Let $C$ be a smooth projective curve (geometrically integral of dimension 1) over a field $k$. Let $K$ be the function field of $C$. Then the set of non-trivial valuation rings of $K/k$ (i.e. $\neq K$ and contains $k$) corresponds bijectively with the set of non-generic points of $C$.
I was told this result is kind of well known. Is there a reference? AI would just give me fake sections.
For completeness I will state the full result and scratch proof below:
Lemma: Let $C$ be a regular proper integral scheme of dimension 1 over a field $k$. Then there exists a bijection between the set of valuation rings of $k(C)/k$ and the set of points of $C$, in particular, the trivial valuation ring $k(C)$ corresponds to the generic point.
Proof: Given a point $v\in C$, it is easy to see that $\mathcal{O}_{C,v}$ is a valuation ring of $k(C)/k$. Reversely given a valuation ring $\mathcal{O}$ of $k(C)/k$, by the Valuative criterion for properness, there exists a unique morphism $\DeclareMathOperator{\Spec}{Spec} f:\Spec \mathcal{O}\to C$ over $k$ s.t. the generic point embedding map $\Spec k(C)\to C$ factors through. Let $v^\prime$ be the closed point of $\Spec \mathcal{O}$ and $v:=f(v^\prime)$. We can see that $f$ factors through $\Spec \mathcal{O}_{C,v}\to C$ and eventually we have $\mathcal{O}_{C,v}\hookrightarrow \mathcal{O}\hookrightarrow K$. So $\mathcal{O}_{C,v}=\mathcal{O}$ by the domination property of valuation ring. The process is clearly mutual.
