Does There Exist an Elliptic Curve E/Q with Mordell-Weil Group Z 2 Z 8 Z 4?

Transcription

1 Does There Exist an Elliptic Curve E/Q with Mordell-Weil Group Z 2 Z 8 Z 4? Edray Herber Goins Department of Mathematics, Purdue University Atkin Memorial Lecture and Workshop: over Q( 5) April 29, 2012

2 Abstract An elliptic curve E defined over the rational numbers Q is an arithmetic-algebraic object: It is simultaneously a nonsingular projective curve with an affine equation Y 2 = X 3 +AX +B, which allows one to perform arithmetic on its points; and a finitely generated abelian group E(Q) E(Q) tors Z r, which allows one to apply results from abstract algebra. The abstract nature of its rank r can be made explicit by searching for rational points (X,Y). The largest possible subgroup of an elliptic curve E is E(Q) tors Z 2 Z 8, and, curiously, these curves seem to have the least known information about the rank r. To date, there are twenty-seven known examples of elliptic curves over Q having Mordell-Weil group E(Q) Z 2 Z 8 Z 3, yet no larger rank has been found. In this talk, we give some history on the problem of determining properties of r and analyze various approaches to finding curves of large rank.

3 Outline of Talk Motivation 1 Motivation Challenge Problem Elliptic Integrals Addition Formulas 2 Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 3 Ranks of y 2 = (1 x 2 )(1 k 2 x 2 )

4 Challenge Problem Motivation Challenge Problem Elliptic Integrals Addition Formulas Modern Language E : y 2 = x 3 + ( 5 5 ) x 2 + 5x The curve has invariant j(e) = The curve has conductor f E = p 6 2 p2 5 in terms of the prime ideals p 2 = 2Z[ϕ] and p 5 = 5 Z[ϕ], where ϕ = This curve is 2-isogeneous to (a quadratic twist of) its Galois conjugate. Theorem (G, 1999) The elliptic curve E is modular. More precisely, there is a modular form f(q) S 2 ( Γ0 (160),ǫ ) and a Dirichlet character χ : Z[ϕ] C such that χ 2 = ǫ N Q( 5)/Q and a p (f) = χ(p)a p (E) for almost all primes p. Challenge Compute the Mordell-Weil group E ( Q( 5) ) before the end of this talk!

5 Challenge Problem Elliptic Integrals Addition Formulas Modern Language My Favorite Elliptic Curve: y 2 = ( 1 x 2)( 1 k 2 x 2)

6 Challenge Problem Elliptic Integrals Addition Formulas Modern Language Theorem (Galileo Galilei, 1602; Christiaan Huygens, 1673) Say we have a mass m attached to a rigid rod of length l that is allowed to swing back and forth at one end. The period of the oscillation, given an initial angle θ 0, is Period = 4 l g K ( sin θ ) [ 0 l = 2π 1+ 1 ] θ 0 2 g 4 sin2 2 + in terms of the complete elliptic integral of the first kind: K(k) = 1 0 dt (1 t2 )(1 k 2 t 2 ) = π 2 [ ] 2 (2n 1)!! k 2n. (2n)!! n=0

7 Challenge Problem Elliptic Integrals Addition Formulas Modern Language Theorem (Jakob Bernoulli, 1694) The circumference of the lemniscus ( x 2 +y 2) 2 ( = a 2 x 2 y 2) is Arc Length = 4a K( 1) = 2πa [ ] 2 (2n 1)!! ( 1) n. (2n)!! n=0

8 Challenge Problem Elliptic Integrals Addition Formulas Modern Language Theorem (Giulio Fagnano, 1718) Define w = w(z) implicitly via z = w 0 dt 1 t 4. Then w(2z) = 2w(z)w (z) 1+w(z) 4 where w (z) = 1 w(z) 4. Theorem (Leonhard Euler, 1751) Fix a modulus k satisfying k < 1, and define w = w(z) implicitly via w dt the incomplete elliptic integral z = (1 t2 )(1 k 2 t 2 ). Then w(z ±ξ) = w(z)w (ξ)±w (z)w(ξ) 1 k 2 w(z) 2 w(ξ) 2 where w (z) = [1 w(z) 2 ][1 k 2 w(z) 2 ]. Remark: w(z) = sn(z) is a Jacobi elliptic function. 0

