The context is that I'm trying to understand the implication that if $C$ is a smooth curve of genus $0$ then $C \cong \mathbb{P}^1$.
Using Riemann-Roch one can see that there exists a non-constant function $f \in k(C)$ (the function field of $C$) with exactly one simple pole at $P \in C$ and no poles anywhere else. If we can show $C \cong \mathbb{P}^1$ we need to show that $f$ induces a map of degree $1$ then we are done.
In for example Hartshorne II 6.10.1 this is done by saying that if $\varphi: C \to \mathbb{P}^1$ is the by $f$ induced morphism then $\varphi^*(0) = P$ and thus by II 6.9. $\deg \varphi = \deg \varphi \deg (0) = \deg \varphi^*(0) = \deg P = 1$.
I don't understand why $\varphi^*(0) = P$?
I have also seen that the degree of $\varphi$ is the degree of the divisors of poles of $f$. This I also have trouble seeing.
My thought process thus far is that we by definition (Hartshorne page 137) we have $$\varphi^* (0) = \sum_{Q \in \varphi^{-1}(0)} = v_Q(t) \cdot (Q)$$ where in this case since $\varphi$ is injective so $\varphi^{-1}(0)$ consists of a single point $Q$. Thus $\deg \varphi^*(0) = \deg v_Q(t) (Q) = v_Q(t)$ but I have no idea how to evaluate $v_Q(t)$. I know it can be seen as the ramification index but this does not seem to be necessary.
Thanks a lot in advance!
My question is answered in detail in these notes, Corollary 19.23.