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This is part of Ex1.7 in Chapter IV of Hartshorne's Algebraic Geometry. Let $X$ be a curve of degree $2$ and genus $2$ over $\mathbb{P}^1$. Show that the canonical divisor defines a complete linear system of degree $2$ and dimension $1$ without base points. I use RR Thm $h^0(D)-h^0(K-D)=d+1-g$. Let $D=K$. Then I can get the degree and dimension easily. But for the base point freeness, I let $D=K-P$ for any closed point $P$ and use the criterion in Proposition3.1 in the same chapter(saying that a complete linear system $|D|$ is base point free iff for every $P\in X$, $\dim|D-P|=\dim|D|-1$). In other words, I get $h^0(K-P)=h^0(P)$ and want to show $h^0(P)=1$. However, I think it should be $2$, because $H^0(X,P)=\{f\in k(X)|div(f)+P\geq 0\}$ and $f$ can be spanned by either constant or functions having one pole at $P$(it cannot have higher degree vanishment at $P$ since $deg(div(f))=0$ and if it did, $div(f)+P$ would not be effective). Can anyone tell me what my mistake is?

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    $\begingroup$ There are no functions with a simple pole at a single point in your case. $\endgroup$ – Mohan Jun 19 at 16:34
  • $\begingroup$ @Mohan Why not? Can you explain a little more? $\endgroup$ – Li Li Jun 19 at 21:13
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    $\begingroup$ If such a function existed, it gives a morphism $X\to\mathbb{P}^1$ of degree one, which says $X$ is isomorphic to the projective line and thus genus zero. $\endgroup$ – Mohan Jun 19 at 21:29

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