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k is an algebraically closed field, X is a smooth, connected, projective curve over k. f: X-->P^1 is a finite morphism. Let t be a parameter of P^1, suppose f is etale outside t=0 and t=\infty, and tamely ramified over these two points. Prove that f is a cyclic cover, i.e., K(X)=k(t)[h]/(h^n-ut), u is a unit in field k.

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By Hurewicz's formula I can prove the ramification indices at these two points are n, n is the degree of f. But I can't see why it must be a cyclic cover. –  Taisong Jing Dec 26 '09 at 21:00
By the way, I think you mean (Adolf) Hurwitz, rather than (Wittold) Hurewicz. –  Pete L. Clark Dec 26 '09 at 23:38
yes, you are right~ –  Taisong Jing Dec 27 '09 at 4:25

2 Answers 2

up vote 9 down vote accepted

Here's an alternative way to think about it:

You can easily deduce from Riemann-Hurwitz that the genus of X is 0, i.e. it is just the projective line. Look at the affine patch: t is not infinity. Above t=infinity there's only one point. So take that point out, and call the parameter of the resulting affine line h. Then we have the inclusion of k[t] in k[h]. So t=g(h) where g is a polynomial. Since over t=0 there is one point with ramification n, g has multiplicity n: g=u(h-c)n. So after change of variables g=u*hn, as required.

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Thank you so much! –  Taisong Jing Dec 27 '09 at 4:25

Take the Galois closure, which satisfies the same hypotheses and is Galois. The hardest part is to see that it is again tamely ramified: for this see Theorem 2.1 of

Then observe that the tame fundamental group of $\mathbb{P}^1$ minus two points is procyclic. This follows from a comparison theorem of Grothendieck, which allows you to reduce to the case $k = \mathbb{C}$.

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