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Division Polynomials and Intersection of Projective Torsion Points

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Division Polynomials and Intersection of Projective Torsion Points European Journal of Mathematics (2016) 2:644–660 DOI 10.1007/s40879-016-0111-7 RESEARCH ARTICLE Division polynomials and intersection of projective torsion points Fedor A. Bogomolov1,2 · Hang Fu1 Received: 29 February 2016 / Revised 13 June 2016 / Accepted: 23 June 2016 / Published online: 19 July 2016 © Springer International Publishing AG 2016 Abstract Given two elliptic curves, each of which is associated with a projection map that identifies opposite elements with respect to the natural group structure, we investigate how their corresponding projective images of torsion points intersect. Keywords Elliptic curve · Division polynomial · Unlikely intersection · Jordan’s totient function Mathematics Subject Classification 14H52 1 Introduction Throughout the article, let K be a field of characteristic 0, (E, O) an elliptic curve defined over K , E[n] the collection of n-th torsion points, E∗[n] the collection of torsion points of exact order n, E[∞] the collection of all torsion points, and π ∈ K (E) an even morphism of degree 2. The first author acknowledges that the article was prepared within the framework of a subsidy granted to the HSE by the Government of the Russian Federation for the implementation of the Global Competitiveness Program. The first author was also supported by a Simons Travel Grant. The second author was supported by the MacCracken Program offered by New York University. B Hang Fu [email protected] Fedor A. Bogomolov [email protected] 1 Courant Institute of Mathematical Sciences, New York University, 251 Mercer Str., New York, NY 10012, USA 2 Laboratory of Algebraic Geometry, NRU HSE, 7 Vavilova Str., Moscow, Russia 117312 123 Division polynomials and intersection... 645 Bogomolov and Tschinkel [1] observed that Theorem 1.1 If π1(E1[2]) and π2(E2[2]) are different, then the intersection of π1(E1[∞]) and π2(E2[∞]) is finite. 1 1 Proof Consider the product map π1×π2 : E1× E2 → P ×P ,let be the diagonal curve. Suppose that # π1(E1[2]) ∩ π2(E2[2]) = 0 (resp. 1, 2, 3), then the preimage −1 (π1×π2) () is a curve of genus 5 (resp. 4, 3, 2), and hence contains only finitely many torsion points of E1× E2, by Raynaud’s theorem [10]. However, we expect not only the finiteness, but also the existence of a universal bound of the cardinality of their intersection. Conjecture 1.2 sup (# π1(E1[∞]) ∩ π2(E2[∞])) < ∞, where supremum is taken over (K, E1, O1,π1, E2, O2,π2) such that π1(E1[2]) = π2(E2[2]). Here the supremum is taken over all K , but clearly we can restrict to K = Q or K = C. In Sect. 3, Theorems 3.1 and 3.2 will indicate that under some mild conditions, the cardinality is small, while Theorem 3.7 will give a construction to show that it can be at least 14. (See also Remark 3.8). The main tool to achieve these results is the explicit calculation of division polynomials, which will be introduced and developed in Sect. 2. Jordan’s totient function, as an ingredient of division polynomials, will be briefly discussed in Appendix. Calculations were assisted by Mathematica 10.0 [14]. 2 Division polynomials 2 3 2 Now let E : y +a1xy+a3 y = x +a2x +a4x +a6 be in the generalized Weierstrass form with the identity element OW = (0:1:0), then it has a canonical projection map π W : E → P1, (x, y) → x. We have the standard quantities = 2 + , = + , b2 a1 4a2 b4 2a4 a1a3 = 2 + , = 2 + − + 2 − 2, b6 a3 4a6 b8 a1a6 4a2a6 a1a3a4 a2a3 a4 = 2 − , =− 3 + − , c4 b2 24b4 c6 b2 36b2b4 216b6 =− 2 − 3 − 2 + , = 3/, b2b8 8b4 27b6 9b2b4b6 j c4 with relations = − 2, = 3 − 2. 4b8 b2b6 b4 1728 c4 c6 Traditionally, the division polynomials ψn [13] are defined by the initial values ψ1 = 1, ψ2 = 2y + a1x + a3, ψ = 3x4 + b x3 + 3b x2 + 3b x + b , 3 2 4 6 8 ψ = ψ · 2x6 + b x5 + 5b x4 + 10b x3 4 2 2 4 6 + 2 + ( − ) + ( − 2) , 10b8x b2b8 b4b6 x b4b8 b6 123 646 F.A. Bogomolov, H. Fu and the inductive formulas ψ = ψ3ψ − ψ ψ3 , , 2n+1 n n+2 n−1 n+1 n 2 ψ ψ = ψ2 ψ ψ − ψ ψ ψ2 , . 