9 Challenge Problem Elliptic Integrals Addition Formulas Modern Language Theorem The Jacobi elliptic function sn : C/Λ C is well-defined modulo the period lattice Λ = {mω 1 +nω 2 m,n Z} in terms of the integrals 1/k ω 1 = 2 ω 2 = 2 1/k 1 1 dt (1 t2 )(1 k 2 t 2 ) = 4 ( ) 1 k K k dt (1 t2 )(1 k 2 t 2 ) = 4 K(k) The map C/Λ C 2 which sends z ( sn(z), sn (z) ) parametrizes all points (x,y) on the quartic curve y 2 = ( 1 x 2)( 1 k 2 x 2). Moreover, 0 (0,1). Say that P = ( sn(z), sn (z) ) and Q = ( sn(ξ), sn (ξ) ) are on the quartic curve. Then P Q = ( sn(z +ξ), sn (z +ξ) ) has coordinate x(p Q) = x(p)y(q)±y(p)x(q) 1 k 2 x(p) 2 x(q) 2.

10 Challenge Problem Elliptic Integrals Addition Formulas Modern Language Proposition y 2 = ( 1 x 2)( 1 k 2 x 2) is a quadric intersection in P 3 and has a Weierstrass model in P 2. It is nonsingular if and only if k 1, 0, 1. ( x1 y 2 = ( 1 x 2)( 1 k 2 x 2) (x,y) =, x 0 x2 2 = ( )( x 3 x 0 k 2 ) x 3 x 0 ) x 2 x 0 (x x1 2 1 : x 2 : x 3 : x 0 ) = x 3 x 0 Y 2 Z = X 3 +AX Z 2 +BZ 3 A = 27 ( k 4 +14k 2 +1 ) B = 54 ( k 6 33k 4 33k 2 +1 ) X Z = 3(5k2 1)x +3(k 2 5) x 1 Y Z = 54(1 k2 )y (x 1) 2

11 Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 More generally, we consider cubic curves E : Y 2 = X 3 +AX +B where the rational numbers A and B satisfy 4A 3 +27B 2 0. Given a field K such as either Q, R, C, or even Q( 5), denote { } E(K) = (X : Y : Z) P 2 (K) Y 2 Z = X 3 +AX Z 2 +BZ 3. Remark: O = (0 : 1 : 0) comes from (x,y) = (1,0) not (x,y) = (0,1)!

12 Mordell-Weil Group Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 Conjecture (Henri Poincaré, 1901) Let E be an elliptic curve over Q. Then E(Q) is a finitely generated abelian group. Theorem (Louis Mordell, 1922; André Weil, 1928) Let E be an elliptic curve over a number field K. There exists a group E(K) tors and a nonnegative integer r such that E(K) E(K) tors Z r. Theorem (Barry Mazur, 1977) The torsion subgroup of an elliptic curve E over Q is one of fifteen types: { Z N for 1 N 10 or N = 12; E(Q) tors Z 2 Z 2N for 1 N 4. Question: What can one say about the Mordell-Weil rank r = r(e)?

13 Rank Conjecture Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 Conjecture Let T be one of the fifteen torsion groups in Mazur s Theorem. For any given nonnegative integer r 0, there exists an elliptic curve E over Q with torsion subgroup E(Q) tors T and Mordell-Weil rank r(e) r 0. Project Given T and r 0, find an elliptic curve E over with torsion subgroup E(Q) tors T and Mordell-Weil rank r(e) r 0. For each torsion group T, define the quantity { } B(T) = sup r Z there exists a curve E with E(Q) T Zr. Question: Is B(T) unbounded?

