(Non)-Uniform Bounds for Rational Preperiodic Points in Arithmetic Dynamics

Transcription

1 (Non)-Uniform Bounds for Rational Preperiodic Points in Arithmetic Dynamics Robert L. Benedetto Amherst College Special Session on The Interface Between Number Theory and Dynamical Systems AMS Spring Sectional Meeting UIUC Friday, March 27, 2009

2 Dynamics over Number Fields K = global field, N 1. φ : P N (K) P N (K) morphism over K, degree d 2. (N = 1: φ K(z) is a rational function.) Preper(φ, K) := {preperiodic points of φ in P 1 (K)}. Example. φ(z) = z , on P1 (Q) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

3 φ(z) = z φ(z) = z ±

4 Theorem (Northcott, 1950): Let K be a global field. Let φ : P N (K) P N (K) be a morphism, defined over K, of degree d 2. Then #Preper(φ, K) <. Dynamical Uniform Boundedness Conjecture (Morton & Silverman, 1994): For any integers d 2, D 1, and N 1, there is a constant C = C(d, D, N) such that for any number field K with [K : Q] = D, and for any morphism φ : P N (K) P N (K) defined over K and of degree d, #Preper(φ, K) C(d, D, N). Conjecture (DUBC Lite): There is a constant C > 0 so that for any quadratic polynomial φ Q[z], #Preper(φ, Q) C. Refined DUBC Lite Conjecture (Poonen, 1998): The DUBC Lite Constant is 9.

5 Definition. Let K be a global field, v M K non-archimedean, and φ K(z) a rational function. We say φ has good reduction at v if φ may be written in homogeneous coordinates as [f, g], i.e., ( ) x f(x, y) φ = y g(x, y) for some f, g O v [x, y] homogeneous of the same degree such that: The reductions f and g modulo v have no common zeros besides (0, 0). In that case, φ maps residue classes into residue classes. Idea: φ still makes sense everywhere modulo v.

6 Example. Any polynomial with v-adic integer coefficients and with a d v = 1 has good reduction: φ(z) = a d z d + a d 1 z d a 0 = a dx d + a d 1 x d 1 y + + a 0 y d y d. Example. φ(z) = z Q[z] has bad reduction at v = 2, 3,. Modulo 2 or 3: φ(z) = 144x2 133y 2 144y 2 y2 0. Example. ψ(z) = 3z2 8z 4z + 11 Q(z) has bad reduction at v = 3, 5, 11, 13,. [Resultant of 3z 2 8z and 4z + 11 is ] E.g. modulo 5: ψ(z) = 3x2 8xy 4xy + 11y 3x2 3xy 3x(x y) = 2 xy + y2 y(x y).

7 Theorem. (Pezda, Morton & Silverman, Zieve, 1990s). Let φ K(z) such that no iterate of φ is the identity, and with good reduction at v. Let x P 1 (K) be a K-rational periodic point of φ. Then the period n of x is at most O(Np 2 v). [Np v is the norm of the prime ideal associated to v.] Idea of Proof: Write q = #k v = p f. So Np v = qp e. WLOG x = 0. D(0, 1) is periodic of period m #P 1 (k v ) = q + 1. Let λ = (φ m ) (0). Good reduction implies λ v 1. If <, then n = m. Otherwise, let r be the order of λ k v. Note r #k v = q 1. Then n = mrp E for some integer E 0. Pezda/Zieve: E e. [In fact, much less.] Note: There is no reference to the degree d = deg φ!!! The proof works entirely in the local field K v. More work gives #Preper(φ, Q) < [ d Np2 ].

8 Definition. Let v M K, and let φ K[z] be a polynomial of degree d 2. Let C v be the completion of an algebraic closure of K v. The filled Julia set of φ at v is K φ,v := {x C v : { φ n (x) v } n 0 is bounded} Note: (1) All preperiodic points (besides ) lie in K φ,v. (2) If φ is good at v, then K φ,v = D(0, 1).

