PREPERIODIC POINTS OF POLYNOMIALS OVER GLOBAL FIELDS

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1 PREPERIODIC POITS OF POLYOMIALS OVER GLOBAL FIELDS ROBERT L. BEEDETTO Abstract. Gien a global field K and a polynomial φ defined oer K of degree at least two, Morton and Silerman conjectured in 1994 that the number of K-rational preperiodic points of φ is bounded in terms of only the degree of K and the degree of φ. In 1997, for quadratic polynomials oer K = Q, Call and Goldstine proed a bound which was exponential in s, the number of primes of bad reduction of φ. By careful analysis of the filled Julia sets at each prime, we present an improed bound on the order of s log s. Our bound applies to polynomials of any degree (at least two) oer any global field K. Let K be a field, and let φ K(z) be a rational function. Let φ n denote the n th iterate of φ under composition; that is, φ 0 is the identity function, and for n 1, φ n = φ φ n 1. We will study the dynamics φ on the projectie line P 1 (K). In particular, we say a point x is preperiodic under φ if there are integers n > m 0 such that φ m (x) = φ n (x). The point y = φ m (x) satisfies φ n m (y) = y and is said to be periodic (of period n m). ote that x P 1 (K) is preperiodic if and only if its orbit {φ n (x) : n 0} is finite. For example, let K = Q and φ(z) = z 2 29/16. Then {5/4, 1/4, 7/4} forms a periodic cycle (of period 3), and 5/4, 1/4, 7/4, and ±3/4 each land on this cycle after one or two iterations. In addition, the point is of course fixed. These nine Q-rational points are all preperiodic under φ. Meanwhile, it is not difficult to see that no other point in P 1 (Q) is preperiodic by showing that the denominator of a rational preperiodic point must be 4, and that the absolute alue must be less than 2. In general, for any global field K, any dimension 1, and any morphism φ : P P oer K of degree at least two, orthcott proed in 1950 that the number of K-rational preperiodic points of φ is finite [25]. More precisely, he showed that the preperiodic points form a set of bounded arithmetic height. Years later, by analogy with the Theorems of Mazur [19] and Merel [20] on K-rational torsion of elliptic cures, Morton and Silerman proposed the following Conjecture [23]. Uniform Boundedness Conjecture. (Morton and Silerman, 1994) Gien integers D, 1 and d 2, there is a constant κ = κ(d,, d) with the following property. Let K be a number field with [K :Q] = D, and let φ : P P be a morphism of degree d defined oer K. Then φ has at most κ preperiodic points in P (K). The analogy between preperiodic points and torsion comes from the fact that the torsion points of an elliptic cure E are precisely the preperiodic points of the multiplication-bytwo map [2] : E E. In fact, taking x-coordinates, the map [2] induces a rational function (known to dynamicists as a Lattès map) φ : P 1 P 1 of degree 4 whose preperiodic points Date: June 21, 2005; reised December 21, Mathematics Subject Classification. Primary: 11G99 Secondary: 11D45, 37F10. Key words and phrases. filled Julia set, transfinite diameter, uniform bounds. The author gratefully acknowledges the support of a Miner D. Crary Research Fellowship from Amherst College and SA Young Inestigator Grant H

2 2 ROBERT L. BEEDETTO are precisely the x-coordinates of the torsion points of E. Thus, Merel s Theorem would follow as a simple corollary of the Morton and Silerman Conjecture for = 1 and d = 4. More generally, Fakhruddin has shown [12] that the full Morton and Silerman Conjecture for D = 1 would imply uniform boundedness of torsion for abelian arieties. The Conjecture seems to be ery far from a proof. Howeer, there is growing eidence that it is alid, at least in the simplest case, that K = Q, = 1, and φ is a polynomial of degree 2. (The problem then reduces to considering φ c (z) = z 2 + c, with c Q.) In particular, the computations in [22] and [13] show that φ c neer has a rational point of period 4 or 5, respectiely. Moreoer, Poonen showed in 1998 that if φ c neer has rational periodic points of period greater than 5, then it neer has more than 9 rational preperiodic points [28]. (That bound, if true, would be sharp, in light of the c = 29/16 example aboe.) Those results all considered moduli spaces, for fixed n > m 0, of pairs (c, x) such that φ n c (x) = φ m c (x), giing cures analogous to modular cures, but with no known structure to take the place of a Hecke ring. Instead, the theorems were proen by delicate ad hoc computations on the particular cures that arose. Other researchers hae found bounds for the longest possible period of a K-rational periodic point by analyzing at a prime of good reduction (see Definition 2.1 below); see, for example, [11, 23, 24, 26, 27, 35]. If s is the total number of primes of bad reduction, then these results lead to bounds on the order of at least d s4d for the number of K-rational periodic points (cf. Corollary B of [23], for example). A different strategy (for the family φ c (z) = z 2 +c for c Q) appeared in a 1997 theorem of Call and Goldstine [9], who showed that φ c has at most 1+2 s+2 rational preperiodic points, where s is the number of primes of bad reduction. They analyzed the dynamics at the primes of bad, not good, reduction, by studying the filled Julia set K (see Definition 2.2 below). All preperiodic points in Q sit inside K, which in turn lies in a union of two -adic disks, each of olume 1. (A slightly different condition holds at = 2,.) For good, a single such disk suffices. The bound of O(2 s ) then follows naturally. In this paper, we will still work only with polynomials and only in dimension 1, but the degree d 2 and global field K of definition will be arbitrary. Like Call and Goldstine, we will study dynamics oer the associated complete alued fields C. We refer the reader to [4, 10, 21] for expositions on complex dynamics (where is archimedean and C = C), and to [5, 6, 8, 17, 30, 31, 33] for papers exploring arious aspects of the newer realm of p-adic and non-archimedean dynamics. By a detailed analysis of the filled Julia sets K C, we will obtain the following substantial improement oer the results of [9]. Main Theorem. Let K be a global field, let φ K[z] be a polynomial of degree d 2, and let s be the number of bad primes of φ in K. Then the number of preperiodic points of φ in P 1 (K) is at most O(s log s). A more precise statement appears in Theorem 7.1; the big-o constant is essentially (d 2 2d + 2)/ log d. The idea of the proof is to consider, for each M K, the product P = i j x i x j, where {x 1,..., x } are finite K-rational preperiodic points of φ. (The product P is related to transfinite diameters and capacities, as discussed in Section 4.) If naiely; but in Lemmas 3.4.a and 4.1, we obtain P r (d 1) log d, with some correction factors for archimedean. The key, howeer, is our treatment of the prime w with filled Julia set K w of the largest diameter. We partition K w into two pieces, and we show in Lemmas 3.4.b, 5.1, and 6.3 that r is the diameter of the filled Julia set K, then P r ( 1)

