GONALITY OF DYNATOMIC CURVES AND STRONG UNIFORM BOUNDEDNESS OF PREPERIODIC POINTS

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1 GONALITY OF DYNATOMIC CURVES AND STRONG UNIFORM BOUNDEDNESS OF PREPERIODIC POINTS JOHN R. DOYLE AND BJORN POONEN Abstrat. Fix d 2 and a field k suh that har k d. Assume that k ontains the dth roots of 1. Then the irreduible omponents of the urves over k parameterizing preperiodi points of polynomials of the form z d + are geometrially irreduible and have gonality tending to. This implies the funtion field analogue of the strong uniform boundedness onjeture for preperiodi points of z d +. It also has onsequenes over number fields: it implies strong uniform boundedness for preperiodi points of bounded eventual period, whih in turn redues the full onjeture for preperiodi points to the onjeture for periodi points. 1. Introdution 1.1. Dynatomi urves. Fix an integer d 2. Let k be a field suh that har k d. View f = f := z d + as a polynomial in z with oeffiients in k[]. Let f n (z) be the nth iterate of f; in partiular, f 0 (z) := z. If n and m are nonnegative integers with n > m, then any irreduible fator of f n (z) f m (z) k[z, ] defines an affine urve over k. By a dynatomi urve over k, we mean any suh urve, or its smooth projetive model. Any k-point on suh a urve yields 0 k equipped with a preperiodi point in k, that is, an element z 0 k that under iteration of x d + 0 eventually enters a yle; the length of the yle is alled the eventual period. We onsider two dynatomi urves to be different if the orresponding losed subshemes of A 2 k are distint. Setion 2 desribes all dynatomi urves in harateristi 0 expliitly Gonality. Let k be an algebrai losure of k. Let µ d = {x k : x d = 1}. For a urve X over k, let X k = X k k. If X is irreduible, define the gonality γ(x) of X as the least possible degree of a dominant rational map X P 1 k. If X is geometrially irreduible, define its k-gonality as γ(x k ). Theorem 1.1. Fix d 2 and k suh that har k d. Suppose that µ d k. (a) Every dynatomi urve over k is geometrially irreduible. (b) If the dynatomi urves over k are listed in any order, their gonalities tend to. Remark 1.2. Part (a) of Theorem 1.1 an fail if µ d k. See Remark 2.2. Remark 1.3. In proving Theorem 1.1 in positive harateristi, we fae the hallenge that we do not know expliitly what the dynatomi urves are, sine we are not sure whether Date: November 11, Mathematis Subjet Classifiation. Primary 37P35; Seondary 11G30, 14H51, 37P05, 37P15. Key words and phrases. Gonality, dynatomi urve, preperiodi point. The seond author was supported in part by National Siene Foundation grants DMS and DMS and Simons Foundation grants # (to Bjorn Poonen) and #

2 the known fators of the polynomials f n (z) f m (z) are irreduible. Some partial results regarding the irreduibility of dynatomi urves in positive harateristi may be found in [DKO + 17], but we will not use them. Let us now introdue notation for our seond main result. Let µ(n) denote the Möbius µ-funtion. Then f n (z) z = e n Φ e(z, ), where (1) Φ n (z, ) := e n (f e (z) z) µ(n/e) k[z, ]. Let Y dyn 1 (n) be the urve defined by Φ n (z, ) = 0 in A 2 k, and let Xdyn 1 (n) be the normalization of its projetive losure. To simplify notation, we omit the supersript dyn from now on. General points of X 1 (n) parametrize polynomials of the form z d + equipped with a point of