Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

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.

share|improve this answer
Thank you so much! –  Taisong Jing Dec 27 '09 at 4:25
add comment

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 $k = \mathbb{C}$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.