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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? –  Martin May 19 '12 at 0:20

1 Answer 1

up vote 2 down vote accepted

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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What's the degree theorem? –  Martin May 19 '12 at 0:50
    
Just that $deg(div(f)) = 0$ –  uncookedfalcon May 19 '12 at 0:53
    
I'm really having a dim moment. $div(f) = P - Q$, so $f$ has exactly one 0 (at $Q$) and exactly one pole (at $P$). So $f$ only hits $0$ once. Now $deg(div(f-\lambda)) = 0$, but why does this mean $f$ hits $\lambda$ exactly once? –  Martin May 19 '12 at 1:05
    
Oh, I think I understand. $f$ just has one pole, so $f - \lambda$ has just one pole (can see this by observing $\nu_R(f - \lambda) \geq min(\nu_R (f), \nu_R(\lambda))$ . So $f- \lambda$ has just one zero, i.e $\lambda$ is hit only once. –  Martin May 19 '12 at 1:11
    
yuuuuuuuuuuuuup :) –  uncookedfalcon May 19 '12 at 1:16

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