# Global section of a closed point as a divisor for a degree 2 curve

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?

• There are no functions with a simple pole at a single point in your case. – Mohan Jun 19 at 16:34
• @Mohan Why not? Can you explain a little more? – Li Li Jun 19 at 21:13
• 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. – Mohan Jun 19 at 21:29