Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Belyi's theorem states that every smooth projective algebraic curve $C$ defined over $\bar{\mathbb{Q}}$ admits a map $C\to\mathbb{P}^1$ ramified only over $0,1,\infty$. Is there an analogue of this theorem with $\mathbb{Q}$ replaced by a global function field (i.e. finite extension $\mathbb{F}_q(t)$)?

I am especially interested in the existence of a tamely ramified map.

share|cite|improve this question
Have you seen Wushi Goldring's "Unifying themes suggested by Belyi's theorem"? See… – Ariyan Javanpeykar Jun 8 '13 at 8:58
@Ariyan Javanpeykar Yes, I've seen it. It discusses the problem over (the algebraic closure of) finite fields and function fields in characteristic zero. It also discusses characteristic p in general but with wild ramification allowed, while I am more interested in tame ramification. – Alex Jun 8 '13 at 11:07
(answering your comment below). I think that it is unlikely that this will work if you want the morphism to be tamely ramified. Galois coverings of the affine line in char. p, which are tamely ramified along $\infty$, are trivial. – Damian Rössler Jun 8 '13 at 20:22
@Damian Rossler But I also allow ramification at 0,1. – Alex Jun 8 '13 at 20:42

2 Answers 2

up vote 6 down vote accepted

Much stronger results are available in positive characteristic. See

Kedlaya, Kiran S.

More étale covers of affine spaces in positive characteristic. J. Algebraic Geom. 14 (2005), no. 1, 187–192.

share|cite|improve this answer
What if I want the map to be tamely ramified? – Alex Jun 8 '13 at 8:25

In a comment on the accepted answer, the OP asks about tame ramification. Saidi shows in Theorem 5.6 here that a smooth projective curve $C$ over a field $k$ of characteristic $p>2$ is defined over $\overline{\mathbb{F}_p}$ if and only if $C$ admits a map $C\to \mathbb{P}^1$ with only tame ramification over $\{0,1,\infty\}$.

The proof is rather easy. I'll first sketch the argument that a curve defined over $\overline{F_p}$ admits a map as claimed. A result of Fulton shows that any curve admits a map $g: C\to \mathbb{P}^1$ with ramification indices at most two, hence a tamely ramified map if the characteristic is different from $2$. Let $q$ be such that the ramification values are defined over $\mathbb{F}_q$. Then composing $g$ with the map $$z\mapsto z^{q-1}$$ gives a map $C\to \mathbb{P}^1$ with the desired property.

To see the other direction, one may use Riemann existence or just observe that maps with the desired property don't deform and the moduli space of such maps is locally finite type, hence any $k$-point comes from an $\overline{\mathbb{F}_p}$-point.

share|cite|improve this answer

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.