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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.

Suzet
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  • The local ring at a generic point must be $K$, right? – asdq May 20 '18 at 08:59
  • Yes, I agree with this. – Suzet May 20 '18 at 09:03
  • This shows that not every local ring is a dvr in your situation. In fact, I think you can prove the claim by using the fact that a ring is a dvr if and only if it is a regular noetherian local ring of dimension 1. – asdq May 20 '18 at 09:09
  • A ring is a DVR iff it's a local PID that is not a field. The stalk of the structure sheaf at the generic point of any subvariety (of a curve or any higher dimension variety) is indeed local, but it will be a PID if and only if the subvariety has codimension 1, i.e., it's a divisor. – user347489 May 20 '18 at 09:25
  • The converse requires completeness, you probably want to use the valuative criteria here. What's happening is that if your curve is not complete then there will be discrete valuations that don't correspond to any divisor on the curve. For example, you can see that the valuation corresponding to the point at infinity of $\mathbb{P}^1$ doesn't correspond to anything when you consider the valutions of $\mathbb{A}^1$. – user347489 May 20 '18 at 09:25
  • Thanks both of you for your replies. They help make the notions clearer in my mind. I now remember about the general fact that the dimension of a stalk at a point is the codimension of its closure. Hence, in our situation, a local ring is a dvr if and only if the closure of the point has codimension 1. This is satisfied for closed points. I need to see that only closed point have closure of codimension 1. That is to say, every non closed point must be dense. In other words, that there is only one non closed point, which is the generic one. Is this statement true? – Suzet May 20 '18 at 10:08
  • I found a solution to this here, so this settles down my interrogations. https://math.stackexchange.com/questions/218928/why-are-prime-divisors-on-curves-just-closed-points

    Again, thank you very much.

     – Suzet May 20 '18 at 10:31

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