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Growth of torsion in quadratic extensions of quadratic cyclotomic fields
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Growth of torsion in quadratic extensions of quadratic cyclotomic fields
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GROWTH OF TORSION IN QUADRATIC EXTENSIONS OF QUADRATIC CYCLOTOMIC FIELDS by Burton Newman A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulllment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (MATHEMATICS) May 2015 Copyright 2015 Burton Newman In loving memory of Darlene Schaefer ii Acknowledgments First and foremost I would like to thank my advisor, Sheldon Kamienny. As my mentor over the past 14 years, Sheldon has been a constant source of support and encouragement. Without his guidance this dissertation would not have been possible. I feel honored to have been able to work so closely with him and to spend invaluable time with his wife Darlene, who encouraged and inspired me to continue my mathematical pursuits. Sheldon also introduced me to the PROMYS program, where I had some of the most exciting and fullling experiences of my life. I would like to give a special thank you to Bob Guralnick, from whom I took my rst course in group theory 15 years ago. Bob played a large role in my return to graduate school, provided generous nancial support and has always made himself available should I need anything. I am also thankful to Aravind Asok for our mathematical chats, as well as Susan Montgomery for her advice and support during my time in graduate school. I would like to thank Werner D appen for the optimism and warmth he showed me while serving on my committee. I would also like to thank Filip Najman for his advice and clearly-written research articles, which were invaluable in helping me navigate my way through this area of mathematics. While attending USC, I have had the privilege of studying math with so many brilliant and interesting classmates that I cannot name them all here. Nevertheless their camaraderie means a great deal to me. A special thanks to Ozlem Ejder for her willingness to always entertain my mathematical ideas and Jared Warner for his advice and perspective. My friends and family have been a tremendous support during this journey. To my UChicago classmates David Coley, Aaron Marcus and Matthew Thibault, thank you for always making the distance between us feel small and being there in spirit, day in and day out. To Brad, Sean, Matthew - thank you for bringing out the kid in me and for all the adventures we have shared. Thank you Noah for your curiosity and for reminding me what is important in life. Many thanks to Alowie, Dory, Devin and Aron for sharing their iii wisdom and outlook on life. Thank you to my roommate and future brother-in-law Adam for always looking out for me and encouraging me to play. To my Grandma Dorothy, thank you for your love and wit. To my Aunts Robin and Bethanne, thank you for all of your love and encouragement. A special thank you to my sisters, Aimee and Arianne, for always being there to listen. Finally, I would like to thank my parents for their unconditional love, support and guidance. I am very fortunate to have two loving parents that are truly dedicated to their children. iv Table of Contents Dedication ii Acknowledgments iii Abstract vii List of Tables 1 Chapter 1: Introduction 2 Chapter 2: Background 9 1 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1 Quadratic Twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.3 Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.4 Weil Pairing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3 Modular Curves X 0 and X 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Chapter 3: Growth in Case E(K)[2]6C 2 C 2 21 1 Restrictions on Growth in Quadratic Extensions . . . . . . . . . . . . . . . 21 2 K-Rational Points on X 0 (N) . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.1 N=21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.2 N=15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.3 N=20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.4 N=24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.5 N=27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.6 N=30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.7 N=35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3 Classication of Twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4 Examples of Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Chapter 4: Growth in Case E(K)[2]C 2 C 2 36 1 Some Diopantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2 Classication of Twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3 Full 4-torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 v 4 Classication of Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.1 E(K) tor C 2 C 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.2 E(K) tor C 2 C 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.3 E(K) tor C 2 C 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.4 E(K) tor C 4 C 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 4.5 E(K) tor C 2 C 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Bibliography 65 Chapter A: Appendix 68 1 Magma Program: Solving a Diophantine Equation Over a Quadratic Field . 68 2 Magma Program: Determining Growth of Torsion in Quadratic Fields . . . 70 vi Abstract LetK =Q( p 3) orQ( p 1) and letC n denote the cyclic group of order n. We study how the torsion part of an elliptic curve over K grows in a quadratic extension of K. In the case E(K)[2] C 2 C 2 we determine how a given torsion structure can grow in a quadratic extension and the maximum number of extensions in which it grows. We also classify the torsion structures which occur as the quadratic twist of a given torsion structure. We complete a similar classication in the case K =Q( p 3) and E(K)[2] =C 1 . vii List of Tables 1.1 Examples of Torsion Structures overQ . . . . . . . . . . . . . . . . . . . . . 3 1.2 Growth of Torsion of Elliptic Curves Dened over Q in Quadratic Fields . . 6 2.1 Models for X 0 (N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.1 (K =Q( p 3)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (21) . . . . . . . . . . . 26 3.2 (K =Q( p 1)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (15) . . . . . . . . . . . 27 3.3 (K =Q( p 1)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (20) . . . . . . . . . . . 28 3.4 K-Rational Points on a Genus 1 Quotient of X 0 (35) . . . . . . . . . . . . . 30 3.5 Gr obner basis data for determination of f 1 (E(K)) . . . . . . . . . . . . . 31 3.6 Examples of Growth of Torsion over K =Q( p 1) in CaseE(K)[2]6C 2 C 2 34 3.7 Examples of Growth of Torsion over K =Q( p 3) in CaseE(K)[2]6C 2 C 2 35 4.1 K-Rational Points on E C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.2 Gr obner basis data for determination of f 1 (E(K)) . . . . . . . . . . . . . 44 4.3 Examples of Growth of Torsion over K =Q( p 1) in CaseE(K)[2]C 2 C 2 50 4.4 Examples of Growth of Torsion over K =Q( p 3) in CaseE(K)[2]C 2 C 2 50 1 Chapter 1 Introduction The theory of diophantine equations, the study of integer and rational solutions to poly- nomial equations, is a subject that dates back to antiquity. Many well-known examples of diophantine equations are curves (e.g. x 2 +y 2 = 1) and geometric methods were some- times used to generate and classify rational solutions to these equations. Today, these methods have been generalized and rened in the eld of algebraic geometry. In modern language, the basic problem is the following: Given a number eld K and a (geometrically irreducible) curve C/K, determine C(K), the set of K-rational points on C. The resolution of this problem often requires both theoretical and computational tools. By replacing C by a (birationally equivalent) smooth model and homogenizing, we may assume WLOG that C is a smooth projective curve. The cardinality of C(K) depends on an invariant of the curve called the genus. Informally, the genus of C is the number of "handles" of the compact Riemann surface C(C). The genus of a smooth projective curve of degree d is (d-1)(d-2)/2. We have the following trichotomy: If the genus is 0 (e.g. C is a line or a conic section) and C has a K-rational point, then C(K) has innitely many points and can be parametrized. If the genus is at least 2, then by Mordell's Conjecture (proven by Faltings), C(K) is nite. If the genus is 1 and C has a K-rational point, then C is an elliptic curve over K, denoted C/K. In this case C(K) can be nite or innite and deciding which can be challenging. An elliptic curve is often denoted by the letter E. If E/K is an elliptic curve, it has an ane model of the form y 2 = x 3 +ax +b for some a;b2 K. There is just one more projective point on E, namely1 := [0; 1; 0]. Furthermore, there is an abelian group law on E(K), and in fact we have the following: Theorem 1.0.1. (Mordell-Weil 1928) Let K be a number eld and E an elliptic curve dened over K. Then E(K) is a nitely-generated abelian group and hence: 2 E(K) E(K) tor Z r where E(K) tor denotes the torsion part of E(K) and r is the rank of E(K). Determining the nite abelian groups (up to isomorphism) which occur as E(K) tor is a nontrivial problem to which much research has been devoted. For K = Q one can compile a table of torsion structures such as the one below. Let C n denote the cyclic group of order n. Table 1.1: Examples of Torsion Structures overQ Elliptic Curve E(Q) tor y 2 =x 3 2 C 1 y 2 =x 3 + 8 C 2 y 2 =x 3 + 4 C 3 y 2 =x 3 + 4x C 4 y 2 y =x 3 x 2 C 5 y 2 =x 3 + 1 C 6 y 2 +xy =x 3 x + 137 C 7 y 2 + 7xy =x 3 + 16x C 8 y 2 +xy +y =x 3 x 2 14x + 29 C 9 y 2 +xy =x 3 45x + 81 C 10 y 2 +xy +y =x 3 x 2 122x + 1721 C 12 y 2 =x 3 x C 2 C 2 y 2 =x 3 + 2x 2 3x C 2 C 4 y2 +xy +y =x3 19x + 26 C 2 C 6 y 2 + 17xy 120y =x 3 60x 2 C 2 C 8 This list of torsion structures overQ is actually complete: Theorem1.0.2. (Mazur [Maz78], Kubert [Kub76]) Let E be an elliptic curve overQ. Then the torsion subgroup E(Q) tor of E(Q) is isomorphic to one of the following 15 groups: C n for 1n 12;n6= 11 C 2 C 2n for 1n 4 Furthermore, each group above occurs innitely often (up to isomorphism). By contrast, when K=Q( p 5), there is only one elliptic curve over K (up to isomorphism) with torsion part C 15 [Naj12]. The work of Mazur was generalized by Kamienny et al. to the case of quadratic number elds. 3 Theorem 1.0.3. (Kamienny [Kam92], Kenku and Momose [KM88]) Let K be a quadratic eld and E an elliptic curve over K. Then E(K) tor is isomorphic to one of the following 26 groups: C n for 1n 18;n6= 17 C 2 C 2n for 1n 6 C 3 C 3n for 1n 2 C 4 C 4 Note that the above result classies the possible torsion structures as one varies over all quadratic number elds. Hence for a given quadratic eld K, it is in general still unknown which torsion structures occur over K among elliptic curves E/K. Two cases in which the answer is known are K =Q( p d), d =1;3: Theorem 1.0.4. (Najman [Naj11]) Let K be a cyclotomic quadratic eld and E an elliptic curve over K. IfK =Q(i) thenE(K) tor is either one of the groups from Mazur's theorem orC 4 C 4 . If K = Q( p 3) then E(K) tor is either one of the groups from Mazur's theorem, C 3 C 3 or C 3 C 6 . For cubic and quartic number elds, we still do not have a complete classication analogous to Theorem 1.0.3. We do know the possible prime divisors of the order of the torsion ([Par00], [Par03],[Sto10]), as well as the torsion structures that occur innitely often (up to isomorphism) ([JKS04], [JKP06]). Furthermore for cubic elds, we have a classication of possible torsion structures if we restrict to elliptic curves dened over Q ([Naj12]). We also have a classication of torsion structures over the compositum of all quadratic elds for elliptic curves dened overQ ([Fuj05]). Theorem 1.0.3 provided evidence for the Strong Uniform Boundedness conjecture, which was conrmed by Merel: Theorem 1.0.5. (Merel [Mer96]) For every positive integer d there exists B d > 0 such that for every number eld K with [K :Q] =d and every elliptic curve E/K we have jE(K) tor jB d 4 With the above classication results in hand, researchers began to study how the torsion part of an elliptic curve grows in a quadratic extension of the base eld. For example, consider the elliptic curve E: y 2 = x 3 x. We have E(Q) tor =f(1; 0); (1; 0); (0; 0);1g and E(Q) tor C 2 C 2 . Over Q( p 1) one has the additional solutions (i;(1i)) and (i;(1 +i)) and E(Q( p 1)) tor C 2 C 4 . OverQ( p 2) one has the additional solutions (1 p 2;( p 22)) and (1+ p 2;( p 2+2)) and E(Q( p 2)) tor C 2 C 4 . One can show these are the only two quadratic extension in which the torsion grows. As we will soon describe, if L is a quadratic extension of K, there is a bound onjE(L) tor j which depends only on the torsion structures that occur over K. Furthermore, as a consequence of Theorem 1.0.5 and Theorem 3.1.1 one can show that for any number eld K, E(K) tor grows in a nite number of quadratic extensions (see [Naj14, Thm 12]). The case of quadratic growth whenK =Q has already been studied by various authors. Previously Kwon completed the case in which E(Q)[2] = C 2 C 2 ([Kwo97]). In order to classify growth, Kwon studied the the distribution of torsion structures among the quadratic twists of an elliptic curve. The d-quadratic twist of E : y 2 = x 3 +ax +b is the elliptic curve with model E d : dy 2 = x 3 +ax +b. Kwon's result was extended by Tornero et al. to complete the classication over Q ([GJT14]) (See Table 1.2). On the other hand, they obtained non-optimal upper bounds on the number of extensions in which a given torsion structure can grow. Najman nally determined the best possible bounds for each torsion structure overQ ([Naj14]). In particular, no curve overQ can grow in more than 4 quadratic extensions. 