3 PREPERIODIC POITS 3 the corresponding product P w on each piece satisfies P w r (d 1)(log d w A+B) for certain simple constants A and B. The product P of all the P s, restricted to preperiodic points in the gien piece of K w, is then bounded by r w (d 1)E, where E = s log d A + B. For slightly larger than (1/A)s log d s (see Lemma 3.5), we get E < 0, so that P < 1, which contradicts the product formula for the global field K. Thus, we get a bound of about (1/A)s log d s on each piece; summing the two bounds gies the Theorem. Of course, the details are complicated. In Sections 1 and 2, we will set terminology and recall fundamental facts concerning local and global fields, bad primes, and filled Julia sets. In Section 3, we will introduce notation for certain expressions that will arise later, and we will bound these expressions in a series of technical but completely elementary Lemmas. In Section 4, we will discuss transfinite diameters and proe our first nontriial bound for P, for general bad primes. In Sections 5 and 6, we will describe the partition of the filled Julia set at a bad prime. Finally, in Section 7, we will state Theorem 7.1 and combine all the results from the preceding sections to proe it. The author would like to thank Laura DeMarco, Andrew Granille, Jonathan Lubin, Bjorn Poonen, Joseph Silerman, and Daniel Velleman for a number of helpful conersations. Many thanks also to Matthew Baker for suggesting an improement to the archimedean case of Lemma 4.1, and for other comments and stimulating discussions. 1. Global Fields and Local Fields In this section we present the necessary fundamentals from the theory of local and global fields. We also set some notational conentions for this paper. Although this material is well known to number theorists, we present it for the conenience of dynamicists. See Section B.1 of [16] or Section 4.4 of [29] for more details on global fields and sets of absolute alues; see [14, 18] for expositions concerning the local fields C Global fields and absolute alues. Throughout this paper, K will denote a global field. That is, K is either a number field (i.e., a finite extension of Q) or a function field oer a finite field (i.e., a finite extension of F p (T ) for some prime p). We will write M K for the set of standard absolute alues on K. That is, M K consists of functions : K R satisfying x 0 (with equality if and only if x = 0), xy = x y, and x + y x + y, for all x, y K. (We will frequently abuse notation and write M K when our meaning is clear.) Moreoer, the absolute alues in M K are chosen to satisfy a product formula, which is to say that for each M K, there is an integer n 1 such that for all x K, (1) = 1. M K x n Implicit in the product formula is the fact that for any x K, we hae x = 1 for all but finitely many M K. All but finitely many M K satisfy the ultrametric triangle inequality x + y max{ x, y }. (ote that n 1 for all n Z.) Such are called non-archimedean absolute alues; the finitely many exceptions are called archimedean absolute alues. The non-archimedean absolute alues in M K correspond to prime ideals of the ring of integers of K. Hence, we often refer to the absolute alues M K as primes of K, een when is archimedean.

4 4 ROBERT L. BEEDETTO If K is a function field, then all absolute alues are non-archimedean. If K is a number field, then there are archimedean absolute alues, each of which, when restricted to Q, is the familiar absolute alue, commonly denoted ; we write, and we hae (2) n = [K :Q]. M K, If is non-archimedean, then K is a discrete subset of R, and we say that is a discrete aluation on K. In that case, choose π K such that π (0, 1) is the largest absolute alue less than 1 attained in K. Then π is called a uniformizer of K at, and we hae K = { π m : m Z}. Moreoer, if K is a number field, then π n = p f for some prime number p Z and some positie integer f, and restricted to Q is the usual p-adic absolute alue on Q. In this case, we say that lies aboe p Local fields. For each M K, we can form the completion K (often called the local field at ) of K with respect to. We write C for the completion of an algebraic closure K of K. (The absolute alue extends in a unique way to K and hence to C.) The field C is then a complete and algebraically closed field. If is archimedean, then K is isomorphic either to R (in which case we call a real prime) or to C (in which case we call a complex prime), and C = C. We will henceforth aoid the notation K, as we will soon introduce the notation K to denote a completely different object in Section 2. If is non-archimedean, then C is not locally compact, but it has other conenient properties not shared by C. In particular, the disk O = {c C : c 1} forms a ring, called the ring of integers, which has a unique maximal ideal M = {c C : c < 1}. The quotient k = O /M is called the residue field of C. The natural reduction map from O to k, sending a O to a = a + M k, will be used to define good and bad reduction of a polynomial in Definition 2.1 below; but after proing a few simple Lemmas about good and bad reduction, we will not need to refer to O, M, or k again Disks. Let C be a complete and algebraically closed field with absolute alue. Gien a C and r > 0, we write D(a, r) = {x C : x a r} and D(a, r) = {x C : x a < r} for the closed and open disks, respectiely, of radius r centered at a. ote our conention that all disks hae positie radius. If is non-archimedean and U C is a disk, then the radius of U is unique; it is the same as the diameter of the set U iewed as a metric space. Howeer, any point b U is a center. That is, if b a r, then D(a, r) = D(b, r), and similarly for open disks. It follows that two disks intersect if and only if one contains the other. Still assuming that is non-archimedean, the set C of absolute alues actually attained by elements of C is usually not all of (0, ). As a result, if r (0, ) \ C, then D(a, r) = D(a, r) for any a C. Howeer, if r C, then D(a, r) D(a, r). 2. Bad Reduction and Filled Julia Sets The following definition originally appeared in [23]. We hae modified it slightly so that bad reduction now means not potentially good, as opposed to not good.

5 PREPERIODIC POITS 5 Definition 2.1. Let C be a complete, algebraically closed non-archimedean field with absolute alue, ring of integers O = {c C : c 1}, and residue field k. Let φ(z) C (z) be a rational function with homogenous presentation φ ([x, y]) = [f(x, y), g(x, y)], where f, g O [x, y] are relatiely prime homogeneous polynomials of degree d = deg φ, and at least one coefficient of f or g has absolute alue 1. We say that φ has good reduction at if f and g hae no common zeros in k k besides (x, y) = (0, 0). We say that φ has potentially good reduction at if there is some linear fractional transformation h PGL(2, C ) such that h 1 φ h has good reduction. If φ does not hae potentially good reduction, we say it has bad reduction at. aturally, for f(x, y) = d i=0 a ix i y d i, the reduction f(x, y) in Definition 2.1 means d i=0 a ix i y d i. By conention, if C = C is archimedean, we declare all rational functions in C (z) to hae bad reduction. In this paper, we will consider only polynomial functions φ of degree at least 2; that is, φ(z) = a d z d + + a 0, where d 2, a i C, and a d 0. If C is non-archimedean, then, it is easy to check that φ has good reduction if and only if a i 1 for all i and a d = 1. In particular, by the product formula, if φ K[z] for a global field K, then there can be only finitely many primes M K at which φ has bad reduction. The main focus of our inestigation will be filled Julia sets. The motiating idea is that for a polynomial φ, all of the interesting dynamics inoles points that do not escape to the attracting fixed point at under iteration. We rephrase the standard definition from complex dynamics more generally to include the non-archimedean setting, as follows. Definition 2.2. Let C be a complete, algebraically closed field with absolute alue, and let φ(z) C [z] be a polynomial of degree d 2. The filled Julia set of φ at is K = {x C : { φ n (x) } n 1 is bounded}. We note four fundamental properties of filled Julia sets. First, K is inariant under φ; that is, φ 1 (K ) = φ(k ) = K. Second, all the finite preperiodic points of φ (that is, all the preperiodic points in P 1 (C ) other than the fixed point at ) are contained in K. Third, if U 0 is a disk containing K, then K = n 0 φ n (U 0 ). Finally, if the polynomial φ C [z] has good reduction, then K = D(0, 1). Filled Julia sets hae been studied extensiely in the archimedean case C = C. If φ d (z) = z d, then the (complex) filled Julia K of φ d is simply the closed unit disk D(0, 1). Meanwhile, since the degree d Chebyshe polynomial ψ d satisfies ψ d h = h φ d, where h(z) = z + 1/z, it follows that the complex filled Julia set of ψ d is the interal [ 2, 2] in the real line. These two examples are misleadingly simple, howeer; most filled Julia sets are complicated fractal sets. For example, for c > 2, the filled Julia set of φ(z) = z 2 + c is homeomorphic to the Cantor set. For more complex examples (sometimes of the Julia set, which is the boundary of the filled Julia set), see [4, 10, 21]. On the other hand, while complex filled Julia sets are always compact, their nonarchimedean counterparts are not usually compact. Fortunately, this technicality will not be an obstacle for our inestigations. For the conenience of the reader, we present a few examples of non-archimedean filled Julia sets here. More examples may be found in [6, 30].

