Projective curve branch point

Question

Could someone explain why the projective curve $y^2=x$ has a branch point at infinity like the one at 0? I don't understand why this is true, and supposedly it helps probe that the curve is topologically a circle, which I don't understand how it follows from having a branch point at infinity.

Answer 1

Assuming you want $x,y \in \mathbb{R}$, let $[u,v,w]\in\mathbb{R}\mathbb{P}^2$. The affine point $(x,y)\in\mathbb{R}^2$ (away from infinity) corresponds to the point $[u/w, v/w,1]$ for $w\neq 0$. Then $y^2=x \Leftrightarrow v^2=uw$. Clearly this curve has a branch point at $w=0$.

(Another easier way to do the computation is, formally, let $x=1/t, y=y/t$ then letting $t\to 0$ corresponds to letting $x\to \infty$. This agrees with the above if you set $u=1$ which is allowed for $u\neq 0$ in projective coordinates.)

The compactified curve is topologically a circle since the graph is "on the positive branch" above the x-axis and switches branches (goes below the x-axis) at both $0$ and $\infty$.

Of course if $x,y\in \mathbb{C}$, everything is the same here except topologically you have something different than a circle, namely you have a Riemann surface which topologically looks like $S^3$.