exat order n. The morphism (z, ) (f(z), ) restrits to an order n automorphism of Y 1 (n), so it indues an order n automorphism σ of X 1 (n). The quotient of X 1 (n) by the yli group generated by σ is alled X 0 (n). irreduible, so X 0 (n) is too. If har k = 0, it is known that X 1 (n) is geometrially Theorem 1.4. Fix d 2 and a field k of harateristi 0. Then ( 1 γ(x 0 (n)) > 2 1 ) 2d o(1) n as n. In partiular, γ(x 0 (n)). Remark 1.5. Our definition of dynatomi urve does not inlude quotient urves suh as X 0 (n), so the onlusion γ(x 0 (n)) of Theorem 1.4 does not follow from Theorem 1.1(b). To prove Theorem 1.4, we use that X 0 (n) already has a morphism to P 1 of degree lower than expeted for its genus, namely X 0 (n) P 1. If it also had a morphism to P 1 of bounded degree, then the Castelnuovo Severi inequality would make the genus of X 0 (n) smaller than it atually is, a ontradition. See Setion 3 for details. To prove Theorem 1.1(b), we use different arguments in harateristi 0 and harateristi > 0. In harateristi 0, we use that eah dynatomi urve dominates X 1 (n) and hene also X 0 (n) for some n, so its gonality is large when n is large. On the other hand, for a fixed small n, the dynatomi urves above X 1 (n) ome in towers and we use the Castelnuovo Severi inequality to work our way up eah tower. See Setion 3. In harateristi p, we prove that the irreduible omponents of f n (z) f m (z) have large degree over the -line, and we use that to prove that over the finite field F q := F p (µ d ) their smooth projetive models have so many F q -points over = that their F q -gonalities must be large. Finally, we use a result ontrolling how gonality of a urve hanges when the base field is enlarged. See Setion Uniform boundedness of preperiodi points. The growth of gonalities of lassial modular urves implies the strong uniform boundedness theorem for torsion points on ellipti urves over funtion fields (the funtion field analogue of Merel s theorem [Mer96]); see [NS96, Theorem 0.3]. Similarly, from Theorem 1.1 we will dedue the following funtion field analogue of a ase of the Morton Silverman onjeture [MS94, p. 100]: 2

3 Theorem 1.6 (Strong uniform boundedness theorem for preperiodi points over funtion fields). Fix d 2 and a field k suh that har k d. Let K be the funtion field of an integral urve over k. Fix a positive integer D. Then there exists B = B(d, K, D) > 0 suh that for every field extension L K of degree D and every L not algebrai over k, the number of preperiodi points of z d + in L is at most B. If k is finite, the same holds with the words not algebrai over k deleted. See [DKO + 17, Setion 4] for some related results and arguments. Another appliation of Theorem 1.1 is the following, whih over number fields provides a uniform bound on preperiodi points having a bounded eventual period: Theorem 1.7. Fix integers d 2, D 1, and N 1. Then there exists B = B(d, D, N) > 0 suh that for every number field K satisfying [K : Q] D and every K, the number of preperiodi points of z d + in K with eventual period at most N is at most B. Theorems 1.6 and 1.7 are proved in Setion 5. Theorem 1.7 implies that the strong uniform boundedness onjeture for periodi points over number fields implies the strong uniform boundedness onjeture for preperiodi points over number fields, as we now explain: Corollary 1.8. Fix integers d 2 and D 1. Suppose that there exists a