6 6 ROBERT L. BEEDETTO Example 2.3. Gien C non-archimedean and d 2 with d 1 = 1, fix c C, and consider φ(z) = z d c d 1 z. If c 1, then φ has good reduction, and hence K = D(0, 1). Thus, we consider c > 1; let r = c and U 0 = D(0, r). ote that for x > r, we hae φ(x) = x d, so that φ n (x). That is, K U 0. The set φ 1 (0) consists of d points, all distance r from one another. Using standard nonarchimedean mapping properties (see, for example, Section 2 of [7]), it is not hard to show that φ 1 (U 0 ) consists of d disks of radius r 2 d, each centered at one of the points of φ 1 (0). Moreoer, each of these smaller disks maps one-to-one and onto U 0, and in fact φ multiplies distances by a factor of r d 1 = φ (0) on each smaller disk. It follows that U n = φ n (U 0 ) is a union of d n disks, each of radius r 1 (d 1)n. (The sets U n are nested so that each disk of U n contains exactly d disks of U n+1, arranged so that any two are the maximal distance r 1 (d 1)n apart.) It is then easy to erify that K = U n is homeomorphic to a Cantor set on d interals (which, incidentally, is homeomorphic to the standard Cantor set). Example 2.4. Gien C non-archimedean and d 3 with d 1 = 1, fix a C, and consider φ(z) = z d az d 1. If a 1, then φ has good reduction, and hence K = D(0, 1). Once again, then, we consider a > 1 and set r = a, so that K U 0 = D(0, r). This time, howeer, φ 1 (U 0 ) consists of only two disks. One, W 1 = D(a, r (d 2) ), is small and maps one-to-one onto U 0 ; but the other, W 2 = D(0, 1), is comparatiely large, and it maps (d 1)-to-1 onto U 0. Because of the fixed critical point at 0, we see that φ 2 (U 0 ) consists of two disks inside W 1 (one mapping to W 1, and the other to W 2 ), and d disks inside W 2 (d 1 mapping to W 1, and the last mapping (d 1)-to-1 to W 2 ). In general, each U n = φ n (U 0 ) will be a union of disks. Each disk of U n 1 has one preimage in U n inside W 1 and (with one exception) d 1 preimages inside W 2. The exception is the disk D n 1 of U n 1 containing 0; it has only one preimage D n inside W 2, mapping (d 1)-to-one onto D n 1. Ultimately K consists of the disk V = D(0, r 1/(d 2) ) and all of its preimages together with a aguely Cantor-like set at which the preimages of V accumulate. Thus, in contrast with Example 2.3, K is neither a disk nor compact. In general, the filled Julia set of a polynomial of bad reduction oer C will look something like this one. Howeer, the dynamics can be een more complicated when there are regions on which φ maps n-to-1 for some integer n diisible by p, the characteristic of the residue field k. The preceding comments and examples made frequent reference to disks U 0 containing K. The smallest such disk will be of particular importance to us. The following Lemma shows the existence of the smallest disk and gies a partial characterization of it. Lemma 2.5. Let C be a complete, algebraically closed field with absolute alue. Let φ C [z] be a polynomial of degree d 2 with lead coefficient a d C. Denote by K the filled Julia set of φ in C. Then: a. There is a unique smallest disk U 0 C which contains K. b. U 0 is a closed disk of some radius r C, with r a d 1/(d 1). c. If is non-archimedean, then φ has potentially good reduction if and only if r = a d 1/(d 1). In that case, K = U 0. Proof. Choose α C such that α d 1 = a d, and let ψ(z) = αφ(α 1 z), which is a monic polynomial with filled Julia set αk. Gien the scaling factors of a d 1/(d 1) = α 1 in parts (b) and (c), we may assume without loss that a d = 1.

