In Hartshorne's proof of proposition II 6.7, there is a point that I don't really understand, and I couldn't find any clarification online. My question may be trivial, but getting an answer to it would be a great enlightenment to me. Here is the statement.
Let $X$ be a nonsingular curve over [an algebraically closed field] $k$ with function field $K$. Then the following are equivalent:
(i) $X$ is projective;
(ii) $X$ is complete;
(iii) $X$ is isomorphic to $t(C_K)$, where $C_K$ is the abstract nonsingular curve of (I,§6), and $t$ is the functor from varietes to schemes of (2.6).
Recall also that Hartshorne just defined curves over $k$ to be integral separated schemes of finite type over $k$, of dimension one. Such a curve is said to be nonsingular if all its local rings are regular.
The point I don't understand lies in the proof of $(ii) \rightarrow (iii)$. It is written that the closed points of $X$ are in $1-1$ correspondence with the discrete valuation rings of $K/k$.
While I agree that all local rings at closed points are discrete valuation rings of $K/k$, why would this only concern closed points? Given that all local rings are assumed to be regular, I don't see any reason why only those at closed points are discrete valuation rings of $K/k$.
Does this have something to do with the fact that closed points are exactly those with residue field isomorphic to $k$?
I thank you in advance for your clarifications.