bound N = N(d, D) suh that for every number field K satisfying [K : Q] D and every K, every periodi point of z d + in K has period at most N. Then there exists a bound B = B (d, D) suh that for every number field K satisfying [K : Q] D and every K, the number of preperiodi points of z d + in K is at most B. Proof. By assumption, if [K : Q] D and K, then the preperiodi points of z d + in K having eventual period at most N are all the preperiodi points in K. Therefore the bound B(d, D, N) of Theorem 1.7 is atually a bound on the total number of preperiodi points in K. Take B = B(d, D, N) = B(d, D, N(d, D)). 2. Classifiation of dynatomi urves For m, n 1, let Y 1 (m, n) be the urve over k whose general points parametrize polynomials z d + equipped with a preperiodi point that after exatly m steps enters an n-yle. This urve is the zero lous in A 2 k of the polynomial Φ m,n (z, ) := Φ n(f m (z), ) Φ n (f m 1 (z), ). For a general point (z, ) Y 1 (1, n), the elements z and f n (z) are distint preimages of f(z), so z = ζf n (z) for some ζ µ d {1}. Suppose that µ d k. For eah ζ µ d {1}, let Y 1 (1, n) ζ be the subsheme of Y 1 (1, n) defined by the ondition z = ζf n (z), so Y 1 (1, n) = Y 1 (1, n) ζ. ζ µ d {1} Both (z, ) (f(z), ) and (z, ) (ζ 1 z, ) define isomorphisms Y 1 (1, n) ζ Y 1 (n). In partiular, Y 1 (1, n) ζ equals the urve Φ n (ζ 1 z, ) = 0 in A 2 k. For m 2, let Y 1(m, n) ζ be the inverse image of Y 1 (1, n) ζ under Y 1 (m, n) Y 1 (1, n) (z, ) (f m 1 (z), ). 3

4 Then for any m, n 1, (2) Y 1 (m, n) = ζ µ d {1} Y 1 (m, n) ζ, and Y 1 (m, n) ζ equals the urve Φ n (ζ 1 f m 1 (z), ) = 0 in A 2 k. The deomposition (2) orresponds to a fatorization (3) Φ m,n (z, ) = Φ n (ζ 1 f m 1 (z), ). ζ µ d {1} The following is a olletion of results from [Bou92, LS94, Mor96, Gao16]. Theorem 2.1. Let k be a field of harateristi 0 suh that µ d k. Then the urves Y 1 (n) for n 1 and the urves Y 1 (m, n) ζ for m, n 1 and ζ µ d {1} are irreduible, so they are all the dynatomi urves over k. Computer experiments (at least for d = 2) suggest that Theorem 2.1 holds for any field k suh that har k d and µ d k, but in positive harateristi this remains unproved. See [DKO + 17], espeially Theorems B and D, for some progress in this diretion. Remark 2.2. If in Theorem 2.1 we drop the hypothesis that µ d k, then the irreduible omponents of Y 1 (m, n) over k are in bijetion with the Gal(k(µ d )/k)-orbits in µ d {1}. An irreduible omponent orresponding to an orbit of size greater than 1 is not geometrially irreduible. 3. Gonality in harateristi 0 Given a geometrially irreduible urve X over k, let g(x) denote the genus of its smooth projetive model. ( Let D 0 (n) be the degree of the morphism X 0 (n) P 1. Similarly, let D 1 (n) = deg X 1 (n) P ). 1 Proposition 3.1. Fix d 2, and fix a field k of harateristi 0. (a) We have D 1 (n) = (1 + o(1))d n as n. (b) We have D 0 (n) = (1 + o(1))d n /n as n. () We have g(x 0 (n)) > ( 1 1 o(1)) d n as n. 2 2d P 1 is the full sym- (d) The Galois group of (the Galois losure of) the overing X 0 (n) metri group S D0 (n). Proof. We may assume k = C. (a) First, D 1 (n) = deg z Φ n deg(f n (z) z) = d n. On the other hand, one an show (e.g., as in the proof of [Doy16, Lemma 2.2]) that D 1 (n) > d n d n/2+1 /(d 1). (b) The first morphism in the tower X 1 (n) X 0 (n) P 1 has degree n, so D 0 (n) = D 1 (n)/n. Substitute (a) into this. () More preisely, g(x 0 (n)) > ( d n) d n + O(nd n/2 ) as n, by [Mor96, Theorem 13(d)]. (d) This is a onsequene of work of Boush [Bou92] for d = 2, and Lau and Shleiher [LS94] for d > 2. See also [Mor98, Theorem B] and [Sh17]. 4