7 PREPERIODIC POITS 7 If C is archimedean, then C = C. In that case, K C is a compact set. (See, for example, Lemma 9.4 of [21].) Since K is bounded, there is a unique smallest disk U 0 containing K. (See, for example, Exercise 3 in Appendix I of [34].) Moreoer, because K is compact, this disk must be closed. The filled Julia set of a monic polynomial oer C has capacity 1; see, for example, Theorem 4.1 of [3]. Meanwhile, the capacity of the disk U 0 is exactly its radius r. Since U 0 K, we must hae r 1, proing the Lemma in the archimedean case. (See Remark 2.6 below for an alternate proof not using capacity theory.) If C is non-archimedean, then let b C be a fixed point of φ. (Such b exists because φ(z) z is a polynomial of degree d 2.) Clearly b K. By the coordinate change z z + b, we may assume that b = 0. Write φ(z) = z d + a d 1 z d a 1 z. Let r = max{ a i 1/(d i) : i = 1,..., d 1} and r = max{ r, 1}; note that r C. If r = 1, then φ is a monic polynomial with coefficients in O. Hence, φ has good reduction; with U 0 = D(0, 1), the Lemma follows. For the remainder of the proof, assume r > 1. Then the ewton polygon (see Section 6.5 of [14] or Section IV.3 of [18]) for the equation φ(z) = 0 shows that there is some c C with c = r and φ(c) = 0. In particular, any disk containing K must contain D(0, r ). Moreoer, if z > r, then the z d term has larger absolute alue than any other term of φ(z), so that φ(z) = z d. By induction, φ n (z) = z dn for all n 1. It follows that K D(0, r ). By the preious paragraph, D(0, r ) is the smallest disk U 0 containing K. On the other hand, for any x C with r < x r, d all preimages of x lie in U 0. Thus, K U 0. Howeer, if φ had potentially good reduction, then K would be a disk, since a polynomial of good reduction has filled Julia set equal to D(0, 1). Since U 0 is the smallest disk containing K, φ does not hae potentially good reduction. Remark 2.6. The fact that r 1 in the archimedean case can also be proen directly, without reference to the power of capacity theory. The following alternate argument was suggested to the author by Laura DeMarco. Suppose that K D(a, r ) for some r < 1; let s = (1 + r )/2. Since φ is a (monic) polynomial of degree at least 2, there is some radius R > 1 such that for all z C with z a > R, we hae φ(z) a > R. Let A = {z C : s z a R}. Eery point of the annulus A is attracted to under iteration of φ. Since A is compact, there is some n 1 such that f(z) = φ n (z) a has f(z) > 1 for all z A. ote that all d n zeros of f lie in D(a, s). Let g(z) = (z a) dn f(z), which is a polynomial of degree strictly less than d n. Howeer, for all z C with z a = s, we hae f(z) + g(z) = z a dn = s dn < 1 < f(z). By Rouché s Theorem (noting that g(z) 0 for z a = s), f and g hae the same number of zeros in D(a, s), counting multiplicity. That is a contradiction; thus, r 1. The next two Lemmas gie slightly more detailed information about the filled Julia set for a polynomial of bad reduction oer a non-archimedean field. Lemma 2.7. With notation as in Lemma 2.5, suppose that C is non-archimedean and r > a d 1/(d 1). Then φ 1 (U 0 ) is a disjoint union of closed disks D 1,..., D l U 0, where 2 l d. Moreoer, there are positie integers d 1,..., d l with d d l = d such that for each i = 1,..., l, φ maps D i d i -to-1 onto U 0. That is, φ(d i ) = U 0, and eery point U 0 has exactly d i pre-images in D i, counting multiplicity.

8 8 ROBERT L. BEEDETTO Proof. By the proof of Lemma 2.5 in the r > 1 case, we hae φ 1 (U 0 ) U 0. We now construct the disks D 1,..., D l inductiely. For each i = 1, 2,..., suppose we already hae D 1,..., D i 1, and choose b i φ 1 (U 0 ) \ (D 1 D i 1 ). (If this is not possible, then skip to the next paragraph.) By Lemmas 2.3 and 2.6 of [7], there is a unique disk D i containing b i which maps onto U 0, and this disk must be closed. Since D j was also unique for each j < i, the new disk D i must be disjoint from D j. In addition, by Lemma 2.2 of [7], φ maps D i d i -to-1 onto U 0, for some integer d i 1. This process must stop with l d, because for any a U 0, φ 1 (a) consists of exactly d points, counting multiplicity, and since each d i 1, at least one must be contained in each D i. Counting elements of φ 1 (a) also shows that d d l = d. Finally, if l = 1, then D 1 = φ 1 (U 0 ) U 0 is a single disk. Thus, K = φ 1 (K ) D 1 ; but U 0 was by definition the smallest disk containing K. Hence, we must hae l 2. Lemma 2.8. Let K be a field with a discrete aluation, and let π K be a uniformizer at. Let C be the completion of an algebraic closure of K. Let φ(z) = a d z d + +a 0 K[z] be a polynomial of degree d 2. Denote by K the filled Julia set of φ in C, and let r > 0 be the radius of the smallest disk in C containing K. Suppose that r > a d 1/(d 1). If K K, then a d 1/(d 1) r { π 1 if d = 2, π 1/[(d 1)(d 2)] if d 3. Proof. Gien b K K, we may replace φ by φ(z + b) b K[z], which is a polynomial with the same degree and lead coefficient as φ, but with filled Julia set translated by b. In particular, the radius r is presered; so we may assume without loss that 0 K. As in the proof of Lemma 2.5, choose α C such that α d 1 = a d, and let d 1 ψ(z) = αφ(α 1 z) = z d + α 1 i a i z i. Then ψ is a monic polynomial with filled Julia set K = αk ; hence, the radius r of the smallest disk containing K satisfies r > 1. Howeer, ψ may not be defined oer K. Let j be the largest index between 0 and d 1 that maximizes λ j = α 1 j a j 1/(d j). ote that λ j > 1; for if not, then ψ has good reduction, contradicting Lemma 2.5.c. The ewton polygon for the equation ψ(z) = 0 shows that there is some β C with ψ(β) = 0 and β = λ j. We hae 0, β K ; hence, r λ j. If j = 0, then a simple induction shows that ψ n (0) = αa 0 dn 1 for n 1. Since αa 0 > 1, this contradicts the hypothesis that 0 K. Thus, 1 j d 1, and we write a d = π e 1 and a j = π e 2 ; note that e 1, e 2 Z. Our assumptions say that r λ j = α 1 j a j 1/(d j) = π f > 1, where f = 1 ( ) (1 j)e1 + e 2 < 0. d j d 1 If j = 1, then f = e 2 /(d 1) 1/(d 1), which proes the Lemma for d = 2. If 2 j d 1, then f 1/[(d 1)(d j)] 1/[(d 1)(d 2)], and we are done. Remark 2.9. The bounds of Lemma 2.8 are sharp. Indeed, one can check that they are attained by φ(z) = z 2 π 1 z for d = 2 and by φ(z) = π d z d π z 2 for d 3. i=0

9 PREPERIODIC POITS 9 3. Elementary Computations We will now define and bound certain integer quantities that will appear as exponents in the rest of the paper. The reader is encouraged to read the statements of Definition 3.1, Lemma 3.4, and Lemma 3.5 but to skip the proofs, which are tedious but completely elementary, until after seeing their use in Theorem 7.1. We will write log d x to denote the logarithm of x to base d. Definition 3.1. Let 0 and d 2 be integers. We define E(, d) to be twice the sum of all base-d coefficients of all integers from 0 to 1. That is, ( 1 M ) M E(, d) = 2 e(j, d), where e c i d i, d = c i, j=0 for c i {0, 1,..., d 1}. Moreoer, if m is an integer satisfying 1 m d, we may write = c 0 +mk for unique integers c 0 {0, 1,..., m 1} and k 0. We then define and e(, m, d) = c 0 + e(k, d) (d m)k and f(, m, d) = c 0 + e(k, d), 1 1 E(, m, d) = 2 e(j, m, d) and F (, m, d) = 2 f(j, m, d). j=0 We declare E(, d) = E(, m, d) = F (, m, d) = 0 for 1. Clearly, E(, d) and F (, m, d) are always positie for 1; but for large and m < d, E(, m, d) is negatie. ote that e(, d, d) = f(, d, d) = f(, 1, d) = e(, d), and therefore (3) E(, d, d) = F (, d, d) = F (, 1, d) = E(, d). We will need the following two auxiliary Lemmas. Lemma 3.2. Let, m, d be integers satisfying 1, d 2, and 1 m d. Write = c + mk with 0 c m 1 and k 0. Then: a. F (, m, d) = (m c)e(k, d) + ce(k + 1, d) + (m 1) c(m c). (d m) b. E(, m, d) = F (, m, d) m [ 2 m + c(m c)]. c. If m, then E(, m, d) = F (, m, d) = ( 1). Proof. Writing an arbitrary integer j 0 as j = i + ml for 0 i m 1, we compute 1 c 1 k m 1 k 1 F (, m, d) = 2 f(j, m, d) = 2 f(i + ml, m, d) + 2 f(i + ml, m, d) j=0 c 1 = 2 i=0 i=0 k (i + e(l, d)) + 2 l=0 l=0 m 1 i=c i=0 k 1 (i + e(l, d)) l=0 c 1 m 1 = [2(k + 1)i + E(k + 1, d)] + [2ki + E(k, d)] i=0 i=c = ce(k + 1, d) + (m c)e(k, d) + (k + 1)c(c 1) + km(m 1) kc(c 1). i=c j=0 l=0 i=0

