Introduction Main Theorem

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1 Abstract: We study the number of rational pre-images of a rational number a under the quadratic polynomial map (x) = x 2 + c. We state the existence of a uniform bound (uniform over the family of maps (x)) on the number of rational pre-images and examine effective bounds for various choices of a. We use methods from rational points on curves, Falting s Theorem, height functions, elliptic curves, and elliptic surfaces. This is a combination oompleted work, work in progress, and supervised student research.

2 On the Number of Under Quadratic Dynamical Systems Amherst College AMS Special Session: January 2010 Arithmetic and Nonarchimedean Dynamics

3 Outline 1 Introduction 1 Quadratic Dynamical Systems and Pre-images 2 Previous Results 2 1 Pre-image Surfaces 2 Geometric Argument

4 The Dynamical System Definition Consider the quadratic polynomial (x) = x 2 + c. Let f n c (x) be the n th iterate of (x). In other words Definition fc n (x) = (fc n 1 (x)). We say that x is an N th pre-image of a fixed number a, if f N c (x) = a and we denote the set of such x as fc N (a).

5 The Question Given an number field K and a quadratic polynomial (x) K [x] and an element a K. How large is the set N 1 f N c (a)(k )?

6 Example Example f 1 (x) = x 2 1 Working over Q we have the following rational pre-images: f 1 ± 2 f 1 1 f 1 f 1 0 f 1 f The set of all Q-rational pre-images of 0 by f 1 is {0, 1, 1}.

7 A Uniform Bound Theorem (Faber,H.,Ingram,Jones,Manes,Tucker,Zieve) Fix an integer d. For all but finitely many a Q, there exists an integer κ(a, d) such that for any number field K /Q of degree d # N 1 fc N (a)(k ) κ(a, d).

8 A Uniform Bound Theorem (Faber,H.,Ingram,Jones,Manes,Tucker,Zieve) Fix an integer d. For all but finitely many a Q, there exists an integer κ(a, d) such that for any number field K /Q of degree d # N 1 fc N (a)(k ) κ(a, d). We will in fact study Definition κ(a) = lim sup # fc N (a)(q). c Q N 1

9 Previous Results 1 Faber, H., Ingram, Jones, Manes, Tucker, and Zieve. Uniform Bounds on Pre-Images Under Quadratic Dynamical Systems, Math. Res. Lett., 16(1):87 101, κ(a) Faber, H. On The Number of Rational Iterated Pre-Images of the Origin Under Quadratic Dynamical Systems, submitted. κ(0) = 6. 3 Hyde. On The Number of Rational Iterated Pre-Images of 1 Under Quadratic Dynamical Systems, submitted. κ( 1) = 6.

10 Theorem (Hyde, H., Krause) For any a Q\{ 1 4 } we have κ(a) = 6.

11 Pre-Image Curves and Surfaces View (x) = x 2 + c as a mapping A 1 Q A1 Q. Definition Define the N th pre-image curve as Y pre (N, a) = V (f N c (x) a) A 2. and its projectivization X pre (N, a). Definition Allowing a to vary we can similarly define fibered surfaces Y pre (N) and X pre (N).

12 Height Argument The height argument used for a = 0 and a = 1 can only show κ(a) = 6 for a values for which X pre (3, a) is a rank 1 elliptic curve, which are (at least) a {0, 1, 1 4, 3 4, 5 4, 1 2 }.

13 Geometric Argument We check that there are only finitely many c for each a where there can be more than 6 rational pre-images, by defining a curve representing each arrangement of rational pre-images. These fall into 2 possibilities: curve curve

14 Introduction Arrangement q Defined by equations or a t s s u q t u q 2 + c = s, s 2 + c = t, t 2 + c = a, u 2 + c = t az 2 t 2 (sz q 2 ) = az 2 t 2 (tz s 2 ) = az 2 t 2 ( tz u 2 ) = 0.

15 Arrangement q Defined by equations or a t t s s q r q 2 + c = s, s 2 + c = t, t 2 + c = a, r 2 + c = s az 2 t 2 (sz q 2 ) = az 2 t 2 ( sz r 2 ) = az 2 t 2 (tz s 2 ) = 0. r

16 Genus: Arrangement Theorem (Faber,H.,Ingram,Jones,Manes,Tucker,Zieve) Suppose N N and a Q is not a critical value of f j c(0) for any 2 j N. Then Y pre (N, a) is geometrically irreducible, and the genus of X pre (N, a) is (N 3)2 N N Genus

17 Genus Theorem (Hyde, H., Krause) If the curve is nonsingular, the genus depends only on the number of points in the tree. And for 2N points the genus is Proof. (N 3)2 N These are complete intersections of N 1 degree 2 hypersurfaces.

18 Singular Points Theorem The only Q-rational singular points on the curve have a values {0, 1 4, 2}. The only Q-rational singular points on the curve have a values {0, 1 4 }. Proof. Jacobian Criterion.

19 a = 2 In this case, we have the curve generated by 2z 2 t 2 (tz s 2 ) = 0 2z 2 t 2 ( tz u 2 ) = 0 2z 2 t 2 (sz q 2 ) = 0 The singular point is (0, 2, 0, 2, 1). We can compute the genus for this curve as 4.

20 Still to do... 1 a = 1 4 (κ( 1 4 ) = 10?). 2 a Q\{ 1/4} (κ(a) = 6?).