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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.

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Consider a degree $4$ cover of $\mathbb P^1$ that is ramified over $4$ points, of degree $3$ at each. It is easy to see via the group theory of $S^4$ that such exists. Then by Riemann-Hurwitz, this is an elliptic curve. The moduli space of such covers is one-dimensional, and one just has to check that the map to the moduli space of elliptic curves is nonconstant, which I think is true but haven't checked. Similar explicit constructions might work for other genuses. –  Will Sawin Apr 15 '13 at 16:30
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Thanks Will. There are a few things I don't understand. What is the link with $S^4$? How do you know the dimension of the moduli space of covers? (Since $M_{0,4}$ has dimension 1, I would have thought it would be bigger, but this will not cause any trouble anyway.) I don't have time to check the details right now but will come back to it next week. –  Jérôme Poineau Apr 16 '13 at 9:17
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@Will: $S^4$ is meant to be the symmetric group on $4$ letters? Anyway, in this characteristic $2$ context without Riemann's existence theorem, it isn't clear to me that this cover exists. I would be more convinced if one would write it down explicitly for an elliptic curve in Weierstrass form. –  Peter Mueller Apr 16 '13 at 9:29
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@Peter: Thanks for your comment. This $S^4$ makes more sense but I don't know how to make the argument work in characteristic 2 either. On the other hand, I may be able to do it using patching techniques from inverse Galois theory (maybe if $k$ is algebraically closed but I don't mind too much). –  Jérôme Poineau Apr 16 '13 at 9:40
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1) The condition that $k$ be algebraically closed is certainly not necessary in the result from Fulton's paper, it suffices to assume that $k$ be infinite. The result probably holds over finite fields as well, but would require some more arguments. 2) An example of an elliptic curve in characteristic $2$ where there is a tamely ramified map is the elliptic curve with an automorphism of order $3$: the quotient of the curve modulo the group generated by this automorphism is $\mathbb{P}^1$. –  ulrich Apr 25 '13 at 10:51

2 Answers 2

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!

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It seems that your answer is correct. The argument is given in [Stefan Schroer, The strong Franchetta conjecture in arbitrary characteristics, Theorem 6.1]. If I understand well, the reason why the argument in your edit is not a problem is that the ramification divisor might be geometrically divisible by 2 but not divisible by 2 if its support contains a point whose residue field is inseparable over the base field. –  Olivier Benoist Mar 28 at 10:06
    
Thank you very much for this clarification and the reference. –  answer_bot Mar 29 at 0:09

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.

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