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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$\begingroup$ 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. $\endgroup$– TJCMCommented Dec 26, 2009 at 21:00
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$\begingroup$ By the way, I think you mean (Adolf) Hurwitz, rather than (Wittold) Hurewicz. $\endgroup$– Pete L. ClarkCommented Dec 26, 2009 at 23:38
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$\begingroup$ yes, you are right~ $\endgroup$– TJCMCommented Dec 27, 2009 at 4:25
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2 Answers
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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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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
http://math.stanford.edu/~conrad/248APage/handouts/tamecomp.pdf.
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 $k = \mathbb{C}$.
answered Dec 26, 2009 at 23:37
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$\begingroup$ For those who like references, the comparison theorem which gives the result about the tame fundamental group of $\mathbf P^1$ minus two points as a corollary is SGA 1 Exp. XIII 2.12. $\endgroup$– TomoCommented Aug 30, 2019 at 21:37