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This is the exercise 19.5B of Ravi Vakil's Foundations of Algebraic Geometry

Verify that a curve of genus $C$ of genus at least $1$ admits a degreee $2$ cover of $\mathbb{P}_k^1$ if and only if it admits a degree $2$ invertible sheaf $\mathscr{L}$ with $h^0(C,\mathscr{L})=2$.

Here a curve is assumed to be geometrically regular, geometrically integral and projective.

The hint is to show that if $C$ is to show that if $h^0(C,\mathscr{L})\ge 2$, then $h^0(C,\mathscr{L})=2$, and that $\mathscr{L}$ is base point free.

I know how to show $\mathscr{L}$ is base point free if $h^0(C,\mathscr{L})\ge 2$: Suppose $s_1,s_2\ne 0$ are two sections of $\mathscr{L}$ with common zero at a point $p$. If $p$ is a degree $2$ point, $s_1/s_2$ has no zeros and no poles and must be a constant, thus $s_1,s_2$ are linearly dependent.

If $p$ is of degree $1$, write $$div(s_1)=p+q_1, div(s_2)=p+q_2$$ then I have $$\mathscr{O}_C(q_1)=\mathscr{L} \otimes \mathscr{O}_C(-p)=\mathscr{O}_C(q_2)$$ therefore $q_1=q_2$ and $s_1,s_2$ are linearly dependent.

Where the last line I have used the following result from the book:

19.4.2 Corollary. If $C$ is a curve not isomorphic to $\mathbb{P}_k^1$, and $q_1, q_2$ are distinct points on $C$, then $\mathscr{O}_C(q_1)\cong\mathscr{O}_C(q_2)$.

What I don't know is how to argue that $h^0(C,\mathscr{L})\le2$, any help or hints are appreciated, thank you so much!

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  • $\begingroup$ Let $p$ be any point. Can you say what $h^0(\mathcal{L}(-p))$ is? $\endgroup$ – Mohan Feb 3 '17 at 15:09
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    $\begingroup$ I have the exact sequence $0 \to \mathscr{L}(-p) \to \mathscr{L} \to \mathscr{O}|_p \otimes \mathscr{L} \to 0$, and the long exact sequence tells me that $h^0(C,\mathscr{L}(-p)) \ge h^0(C,\mathscr{L}(-p)) -1$, so if $h^0(C,\mathscr{L})\ge 3$, we must have $h^0(C,\mathscr{L}(-p))\ge 2$, which is impossible unless $C\cong \mathbb{P}^1_k$, am I right? Thank you so much $\endgroup$ – chan kifung Feb 4 '17 at 3:17
  • $\begingroup$ Good, there is a typo in your comment above, but conclusion is right. $\endgroup$ – Mohan Feb 4 '17 at 14:40

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