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Let $X$ be a smooth projective curve. I'd like to prove $X$ has genus $0$ if and only if $X = \mathbb P^1$. The proof I have goes as follows:

Let $p \in X$. Then $p \in \mathrm{Div}(X)$ with $\mathrm{deg}(p)=1$. By Riemann Roch, $l(p) = 1$. So $\mathcal{L}(p) \supsetneq \mathcal{L}(0) = k$, i.e. $\exists f \in \mathcal{L}(p) \backslash k$. By definition of $\mathcal{L}(p)$, $\mathrm{div}(f) + (p) \geq 0$. So $\mathrm{div}(f) + (p) = (q)$, for some $q \in X$, i.e. $\mathrm{div}(f) = (q) - (p)$, and $q \neq p$ as $f$ is not a constant function. So $\alpha = (f:1) : X --> \mathbb P^1$ is a non-constant rational map of degree 1.

My problem is: why is $\mathrm{deg}(\alpha) = 1$?

Thanks!

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    @Andrea Are you using the Finiteness Theorem? Why does $\mathrm{deg}(\mathrm{div}(f)) = 0$ imply $\alpha$ has degree 1?2012-05-19

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By construction, your $f$ is a rational function, with exactly one pole of order 1. By the degree theorem, $deg(div(f)) = 0$, i.e. number of zeroes = number of poles, so $f$ attains every value in $\mathbb{P}^1 = k \cup \{\infty\}$, $k$ your ground field, with multiplicity one!

Well, it certainly hits 0 only once, and to see it hits $\lambda$ only once, apply the degree theorem to $f - \lambda$; still just one pole!

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    yuuuuuuuuuuuuup :)2012-05-19