5 Table 1.2: Growth of Torsion of Elliptic Curves Dened over Q in Quadratic Fields E(Q) tor E(L) tor in some quadratic eld L (depending on E) C 1 C 1 ;C 3 ;C 5 ;C 7 ;C 9 C 2 C 2 ;C 4 ;C 6 ;C 8 ;C 10 ;C 12 ;C 16 ;C 2 C 2 ;C 2 C 6 ;C 2 C 10 C 3 C 3 ;C 15 ;C 3 C 3 C 4 C 4 ;C 8 ;C 12 ;C 2 C 4 ;C 2 C 8 ;C 2 C 12 ;C 4 C 4 C 5 C 5 ;C 15 C 6 C 6 ;C 12 ;C 2 C 6 ;C 3 C 6 C 7 C 7 C 8 C 8 ;C 16 ;C 2 C 8 C 9 C 9 C 10 C 10 ;C 2 C 10 C 12 C 12 ;C 2 C 12 C 2 C 2 C 2 C 2 ;C 2 C 4 ;C 2 C 6 ;C 2 C 8 ;C 2 C 12 C2C 4 C 2 C 4 ;C 2 C 8 ;C 4 C 4 C 2 C 6 C 2 C 6 ;C 2 C 12 C 2 C 8 C 2 C 8 Our research is a rst step toward classifying growth of torsion in quadratic extensions of base elds larger thanQ. We choose to work over the base eldsQ( p 3) andQ( p 1), as these are the only two number elds besidesQ for which a classication of torsion structures is known. Let K = Q( p 3) or Q( p 1). In the case E(K)[2] = C 2 C 2 , we generalize Kwon's work, determining (i) a classication of the torsion structures which occur as the quadratic twists of a given torsion structure (Theorem 4.2.1) (ii) a classication of the torsion structures which occur as the growths of a given torsion structure (Theorem 4.4.1, Theorem 4.4.2) (iii) Tight bounds on the number of quadratic extensions in which a given torsion structure can grow (Theorem 4.4.1, Theorem 4.4.2). Kwon's methods relied upon the fact thatZ was a UFD and had nite unit group. Our methods don't depend explicitly on these properties and hence generalize more easily to arbitrary number elds. To make progress on the caseE(K) tor 6=C 2 C 2 , we (iv) classify N-isogenies dened over quadratic extensions of K for N=15,20,21,24,27,30,35 (Theorem 3.2.3). In particular, (v) in the case K=Q( p 3), E(K)[2] =C 1 this allows us to accomplish the same three tasks as in the full 2-torsion case above (Theorem 3.3.1). In order to seek out examples of growth (vi) we used division polynomials to write a program in Magma which on input an elliptic curve E/K, outputs a list of quadratic extensions of K where the torsion grows and the torsion in each case. 6 In Chapter 2 we describe some background necessary to understand the rest of the paper. In Section 1 we dene the genus of a smooth projective curve C, a birational invariant of C which in uences the structure of C(K). Given any curve C, we also introduce an abelian variety called the Jacobian of C, which can often be used to obtain bounds on the size of C(K). In Section 2 we describe algorithms for computing the rank and torsion of E(K). The calculation of torsion makes use of division polynomials, a class of polynomials of great use computationally. In Section 3.2 we describe the Weil pairing, which puts certain constraints on E(K) tor depending on K. It will be used to show certain types of growth cannot occur (see Theorem 3.1.2). In Section 3.3 we describe the modular curves X 0 and X 1 , which can be used to classify isogenies over K and parametrize elliptic curves with certain torsion structures over K. In Section 2.1 we introduce quadratic twists, which play a crucial role in the study of growth. Suppose L =K( p d) is a quadratic extension of K and E/K is an elliptic curve. To studyE(L) tor , one can show it is nearly the direct product of two torsion subgroups which occur over K (Theorem 2.2.4). These two groups areE(K) tor andE d (K) tor . If L =K( p d), andE(K) tor is given thenE d (K) tor determines the odd part ofE(L) tor and yields a bound for the 2-part of E(L) tor (see Corollary 2.2.5, Proposition 2.2.6). Hence if one wants to study the growth of the torsion structure E(K) tor the rst task is to classify which torsion structures occur as E d (K) tor . In Chapter 4, we study growth for arbitrary E/K. In particular when E(K)[2] = C 1 , we have E(L)[2] =C 1 (Theorem 3.1.2). Hence in order to complete tasks (i), (ii) and (iii), it suces to complete (i). But by Proposition 2.2.7, if E(K) tor 6= C 1 and E d (K) tor 6= C 1 then one can often show E has an N-isogeny dened over K( p d) for some large value of N, and these are rare over K (Theorem 3.2.3). The classication of N-isogenies leads to the Diophantine problem of determining X 0 (N)(K). In Chapter 5 we study growth in the caseE(K)[2] =C 2 C 2 . First we classify the twists (Theorem 4.2.1). The proof of this result uses a parametrization of elliptic curves containing a given torsion structure C 2 C 2n (1 n 4) (Theorem 4.0.2). Explicitly twisting and equating parametrized models of curves yields systems of diophantine equations. These systems are solved in Chapter 4 Section 1. Now, given a torsion structure E(K) tor , the classication of twists yields a nite list of possible torsion structures for the 2-part of E(L). Furthermore, the 2-part of E(K) grows in L if and only if for some point P2 E(K) tor , P 2 2E(L) but P62 2E(K). Given E/K such that E(K)[2] =C 2 C 2 and P2E(K), Lemma 4.0.1 provides a criteria for when P 7 2 2E(K). Applying this criteria to parametrized models as in Theorem 4.0.2 yields systems of Diophantine equations. Finding all the K-rational solutions to the Diophantine equations arising in our research would have been quite dicult without computer assistance. We used the software packages Magma and Sage. Given a curve C/K, Magma can determine the genus of C. If the genus is 1 and we can nd just one K-rational point P on C, then Magma can nd a Weierstrass model E for C and a birational map f: C!E.Given the Weierstrass equation for E, Magma can now try to compute the rank of E(K). If the rank is 0, then E(K) =E(K) tor is nite and Magma can carry out can carry out standard algorithms (e.g. Nagell-Lutz or reduction mod p) to nd all of E(K). Finally one can use the map f to recover all of C(K). If the curve C has genus at least 2, then C(K) does not form a group as above, but the K-points of its Jacobian J(C)(K) do. We have an injective morphism f:C! J(C). If C is genus 2 hyperelliptic, Magma has functionality analogous to the the tools above for elliptic curves: it can tell you if J(C)(K) is nite and can help compute the torsion points on J(C)(K). Finally one can often use the morphism f to recover C(K). The algorithm we wrote to nd examples of growth of torsion also relied upon Magma to factor division polynomials over quadratic elds, as well as compute the torsion over quartic number elds. Finally, the classication of N-isogenies relied upon the Small Modular Curves package in Magma. 8 Chapter 2 Background 1 Curves Let K be a number eld. Given a smooth projective curve C/K, we can attach an abelian group to C as follows. A divisor is a formal sum X p2C(K) n p P withn p 2Z andn p = 0 for almost all p2C(K). Let Div(C) denote the (additive) group of divisors. Given a divisor D = P p2C(K) n p P dene the degree of D, denoted Deg(D), to be P p2C(K) n p . Since C is smooth, for each point P on C the localization of the coordinate ring K[C] at P is a DVR. Hence iff2K(C), we can dene the orderord p (f) of a zero or pole at P. So to each f2K(C) we can associate the principal divisor div(f)= P p2C(K) ord p (f)P . Note that this is well-dened as f has only a nite number of zeroes and poles [Sil09, Prop. 1.2]. For each divisor D, we can construct the space of functions on C with `prescribed zeroes and allowed poles' as follows: L(D) :=f0g[ff2C(K)j ord P (f)n P for all P2Cg This is a nite dimensional K-vector space and we let l(D) denote its dimension. Dene the genus g of C to be g :=max D Deg(D) + 1l(D) where the maximum is taken over all divisors D on C. By Riemann's inequality, the maxi- mum exists and is a nonnegative integer. The set of all principal divisors of C forms a subgroup denoted Princ(C). If f2K(C) then Deg(div(f)) = 0 ([Sil09, Prop. 3.1]) so Princ(C) is a subgroup of the group of degree 0 divisors on C, denoted Div 0 (C). Hence we can form the quotient Jac(C) = Div 0 (C)=Prin(C), the Jacobian of C. If C is dened over K then the Jacobian of C can be realized as a projective variety (in fact, an abelian variety) dened over K, of dimension 9 equal to the genus of C. Furthermore if the genus of C is not 0 and we choose a K-rational point Q on C, there is an embedding dened over K: C! Jac(C) P7!PQ The absolute Galois group Gal(K=K) acts on C as follows: if P= [x;y;z]2 C(K) and 2 Gal(K=K) then (P) = [(x);(y);(z)]. This action can be extended linearly to Div 0 (C) and induces a well-dened action on Jac(C). Hence if D is divisor in Div 0 (C), we say D is dened over K if(D) =D in Jac(C) and let Jac(C)(K) denote the set of all points in Jac(C) dened over K. Note that if the divisor D is dened over K, it is not required that the points in the support of D are dened over K. Now let K be a number eld. By the Mordell-Weil Theorem, Jac(C)(K) is a nitely- generated abelian group. Suppose C has good reduction at a prime ideal ofO K lying over the prime p in Z and let C denote the reduction of C mod . The reduction map Jac(C)(K) tor !Jac(C ) induces an injection on the prime-to-p part of Jac(C)(K) tor . 2 Elliptic Curves 2.1 Quadratic Twists Let K be a number eld and E/K a smooth projective curve of genus 1 dened over K with at least one K-rational point. Then E is isomorphic over K to a curve in Weierstrass form Y 2 +a 1 XY +a 3 Y =X 3 +a 2 X 2 +a 4 X +a 6 with a i 2 K (i=1,..,5). If char(K)6= 2; 3 then by completing the square and translating we see E is isomorphic over K to a curve in short Weierstrass form: y 2 =x 3 +Ax +B :=16(4A 3 + 27B 2 ) j :=1728(4A) 3 = Short Weierstrass form is useful as isomorphisms between curves in this form are simple to describe: 10 Proposition 2.2.1. Suppose E;E 0 are elliptic curves with models: E :y 2 =x 3 +Ax +B E 0 :y 2 =x 3 +A 0 x +B 0 If E;E 0 are isomorphic (over K) then there exists a d2 K such that A =d 4 A 0 B =d 6 B 0 and in this case =d 12 0 In particular, this allows for easy calculation of the automorphism group (over K) of an elliptic curve. Corollary 2.2.2. Let K be a eld with char(K)6= 2; 3, E/K an elliptic curve and Aut(E) the automorphism group of E: 1. if j6= 0; 1728 then Aut(E)C 2 2. if j = 1728 then Aut(E)C 4 3. if j = 0 then Aut(E)C 6 Corollary 2.2.3. Let K be a eld with char(K)6= 2; 3 and E/K,E 0 /K elliptic curves. Then E andE 0 are isomorphic over K if and only ifj =j 0 . Ifj =j 0 then E andE 0 are isomorphic over an extension L/K of degree at most 6. Ifj =j 0 6= 0; 1728, thenE andE 0 are actually isomorphic over a quadratic extension of K. In this case we say E is a quadratic twist of E 0 . If E is an elliptic curve with model: E :y 2 =x 3 +Ax +B Then its d-quadratic twist is: E d :y 2 =x 3 +d 2 Ax +d 3 B Note this curve is isomorphic over K to the curve with model dy 2 =x 3 +Ax +B 11 so we often use them interchangeably. On the other hand the map T :E d !E (x;y)7! (x; p dy) denes an isomorphism over K( p d). Hence if d is a square in K then E and E d are isomorphic over all extensions of K and if d is not a square then we have E(K( p d)) E d (K( p d)). Let L be a quadratic extension of K. The following theorem tell us thatE(L) tor is nearly the direct product of two torsion structures which occur over K. Theorem 2.2.4. [GJT14, Thm 3] Let K be a number eld, d a non-square in K and E/K an elliptic curve. There are homomorphisms E(L) f !E(K)E d (K) E(L) f 0 E(K)E d (K) such that: 1. f 0 f = [2] 2. ff 0 = [2] [2] 3. ker(f) = E(K)[2] 4. ker(f')E(K)[2]E d (K)[2] 5. coker(f) and coker(f') have exponent 2 Corollary 2.2.5. [GJT14, Cor. 4] If n is an odd positive integer we have E(K( p d))[n]E(K)[n]E d (K)[n] . We also have the following proposition of Kwon, which allows us to bound the growth (e.g. of the 2-part of E(K) tor ) in a quadratic extension. 12 Proposition 2.2.6. [Kwo97] Let K be a quadratic number eld, L = K( p d) a quadratic extension of K and let denote the generator of Gal(L/K). The map E(L) tor !E d (K) tor P7!P(P ) is a homomorphism with kernel E(K) tor and hence induces an injection E(L) tor =E(K) tor ,!E d (K) tor Proposition2.2.7. Let K be a number eld and E/K an elliptic curve in short Weierstrass form. Let d2 K be a nonsquare and let L = K( p d). If H is a subgroup of E d (K) tor of odd order, then there is a Gal(K=K)-invariant subgroup J of E(L) tor such that JHE(K) tor . Proof. Recall we have an isomorphism: T :E d !E (x;y)7! (x; p dy) so T(H) H and T(H) is Gal(K=K)-invariant since H is a subgroup dened over K (so closed under inverses). Since H has odd order, it has no points of order 2 so T(H)\E(K) tor = f0g. Hence J := T(H)+E(K) tor HE(K) tor is a subgroup of E(L). As J is the sum of Gal(K=K)-invariant subgroups, it is invariant as well. 2.2 Torsion Let K be a number eld and E/K an elliptic curve. There are several algorithms (guaranteed to halt) which will produceE(K) tor . We begin with the most elementary algorithm, which is most easily applied when K = Q. We will later learn a more ecient algorithm. The algorithm is a consequence of the following theorem. There is an analogous result for arbitrary number elds (see Silverman Rmk 7.3.1, Exercise 8.11 for the general case). Theorem 2.2.8. Let E=Q be an elliptic curve in short weierstrass form y 2 =x 3 +ax +b (a;b2Z) If (x 0 ;y 0 )2 E(Q) tor then x 0 ;y 0 are in Z and y 0 = 0 or y 2 0 j 4a 3 + 27b 2 . 13 Given a curve E in short weierstrass formy 2 =f(x) as above, we can compute its torsion overQ as follows: 1)Compute disc(E)=16(4a 3 + 27b 2 ) 2)Factor disc(E) inZ 3)For each square divisor d of disc(E) in Z, use the rational roots theorem to nd the integer roots of f(x)d. 4) Let R =f(c;d)2 E(Q)jc;d2 Z;d = 0 or d 2 j4a 3 + 27b 2 g. Then E(Q) tor R so the order of any P2 E(Q) tor is bounded byjRj. If nP =1 for some n, 1 njRj, then P2E(Q) tor . Otherwise it is not. Note that bound in step (4) on the order of torsion points can be improved by Theo- rem 1.0.2. Although Theorem 2.2.8 yields a guaranteed algorithm, it is not often the fastest way to compute the torsion. We will now study the most common way the torsion of an elliptic curve over a number eld is computed. Let L be a local eld,O the ring of integers of L, P the (principal) maximal ideal ofO and p a generator of P. If m is a positive integer, let E(K)[m] denote the K-rational torsion points of order prime to m. Since K is local, each nonzero x in K can be written uniquely as x = up n for some unit u inO and integer n. Hence if (x;y;z) is inP 2 (L) then we can nd a,b,c inO with at least one of a,b,c a unit in O such that (a;b;c) = (x;y;z). Hence we obtain a well-dened reduction map P 2 (L)!P 2 (O=P ) (x;y;z)7! (a;b;c) For an elliptic curve E/O, let e E denote the reduction of E mod p. The map above induces a map E(L)! e E(O=P ) Theorem2.2.9. [Sil09, Prop. 3.1] With notation as above, suppose E/L is an elliptic curve with good reduction at P and let n = char(O=P ). We have an injection: E(L)[n],! e E(O=P ) We apply this result as follows: Let K be a number eld and v the valuation on K corresponding to P. The completion of K with respect to v, denoted K v , is a local eld 14 with ring of integersO v and unique maximal ideal P v . Clearly E(K) ,! E(K v ) and by Theorem 2.2.9 E(K v )[n] ,! E(O v =P v ). ButO v =P v is a eld, and in fact a nite eld of orderjO=Pj so the order (and group structure) E(O v =P v ) can be determined in a nite amount of time. Using this result, one can usually get very good bounds on the order of E(K) tor (and a short list of possible group structures). In order to determineE(K) tor itself we will need one more tool: division polynomials. Theorem 2.2.10. Let E/K be an elliptic curve. For each positive integer n, there exists an f2K[x;y] such that div(f) = X P2E[n]=1 (P1) Furthermore if n is odd then f2K[x] and if n is even then f=y2K[x]. If is E/K an elliptic curve in short Weierstrass form then we can construct the division polynomials inductively as follows: 0 = 0 1 = 1 2 = 2y 3 = 3x 4 + 6Ax 2 + 12BxA 2 4 = 4y(x 6 + 5Ax 4 + 20Bx 3 5A 2 x 2 4ABx 8B 2 A 3 ) 2m+1 = m+2 3 m m1 3 m+1 for m 2 2m = (2y) 1 ( m )( m+2 2 m1 m2 2 m+1 ) for m 3 We can now compute E(K) tor as follows: 1) Obtain a bound n = Q k i=1 p e i i onjE(K) tor j as described after Theorem 2.2.9. Alter- natively, when K= Q Theorem 1.0.2 yields a bound and when K is a quadratic eld, Theorem 1.0.3 yields a bound. 2) For each i, 1in, compute p e i i and nd its roots over K. 3) For each root r found above, computer 3 +Ar +B and determine if it is a square s in K. If so, then (r; p s)2E(K) tor and otherwise r is not the x-coordinate of a point inE(K) tor . 