14 Competing Points of View Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 Conjecture (Taira Honda, 1960) If E is an elliptic curve defined over Q, and K is a number field, then the ratio of the Mordell-Weil rank of E(K) to the degree [K : Q] should be uniformly bounded by a constant depending only on E. Remark: If true, this would imply that there are infinite families of elliptic curves over the rational numbers which have a uniformly bounded rank. Theorem (Igor Shafarevich and John Tate, 1967) The ranks are not uniformly bounded for elliptic curves defined over function fields F q (t).

15 Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 E(Q)tors Highest Known Rank r Found By Year Discovered Trivial 28 Elkies Z 2 19 Elkies Z 3 13 Eroshkin 2007, 2008, Z 4 12 Elkies Z 5 8 Dujella, Lecacheux Eroshkin Z 6 8 Z 7 5 Eroshkin Dujella, Eroshkin Elkies Dujella Dujella, Kulesz Elkies Eroshkin Dujella, Lecacheux Dujella, Eroshkin Z 8 6 Elkies Z 9 4 Fisher Z 10 4 Dujella 2005, 2008 Elkies Z 12 4 Fisher 2008 Z 2 Z 2 15 Elkies Z 2 Z 4 8 Elkies Eroshkin Dujella, Eroshkin Z 2 Z 6 6 Elkies Z 2 Z 8 3 Connell Dujella Campbell, Goins Rathbun Flores, Jones, Rollick, Weigandt, Rathbun Fisher , 2001,, , 2007

16 Classification Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 Theorem Fix a rational k 1, 0, 1 for the curve E k : y 2 = (1 x 2 )(1 k 2 x 2 ). Z 2 Z 8 if k = t4 6t 2 +1 E k (Q) tors (t 2 +1) 2 for some rational t, Z 2 Z 4 otherwise. Conversely, if E is an elliptic curve over K with torsion subgroup E(Q) tors Z 2 Z 4 or Z 2 Z 8, then E E k for some k K. The modular curve X 0 (24) : Y 2 = X 3 +5X 2 +4X has Mordell-Weil group X 0 (24) ( Q ) Z 2 Z 4, and so corresponds to k = 1/3. The modular curve X 1 (15) : Y 2 +X Y +Y = X 3 +X 2 10X 10 has X 1 (15) ( Q ) Z 2 Z 4, and so corresponds to k = 1/9. Moreover, X 1 (15) ( Q( 5) ) Z 2 Z 8, and so t = (3 5)/2.

17 Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 H X(2,8) = Γ(2) Γ 1 (8) 4 H X(2,4) = Γ(2) Γ 1 (4) 2 X(2) = H Γ(2) 6 2 X 1 (8) = H Γ 1 (8) 4 2 X 1 (4) = H Γ 1 (4) 2 2 X 1 (2) = H Γ 1 (2) 3 2 X 0 (8) = H Γ 0 (8) 2 1 X 0 (4) = H Γ 0 (4) 2 1 X 0 (2) = H Γ 0 (2) 3 X(1) = H SL 2 (Z) 1 X 1 (1) = H SL 2 (Z) 1 X 0 (1) = H SL 2 (Z)

18 Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 [ ] η(q) 4 k(q) = 4 η(q 2 η(q4 8 ) ) η(q 2 ) = t(q)4 6t(q) ( ) t(q) λ(q) = 1 η(q) η(q 1/2 ) η(q 2 ) 2 = 4 k(q) ( k(q) + 1 ) 2 ( λ(q) 2 ) λ(q) j(q) = 256 λ(q) 2 ( λ(q) 1 ) 2 µ 4 (q) = η(q2 8 ) η(q2 16 ) η(q) η(q 4 ) = 16 k(q) 2 [ ] η(q) 24 µ 2 (q) = η(q 2 ) = 256λ(q) ( λ(q) 1 ) ν 4 (q) = ν 2 (q) = [ ] η(q) 8 η(q 4 ) = µ 4 (q) 16 [ ] η(q) 24 η(q 2 ) = µ 2 (q) ( µ4 (q) 16 ) 2 = = ν 4 (q)2 µ 4 (q) ν 4 (q) ( µ2 (q) ) σ 3 3 (n)q n n=1 j(q) = µ 2 (q) 2 j(q) = q (1 q n ) 24 n=1

