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Let $C$ be a proper smooth geometrically connected curve over a field $K$, and let $P\in C(K)$ be a point. Must $C - P$ be affine?

EDIT: By Riemann-Roch, you can definitely find functions $f_1,\ldots,f_r : C-P\longrightarrow\mathbb{A}^n_K$, but how do you guarantee that for some $n$, you can find enough such $f_i$'s such that this gives you an embedding?

EDIT: Is the same true with $C$ not smooth?

oxeimon
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    Yes. And the original curve $C$ needn't even be complete! – Georges Elencwajg Dec 30 '15 at 18:46
  • @GeorgesElencwajg That's what I thought! Can you explain the reasoning? – oxeimon Dec 30 '15 at 18:46
  • @GeorgesElencwajg The assumption that it's not complete is not a strengthening right? (Since you can always take the completion?) – oxeimon Dec 30 '15 at 18:53
  • But yes it is a strengthening! If you take the completion $\bar C$ of $C$ you might have to add 666 points and I'm telling you that $\bar C$ minus 667 points is affine! This cannot be deduced from your statement. – Georges Elencwajg Dec 30 '15 at 18:57
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    Obviously you are not asking for an explanation like you're 5, since you're happy to assume the Riemann-Roch theorem... – Qiaochu Yuan Dec 30 '15 at 19:09
  • I have now also proved the result for singular curves in an edit. – Georges Elencwajg Dec 30 '15 at 20:24
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    @QiaochuYuan Balamurali Ambati was allegedly doing calculus when he was four years old. I am not sure if he was doing classical algebraic geometry by age of 5 but it is not entirely unreasonable. So your usage of the word "obviously" seems unwarranted. –  Apr 16 '19 at 09:42

1 Answers1

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Theorem
If a nonzero finite number of points $p_1,\dots, p_r $ are deleted from $C$ the resulting curve will be affine.

Indeed consider the divisor $D=p_1+\dots+ p_r $ on $C$.
Since it has positive degree some positive multiple $nD$ of it will be very ample.
Thus we get an embedding of $j:C\to \mathbb P^N$ (for some huge $N$) and a hyperplane section divisor $\Delta =H\cap j(C)$ on $j(C)$ such that $j^*\Delta=nD$.
But then $C\setminus \{p_1,\dots, p_r\}$ is isomorphic to $j(C)\cap (\mathbb P^N\setminus H)\cong j(C)\cap \mathbb A^N$ (the complement of a hyperplane in projective space is affine space) and since this last variety $j(C)\cap \mathbb A^N$ is clearly affine, so is $C\setminus \{p_1,\dots, p_r\}$.

Edit
The theorem is valid even if $C$ is singular.
To see that, consider the finite normalization morphism $n:\tilde C\to C$ and delete the inverse image of $\{p_1,\dots, p_r\}$, obtaining the smooth curve $C'=\tilde C\setminus n^{-1}(\{p_1,\dots, p_r\})$ which is affine by the result already proved for smooth curves.
Now consider the restricted finite morphism $n':C'\to C\setminus \{p_1,\dots, p_r\}$.
Since $C'$ is affine and the finite morphism $n'$ is surjective the curve $C\setminus \{p_1,\dots, p_r\}$ will also be affine by Chevalley's Theorem (EGA $_{II}$, Théorème (6.7.1), page 136), and we are done.

Georges Elencwajg
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