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Let $X$ be an abstract curve in the following sense: $X$ is a scheme, proper over $k$ which is noetherian, integral, dimension 1, and normal. The important thing to point out is that I am not assuming any fixed embedding into affine or projective space.

Question: How do we conclude that in fact $H^{0}(X,\mathcal{O}_{X}^{*})=k$ as you would expect from the analogous statement for Riemann surfaces? Go ahead and assume $k$ is algebraically closed if it helps you.

I'll say some more about it. From the definitions it follows that for each $p$ in $X$ we have that $\mathcal{O}_{X,p}$ is a D.V.R., which is the valuation ring of $k(X) \simeq \mathcal{O}_{X,\eta}$ where $\eta$ is of course the generic point of $X$.

Using the general machinery one can prove that we can define a divisor of a nonzero element $f$ of $k(X)$ to be

$$Div(f) = \sum_{p\in{X}}v_{p}(f)[p]$$

where $v_{p}$ is the obvious valuation of $k(X)$ at $p$. (The only thing to check would be that this sum is actually finite). Thus to rephrase the question:

Rephrasing Why does $Div(f) = 0$ imply that $f$ is in $k$?

This should probably be simple enough given how nice everything is, but I'm stuck.

note: I'm already aware of any kind of answer which includes something like [GAGA] in it.

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Are you sure your statement is true? The affine line with a point removed has a lot of nowhere-vanishing regular functions. – Piotr Pstrągowski May 5 '13 at 23:21
you're right, maybe I should add proper over k? – martin May 5 '13 at 23:25
Hartshorne, Theorem 3.4 (a), page 18. Normal and dimension one are irrelevant. Be sure to remember that proper over $k $ =complete. – Georges Elencwajg May 6 '13 at 8:30
@GeorgesElencwajg I'm aware of the result if we're talking about projective varieties. In fact after spending some time looking today, the answer is quite clear if we assume $k$ is algebraically closed by Hartshorne II.4 excercise 4.5, and with a little more work the general case follows from proposition II.6.7. – martin May 6 '13 at 21:10

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