10 10 ROBERT L. BEEDETTO Part (a) of the Lemma now follows by rewriting the last three terms as c(c 1) + mk(m 1) = c(c m) + (c + mk)(m 1) = (m 1) c(m c). ext, we compute [ ] 1 1 k 1 c 1 E(, m, d) = 2 e(j, m, d) = 2 f(j, m, d) 2(d m) m l + k j=0 j=0 = F (, m, d) k(d m)[m(k 1) + 2c] Writing k = ( c)/m, the last term becomes ( c) m) (d m)( + c m) = (d m m [ 2 m + c(m c)], proing part (b). Finally, part (c) is immediate from the fact that e(j, m, d) = f(j, m, d) = j for 0 j m 1. Lemma 3.3. Let, m, d be integers satisfying 1, d 2, and 1 m d. Write = c + mk with 0 ( c m ) 1 and ( k 0. Then: ) mk mk + m a. (m c) log d + c log d 0. ( ) mk + m b. If d, then (d 1) log d (m c) 0. Proof. The function log d (x) is of course concae down. Letting x 1 = mk/ and x 2 = (mk + m)/, then, we hae x 1 1 < x 2, and therefore log d (1) L(1), where 1 L(x) = [(x 2 x) log x 2 x d (x 1 ) + (x x 1 ) log d (x 2 )] 1 is the line through (x 1, log d (x 1 )) and (x 2, log d (x 2 )). That is, 0 1 [ ( ) ( )] mk mk + m (m c) log m d + c log d, proing part (a). For part (b), we hae ( ) mk + m (d 1) (d 1) log d = log d ( log 1 + m c ) l=0 (d 1) log d j=0 (m c). Howeer, log d = log[1 (d 1)/d] (d 1)/d, and since d, ( ) mk + m d (d 1) log d (d 1) d 1 m c = d (m c) (m c). Lemma 3.4. Let, m, d be integers satisfying 1, d 2, and 1 m d 1. Then: a. E(, d) (d 1) log[ d, with equality if is a power of ] d. b. E(, m, d) (d 1) log d + 1 log d m, with equality if /m is a power of d. c. F (, m, d) (d 1) log d. d. For m, F (, m, d) (d 1) /m is a power of d. (d m) m(d 1) [ log d log d m + m 1 ], with equality if d 1

11 PREPERIODIC POITS 11 Proof. Fix d 2. If = 1, then both sides of part (a) are clearly zero. If 2 d, then E(, d) = ( 1) by Lemma 3.2.c (with m = d) and equation (3). Because (log x)/(x 1) is a decreasing function of the real ariable x > 1, we hae (log d)/(d 1) (log )/( 1), with equality for = d. Part (a) then follows for 1 d. For d + 1, we proceed by induction on, assuming part (a) holds for all positie integers up to 1. Write = c + dk, where 0 c d 1, so that 1 k 2. By Lemma 3.2.a (with m = d) and equation (3), we hae E(, d) = (d c)e(k, d) + ce(k + 1, d) + (d 1) c(d c) (d c)(d 1)k log d k + c(d 1)(k + 1) log d (k + 1) + (d 1) c(d c) = (d c)(d 1)k log d (dk) + c(d 1)(k + 1) log d (dk + d) c(d c) where the final equality is because = (d c)k + c(k + 1), and the inequality (which is equality if is a power of d) is by the inductie hypothesis, since k, k More generally, adding and subtracting (d 1) log d, we hae ( ) dk E(, d) (d 1) log d + (d c)(d 1)k log d ( ) dk + d + c(d 1)(k + 1) log d c(d c) [ ( ) ] dk + d = (d 1) log d + c (d 1) log d (d c). [ ( ) ( )] dk dk + d + (d 1)k (d c) log d + c log d By Lemma 3.3 with m = d, the quantities in brackets are nonpositie, and part (a) follows. If m = 1, then parts (c d) are immediate from part (a) and equation (3). Moreoer, by Lemma 3.2.a b (with m = 1) and part (a), E(, 1, d) = E(, d) (d 1)( 1) (d 1)[log d + 1 ], with equality if is a power of d. This is exactly part (b) for m = 1. Thus, we may assume for the remainder of the proof that 2 m d. We now turn to part (d). If = m, then by Lemma 3.2.c, we hae F (m, m, d) m(m 1), which exactly equals the desired right hand side. For m + 1, write = c + mk, where k 1 and 0 c m 1. By Lemma 3.2.a, (4) F (, m, d) = (m c)e(k, d) + ce(k + 1, d) + (m 1) c(m c). If m + 1 d 1, then k d 1, so that by Lemma 3.2.c, equation (4) becomes F (, m, d) = (m c)k(k 1) + ck(k + 1) + (m 1) c(m c) = mk 2 mk + 2ck + (m 1) c(m c) = (m + k 1) (m c)(k + c) = m 1 [ ( + m 2 c m) (m c)( c + cm) ] = m 1 [ ( + m 2 2m) c(m c)(m 1) ] [(/m) + m 2], where we hae substituted k = ( c)/m along the way. Thus, we must show [(/m) + m 2] [(d 1) log d (/m) + m 1].

