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Journal of Number Theory 135 (2014) 151–154

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Journal of Number Theory www.elsevier.com/locate/jnt

A new proof of Belyi’s Theorem Wushi Goldring 1 Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544-1000, USA

a r t i c l e

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Article history: Received 12 June 2013 Received in revised form 2 August 2013 Accepted 4 August 2013 Available online 18 October 2013 Communicated by David Goss

a b s t r a c t A new proof of Belyi’s Theorem is given. While Belyi’s two proofs use different methods to first reduce the branch locus from Q to Q and then from Q to {0, 1, ∞}, our argument uses a single, new method to reduce directly from Q to {0, 1, ∞}. © 2013 Elsevier Inc. All rights reserved.

MSC: primary 11G32 secondary 14H57 Keywords: Belyi’s Theorem Covers of curves

The purpose of this short note is to describe a new proof of Belyi’s Theorem. Given a map ϕ : X → Y of connected, smooth projective curves, denote by B(ϕ) ⊂ Y the branch locus of ϕ. Belyi’s Theorem states: Theorem 1. (Belyi [Bel80,Bel02].) Let X be a connected, smooth, projective curve defined over the field of algebraic numbers Q. Then there exists a morphism ϕ : X −→ P1

1

E-mail address: [email protected]. The author gratefully acknowledges support from NSF MSPRF.

0022-314X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jnt.2013.08.017

(1)

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W. Goldring / Journal of Number Theory 135 (2014) 151–154

with B(ϕ) ⊂ {0, 1, ∞}.

(2)

A map ϕ which satisfies (1) and (2) is called a Belyi map for X. See [Gol11] for an exposition of Belyi’s two proofs of Theorem 1. Both of Belyi’s proofs and the one of this note deduce Theorem 1 from the following: Theorem 2. Suppose S is a finite subset of P1 (Q). Then there exists a map fS : P1 → P1 defined over Q satisfying the following two conditions: S-Bel1. B(fS ) ⊂ {0, 1, ∞}. S-Bel2. fS (S) ⊂ {0, 1, ∞}. It is immediate that Theorem 2 implies Theorem 1. Indeed, let ϕ0 : X → P1 be any map defined over Q and take S = B(ϕ0 ). If fS satisfies (S-Bel1) and (S-Bel2) of Theorem 2, then the composite fS ◦ ϕ0 is a Belyi map for X. We give a new proof of Theorem 2. Proof of Theorem 2. Since S is a finite set and every element of S is contained in a finite extension of Q, we may assume without loss of generality that (i) S is stable under the action of Gal(Q/Q) and (ii) {0, 1, ∞} is a subset of S. By induction, it suffices to construct a map hS : P1 → P1 such that (i) the cardinality of B(hS ) ∪ hS (S) is strictly less than that of S, (ii) B(hS ) ∪ hS (S) is Gal(Q/Q)-stable and (iii) {0, 1, ∞} ⊂ B(hS ) ∪ hS (S). Put S = S − {0, 1, ∞} and let d be the cardinality of S . Let p(x) = α∈S (x − α); since S is Galois stable, p(x) has rational coefficients. Our function hS has the form hS (x) = xm (x − 1)n p(x)k

(3)

for some integers m, n, k to be determined. Taking the logarithmic derivative gives n kp (x) q(x) m hS + + = (x) = hS x x−1 p(x) x(x − 1)p(x)

(4)

q(x) = m(x − 1)p(x) + nxp(x) + kx(x − 1)p (x).

(5)

where

So deg q d + 1. Write p(x) = xd + ad−1 xd−1 + · · · + a1 x + a0


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and q(x) = bd+1 xd+1 + bd xd + · · · + b1 x + b0 . Then one has

bd+1 = m + n + dk,

bd = (ad−1 − 1)m + ad−1 n + (d − 1)ad−1 − d k.

(6)

Setting bd+1 = bd = 0, one gets from (6) two linear equations with rational coefficients in the three variables m, n, k. Therefore there exists a solution consisting of integers that are not all zero. Let m, n, k denote such a solution. We claim that then hS (x) defined in (3) has the properties alluded to above. Since bd+1 = bd = 0, one has deg q d − 1. Also, it follows from (6) that k = 0 and that hS (∞) = 1, so that {0, 1, ∞} ⊂ hS (S). Assume first m = 0 and n = 0. Then hS (S) = {0, 1, ∞}. The set of ramification points of hS is contained in the set of zeroes and poles of the logarithmic derivative hS /hS . The poles of hS /hS are at elements of S, which map to {0, 1, ∞} under hS . The zeroes of hS /hS besides ∞, which maps to 1 under hS , are the roots of q(x), so there are at most d − 1 of them. These correspond to at most d − 1 branch points of hS . Hence the cardinality of hS (S) ∪ B(hS ) is at most (d − 1) + 3 = d + 2, which is one less than that of S. Moreover, since q has rational coefficients, the set of its roots is Gal(Q/Q)-stable and since hS has rational coefficients, applying hS to the set of roots of q also gives a Gal(Q/Q) stable set. Therefore B(hS ) ∪ hS (S) is Gal(Q/Q) stable. It remains to treat the case when either m = 0 or n = 0. Suppose that m = 0 (a completely analogous argument handles the case n = 0). Then n = 0 and q(x) is divisible by x, so we can divide both q(x) and x(x−1)p(x) by x to simplify the expression for hS /hS in the right-hand side of (3). Let q0 (x) = q(x)/x. Then deg q0 (x) d − 2. Also, hS (S − {0}) = {0, 1, ∞} and hS (0) is rational since hS has rational coefficients. So hS (S) has cardinality 4 and the cardinality of hS (S)∪B(hS ) is at most (d−2)+4 = d+2 again. Since hS (0) is rational the same argument as before shows that hS (S) ∪ B(hS ) is Gal(Q/Q) stable. 2 Acknowledgments It is a pleasure to thank Joseph Oesterlé, Barry Mazur, Pierre Deligne and Leonardo Zapponi for helpful discussions about Belyi’s Theorem. I am also grateful to Anders Karlsson and the University of Geneva for inviting me to speak about this topic in January 2013. Finally I would like to thank the referee for his/her helpful comments. References [Bel80] G.V. Belyi, On Galois extensions of a maximal cyclotomic field, Math. USSR Izv. 14 (1980) 247–256.

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[Bel02] G.V. Belyi, A new proof of the three point theorem, Sb. Math. 193 (3–4) (2002) 329–332. [Gol11] W. Goldring, Unifying themes suggested by Belyi’s theorem, in: Ramakrishnan, et al. (Eds.), Number Theory, Analysis and Geometry, Springer-Verlag, 2011 (S. Lang memorial volume).