Quick question: a 2:1 map onto the projective line

Question

Given a line $L$ in $\mathbb{P^2}$. How do we see that a surjective map $\mathcal{O}_\mathbb{P^2}^{\oplus2}\rightarrow j_{*}{\mathcal{O}_L(2)}$ ($j$ is the inclusion of $L$ to $\mathbb{P^2}$) corresponds to a $2$ to $1$ map $L\rightarrow \mathbb{P^1}$ ramified at two points? Thank you very much.

Answer 1

If $\mathcal{O}_{\mathbb{P}^1}\to j_*\mathcal{O}_L(2)$ is surjective, then $L$ embeds as a degree 2 curve in $\mathbb{P}^2$ (this means that $j$ is a closed immersion). Take a point outside of the curve and project to $\mathbb{P}^1$. This will give a 2-1 map, and the ramification points are the points where the map fails to be injective, of which there are two. This is easy to check. Say your embedded curve is $C:y_0^2+y_1^2+y_2^2=0$ and you are projecting from $(0:0:1)$. Take a point $(x_0:x_1)$ in $\mathbb{P}^1$ and note that the points above it in $P^2$ lie on $y_0x_1-y_1x_0=0$. This meets the curve $C$ generically in two points and to say that it meets it tangentially (i.e., to have a ramification) is a degree 2 condition on $x_0$ and $x_1$.