9 Theorem. (Call & Goldstine, 1997.) Let c Q and let φ(z) = z 2 + c. Let s be the number of bad primes (i.e., one plus the number of distinct primes dividing the denominator of c). Then #Preper(φ, Q) s+2 = O(2 s ) except for c = 2, with #Preper(φ, Q) = 6. Idea of Proof: 1. (p-adic dynamics step): (a.) For good primes p, we know K φ,p = D(0, 1) is disk of radius 1. (b.) For bad primes p, prove that K φ,p sits inside a union of two disks of radius 1. (Slightly different for p = 2,.) 2. (global step): In each choice of one unit disk at each prime (or interval length 1 at v = ), there is only one rational number.

10 Theorem. (RB, 2004.) Let K be a global field, and let φ(z) K[z] be a polynomial of degree d 2. Let s be the number of bad primes (i.e, not potentially good) of φ. Then ( ) d 2 #Preper(φ, K) O log d s log s. More properties of K φ,v : (1) If φ is monic, then the smallest disk U 0 := D(a, r) containing K φ,v has radius r 1. (2) If v is non-archimedean, r = 1 potentially good. (3) Define U n := φ n (U 0 ). Then U 0 U 1 U 2 and K φ,v = n 0U n. (4) If v is non-archimedean, then U n is a disjoint union of closed disks.

11 Lemma 1. Let φ C v [z] be a polynomial of degree d 2. Assume φ is monic, and let r φ,v be the radius of the smallest disk U 0 C v containing K φ,v. Given N 2, let x 1,..., x N K φ,v. Then where i j B v (N) = x i x j v B v (N) r (d 1)N log d N φ,v, { N N if v is archimedean, 1 if v is non-archimedean. Note: (1) If φ not monic, you also get a correction factor of a d N(N 1)/(d 1) on the right. (2) The N N can probably be substantially reduced (but not eliminated). (3) Otherwise, these bounds are sharp: φ(z) = z d + c.

12 Proof. Let U 0 = D(a, r φ,v ) be the smallest disk containing K φ,v. For any integer j 0, write j = c 0 + c 1 d + c 2 d c M d M in base d. (0 c i d 1.) Let M f j (z) = [φ i (z) a] c i, i=0 so that f j is a monic polynomial of degree j with for x K φ,v. f j (x) v r c 0+c 1 +c 2 + +c M φ,v Meanwhile, i j(x i x j ) = ±(det V ) 2, where V = 1 x 1 x x N x 2 x x N x N x 2 N... xn 1 N

13 Column operations (f j monic) gives det V = det A, where 1 f 1 (x 1 ) f 2 (x 1 )... f N 1 (x 1 ) A = 1 f 1 (x 2 ) f 2 (x 2 )... f N 1 (x 2 ) f 1 (x N ) f 2 (x N )... f N 1 (x N ) Hadamard: det A v N 1 j=0 (j th column) (v arch: L 2 -norm; v non-arch: L -norm.) But by properties of f j : (j th column) b v (N)r c 0+ +c M φ,v { 1 v non-archimedean, where b v (N) = N v archimedean. Hence i j x i x j v = det A 2 v B v (N) N 1 j=0 r 2(c 0+ +c M ) φ,v.

14 That is, i j x i x j v B v (N) r E(N,d) φ,v, where N 1 E(N,d) = 2 [c 0 (j) + c 1 (j) + + c M (j)] j=0 = twice the sum of all base-d coefficients of all integers from 0 to N 1. Finally, it is elementary to show that E(N, d) (d 1)N log d N.

15 Lemma 2. Let K,v, φ, d, and r φ,v be as in Lemma 1. Assume that { 4d if v is archimedean, r φ,v > 1 if v is non-archimedean. Then there are disjoint sets V 1, V 2 C v and an integer 1 m d 1 such that K φ,v = V 1 V 2, φ : V 1 K φ,v is m-to-1, φ : V 2 K φ,v is (d m)-to-1, and For any y 1 V 1 and y 2 V 2, y 1 y 2 v r φ,v { 2/d if v is archimedean, 1 if v is non-archimedean. Idea: If the diameter of K φ,v is big, then it splits into (at least) two pieces which are very far apart.