5 Proposition 3.2 (Castelnuovo Severi inequality). Let F, F 1, F 2 be funtion fields of urves over k, of genera g, g 1, g 2, respetively. Suppose that F i F for i = 1, 2 and the ompositum of F 1 and F 2 in F equals F. Let d i = [F : F i ] for i = 1, 2. Then Proof. See [Sti93, III.10.3]. g d 1 g 1 + d 2 g 2 + (d 1 1)(d 2 1). h Proof of Theorem 1.4. Let X 0 (n) P 1 be a dominant rational map of minimal degree. Case I: h fators through X 0 (n) P 1. Then deg h D 0 (n) = (1 + o(1)) dn n by Proposition 3.1(b), so deg h is muh larger than n when n is large. Case II: h does not fator through X 0 (n) P 1. Then the ompositum of k() and k(h) in the funtion field k(x 0 (n)) is stritly larger than k(). Beause of the Galois group (Proposition 3.1(d)), the only nontrivial extension of k() in k(x 0 (n)) is the whole field k(x 0 (n)). Thus k() and k(h) generate k(x 0 (n)). By Proposition 3.2, Thus g(x 0 (n)) (D 0 (n) 1)(deg h 1). deg h 1 + g(x 0(n)) D 0 (n) 1 = as n, by Proposition 3.1(b,). ( ) 2d o(1) n Lemma 3.3. Let k be a field of harateristi 0 suh that µ d k. Let m and n be positive integers, and let ζ µ d {1}. Then the polynomial Φ n (ζ 1 f m 1 (0), ) k[] has only simple roots, and their number is d m 2 D 1 (n) if m 2. Proof. First suppose that n does not divide m 1. In this ase, the roots of Φ m,n (0, ) are distint by [HT15, Theorem 1.1], and therefore the roots of Φ n (ζ 1 f m 1 (0), ) are distint by (3). Now suppose that n divides m 1. By [Eps12, Proposition A.1], the roots of Φ n (0, ) are simple. If is suh a root, then the polynomial f = f satisfies f n (0) = 0, so f m 1 (0) = 0 and Φ n (ζ 1 f m 1 (0), ) = 0. Thus Φ n (0, ) divides Φ n (ζ 1 f m 1 (0), ). The fatorization (3) yields (4) Φ m,n (0, ) Φ n (0, ) d 1 = ζ µ d {1} Φ n (ζ 1 f m 1 (0), ). Φ n (0, ) By [HT15, Theorem 1.1], Φ m,n (0, )/Φ n (0, ) d 1 has only simple roots, none of whih are also roots of Φ n (0, ). Combining this with (4) shows that Φ n (ζ 1 f m 1 (0), ) has only simple roots. It remains to prove deg Φ n (ζ 1 f m 1 (0), ) = d m 2 D 1 (n). In fat, this is [Gao16, Lemma 4.8]. In our notation, the argument is as follows. By indution on m, the degree of the polynomial f m 1 (0) k[] is d m 2 if m 2. Hene, by indution on e, we have deg f e (ζ 1 f m 1 (0)) = d e d m 2 for eah e 0. Thus the -degree of f e (ζ 1 f m 1 (0)) ζ 1 f m 1 (0) is d m 2 times the z-degree of f e (z) z for eah e 1. Substituting ζ 1 f m 1 (0) for z in (1) shows that deg Φ n (ζ 1 f m 1 (0), ) = d m 2 deg z Φ n (z, ) = d m 2 D 1 (n). 5

6 Proof of Theorem 1.1 in harateristi 0. (a) By Theorem 2.1, the dynatomi urves over k are the urves Y 1 (n) and Y 1 (m, n) ζ, and they are geometrially irreduible. (b) The urves Y 1 (n) and Y 1 (m, n) ζ dominate X 0 (n), so their gonalities are at least the gonality of X 0 (n), by [Poo07, Proposition A.1(vii)]. In light of Theorem 1.4, it remains to prove that in eah tower Y 1 (m, n) ζ Y 1 (m 1, n) ζ Y 1 (1, n) ζ for fixed n and ζ, the gonality tends to as m. Let g m = g(y 1 (m, n) ζ ) and γ m = γ(y 1 (m, n) ζ ). Let Y 1 (m, n) ζ h P 1 be a dominant rational map of degree γ m. Case I: h fators through a urve Z with k(y 1 (m 1, n) ζ ) k(z) k(y 1 (m, n) ζ ). Then γ m 2γ(Z) 