An algebraic curve $C$ as a cover of $\mathbb{P}^1$ over $\bar{\mathbb{Q}}$

Question

Could someone explain what it means for an algebraic curve $C$ to be a cover of $\mathbb{P}^1$ over $\bar{\mathbb{Q}}$, ramified over $n$ points?

Answer 1

I don't know what language you're using (classical or modern). In the modern sense, it means that you have a curve $C$ over $\overline{\mathbb{Q}}$ (perhaps it was originally over some subfield of $\overline{\mathbb{Q}}$, but we just base change up to $\overline{\mathbb{Q}}$) and a surjective map $\varphi:C\to\mathbb{P}^1$.

What it means for a point $x\in C$ to be ramified relative to this morphism is that the sheaf $\Omega^1_{C/\mathbb{P}^1}$ has non-zero stalk there. Or said differently (although equivalently, since I'm assuming $C/\overline{\mathbb{Q}}$ is smooth) that if $y=\varphi(x)$, and $\mathcal{O}_{\mathbb{P}^1,y}\to\mathcal{O}_{C,x}$ is the natural map, that a uniformizer $\pi$ of $\mathcal{O}_{\mathbb{P}^1,y}$ doesn't map to a uniformizer $\mathcal{O}_{C,X}$.

There is an easy way to understand the images of the points of ramification. If $[K(C):k(t)]=n$ (where $K(C)$ is the function field of $C$, and $k(t)$ the function field of $\mathbb{P}^1$) a point $y\in\mathbb{P}^1$ is the image of a point of ramification if the fiber contains less than $n$ points.

Intuitively, ramification means that $C\to\mathbb{P}^1$ is generally $n$-to-$1$, with $n$ "sheets", and the points of ramification are where these sheets come together. This can be made entirely formal if you further base change to $\mathbb{C}$, and look at the analytification of the curves. Then this becomes a finite map of Riemann surfaces (real surfaces, if that makes you happier), which is generically a covering map, except at the points of ramification!

Let me know if I can further clarify!

(PS: Here's a shameless plug for a post on my blog talking about unramified morphisms. Check it out :) )