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?

  • 2
    $\begingroup$ Yes. And the original curve $C$ needn't even be complete! $\endgroup$ – Georges Elencwajg Dec 30 '15 at 18:46
  • $\begingroup$ @GeorgesElencwajg That's what I thought! Can you explain the reasoning? $\endgroup$ – oxeimon Dec 30 '15 at 18:46
  • $\begingroup$ @GeorgesElencwajg The assumption that it's not complete is not a strengthening right? (Since you can always take the completion?) $\endgroup$ – oxeimon Dec 30 '15 at 18:53
  • 4
    $\begingroup$ Obviously you are not asking for an explanation like you're 5, since you're happy to assume the Riemann-Roch theorem... $\endgroup$ – Qiaochu Yuan Dec 30 '15 at 19:09
  • 1
    $\begingroup$ @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. $\endgroup$ – user251240 Apr 16 at 9:42

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 $C\setminus \{p_1,\dots, p_r\}$.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.