Ramification at infinity for a cyclic cover of $\Bbb P^1$

Solution 1:

If $\sum r_i = 3d$ one can construct the cyclic covering of $\mathbb{P}^1$ as $$ C = \mathrm{Spec}_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(-d) \oplus \mathcal{O}(-2d)), $$ where the algebra struture is induced by the morphism $$ \mathcal{O}(-3d) \to \mathcal{O} $$ that has zero of multiplicity $r_i$ at point $\alpha_i$. The induced field extension is exactly the same and by construction there is no ramification at infinity.

Solution 2:

Here is a topological reason why $\infty$ is not ramified.

By removing the ramification points over $\alpha_1,\ldots,\alpha_s$, the curve defines a 3-sheeted covering space of $U=\mathbb C-\{\alpha_1,\ldots,\alpha_s\}$. We pick a base point $\alpha_0\in U$, then any loop $l\subseteq U$ based at $\alpha_0$ defines a permutation action of the three points on the fiber. In other words, we have a representation

$$\rho:\pi_1(U,\alpha_0)\to S_3.$$

We can choose a loop $l_{\infty}$ large enough to enclose all the $\alpha_i$'s and counter-clockwise oriented.

Claim: The infinity point $\infty$ is not ramified if and only if $\rho(l_{\infty})$ is identity element.

The problem reduces to prove the latter statement. We perturb the equation a little to separate the branching points whenever $\alpha_i$ is not simple. So we can assume the equation is

$$y^3=\Pi_{i=1}^{3n}(x-\beta_i),$$

where $n$ is an integer and all $\beta_i$'s are distinct and contained in $l_{\infty}$. Of course, this doesn't change the topology far away from the branchings.

Now, we can choose $l_i$ is a loop based at $\alpha_0$ and only enclose $\beta_i$ and counter-clockwise oriented. Then up to relabelling the index, $l_{\infty}$ is homotopic to the concatenation $l_{3n}*l_{3n-1}*\cdots*l_1$. See picture below.

Finally, each $\rho(l_i)$ is the same generator $\sigma:1\mapsto 2\mapsto 3\mapsto 1$ of the cyclic subgroup permutating the three points (this is because topologically along $l_i$, the curve looks like a small circle around zero on $y^3=x$). Therefore $$\rho(l_{\infty})=\rho(l_{3n})\cdot\rho(l_{3n-1})\cdots\rho(l_1)=\sigma^{3n}=e$$ is the identity element.

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