2.3 Rank Let K be a number eld. Suppose that E/K is an elliptic curve and E(K) has a point P of order 2. By translating we may assume E has a model of the form y 2 =x 3 +ax 2 +bx and 15 P=(0,0). We will describe a method for computing the rank r of E(K). By Theorem 1.0.1 it follows that [E : 2E] = 2 r+t where t is dened by E(K)[2] = 2 t . Since P is in E(K), the subgroupH :=f1;Pg is Gal(K=K)-invariant, so by [Sil09, Rmk 4.13.2] there is an elliptic curve E'/K and an isogeny :E!E 0 dened over K such that Ker() =H. In fact, we can take E' to have model y 2 = x 3 2ax 2 + (a 2 b)x and (x;y) = (( y x ) 2 ; y(bx 2 ) x 2 ). By [Sil09, Thm 6.2] there is an isogeny ^ :E 0 !E such that ^ = [2] on E and ^ = [2] on E'. Elementary group theory shows 2 r = [E 0 :(E)][E : ^ (E 0 )] 4 Hence to compute the rank it suces to compute the two indices in the equation above. There is a homomorphism :E 0 (K)!K =(K ) 2 17! 1 (0; 0)7!a 2 4b (x;y)7!x(x6= 0) and Ker() =(E), so [E 0 :(E)] =jIm()j. To each d inK =(K ) 2 we associate a smooth genus 1 curve C d : dw 2 =d 2 2adz 2 + (a 2 4b)z 4 The curveC d has the property thatC d (K)6=; if and only if d2Im(). Furthermore, Im() is contained in a nite, computable set denoted Sel (E=K), the -Selmer group for E/K. Let S be the set of prime ideals P ofO K dividing 2, b or a 2 4b. K(S; 2) :=fd2K =(K ) 2 j ord v (d) 0 mod 2 for all prime ideals P 62Sg Sel (E=K) :=fd2K(S; 2)j C d (K v )6=; for all places v of Kg Now K(S,2) is nite and for each d in K(S,2) and each v in S, one can check whether C d (K v )6=; by Hensel's lemma. Hence Sel (E=K) is computable as well. The map induces a short exact sequence: 0!E 0 =(E)!Sel (E=K)!X(E=K)[]! 0 16 where the nal termX(E=K) denotes the Tate-Shafarevich group. Hence we obtain the bound [E 0 :(E)]jSel (E=K)j. If for each d2Sel (E=K) we have C d (K)6=;, then Im() = Sel (E=K) and hence [E 0 : (E)] =jSel (E=K)j. Applying the method above with E replaced E 0 also yields a bound [E : ^ (E 0 )]jSel ^ (E 0 =K)j and so we obtain a bound on the rank. 2.4 Weil Pairing Let K be a number eld and E/K an elliptic curve. We relate K and E(K) tor via the Weil pairing. We will use this pairing to deduce restrictions on the sort of growth of torsion that can occur. Proposition 2.2.11. [Sil09] Let E/K be an elliptic curve, n a positive integer and U n the set of nth roots of unity in K. There exists a map h;i :E[n]E[n]!U n such that: 1. (bilinear)ha;b +ci =ha;biha;ci andhb +c;ai =hb;aihc;ai for all a,b,c2 E[n]. 2. (alternating)ha;ai = 1 for all a2 E[n] and henceha;bi =hb;ai 1 for all a,b2 E[n]. 3. (nondegenerate) If a2 E[n] andha;bi = 1 for all b2 E[n] then a = 0. 4. (Galois-invariant) (ha;bi) =h(a);(b)i for all a,b2 E[n] and 2 Gal(K=K). If we x a basisfv 1 ;v 2 g of E[n] then we can obtain a more concrete description of this pairing. Using (1) above, for any a 1 ;a 2 ;b 1 ;b 2 2Z=nZ we have ha 1 v 1 +a 2 v 2 ;b 1 v 1 +b 2 v 2 i =hv 1 ;v 2 i det(M) where M = " a 1 b 1 a 2 b 2 # Furthermorehv 1 ;v 2 i must be a primitive nth root of unity n , as otherwise using (1) above we can contradict (3). With respect to the basis above, we have a Galois representation : Gal(K/K)! Gl 2 (Z=nZ) 17 7!M and hence ( n ) =h(v 1 );(v 2 )i = det(M) n Corollary 2.2.12. Let E/K be an elliptic curve, n a positive integer and K(E[n]) the eld of denition of E[n]. Then n 2K(E[n]). Proof. If 2G :=Gal(K=K(E[n])) then xes K(E[n]) so M is the identity and hence ( n ) = det(M) n = 1 n = n . Therefore G xes K( n ) so by the Galois correspondence, we have K( n )K(E[n]). Corollary2.2.13. Let K=Q( p D) (D =1;3), E/K and elliptic curve and L a quadratic extension of K. Then the only odd prime power n such that C n C n E(L) is n=3. Proof. Let denote Euler's totient function. If n =p t where p is a prime and C n C n E(L) then by Corollary 2.2.12, L contains an nth root of unity n . Hence (p 1)p t1 = (n) [L : Q] = 4 so n=2,3,4,5 or 8. Note there is either a 3rd or 4th root of unity in K, so if 5 is in L, then there is a 15th or 20th root of unity in L. But (15) = 8> 4 and (20) = 8> 4, a contradiction. On the other hand, there is an elliptic curve E/Q (namely [0,-1,1, 217, -282]) which has full 3-torsion over K=Q( p 3) and hence provides examples in each case with L=Q( p 1; p 3). 3 Modular Curves X 0 and X 1 Let H =fx2 CjIm(x) > 0g and let = SL 2 (Z) denote the modular group. If M2 , M = " a b c d # then M induces an automorphism of H via z7! az+b cz+d and hence (and its subgroups) act onH. The following subgroups are of particular importance: 0 (N) =f " a b c d # 2 j " a b c d # " 0 # modNg 1 (N) =f " a b c d # 2 j " a b c d # " 1 0 1 # modNg (N) =f " a b c d # 2 j " a b c d # " 1 0 0 1 # mod Ng 18 The quotientH= is an open Riemann surface and by adding a nite number of points (called cusps) one obtains a compact Riemann surface X(N). Hence X(N) is an algebraic curve. In fact, this curve has a model over Q. Similarly, X 0 (N) and X 1 (N) have models overQ. This is one of the main reasons one can get so much information out of these curves: because they are dened overQ, if K is any number eld, we have a wealth of tools to study their K-rational points. For a concrete construction of the curve X 1 (N) see [Rei86]. The cusps of X i (N) can be described as follows: i (N) acts on P 1 (Q) via M [e;f] = [ae +bf;ce +df] (e,f2Q). Furthemore if we identifyP 1 (Q) withQ[1 then the action of onQ is compatible with its action on H. The cusps of X i (N) can be identied with the orbits ofP 1 (Q) under this action. The curves are related to the torsion part of an elliptic curve due to the following interpretation: The non-cuspidal K-rational points of X 1 (N) classify isomorphism classes of pairs (E,P) where E/K is an elliptic curve and P is a K-rational point of order N on E. Two pairs (E,P),( ~ E; ~ P ) are equivalent i there is an isomorphism f:E7! ~ E such that f(P)= ~ P . The non-cuspidal K-rational points of X 0 (N) classify isomorphism classes of pairs (E,C) where E/K is an elliptic curve and C is a cyclic subgroup of E of order N such that C is invariant under the action of Gal( K=K). Two pairs (E,C),( ~ E; ~ C) are equivalent i there is an isomorphism f:E7! ~ E such that f(C)= ~ C. A smooth projective curve X of genus at least 2 is hyperelliptic if there is a degree 2 map X! P 1 . The curve X is bielliptic if there is an elliptic curve E and a degree 2 map X ! E. The existence of such maps may make it easier to nd quadratic points on X. For convenience we have organized this information in Table 2.1. 19 Table 2.1: Models for X 0 (N) Model Genus N P 1 0 1-10,12,13,16,18,25 Elliptic 1 11,14,15,17,19-21,24,27,32,36,49 Hyperelliptic 2 22,23,26,28,29,31,37, 50 3 30,33,39,40,41,48 4 47 5 46,59 6 71 Bielliptic 2 22,26,28,37,50 3 30,33,34,35,39,40,43,45,48,64 4 38,44,53,54,61,81 5 42,51,55,56,63,65,72, 75 6 79 7 60, 62,69, 83, 89, 8 101 9 95 10 92 11 94, 119, 131 y The bolded numbers are hyperelliptic and bielliptic. 20 Chapter 3 Growth in Case E(K)[2]6C 2 C 2 1 Restrictions on Growth in Quadratic Extensions The structure of the n-torsion of an elliptic curve is well-understood over K: Theorem 3.1.1. [Sil09, Cor. 6.4] If K is a eld with char(K)=0, E/K is an elliptic curve and n is a positive integer then E(K)[n]C n C n The following theorem lists various restrictions on growth in quadratic extensions. It is a generalization of Theorem 5 in [GJT14] (where it is assumed K=Q). Theorem 3.1.2. Let K be a number eld, E/K an elliptic curve, L a quadratic extension of K and p an odd prime. 1. If E(K)[2]C 1 then E(L)[2]C 1 . 2. If d2 K, d6= 0, then E d (K)[2]E(K)[2]. 3. If E(K) tor C 2k with k odd then C 2 C 4 6 E(L) tor . 4. If E(K) tor C 4 then E(L) tor 6C 16 . 5. [Naj12] Let k be a positive integer. If E(L) tor C 2p k then E d (K) tor C 2p k for some d2 K, d6= 0. 6. If E(K)[p]C 1 and E(L)[p]C p C p then K contains a primitive pth root of unity. 7. If E(K)[p]C p and E(L)[p 1 ]6= E(K)[p 1 ] then E(L)[p]C p C p . 8. If E(K)[p]C p and E(L)[p]C p C p then K does not contain a primitive pth root of unity. 9. If E(K)[p]C p C p then E(L)[p 1 ] = E(K)[p 1 ]. 21 Proof. 1) We may assume E has a model of the formy 2 =f(x) (deg(f)=3) with identity1. Hence the order 2 points are of the form (r; 0) where f(r) = 0. If E(K)[2]C 1 then f(x) is irreducible over K so 3 is the smallest degree of an extension M of K such that E(M)[2] 6C 1 . As [L:K] =2, E(L)[2]C 1 . 2) Suppose f(x) = x 3 +ax 2 +bx +c for some a,b,c2 K and for d2 K, d6= 0 let f d (x) =x 3 +dax 2 +d 2 bx +d 3 c. If r2K and f(r)=0, then f d (dr) = 0 and this induces a bijection between the K-roots of f and f d . 3) Suppose E(K) tor C 2k with k odd. Then WLOG we may assume E has a model of the form y 2 =x(x +r)(x +r) where r;r are roots of a quadratic polynomial that is irreducible over K. The unique quadratic extension L of K in which C 2 C 2 L is L =K(r) =K(r). We now claim that E(L) has no point P of order 4. If such a point P exists then by Lemma 4.0.1 one of the following pairs consists of squares in K:(i) r;r (ii)r;rr (iii)r;rr. Furthermore if (i) consists of squares, we can write r =u 2 ;r =u 2 for some u2K and the following point P satises 2P = (0; 0): P = (uu;uu(u +u)) Since the coordinates of P are symmetric in u and u, P is actually dened over K, so E(K) has a point of order 4, a contradiction to the hypothesis. Hence (i) does not hold. Furthermore (ii) consists of squares i (iii) consists of squares. Let L = K( p d) for some non-square d2 K and suppose (iii) holds. If r = a +b p d for some a;b2 K, b6= 0 then rr = 2b p d is a square in K. But if v;w2 K and (v +w p d) 2 = 2b p d then w6= 0, v 2 +dw 2 = 0 and 2vw = 2b. From the rst equality we conclude d =s 2 for some s2K so WLOG we can assume d =1. But then1 is a square in L so asr is a square by assumption, we conclude r and hence r are squares as well. Hence (i) holds, but we have shown this case cannot occur. Hence E(L) has no point of order 4. 4) See [GJT14, Thm 5]. 5) Let L = K( p d). If E(L) tor C 2p k then by Corollary 2.2.5 it follows that E(K) tor or E d (K) tor contains a point of order p k . But by (1), E(K) tor must have a point of order 2 and hence E d (K) tor must as well by (2). Therefore, some twist of E has a point of order 2p k dened over K. 6) Suppose E(K)[p] is trivial. By Corollary 2.2.5, it follows that E d (K)[p]C p C p so by Corollary 2.2.12 we conclude K contains a primitive pth root of unity. 22 7) Let m the largest positive integer such that there is an element of orderp m inE(L) tor . We have E(L)[p m ] = E(K)[p m ]E d (K)[p m ] by Corollary 2.2.5. If E(L)[p m ]6= E(K)[p m ] then E d (K)[p m ]6 C 1 so E d (K)[p]6 C 1 . Hence E d (K)[p] C p or C p C p by Theorem 3.1.1. In the latter case this would yield E(L)[p]C p C p C p which contradicts Theorem 3.1.1, so E(L)[p]C p C p . 8) Let p be a primtive pth root of unity. Suppose E(K)C p , and E(L)[p]C p C p . Let 2 Gal(L/K) be nontrivial. We can choose a basis for E(L)[p] such that the induced Galois representation satises : Gal(L/K)! Gl 2 (Z=pZ) 7! " 1 0 # for some 2 (Z=pZ) , 2Z=pZ. If p 2K then p =( p ) = ( p ) det(()) = ( p ) by the comment before Corollary 2.2.12 so = 1 mod p. As 2 = 1, (()) 2 = 1 so 2 = 0. As p is odd, we conclude = 0, so () is the identity. This means acts trivially on the p-torsion, so E(K)[p]C p C p , contradicting our hypothesis. 9) Suppose E(K)[p] C p C p . By Corollary 2.2.5, if E(L)[p 1 ] 6= E(K)[p 1 ] then E d (K)[p 1 ]6 C 1 so E d (K)[p]6 C 1 . Hence C p C p C p E(L)[p], contradicting Theo- rem 3.1.1. Note that there are growths which occur overQ but not over some quadratic extension: C 3 toC 3 C 3 occurs overQ but by Theorem 3.1.2(8) not overQ( p 3). On the other hand, there are growths which occur over a quadratic eld but not overQ: C 1 toC 3 C 3 cannot overQ because if it did,Q would contain a primitive 3rd root of unity by Theorem 3.1.2(6). On the other hand, C 1 to C 3 C 3 occurs overQ( p 3) (see table below). Also, C 1 to C 15 cannot occur overQ by Corollary 2.2.5 since X 1 (15) has no noncuspidalQ-rational points. On the other hand, this growth does occur over K =Q( p 5)): By [Naj12, Thm 2] there is an elliptic curve E/K withE(K) tor C 15 . Choose d2K,d6= 0, such thatE d (K) tor =C 1 . Then E(K( p d)) tor C 15 by Corollary 2.2.5 and Theorem 3.1.2(1). 23 2 K-Rational Points on X 0 (N) Lemma 3.2.1. Let p be a prime and E=F p an elliptic curve with model y 2 =x 3 +Ax +B. 1. If A=0 (i.e. j(E)= 0) and p 2 mod 3 thenjE(F p )j =p + 1 andjE(F p 2)j = (p + 1) 2 . 2. If B=0 (i.e. j(E)=1728) and p 3 mod 4 thenjE(F p )j =p+1 andjE(F p 2)j = (p+1) 2 . Proof. If A=0 and p 2 mod 3 thenjE(F p )j = p + 1 by [Was08, Prop 4.33]. If B=0 and p 3 mod 4 thenjE(F p )j = p + 1 by [Was08, Thm 4.23]. Now in either case above, jE(F p 2)j = p 2 + 1 ( 2 + 2 ) by [Was08, Thm 4.12], where and are roots of x 2 +p. Hence jE(F p 2)j =p 2 + 1 ( 2 + 2 ) =p 2 + 1 (pp) =p 2 + 2p + 1 = (p + 1) 2 Theorem 3.2.2. Let K be a quadratic eld and E/K an elliptic curve. If j(E)=0 and p> 3 is a prime thenE(K) tor has no element of order p. If j(E)=1728 and p> 2 is a prime then E(K) tor has no element of order p. Proof. Suppose j(E)=0. Twisting by a square inO K if necessary, we may assume E has a model of the form y 2 =x 3 +AX +B with A,B2O K . Note that sinceO K is a Dedekind domain, the principal ideal (disc(E)) has only a nite number of prime ideal divisors, and hence disc(E) lies in only a nite number of prime ideals ofO K . Let q > 3 be a prime in Z. Since q6= 3, by the Chinese remainder theorem there exists an integer n satisfying: n + 1 2 mod q n 2 mod 3 Furthermore, n+3qk satises the congruences above for every integer k, and (n; 3q) = 1 by the congruences above. Hence by Dirichlet's theorem on arithmetic progressions, there are innitely many primes in this arithmetic progression. In particular, there is a prime 24 p satisfying the congruences above such that E has good reduction modulo a prime ideal above p. As [K : Q] = 2, we haveO K = F p or F p 2. By the comments following Theorem 2.2.9 we have an injection of the group E(K)[p] into E(F p ) or E(F 2 p ). But by Lemma 3.2.1 we have: jE(F p )j =p + 1 26 0 mod q jE(F p 2)j = (p + 1) 2 46 0 mod q as q6= 2. Hence in either case (noting p6= q), we conclude there is no point of order q in E(K) tor . Now suppose j=1728. If q is an odd prime, then one can argue just as in the j=0 case that there is no point of order q. 2.1 N=21 By an n-isogeny of an elliptic curve E/K we mean a cyclic subgroup of E(K) which is invariant under the action of Gal(K/K). Magma describes an n-isogeny C by providing a polynomialf C whose roots are precisely the x-coordinates of the points in C. LetK C denote the eld of denition of C and for a polynomial f, let K(f) denote the splitting eld of f over K. If an n-isogeny C is dened over a eld L then f C should split completely over L. In particular if L is a quadratic extension of K, then f C must have irreducible factors of degree at most 2 over K. We will now argue that no elliptic curve overK =Q( p 3) has a 21-isogeny dened over a quadratic extension of K. Magma tells us the modular curve X 0 (21) has rank 0, torsion C 2 C 8 over K and 4 cusps over K. The 12 non-cuspidal points correspond to isomorphism classes (E;C) and using Magma we found representatives of each class (see Table 3.1). As one can see from the table, for each representative (E,C) with j6= 0, f C has an irreducible factor of degree at least 3 and hence [K C :K] [K(f C ) :K] 3. So C will not be dened over a quadratic extension of K. Now since in each of the cases just mentioned, j6= 0; 1728, by Proposition 2:2:1 the isomorphism class of (E,C) just consists of (E d ;C d ) for nonzero d2 K, where C d denotes the image of C under quadratic twist by d. If (x;y)2 C then (dx;d 3=2 y)2C d . Asd2K,K(f C ) =K(f C d). Hence the 8 isomorphism