Function field of a curve over $\mathbb F_q$: why do valuation rings correspond to closed points?

Question

Let $C$ be a smooth projective curve over $\mathbb F_q$ given by $f\in \mathbb F_q[x,y]$ a prime polynomial. Then we can consider the function field in one variable $\mathbb F_q(C)/\mathbb F_q$, where $\mathbb F_q(C)$ is the fraction field of the domain $\mathbb F_q[C]$. How can we prove the correspondence between the closed points of $C$ (orbits of points of $C(\overline{\mathbb F_q})$ by $\operatorname{Gal}(\overline{\mathbb F_q}/\mathbb F_q)$) and the places/valuation rings of the function field? By a valuation ring over $\mathbb F_q(C)/\mathbb F_q$ I mean an intermediate ring $\mathcal O$ such that there exists $t\in\mathcal O$ such that for every $z\in \mathbb F_q(C)$ there exists $u\in\mathcal O^\times$ and $n\in\mathbb Z$: $z=ut^n$; furthermore, $\mathcal O=\{ut^n:\;u\in\mathcal O^\times,\;\;n\geq 0\}$.

I've been looking for proofs of this but I can't seem to find them, or they have slightly different hypothesis.

Answer 1

Over a more interesting base field $k$, one should require that the base field $k$ is inside the valuation ring $\mathcal{O}$. Here, as Jyrki Lahtonen points out in the comments, it is automatic. The following will actually show that for a smooth projective curve $C\to\operatorname{Spec} k$, the valuation rings of $k(C)$ containing $k$ are precisely the local rings of closed points. (Technicality for those who like them: smooth can be weakened to regular.)

In one direction, if $c\in C$ is a closed point, then $\mathcal{O}_{C,c}$ is a regular noetherian local domain of dimension one, and therefore a DVR. It contains $\Bbb F_q$, and has field of fractions $\Bbb F_q(C)$, so it is a valuation ring of $\Bbb F_q(C)/\Bbb F_q$.

In the other direction, since $C\to \operatorname{Spec} \Bbb F_q$ is projective, hence proper, we may use the valuative criteria of properness and a characterization of morphisms from a valuation ring:

Valuative Criteria for Properness (Hartshorne theorem II.4.7) Let $f:X\to Y$ be a morphism of finite type, with $X$ noetherian. Then $f$ is proper iff for every valuation ring $R$ with field of fractions $K$ and every diagram of the form $$\require{AMScd} \begin{CD} \operatorname{Spec} K @>>> X\\ @VVV @VV{f}V \\ \operatorname{Spec} R @>>> Y \end{CD}$$ there is a unique morphism $\operatorname{Spec} R\to X$ making the diagram commute.

Characterization of morphisms from a valuation ring (Hartshorne lemma II.4.4) Let $R$ be a valuation ring of a field $K$. To give a morphism of $\operatorname{Spec} K$ to a scheme $X$ is equivalent to giving a point $x_1\in X$ and an inclusion of fields $k(x_1)\subset K$; to give a morphism of $\operatorname{Spec} R\to X$ is equivalent to giving two points $x_0,x_1$ in $X$ with $x_0$ a specialization of $x_1$, and an inclusion of fields $k(x_1)\subset K$, such that $R$ dominates the local ring of $x_0$ on the subscheme $\overline{\{x_1\}}$ equipped with the reduced induced structure.

Taking $R=\mathcal{O}$, $Y=\operatorname{Spec} \Bbb F_q$, $X=C$, and $f$ the structure morphism, as $\Bbb F_q \subset \mathcal{O}$ and $\operatorname{Frac} \mathcal{O}=\Bbb F_q(C)$, we can take $\operatorname{Spec} R\to\operatorname{Spec}\Bbb F_q$ to be the spectrum of the natural inclusion and $\operatorname{Spec} K\to X$ as the natural inclusion of the generic point to get the following diagram: $$\require{AMScd} \begin{CD} \operatorname{Spec} K @>>> C\\ @VVV @VV{f}V \\ \operatorname{Spec} R @>>> \operatorname{Spec} \Bbb F_q \end{CD}$$ Since a projective morphism is proper, there exists a unique morphism $\operatorname{Spec} R\to C$ making the diagram commute. Since $\operatorname{Spec} K$ picks out the generic point $\eta$ of $C$, the inclusion $k(\eta)\subset K$ is the identity, and $C$ is exactly the closure of $\eta$ with the reduced induced structure, this morphism $\operatorname{Spec} R\to X$ is equivalent to a choice of closed point $c\in C$ so that $R$ dominates $\mathcal{O}_{C,c}$. But valuation rings are maximal elements for local rings contained in a field under the relation of domination, so since $\mathcal{O}_{C,c}$ is also a valuation ring, we must have $R=\mathcal{O}_{C,c}$.