19 Example Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 On the quartic curve y 2 = (1 x 2 )(1 k 2 x 2 ), the rational point (x,y) has order 2 if and only if [2](x,y) = (1,0). There are only four: ( ) 1 k, 0, (1,0), ( 1,0), and ( 1k ),

20 Example Motivation Mordell-Weil Group Are the ranks unbounded? Z 2 Z 4 and Z 2 Z 8 On the quartic curve y 2 = (1 x 2 )(1 k 2 x 2 ), the rational point (x,y) has order 4 if and only if [2](x,y) = (,0). There are only four: (0, 1), (0, 1), and (two points at infinity)

21 E(Q) Z 2 Z 4 Z r Rank r = 8: Rank r = 7: Author(s) Fiber k Year Discovered Elkies / Eroshkin / / Dujella, Eroshkin / / Author(s) Fiber k Year Discovered Dujella Eroshkin Dujella, Eroshkin / / / / / / / / / / / / / / / / Rank r = 6: Author(s) Fiber k Year Discovered Ansaldi, Ford, George, Mugo, Phifer / /

22 E(Q) Z 2 Z 8 Z 3 Author(s) Fiber t Year Discovered Connell, Dujella 5/ Dujella 18/47 87/ / / / Dujella, Rathbun 230/ /1012 Campbell, Goins 15/ Campbell, Goins (with Watkins) 19/ Rathbun Flores - Jones - Rollick - Weigandt (with Rathbun) Fisher 47/219 74/207 17/ /569 86/ /299 65/337 47/ /321 97/ / / / / /815 76/ /

23 Example Motivation In, Dujella discovered the elliptic curve E : Y2 +X Y = X X with conductor N E = = has Mordell-Weil group E(Q) Z 2 Z 8 Z 3. Using the substitutions 6240( x ) X =, x 1 Y = y (x 1) x x we see that it is birationally equivalent to the quartic curve with k = = t4 6t 2 +1 (t 2 +1) 2 where t =

24

25 Can we do better than E(Q) Z 2 Z 4 Z 8 or E(Q) Z 2 Z 8 Z 3?

26 Elliptic Surfaces Motivation We will focus on the cases where the quartic curve E k : y 2 = (1 x 2 )(1 k 2 x 2 ) has torsion subgroup E k (Q) tors Z 2 Z 8. We express our results in terms of elliptic surfaces. Consider the affine curve { C = t = (a : b) P 1 ab ( a 4 b 4)( a 4 6a 2 b 2 +b 4) } 0. Fix the rational functions A, B : C P 1 defined by A(t) = 27 ( k 4 +14k 2 +1 ) B(t) = 54 ( k 6 33k 4 33k 2 +1 ) where k = t4 6t 2 +1 (t 2 +1) 2 and consider the surface E = { [(X : Y : Z), t ] P 2 C } Y 2 Z = X 3 +A(t)X Z 2 +B(t)Z 3.

27 Theorem (G, 2008) With respect to E C which sends [ (X : Y : Z), t ] t, the variety E is an elliptic surface. Each of the fibers E t is semistable. We have two sections P : t 12 4t 6 26t 4 4t (t 2 + 1) 4 : 0 : 1,t Q : t 12 4t 6 12t 5 2t t t + 1 (t 2 + 1) 4 : 864 t7 5t5 4 t4 + 3t3 + 4t2 + t (t 2 + 1) 5 : 1,t All elliptic curves E over a number field K with torsion subgroup P(t), Q(t) Z 2 Z 8 arise from such a fiber, i.e., are birationally equivalent to E t for some t C(K). The automorphisms σ : (a : b) (a b : a+b) and τ : (a : b) ( a : b) act on C, yet leave A and B invariant. Moreover, D 8 = σ, τ Aut(C) is the dihedral group.