classes with j6= 0 in Table 3.1 do not contain an example of an elliptic curve E/K with a 21-isogeny. In the j = 0; 1728 case, Magma is not yet able to describe the isomorphism class, so we instead argue as follows: If there is an elliptic curve E/K with a 21-isogeny dened over a quadratic 25 extension L of K, then by Corollary 2.2.5 there is an elliptic curve E/K with a point of order 7. But this is impossible by Theorem 3.2.2. Table 3.1: (K =Q( p 3)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (21) Point j(E) E f C (1=4; 1=8) 3375=2 [20=441;16=27783] (1; 3; 3; 3) (2;1) 189613868625=128 [1915=36;48383=324] (1; 3; 6) (1; 2) 1159088625=2097152 [505=192;23053=6912] (1; 3; 6) (5; 13) 140625=8 [1600=147;134144=9261] (1; 3; 3; 3) ( (+1) 2 ; 1) 12288000 [ (40+10) 49 ; (25306831) 12348 ] (1; 3; 6) ( (+1) 2 ; 1) 12288000 [ (40+10) 49 ; (25306831) 12348 ] (1; 3; 6) ( (+1) 2 ; (3+1) 2 ) 54000 [ 135585 98 ; 6601782 343 ] (1; 3; 6) ( (+1) 2 ; (3+1) 2 ) 54000 [ 135585 98 ; 6601782 343 ] (1; 3; 6) ( 35 2 ; 8) 0 See Theorem 3.2.2 ( 35 2 ; 8) 0 See Theorem 3.2.2 ( 35 2 ; 311 2 ) 0 See Theorem 3.2.2 ( 35 2 ; 311 2 ) 0 See Theorem 3.2.2 y In the last column we list the degrees of the irreducible factors of f C over K. z We use the model y 2 +xy =x 3 4x 1 for X 0 (21). On the other hand, over K =Q(i), X 0 (21) has rank 1. A search of points has not yet produced an example of a 21-isogeny dened over a quadratic extension of K. 26 2.2 N=15 Table 3.2: (K =Q( p 1)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (15) Point j(E) E Deg(f C ) (8;27) 121945=32 [87=20;421=100] (1,1,1,2,2) (-2, -2) 46969655=32768 [633=54080; 239=1081600] (1,1,1,2,2) (-13/4, 9/8) -25/2 [12=25;944=625] (1,2,4) (3,-2) -349938025/8 [ 46272 4225 ; 1473536 105625 ] (1,2,4) (1=2; 15i3 4 ) 198261i62613 2 [ 6846i+9528 105625 ; 22652i+30164 2640625 ] (1,2,4) (1=2; 15i3 4 ) 198261i62613 2 [ 6846i+9528 105625 ; 22652i+30164 2640625 ] (1,2,4) (3i 1;6i + 6) 15363i47709 256 [ 3i+96 200 ; 3989i373 10000 ] (1,2,4) (3i 1; 6i + 6) 15363i47709 256 [ 3i+96 200 ; 3989i373 10000 ] (1,2,4) (3i 1; 3i 6) 13670181i+19928133 8 [ 2583i+9444 8450 ; 93373i+39511 211250 ] (1,2,4) (3i 1;3i 6) 13670181i+19928133 8 [ 2583i+9444 8450 ; 93373i+39511 211250 ] (1,2,4) (7; 15i + 3) 86643i1971 4 [ 216i2688 625 ; 8608i53344 15625 ] (1,2,4) (7;15i + 3) 86643i1971 4 [ 216i2688 625 ; 8608i53344 15625 ] (1,2,4) y In the last column we list the degrees of the irreducible factors of f C over K. z We use the model y 2 +xy +y =x 3 +x 2 10x 10 for X 0 (15). Now we will classify 15-isogenies overK =Q(i). We see the rst two entries in Table 3.2 indicate the only potential isomorphism classes in which we could nd a 15-isogeny dened over a quadratic extension of K. Hence if a pair (E,C) exists with E/K, C Gal(K/K)- invariant and C dened over a quadratic extension of K then in fact E is dened over Q and C is Gal(Q=Q)-invariant. The point (8,-27) corresponds to (E,C) with f C = (x 7=10)(x + 1=2)(x + 17=10)(x 2 +x 139=20)(x 2 + 13x + 269=20) A brief computation yields K(f C ) = Q( p 5). Since j 6= 0; 1728, any pair (E 0 ;C 0 ) equivalent to (E;C) is of the formE 0 =E d ,C 0 =C d for some d in K. AsK(f C ) =K(f C d), if C is dened over a quadratic extension of K then it must be dened over K( p 5). The point (1=2; 3 p 6=5) is in C, so the only potential d-twists (up to a square in K) in which C d is dened over a quadratic extension of K (namely K( p 5)) are d =6;6 5. Magma now tells us that for these two values of d, E d (K( p 5)) tor C 15 . 27 The point (-2,-2) corresponds to (E,C) with f C = (x3=104)(x+17=520)(x+113=520)(x 2 (11=52)x+2333=54080)(x 2 +(1=52)x+437=54080) A brief computation yields K(f C ) = Q( p 15), so as above, if C is dened over a quadratic extension of K then it must be dened over K( p 15). The point (3=104; 4 p 26=845) is in C, so the only potential d-twists in which C d is dened over a quadratic extension of K are d = 26; 26 (15). Magma now tells us that for these two values of d,E d (K( p 15)) tor C 15 . Hence there are exactly four elliptic curves over K (up to isomorphism over K) with a 15-isogeny dened over a quadratic extension of K. Similarly, when K =Q( p 3) we nd the same four elliptic curves are the only elliptic curves over K with a 15-isogeny dened over a quadratic extension of K. 2.3 N=20 Table 3.3: (K =Q( p 1)) Representatives (E,C) of isomorphism classes corresponding to non-cuspidal K-rational points on a model of X 0 (20) Point j(E) E f C (2i; 0) 287496 [ 264i+77 625 ; 616i+1638 15625 ] (1; 1; 2; 2; 4) (2i; 0) 287496 [ 264i+77 625 ; 616i+1638 15625 ] (1; 1; 2; 2; 4) (2i 2;2i 4) 287496 [ 264i+77 625 ; 616i+1638 15625 ] (1; 1; 2; 2; 4) (2i 2; 2i 4) 287496 [ 264i+77 625 ; 616i+1638 15625 ] (1; 1; 2; 2; 4) (2i 2; 2i + 4) 1728 See Theorem 3.2.2 (2i 2;2i + 4) 1728 See Theorem 3.2.2 y In the last column we list the degrees of the irreducible factors of f C over K. z We use the model y 2 =x 3 +x 2 + 4x + 4 for X 0 (20). Over K =Q( p 3), X 0 (20) has rank 0, torsion C 6 and these points are all cusps. Over K = Q( p 1), X 0 (20) has rank 0, torsion C 2 C 6 and 6 cusps. Table 3.3 shows that 4 non-cuspidal K-points correspond to 20-isogenies dened over extensions of K of degree at least 4. If E/K has j(E) = 1728 and a 20-isogeny dened over a quadratic extension of K then some quadratic twist of E has a point of order 5 over K by Corollary 2.2.5. But this contradicticts Theorem 3.2.2. Therefore there are no 20-isogenies dened over quadratic extensions of K =Q( p 3) orQ(i). 28 2.4 N=24 Over Q, X 0 (24) has rank 0, torsion C 2 C 4 and 8 cusps. The torsion and rank do not grow upon extension to K =Q( p 3) orQ(i) so there are no elliptic curves over K with a 24-isogeny (dened over any extension of K). 2.5 N=27 The curveX 0 (27) is an elliptic curve with modely 2 +y =x 3 7. OverQ,X 0 (27) has rank 0, torsion C 3 and 2 cusps. The torsion and rank do not grow upon extension to K = Q(i) The one non-cuspidal point (3,-5) corresponds to a pair (E,C) with j(E) = -12288000 and the degrees of the irreducible factors of f c over Q(i) are (1,3,9). Because j(E)6= 0; 1728, the isomorphism class of (E,C) just consists of quadratic twists of this pair, and hence will yield the same degrees of irreducible factors. Over K =Q( p 3), X 0 (27) has 6 cusps and X 0 (27)(K) = C 3 C 3 . As in the case K = Q(i), the 3 non-cuspidal points do not yield 27-isogenies dened over a quadratic extension of K. Therefore in either case, there are no elliptic curves over K with a 27-isogeny dened over a quadratic extension of K. 2.6 N=30 Let K = Q( p 3) or Q(i). If E/K possesses a cyclic Gal(K=K)-invariant subgroup C of order 30, then C has a unique cyclic subgroup of order 15 and hence this subgroup is Gal(K=K)-invariant as well. So if C is dened over a quadratic extension of K then E possesses a 15-isogeny dened over a quadratic extension of K. But there are only four such pairs (E;C 0 ), and we found that in each case C 0 was dened over a unique quadratic extension (i.e. C 0 was not dened over K) so C would have to be dened over that same extension. But as already noted,the torsion over the extension was C 15 in each case, so there are no 30-isogenies over K (dened over a quadratic extension of K). 2.7 N=35 Magma tells us X 0 (35) is genus 3 with ane model y 2 + (x 4 x 2 1)y =x 7 2x 6 x 5 3x 4 +x 3 2x 2 +x 29 Furthermore Magma found an automorphism of X 0 (35) such that the quotient curve E is genus 1 with ane model: y 2 +y =x 3 +x 2 + 9x + 1 The quotient map (dened between the projective closures) is given by: f :X 0 (35)!E (x;y;z)7! (p f 1 ;p f 2 ;p f 3 ) p f 1 =x 4 5x 3 z 8x 2 z 2 + 5xz 3 +z 4 p f 2 = 3x 4 x 3 z + 4x 2 z 2 +xz 3 7yz 3 + 3z 4 p f 3 =x 4 + 2x 3 zx 2 z 2 2xz 3 +z 4 Because f is a rational map, the only potential K-rational points of X 0 (35) are the non- regular points of f andf 1 (E(K)). In order to computef 1 (E(K)) we must rst compute E(K). Magma/Sage give us the following information: Table 3.4: K-Rational Points on a Genus 1 Quotient of X 0 (35) K rk(E(K)) E(K) tor Points of E(K) tor Q( p 1) 0 C 3 [0,1,0], [1,3,1],[1,-4,1] Q( p 3) 0 C 3 C 3 [0; 1; 0]; [1; 3; 1]; [1;4; 1]; [ 1 2 (5 1); 1 2 (5 + 9); 1]; [ 1 2 (5 1); 1 2 (5 + 9); 1]; [ 1 2 (5 1); 1 2 (5 11); 1]; [ 1 2 (5 1); 1 2 (5 11); 1]; [ 4 3 ; 1 18 (35 9); 1]; [ 4 3 ; 1 18 (35 9); 1] To computef 1 ([x;y;z]), we form the ideal<C;p f 1 wx;p f 2 wy;p f 3 wz> (C denotes the model of X 0 (35) above) and compute it's Gr obner Basis (with respect to the ordering x,y,z,w). Often, one can nd basis elements that allow the system to be solved by hand. We can assumez6= 0 as the only point on our model ofX 0 (35) with this property is [0,1,0] and f is not dened at this point. 30 Table 3.5: Gr obner basis data for determination of f 1 (E(K)) Point P of E(K) f 1 (P ) Gr obner basis elements [0,1,0] ; w 2 [1,3,1] [0,0,1] xw 2 ;yw 2 [1,-4,1] [0,1,1] xw 2 ;yw 2 zw 2 For each of the six extra points over K =Q( p 3) the Gr obner basis contains a poly- nomial g(w). Using Magma one can check that in each case the only root of g(w) over K is 0. Hence the (K-rational) inverse image of these points under f is empty. Finally, using a Gr obner basis for the ideal <C;p f 1 ;p f 2 ;p f 3 > we can determine the non- regular points of f. Ifz6= 0 then the Groebner basis containsy 2 6y + 4. This has no roots over K. Hence the only K-rational points on X 0 (35) are [0,0,1], [0,1,0] and [0,1,1]. These points are all cusps so there are no 35-isogenies dened over K =Q( p 1);Q( p 3). We summarize our ndings in the following theorem. Theorem3.2.3. If K =Q( p 3) and E/K is an elliptic curve, E has no N-isogenies dened over a quadratic extension of K for N= 20,21,24,27,30,35,45. If K = Q( p 1) and E/K is an elliptic curve, E has no N-isogenies dened over a quadratic extension of K for N= 20,24,27,30,35,45. The curve X 0 (21) is genus 1 and rank 1 over K. In either case above, there are exactly four elliptic curves over K (up to isomorphism over K) with a 15-isogeny dened over a quadratic extension of K. 3 Classication of Twists Theorem 3.3.1. Let K = Q( p 3), d2O K , d a nonsquare, and E/K an elliptic curve with full 2-torsion. 1. If E(K) tor C 7 ;C 9 or C 3 C 3 ; then E d (K) tor C 1 2. If E(K) tor C 3 then E d (K) tor C 1 or C 5 3. If E(K) tor C 5 then E d (K) tor C 1 or C 3 4. If E(K) tor C 1 then E d (K) tor C 1 ;C 3 ;C 5 ;C 7 ;C 9 or C 3 C 3 Hence the torsion structuresC 7 ;C 9 andC 3 C 3 do not grow in any quadratic extension of K. The torsion structures C 3 and C 5 grow in at most 1 extension, and C 1 grows in at most 2 extensions. 31 Proof. Let d2 K be a non-square. Note that if E 0 is a quadratic twist of E then E is a quadratic twist ofE 0 (up to isomorphism over K). Also by Theorem 3.1.2, all quadratic twists of a curve with odd order torsion will be odd order. By Theorem 1.0.4 the only odd-order torsion structures occurring over K areC 1 ;C 3 ;C 5 ;C 7 ;C 9 andC 3 C 3 . Now ifE(K)[3]6=C 1 and E d (K)[3]6= C 1 then by Corollary 2.2.5, E(K( p d))[3] = C 3 C 3 , C 3 C 3 C 3 or C 3 C 3 C 3 C 3 , contradicting Theorem 3.1.2 Item 8 or Theorem 3.1.1 respectively. If m = 5 or 7,E(K) tor C m andE d (K) tor C m , then by Proposition 2.2.7,C m C m E(L), contradicting Corollary 2.2.13. IfE(K)[3]6=C 1 andE d (K)C 7 , then by Proposition 2.2.7 E d has a 21-isogeny dened over a quadratic extension of K. But no such isogeny exists by Theorem 3.2.3. If E(K)[5] =C 5 and E d (K)C 7 , then by Proposition 2.2.7 E d has a 35-isogeny dened over a quadratic extension. But no such isogeny exists by Theorem 3.2.3. If E(K)[3]6= C 1 and E d (K) C 5 , then by Proposition 2.2.7 E d has a 15-isogeny dened over a quadratic extension. There are four elliptic curves (two pairs of quadratic twists) over K (up to isomorphism over K) with a 15-isogeny dened over a quadratic eld. For each such curve E (we actually need only check one member of each pair), the factorization of the 3-division polynomial of E indicates that the nontrivial torsion structures occurring among the quadratic twists of E are C 3 and C 5 and each occurs exactly once. 4 Examples of Growth Given an elliptic curve E/K and a list of all possible torsion sructures over K (as in The- orem 1.0.2), one can describe all quadratic extensions L/K in which E(K) tor grows and describe E(L) tor in each such case. See the appendix (Section 2) for such an implementa- tion in Magma. Below we give an informal description of the algorithm. Input: The integer d =1 or3 and an elliptic curve E/K where K =Q( p d) Output: List of quadratic extensions where torsion grows and the torsion structure in each case 1. Let S :=f2; 3; 5; 7g be the set of primes p for which there is an elliptic curve E/K with pjjE(K) tor j. 2. Now for each prime p in S, determine the division polynomial f p of smallest degree necessary to detect growth of the p-part of E(K) tor in a quadratic extension of K: (a) Find the primary decomposition of E(K) tor . 32 (b) For each prime p in S, count the number S p of p-summands in the p-part of E(K) tor . (i) For p = 2: If S 2 = 0 then E(K)[2 1 ] does not grow in a quadratic extension by Theorem 3.1.2. If S 2 6= 0 and the E(K)[2 1 ] [2 a ; 2 b ] with a<b then let f 2 = 2 b. (ii) For p=3: If S p = 0 or 1 then if any growth occcurs, by Corollary 2.2.5, E(K)[p] must grow (Theorem 3.1.2), so we let f p := p . IfS p = 2 then the p-part cannot grow in a quadratic extension by Corol- lary 2.2.5 and Theorem 3.1.1, so let f p = 1. (iii) For p > 3: If S p = 0 then if any growth occcurs, by Corollary 2.2.5, E(K)[p] must grow by Theorem 3.1.2, so we let f p := p . IfS p = 1 then the p-part cannot grow in a quadratic extension by Corol- lary 2.2.13, so let f p = 1. IfS p = 2 then the p-part cannot grow in a quadratic extension by Corol- lary 2.2.5 and Theorem 3.1.1, so we ignore it. 3. For each p in S: (a) Factor f p over K. (b) For each factor g of f p over K: (i) If degree(g) = 1: We may write g = x-c. Compute (c,d) on E. If d62 K, then the torsion grows in L=K(d) and Magma can compute E(L) tor . (ii) If degree(g) = 2: Construct the splitting eld L of g over K, and let c be a root of g in L. Compute (c,d) on E. If d2 L, then the torsion grows in L and Magma can compute E(L) tor . 