# Ramification of covering maps of curves and the étale fundamental group

I'm interested in finite galois covers $\varphi: Y \rightarrow X$ between smooth proper curves over an algebraically closed field of characteristic zero, which are étale outside a prescribed finite set $D$ of points of $X$.

More precisely, I'd like to understand how the étale fundamental group $\widehat\pi_1(U_{X})$ of $U_{X}:=X \setminus D$ determines the ramification indices $e_{y}$ of points $y$ of $Y$.

As a starting point I have the following more specific question:

Suppose that $X=\mathbb{P}^{1}_{k}$, where $k$ is an algebraically closed field of characteristic zero, and that $D=\{0,1,\infty\}$. The étale fundamental group of $U_{X}=\mathbb{P}^{1}_{k}\setminus \{0,1,\infty\}$ is then the profinite group generated by three elements $\sigma_1,\sigma_2,\sigma_3$, subject to the only relation $$\sigma_1 \cdot \sigma_2 \cdot \sigma_3=1$$ Finite quotients of this group classify finite galois coverings $\varphi: Y \rightarrow \mathbb{P}^{1}_{k}$, which are étale outside of $\{0,1,\infty\}$.

Is it possible to choose the covering $\varphi:Y \rightarrow \mathbb{P}^{1}_{k}$ such that every point $y \in \varphi^{-1}(\{0,1,\infty\})$ has ramification index $e$, where $e$ is a fixed positive integer, $e \geq 2$? If so, which are the properties of the finite quotient of $\widehat{\pi}_1(\mathbb{P}^{1}_{k}\setminus\{0,1,\infty\})$, associated to $\varphi$, which reflect the order of the ramification indices?

I think that at least my first question might be related to (or possibly even be answered by) the Riemann existence theorem, but I was unable to figure out how.

• I think by Riemann-Hurwitz you can numerically rule out some cases of being able to find a covering $X\to \mathbb{P}^1$ so that every point has the same ramification index. I should probably wait until morning to double check this, but it seems you'd get $2g(X)-2=-2d+3(e-1)$ where $d$ is the degree of the covering. So maybe for even $e$ no such cover can exist?
– Matt
Commented Oct 15, 2012 at 5:29
• Dear @Matt, how do you get the $3$ in your formula using Riemann-Hurwitz? Commented Oct 15, 2012 at 8:52
• Ah. You're right. I was making the simplifying assumption that there was only one "point" over each of $0, 1$ and $\infty$. You could of course have 2 over each, and that fixes the problem because there would be a 6 in front.
– Matt
Commented Oct 15, 2012 at 23:53
• @Matt: yes, that's what I was thinking. Thanks for your response, nonetheless. I wish there were more of them. Commented Oct 16, 2012 at 8:06

The relevant reference seems to be Exposé XIII, Corollaire 2.12. of SGAI. For the specific example $\Bbb{P}^1_k \setminus \{0,1,\infty\}$, with $k$ algebraically closed of characteristic zero, it says that giving a finite Galois cover
$$\varphi: Y \rightarrow \Bbb{P}^1_k$$
with (finite) Galois group $G$, which is étale outside of $\{0,1,\infty\}$, is equivalent to giving a surjective group homomorphism
$$\phi: \widehat{\pi}_1(\Bbb{P}^1_k \setminus \{0,1,\infty\}) \rightarrow G$$ Also it is shown in loc.cit. that $\widehat{\pi}_1(\Bbb{P}^1_k \setminus \{0,1,\infty\})$ is the pro-finite group generated by elements $\sigma_1,\sigma_2,\sigma_3$, subject only to the single relation $$\sigma_1 \cdot \sigma_2 \cdot \sigma_3=1$$ Moreover, if we set $x_1=0$, $x_2=1$, $x_3=\infty$ (the actual ordering of the three points $0,1,\infty$ is irrelevant) then the ramification index of every point in $\varphi^{-1}(x_i)$ equals the order of $\phi(\sigma_i)$ in $G$.
So putting everything in context of the question, if I want a finite Galois cover $$\varphi: Y \rightarrow \Bbb{P}^1_k$$ which is étale outside of $D=\{0,1,\infty\}$ and has ramification index equal to $e$ at every point of $D$, I just have to find a finite group $G$, which can be generated by three elements $\overline{\sigma}_1,\overline{\sigma}_2,\overline{\sigma}_3$, each of order $e$ such that $$\overline{\sigma}_1 \cdot \overline{\sigma}_2 \cdot \overline{\sigma}_3 = 1$$ It is clear that for a prime number $e \geq 3$, $G=\Bbb{Z}/e\Bbb{Z}$ does the job. This is in fact sufficient for my purposes.