28 Proposition (A. O. L. Atkin and François Morain, 1993) The elliptic curve C 1 : v 2 = u 3 8u 32 has Mordell-Weil group C 1 (Q) Z 2 Z as generated by (u : v : 1) = (12 : 40 : 1). One can construct infinitely many fibers E t having positive rank via the map C 1 C defined by (u : v : 1) 2(u 9)/(3u +v 2). Theorem (Garikai Campbell and G, 2003) The elliptic curve C 2 : v 2 = u 3 u 2 9u +9 has Mordell-Weil group C 2 (Q) Z 2 Z 2 Z as generated by (u : v : 1) = (5 : 8 : 1). One can construct infinitely many fibers E t having positive rank via the map C 2 C defined by (u : v : 1) t = (u +v 3)/(2u). Indeed, upon setting w = 3(u 2 2u +4v +9)/(u 2 18u +9), we have a section R : (u : v : 1) 3(w2 2w 3) (w 2 w 3) (w 2 + 2w 3) 3 (w 4 2w 2 + 9) 2 54(w 4 9)(w 2 2 w 3)(w w 3) 3 : 1, u + v 3 (w 4 2w 2 + 9) 3 : 2u

29 Infinite Families Motivation There are infinitely many choices of rational t such that E t : y 2 = (1 x 2 )(1 k 2 x 2 ) where k = t4 6t 2 +1 (t 2 +1) 2 has torsion subgroup E t (Q) Z 2 Z 8 and rank r 1. These choices of t correspond to rational points on elliptic curves. Open Questions Are there other elliptic curves besides C 1 and C 2 which work? Is there a curve of genus 0 which gives E t having rank r 1? Are there infinitely many rational t which give E t having rank r 2?

30 Finding Curves of High Rank Approach #1 Fix a square-free integer D, and consider the quadratic twist E (D) : Y 2 = X 3 +D 2 AX +D 3 B. This is very efficient (i.e., no redundant curves), but E (D) (Q) tors changes with each D. Approach #2 Fix polynomials A = A(t) and B = B(t) such that (t) = 16 ( 4A 3 +27B 2) 0, and consider the elliptic surface E t : Y 2 = X 3 +A(t)X +B(t). This is not very efficient (i.e., different t s may give the same curves), polynomials can be chosen to fix E t (Q) tors for all t.

31 Algorithm Motivation #1. Classify those elliptic curves E over Q with torsion subgroup E(Q) tors Z 2 Z 8. Express these curves as an elliptic surface E t. #2. Find a criterion on t such that any t Q may be associated to an element from a fundamental region α < t < β. #3. Create a list of candidate elliptic curves E t for this fundamental region. #4. Compute the 2-Selmer ranks to find upper bounds on the Mordell-Weil ranks. #5. Compute the Mordell-Weil ranks.

32 E(Q) Z 2 Z 8 Z 3 Author(s) Fiber t Year Discovered Connell, Dujella 5/ Dujella 18/47 87/ / / / Dujella, Rathbun 230/ /1012 Campbell, Goins 15/ Campbell, Goins (with Watkins) 19/ Rathbun Flores - Jones - Rollick - Weigandt (with Rathbun) Fisher 47/219 74/207 17/ /569 86/ /299 65/337 47/ /321 97/ / / / / /815 76/ /

33 Fundamental Region Theorem (G, ) Fix a rational number t 1,0,1 and consider E t : y 2 = ( 1 x 2)( 1 k 2 x 2) where k = t4 6t 2 +1 (t 2 +1) 2. D 8 = σ, τ σ 4 = τ 2 = 1, τ στ = σ 1 in terms of σ : t t 1 t +1 and τ : t t. We may assume that t satisfies 0 < t < 2 1. Remark: Given a bound N, choose coprime integers a and b satisfying ( 0 < 1+ ) 2 a < b < N and set t = a b.