33 Table 3.6: Examples of Growth of Torsion over K =Q( p 1) in Case E(K)[2]6C 2 C 2 E(K) tor E(L) tor Model for E C 1 C 3 [ 0, 1, 1, -769, -8470 ] C 5 [ 1, 0, 1, 549, -2202 ] ] C 7 [ 1, -1, 0, -24, -64 ] C 9 [ 1, -1, 0, -123, -667 ] C 2 C 4 [ 0, 1, 0, -41, -116 ] C 6 [ 1, 0, 1, -171, -874 ] C 8 [ 1, -1, 0, 0, -5 ] C 10 [ 1, 1, 0, -700, 34000 ] C 12 [ 1, 0, 1, -14, -64 ] C 16 [ 1, -1, 1, 47245, -2990253 ] C 2 C 2 [ 1, 0, 1, -171, -874 ] C 2 C 6 [ 0, 0, 0, 0, -27 ] C 2 C 10 [ 1, -1, 1, -6305, -924303 ] C 3 C 3 C 5 [ 1, 0, 1, -76, 298 ] C 3 C 3 [ 0, 1, 1, -9, -15 ] C 4 C 8 [ 1, 1, 1, -2160, -39540 ] C 12 [ 1, -1, 1, 13, -61 ] C 2 C 4 [ 1, 1, 1, -2160, -39540 ] C 2 C 1 2 [ 1, 1, 1, -338, -7969 ] C 2 C 8 [ 0, 0, 0, 6, -7 ] C 5 C 3 C 5 [ 1, 1, 1, -3, 1 ] C 6 C 12 [ 0, 1, 0, -1, 0 ] C 2 C 6 [ 1, 0, 1, 4, -6 ] C 3 C 6 [ 1, 0, 1, 4, -6 ] C 8 C 16 [ 1, 0, 0, 210, 900 ] C 2 C 8 [ 1, 0, 0, -39, 90 ] C 10 C 2 C 10 [1, 0, 0, -45, 81] C 12 C 2 C 12 [ 1, -1, 1, -122, 1721 ] 34 Table 3.7: Examples of Growth of Torsion over K =Q( p 3) in Case E(K)[2]6C 2 C 2 E(K) tor E(L) tor Model for E C 1 C 3 [ 1, 1, 1, -13, -219 ] C 5 [ 0, -1, 1, 42, 443 ] C 7 [ 1, 0, 1, -975873773, 11746188793640 ] C 9 [ 0, 0, 0, -1971, 44658 ] C 3 C 3 [ 0, -1, 1, 217, -282 ] C 2 C 4 [ 1, 1, 1, -2160, -39540 ] C 6 [ 0, -1, 0, -1, 0 ] C 8 [ 0, -1, 0, -384, -2772 ] C 10 [ 1, 1, 0, -20700, 1134000 ] C 12 [ 0, -1, 0, 24, -144 ] C 16 [ 1, -1, 1, 47245, -2990253 ] C 2 C 2 [ 1, 1, 1, -2160, -39540 ] C 2 C 6 [ 0, -1, 0, 4, -4 ] C 2 C 10 [ 1, 1, 0, -700, 34000 ] C 3 C 6 [ 1, 1, 0, -1740, 22184 ] C 6 C 6 [ 1, 1, 0, 220, 2192 ] C 3 C 3 C 5 [ 1, 0, 1, -76, 298 ] C 4 C 8 [ 1, 1, 1, -80, 242 ] C 12 [ 0, -1, 0, -4616, -119184 ] C 2 C 4 [ 1, 1, 1, -80, 242 ] C 2 C 12 [ 1, 1, 1, 37, 281 ] C 2 C 8 [ 1, 0, 1, -1, 23 ] C 4 C 4 [ 0, 0, 0, 13, -34 ] C 5 C 3 C 5 [ 1, 1, 1, -3, 1 ] C 6 C 12 [ 1, 0, 1, -69, -194 ] C 2 C 6 [ 1, 0, 1, -171, -874 ] C 8 C 16 [ 1, 0, 0, -39, 90 ] C 2 C 8 [ 1, 1, 1, 35, -28 ] C 10 C 2 C 10 [ 1, 0, 0, -45, 81 ] C 12 C 2 C 12 [ 1, 0, 1, 1, 2 ] C 3 C 6 C 6 C 6 [ 1, 0, 1, 4, -6 ] 35 Chapter 4 Growth in Case E(K)[2]C 2 C 2 The arguments in this chapter were adapted from [Kwo97], who studied the case K =Q. The following lemma plays a crucial role in studying the 2-part of the torsion of an elliptic curve: Lemma 4.0.1. [Kna92] Let E be an elliptic curve over a eld F with char(F) 6= 2; 3. Suppose E is given by y 2 = (x)(x)(x ) with ;; 2 F . For P = (x 0 ;y 0 ) in E(F) there exists Q in E(F) such that 2Q = P i x 0 ;x 0 ;x 0 are squares in F. If a point P meets the criteria in the theorem above, we will say the point P lifts in K. Below we have a variant of a result of Ono. His result held for elliptic curves over Q. Note the result below does not depend on the class group nor the unit group of the number eld K. Theorem 4.0.2. [Ono96] Let K be number eld and E/K an elliptic curve with full 2- torsion. Then E has a model of the form y 2 =x(x +M)(x +N) where M,N2O K . 1. E(K) has a point of order 4 i M, N are both squares (inO K ), or -M, N-M are both squares, or -N,M-N are both squares. 2. E(K) has a point of order 8 i there exists a d2O K , d6= 0 and a pythagorean triple (u,v,w) (i.e. u,v,w2O K with u 2 +w 2 = v 2 ) such that M = d 2 u 4 ;N = d 2 v 4 or M =d 2 u 4 ;NM =d 2 v 4 , orN =d 2 u 4 ;MN =d 2 v 4 . 3. E(K) contains a point of order 3 i there exists a;b;d2O K with d6= 0 and a b 62 f2;1; 1 2 ; 0; 1g such that M =a 3 (a + 2b)d 2 and N =b 3 (b + 2a)d 2 . Proof. 1) E has a point of order 4 over K i (0,0), (-M,0) or (-N,0) lifts over K. But by the Lemma 4.0.1, (0,0) lifts i M,N are squares in K, (-M,0) lifts iM;NM are squares in K and (-N,0) lifts iN;MN are squares in K. Also note that sinceO K is integrally closed, if x2O K is a square in K then it is a square inO K . 36 2) Suppose E(K) has a point P of order 8. By translating if necessary, we may assume WLOG that 4P = (0; 0). Hence by lemma 4.0.1 there exist a;b2O K such that M = a 2 and N =b 2 and the four order 4 points above (0,0) are: (ab;ab(a +b)) (ab;ab(ab)) If [2] denotes the multiplication-by-2 map on E, note thatjKer([2]j = 4 so we expect 4 points above. Replacinga bya if necessary, we may assume WLOG that (ab;ab(ab)) lifts over K. By lemma 4.0.1ab;ab+a 2 andab+b 2 are squares in K so there existsu;w2O K such that ab =u 2 and ab +b 2 =w 2 . Hence if we let d =b 1 then: M =a 2 = (b 1 ) 2 (ab) 2 =d 2 u 4 N =b 2 =d 2 b 4 Furthermore, u 2 +b 2 =ab +b 2 =w 2 The converse is easy to verify by lemma 4.0.1. 3) A point (x;y) on our curve is order 3 i x is a root of the 3-division polynomial: 3x 4 + 4(M +N)x 3 + 6MNx 2 M 2 N 2 = 0 As Ono notes, this curve (in M,N,x) has a rational parametrization (due to Nigel Boston) of the form: M x = (1 +t) 2 1 N x = (1 +t 1 ) 2 1 Hence M =x((1 +t) 2 1) N =x((1 +t 1 ) 2 1) 37 and as (x;y) lies on our curve, we have have: y 2 =x(x +M)(x +N) =x(x +x((1 +t) 2 1))(x +x((1 +t 1 ) 2 1)) =x 3 (1 + (1 +t) 2 1)(1 + (1 +t 1 ) 2 1) =x 3 (1 +t) 2 (1 +t 1 ) 2 and hence x is a square: x = [ y x(1 +t)(1 +t 1 ) ] 2 Let c = y x(1+t)(1+t 1 ) so we have x =c 2 . Then: M =c 2 ((1 +t) 2 1) =c 2 (t 2 + 2t) = ( c t ) 2 t 3 (t + 2) Similarly: N =c 2 ((1 +t 1 ) 2 1) = ( c t ) 2 (2t + 1) Writing t := a b for some a,b2O K and d := c tb 2 , the result follows. 1 Some Diopantine Equations The curve x 2 +y 2 =z 2 is genus 0 and hence can be parametrized. Proposition 4.1.1. Let K be a number eld and let C be the curve x 2 +y 2 =z 2 38 For every K-rational projective point P6= [1; 0; 1] on C there exists an m2 K such that P = [1m 2 ; 2m; 1 +m 2 ]. Proposition 4.1.2. Let K= Q( p D) (with D a squarefree integer) be any imaginary quadratic number eld with class number 1 except D=-7. The equation u 4 v 4 =w 2 has no solutions (u;v;w) with u;v;w2O K and (u;v;w)6= (0; 0; 0). Proof. Let C be the (ane) curve with equation: u 4 v 4 =w 2 and let E be the elliptic curve: y 2 =x 3 + 4x The curves E and C are birational via: f :C!E (u;v)7! (x;y) = 2v 2 u 2 w ; 4uv wu 2 Hence, in order to determine the K-rational points on C, it suces to consider the non-regular points of f and f 1 (E(K)). At a non-regular point (u,v,w) we have w = u 2 and hence v = 0. Now using Magma to nd the torsion subgroup and rank: Rank(E(K)) = 8 < : 1 D =7 0 otherwise E(K) tor = 8 < : C 2 C 4 D =1 C 4 otherwise Furthermore, the torsion points over K=Q(i) are (0; 0); (2;4); (2i; 0); (2;4i). If y = 0 then we see u=0 or v=0. Ifx =2 andy = 4u 0 whereu 0 is a unit in K, then it follows that either v=0 or v and u dier by a unit, in which case w 2 =u 4 v 4 = 0 so w = 0. In more detail, if 2v 2 u 2 w =2 and 4uv wu 2 = 4u 0 thenv 2 =u 2 w andu 1 uv =u 2 w for some 39 unitu 1 in K. Hencev 2 =u 1 uv sov(vu 1 u) = 0. Hencev = 0 orv =u 1 u. Therefore the curve C has no nontrivial solutions over K. For D6= -1,-7 we have E(K)=f(0; 0); (2;4)g and the same argument shows these points do not yield nontrivial solutions over K. The extra points in the D =7 case give rise to growth that doesn't occur in the D =1;3 cases. For example, the curve E dened over K =Q( p 7) by: y 2 =x(x +u 4 )(x +v 4 ) where u = 1 2 3 2 p 7 and v = 3 p 7 has E(K) C 2 C 8 but over L = K( q 465 2 + 45 2 p 7) we have E(L)C 2 C 16 . Proposition 4.1.3. Let K = Q( p D) (D =1;2;3;7;11) and let s,t,d2O K with s;t;d6= 0 and s 2 6=t 2 . Then there do not exist a,b2O K such that ds 2 =a 3 (a + 2b)c 2 dt 2 =b 3 (b + 2a)c 2 Proof. It's easy to check a b 62f2;1; 1 2 ; 0; 1g as otherwise a hypothesis is contradicted. Let K = Q( p D) (D=-1,-2,-3,-7,-11) and let s,t,d2 K with s,t,d6= 0, s 2 6= t 2 . Suppose there exists a,c2K such that ds 2 =a 3 (a + 2b)c 2 dt 2 =b 3 (b + 2a)c 2 Taking products and dividing both sides by a 2 c 4 b 4 yields: ( dst ab 2 c 2 ) 2 =w(w + 2)(1 + 2w) where w := a b . Multiplying by 4 yields: ( 2dst ab 2 c 2 ) 2 = 2w(2w + 4)(2w + 1) Hence we a get a K-point on the elliptic curve E with equation y 2 =x(x + 4)(x + 1) =x 3 + 5x 2 + 4x 40 Magma tells us E(K) has rank 0 and Sage tells us E(K) tor C 2 C 4 . so E(K)= E(K) tor =f(4; 0); (2;2); (1; 0); (0; 0); (2;6);1g. Hence 2w =4;2;1 or 0 so w =2;1 or 0, a contradiction. Proposition 4.1.4. Let K = Q( p D) (D=-1,-3) and let s,t2O K with s,t,d6= 0, s 2 6= t 2 . Then there do not exist a,b2O K such that s 2 =a 3 (a + 2b)c 2 t 2 =b 3 (b + 2a)c 2 Proof. Note that when D =1,1 is a square in K so this system has no solutions by Proposition 4.1.3 (with d=1). Hence we may assume D =3. It's now easy to check a b 62f2;1; 1 2 ; 0; 1g as otherwise a hypothesis is contradicted (or1 is a square in K). Let s,t2K with s,t6= 0, s 2 6=t 2 . Suppose there exists a,b,c2K such that s 2 =a 3 (a + 2b)c 2 t 2 =b 3 (b + 2a)c 2 As in the proof of Proposition 4.1.3 we obtain ( 2st ab 2 c 2 ) 2 =2w(2w 4)(2w 1) where w := a b . Hence we a get a K-point on the elliptic curve E with equation y 2 =x(x 4)(x 1) =x 3 5x 2 + 4x Magma tells us E(K) has rank 0 and E(K) tor C 2 C 4 so E(K) =f(1;3 ); (1;3 +); (1 +;3 +); (1 +; 3); (0; 0); (4; 0); (1; 0)g where 2 =3. Hence 2w = 0; 1; 4; 1 sow = 0; 1 2 ;2; or 2 where = 1 2 + p 3 2 is a primitive cube root of unity. We have already noted that w6= 0; 1 2 ;2. If w = then a =b so s 2 =b 3 (b + 2a) =b 3 (b + 2b) =b 4 (1 + 2) Hence 1 + 2 must be a square in K, but N(1 + 2) = (1 + 2)(1 + 2 2 ) = 3, which is not a square inQ. Hence there is no solution to the equation above with w = and as = 2 the case w = 2 follows similarly. 41 Proposition 4.1.5. Let K = Q( p D) (D =1;3) and let a;b;c;a 0 ;b 0 ;c 0 ;d2O K with d a non-square in K and a b ; a 0 b 0 62f2;1;1=2; 0; 1g. Then the following system of equations has noO K -solutions: da 3 (a + 2b)c 2 =a 3 0 (a 0 + 2b 0 )c 2 0 db 3 (b + 2a)c 2 =b 3 0 (b 0 + 2a 0 )c 2 0 Proof. Suppose a solution to the following system of equations exists satisfying the hypothe- ses above: da 3 (a + 2b)c 2 =a 3 0 (a 0 + 2b 0 )c 2 0 db 3 (b + 2a)c 2 =b 3 0 (b 0 + 2a 0 )c 2 0 Note that we must have a;b;a + 2b6= 0 so we may divide: a b 3 a + 2b b + 2a = a 0 b 0 3 a 0 + 2b 0 b 0 + 2a 0 and letting x = a 0 b 0 and y = a b we get y 3 y + 2 1 + 2y =x 3 x + 2 1 + 2x x 4 + 2x 3 + 2yx 4 + 4yx 3 y 4 2y 3 2xy 4 4xy 3 = 0 Magma factored the polynomial on the l.h.s. into absolutely irreducible polynomials: (xy)(2x 3 y + 2xy 3 +x 3 +y 3 + 5x 2 y + 5xy 2 + 2x 2 + 2y 2 + 2x 2 y 2 + 2xy) Now if x =y, it follows that there is an e2K such that a =be;a 0 =b 0 e.Substituting this into the system of equations above yields that d is a square (a contradiction). Hence there exists a K-rational point (x;y) (withx;y62f2; 1 2 ;1; 0; 1g) on the curve C dened by p C (x;y) = 0 where p C = 2x 3 y + 2xy 3 +x 3 +y 3 + 5x 2 y + 5xy 2 + 2x 2 + 2y 2 + 2x 2 y 2 + 2xy 42 Table 4.1: K-Rational Points on E C K rank(E C (K)) E C (K) tor points of E C (K) tor Q( p 1) 0 C 6 [0,1,0], [-1,0,1],[0,-2,1],[0,0,1], [2,-6,1],[2,0,1] Q( p 3) 0 C 2 C 6 [0,1,0],[-1,0,1],[0,-2,1],[0,0,1],[2,-6,1],[2,0,1], [1;3+; 1]; [1; 3+; 1]; [1+;3 ; 1]; [1 +; 3; 1]; [ 1 2 (1); 1 2 (3 +); 1]; [ 1 2 (1 +); 1 2 (3); 1] Let C denote the projective closure of C. Magma tells us C has genus 1 and is birational to the elliptic curve E C with dening equation y 2 + 2xy + 2y =x 3 x 2 2x (In fact, E C is birational to the curve with equation y 2 =x 3 + 1). Magma also provided a birational map: f :C!E C (x;y;z)7! (p f 1 ;p f 2 ;p f 3 ) p f 1 = 2x 2 y 2 + 3x 2 yz + 4xy 2 zy 3 z +x 2 z 2 + 6xyz 2 y 2 z 2 + 2xz 3 p f 2 = 2x 2 y 2 + 4xy 3 x 2 yz + 10xy 2 z + 3y 3 zx 2 z 2 + 7y 2 z 2 2xz 3 + 2yz 3 p f 3 =y 4 + 3y 3 z + 3y 2 z 2 +yz 3 Because f is a rational map, the only potential K-rational points of E C are the non- regular points of E C and f 1 (E C (K)). Suppose p f 3 = 0. If z=0 then y=0 and if z=1 then y = 0 or -1. Hence any non- regular point of f is of the form [x,0,0],[x,0,1] or [x,-1,1]. Evaluating p f 2 = 0 at the last two points yields the following possibilities for non-regular points of f: [1,0,0], [0,0,1], [-2,0,1], [-1,-1,1]. All such points do not give 'non-trivial' points on the curve C: [1,0,0] does not correspond to a point on C and the other points fail to produce points (x,y) on C with x;y62f2;1;1=2; 0; 1g. In order to compute f 1 (E C (K)) we must rst compute E C (K). Magma gives us the following information: 43 Table 4.2: Gr obner basis data for determination of f 1 (E(K)) Point P of E(K) f 1 (P ) Gr obner basis elements [0,1,0] [1,-1,1] xwzw;yw +zw [-1,0,1] [-1,1,1] yw 2 zw 2 ;xw +zw; [0,-2,1] ; w 2 [0,0,1] [0,1,0] xw;yw [2,-6,1] ; w 2 [2,0,1] [0,-2,1] xw;yw 2 + 2zw 2 [1;3 +; 1] [ 1; 2; 0] xw + 1 2 (q + 1)yw;zw [1 +;3; 1] [ 1; 2; 0] [1; 3 +; 1] ; w 2 [1 +; 3; 1] ; [ 1 2 (1); 1 2 (3 +); 1] ; w 2 [ 1 2 (1 +); 1 2 (3); 1] ; To compute f 1 ([x;y;z]), we form the ideal < p C ;p f 1 wx;p f 2 wy;p f 3 wz > and compute it's Gr obner basis. Often, one can nd basis elements that allow the system to be solved by hand. Now we see each point of f 1 (E C (K)) either does not correspond to a point on C, or fails to satisfy x;y62f2;1;1=2; 0; 1g. 2 Classication of Twists Theorem 4.2.1. Let K =Q( p D) (D =1;3), d2 K a non-square, and E/K an elliptic curve with full 2-torsion. 1. If E(K) tor C 2 C 8 then E d (K) tor C 2 C 2 . 2. If E(K) tor C 2 C 6 then E d (K) tor C 2 C 2 . 3. If E(K) tor C 4 C 4 then K =Q( p 1) and E d (K) tor C 2 C 2 . 4. If E(K) tor C 2 C 4 then WLOG we may write M = s 2 and N = t 2 . We have E d (K) tor C 2 C 2 unless K =Q( p 3);d =1; and9 v2O K such that s 2 t 2 = v 2 , in which case E d (K) tor C 2 C 4 . 5. If E(K) tor C 2 C 2 then E d (K) tor C 2 C 2 for almost all d. If for some d-twist, E d (K) tor C 2 C 8 , C 2 C 6 or C 4 C 4 then it is the unique nontrivial twist of E. If 44 for some d-twist, E d (K) tor C 2 C 4 then there is at most one other nontrivial twist of E and it is also isomorphic to C 2 C 4 . Proof. Before we begin we note that in general, if E and E 0 are isomorphic elliptic curves and d 2 K, then E d and E 0d may only be isomorphic over K( p d). But if E and E 0 are isomorphic via twist by a square in K, or by translation of the x-coordinate by an element of K, then one does in fact get an isomorphism of E d and E 0d over K (and hence E d (K)E 0d (K)). 1) Suppose E(K) tor C 2 C 8 . Then WLOG we may assume there exists x,y,z2O K such that M = x 4 , N = y 4 and x 2 +y 2 = z 2 . If E d (K) has a point of order 4 then by Theorem 4.0.2 either (i) dx 4 ;dy 4 are both squares, or (ii)dx 4 ;dy 4 dx 4 are squares, or (iii)dy 4 ;dx 4 dy 4 are squares. In case (i) it follows that d is a square, a contradiction. In case (ii), ifdx 4 is a square, thend =u 2 for some u2O K . Ifdy 4 dx 4 =w 2 for some w2O K , then dividing by d yields y 4 x 4 =(wu 1 ) 2 and multiplying by -1 we obtain x 4 y 4 = v 2 for some v2O K . Note we have x 4 ;y 4 ;x 4 y 4 6= 0 (as otherwise E is not smooth) and hence x;y;v6= 0. But this contradicts Prop 4.1.2, so E d (K) does not have a point of order 4. Now looking above (or applying Theorem Theorem 4.0.2) we can write M =s 2 ;N =t 2 for some s,t2O K . Hence if E d (K) has a point of order 3, there exists a;b;c2O K , c6= 0 and a b 62f2;1; 1 2 ; 0; 1g such that ds 2 = a 3 (a + 2b)c 2 and dt 2 = b 3 (b + 2a)c 2 . But this contradicts Proposition 4.1.3 so E d (K) does not have point of order 3. Hence by Theorem 1.0.4 we conclude E d (K)C 2 C 2 . 2) Suppose E(K) tor C 2 C 6 . Then we may assume there exists a;b;c2O K with c6= 0 and a b 62f2;1; 1 2 ; 0; 1g such that M = a 3 (a + 2b)c 2 and N = b 3 (b + 2a)c 2 . If E d (K) has a point of order 4, then by Theorem 4.0.2 (i) M, N are both squares (inO K ), (ii) -M, N-M are both squares, or (iii) -N,M-N are both squares. Hence WLOG we can assume case (i) holds, so dM =s 2 ;dN =t 2 for some s;t2O K (s;t6= 0;s 2 6=t 2 ). Hence s 2 =dM =da 3 (a + 2b)c 2 and t 2 =dN =db 3 (b + 2a)c 2 and multiplying by d: ds 2 =a 3 (a + 2b)(cd) 2 and dt 2 =b 3 (b + 2a)(cd) 2 But this contradicts Prop 4.1.3. Hence E d (K) does not have a point of order 4. 45 Let K = Q( p D) (D =1 or3) and let d2O K be a nonsquare. Suppose that E(K) tor C 2 C 6 . We will argue that the d-twist E d (K) tor 6 C 3 . We may assume (WLOG) that E is given by an equation of the form y 2 =x(x +M)(x +N) where M,N2O K . By Theorem 4.0.2, there exists a;b;c2O K with a b 62f2; 1 2 ;1; 0; 1g such that M =a 3 (a + 2b)c 2 and N =b 3 (b + 2a)c 2 Hence if E d T (K)C 3 then there existsa 0 ;b 0 ;c 0 2O K and a 0 b 0 62f2; 1 2 ;1; 0; 1g such that da 3 (a + 2b)c 2 =dM =a 3 0 (a 0 + 2b 0 )c 2 0 db 3 (b + 2a)c 2 =dN =b 3 0 (b 0 + 2a 0 )c 2 0 But this system has no solution by Proposition 4.1.5. Hence we reach a contradiction and conclude E d (K) tor 6C 3 . Therefore by Theorem 1.0.4 we conclude E d (K)C 2 C 2 . 