2γ m 1 by [Poo07, Proposition A.1(vii)], so γ m grows by indution on m. Case II: h does not fator through any suh urve Z. Denote by π m the degree d map π m : Y 1 (m, n) ζ Y 1 (m 1, n) ζ (z, ) (f(z), ), and let π m : X 1 (m, n) ζ X 1 (m 1, n) ζ be its extension to the smooth projetive models. Let R m be the ramifiation divisor of π m. Applying Proposition 3.2 to π m and h yields g m dg m 1 + (d 1)(γ m 1), so it suffies to show that g m dg m 1 as m. By Riemann Hurwitz, this is equivalent to showing that deg R m as m. In the fiber produt diagram Y 1 (m, n) ζ π m Y 1 (m 1, n) ζ z A 1 f both vertial morphisms are étale above 0 by Lemma 3.3, while f has ramifiation index d at 0, so π m has ramifiation index d at eah point of Y 1 (m, n) ζ where z = 0. For m 2, the number of suh points is d m 2 D 1 (n) by Lemma 3.3. Thus deg R m (d 1)d m 2 D 1 (n), whih tends to as m. z A 1, 4. Gonality in harateristi p 4.1. Redution to the ase of a finite field. Let F q = F p (µ d ). Lemma 4.1. Theorem 1.1 for F q implies Theorem 1.1 for any harateristi p field k ontaining µ d. Proof. Theorem 1.1(a) for F q implies that the dynatomi urves over k are just the base extensions of the dynatomi urves X over F q, and that they are geometrially irreduible too. We will prove in Setion 4.4 that the smooth projetive model of eah X has an F q - point. Then Theorem 2.5(iii) and Proposition A.1(ii) of [Poo07] imply γ(x k ) γ(x), whih implies Theorem 1.1(b) for k. 6

7 4.2. Symboli dynamis. View f(z) as a polynomial in z over the loal field F q (( 1 )). Normalize the valuation v on F q (( 1 )) so that v( 1 ) = 1, and extend v to an algebrai losure. Let t = 1/d, so F q ((t)) is a degree d totally tamely ramified extension of F q (( 1 )). Lemma 4.2. (a) For any nonnegative integers n > m, the polynomial f n (z) f m (z) over F q () is separable and splits ompletely over F q ((t)). (b) Eah zero of f n (z) f m (z) has valuation 1/d and generates F q ((t)) over F q (( 1 )). Proof. The ideas in the following argument are well known; f. [Mor96, Lemma 1]. (a) For eah dth root of, interpreting ( + z) 1/d as ( ) 1/d (1 1 z) 1/d and expanding (1 1 z) 1/d in a binomial series defines a branh of the inverse of z d + on the open disk D := {z F q ((t)) : v(z) > v()}. Taking the derivative of ( + z) 1/d shows that eah branh is a ontrating map D D. These branhes have disjoint images, eah a smaller open disk around a different dth root of. Let S be the set of these d funtions. Eah finite sequene of elements of S defines a omposition of funtions, and for eah m, the images of the different m-fold ompositions are disjoint open disks. For eah infinite sequene s 1, s 2, of elements of S, the images of s 1 s m for m 1 are nested open disks whose radii tend to 0, so they have a unique point in their intersetion; denote it [s 1 s 2 ]. Any two distint infinite sequenes yield two points in disjoint disks, so these points are distint. Sine f s 1 is the identity, f maps [s 1 s 2 ] to [s 2 s 3 ]. For fixed nonnegative integers n > m, any n-long sequene s 1,..., s n in S extends uniquely to an infinite sequene (s i ) satisfying s i+n = s i+m for all i 1, and then [s 1 s 2 ] is a zero of f n (z) f m (z). There are d n of these, so they are all the zeros. In partiular, these zeros are distint elements of F q ((t)). This implies that f n (z) f m (z) is separable. (b) The image of eah s S onsists of elements