12 12 ROBERT L. BEEDETTO Equialently, we must show log d d 1 log(/m) (/m) 1, which is true because (log x)/(x 1) is a decreasing function of x > 1, and 1 < /m < d. If d in part (d), then k 1. Applying part (a) to equation (4), we obtain (5) F (, m, d) (m c)(d 1)k log d k + c(d 1)(k + 1) log d (k + 1) + (m 1) c(m c), with equality if c = 0 and k is a power of d, whence we immediately obtain the statement of the Lemma for = md i. More generally, (5) becomes ( F (, m, d) (d 1) log d m + m 1 ) ( ) mk + (m c)(d 1)k log d 1 d ( ) mk + m + c(d 1)(k + 1) log d c(m c) ( = (d 1) log d m + m 1 ) [ ( ) ] mk + m + c (d 1) log d 1 d (m c) [ ( ) ( )] mk mk + m + (d 1)k (m c) log d + c log d. Part (d) now follows from Lemma 3.3, as before. For part (c), if 1 m, then by part (a) and Lemma 3.2.c, F (, m, d) = ( 1) = E(, d) (d 1) log d, as desired. The remaining case, that m, will follow from part (d) proided m 1 (d 1) log d m. Howeer, this is the same as showing that log d/(d 1) log m/(m 1), which once again follows from the fact that (log x)/(x 1) is decreasing for x > 1. Last, we turn to part (b). If 1 m, then E(, m, d) = ( 1) by Lemma 3.2.c. Thus, we wish to show that ( ) 1 (d 1) 1 + log d (d m) m m, which is to say ( ) d d m 1 (d 1) log d, m with equality when = m. Yet again, this inequality follows immediately from the facts that (log x)/(x 1) is decreasing for x > 1 and that 1 d/m d. It only remains to consider m + 1. Writing = c + mk, where k 1 and 0 c m 1, and inoking Lemma 3.2.b, we hae (d m) E(, m, d) = F (, m, d) m [ 2 m + c(m c)] (d m) F (, m, d) m 2 + (d m),

13 PREPERIODIC POITS 13 with equality for c = 0. By part (d), we obtain [ E(, m, d) (d 1) log d log d m + m 1 ] (d m) d 1 m 2 + (d m) [ = (d 1) log d log d m + 1 d m ] m(d 1), with equality if is of the form = md i. Besides the preceding integer quantities and their bounds, we will need the following bound inoling a certain family of real-alued functions. Lemma 3.5. Let d 2 be an integer, and let A, B, t be positie real numbers such that Define η : (0, ) R by Set the real number M(A, B, t) to be Then η(x) < 0 for all x M(A, B, t). (d 1)A d B 1 and t 1. η(x) = t log d x Ax + B. M(A, B, t) = t A (log d t + log d (max{1, log d t}) + 3). Proof. By differentiating, we see that η is decreasing for x t/(a log d), and hence for x M(A, B, t). Thus, it suffices to show that η(m(a, B, t)) < 0. First, suppose that t < d. Then η(m(a, B, t)) = t log d t + t log d [ A 1 (log d t + 3) ] t log d t 3t + B = t log d [ A 1 d B 3 (log d t + 3) ] B(t 1) t log d [ (d 1)(logd t + 3)/d 2], where the inequality is because A 1 d B 1 (d 1) and B > 0, by hypothesis. Since t < d, the quantity inside square brackets is strictly less than 4(d 1)/d 2 1. Thus, η(m(a, B, t)) < t log d (1) = 0, and we are done. Second, if t d, then by a similar computation, η(m(a, B, t)) = t log d [ A 1 d B 3 u 1 (u + log d u + 3) ] B(t 1) < t log d [ (d 1)(u + logd u + 3)/(d 2 u) ] where u = log d t 1. Writing H(u) = (d 1)(u + log d u + 3)/(d 2 u), it suffices to show that H(u) 1 for u 1. Differentiating, it is easy to see that H is decreasing for such u. Since H(1) = 4(d 1)/d 2 1, we are done. 4. Transfinite Diameters and Bad Primes Gien a metric space X and an integer 2, the th diameter of X is defined to be d (X) = sup x 1,...,x X d X (x i, x j ) 1/[( 1)], i j

14 14 ROBERT L. BEEDETTO which measures the maximal aerage distance between any two of points in X. (See [15], for example, for a computation of the th diameter of the interal [0, 1].) This quantity is usually used to define the transfinite diameter of X, d(x) = lim d (X), which conerges because {d (X)} 2 is a decreasing sequence. If X is a nice enough (e.g., compact) subset of a alued field, then the transfinite diameter coincides with the Chebyshe constant and the logarithmic capacity of X; see Section 5.4 of [1], or Chapters 3 and 4 of [32]. Baker and Hsia used this equality in [3] to compute the transfinite diameter of filled Julia sets of polynomials, een when those sets are not compact. (Their result of a d 1/(d 1), where d is the degree and a d the lead coefficient of the polynomial, was already well known for C = C.) See [32] for more on transfinite diameters and capacities in C. Howeer, in this paper we will be interested in the th diameters d (X) themseles, rather than the transfinite diameter. In particular, the following Lemma contains our main bound for d (K ) ( 1), where K is the filled Julia set of a polynomial φ C [z]. The proof uses an estimate inoling an der Monde determinants similar to a bound that appears in the proof of Lemme in [1]. Lemma 4.1. Let C be a complete, algebraically closed field with absolute alue. Let φ C [z] be a polynomial of degree d 2 with lead coefficient a d C. Denote by K the filled Julia set of φ in C, and let r be the radius of the smallest disk that contains K. Set r = a d 1/(d 1) r. Then for any integer 2 and any set {x 1,..., x } K of points in K, i j x i x j a d ( 1)/(d 1) max{1, }r E(,d), where E(, d) is twice the sum of all base-d coefficients of all integers from 0 to 1, as in Definition 3.1. Proof. Choose α C such that α d 1 = a d, and let ψ(z) = αφ(α 1 z). Then ψ is a monic polynomial with filled Julia set K = αk, and the smallest disk containing K has radius r. If the Lemma holds for ψ, then for x 1,..., x K, we hae αx i K, and therefore i j x i x j = α ( 1) i j αx i αx j α ( 1) max{1, }r E(,d), as desired. Thus, it suffices to proe the Lemma in the case that φ is monic. We will now construct a sequence {f j } j=1 of monic polynomials oer C such that each f j has degree j and such that f j (x) is not especially large for any x K. First, let D(a, r ) be the smallest disk containing K, where a C and r is as in the statement of the Lemma. For any integer j 0 written in base-d notation as with c i {0, 1,..., d 1}, define j = c 0 + c 1 d + c 2 d c M d M, f j (z) = M [φ i (z) a] c i. i=0