2 2n n−1 n n+2 n−2 n n+1 n 3 Notice that ψ2 = 3 + 2 + + . 2 4x b2x 2b4x b6 Since char(K ) = 0, we can eliminate b8 and the leading coefficients. Definition 2.1 Let n > 1, (A) the normalized n-th division polynomial is W fn(x) = (x − π (P)); P∈E[n]\{OW} (B) the normalized n-th primitive division polynomial is W Fn(x) = (x − π (P)). {πW(P):P∈E∗[n]} Theorem 2.2 We have the following explicit formulas: (A) ( ) = ( ) r s t d(n)−(r+2s+3t), fn x cr,s,t n b2 b4 b6 x (r,s,t) r+2s+3td(n) the degree and the first three coefficients are d(n) = n2 − 1, 2 − n 1 1 2 1 c , , (n) = = n − , 1 0 0 12 12 12 ( 2 − )( 2 + ) n 1 n 6 1 4 1 2 1 c , , (n) = = n + n − , 0 1 0 60 60 12 10 ( 2 − )( 4 + 2 + ) n 1 n n 15 1 6 1 2 1 c , , (n) = = n + n − ; 0 0 1 420 420 30 28 (B) ( ) = ( ) r s t D(n)−(r+2s+3t), Fn x Cr,s,t n b2 b4 b6 x (r,s,t) r+2s+3tD(n) 123 Division polynomials and intersection... 647 the degree and the first three coefficients are 1 D(n) = J (n) I (n), 2 2 1 C1,0,0(n) = J2(n) I (n), 24 1 1 C0,1,0(n) = J4(n) + J2(n) I (n), 120 24 1 1 C , , (n) = J (n) + J (n) I (n), 0 0 1 840 6 60 2 where 2ifn = 2, I (n) = 1ifn > 2, and 1 J (n) = nk 1 − k pk p|n is Jordan’s totient function. ( ) = ψ2( )/ 2 ψ Proof (A) Notice that fn x n x n , so the inductive formulas for n can be transformed to 3( + ) ( − )( + )3 2 n n 2 3 n 1 n 1 3 f n+ = f fn+ − fn− f 2 1 2n + 1 n 2 2n + 1 1 n+1 for n 2, and ( − )2( + ) 2 ( − )( + )2 2 2 n 1 n 2 fn−1 fn fn+2 n 2 n 1 fn−2 fn fn+1 f2n = − 4 f2 4 f2 for n 3. In order to use induction to prove the formulas for t(n) = d(n), c1,0,0(n), c0,1,0(n), and c0,0,1(n), we need to check the initial values t(n) for 1 n 4, and verify that they all satisfy n3(n + 2) (n − 1)(n + 1)3 t(2n +1) = (3t(n) + t(n +2)) − (t(n −1) + 3t(n +1)) 2n + 1 2n + 1 123 648 F.A. Bogomolov, H. Fu for n 2, and (n − 1)2(n + 2) t(2n) = 2t(n −1) + t(n) + t(n +2) − t(2) 4 (n − 2)(n + 1)2 − t(n −2) + t(n) + 2t(n +1) − t(2) 4 for n 3. All of these can be easily done. (B) By definition, 2/I (d) fn(x) = Fd (x) , d|n d =1 so for T (n) = D(n), C1,0,0(n), C0,1,0(n), and C0,0,1(n),wehave 2 t(n) = T (d). I (d) d|n d =1 Then the rest is a standard application of the Möbius inversion formula. Theorem 2.3 Let E1 and E2 be two elliptic curves in the generalized Weierstrass form, then the following are equivalent: (A) bi (E1) = bi (E2),i = 2, 4, 6; π W( ∗[ ]) = π W( ∗[ ]) (B) for any n, E1 n E2 n ; W W (C) π (E1[∞]) = π (E2[∞]); W W (D) π (E1[∞]) ∩ π (E2[∞]) is infinite; > π W( ∗[ ]) = π W( ∗[ ]) (E) for some n 1, E1 n E2 n ; ,..., > π W k ∗[ ] = π W k ∗[ ] (F) for some n1 nk 1, i=1 E1 ni i=1 E2 ni . Proof (A) ⇒ (B) ⇒ (C) ⇒ (D) ⇒ (E) ⇒ (F) is clear, where (D) ⇒ (E) is given by The- k ( ) ( ) orem 1.1. Assume (F), then E1 and E2 share the same i=1 Fni x . Since C1,0,0 n , ( ) ( ) , C0,1,0 n , and C0,0,1 n are all strictly positive, the coefficients of b2 b4, and b6 in k ( ) , the product i=1 Fni x will always be nonzero, then b2 b4, and b6 can be solved, which proves (A). Now let us go back to the general (E, O,π), and assume that π(O) =∞.Bythe Riemann–Roch theorem, there exists an isomorphism φ : E → E such that E is in the generalized Weierstrass form. Then φ induces φ ∈ Aut(P1) fixing ∞, which must be a linear function [5]. Its inverse φ −1 can be lifted to φ : E → E such that E remains in the generalized Weierstrass form. Thus the general case can be reduced to the Weierstrass case. Note that we can make everything above defined over K , except possibly φ has to be defined over a quadratic extension of K . 123 Division polynomials and intersection... 649 φ φ / / (E, O) (E , OW) (E , OW) π πW πW φ φ −1 1 / 1 / 1 (P , ∞) (P , ∞) 5(P , ∞). id Corollary 2.4 Given (E1, O1,π1) and (E2, O2,π2) such that π1(O1) = π2(O2), then the following are equivalent: π ( ∗[ ]) = π ( ∗[ ]) (A) for any n, 1 E1 n 2 E2 n ; > π ( ∗[ ]) = π ( ∗[ ]) (B) for some n 1, 1 E1 n 2 E2 n ; ,..., > π k ∗[ ] = π k ∗[ ] (C) for some n1 nk 1, 1 i=1 E1 ni 2 i=1 E2 ni ; (D) π1(E1[∞]) = π2(E2[∞]); (E) π1(E1[∞]) ∩ π2(E2[∞]) is infinite.
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