4) SupposeE(K) tor C 2 C 4 . Then we may assume WLOG that there existss;t2O K such that M = s 2 and N = t 2 . If E d (K) has a point of order 4, then either (i) dM, dN are both squares, (ii) -dM, dN-dM are both squares, or (iii) -dN,dM-dN are both squares. Case (i) is not possible since d is a nonsquare. In case (ii) there exists u,v2O K such that ds 2 = u 2 and dt 2 ds 2 = v 2 so d =w 2 for some w2O K Furthermore D6=1 as otherwise d is a square, so K =Q( p 3). Hence t 2 s 2 =z 2 for some z2O K . Case (iii) similarly yields that s 2 t 2 = z 2 for some z2O K and d =w 2 for some unit w2O K . On other hand, if those condition are fullled then there is a point of order 4. Note that the only quadratic extension over which we can have E(K) tor C 4 C 4 is K = Q( p 1). Hence if E(K) tor C 2 C 4 then the 2-part of E d (K) can only be C 2 C 2 or C 2 C 4 (clearly E d (K) C 2 C 2 ). Now applying Prop 4.1.3 as in Case 2 shows thatE d (K) has no point of order 3 and by Theorem 1.0.4 it has no point of order 5. 3) Suppose E(K) tor C 4 C 4 . Then we must have K = Q( p 1). On other other hand, as in the proof of Case 4 above, E d (K) cannot have a point of order 4. Also as in the proof above, E d (K) cannot have a point order 3 or 5 so E d (K)C 2 C 2 . 46 5) Note that for nonzerod;e2K,E d is isomorphic (over K) to (E e ) ed . Suppose that E has a quadratic twist that is not isomorphic to C 2 C 2 . IfE d (K) tor C 2 C 8 ;C 2 C 6 or C 4 C 4 for some nonsquare d2K, then E has only one quadratic twist not isomorphic to C 2 C 2 by Cases 1,2 and 3. IfE d (M 0 ;N 0 ) tor C 2 C 4 for some nonsquare d, then WLOG (translating E if necessary) we may assume M 0 =s 2 and N 0 =t 2 for some s,t inO K .Now applying Case 4, we see that if K =Q( p 3), e =1 and there exists v2O K such that s 2 t 2 =v 2 , then E ed (K) tor C 2 C 4 so we have two twists not isomorphic to C 2 C 2 (byd). 3 Full 4-torsion Lemma 4.3.1. Let E be an elliptic curve dened over K =Q( p D)(D =1;3), d2O K a nonsquare and let L =K( p d). If E(K) tor C 2 C 8 , then If D = -3, C 4 C 4 6E(L). If D = -1, there is a unique quadratic extension L of K such that C 4 C 4 E(L). WLOG, writing M = x 4 and N = y 4 where x 2 +y 2 = z 2 , the extension is L = K( p x 2 y 2 ). Proof. By Thm 4.0.2, WLOG we can assume there exists x;y;z2O K such that M =x 4 , N =y 4 andx 2 +y 2 =z 2 : IfC 4 C 4 E(L), then every point of order 2 lifts, so in particular there exists a point P inE(L) such that 2P = (x 4 ; 0). Hence by Thm 4.0.1,x 4 andy 4 x 4 are squares in L, so there exist a;b2O L such thatx 4 =a 2 and y 4 x 4 =b 2 . Therefore 12L so either (i) D =1 (if12K) or (ii) D =3 and d =1(L =K( p 1)). Also, b 2 =y 4 x 4 = (y 2 x 2 )(y 2 +x 2 ) = (y 2 x 2 )z 2 so c 2 = y 2 x 2 2 K for some c2 L. Writing c = e +f p d (e;f2 K) since c 2 2 K, we conclude thatc 2 =e 2 ordf 2 . Hence we get a solution to eitherg 2 =y 2 x 2 ordg 2 =y 2 x 2 (g2O K ). In case (ii), we obtain either way y 4 x 4 = z 2 (or x 4 y 4 = z 2 ) for some z2O K . Furthermore we must have x;y;x 4 y 4 6= 0, so z6= 0 because E is nonsingular. But this contradicts Prop. 4.1.2. In case (i), we obtain eithery 4 x 4 =dz 2 ory 4 x 4 =z 2 for some d;z2O K where K =Q(i). The latter cannot occur by Prop.4.1.2. 47 On the other hand, if we choose any pythagorean triple (x;y;z) inO K with x;y6= 0 andx 4 6=y 4 and setM =x 4 ;N =y 4 then writingy 4 x 4 asdz 2 with d squarefree, we nd (by Thm 4.0.1) that L =K( p d)) is the unique quadratic extension of K (in K) such that C 4 C 4 E(L). Note that the argument in case (i) above implies that L/K is a nontrivial extension. Lemma 4.3.2. Let E be an elliptic curve dened over K =Q( p D)(D =1;3), d2O K , d a nonsquare and let L = K( p d) . Suppose E(K) tor C 2 C 4 . Then WLOG we may write M =u 2 ;N =v 2 for some u;v2O K . We have E(L) tor C 4 C 4 i: D = -3, d=-1 and there exists a w2O K such that v 2 u 2 =w 2 D = -1 and L=K( p v 2 u 2 ) Proof. WLOG we may assume (0,0) lifts over K and write M = u 2 ;N = v 2 for some u;v2O K . IfE(L) tor C 4 C 4 then some other order 2 point lifts. But by lemma 4.0.1, if (u 2 ; 0) lifts over L thenu 2 andu 2 v 2 are squares in L, or equivalently,1 andu 2 v 2 are squares in L. As -1, u 2 v 2 2 K, if1 and u 2 v 2 are squares in L then each of them is a square in K or d times a square in K (and both will not be squares in K, as otherwise E(K)C 4 C 4 ). Now ifK =Q(i), then -1 is a square in K sou 2 v 2 is not, and henceLK( p u 2 v 2 ). But K( p u 2 v 2 ) is quadratic over K so we must have L = K( p u 2 v 2 ). Conversely, if M =u 2 ;N =v 2 for someu;v2O K and we deneL =K( p u 2 v 2 ) thenu 2 = (iu) 2 and u 2 v 2 are squares, so the result follows by the lemma 4.0.1. If K =Q( p 3) then as1 is not a square in K, as above we must have L =K( p 1). But if u 2 v 2 is a square in L, as u 2 v 2 2K, we conclude u 2 v 2 =w 2 orw 2 for some w2K. Conversely, , if M =u 2 ;N =v 2 for some u;v2O K and we dene L =K(i) then u 2 = (iu) 2 and u 2 v 2 = w 2 or (iw) 2 are squares, so the result follows by the lemma 4.0.1. 4 Classication of Growth Now we prove Theorem 4.4.1 and Theorem 4.4.2. For clarity of proof, we prove each case of the main theorems simultaneously. In the results below, the elliptic curves have full 2-torsion over the base eld K so each has a model of the form: y 2 =x(x +M)(x +N) 48 for some M;N2K. Our main results are the following: Theorem 4.4.1. Let K =Q( p 3) and E/K an elliptic curve with full 2-torsion. 1. If E(K) tor C 2 C 8 then E(L) tor C 2 C 8 for all quadratic extensions L of K. 2. If E(K) tor C 2 C 6 then there is at most one quadratic extension L 0 such that E(L 0 ) tor C 4 C 6 and E(L) tor C 2 C 6 for all other quadratic extensions L of K. 3. IfE(K) tor C 2 C 4 then WLOG we may writeM =u 2 ;N =v 2 . There is a quadratic extension L of K such E(L)C 4 C 4 i u 2 v 2 =w 2 for some w2K and in this case L =K(i) and E(L)C 4 C 4 . There are at most two quadratic extensions L in which E(L) has a point of order (at least) 8 and in this case E(L)C 2 C 8 . 4. If E(K) tor C 2 C 2 and L is a quadratic extension of K then one may have E(L) C 2 C 2 ;C 2 C 4 ;C 2 C 6 ;C 2 C 8 ;C 4 C 6 or C 4 C 4 . If L = K( p d) for some d2 K then this can be determined by replacing E by E d . The group E(K) tor grows in at most 3 extensions. Theorem 4.4.2. Let K =Q( p 1) and E/K an elliptic curve with full 2-torsion. 1. If E(K) tor C 2 C 8 then there is a unique quadratic extension L 0 of K in which E(L 0 ) tor C 4 C 8 and E(L) tor C 2 C 8 for all other quadratic extensions L of K. 2. If E(K) tor C 2 C 6 then there are at most two quadratic extension L' such that E(L 0 ) tor C 4 C 6 and E(L) tor C 2 C 6 for all other quadratic extensions L of K. 3. If E(K) tor C 2 C 4 then WLOG we may write M = u 2 ;N = v 2 . The quadratic extensionL =K( p u 2 v 2 ) is the unique quadratic extension of K such E(L)C 4 C 4 and in this caseE(L)C 4 C 4 . There are at most two other quadratic extensions L 0 in which a point of order 4 (over K) lifts and in this case E(L 0 )C 2 C 8 . 4. If E(K) tor C 2 C 2 and L is a quadratic extension of K then one may have E(L) C 2 C 2 ;C 2 C 4 ;C 2 C 6 ;C 2 C 8 ;C 4 C 6 ,C 4 C 4 or C 4 C 8 . If L =K( p d) then this can be determined by replacing E by E d . The group E(K) tor grows in at most 3 extensions. 5. If E(K) tor C 4 C 4 then there is at most one quadratic extension L of K in which E(K) tor grows and in this case E(L) =C 4 C 8 . Let denotes a primitive 3rd root of unity. 49 Table 4.3: Examples of Growth of Torsion over K =Q( p 1) in Case E(K)[2]C 2 C 2 (M,N) E(K) tor E(K( p d)) tor d (-2,2) C 2 C 2 C 4 C 4 2 (5 64; 5 89) C 2 C 2 C 4 C 6 5 (7 3 4 ; 7 4 4 ) C 2 C 2 C 4 C 8 7 (1,4) C 2 C 4 C 4 C 4 -3 (64,189) C 2 C 6 C 4 C 6 5 (3 4 ; 4 4 ) C 2 C 8 C 4 C 8 7 (4 2 ; 5 2 ) C 4 C 4 C 4 C 8 5 Table 4.4: Examples of Growth of Torsion over K =Q( p 3) in Case E(K)[2]C 2 C 2 (M,N) E(K) tor E(K( p d)) tor d (-1, ) C 2 C 2 C 4 C 4 -1 (21 64, 21 189) C 2 C 2 C 4 C 6 21 (4 2 ; 5 2 ) C 2 C 4 C 4 C 4 -1 (64,189) C 2 C 6 C 4 C 6 21 Note in only one case our example is not dened over Q. In that case there is in fact no example dened overQ. 4.1 E(K) tor C 2 C 8 Proof. 1) SupposeE(K) tor C 2 C 8 WLOG we can assume there existsu;v;w2O K such that M =u 4 and N =v 4 and u 2 +v 2 =w 2 . Let L = K( p d) where d2 K is a nonsquare. By Theorem 4.2.1, it follows that E(L) C 2 C 8 ;C 4 C 8 ;C 2 C 16 or C 4 C 16 . By Lemma 4.0.1, we nd the following point of order 8: P = (uv(u +w)(v +w);uvw(u +v)(v +w)(u +w)) This point satises 4P = (0; 0) and hence E(K) =<P 0 ><P > where P 0 = (u 4 ; 0). Let [2] : E(L) tor ! E(L) tor denote the multiplication-by-2 map. Supppose E(L) C 2 C 16 and Q2 E(L) is a point of order 16. Then E(L) has 8 points of order 8 and Im([2]) is a cyclic subgroup of order 8. AsE(K) has 8 points order 8, the image above must be contained in E(K). But E(K) has only 2 cyclic subgroups of order 8, namely < P > 50 and <P 0 +P >. In particular there is a point Q of order 16 in E(L) such that 2Q =P or P 0 +P where explicit calculation shows P 0 +P = (uv(u +w)(vw);uvw(uv)(vw)(u +w)) Suppose 2Q =P (the case 2Q =P 0 +P case similar). By Thm 4.0.1, P x , P x +u 4 and P x +v 4 are squares in L. Since each of them lies in K, it follows that each of the above is either of the form c 2 or dc 2 for some c2K. We claim that one of them must be a square in K (In fact exactly one, as if they were all squares in K, then E(K) would have a point of order 16). Note that P 2 y = (P x )(P x +u 4 )(P x +v 4 ) so as d is a nonsquare and P y lies in K, the claim follows. We will now argue that none of them are squares in K. Explicit calculation shows: P x =uv(u +w)(v +w) P x +u 4 =uw(u +v)(w +v) P x +v 4 =vw(v +u)(w +u) If P x is a square then we have u;v;w;z2O K such that z 2 =uv(u +w)(v +w) u 2 +v 2 =w 2 Note that if (u,v,w)=(-1,0,1) then v = 0 and hence our curve is not smooth, contradicting our hypothesis. By Proposition 4.1.1 we can write any other solution to the above system (twisitng by a square if necessary) (u;v;w) = (1m 2 ; 2m; 1 +m 2 ). Again since our curve is smooth by hypothesis, we see that m6= 0;1;i. Evaluating the above equation yields the following curve C: z 2 = (1m 2 )(2m)(2)(m + 1) 2 =m(1m 2 )(2m + 2) 2 Letting E 0 denote the elliptic curve with equation y 2 = x 3 x, we have a birational map f : C ! E 0 given by (m;z) 7! (m; z 2m+2 ). Magma tells us E 0 (K) has rank 0 and E 0 (K) tor C 2 C 4 . In fact E(K) tor =f1; (0; 0); (1; 0); (i;(1i)); (i;(1 +i))g. Hence all rational points (m,z) on C satisfy m = 0;1 ori, a contradiction. Similarly whenK =Q( p 3) we haveE 0 (K)C 2 C 2 (the rank is 0), som = 0;1, a contradiction. If P x +u 4 is a square, then we have u;v;w;z2O K such that 51 z 2 =uw(u +v)(w +v) u 2 +v 2 =w 2 Again by Proposition 4.1.1 (noting that (-1,0,1) does not yield a solution satisfying the hypotheses), WLOG we can write any solution to the above system (u;v;w) = (1 m 2 ; 2m; 1 +m 2 ). By our hypotheses concerning u,v,w we see that m6= 0;1. Evaluat- ing the above equation yields the following curve C: z 2 = (1m 2 )(1 +m 2 )(1 + 2mm 2 )(m + 1) 2 Let E 1 denote the curve y 2 = (1x)(1 +x)(1 +x 2 )(1 + 2xx 2 ) and let J denote its Jacobian. We wish to classify all K-rational points on this curve. Magma tells us E 1 is smooth (hence hyperelliptic) of genus 2. Suppose K= Q( p 1) (the case K=Q( p 3) is simpler). Magma tells us J(K) has rank 0 and J(K) 2part C 2 C 2 C 2 J(Q) tor C 2 C 2 C 5 and furthermore Magma gives us a list of generators. Together this allows us to generate 40 points on J(K). On the other hand, we will argue that J(K) has size at most 40. Note that 3 remains prime and 5 splits inO K . LetB denote a prime above 5 inO K . The curve E 1 remains non-singular mod 3 andB and hence J has good reduction at those primes. Magma tells us jJ(F 9 )j = 80 andjJ(F 5 )j = 40. Since the reduction map J(K) tor ! J(F 5 ) (moduloB) is injective on the prime-to-5 part, we see the prime-to-5-part has size at most 40 (and hence is a 2-group of size at most 8). On the other hand, under reduction mod 3 the 5-part of J(K) tor injects into J(F 9 ) and so can be of size at most 5 . It follows thatjJ(K)j = 40. Magma embeds E 1 in (1,3,1)-weighted projective space, yielding two points at innity 1 1 = (1;1; 0) and1 2 = (1; 1; 0). Now E 1 embeds into J via the Abel-Jacobi map: E 1 !J 52 P7!P1 1 Note that the image of P is P1 1 = P +1 2 1 1 1 2 and the latter expression is the unique coset representative that Magma selects. Among the 40 coset representa- tives that Magma found, 7 are of the above form and the corresponding points P are (0;1; 1); (1; 1; 0); (0; 1; 1); (1; 0; 0); (i; 0; 1). Again, any such point corresponds to u,v,w which contradict the hypothesis that our curve is smooth. IfP x +v 4 is a square then one obtains the same system as in the previous case but with u and v interchanged. Hence using the parametrization (u;v;w) = (2m; 1m 2 ; 1 +m 2 ) we obtain the same hyperelliptic curve above. Note thatu 2 +(v) 2 =w 2 and P'+P is related to P by the substitution -v for v. Hence if 2Q=P'+P then one obtains the same (hyper)elliptic curves as in the case 2Q =P . We conclude that we cannot have E(L)C 2 C 16 . Now ifE(L)C 4 C 16 , then in particularC 4 C 4 E(L), so by lemma 4.3.1,K =Q(i) and d = p v 2 u 2 . In thise case E(L) has 16 points of order 8. Applying Thm 4.0.1 to obtain all 16 points of order 8, we nd that (replacing w byw or u byiu if necessary) we may asssume any point P of order 8 has P x = uv(u +w)(v +w) or uv(u +w)(vw). But in the case E(L) C 2 C 16 we already showed that points P of the above form do not lift. 4.2 E(K) tor C 2 C 6 Proof. 2) Now supposeE(K) tor C 2 C 2 C 3 . Again by Theorem 4.2.1 we have the only possibilities are E(L) tor C 2 C 2 C 3 or C 2 C 4 C 3 or C 4 C 4 C 3 . If E(L) has a point of order 4, then either (i) M, N are squares in L (if (0,0) lifts), (ii) -M,N-M are squares in L (if (-M,0) lifts) or (iii) -N, M-N are squares in L (if (-N,0) lifts). It follows that each order 2 point lifts in at most one quadratic extension. Furthermore note that if at least two order 2 points lift, then all three lift. If (0,0) lifts in L =K( p d) then M and N are squares in L and hence WLOG M =s 2 or ds 2 and N = t 2 or dt 2 for some s;t2O K . Now if M = s 2 and N = t 2 then E(K) has a point of order 4, contradicting our hypothesis. Also, if M = ds 2 and N = dt 2 then as E(K) has a point order 3, we get a contradiction by Proposition 4.1.3. Hence WLOG we can writeM =ds 2 andN =t 2 andL =K( p d) is the unique quadratic extension in which (0,0) lifts. Noting that E(M;NM)E(M;N) we can apply the same argument and conclude that if -M, N-M are square in L (that is, (-M,0) lifts) that one is a square (and 53 the other d times a square). Similar reasoning applies to -N, M-N. Suppose K =Q( p 3). We will argue that: (I) If (0,0) lifts in a quadratic extension L = K( p d), then none of -M,N-M, -N, M-N are squares in K. If this holds, then by the argument above, it must be that (-M,0) and (-N,0) dont lift in any quadratic extension of K, Hence L is the unique quadratic extension in which the torsion grows and E(L)C 2 C 4 C 3 . In particular, E(L) cannot achieve full 4-torsion. We would also like to show that: (II) If (-M,0) lifts in some quadratic extension L of K then (0,0) and (-N,0) dont lift in any quadratic extension of K. (III) If (-N,0) lifts in some quadratic extension L of K then (0,0) and (-M,0) dont lift in any quadratic extension of K. But these latter two claims ((II) and (III)) follow from the rst (I). Let's prove (II) as (III) follows similarly. If the rst claim holds and (-M,0) lifts then (0,0) lifts onE(M;N M) so applying (I) to the curve E(M;NM) we conclude that the following are not squares in K: (M) =M; (NM) (M) =N;(NM) =MN;M (NM) =N Hence it must be that (0,0) and (-N,0) do not lift in any quadratic extension of K. Now let's prove the rst claim. First ifM = v 2 for some v2 K then ds 2 =v 2 so WLOG d=-1. As E has a point of order 3, we obtain a nontrivial solution to: s 2 =M =a 3 (a + 2b)c 2 t 2 =N =b 3 (b + 2a)c 2 contradicting Prop 4.1.4 . If MN = v 2 for some v2 K. Then as E(N;MN) E(M;N) has a point of order 3, we obtain a nontrivial solution to: t 2 =N =a 3 (a + 2b)c 2 v 2 =MN =b 3 (b + 2a)c 2 54 contradicting Prop 4.1.4. If NM = v 2 for some v2 K. Then as E(N;MN) E(M;N) has a point of order 3, we obtain a nontrivial solution to: t 2 =N =a 3 (a + 2b)c 2 v 2 =MN =b 3 (b + 2a)c 2 contradicting Prop 4.1.3. Finally if -N is a square in K then it follows -1 is a square in K, a contradiction. Now suppose K = Q(i) and some point of order 2 lifts. WLOG we may assume this point is (0,0) and the same proof as above shows (I) holds with the exception that -N is of course a square. As (I) implies (-M,0) never lifts, we still cannot have E(L)C 4 C 4 in any quadratic extension L, but (-N,0) lifts in the extension K( p MN). Hence we have at most two quadratic extensions L of K such thatE(L)C 4 C 6 . For example, the curve E(64,189) satises E(K) tor C 2 C 6 but E(K( p d)) tor C 2 C 12 for d = 5; 21. 