15 PREPERIODIC POITS 15 Clearly, f j is monic of degree j. Moreoer, for x K, we hae φ i (x) K, and therefore M f j (x) r c i = r e(j,d), i=0 where e(j, d) is as in Definition 3.1. Gien x 1,..., x K, denote by V (x 1,..., x ) the corresponding an der Monde matrix (i.e., the matrix with (i, j) entry x j 1 i ). Recall that x i x j = det V (x 1,..., x ) 2. i j Because f 1 is monic, we may replace the last column of the matrix by a column with entry f 1 (x i ) in the ith row, without changing the determinant. We may then replace the second to last column by a column with entry f 2 (x i ) in the ith row, and so on. Thus, if we denote by A(x 1,..., x ) the matrix with (i, j) entry f j 1 (x i ), then det V (x 1,..., x ) = det A(x 1,..., x ). If C = C is archimedean, then by Hadamard s inequality applied to the columns of A, det A(x 1,..., x ) 2 1 j=0 ( fj (x 1 ) f j (x ) 2) 1 j=0 r 2e(j,d) = r E(,d). Similarly, if C is non-archimedean, then by the non-archimedean ersion of Hadamard s inequality (see, for example, [1], Preue du Lemme 5.3.4), we hae det A(x 1,..., x ) 2 1 j=0 max f j(x i ) 2 i=1,..., 1 j=0 r 2e(j,d) = r E(,d). Remark 4.2. We can recoer the Baker and Hsia bound d(k ) a d 1/(d 1) immediately from Lemmas 4.1 and 3.4.a. (The opposite inequality is more subtle, howeer.) Remark 4.3. There are many cases for which the bound of Lemma 4.1 is sharp. In particular, for non-archimedean, degree d 2 with d 1 = 1, and c C with c > 1, recall that the function φ(z) = z d c d 1 z of Example 2.3 has K homeomorphic to a Cantor set on d pieces. For arbitrary 2, one can distribute points in K in the following way. Write = M i=0 c id i, and put c M points in each of the d M pieces at leel M, maximally far apart in each piece; then put c M 1 in each of the d M 1 pieces at leel M 1, each as far as possible from the existing points; and so on. Keeping track of the radii of the disks at each leel, one can show that i j x i x j = r E(,d) exactly. In many other cases, howeer, the bound is not quite sharp, though it appears to be approximately the right order of magnitude. In the archimedean case, of course, the Hadamard inequality introduces some error. Still, the greater factor seems to be the choice of the monic polynomial f j. When j is a power of d, computations suggest that our choice of f j is ery close to sharp, if not actually sharp. Howeer, when j is not a power of d, our construction of f j as a product of smaller factors is in general not optimal, een in the non-archimedean setting. For example, if φ(z) = z 3 az 2 is the map of Example 2.4 (nonarchimedean, with d = 3, a > 1, and 2 = 1), then the function f 6 (z) = (φ(z)) 2 of the proof has f 6 (z) growing as large as r 2 on K ; but the function f 6 (z) = (φ(z)) (φ(z) a)

16 16 ROBERT L. BEEDETTO has f 6 (z) r. Ultimately, while the exponent E(, 3) of Lemma 4.1 is essentially 2 log 3, the actual exponent for this φ should be something more like (4/3) log 3. In the archimedean case, the Chebyshe polynomials {ψ j } j 1 proide an een stronger example of this phenomenon. More precisely, if C = C and φ(z) = ψ 2 (z) = z 2 2, then K is simply the interal [ 2, 2] in the real line. For j 1, the j th Chebyshe polynomial ψ j has ψ j 2 on K, as compared with the proof s bound of 2 c 0+c 1 + for f j. In general, howeer, knowing nothing about the polynomial other than its degree and the radius r, we cannot substantially improe on Lemma A Partition of the Filled Julia Set: on-archimedean Case The key to the Main Theorem, as described in the introduction, is to diide the filled Julia set at a particular bad prime into two smaller pieces X 1 and X 2. As a result, the product i j x i x i, when restricted to {x i } X k (for fixed k = 1, 2), will be substantially smaller than the bound of Lemma 4.1. We begin with non-archimedean primes. Lemma 5.1. Let C be a complete, algebraically closed field with non-archimedean absolute alue. Let φ C [z] be a polynomial of degree d 2 with lead coefficient a d C. Denote by K the filled Julia set of φ in C, and let r be the radius of the smallest disk U 0 that contains K. Set r = a d 1/(d 1) r, and suppose that r > 1. Then there are disjoint sets X 1, X 2 K and positie integers m 1, m 2 with the properties that X 1 X 2 = K, that m 1 + m 2 = d, that for k = 1, 2, φ : X k K is m k -to-1, and that for k = 1, 2, for any integer 2, and for any set {x 1,..., x } X k of points in X k, x i x j a d ( 1)/(d 1) r E(,m k,d), i j where E(, m k, d) is as in Definition 3.1. Proof. As in the proof of Lemma 4.1, we may assume that φ is monic. By Lemma 2.5, U 0 is a closed disk of radius r C. We may write U 0 = D(a, r ) for some point a K, since K is nonempty, and since any point of a non-archimedean disk is a center. Pick b φ 1 (a). ote that b K U 0. Write U 1 = φ 1 (U 0 ). By Lemma 2.7, U 1 = D 1 D l for some disjoint closed disks {D i }, with 2 l d. Moreoer, φ : D i U 0 maps d i -to-one for some positie integers {d i } with d d l = d. Define W 1 = {x U 1 : x b < r }, and W 2 = U 1 \ W 1, so that W 1 W 2 = and W 1 W 2 = U 1. If W 2 =, then K D(b, r ) U 0, contradicting the minimality of U 0. (The second inclusion is strict because r C.) Thus, since b W 1, both W 1 and W 2 are nonempty. Furthermore, W 1 and W 2 are both finite unions of disks D i aboe. Hence, there are integers m 1, m 2 1 so that each W k maps m k -to-one onto U 0, with m 1 + m 2 = d. Let X k = W k K for k = 1, 2. Since φ 1 (K ) = K, φ must map X k m k -to-one onto K. For any integer i 1, obsere that the polynomial φ i (z) a is monic of degree d i. Moreoer, since the equation φ i 1 (z) = a has exactly d i 1 roots (counting multiplicity), all of which lie in U 0, it follows that φ i (z) = a has m 1 d i 1 roots in W 1 and m 2 d i 1 roots in W 2, counting multiplicity. Thus, we may write φ i (z) a = g i (z)h i (z)

17 PREPERIODIC POITS 17 where g i is monic of degree m 1 d i 1 with all its roots in W 1, and h i is monic of degree m 2 d i 1 with all its roots in W 2. In addition, define g 0 (z) = h 0 (z) = z a. We will now use the polynomials g i to compute the bounds gien in the Lemma for X 1 ; the proof for X 2 is similar, using h i. To simplify notation, write X = X 1 and m = m 1. For any integer j 0, write j = c 0 + mk, and write k in base-d notation, so that j = c 0 + m(c 1 + c 2 d + c 3 d c M d M 1 ), with c 0 {0, 1,..., m 1}, and with c i {0, 1,..., d 1} for i 1. Define M f j (z) = [g i (z)] c i. i=0 Clearly, f j is monic of degree j. Meanwhile, for x X and i 1, obsere that φ i (x) K, and therefore φ i (x) a r. On the other hand, all roots of h i lie in W 2, which is distance r from x; therefore, h i (x) = r (d m)di 1. It follows that g i (x) r 1 (d m)di 1 for all i 1. In addition, since X U 0, we hae g 0 (x) r. Thus, M f j (x) r c 0 r c i(1 (d m)d i 1 ) = r, e i=1 where e = e(j, m, d) in the notation of Definition 3.1. By the same an der Monde determinant argument as in the proof of Lemma 4.1, it follows that if 2 and x 1,..., x X, then x i x j r E(,m,d). i j Remark 5.2. In some cases, K can be split into more than two pieces, each much smaller than the X 1, X 2 of Lemma 5.1. For example, the filled Julia set of the map φ(z) = z d c d 1 z of Example 2.3 breaks naturally into d pieces. Adapting the method of the Lemma for each piece, we could ultimately replace the coefficient d 2 2d + 2 in Theorem 7.1 by d. Howeer, as preiously noted, most polynomials are not so simple. Indeed, the filled Julia set of φ(z) = z d az d 1 from Example 2.4 splits into only two pieces. (Of course, if we take a higher preimage U n in that example, we get more than two pieces; but because of the large radii, there appears to be no improement gained by using n > 1.) Een an application of the arguments of Remark 4.3 would result in only a slight decrease in the coefficient of log d in the exponent (cf. Lemma 3.4.b). Unfortunately, a real improement would require an increase in the size of the (negatie) coefficient of 2, not the log d term. 6. A Partition of the Filled Julia Set: Archimedean Case The final tool needed for Theorem 7.1 is an archimedean analogue of Lemma 5.1. Roughly the same argument works, but only if the diameter of the filled Julia set K is large enough. This phenomenon is familiar to complex dynamicists. For example, gien φ(z) = z 2 + c C[z], if the diameter of K is small, then c lies in the Mandelbrot set, in which case K is connected. Howeer, once the diameter is large enough, c leaes the Mandelbrot set and K becomes disconnected. In fact, as the diameter grows, the arious pieces of K shrink.

