Variety isomorphic to Riemann sphere

Question

I have read that the complex solutions to $y^2=p(x)$ where $p(x)$ has degree 1 or 2 (and distinct roots) is topologically and complex analytically isomorphic to the Riemann sphere $\mathbb{P}^1$. How would one prove such statements?

Answer 1

There are several abstract ways to show this (Riemann-Roch, Riemann-Hurewitz, Bezout), but I think the direct geometrical construction of a map is very instructive.

Let $P = (x,y)$ be any solution of the equation in $\mathbb P^2$ (you have to allow for solutions at infinity). Now let $[a:b]$ be any point of $\mathbb P^1$. This defines a line $P + aX + bY$ in $\mathbb C^2$. Find the other point of intersection of this line with the solution set (one point is $P$ of course). This is easy: simply substitute either $X$ or $Y$ to be left with a quadratic equation in the other, one of whose roots you know (note that there is exactly one such point because the resulting equation is quadratic, but you will have to consider points at infinity). This will be the image of $[a:b]$. This essentially is the desired isomorphism.

There are several details you'll have to fill in, to show that this is a biholomorphic/regular map, take care of the point at infinity, etc.