4.3 E(K) tor C 2 C 4 Proof. 3)Now suppose E(K) C 2 C 4 . First suppose K = Q( p 3);L = K(i);M = u 2 ;N =v 2 andu 2 v 2 =w 2 for someu;v;w2O K .By lemma 4.3.2, we haveE(L)C 4 C 4 and we will argue that in this case there is no point of order 8. Q = (uv;uv(u +v)) Q 0 = (u(u +w);iuw(u +w)) Q +Q 0 = (v(v +iw);vw(v +iw)) Note 2Q = (0; 0) and 2Q 0 = (u 2 ; 0) so Q, Q 0 are independent order 4 points and hence form a basis for C 4 C 4 E(L). Now if some point of order 4 lifts in L, then choosing coset representatives inC 4 C 4 =(2(C 4 C 4 )) and pulling back to E(L), we can say WLOG that Q;Q 0 or Q +Q 0 lifts. We will consider each case: If Q lifts, then by Lemma 4.0.1 uv;u(u +v);v(u +v) are squares in L. As they lie in K, it follows that each is a square in K. If uv =c 2 note (-1,0,1) is not a solution to our problem as our curve is smooth. By Proposition 4.1.1, 55 WLOG we may use the parametrization (u;v;w) = (1 +m 2 ; 2m; 1m 2 ) of u 2 = v 2 +w 2 (note m6= 0;1 as our curve is nonsingular) yielding c 2 = (1 +m 2 )(2m) And multiplying by 4 yields (2c) 2 = (4 + (2m) 2 )(2m) So we get a K-point on the elliptic curve E 0 with model: y 2 =x 3 + 4x Magma tells us E 0 (K) tor C 4 and E 0 has rank 0, so the ane points of E 0 (K) are (0; 0); (2;4). Hence2m = 0; 2 so m = 0 or1, a contradiction. If uv = c 2 then one obtains the same curve E 0 as above, hence no nontrivial solutions. If Q 0 lifts then by Lemma 4.0.1u(u +w);uw;2(u +w) are squares in L and as above we concludeuw is a square in K. Ifuw =c 2 then the parametrization above yields: c 2 = (1 +m 2 )(1m 2 ) = 1m 4 and multiplying by m 2 yields: (mc) 2 =m 2 m 6 = (m 2 ) (m 2 ) 3 = (m 2 ) 3 (m 2 ) Letting E 0 denote the elliptic curve: y 2 =x 3 x Magma tells us E 0 has rank 0 and E(K) tor C 2 C 2 so the only ane K-points are (1; 0); (0; 0) and (1; 0) . Hence we getm 2 = 0;1 som = 0;1, a contradiction. Finally, suppose Q +Q 0 lifts. By Thm 4.0.1, it follows that -iwv is a square in L. iwv = (a +bi) 2 iwv =a 2 b 2 + 2abi 56 From this it follows thata =b and sowv =2a 2 . Using the parametrization above yields a 2 =m(1m 2 ) Let E denote the elliptic curve y 2 =x 3 x We wish to classify the points on E(K) and E 1 (K) but note that E 1 E. Magma tells us E has rank 0 and E(K) tor C 2 C 2 so the only ane K-points are (1,0),(0,0) and (-1,0) . In any case, we geta = 0 which impliesw = 0 orv = 0, yielding a contradiction. Now suppose K =Q( p 1);M =u 2 ;N =v 2 , L = K(w) where u 2 v 2 =w 2 for some u;v2O K with [L:K]=2. By Lemma 4.3.2, we have E(L)C 4 C 4 and so by Prop 2.2.6 and Thm 4.2.1 E(L)C 4 C 4 or C 4 C 8 . Q = (uv;uv(u +v)) Q 0 = (u(u +w);iuw(u +w)) Q +Q 0 = (v(v +iw);vw(v +iw)) Note 2Q = (0; 0) and 2Q 0 = (u 2 ; 0) so Q, Q 0 are independent order 4 points and hence form a basis for C 4 C 4 E(L). Now if some point of order 4 lifts in L, then as in the previous case we can say WLOG that Q;Q 0 or Q +Q 0 lifts. Two cases can be dismissed quickly: If Q 0 lifts then by Thm 4.0.1u(u +w) +u 2 =uw is a square in L and if Q +Q 0 liftsv(v +iw) +v 2 =ivw is a square in L. In both cases, we nd a square in L of the form cw with c2 K (note i2 K). Now we make the following observation: if F F 0 is a quadratic extension of elds and F contains a square root of1, then the (nonzero) squares in F cannot be of the above form: if F 0 = F (w) then a typical element in F 0 can be written a +bw with a;b2 F . So we nd (a +bw) 2 = a 2 +b 2 w 2 + 2abw will be of the form mentioned above only if a 2 +b 2 w 2 = 0. But in this case it follows that either b = 0 (and hence a = 0) or w = ia b 2K, a contradiction. Therefore neither Q 0 nor Q +Q 0 lift over L. If Q lifts, then by Thm 4.0.1 uv;u(u +v);v(u +v) are squares in L. As they lie in K, it follows that each is a square in K or u 2 v 2 times a square in K. As they cannot all be squares in K, it follows that exactly one is a square in K and hence the other two have the 57 same square free part (as their product is square). Hence there is at most one extension in which E(L)C 4 C 8 . We will prove that Q does not lift in L (so there are in fact no extensions in which E(L) C 4 C 8 ). If Q lifts, then by Thm 4.0.1 uv;u(u +v);v(u +v) are squares in L. As they cannot all be squares in K (since otherwise E(K) would contain a point of order 8) they lie in K and their product is a square, it follows that one is a square in K and the other two are u 2 v 2 times a square in K. Hence one of the three sets are squares in K: (i)uv;u(u +v)(u 2 v 2 );v(u +v)(u 2 v 2 ) (ii)uv(u 2 v 2 );u(u +v);v(u +v)(u 2 v 2 ) (iii)uv(u 2 v 2 );u(u +v)(u 2 v 2 );v(u +v) Modulo squares, this is equivalent to: (i)uv;u(uv) (ii)u(u +v);v(uv) (iii)u(uv);v(u +v) If (i) holds then by Lemma 4.0.1 the point P of order 4 in E(K) with P x =uv lifts in E(K) (a contradiction). If (ii) holds then letting z = u v we obtain a 2 =z(z + 1) b 2 =z 1 and taking products yields (ab) 2 =z 3 z . Likewise if (iii) holds a 2 =z(z 1) b 2 =z + 1 and taking products yields (ab) 2 =z 3 z 58 Hence in either case we obtain a point on the elliptic curve y 2 =x 3 x But as noted in the case E(K) tor C 2 C 8 , the only K-rational points (x;y) on this curve satisfy x = 0;1;i. If u v = 0;1 then E is not smooth, a contradiction. If u v =i then u 2 =v 2 . But for any nonzero v2K, the curve E(v 2 ;v 2 ) is isomorphic over K to E(1;1) and the latter curve does not have a point of order 8 in K( p 2). Now suppose E(K) C 2 C 4 (K = Q( p 1) or Q( p 3)) and the conditions in Lemma 4.3.2 do not hold (i.e. E(L)6C 4 C 4 ). Again by Theorem 4.2.1 (and Lemma 4.3.2) we have the only possibilities are E(L) C 2 C 4 or C 4 C 4 or C 2 C 8 and as noted, two of these cases (full 4-torsion) cannot occur. So now we just need determine in which quadratic extensions of K a point of order 4 lifts. WLOG assume M = u 2 , N = v 2 . There are 4 order 4 points in E(K) (each a lift of (0,0)) and we can describe them explicitly: (uv;uv(u +v)); (uv;uv(uv)). Hence if P is an order 4 point in E(K) thenP x =uv oruv and furthermore these cases cannot occur simultaneously (as if E(L)C 2 C 8 then Im([2]) contains only 2 order 4 points, and it is closed under inverses). IfP x =uv and P lifts in L=K( p d) with d a nonsquare inO K thenuv;u(u+v);v(u+v) are squares in L, so each is a square in K or d times a square in K. Since their product is a square in K, and not all of them can be squares in K (since otherwise we have a point of order 8 in E(K)), it follows that exactly one of uv;u(u +v);v(u +v) is a square in K. If uv is a square then all three expressions are squares in K( p u(u +v)) =K( p v(u +v)). Likewise ifu(u +v) is a square in K then all three expressions are squares in K( p uv) (and similarly ifv(u +v) is a square in K). IfP x =uv then one obtains the same results above but with v replaced by -v. Hence the growth C 2 C 4 to C 2 C 8 occurs in at most two quadratic extensions. If K = Q(i), then uv is a square i -uv is a square. Hence if we choose uv to be a square and ifE(K) tor C 2 C 4 then each order 4 points will lift in some extension. When u = 1;v = 4 then E(1; 16) tor C 2 C 4 and E(K( p d)) tor C 2 C 8 for d =3; 5. We also have E(K( p 15)) tor C 4 C 4 so E(1; 16) grows in 3 quadratic extensions. If K = Q( p 3), suppose E(K) tor C 2 C 4 and E obtains full 4-torsion in a quadratic extension. By Lemma 4.3.2 there is a w2 K such that u 2 v 2 = w 2 . Hence 59 (u +v)(uv) = w 2 so u(u +v) is a square i u(uv) is a square (and in this case each order 4 point lifts in some extension). Consider the system a 2 =u(u +v) u 2 v 2 =w 2 Using the parametrization (1+m 2 ; 2m; 1m 2 ) for (u,v,w) (note we must have u,v,w nonzero above) yields b 2 = 1 +m 2 for someb2K. Using the parametrization (b;m) = ( 1+n 2 2n ; 1n 2 2n ) and letting say n=2 yields (after twisting) E(25,-24). We have E(K) tor C 2 C 4 and E(K( p d)) tor C 2 C 8 for d =6. We also have E(K(i)) tor C 4 C 4 so E(K) tor grows in 3 quadratic extensions. 4.4 E(K) tor C 4 C 4 Proof. 5) If E(K)C 4 C 4 where K is a quadratic extension then it follows from Corol- lary 2.2.12 that we must have K =Q(i). It follows from Thm 4.2.1 that E(L)C 4 C 4 , C 4 C 8 or C 8 C 8 . WLOG we may write M = u 2 ;N = v 2 , u 2 v 2 = w 2 for some u;v;w2O K . Q = (uv;uv(u +v)) Q 0 = (u(u +w);iuw(u +w)) Q +Q 0 = (v(v +iw);vw(v +iw)) As aboveQ,Q 0 are independent order 4 points and hence form a basis forC 4 C 4 E(L). Also, if some point of order 4 lifts in L then we can say WLOG that Q;Q 0 or Q +Q 0 lifts. WLOG we may now assume the parametrization of w 2 +v 2 = u 2 of the form (1 m 2 ; 2m; 1 +m 2 ). Note in this case thatu +v = (1 +m) 2 andwiv = (im 1) 2 are squares in K andu +w = 2. First we claim thatuv,uw andivw are never squares in K. Ifuv =c 2 for some c2K then: c 2 =uv = (1 +m 2 )2m And multiplying by 4 yields: (2c) 2 = (4 + (2m) 2 )(2m) 60 So we get a K-point on the elliptic curve E 0 : y 2 =x 3 + 4x Magma tells us E 0 has rank 0 and E(K) tor C 2 C 4 so the only ane K-points are (0; 0); (2i; 0); (2;4); (2;4i). Hence we get 2m = 0;2;2i so m = 0;1;i, a con- tradiction. If uw =c 2 for some c2K, then c 2 =uw = (1 +m 2 )(1m 2 ) = 1m 4 And multiplying by m 2 yields: (mc) 2 = (m 2 ) (m 2 ) 3 = (m 2 ) 3 (m 2 ) So we get a K-point on the elliptic curve E 0 : y 2 =x 3 x Magma tells us this curve has has rank 0 andE(K) tor C 2 C 4 and list of 8 points shows that the x-coordinates of points in E(K) are 0;1;i. Hence m satisesm 2 = 0;1;i so m = 0;1;i, a contradiction (as our curve E is nonsingular by hypothesis). If ivw =c 2 for some c2K then c 2 = 2im(1m 2 ) and multiplying by -4 yields (2ic) 2 = (2im)(4 + (2im) 2 ) So we get a K-point on the elliptic curve E 0 : y 2 =x 3 4x Magma tells us this curve has has rank 0 and E(K) tor C 2 C 2 and list of 4 points shows that the x-coordinates of points in E(K) are 0;2. Hence m satises 2im = 0;2 so m = 0;i, a contradiction (as E nonsingular by hypothesis). 61 Now if Q lifts (in some extension) it follows that exactly one of uv;u(u +v);v(u +v) is a square in K and hence one of uv;u;v is a square in K (as u +v is a square). If Q 0 lifts in some extension then exactly one of uw; 2u; 2w is a square in K. Finally if Q 00 lifts in some extension then exactly one of ivw;w;iv is a square in K. We have already argued that uv;uw and ivw are never squares in K. Suppose thatQ lifts in some extension andQ 0 lifts in some (possibly dierent) extension. The only way this can occur is if one of the following pairs consists of squares in K: (i) u and 2u (ii) u and 2w (iii) v and 2u or (iv) v and 2w. In case (i) we nd 2 is a square in K, a contradiction. In case (ii), we get K-solutions to the system a 2 = 1 +m 2 b 2 = 2(1m 2 ) Taking products yields: (ab) 2 = 2(1m 4 ) Multiplying both sides by 4m 2 : (2mab) 2 = 8(m 2 m 6 ) (2mab) 2 = (2m 2 ) 3 4(2m 2 ) So we get a point on the curve y 2 =x 3 4x Magma tells us this curve has has rank 0 and E(K) tor C 2 C 2 and list of 4 points shows that the x-coordinates of points in E(K) are 0;2. Hence m satises2m 2 = 0;2 som 2 = 0;1 and m = 0;1;i. But then m = 0;1;i and each such choice for m yields u;v or w = 0 (contradiction). In case (iii), if v and 2u are squares a 2 = 2m b 2 = 2(1m 2 ) Taking products yields: (ab) 2 = 4(mm 3 ) 62 (ab=2) 2 = (m) 3 (m) So we get a point on the curve y 2 =x 3 x Magma tells us this curve has has rank 0 andE(K) tor C 2 C 4 and list of 8 points shows that the x-coordinates of points in E(K) are 0;1;i. As above we reach a contradiction. Finally in case (iv) if v and 2u are squares a 2 = 2m b 2 = 2(1 +m 2 ) Taking products yields: (ab) 2 = 4(m +m 3 ) (ab=2) 2 =m 3 +m So we get a point on the curve y 2 =x 3 +x Magma tells us this curve has has rank 0 andE(K) tor C 2 C 2 and list of 4 points shows that the x-coordinates of points in E(K) are 0;i. As above we reach a contradiction. Therefore if Q lifts in some extension then Q 0 does not (and vice versa). In particular, as both Q and Q 0 would lift in an extension L such that E(L) tor C 8 C 8 (as any two order 4 points would lift) we conclude that there is no quadratic extension L of K such that E(L) tor C 8 C 8 . If Q and Q 00 both lift in some (possibly dierent) extension(s), then one of these pairs consists of squares in K : (i) u;w (ii) u;iv (iii) v;w (iv) v;iv. Each of the rst three cases contradict Prop 4.1.2 and the last case can't hold since i has no square root in K. Finally if Q 0 and Q 00 both lift then one of these pairs consists of squares in K: (i) w; 2u (ii) w; 2w (iii) 2u;iv (iv) 2w;iv. Cases (i), (ii) have been addressed above. In case (iii) we obtain c 2 = 4im(1 +m 2 ) and multiplying by1 yields (ic=2) 2 = (im) 3 (im) 63 So we get a point on the elliptic curve y 2 =x 3 x As noted above, the x-coordinates of points inE(K) are 0;1;i so m satisesim = 0;1 ori. Hence m = 0;1, ori which leads to u,v or w = 0 (a contradiction). In case (iv) we obtain c 2 = 4im(1m 2 ) (c=2) 2 = (im) 3 + (im) So we get a point on the elliptic curve y 2 =x 3 +x As noted above, the x-coordinates of points in E(K) are 0;i so m satises im = 0 ori. Hence m = 0;1, which leads to u,v or w = 0. We conclude that there is at most one quadratic extension L of K in which E tor grows and in this case E(L) =C 4 C 8 . 4.5 E(K) tor C 2 C 2 Proof. 