Tame morphism from a curve to $\mathbb{P}^1$ Let $k$ be an algebraically closed field of characteristic $p\ge 0$. Let $C$ be a smooth projective curve over $k$. Is it possible to find a map $C \to \mathbb{P}^1$ that is tamely ramified at every point of $C$, i.e. such that the ramification index at every point of $C$ is prime to $p$?
A result of Fulton says that, if $k$ is (algebraically closed) of characteristic $p\ne 2$, then it is possible to find a morphism $C \to \mathbb{P}^1$ that is a simple cover: only double points may appear and at most one in every fiber. (This is theorem 8.1 in "Hurwitz schemes and the irreducibility of moduli of algebraic curves", Ann. of Math. 90, 1969. He says it is classical and dates back to Severi.)
Fulton's result gives a positive answer for fields of characteristic $p\ne 2$. But what about characteristic 2? Does the result still hold? I would already be interested in answers in particular cases (elliptic curves for instance).
EDIT: I added the hypothesis that the field is algebraically closed in order to focus on what I am really interested in. Still, I would also appreciate comments on how relevant this hypothesis is.
 A: this is silly but if you do not assume the ground field is algebraically closed, then the answer is no. Namely, suppose that C is the generic curve of genus g in char 2 where g is large. Then every divisor class on C is a multiple of K_C. (This is a highly nontrivial theorem.) Now if we had a tame morphism C ---> P^1 then the ramification indices would all be even, hence the ramification divisor would be 2E for some effective divisor. Then K_C = -2H + 2E where H is the pullback of the ample divisor from P^1. Contradiction.
Edit: Actually, now I just got a little bit worried about the difficult thing referenced above, as I don't know a reference and it is possible that one can take the square root of the canonical divisor on a general curve in characteristic 2. Namely, there is that weird thing where d(x^2 + x^3) = x^2dx in characteristic 2. So if you take a general Lefschetz pencil on that curve then it seems that the ramification has degree 2 everywhere and the x^3 term is also present at every point and then you'd actually get a square root of K_C. So I am sorry but my argument is faulty! Please take away the upvotes for this answer! Thanks!
A: In [Stefan Schroer, Curves with only triple ramification], the author gives lower bounds on the dimension of the subset of the moduli space of curves for which the question has a positive answer.
