Let $C$ be a projective curve over $k$ (geometrically integral, nonsingular).

I am confused about the following argument in Vakil's notes on algebraic geometry:

(1) In (20.6.2), he writes:

Fix a curve $C$ of genus $g = 2$. Then $\omega_X$ is degree $2g-2 = 2$, and has 2 sections (Exercise 20.2.A). I claim that $\omega_X$ is base-point-free. We may assume $k$ is algebraically closed [...]. If $p$ is a base point of $\omega_X$, then $\omega_X(-p)$ is a degree 1 invertible sheaf with two sections, which Proposition 20.4.1 shows is impossible.

I am not completely sure what he means by $\omega_X$ having two sections, but I assume he just means $h^0(X, \omega_X) = 2$. Now indeed $\deg(\mathscr{L}(-p)) = \deg(\mathscr{L} \otimes \mathscr{O}_X(-p)) = \deg(\mathscr{L}) - 1$ since $p$ has degree 1. However, since $p$ is a base point of $\omega_X$, we have $$ h^0(X, \omega_X) - h^0(X, \omega_X(-p)) = 1 $$ which implies that $\omega_X(-p)$ has 1 section, not two -- unless I misunderstand the meaning of this phrase.

  • 2
    $\begingroup$ what you said makes sense to me: two "linearly independent" sections yeah $\endgroup$ – uncookedfalcon Jan 10 '13 at 9:57

Note that $p$ is a base point if and only of the inclusion $H^0(X, \omega_X(-p))\subseteq H^0(X, \omega_X)$ is an equality. This is essentially the definition of base point.

So if $p$ is a base point, then $h^0(X, \omega_X(-p))=2$.


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