34 Isogeny Graph Motivation D t D t C t E t E t φ φ ˆφ ˆφ E t C t C t

35 Isogeny Graph Motivation Curve Weierstrass ( Model Y 2 = X 3 ) + A X + B Torsion A = 27 k k Et ( ) Z 2 Z 8 B = 54 k 6 33k 4 33k ( ) A = 27 k 4 k E t ( B = 27 2k 6 3k 4 3k 2 ) Z 2 Z ( ) A = 27 k 4 60k k 2 60k + 1 C t ( B = 54 k k k k k 2 ) Z k + 1 ( ) A = 27 k k k k + 1 D t ( B = 54 k 6 126k k k k 2 ) 126k + 1 Z 8 ( ) A = 27 k 4 16k E t ( B = 54 k k 4 96k 2 ) Z 2 Z ( ) A = 27 16k 4 16k C t ( B = 54 64k 6 96k k 2 ) + 1 Z 4 y 2 = x 3 ( t + 20 t t 3 26t 4 24 t t 6 24t 7 + t 8) x 2 C t D t + (1 2t t 2) Z 8 2 x y 2 = x 3 ( t + 20t 2 24t 3 26t t t t 7 + t 8) x 2 + (1 + 2t t 2) Z 8 8 x

36 Motivation Define the curves and homogeneous spaces E t : y 2 = ( 1 x 2)( 1 k 2 x 2) E t : y 2 = ( 1 x 2)( 1 κ 2 x 2 ) E t : y 2 = ( 1+x 2)( 1+k 2 x 2 ) where C d : d w 2 = ( 1 d z 2)( 1 d k 2 z 2) C d : d w 2 = ( 1+d z 2)( 1+d κ 2 z 2) C d : d w 2 = ( 1+d z 2)( 1+d k 2 z 2 ) κ = 1 k 1+k, κ = 1 k 1+k, and k2 +k 2 = 1. D t C d Φ C d ˆφ E t E t ψ Φ C d ˆφ ψ E t

37 Descent via 4-Isogeny Theorem (G, ) There are 2-isogenies φ : E t E t and φ : E t E t. If E E t and E E t, then E(Q) 2E(Q) = E (Q) φ ( E(Q) ) E(Q) ˆφ ( E (Q) ). Write k = p/q for relatively prime integers p and q. The image of δ φ (of δˆφ, respectively) is the set of those square-free divisors d of pq (of p 2 q 2, respectively) such that C d (C d, respectively) has a Q-rational point. ( δˆφ ψ ) (z,w) ( δ φ ψ ) (z,w) d mod (Q ) 2 for the maps ψ : C d E t ψ : C d E t ( ) 1 d κz 2 4d κz w (z,w) 1+d κz 2, (1+κ)(1+d κz 2 ) 2 ( 1 d k z 2 4d k ) z w (z,w) 1+d k z 2, (1+k )(1+d k z 2 ) 2

38 Example Motivation Proposition (Samuel Ivy, Brett Jefferson, Michele Josey, Cheryl Outing, Clifford Taylor, and Staci White, 2008) When t = 9/296 we have 1, , 2, 7 δˆφ 1, , 2, 7, 37. Hence E t has Mordell-Weil group E t (Q) Z 2 Z 8 Z 3 if and only if at least one of the following homogeneous spaces corresponding to d = 37 contains a rational point (z,w): C 37 : w 2 = z z ; C 37 : w2 = z z

39 Challenge Problem Revisited E : y 2 = x 3 + ( 5 5 ) x 2 + 5x The curve has invariant j(e) = The curve has conductor f E = p 6 2 p2 5 in terms of the prime ideals p 2 = 2Z[ϕ] and p 5 = 5 Z[ϕ], where ϕ = This curve is 2-isogeneous to (a quadratic twist of) its Galois conjugate. Theorem (G, 1999) The elliptic curve E is modular. More precisely, there is a modular form f(q) S 2 ( Γ0 (160),ǫ ) and a Dirichlet character χ : Z[ϕ] C such that χ 2 = ǫ N Q( 5)/Q and a p (f) = χ(p)a p (E) for almost all primes p. Question Did you compute the Mordell-Weil group E ( Q( 5) )?

40 Questions?