Isomorphism between $Z(x_1^2-x_0x_2) \subseteq \mathbb{P}^2$ and $\mathbb{P}^1$

Question

I want to show that if I have morphism $\varphi_0 : C_0 \rightarrow \mathbb{P}^1, \varphi_0([a_0:a_1:a_2])=[a_0:a_1]$ then I can extend it to be an isomorphism $\varphi : C:=Z(x_1^2-x_0x_2) \rightarrow \mathbb{P}^1, C \subseteq \mathbb{P}^2$ with $C_0:=\{ [a_0:a_1:a_2] \in C | a_0 \neq 0 \}$. My idea is, since in $C$ we have $\frac{x_1}{x_0}=\frac{x_2}{x_1}$, we define for $a_0=0, \varphi([a_0:a_1:a_2])=[a_1:a_2]$, that is, $\varphi([0:0:1])=[0:1]$ since $a_0=0$ implies $a_1=0$. Is this the correct extension? If so, how can I show it is an isomorphism? Thank you.

Answer 1

Given $[a_1:a_2] \in \mathbb P^1$, plug into the equation of $C$: you get $a_1^2 - x_0a_2$, so the unique point mapping to $[a_1:a_2]$ is $[a_1^2/a_2:a_1:a_2] = [a_1^2:a_1a_2:a_2^2]$. Thus $\varphi$ is a bijective morphism between smooth curves which is enough to conclude that it is an isomorphism, or you can explicitly check that $[a_1:a_2] \mapsto [a_1^2:a_1a_2:a_2^2]$ is the inverse map.