16 Lemma 3. Under the hypotheses of Lemma 2, for any x 1,..., x N V 1, where and i j B v(n) = x i x j v B v(n)r (d 1)N[log d N P m (N)] φ,v, P m (N) = d m m(d 1) N (1 log d m) { N N (d 1) P m(n) if v is archimedean, 1 if v is non-archimedean. Similar bound for V 2, with m d m. Throw in the same correction factor a d N(N 1)/(d 1) v if φ is not monic.

17 Summary of the three Lemmas: Lemma 1. For x 1,..., x N K φ,v, x i x j v rv N log N, essentially. i j That is: On average, x i x j v is only a little bigger than 1. Lemma 2. If r v is big enough, then K φ,v splits into two small pieces, V 1 and V 2, that are far apart. Lemma 3. For x 1,..., x N V 1 (or in V 2 ), x i x j v r a mn 2 v, essentially. i j That is: On average, x i x j v r a m v, which is small.

18 Theorem: Sketch of Proof. [For simplicity, assume φ is monic.] For each v M K, let R v = r n v φ,v. [Actually, adjust R v slightly at archimedean v.] Let w M K be the place for which R w is largest. Split K φ,w into V 1 and V 2. If V 1 contains N distinct rational preperiodic points x 1,..., x N V 1 C w, then by the product formula, 1 = i j x i x j nv v v M K v bad i j x i x j n v v [ [ R P m(n) w v bad R s log d N P m (N) w R log d N v ] (d 1)N v arch ] (d 1)N (B v or B v) v arch (B v or B v)

19 Since R w > 1, we only need to choose N large enough so that s log d N d m m(d 1) N + (1 log d m) < 0 to get a contradiction. Letting N be slightly bigger than m(d 1) N m = d m s log d s does the trick. Do the same for V 2. So if the total number of rational preperiodic points is at least N m + N d m, we get a contradiction. The worst case is m = 1, which gives a total number of points on the order of at most [1 + (d 1) 2 ]s log d s = (d 2 2d + 2)s log d s.

20 Let K be a number field. Heights The standard height on P N (K) is h([x 0,..., x N ]) := 1 [K v : Q v ] log max{ x 0 v,..., x N v }. [K : Q] v M K (Analogous definition for function fields.) For K = Q and for x i Z with gcd(x 0,..., x N ) = 1, we can write h([x 0,..., x N ]) = log max{ x 0,..., x N }. Key properties: If φ : P N P N is any morphism of degree d, then h(φ(x)) d h(x) is a bounded function of x. (Non-degeneracy) For global fields K and any real number B, the set of K-points of height at most B is finite.

21 Canonical Heights Given a morphism φ : P N P N defined over K of degree d 2, the canonical height for φ on P N (K) is We have: The limit converges. ĥφ h is bounded. ĥφ(φ(x)) = d ĥφ(x). ĥφ(x) 0. 1 ĥ φ (x) = lim n d nh (φn (x)). For N = 1, φ a polynomial, and x : ĥ φ (x) = 0 x K φ,v for all v M K. If x is preperiodic, then ĥ(x) = 0. For global fields, if ĥ(x) = 0, then x is preperiodic.

22 Points of Small Canonical Height a.k.a. Almost Preperiodic Points Example. φ(z) = z ĥ φ (7/) = 2 5 log 3 = , vs. h(φ) = h(181/144) = log 181 = Ratio is ĥφ(7/)/h(φ) = Example. φ(z) = z ĥ φ (153/140) = 2 10 log log 7 = , vs. h(φ) = h(36989/19600) = log = Ratio is ĥφ(153/140)/h(φ) =

23 Another Point of Small Canonical Height Example. φ(z) = 1 24 z z ĥ φ ( 7) = , vs. h(φ) = log(97) = Ratio is ĥφ( 7)/h(φ) = Conjecture. (Silverman) Let K be a number field and d 2. There is a constant C = C(K,d) such that if φ K(z) with deg φ = d, then for any non-preperiodic P P 1 (K), ĥ φ (P) Ch(φ).