4)Finally, suppose E(K)C 2 C 2 . Fix d2K, d a nonsquare. We wish to study E((K p d)). But as E is isomorphic toE d overK(( p d)), we haveE(K( p d))E d (K p d) so it suces to study the growth ofE d (K) overK( p d). IfE d (K) is not isomorphic toC 2 C 2 then we may apply a previously proven case. On the other hand, if E d (K)C 2 C 2 then it follows that E(K( p d))C 2 C 2 ;C 2 C 4 or C 4 C 4 . Now we claim that if K = Q( p d) (d =1; 3) and E(K) C 2 C 2 then E(K) tor grows in at most 3 quadratic extensions. Note that E(K)[2 1 ] grows in a quadratic exten- sion L if and only if one of the three points (0,0), (-M,0), (-N,0) lifts in L. But the point (0,0) lifts in at most one quadratic extension, namely K( p M; p N) (when it is quadratic over K) by Lemma 4.0.1. Similarly, (-M,0) (resp. (-N,0)) can only lift in the extension K( p M; p NM) (resp. K( p N; p MN). Hence E(K)[2 1 ] grows in at most 3 exten- sions. On the other hand, by Corollary 2.2.5, the number of quadratic extensions where E(K)[2] grows equals the number of d-quadratic twists (d dened up to a square) with 64 E d (K)[2] nontrivial. But by Theorem 4.2.1 there can be at most one such twist (namely C 2 C 6 , so E(K)[2] grows in at most one extension. Now if E(K)[2 1 ] grows in 3 quadratic extensions then K( p M; p N) is quadratic over K, so either (i) M =ds 2 ;N =dt 2 for some s;t;d2K, d a nonsquare or (ii) (WLOG) M =s 2 for some s2 K and N is a nonsquare. If (i) holds, then E d (K) has a point of order 4 by Lemma 4.0.1 so E cannot have a twist with a point of order 3 by Theorem 4.2.1 and hence E(K)[2] cannot grow in a quadratic extension by Corollary 2.2.5. Now if -M is a nonsquare in K (i.e. d6=1 modulo squares) then (-M,0) must lift in K( p M). But in this case -N is not a square as well, so (-N,0) must lift in K( p N) = K( p M) (a contradiction, as we assume the three order two points lift in pairwise dierent extensions). Hence it must be the case that -M is a square, so WLOG d=-1. As this is a nonsquare in K by assumption, this cannot happen over K= Q(i). On the other hand, over K( p 3) let E:y 2 =x(x 1)(x 9). Then E(K)[2 1 ] grows in K(i), K( p 2) and K( p 2). If (ii) holds then as N is a nonsquare in K, (0,0) lifts in K( p N). But if K= Q(i), -N is a nonsquare as well so (-N,0) lifts in K( p N) = K( p N) (a contradiction). If K( p 3) then -M is not a square so (-M,0) lifts in K( p M)=K( p 1). But then by Lemma 4.0.1 NM =v 2 for some v2 K. But ifNM =v 2 thenMN =v 2 so (-N,0) must lift in K( p MN) = K( p 1) as well (a contradiction, as we assumed the points lift in dierent extensions). Hence NM =v 2 . As noted in the proof of Theorem 4.2.1, for all nonzero d2 K we have E d (M;N) E d (M;NM) = E d (a 2 ;v 2 ) and by case (i), the latter curve has a no twist with a point of odd order greater than 1. In conclusion, if K = Q(i) then E(K)[2 1 ] never grows in 3 distinct extensions. The curve E(5 2 ; 5 32) grows in K( p 5),K( p 10),K( p 15).If K = Q( p 3) then E(K)[2 1 ] can grow in 3 dierent extensions but for such a curve E(K)[2] will not grow in any extension. The curve E(-1,-9) grows in K( p 2),K( p 2),K( p 1). 65 Bibliography [Fuj05] Yasutsugu Fujita, Torsion subgroups of elliptic curves in elementary abelian 2-extensions of Q, J. Number Theory 114 (2005), no. 1, 124{134, DOI 10.1016/j.jnt.2005.03.005. MR2163908 (2006h:11055) [GJT14] Enrique Gonzalez-Jimenez and Jose M. Tornero, Torsion of rational elliptic curves over quadratic elds, Revista de la Real Academia de Ciencias Exactas, Fisicas y Naturales. Serie A. Matematicas 108 (2014), no. 2, 923-934, DOI 10.1007/s13398-013-0152-4 (English). [JKS04] Daeyeol Jeon, Chang Heon Kim, and Andreas Schweizer, On the torsion of elliptic curves over cubic number elds, Acta Arith.113 (2004), no. 3, 291{301, DOI 10.4064/aa113-3-6. MR2069117 (2005f:11112) [JKP06] Daeyeol Jeon, Chang Heon Kim, and Euisung Park, On the torsion of elliptic curves over quartic number elds, J. London Math. Soc. (2)74 (2006), no. 1, 1{12, DOI 10.1112/S0024610706022940. MR2254548 (2007m:11079) [Kam92] S. Kamienny, Torsion points on elliptic curves and q-coecients of modular forms, Invent. Math. 109 (1992), no. 2, 221{229, DOI 10.1007/BF01232025. [KN12] Sheldon Kamienny and Filip Najman, Torsion groups of elliptic curves over quadratic elds, Acta Arith.152 (2012), no. 3, 291{305, DOI 10.4064/aa152-3-5. [KM88] M. A. Kenku and F. Momose, Torsion points on elliptic curves dened over quadratic elds, Nagoya Math. J.109 (1988), 125{149. [Kna92] Anthony W. Knapp, Elliptic curves, Mathematical Notes, vol. 40, Princeton University Press, Princeton, NJ, 1992. MR1193029 (93j:11032) [Kub76] Daniel Sion Kubert, Universal bounds on the torsion of elliptic curves, Proc. London Math. Soc. (3)33 (1976), no. 2, 193{237. [Kwo97] Soonhak Kwon, Torsion subgroups of elliptic curves over quadratic extensions, J. Number Theory 62 (1997), no. 1, 144{162, DOI 10.1006/jnth.1997.2036. [Kub79] Daniel Sion Kubert, Universal bounds on the torsion of elliptic curves, Compositio Math.38 (1979), no. 1, 121{128. [LL85] Michael Laska and Martin Lorenz, Rational points on elliptic curves over Q in elementary abelian 2-extensions of Q, J. Reine Angew. Math.355 (1985), 163{172, DOI 10.1515/crll.1985.355.163. [Maz78] B. Mazur, Rational isogenies of prime degree (with an appendix by D. Goldfeld), Invent. Math.44 (1978), no. 2, 129{162, DOI 10.1007/BF01390348. 66 [Mer96] Lo c Merel, Bornes pour la torsion des courbes elliptiques sur les corps de nombres, Invent. Math. 124 (1996), no. 1-3, 437{449, DOI 10.1007/s002220050059 (French). MR1369424 (96i:11057) [Naj10] Filip Najman, Complete classication of torsion of elliptic curves over quadratic cyclotomic elds, J. Number Theory 130 (2010), no. 9, 1964{1968, DOI 10.1016/j.jnt.2009.12.008. [Naj11] , Torsion of elliptic curves over quadratic cyclotomic elds, Math. J. Okayama Univ. 53 (2011), 75{82. [Naj12] , Torsion of elliptic curves over cubic elds, J. Number Theory 132 (2012), no. 1, 26{36, DOI 10.1016/j.jnt.2011.06.013. [Naj14] , The number of twists with large torsion of an elliptic curve, Revista de la Real Academia de Ciencias Exactas, Fisicas y Naturales. Serie A. Matematicas, posted on 2014, 1-13, DOI 10.1007/s13398-014-0199-x (English). [Ono96] Ken Ono, Euler's concordant forms, Acta Arith.78 (1996), no. 2, 101{123. MR1424534 (98c:11051) [Par00] Pierre Parent, Torsion des courbes elliptiques sur les corps cubiques, Ann. Inst. Fourier (2000), 723{749. [Par03] , No 17-torsion on elliptic curves over cubic number elds, Journal de thorie des nombres de Bordeaux15 (2003), no. 3, 831-838 (eng). [Sil94] Joseph H. Silverman, Advanced topics in the arithmetic of elliptic curves, Graduate Texts in Math- ematics, vol. 151, Springer-Verlag, New York, 1994. [Sil09] , The arithmetic of elliptic curves, Second, Graduate Texts in Mathematics, vol. 106, Springer, Dordrecht, 2009. [Sto10] Michael Stoll, Torsion points on elliptic curves over quartic number elds, 2010. Algorithmic Num- ber Theory Symposium ANTS IX, 2010. [Rab10] F. Patrick Rabarison, Structure de torsion des courbes elliptiques sur les corps quadratiques, Acta Arith.144 (2010), no. 1, 17{52, DOI 10.4064/aa144-1-3. [Rei86] Markus A. Reichert, Explicit determination of nontrivial torsion structures of elliptic curves over quadratic number elds, Math. Comp. 46 (1986), no. 174, 637{658, DOI 10.2307/2008003. MR829635 (87f:11039) [Was08] Lawrence C. Washington, Elliptic curves, 2nd ed., Discrete Mathematics and its Applications (Boca Raton), Chapman & Hall/CRC, Boca Raton, FL, 2008. Number theory and cryptography. MR2404461 (2009b:11101) 67 Chapter A Appendix 1 Magma Program: Solving a Diophantine Equation Over a Quadratic Field //Suppose C is the curve defined by f=0 where f is the polynomial below. //We wish to find all K-rational points on C for K = Q(i). R<x, y> := PolynomialRing(Rationals(), 2); f := x^4 + 2*y*x^4+2*x^3+4*x^3*y-y^4 -2*x*y^4-2*y^3-4*x*y^3; //First we factor f over Q. Fac := Factorization(f); //Now we try to find the K-rational points on the irreducible factor of f of degree > 1. g := Fac[2][1]; A<x,y> := AffineSpace(Rationals(),2); C := Curve(A, g); IsAbsolutelyIrreducible(C); // next 4 lines just to homogenize the polynomial g I := Ideal(g); I_hom := Homogenization(I); B := Basis(I_hom); g_hom := B[1]; g_hom; //Now that g_hom is homogenous we can define a curve in projective space. P<x,y,H> := ProjectiveSpace(Rationals(),2); 68 CP := Curve(P,g_hom); Genus(CP); //The genus is 1, so it is an elliptic curve over Q if we can find one rational point! //We search for rational points on CP. points := Points(CP : Bound := 10); // There was a rational point (we chose points[1] = [0,1,0]). Magma can now find //a Weierstrass model E for C and a birational map phi:C --> E E, phi := EllipticCurve(CP, points[1]); //E; //phi; K := QuadraticField(-1); //Now we tell Magma to view E as a curve over K. EK := BaseExtend(E,K); Torsion := TorsionSubgroup(EK); Invariants(Torsion); RankBound(EK); //The curve has rank 0 over K and 6 torsion points, so let's find them. points := Points(EK: Bound := 100); points; //Now by hand, one can compute the points phi^(-1)(E(K)) and the points where phi //is undefined. These are the only candidates for K-rational points on C. //Here we illustrate how to compute phi^(-1)([-5 : 2 : 1]) using a Groebner basis. R<x,y,H,w> := PolynomialRing(K,4); g_hom := H^2*x^2 + 1/2*H*x^3 + H^2*x*y + 5/2*H*x^2*y + x^3*y + H^2*y^2 + 5/2*H*x*y^2 + x^2*y^2 + 1/2*H*y^3 + x*y^3; phi_x := 2*x*y^2*H + y^3*H + 2*x*y*H^2 + y^2*H^2 + 2*x*H^3 - 2*y*H^3 + 6*H^4; phi_y := 2*x*y^3 + y^4 + 2*x*y^2*H + y^3*H + 2*x*y*H^2 + y^2*H^2 + 3*y*H^3; 69 phi_H := y*H^3 - H^4; // The zero set of this ideal is phi^(-1)([-5 : 2 : 1]) I := Ideal([g_hom, phi_x - w*(points[3][1]), phi_y - w*(points[3][2]), phi_H - w*(points[3][3])]); GroebnerBasis(I); // Note w is not 0, and two Groebner basis elements are (y+H)w =0 and (x-H)w=0. Hence y = -H and x = H, and we can now easily check phi([1,-1,1]) = [-5,2,1]. have := [Evaluate(phi_x, [1,-1,1,0]) , Evaluate(phi_y, [1,-1,1,0]) , Evaluate(phi_H, [1,-1,1,0])]; have; // This is the same projective point as [-5,2,1], //but we can also ask Magma to check it for us: // We must first tell Magma to view 'have' and 'want' as points in projective space. have := P ! have; want := P ! [-5,2,1]; want eq have; 2 Magma Program: Determining Growth of Torsion in Quadratic Fields //Inputs: a non-square integer d (-1 or -3) and an elliptic curve over Q(\sqrt{d}) //Output: List of quadratic extensions where torsion grows and the torsion structures over those fields d := -3; K<a>:= QuadraticField(d); "K:",K; R<x> := PolynomialRing(K); E := EllipticCurve( x*(x+1)*(x+4)); 70 "E:",E; I := Integers(); /////////////////////////////// Defining Function //Gives integer square-free part of rational number squarefree := function(D) ND := I ! Numerator(D); DD := I ! Denominator(D); SFN := SquarefreeFactorization(ND); SFD:= SquarefreeFactorization(DD); output := SFD*SFN; // numerator, denominator relatively prime return output; end function; /////////////////////////////////// /////////////////////////////////////////////////// Defining another function // Given element of quadratic field this function gives a rational number //congruent to it mod squares (if such a number exists) the element of the quadratic field should be of the form //f=u+v*a RationalModSquares := function(f) //f is of the form v+w*a for rationals v,w; f := K!f; m :=f[1]; n:= f[2]; R<x> := PolynomialRing(Rationals()); if (m eq 0) and (n ne 0) then if d eq -1 then output := n*2; else output := f; end if; 71 elif (m ne 0) and (n ne 0) then pol := x^2-(2*m/n)*x+ d; roots := Roots(pol); if #roots ge 1 then //f is congruent to a square! c := roots[1][1]; output := ((c+a)^2)/f; else output := f;//f is not congruent to a square end if; else // f is rational output := f; end if; return output; end function; ////////////////////////////////////// //First we list the primes which occur as the order of points in E(K)_{tor} (over the BASE FIELD k) primes := [2,3,5,7]; //Now we will collect the relevant division polynomials to detect growth. //We will grab one division polynomial f_p for each prime p in Primes. T:=TorsionSubgroup(E); two_invariants := pPrimaryInvariants(T, 2); invariants := Invariants(T); "Torsion over base field K:"; invariants; count_summands := [0,0,0,0]; for n in [1..#primes] do for i in [1..#invariants] do 72 rem := invariants[i] mod primes[n]; if rem eq 0 then count_summands[n] := count_summands[n]+1; end if; end for; end for; two_length := #two_invariants; /// chose last entry x so that Magma would accept polynomial entries in this list DivPols := [0,0,0,0,0,x]; //Defining division polynomial for p=2 if two_length ne 0 then twodivpol := DivisionPolynomial(E,2*(two_invariants[two_length]) ); DivPols[1] := twodivpol; end if; //Defining division polynomials for odd primes for n in [2..#primes] do if count_summands[n] lt 2 then DivPols[n] :=DivisionPolynomial(E,primes[n]); end if; end for; ////////////// Now for each p, we will study the roots of the associated division //polynomial to determine where the p-part of the torsion grows. for n in [1..#primes] do if DivPols[n] ne 0 then Fac := Factorization(DivPols[n]); 73 size := #Fac; // now i'm gonna run through each factor of the dvision polynomial. //For each one, i first check it's degree and then do something different depending on it's degree... for s in [1..size] do /////////////////////////////// Degree 1 Case if Degree(Fac[s][1]) eq 1 then pol := Fac[s][1]; root := Roots(pol); a := root[1]; b := a[1]; //root of the linear factor // The function Points below only detects the y-values defined over the SAME base field P := Points(E,b); //////////////////////////////////// / if #P eq 0 then ai := aInvariants(E) ; R<y> := PolynomialRing(K); f := y^2 + ai[1]*b*y + ai[3]*y - (b^3 + ai[2]*b^2+ai[4]*b+ai[5]); L<e> := ext<K | f>; Disc := Discriminant(f); RR := RationalModSquares(Disc); if RR[2] eq 0 then dd := squarefree(RR); // my user-defined functions else dd := RR; end if; "Torsion grows in L=K(\sqrt{d}) where d =",dd; EL := BaseExtend(E,L); T := TorsionSubgroup(EL); 74 "The invariants of E(L)_{tor} are:"; Invariants(T); end if; end if; //////////////////////////////////////////////////////////////////////////Degree 2 case if Degree(Fac[s][1]) eq 2 then ////////////////////////////////// R<y> := PolynomialRing(K); f := Fac[s][1]; L<e> := ext<K | f>; EL := BaseExtend(E,L); P := Points(EL, e) ; if #P gt 0 then Disc := Discriminant(f); RR := RationalModSquares(Disc); if RR[2] eq 0 then dd := squarefree(RR); // my user-defined functions else dd := RR; end if; "Torsion grows in K(\sqrt{d}) where d =",dd; EL := BaseExtend(E,L); T := TorsionSubgroup(EL); "The invariants of E(L)_{tor} are:"; Invariants(T); end if; end if; end for; end if; end for; 75
Abstract (if available)
Abstract
Let K = ℚ(√(-3)) or ℚ(√(-1)) and let C_n denote the cyclic group of order n. We study how the torsion part of an elliptic curve over K grows in a quadratic extension of K. In the case E(K)[2] = C₂ ⊕ C₂ we determine how a given torsion structure can grow in a quadratic extension and the maximum number of extensions in which it grows. We also classify the torsion structures which occur as the quadratic twist of a given torsion structure. We complete a similar classification in the case K = ℚ(√(-3)) and E(K)[2] = C₁.
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Newman, Burton
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Growth of torsion in quadratic extensions of quadratic cyclotomic fields
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Doctor of Philosophy
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Mathematics
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04/14/2015
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