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!