chart ring $R_0=k[t]$ or $R_1=k[s]$. Each preimage contains $\eta$, since $\eta$ maps to the generic point of ${\mathbb{P}}_k^1$, which belongs to both charts. Thus $B_i$ is a domain with fraction field $K$ by [F3]. Dominance embeds $R_i$ into $B_i\subseteq K$, so $B_i$ is torsion-free over $R_i$. By [F18], each $B_i$ is a free $R_i$-module. Its generic localization is a finite-dimensional domain over $k(t)$, respectively $k(s)$, hence a field. Since its fraction field is $K$, it equals $K$; under $s=f^{-1}$ we have $k(s)=k(f)$. Therefore each free module has rank ${\dim }_{k(f)}K=[K:k(f)]=d$. The two target charts cover ${\mathbb{P}}_k^1$, so ${\phi }_f$ is finite locally free of degree $d$.

The target coordinate-ring maps in step 4.2 have images $k[f]$ and $k[f^{-1}]$ in $K$. These are not in general the full coordinate rings $B_0,B_1$ of the affine preimages; step 9.1 proves that the latter are finite free of rank $d$ over the respective target chart rings. This completes the transcendental case.