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This fact without a proof is mentioned on the page 147 of Bogomolov's book "Algebraic curves and one-dimensional fields", but the statement is not evident to me. Can anyone explain to me why is it true? Thank you very much.

flag	asked Nov 17 at 19:02 D.V. 21●1●1●1

rita · Answer 1 · 2012-11-17 19:15:34Z UTC

If $X$ is a smooth projective curve and $P\in X$ is any point, then $X\setminus P$ is affine. Indeed, by Riemann-Roch for $m$ large enough the linear system $|mP|$ embeds $X$ in some projective space $\mathbb P^n$, so there is a hyperplane $H_0$ of $\mathbb P^n$ that intersects $X$ only at $P$. So, if $P\ne Q$ then $X_1=X\setminus{P}$ and $X_2=X\setminus{Q}$ are affine curves such that $X=X_1\cup X_2$.

algori · Answer 2 · 2012-11-17 19:15:21Z UTC

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One way to see this would be as follows: if $X\subset\mathbb{P}^n$ is a curve one can find hyperplanes $H_1,H_2$ such that $X\cap H_1\cap H_2=\varnothing$. Then $X$ will be the union of two affine curves $X\setminus H_1$ and $X\setminus H_2$.

answered Nov 17 at 19:15

algori
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Very nice ! Over a finite field, one can use hypersurfaces instead of hyperplanes. – Qing Liu Nov 17 at 22:09

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Qing Liu -- yes, indeed. By the way, slightly generalizing the argument one sees that a projective variety $X$ can be covered by $\dim X+1$ affine sets. – algori Nov 17 at 23:52

Is it true that every projective curve is a union of two affine curves?

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