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^nut), u is a unit in field k.

Here's an alternative way to think about it: You can easily deduce from RiemannHurwitz 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(hc)^{n}. So after change of variables g=u*h^{n}, as required. 


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 $k = \mathbb{C}$. 

