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.
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2 Answers
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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 $X=X_1\cup X_2$.
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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 $X\setminus H_2$.
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2$\begingroup$ Very nice ! Over a finite field, one can use hypersurfaces instead of hyperplanes. $\endgroup$– Qing LiuNov 17, 2012 at 22:09
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1$\begingroup$ 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. $\endgroup$– algoriNov 17, 2012 at 23:52