Let $k$ be an algebraically closed field. Throughout, by curve I mean integral, nonsingular, dimension $1$ scheme proper over $k$. In particular I'm assuming we're dealing with complete curves in the scheme theoretic sense. For a curve $X$, denote by $k(X)$ the function field which is a finitely-generated transcendence degree $1$ extension of $k$.
For a pair of finitely-generated transcendence degree $1$ extensions $K$ and $L$ of $k$, we must have that the extension $L / K$ is finite. I would like to use this to construct a morphism of curves. Assume for the time being that every such extension corresponds to a curve so I'm only concerned with constructing the actual morphism. I'm wondering if someone can point me in the right direction since I am fairly stuck at the moment. Here's what I've done so far, which is mostly what I've managed to piece together from Hartshorne I.6 and II.4,6:
Say $K(Y) \subseteq K(X)$ is a finite extension of fields of functions of curves $X$ and $Y$ respectively. Since these are proper. For a point $p \in X$ the local ring $\mathcal{O}_{X, p}$ is a DVR for a discrete valuation on $K(X)$ and the intersection $\mathcal{O}_{X, p} \cap K(Y)$ is a DVR for a discrete valuation on $K(X)$. Using the valuative criterion for properness, I have determined that $\mathcal{O}_{X, p} \cap K(Y)$ must dominate precisely one local ring for a closed point on $Y$, say $\mathcal{O}_{Y, q}$. Since valuation rings are characterised by the property of being maximal with respect to the relation of domination, we have that $$ \mathcal{O}_{X, p} \cap K(Y) = \mathcal{O}_{Y, q} $$ I would like my morphism $f: X \longrightarrow Y$ to be defined by the assignment $f(p) = q$. Is this the correct morphism?
I'm fairly confident this is the morphism I want to construct, so I would like to first show it is continuous. This is where I hit my first major problem. So to do this let $V = \text{spec } B$ be an affine open set in $Y$. Since $B$ must have Krull dimension $1$ and is an integral domain, it must be that every non-zero prime is maximal, and so every localization is a DVR since it corresponds to a closed point in the curve $Y$. So $B$ is a Dedekind domain. I'd like to show that the following set is open, $$ f^{-1}(\text{spec } B) = \{ p \in X : \mathcal{O}_{X, p} \cap K(Y) = B_{\mathfrak{q}} \text{ for some prime ideal } \mathfrak{q} \subseteq B \}. $$ From what I've found online, this should be given by the spectrum of the integral closure of $B$ in $K(X)$, but I haven't been able to show this. Hartshorne claims it is "clear". But so far I haven't even been able to show it is open. I suspect that it will have something to do with the result from commutative algebra that says that since $B$ is an integrally closed noetherian domain (indeed it is a Dedekind domain) that it is equal to the intersection of the localisations at all prime ideals of height $1$ taken in $K(Y)$. Is anyone able to point me in the right direction? Ideally a suggestion on how to show that the set is open, or even better how to show it's actually the spectrum of the integral closure of $B$ in $K(X)$.