of valuation exatly 1/d. Thus eah element [s 1 s 2 ] has valuation 1/d. In partiular, eah zero of f n (z) f m (z) generates an extension field of F q (( 1 )) of ramifiation index divisible by d; this extension field an only be the whole field F q ((t)) Dynatomi urves of low degree. Lemma 4.3. For eah e 1, the set of dynatomi urves X over F q suh that deg(x P 1 ) = e is finite. Proof. Suppose that Q(z) = e r=0 q rz e r is a moni degree e fator of f n (z) f m (z) over F q () for some n and m. For eah r, the oeffiient q r is the rth elementary symmetri polynomial evaluated at the negatives of the zeros of Q; those zeros have valuation 1/d by Lemma 4.2(b), so v(q r ) r/d. On the other hand, by Gauss s lemma, q r F q [], so deg q r r/d. Thus there are only finitely many possibilities for eah q r, and hene finitely many possibilities for Q, eah of whih yields one dynatomi urve Gonality of dynatomi urves. Proof of Theorem 1.1. By Lemma 4.1, we may assume that k = F q. (a) Let X be the smooth projetive model of a dynatomi urve, orresponding to a fator of f n (z) f m (z) for some n and m. By Lemma 4.2(b), f n (z) f m (z) splits ompletely over F q ((t)), and the preimage of under X P 1 onsists of F q -points, eah of ramifiation 7

8 index d. Every irreduible omponent Z of X Fq dominates P 1 and hene must ontain one of those F q -points, say x. Then eah Gal(F q /F q )-onjugate of Z ontains x. On the other hand, sine X Fq is smooth, its irreduible omponents are disjoint. Thus Z equals eah of its onjugates, so Z desends to an irreduible omponent of X, whih must be X itself. (This proves also that X has an F q -point, as promised in the proof of Lemma 4.1.) (b) To bound gonality from below, we use Ogg s method of ounting points over a finite field; f. [Ogg74], [Poo07, Setion 3], and [DKO + 17, Setion 4]. Let X be the smooth projetive model of a dynatomi urve. Let e = deg(x P 1 ). Eah preimage of in X is an F q -point of ramifiation index d, so there are e/d suh points. On the other hand, P 1 has only q +1 points over F q, so any nononstant morphism X P 1 has degree at least e/(d(q + 1)). As X varies, e by Lemma Strong uniform boundedness of preperiodi points Proof of Theorem 1.6. Without loss of generality, K = k(u) for some indeterminate u. Given L, let Y be the smooth projetive integral urve over k with funtion field L. The ondition [L : K] d implies that Y has gonality at most D, so eah irreduible omponent of Y k has gonality at most D. For L not algebrai over k, if z L and n > m satisfy f n (z) f m (z) = 0, then (z, ) is a nononstant and hene smooth L-point of the urve f n (z) f m (z) = 0 in A 2, so it yields an L-point on a dynatomi urve X orresponding to a fator of f n (z) f m (z). This L-point defines a nononstant k-morphism Y X, so γ(x) γ(y ) D. By Theorem 1.1(b), this plaes a uniform bound on n. For eah n, the number of preperiodi points of z d + orresponding to that value of n is uniformly bounded by n 1 m=0 deg(f n (z) f m (z)) = nd n, so bounding n bounds the number of preperiodi points too. If k is finite and lies in the maximal algebrai extension l of k in L, then all the preperiodi points of x d + in L are in l, but [l : k] D, so the number of preperiodi points is uniformly bounded by (#k) D. To prove Theorem 1.7, we need the following result of Frey [Fre94, Proposition 2]. Lemma 5.1. Let C be a urve defined over a number field K. Let