18 18 ROBERT L. BEEDETTO We begin with the following preliminary result. Lemma 6.1. Let φ C[z] be a polynomial of degree d 2 with lead coefficient a d C. Denote by K the filled Julia set of φ in C, and let U 0 = D(a, r ) be the smallest disk that contains K. Set r = a d 1/(d 1) r, and suppose that r > { 3, if d = 2, or 2 + 3, if d 3. Then K is contained in a union of d open disks of radius a d 1/(d 1). Proof. As in the proof of Lemma 4.1, we may assume that φ is monic. Denote by b 1,..., b d the (possibly repeated) roots of φ(z) = a, and let D(c, s) be the smallest disk containing b 1,..., b d. (Here, we break our conention and allow s = 0 if b 1 = = b d.) Our first goal is to show that s r 1. Because K is not contained in D(c, r), there must be some y 0 K such that y 0 c r. Let Y = D(c, s) D(y 0, y 0 c ). We claim that Y is contained in a disk of radius strictly less than s (or that Y is empty, if s = 0). Indeed, if y 0 c < s, then Y D(y 0, y 0 c ) triially. Otherwise, y 0 c s, and since the center c of the first disk lies on the boundary of the second, the intersection Y is contained in a strictly smaller disk. (For example, center the new disk at the midpoint of the two intersection points of the two boundary circles.) By the minimality of s, then, not all of b 1,..., b d can be in Y. Thus, there is some 1 i d such that y 0 b i y 0 c r. Without loss, assume that y 0 b 1 r. For all i 2, we hae y 0 b i r s, because b i D(c, s). Since y 0 K, we hae φ(y 0 ) K, and therefore φ(y 0 ) a r. If r s 0, then, we hae r φ(y 0 ) a = d d y 0 b i = y 0 b 1 y 0 b i r (r s) d 1, i=1 i=2 from which we obtain r s 1. Regardless of the sign of r s, then, we hae s r 1. Re-index {b 1,..., b d } (possibly changing the preious role of b 1 ) so that b 1 and b d are distance max{ b i b j } apart, and so that for all i = 1,..., d 1, we hae b i+1 b 1 b i b 1. ote that b d b 1 3s 3(r 1); see, for example, [34], Exercise 6-1. If d = 2, we can improe this lower bound. In that case, the smallest disk containing b 1 and b 2 is the closed disk centered at (b 1 +b 2 )/2 of radius b 1 b 2 /2. That is, s = b 1 b 2 /2. It follows that b 1 b 2 = 2s 2(r 1). For all degrees d 2, we hae r > 3, so that s > 2, and therefore the two disks D(b 1, 1) and D(b d, 1) are disjoint. Moreoer, as y ranges through C \ [D(b 1, 1) D(b d, 1)], the minimum alue of y b 1 y b d is b 1 b d 1, attained at only two points, namely the point on the boundary of each disk closest to the other disk. Let U 1 = φ 1 (U 0 ). Since K = φ 1 (K) U 1, it suffices to show that (6) U 1 d D(b i, 1). i=1

19 PREPERIODIC POITS 19 If not, then there is some y U 1 \ D(b i, 1). If d 3, then by the aboe computations, we hae y b 1 y b d 3(r 1) 1. Since φ(y) U 0, we obtain d r φ(y) a = y b i ( d 1 3(r 1) 1) y b i ( 3(r 1) 1), i=1 contradicting the hypothesis that r > Similarly, if d = 2, then i=2 r φ(y) a = y b 1 y b 2 > 2(r 1) 1, contradicting the hypothesis that r > 3, and proing the Lemma. Remark 6.2. Because K is compact for archimedean, the conclusion of Lemma 6.1 implies that K is in fact contained in d closed disks of radius strictly less than a d 1/(d 1). This fact will be useful in Cases 2 and 3 of the proof of Theorem 7.1. We are now prepared to present our archimedean ersion of Lemma 5.1. Lemma 6.3. Let φ C[z] be a polynomial of degree d 2 with lead coefficient a d C. Denote by K the filled Julia set of φ in C, and let r be the radius of the smallest disk U 0 that contains K. Set r = a d 1/(d 1) r and Suppose that C d = d (d 2)/(d 1) min { 1, } 1.2. d 1 4 if d = 2 r 3 + 2(d 1), if d 3. 3 (d 1)Cd Then there are disjoint sets X 1, X 2 K and positie integers m 1, m 2 with the properties that X 1 X 2 = K, that m 1 + m 2 = d, that for k = 1, 2, φ : X k K is m k -to-1, and that for k = 1, 2, for any integer 2, and for any set {x 1,..., x } X k of points in X k, x i x j a d ( 1)/(d 1) C F (,m k,d) d (C d r) E(,mk,d), i j where E(, m k, d) and F (, m k, d) are as in Definition 3.1. Proof. As in the proof of Lemma 4.1, we may assume that φ is monic. It is easy to check that C d min{1, 1.2/(d 1)} (the closest approach for d 3 occurs at d = 5), and that the lower bound ( 3 + 2(d 1))/( 3 (d 1)C d ) (respectiely, 4) for r is greater than (respectiely, 3), so that we may inoke Lemma 6.1. Write U 0 = D(a, r), and define and order b 1,..., b d as in the proof of Lemma 6.1, so that b 1 b d 3(r 1) (or b 1 b d 2(r 1), if d = 2). If d 3, obsere that for some m = 1,..., d 1, we hae b m+1 b 1 b m b C d r. For if not, then 3(r 1) bd b 1 < (d 1) [2 + C d r], so that [ 3 (d 1)C d ]r < 3 + 2(d 1), contradicting the hypotheses. If d = 2, we hae b 2 b 1 2r r, since r 4. Let m = 1 in this case.