D 1. If there are infinitely many points P C(K) of degree D over K, then γ(c) 2D. Proof of Theorem 1.7. If the onlusion fails, then there exists n N suh that there are infinitely many m 1 suh that Y 1 (m, n) has a point of degree D over Q. On the other hand, by Northott s theorem, for any given Q, the set of preperiodi points of x d + is finite. Thus for any m 0, the points on Y 1 (m, n) in the first sentene above for all m m 0 together must inlude infinitely many distint -oordinates. In partiular, their images in Y 1 (m 0, n) inlude infinitely many distint points. Thus some Q(µ d )-irreduible omponent Y 1 (m 0, n) ζ ontains infinitely many points of degree D over Q(µ d ). By Lemma 5.1, γ(y 1 (m 0, n) ζ ) 2D. This argument applies for every m 0, ontraditing Theorem 1.1(b). Aknowledgements We thank Joseph Gunther, Holly Krieger, Andrew Obus, Padmavathi Srinivasan, and Isabel Vogt for omments and disussions. 8

9 Referenes [Bou92] Thierry Boush, Sur quelques problèmes de dynamique holomorphe, PhD thesis, Université de Paris-Sud, Centre d Orsay, [Doy16] John R. Doyle, Preperiodi portraits for uniritial polynomials, Pro. Amer. Math. So. 144 (2016), no. 7, MR [DKO + 17] John R. Doyle, Holly Krieger, Andrew Obus, Rahel Pries, Simon Rubinstein-Salzedo, and Lloyd West, Redution of dynatomi urves, Marh 29, Preprint, arxiv: v2. 2, 3, 4, 8 [Eps12] Adam Epstein, Integrality and rigidity for postritially finite polynomials, Bull. Lond. Math. So. 44 (2012), no. 1, With an appendix by Epstein and Bjorn Poonen. MR [Fre94] Gerhard Frey, Curves with infinitely many points of fixed degree, Israel J. Math. 85 (1994), no. 1-3, MR [Gao16] Yan Gao, Preperiodi dynatomi urves for z z d +, Fund. Math. 233 (2016), no. 1, MR , 5 [HT15] Benjamin Hutz and Adam Towsley, Misiurewiz points for polynomial maps and transversality, New York J. Math. 21 (2015), MR [LS94] Eike Lau and Dierk Shleiher, Internal addresses in the Mandelbrot set and irreduibility of polynomials, SUNY Stony Brook Preprint 19 (1994). 4 [Mer96] Loï Merel, Bornes pour la torsion des ourbes elliptiques sur les orps de nombres, Invent. Math. 124 (1996), no. 1-3, (Frenh). MR (96i:11057) 2 [Mor96] Patrik Morton, On ertain algebrai urves related to polynomial maps, Compositio Math. 103 (1996), no. 3, MR , 7 [Mor98] Patrik Morton, Galois groups of periodi points, J. Algebra 201 (1998), no. 2, , DOI /jabr MR [MS94] Patrik Morton and Joseph H. Silverman, Rational periodi points of rational funtions, Internat. Math. Res. Noties 2 (1994), , DOI /S MR (95b:11066) 2 [NS96] Kha Viet Nguyen and Masa-Hiko Saito, d-gonality of modular urves and bounding torsions (Marh 29, 1996). Preprint, arxiv:alg-geom/ [Ogg74] Andrew P. Ogg, Hyperellipti modular urves, Bull. So. Math. Frane 102 (1974), MR (51 #514) 8 [Poo07] Bjorn Poonen, Gonality of modular urves in harateristi p, Math. Res. Lett. 14 (2007), no. 4, , DOI /MRL.2007.v14.n4.a14. MR , 8 [Sh17] Dierk Shleiher, Internal addresses of the Mandelbrot set and Galois groups of polynomials, Arnold Math. J. 3 (2017), no. 1, MR [Sti93] Henning Stihtenoth, Algebrai funtion fields and odes, Universitext, Springer-Verlag, Berlin, MR (94k:14016) 5 Department of Mathematis & Statistis, Louisiana Teh University, Ruston, LA 71272, USA address: jdoyle@lateh.edu URL: Department of Mathematis, Massahusetts Institute of Tehnology, Cambridge, MA , USA address: poonen@math.mit.edu URL: 9